uKs m J(MJ i ^^t^mmig tiM ^w i e^ im» i >mti PLANED SOLID GEOMETRY SLAUGHT AND MMjH i »w«Bi*Mi i ^ a i<««a *'' ^ M *'» w « ' M ' flt^ (!}arnell Hititteratty ffitbrarij Cornell University Library arV19325 Plane and solid geometry 3 1924 031 226 263 olin.anx Cornell University Library The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031226263 PLANE AND SOLID GEOMETRY WITH PROBLEMS AND APPLICATIONS BY H. E. SLAUGHT, Ph.D. ASSOCIATE PROFESSOR OF MATHEMATICS IN THE UNIVTERSITY OF CHICAGO AND N. J. LENNES, Ph.D. INSTRUCTOR IN MATHEMATICS IN COLUMBIA UNIVERSITY iSoston ALLYN AND BACON 191 1 COPYRIGHT,' 1911. BY H. E. SLAUGHT AND N. J. LENNES. Nartnooti ^reiiis .1. S. Gushing Co, — Berwick & Smith Oi Norwood, Muf-s., U.S.A. PREFACE. In writing this book the authors have been guided by two main purposes : (a) That pupils may gain by gradual and natural processes the power and the habit of deductive reasoning. (b) That pupils may learn to know the essential facts of elementary geometry as properties of the space in which they live, and not merely as statements in a book. The important features by which the Plane Geometry seeks to accomplish these purposes are : 1. The simplification of the first five chapters by the exclusion of many theorems found in current books. These five chapters correspond to the usual five books, and the most important omissions are the formal treatment of the theory of limits, the incommensurable cases, maxima and minima, and numer- ous other theorems, together with the deduction of complicated algebraic formulae. Chapter VI contains a graphic representation of certain im- portant theorems and an informal presentation of incommen- surable cases and ■ limits. The treatment of limits is based upon the graph, since the visual or graphic method appeals more directly to the intuition than the usual abstract processes. Chapter VII is devoted to advanced work and to a review of the preceding chapters. 2. T7ie subject has been enriched by including many applica- tions of special interest to pupils. Here an effort has been made to include only such concrete problems as come fairly within iv PREFACE. the observation and comprehension of the average pupil. This led to the omission, for example, of problems relating to machinery and technical industries, which might appeal to an exceptional boy, but which are entirely inappropriate for the average student. On the other hand, free use is made of certain sources of problems which may be easily compre- hended without extended explanation and which involve varied and simple combinations of geometric forms. Such problems pertain to decoration, ornamental designs, and architectural forms. They are found in tile patterns, parquet floors, lino- leums, wall papers, steel ceilings, grill work, ornamental windows, etc., and they furnish a large variety of simple exercises both for geometric construction and proofs and for algebraic computation. They are not of the puzzle type, but require a thorough acquaintance with geometric facts and develop the power. to use mathematics. These problems form an entirely new type of exercises, and while they require more space in the text-book than the more difficult " originals " stated in the usual abstract terms, they excel the latter in interest for the pupil and in helping to train his mathematical common sense. Jlany of these exer- cises are simple enough to be solved at sight, and such solution should be encouraged whenever possible. All the designs are taken from photographs or from actual commercial patterns now in use. 3. Persistent effort is made to vitalize the content of the defini- tions and theorems. It is well known that pupils often study and recite definitions and theorems without really comprehend- ing their meaning. It is sought to check this tendency by giving definitions only when tlioy are to be used, and by imme- diately verifying both definitions and theorems in concrete cases. The figure on page 4 is the basis for a large number of questions of this type. For example, see § !.'."), E.\.. o ; § 30, Ex. 1 ; § 34, Exs. 1, 2; § 36, Ex. 3; §§ 321.', ;i24. PREFACE. V In this connection special attention is called to the emphasis placed upon those theorems which are of fundamental impor- tance both in the logical chain and in their immediate use in effecting constructions and indirect measurements otherwise difficult or impossible. For example, see the theorems on con- gruence of triangles, §§ 31-43, the constructions of §§ 44^58, and the theorems on proportional segments, §§ 243-254. Com- pare especially § 34, Ex. 6, § 244, Ex. 2, and § 254, Exs. 4, 5. The summaries at the close of the chapters, which are to be made by the pupil himself, will vitalize the theorems as no made-to-order summaries can possibly do. 4. The student is made to approach the formal logic of geometry by natural and gradual processes. He is expected to grow into this new, and to him unusual, way of thinking. The treat- ment is at the start informal, leading through the congruence theorems directly to concrete applications and geometric con- structions. The formal development then follows gradually and is characterized by a judicious guidance of the student, by questions, outlines, and other devices, into an attitude of mental independence and an appreciation of clear reasoning. There are certain terms whose general meaning in mathe- matics is inconvenient in some parts of a book like the present. Thus we say that two circles (meaning by circle the curved line) meet in two points, and it is also convenient to say that the base of a certain cylinder is a circle. In such cases the generally accepted mathematical meanings of the terms have been given in the definitions, while notes have been added that different meanings are sometimes to be understood, depending on the context to make the meaning clear. See §§ 623, 695. This avoids learning definitions which must be discarded later, and also the clumsiness resulting from a strict adherence to the same meaning throughout. Experience has shown that no confusion whatever results from this, while the resulting sim- plicity of statement is of considerable importance. VI PREFACE. By its arrangement the Plane Geometry is adapted to three different courses : (a) A minimuTn course, consisting of Chapters I to VI, ■without the problems and applications at the end of each chapter. This would provide about as much material, theo- rems, constructions, and originals as is found in the briefest books now in use. (6) A medium course, consisting of Chapters I to VI, includ- ing a reasonable number, say one half or two thirds, of the applications at the end of each chapter. This would fully cover the college entrance requirements. (c) An extended course, including Chapter VII, which eon- tains a complete review, together with many additional theo- rems and a large number of further applications. This would provide ample work for the strongest high schools, and for normal schools in which more mature students are found or more time can be given to the subject. Chapter VII gives a complete treatment of the incommen- surable cases, though not based on the formal theory of limits. It is believed that for high school pupils the notion of a limit is best studied as a process of approximation, and that the best preparation for the later understanding of the theory is by a preliminary study of what is meant by " approaches," such as is given in Chapters III and IV. The important features by which the Solid Geometry seeks to accomplish the two main purposes stated with reference to the Plane Geometry are : 1. The concepts of three-dimensional space are made clear by many simple iRustratioDS and exercises. At best the average pupil comes but slowly and gradually to a full comprehension of space forms and relations. Pages L'SL', L'S.S, L'SG, 2SS. L\S;1, 295, 296, etc., will show how the pupil is aided in understand- ing and appreciating these forms and relations at the outset. In this connection, Chapter XIII on Graphic Representation PREFACE. vii is of fundamental importance, since it exhibits to the eye the functional relations among varying geometric forms. 2. Interesting concrete applications are interspersed throughout the text. It is the aim profoundly to impress upon the mind of the pupil the practical significance of certain fundamental theorems in solid geometry. Eor instance, how many pupils in the ordinary study of the theorems on the ratios of surfaces and volumes of similar figures realize that there is in them any connection with the possibility of successfully launching a steamship a mile long ? See the exercises on pages 390, 391, 444, 445. For illustrations of other interesting and useful applications, see pages 337, 344, 345, 434, 435. 3. The logical structure is made more complete and more prominent than in the Plane Geometry. Solid Geometry is studied by more mature pupils who have been led by gradual stages in the Plane Geometry to a knowledge and apprecia- tion of deductive reasoning. Hence the axioms are stated and applied in strictly scientific form and at the precise points where they are to be used. For instance, see §§ 460, 462, 464, 538, 560, 562, 593, 612, etc. Note also the consistent and scientific definitions of all solids, not as bounded portions of space, but as configurations in space, the uniform conception in all higher mathematical usage. See §§ 68, 70, 99, 132, 149, 200, 523, 525, 554, 587, 604, 655. 4. Throughout both parts of this Geometry a consistent scheme has been followed in the presentation of incommensurables and the theory of limits. In Chapters I to VI the idea of "ap- proach " is made clear by many concrete illustrations, and the theorems involving this idea are shown to hold for all possible approximations. In Chapter VII rigorous proofs of these theo- rems are given and in far simpler terminology than is found in current text-books. This latter method is followed throughout VIU PREFACE. Chapters VIII to XIII, thus giving a complete and scientific treatment up to that point. In Chapter XIV the theory of limits is presented in such a way as to leave nothing to be unlearned or compromised in later mathematical work. This chapter may be omitted without affecting the logical complete- ness of the book. Acknowledgment is due to Miss Mabel Sykes, of Chicago, for the use in the Plane Geometry of a large number of draw- ings and designs from her extensive collection ; also to numer- ous commercial and manufacturing houses, both in this country and in Europe, through whose courtesy many of the patterns were obtained. H. E. SLAUGHT. N. J. LENNES. Chicago and New York, May, 1911. CONTENTS. Chaptek I. Rectilinear Figures. PASS Introduction .... 1 Notation for Points and Lines 5 Angles and their Notation . . 7 Triangles and their Notation 10 Congruence of Geometric Figures 12 Tests for Congruence of Triangles 14 Construction of Geometric Figures 20 Theorems and Demonstrations 26 Inequalities of Parts of Triangles 32 Theorems on Parallel Lines 35 Applications of -Theorems on Parallels 40 Other Theorems on Triangles 43 Determinations of Loci 50 Theorems on Quadrilaterals 54 Polygons 66 Symmetry 68 Methods of Attack 72 Summary of Chapter I 75 Problems and Applications 76 Chapter II. Straight Lines and Circles. Theorems on the Circle 86 Measurement of Angles 88 Problems and Applications 92 Angles formed by Tangents, Chords, and Secants .... 95 Summary of Chapter 11 106 Problems and Applications 107 Chapter III. Measurement of Straight Line-segments. Approximate Measurement 112 Katies of Line-segments 114 Theorems ou Proportional Segments 116 ix CONTENTS. PAGE Similar Polygons 124 Computations by Means of Sines of Angles .... 136 Problems of Construction 140 Summary of Chapter III . 145 Problems and Applications 146 Chapter IV. Areas of Polygons. Areas of Rectangles . Areas of Polygons Problems of Construction . Summary of Chapter IV Problems and Applications Chapter V. Regular Polygons and Circles. Regular Polygons Problems and Applications Measurement of the Circle Summary of Chapter V Problems and Applications Chapter VI. Variable Geometric Magnitudes. Graphic Representation ....... Dependence of Variables Limit of a Variable . . ... 1.54 156 166 169 170 176 18.3 188 104 195 207 215 217 Chapter VII. Review and Further Applications. On Definitions and Proofs . . . . ilO Loci Considerations ... .... 226 Problems and Applications 22ii Further Data concerning Triangles 2:50 Problems and Applications 233 The Incommensurable Cases 238 Problems and Applications 245 Further Applications of Proportion 2,i0 Further Properties of Triangles 2,V,1 Problems and Applications jiU Maxima and Minima 208 CONTENTS. xi Chapter VTII. Lines and Planes in Space. PAGE Determinations of a Plane 279 Theorems on Perpendiculars 284 Theorems on Parallels 290 Dihedral Angles 296 Polyhedral Angles 304 Summary of Chapter VIII 306 Problems and Applications 306 Chaptek IX. Prisms and Cylinders. Definitions 310 Theorems on Prisms . . 312 Volumes of Rectangular Parallelepipeds ...... 317 Volumes of Prisms in General 320 Cylinders. Definitions . 324 Measurement of the Surface and Volume of a Cylinder . . . 328 Problems and Applications 331 Theorems on Projection ... 333 Projection of a Plane Segment 340 Summary of Chapter IX 343 Problems and Application 344 Chapter X. Pyramids and Cones. Definitions and Theorems on Pyramids Definitions and Theorems on Cones . Measurement of the Surface and Volume of a Cone Summary of Chapter X Problems and Applications .... 316 358 362 369 Chapter XI. Regular and Similar Polyhedrons. Regular Polyhedrons 378 Inscription of Regular Polyhedrons 377 Similar Polyhedrons 380 Applications of Similarity 388 Summary of Chapter XI 390 Problems and Applications 390 XU CONTENTS. Chapter XII. The Sphere. PAGE Plane Sections of the Sphere 392 Trihedral Angles and Spherical Triangles 402 Polar Triangles . 410 Polyhedral Angles and Spherical Polygons ... . 418 Areas of Spherical Polygons 419 Area and Volume of the Sphere . 423 Summary of Chapter XII . 433 Problems and Applications 434 Chaptek XIII. Variable Geometric Magnitudes. Graphic Representation .... ... 436 Summary of Chapter XIII . . .... 443 Problems and Applications . 444 Chapter XIV. Theory of Limits. General Principles . . ... 446 Application of Limits to Geometry 460 Volume of the Sphere . . 460 Summary of Chapter XIV 463 Index ............ 465 PLAICE GEOMETRY. CHAPTER I. RECTILINEAR FIGURES. INTRODUCTION. , 1. Elementary geometry is a science which deals with the space in which we live. It begins with the consideration of certain elements of this space which are called points, lines, planes, solids, angles, triangles, etc. Some of these terms, such as point, line, plane, are here used without being defined in a strictly logical sense. Their meaning is made clear by description and by con- crete illustrations like the following. 2. Certain portions of space are occupied by objects which we call physical solids, as, for instance, an ordinary brick. That which separates a solid from the surrounding space is called its surface. This may be rough or smooth. If a surface is smooth and flat, we call it a plane surface. A pressed brick has six plane surfaces called faces. Two adjoining faces meet in an edge. Three edges meet in a corner. The brick is bounded by its six faces. Each face is bounded by four edges, and each edge is bounded by two corners. 1 2 PLANE GEOMETRY. 3. If instead of the brick we think merely of its form and magnitude, we get a notion of a geometrical solid, which has the three dimensions, length, breadth, and thickness. The faces of this ideal solid are called planes. These are flat and have length and breadth, but no thickness. The edges of this solid are called lines. They are straight and have length, but neither breadth nor thick- ness. The corners of this solid are called points. They have position, but neither length, breadth, nor thickness ; that is, they have no magnitude. 4. It is possible to think of these concepts quite inde- pendently of any physical solid. Thus we speak of the line of sight from one point to another ; and we say that light travels in a straight line. The term straight line is doubtless connected with the idea of a stretched string. Of all the lines which may be con- ceived as passing through two fixed points that one ^^< y^ is said to be straight between these points which corresponds most nearly to a stretched string. Likewise a plane may be thought of as straight or stretched in every direction, so that a straight line passing through any two of its points lies wholly in the plane. 5. If one of two intersecting straight lines turns about their common point as a pivot, the lines will continue to have only one point in ^.^ common until all at once they will coincide ~~^ throughout their whole length. Hence, Two strniijlit lines cannot have more than one point in common unless tlieii coincide and are the same line; that is, two points determine a straight line. This would not be so if the lines liad width, as may be seen by examining the figures. RECTILINEAR FIGURES. 3 6. EXERCISES. 1. How does a carpenter use a straight-edge to determine whether a surface is a plane? Do you know of any surface to which this test will apply in one direction but not in all directions ? 2. What tool does a carpenter use in reducing an uneven surface to a plane surface ? Why is the tool so named ? 3. If two points of a straight line lie in a plane, what can be said of the whole line ? 4. How many points of a straight line can lie in a plane if it con- tains at least one point not in the plane ? 5. If two straight lines coincide in more than one point, what can you say of them throughout their whole length ? Do you know of any lines other than straight lines of which this must be true ? 6. How do the material points and lines made by crayon or pencil differ in magnitude from the ideal points and lines of geometry ? 7. A machine has been made which rules 20,000 distinct lines side by side within the space of one inch. Do such lines have width ? Are they geometrical lines ? 8. Of all the lines, straight or curved, through two points, on which one is the shortest distance measured between the two points ? See the figure of § 4. Historical Note. The Egyptians appear to have been the first people to accumulate any considerable body of exact geometrical facts. The building of the great pyramids (before 3000 b.c.) re- quired not a little knowledge of geometric relations. They also used geometry in surveying land. Thus it is known that Eameses II (about 1400 B.C.) appointed surveyors to measure the amount of land washed away by the Nile, so that the taxes might be equalized. The Greeks, however, were the first to study geometry from a logi- cal point of view. Between 600 b.c, when Thales, a Greek from Asia Minor, learned geometry from the Egyptians, and 300 b.c, when Euclid, a Greek residing in Alexandria, Egypt, wrote his Elements of Geometry, the crude, practical geometric information of the Egyp- tians was transformed into a well-nigh perfect logical system. Euclid's " Elements " contains the essential facts of every text- book on elementary geometry that has been written since his time. PLANE QEOMETRT. RECTILINEAR FIGURES. 6 NOTATION FOR POINTS AND LINES. 7. A point is denoted by a capital letter. A straight line is denoted by two capital ' letters marking two of its points or by — ^ . — one small letter. The word line alone ; usually means straight line. Thus, the point A, the line AB, or the line I. 8. A straight line is usually understood to be un- limited in length in both directions, while that part of it which lies between two of its points is a a B called a line-segment, or simply a segment. " These points are called the end-points of the segment. Thus, the segment AB or the segment a. Two segments with the same end-points are coincident. 9. A part of a straight line, called a ray or half-line, may be thought of as generated by a point ^ ^ starting from a fixed position and mov- ' ' * ing indefinitely in one direction. The starting point is called the end-point or origin of the ray. If A is the origin of a ray and B any other point on it, then it is read the ray AB, not the ray BA. 10. Two line-segments are said to be added if they are placed end to end so as 4 £ ? to form a single segment. Thus, segment ^C = segmeDt^B + segment BC, otAC = AB+BC. If AC = AB + BC, then AC is greater than either AB or BC, and this is written AC > AB and AC > BC. A segment may also be subtracted from a greater or from an equal segment. Thus, itAC = AB + BC, then AB =AC - BCandBC = AC -. AB. 6 PLANE GEOMETRY. A segment is multiplied by an integer n by taking the sum of n such segments. - Thus, if A C is the sum of n segments each AB, then AC = n ■ AB. If AG is n times AB, then AC may be divided by w. Thus, ^S = ^C-Hn or ^B = i.^C. 11. A broken line is composed of connected line-segments not all lying ^ in the same straight line. A curved line, or simply a curve, is a line no part of which is straight. A curved line or a broken line may inclose a portion of a plane, while a straight line cannot. 12. A circle is a plane curve containing all points equally distant from a fixed point in the plane, and no other points. The fixed point is called the center of the circle. Any line-segment joining the center to a point on the circle is a radius of the circle. Any portion of a circle lying between two of its points is called an arc. Evidently all radii of the same circle are equal. Any combination of points, segments, lines, or curves in u plane is called a plane geometric figure. Plane Geometry deals with plane geometric figures. 13. EXERCISES. 1. How many end-points has a straight line? How many has a line-scgmont? How many lias a ray ? A circle? 2. Can you inclosi' a portion of ,i plane with two line-segments? With three? With four? RECTILINEAR FIGURES. ANGLES AND THEIR NOTATION. 14. An angle is a figure formed by two rays proceeding from the same point. The point is the vertex of the angle and the rays are its sides. The angle formed by two rays is ^='*™'^ — siai" said to be the angle between them or simply their angle. Two line-segments having a common end-point also form an angle, namely, the angle of the rays on which the segments lie. An angle is determined entirely by the relative directions of its rays and not by the lengths of the segments laid off on them. 15. An angle is denoted by three letters, one at its vertex and one marking a point on each of its sides. The one at the vertex is. read between the other two, as the angle CAB, or the angle BAG, not ABC. The one letter at the vertex is also used alone to denote an angle in case ^' no other angle has the same vertex, as, for instance, the angle A. In case several angles have the same vertex, a small letter or figure placed within each angle, together with an arc connecting its sides, is a convenient notation. The sign Z. is used for the word angle. Thus in the figure we have Zl,/i2, Z'd, read, angle one, angle two, angle three. A 16. If two rays lie in the same straight line and ex- tend from the same end-point in opposite ^ directions they are said to form a straight | angle. If they extend in the same direc- S'de 'f side tion, they coincide and form a zero angle. 8 PLANE GEOMETRY. 17. An angle may be thought of as generated by a ray turning about its end-point as a pivot. Thus Z BA C is generated by a ray rotating from the position ABto the position A C. The rotating ray is usually conceived as moving in the direction opposite to the hands of a clock and the sides of the angle should usu- ally be read in this order. Thus Z BA C, not / CAB. If the ray continues to rotate until ^_^ it lies in a direction exactly opposite — — * • ' to its original position, it generates a straight angle,, as the straight angle BAG. If the ray rotates until it reaches its original position, the angle generated is called a peri- gon, that is, an angle of complete rotation. The position from which the rotating ray starts is called the origin of the angle. From this point of view two rays from the same point form two angles according as one or the other of the rays is regarded as the origin. In elementary geometry only angles less than or equal to a straight angle are usually considered. The units of measure for angles are one three-hnn- dred-sixtieth of a perigon which is called a degree, one sixtieth of a degree called a minute, and one sixtieth of a minute called a second. These are denoted respectively by the symbols °, ', ". Thus, an angle of 20° 45' 30". A straight angle is therefore an angle of 180° and a perigon is an angle of 360°. 18, Angles are measured by means of an instrument called a protractor, which consists of a semicircular scale with degrees from 0° to 180° marked upon it. An inexpensive protractor made of cardboard or brass may be had at any stationery store. See the figure of § 33. RECTILINEAR FIGURES. 9 19. Two angles are said to be equal if they can be made to coincide without changing the form of either. If a ray is drawn from a point in a gc- straight line so that the two angles thus formed are equal, each angle is called a right angle, and the ray is said to be d~ perpendicular to the line. Thus, if Z 1 = Z 2, each angle is a right angle and ^ C is then said to be perpendicular to BD. Since the straight angle BAD is composed of Z 1 and Z 2, each of which is a right angle, it appears that a straight angle equals two right angles. See § 39 for the addition of angles in general. An acute angle is less than a right angle. ^«. An obtuse angle is greater than a right angle and less than a straight angle. Acute and obtuse angles are called oblique angles. A reflex angle is greater than a straight angle and less than a perigon. One line is oblique to another if the angles between them are oblique. A ray which divides an angle into two equal angles is called its bisector. Thus a perpendicular is the bisector of a straight angle. The angles considered in this book are greater than the zero angle and less than or equal to a straight angle. 20. EXERCISES. 1. Since we can. always place two straight angles so as to make them coincide, what can we say as to whether or not they are equal? What of two right angles? 2. What part of a straight angle is a right angle ? 10 PLANE GEOMETRY. 3. Suppose that in the figure the ray AC rotates about the point A from the position AB to the position AD. What change takes place in Zl? What in Z2? Can there be q more than one position of ^ C for which Z 1 = Z 2 ? In this way it may be made clear that any angle has one and only one bisector. jy -^^ g 4. How many rays perpendicular to BD at the point A can be drawn on the satne side of BD"! Does the answer to this question depend upon the answers to the questions in Ex. 4? How? 5. Pick out three acute angles, three right angles, and three obtuse angles in the figure on page 4. TRIANGLES AND THEIR NOTATION. 21. If the points A, B, C do not lie in the same straight line, the figure formed by the three segments, AB, BC, and CA, is called a triangle. The segments are the sides of the triangle, and the points are its vertices. The symbol A is b used for the word triangle. e^ Each angle of a triangle has one side opposite and two sides adjacent to it. b Similarly each side of a triangle has one angle opposite and two angles adjacent to it. The side opposite an angle is often denoted by the corresponding small letter. Thus, in the figure the side a and the angle A are opposite parts, us are angle B and side h and angle C and side c. The three sides and three angles of a triangle are called the parts of the triangle. These six parts are considered as lying in order around the figure, as Z 4, side b, Z. C, side a, etc. An angle of a triangle is said to be ind>tiJed between its two adjacent sides, and a side is said to be included be- tween the two angles adjacent to it. Thus, in the llnure thi' side it is included between Z B and Z C, and Z. A is included l)et\veen the sides b and c. RECTILINEAR FIGURES. 11 22. A triangle is called equilateral if it has its three sides equal, isosceles if it has at least two sides equal, scalene if it has no two sides equal, equiangular if it has its three angles equal. Select each kind from the figures on this page. 23. A triangle is called a right triangle if it has one right angle, an obtuse triangle if it has one obtuse angle, an acute triangle if all its angles are acute. Select each kind from the figures on this page. The side of a right triangle opposite the right angle is called the hypotenuse in dis- tinction from the other two sides, which are sometimes called its legs. / 24. The side of a triangle on which it is supposed to stand is called its base. The angle opposite the base is called the vertex angle, and its vertex is the vertex of the triangle. The altitude of a tri- angle is the perpendic- ular from the vertex to the base or the base produced. Evidently any side may be taken as thebase,.and.hence.a triangle hasthree different altitudes.. Vertex Vertex Base Base 12 PLANE GEOMETRY. 26. EXERCISES. 1. la every equilateral triangle also isosceles? Is every isosceles triangle also equilateral ? 2. Is a right triangle ever isosceles? Is an obtuse triangle ever isosceles? Draw figures to illustrate your answers. 3. In the figure on page i determine by measuring sides which of the triangles HNP, LKW, IHN, MIJ, KVU, OKJ, LVW, are isos- celes, which are equilateral, and which are scalene. 4. Determine whether /, K, V of the same figure may be the vertices of a triangle ; also whether /, 0, G may be. 5. Pick out ten obtuse triangles in this figure; also ten acute triangles. CONGRUENCE OF GEOMETRIC FIGURES. 26. In comparing geometric figures it is assumed that thet/ may he moved about at tvill, either in the game plane or out of it, without changing their shape or size. 27. Two figures are said to be similar if they have the same shape. This is denoted by the symbol ~, read is similar to. For a more precise definition see §§ 255, 2.5fi. Two figures are said to be equivalent or simply equal if they have the same size or magnitude. This is denoted by the symbol =, read is i r equivalent to or is equal to. I I Two fig-urets are said to be congruent if. , , without changing the sliape or size of either, I I ^ ' ' they may be so placed as to coincide | ._| . throughout. This is denoted by the symbol ^ ^, read is congruent to. In the casi' of linr-scgmonts and angles, congruence is determined by siz<> alone. Hence in these oases we use tlie symbol = to denote congruence, and read it equals or is equal to. RECTILINEAR FIGURES. 13 28, It is clear that if each of two figures is congruent to the same figure they are congruent to each other. Hence if we make a pattern of a figure, say on tracing paper, and then make a second figure from this pattern, the two figures are congruent to each other. 29. If AABC'^AA'b'c', the notation of the triangles may be so arranged that AB = A'b', bc=b'c', ca = c'a', za=Za', zb=Zb' and Ac = Zo'. In this case AB is said to correspond to a'b', bc to b'c', CA te c'a', ZAto Za', etc. Hence, we say that corresponding parts of congruent triangles are equal. 30. EXERCISES. 1. Using tracing paper, draw triangles congruent to the triangles MIN, NHP, OAB, OFE, OKL, UKV, OGL on page 4, and by ap- plying the pattern of each triangle to each of the others determine whether any two are congruent. 2. Find as in § 28 whether any two of three accompanying triangles are congruent, and if so arrange the notation so as to show the corre- sponding parts. 3. Give examples of figures which are similar, equal, or con- gruent, different from those in § 27. 4. If two figures are congruent, does it follow that they are equal ? Similar? 5. If two figures are similar, does it follow that they are equal? Congruent? 6. If two figures are equal, are they similar ? Congruent ? 14 PLANE GEOMETRY. TESTS FOR CONGRUENCE OF TRIANGLES. 31. The method of determining whether two triangles are congruent by making a pattern of one and applying it to the other is often inconvenient or impossible. There are other methods in which it is necessary only to determine whether certain sides and angles are equal. These methods are based upon three important tests for congruence of triangles. 32. First Test for Congruence of Triangles. If tioo triangles have tioo sides and the included angle of one equal respectively to two sides and the included angle of the other, the triangles are congruent. This may be shown by the following argument: c G' Let ABC and A'ffC be two triangles in which AB = A'B', AC = A'C, a.ni ZA = ZA'. We are to show that AABC^Aa'b'c'. Place AABO upon AA'b'c' so that Z.A coincides wiiii Za', which can be done since it is given that Za=Za'. Then point B will coincide with B' and C with c', since it is given that AB = A'b' and Ac = A'C. Hence, side BC will coincide with b'c' (§ ^i). Thus, the two triangles coincide throughout and hence are congruent (§ 27). The process just used is called superxrasition. It /p may sometimes be necessary to move :i figure out of its plane in order to superpose it upon another, as in the case of the accompanying triangles^ liECTILlNEAR FIGURES. 15 33. The equality of short line-segments is conveniently- tested by means of the dividers or compasses. Place the divider points on the end-points of one segment ^B and then see whether they will also coincide with the end-points of the other segment A'B'. If so, the two segments are equal. The equality of two angles may be tested by means of the protractor. Place the protractor on one angle BOC as shown in the figure and read the scale where OC crosses it. Then place the protractor on the other angle B'O'C and see whether 0' C" crosses the scale at the same point. If so, the two angles are equal. . 34, EXERCISES. 1. Using the protractor determine which pairs of the following angles on page 4 are equal : HPG, LGW, GWL, AOB, VLW, LVW. 2. By the test of § 32 determine whether, on page 4, /\JKU SAG WL, also whether A MIH ■^AK VW. First find whether two sides of one are equal respectively to two sides of the other, and if so compare the included angles. 3. Could two sides of one triangle be equal respectively to two sides of another and still the triangles not be congruent ? Illustrate by constructing two such triangles. 4. Show by the test of § 32 that two right triangles are congruent it the legs of one are equal respectively to the legs of the other. Can this be shown directly by superposition ? 5. Find the distance AB when, on account of some obstruction, it cannot be measured directly. Solution. To some convenient point C measure the distances A C and BC. Continuing in the direc- tion .dC lay off CA' = AC, and in the direction BC lay off CB' = BC. Then Z 1 = Z 2 (see § 74) . Test this with the protractor. Show that the length AB is found by measuring A 'B'. 16 PLANE GEOMETRY. 35. Second Test for Congruence of Triangles. If tivo triangles have tioo angles and the included side of one equal respectively to two angles and the in- cluded side of the other, the triangles are congruent. This is shown by tlie following argument : C C A B A' B' Let ABC and A'B'C be two triangles in which /iA=Z.A', AB = Z.B, and AB=A'^. We are to show that A ABC ^ A a'b' c' . Place A ABC upon AA'b'c' so that AB coincides with its equal A'b' , making C fall on the same side of a'b' as C'. Then AC will take the direction of A' c' , since AA= Z.A', and the point C must fall somewhere on the ray a'c'. Also BC will take the direction of b'& (Why?), and hence C must lie on the ray B'c'. Since the point C lies on both of the rays A'cf and b'c', it must lie at their point of intersection c' (§ 5). Hence, the triangles coincide and are, therefore, congruent (§ "27). r 36. EXERCISES. 1. Tn the figure of § '^o is it necessary to move l\ABC out of the plane in wliicli tlie triangles lie? Is it necessary in the figure here given ? 2. Show how to measure the height of a tree by using the second test for congruence. Su(iGF,s iiiiN. Lay out a triangle on the ground which is congruent to A ABC, using § '.i'l. 3. By the second test determine whether A Ofia = A O/A' on page 4. 4. Draw any triangle, (.'oustruct another tri- BECTILINEAR FIGURES. 17 angle congruent to it. Use § 35 and also § 32. Use the protractor to construct the angles. 5. Find the distance A C, when C is inaccessible. Let 5 be a convenient point from which A and C are visible. Lay out a triangle ABC mak- ing ^ 3 = Z 1 and Z 4 = /^ 2. Show that the dis- tance A C may be found by measuring A f. 6. Show how to find the distance between two inaccessible points A and B. Solution. Suppose that both A and B are visible from C and D. (1) Using the triangle CDA, find the length of AD as in Ex. 5 above. (2) Using the triangle CBD, find DB in the same manner. (3) Using the triangle DBA, find AB as in Ex. 5, § 34. 37. The proof of the third test for congruence of triangles involves the following : The angles opposite the equal sides of an Isosceles tri- angle are equal. Let ABC be an isosceles triangle having AC = BC We are to show that Z. A = /^ B. Suppose CD divides /. ACB so that Z 1 = Z 2. By means of § 32 show that Aacd ^Abcd. Then Z 4 = Z B by § 29. The theorems § 35 and § 37 are due to Thales. It is said he used § 35 in calculating the distance from the shore to a ship at sea. 38. EXERCISE. On page 4 pick out as many pairs of angles as possible which may be shown to be equal by § 37. Test these by using the protractor. 18 PLANE GEOMETRY. 39, Definitions. Two angles which have a common vertex and a common side are said to be adjacent if neither angle lies within the other. 1/ i\ Thus, Z 1 and Z 2 are adjacent, while Z 1 and Z 3 are not adjacent The sum of two angles is the angle formed by the sides not common when the two angles are placed adjacent. Thus, Z3 = Z1 + Z2. If Z 3 = Z 1 + Z 2, then we say that Z 3 is greater than either Z 1 or Z 2. This is written Z 3 > Z 1 and Z 3 > Z 2. An angle may also be subtracted from a greater or equal angle. ThusifZ3 = Zl + Z 2, then Z3 -Zl = Z2and Z3-Z2 = Z1. It is clear that : If equal angles are added to equal angles, the sums are equal angles. Angles may be multiplied or divided by a positive in- teger as in the case of line-segments. See § 10. 40, We may now prove the third test for congruence of triangles, namely : // two triangles have three sides of one equal respec- tively to three sides of the other, the triangles are con- gruent. c' Let ABC and A'B'C' be two triangles in which AB= A'B', BC=B'C', CA^CA'. We are to show that A A nc ^ A a'b'c'. RECTILINMAM FIGURES. 19 Place Aa'b'c' so that a'b' coincides with AB and so that c' falls on the side of AB which is opposite C. (Why is it possible to make A 'B' coincide -with AB ?) Draw the segment OC'. From the data given, how can § 37 be used to show that in A ACc' Z 1 = Z 2 ? Use the same argument to show that Z 3 = Z4. But if Z 1 = Z 2 and Z 3 = Z 4, then Z 1 + Z 8 = Z 2 + Z 4. (§39) That is, Z.ACB = Z BC' A. How does it now follow that A ABC ^Aabc' ? (§ 32) But AABC' ^aa'b'c'. (§26) Hence, A ABC ^Aa'b'c'. (§28) Make an outline of the steps in the above argument, and see that each step is needed in deriving the next. 41, Definition. If one triangle is congruent to another because certain parts of one are equal to the corresponding parts of the other, then these parts are said to determine the triangle. That is, any other triangle constructed with these given parts will be congruent to the given triangle. 42. EXERCISES. 1. In § 37 show that CD is perpendicular to AB and that AD = DB. State this fully in words. 2. Using § 40, determine which of the following triangles on page 4 are congruent : OJK, HNP, OIH, PHG, JKU. 3. Do two sides rfeiermine a triangle ? Three sides? Two angles? Three angles ? Illustrate by figures. 4. A segment drawn from the vertex of an isosceles triangle to the middle point of the base bisects the vertex angle and is perpendicular to the base. 5. What parts of a triangle have been found sufficient to determine it? In each case how many parts are needed? 20 PLANE GEOMETRY. 43. The three tests for congruence of triangles, §§ 32, 35, 40, lie at the foundation of the mathematics used in land surveying. The fact that certain parts of a triangle determine it shows that it may be possible to compute the other parts when these parts are known. Rules for doing this are found in Chapter III. CONSTRUCTION OF GEOMETRIC FIGURES. 44. The straight-edge ruler and the compasses are the instruments most commonly used in the construction of geometric figures. By means of the ruler straight lines are drawn, and the compasses are used in laying off equal line-segments and also in constructing arcs of circles (§ 12). Other common instruments are the protractor (§ 33) and the triangular ruler with one square corner or right angle. The three tests for congruence of two triangles are of constant use in geometrical constructions. 45. Problem. To find a point luliose distances frmn the extremities of a given seg- \c/n ment are specified. / \ ^ Solution. Let AB be the given seg- ment and let it be required to find a -4 -B point C which shall be one inch from each extremity of AB. Set the points of the compasses one inch apart. With A as 'A center draw an arc m, and with B as a center draw an arc n meeting the arc m in the point C. Then every point in the arc m is one inch from .1 and every point in the arc n is one inch from B (i^ 1'2). Hence C, which lies on botli m and «, is one inch from A and also from B. RECTILINEAR FIGURES. 21 46. EXERCISES. 1. In the preceding problem is there any other point in the plane besides C which is one inch distant from both .4 and B'l If so, show how to find it. 2. Could AB be given of such length as to make the construction in § 45 impossible ? 3. Is there any condition under which one point only could be found in the above construction ? If so, what would be the length of AB'/ 4. Find a point one inch from .4 and two inches from B and dis- cuss all possibilities as above. • 5. Given three segments a, b, e, construct a triangle having its sides equal to these segments. Discuss all possibilities depending upon the relative lengths of the given segments. 47, Problem. To construct an angle equal to a given angle, unthout using the protractor. Solution. Given the angle A. Lay off any distance AB on one of its sides and any dis- tance AC on the other. Draw the segment BC forming the triangle ABC. As in Ex. 5 above, construct a triangle a'b'c' so that A'b' = AB, b'c' = BC, A'C' = AC. Show that A ABC ^ A a'b'c' by one of the tests, and hence that Z A = Z. A', being corresponding angles of con- gruent triangles, § 29. In the above construction, would it be wrong to make AB = AC? Is it necessary to do so ? 22 PLANE GEOMETRY 48. Problem. To construct the ray dividing a given angle into two equal angles, that is, to bisect the angle- Solution. Given the angle A. To construct the riiy bisecting it. On the sides of the angle lay off segments AB and AC so that AB = AC. With B and C as centers and ^ '^ with equal radii construct arcs m and n meeting at D. Draw the segments CD, BD, and AD. Now show that one of the tests for congruence is applicable to make AACD ^ A ABD. Does it follow that Z 1 = Z 2 ? Why ? 49. EXERCISES. 1. Is it necessary in § 48 to make AB = AC^ In this respect com- pare with the construction in § 47. 2. Is any restriction necessary in choosing the radii for the ares m and n ? Ls it possible to so construct the arcs m and n, ^till usiiii;- equal radii for both, that the point D shall not lie within the angle BA a In that case does the ray AD bisect ZjB.l C? 3. By means of § 48 bisect a straight angle. \\'hat is the ray called which bisects a straight angle? In this case what restriction is necessary on the radii used for the arcs m and n '! 4. By Ex. 3 construct a perpendicular to a line at a given point in it. 5. Construct a perpendicular to a segment at one end of it without prolonging the segment and without using the square ruler. SuG(iKsi ION. Let AB be the given segment. Constrnot a right angle A' I" C^' asm Ex. \. Then as in § 17 construct ZABC— /LA'B'C. 50. Definition. A lino wliich is ptM peiidicular to a line- segiiiciit at its middle point is called tlii' perpendicular bisector of the seoinent. RECTILINEAR FIGURES. 23 51. Problem. To construct the perpendicular bisec- tor of a given line-segment. \U, Solution. Let AB be the given segment. ^^ l As in § 45, locate two points, C \, and D, each of which is equally -i'B distant from A and B. /fP Draw the segment CD meeting ^B in O. Then CD is the required perpendicular bisector of AB. To prove this, show that A ACD ^ A BCD. Hence Z3 = Z4. (Why?) By what test can it now be shown that AAOC^ABOC? Hence Z 1 = Z 2. (Why ?) Therefore CO (or CD} is perpendicular to ^B (Why?) and also AO = OB (Why?). It has thus been shown that CD is perpendicular to AB and bisects it, as was required. 52. The steps proved in the above argument are : (ay A ACD ^ A BCD. (6)Z3 = Z4. (o) AAOC^ABOC. (c?) Z 1 = Z 2, and AO = BO. Study this outline with care. What is wanted is the last result (rf). Notice that (d) is obtained from (c), (c) from (6), and (6) from (o). Thus each step depends on the one preceding, and would be impos- sible without it. To understand clearly the order of the steps in a proof as shown by such an outline is of great importance in master- ing it. 53, EXERCISES. 1. In the construction of § 51, is it necessary to use the same radius in locating the points C and D ? 2. Name the isosceles triangles in the figure §51: (a) if the same radius is used for locating C and D, (J) if different radii are used. 24 PLANE GEOMETRY. 54. Problem. To construct a perpendicular to a given straight line from a given point outside the line. / / \ v/ ^ \^ \ / \ / fS^ ^ Solution. Let / be the given line, and P the given point outside it. With P as a center draw an arc cutting the line I in two points, A and B. With A and U as centers, and with equal radii, draw the arcs m and n intersecting in C. Draw the line PC cutting I in the point H. Then the line PC is the perpendicular sought. To prove this, draw the segments PA, PB, CA, CB. Complete the proof by showing that PC is the perpen- dicular bisector of AB, and hence is perpendicular to I from the point P. 55. Problem. To consfrurt a triamjle ichen two silica and the included angle ^ , are nivcn. c v*'- Solution. Let b and c be the / \y \ given sides, and A the giVen angle. / ,. \ As in § 47, construct an angle ^ A' equal to Z.A. On the si(U\s of Z .l' l;iy off AB = c and A'(' = b. Connect B and r. Then A'bc is the required triangle. (Wliy ?) RECTILINEAR FIGURES. 25 56. Problem. Construct a triangle when tioo and the included side are given. Solution. Let /LA and ZS be the given angles, and c the given side. Construct Z.A'=/.A. On one side of Z a' lay off A'b' = c. At b' construct Za'b'k equal to Zb. Let b'k meet the other side of ZA' at C. Then A'b'c is the required triangle. (Why ?) angles 57. EXERCISES. 1. If in the preceding problem two different triangles are con- structed, each having the required properties, how will these triangles be related? Why? 2. If in the problem of § 55, two different triangles are constructed, each having the required properties, how will these triangles be related ? Why? 3. If two triangles are constructed so that the angles of one are equal respectively to the angles of the other, will the triangles neces- sarily be congruent ? 4. If two different triangles are constructed with the same sides, how will they be related ? Why ? 5. Construct an equilateral triangle. Use § 37 to show that it is also equiangular. 58. We have now seen that the three tests for the con- gruence of triangles are useful in making indirect measure- ments of heights and distances when direct measurement is inconvenient or impossible, and also in making numerous geometric constructions. It will be found, as we proceed, that these tests are of increasing usefulness and importance. 26 PLANE GEOMETRT. THEOREMS AND DEMONSTRATIONS. 59. A geometric proposition is a statement af&rming cer- tain properties of geometric figures. Thus : " Two points determine a straight line " and " The base angles of an isosceles triangle are equal " are geometric propositions. A proposition is proved or demonstrated when it is shown to follow from other known propositions. A theorem is a proposition which is to be proved. The argument used in establishing a theorem is called a proof. 60. In every mathematical science some propositions must be left unproved, since every proof depends upon other propositions which in turn require proof. Propositions which for this reason are left unproved are called axioms. While axioms for geometry may be chosen in many dif- ferent ways, it is customary to select such simple propo- sitions as are evident on mere statement. 61. Among the axioms thus far used are the following : Axioms. I. A figure may he moved about in space without changing its shape or size. See § 26. II. Through two points one and only one straight line can be drawn. See §§ 8, 32. III. The shortest distance between tico points is meas- ured along the straight linc-segnicnt coyinccting them. Thus one side of a triangle is less than the sum of tlie other two. IV. // each of two Jigurcs is congruent to the same figure, they are congruent to each other. See §§ 2S, 40. V. // (I, h, c, d are line-segments {or angles) such that (t = b and c = d, then a +c=b + d and a — c = b — d. In the latter casi; \vu suppose u >c, i>(/. S^e §§ 10, 39. RECTILINEAR FIGURES. 27 VI. // a and b are line-segments (or angles) such that a = b, then a xn=:b xn and a-i-n=b-i-n; and if a>b, then a xn>b xn and a-r-n>b-i-n, n being a positive integer. See §§ 10, 39. Note. An equality or an inequality may be read from left to right or from right to left. Thus, a > 6 may also be read ft < a. Other axioms are given in §§ 82, 96, 119, and in Chapter VII. Certain other simple propositions may be assumed at present without detailed proof. These are called prelimi- nary theorems. PRELIMINARY THEOREMS. 62. Two distinct lines can meet in only one point. For if they have two points in common, then by Ax. II they are the same line. 63. All straight angles are equal. § 20, Ex. 1. 64. All right angles are equal. See Ax. VI. 65. Every line-segment has one and only one middle point. See § 51, where the middle point is found by construction. 66. Every angle has one and only one bisector. See § 48, where the bisector is constructed. 67. One and only one perpendicular can be drawn to a line through a point whether that point is on the line or not. See § 20, Exs. 4, 5; § 49, Ex. 4; § 54. 68. The sum of all the angles about a point in a straight line and on one side of it is two right angles. 69. The sum of all the angles about a point in a plane is four right angles. In §§ 68, 69 no side of one angle is to lie inside another. PLANE GEOMETRY. 70. Definitions. Two angles are said to be complementary if their sum is one right angle. Each is then called the complement of the other. Thus, /. a and /. b are complementary angles. Two angles are said to be supple- mentary if their sum is two right angles. Each is then said to be the supplement of the other. Thus, Z 1 and Z 2 are supplementary angles. Two angles are called vertical angles if the sides of one are pro- longations of the sides of the other. Thus, Z 1 and Z 3 are vertical angles, and also Z 2 and Z 4. 71, EXERCISES. 1. What is the complement of 45°V the supplement? 2. I£ the supplement of an angle is liO", find its complement. 3. If the complement of an angle is 21°, find its supplement. 4. Find the supplement of the complement of 30°. 5. Find the angle whose supplement is five times its complement. 6. Find the angle whose supplement is n times its complement. 7. Find an angle whose complement plus its supplement is 110°. 8. If in the first figure Zb = 2 Za, and Zc = Za + Zb, find each angle. 9. If in the second figure Zb = ^Za, Zc—Za + Zb, and Z, is greater than. <, is less than. ^, is less than or equal to. ^, is greater than or equal to. II, parallel, or is parallel to. . Tperpendicular, or I is perpendicular to. lis, parallels. Js, perpendiculars. .'. , therefore or hence, ax., axiom, th., theorem, def., definition, cor., corollary, alt., alternate, ext., exterior, int., interior, hyp., by hypothesis. 32 PLANE GEOMETRY. INEQUALITIES OF PARTS OF TRIANGLES. 81. Definition. If one side of a triangle is produced, the angle thus formed is called an exterior angle of the triangle. Thus, .^1 is an exterior angle of the triangle ^BC. A 82. Axiom VII. If a, b, c are line-sefjments (or angles) such that a>b and h^c,or such that a^b and b>c, then a>c. The proof of the following theorem is shown in full detail as it should be written by the pupil or given orally, except that the numbers of paragraphs should not be required. 83. Theorem. An exterior angle of a triangle is greater than either of the opposite interior angles. B\ \ o Given the A ABC with the exterior angle DBC formed by pro- ducing the side AB. To prove that Z DBC > Z c and also Z DBC > Z A. Proof: Let E be the middle point of BC. Find E by the construction for bisecting a line-segment (§ 51). Draw AE and prolong it, making EF= AE, and draw BF. In tlie two AACK luid FBE, we have by ronstruction CE = EB ami AE = EF. RECTILINMAB FIGURES. 33 Also Z GEA = Z BEF. (Vertical angles are equal, § 74.) .-. AACE^AFBE. (Two triangles which have two sides and the included angle of the one equal respectively to two sides and the included angle of the other are congruent, § 32.) .-. ZC = Z FBE. (Being angles opposite equal sides in congruent triangles, § 29.) But Z DBC > Z FBE. (If an angle is the sum of two angles it is greater than either of them, §39.) .-. Zdbc> Z C. (Since ZDBOZFBE and A FBE = ZC, Ax. VII, § 83.) In order to prove Z DBC > Z A, prolong CB to some point G. Then Z ABG = Z dbc. (Vertical angles are equal, § 74.) Now bisect AB, and in the same manner as before we may prove Zabg>ZA. .-. Zdbc> Za. (Since ZDBC = Z ABG and Z ABG >Z A, Ax. VII, § 82.) For the second part of the proof let H be the middle point of AB. Draw CH and prolong it to K, making Off = HK. Let the student draw the figure for the second part of the proof and give it in full. Hereafter more and more of the details of the proofs will be left for the student to fill in. When reference is made to a paragraph in the text or when the reason for a step is called for, the complete state- ment of the definition, axiom, or theorem should be given by the student. 34 PLANK GEOMETRY 84. Theorem. If tivo sides of a triangle are un- equal, the angles opposite these sides are unequal, the greater angle being opposite the greater side. Given AjliSC in which AC>BC. . To prove that Z ABC > /.A. Proof : Lay off CD= CB and draw BO. Now give the reasons for the follow- ing steps: (1) Z ABC > Z DBC. r§39) (2) Z DBC = Z CDB. (§37) (3) ZCDB > ZA. (§83) (i) .-. Z ABO ZA. (§82) 85. Theorem. // two angles of a triangle arc un- equal, the sides opposite them are unequal, the greater side being opposite the greater angle. Given A ABC in which ^B>ZA. To prove that b > a. Proof: One of the following three statements must be true: (1) 5 = a, (2) ba. But it cannot be true that b = d. for in that case Zb =Z a. contrary to the hypothesis that Z n > Z A. And it cannot be true that b < a, for in that case Zb a. (§-) (§ 84) RECTILINEAR FIGURES. 35 86. The above argument is called proof by exclusion. Its success depends upon being able to enumerate all the possible cases, and then to exclude all but one of them by showing that each in turn leads to some contradiction. 87. EXERCISES. 1. The hypotenuse of a right triangle is greater than either leg. 2. Show that not more than two equal line-segments can be drawn from a point to a straight line. Suggestion. Suppose a third drawn. Then apply §§ 37, 83, 84. 3. Show by joining the vertex A of the triangle ABC to any point of the side BC that ZB + ZC<2 rt.A. Use § 83. 4. If two angles of a triangle are equal, the sides opposite them are equal. Use §§ 84, 86. 5. Either leg of an isosceles triangle is greater than half of the base. 6. Show that an equiangular triangle is equilateral, and conversely. THEOREMS ON PARALLEL LINES. 88. A straight line which cuts two straight "lines is called a transversal. The various angles formed are named as follows : Z 4 and Z 5 are alternate-interior angles ; also Z3 and Z6. Z 2 and Z 7 are alternate-exterior angles ; also Z 1 and Z8. Z 1 and Z 5 are corresponding angles ; also Z 3 and Z 7, Z2 andZ6, Z4 and Z 8. Z 3 and Z 5 are interior angles on the same side of the transversal; also Z 4 and Z 6. 89. Definition. Two complete lines which lie in the same plane and which do not meet are said to be parallel. Two line-segments are parallel if they lie on parallel lines. 36 PLANE GEOMETRY. 90. Theorem. // two lines cut by a transversal have equal alternate interior angles, the lines are parallel. Given the lines /i and l^ cut by t so that Z 1 = Z 2. To prove that Zj II l^. Proof : Suppose the lines Zj and l^ were to meet on the right of the trans- versal. Then a triangle would be formed of which Z 1 is an exterior angle and Z 2 an opposite interior angle. This gives an exterior angle of a triangle equal to an opposite interior angle, which is impossible. (Why ?) Repeat this argument, supposing l^ and l^ to meet on the left side of the transversal. Hence Zj and l^ cannot meet and are parallel (§ 89). 91. The type of proof used here is called an indirect proof. It consists in showing that something impossible or contradictory results if tlie theorem is supposed not true. 92. Theorem. If two lines cut by a transversal have equal corresponding angles, the lines are parallel. Given the lines 7, and L cut by t so that A Z2 = Z3. . A^ /. To prove that 1^ \\ l^. Proof : Quote the authority for each of the following steps : Z :5 = Z 2. ^;! = ^1. (§74^ .-. ^1 =^2. (Ax. IV) ••• h II h- (§90) RECTILINEAR FIGURES. 37 93. Theorem. // two lines cut by a transversal have the sum of the interior angles on one side of the trans- versal equal to two right angles, the lines are parallel. A Given /i and fa cut by t so that Z.i + /.2=2rt. A. To prove that ?j II \. Proof: Z 4 is supplementary to Z 1 and also to Z 2. (Why ?) .-. Z1 = Z2. (Why?) .-. \\\l^. (Why?) 94, EXERCISES. 1. Show that if each of two lines is perpendicular to the same line, they are parallel to each other. 2. Let ABC be any triangle. Bisect BC at D. q -g. Draw AD and prolong it to make DE = AD. Draw CE. Prove CE II AB. 3. Use Ex. 2 to construct a line througb a A given point parallel to a given line. Suggestion. Let AB be a segment of the given line and let C be the given point. Draw CA and CB and proceed as in Ex. 2. 95. Exs. 2 and 3 above show that through a point P, not on a line I, at least one line ?j can be drawn parallel to I. It seems reasonable to suppose that no other line Zg can be drawn through P parallel to I, al- though this cannot be proved from the preceding theorems. See § 60. Hence we assume the following : 38 PLANE GEOMETRY. 96. Axiom VIII. Through a 'point not on a line only one straight line can he drawn 'parallel to that line. Historical Note. This so-called axiom of parallels has attracted more attention than any other proposition in geometry. Until the year 1829 persistent attempts were made by the world's most eminent mathematicians to prove it by means of the other axioms of geometry. In that year, however, a Russian, Lobachevsky, showed this to be im- possible and hence it must forever remain an axiom unless some other equivalent proposition is assumed. 97. Theorem. // two parallel lines are cut by a transversal, the alternate Interior angles are equal. Given /j II l^ and cut by t yf To prove that Z 1 = Z 2. "' — r--,f/ z, Proof : Suppose Z 2 not equal to Z 1. / ' Through P draw ?g, making Z 3 = Z1. — -/— ?, .-. ZgllV (Why?) / But by hypothesis l^ \\ l^ and thus we have through P two lines parallel to ?j, which is contrary to Ax. VIII. Therefore, the supposition that Z "2 is not equal to Z 1 leads to a contradiction, and hence Z 1 = Z2. 98. Compare this theorem with that of § 90. The hy- pothesis of either is seen to be the conclusion of the other. When two theorems are thus related, each is said to be the converse of the other. Other pairs of converse theo- rems thus far are those in § -37 and Ex. 4. § 87; §§ 75 and 76, and §§ 84 and 85. The converse of a theorem is never to be taken for granted without proof, since it does not follow tluit a state- ment is true because its converse is true. Thus, it is true that if a Iriamjle is i i/uilaleral, it is also isosceles, but the converse, if a triangle is isosceles, it is also equilateral, is not true. RECTILINEAR FIGURES. 39 99. Theorem. If two parallel lines are cut by a transversal, the sum of the interior angles on one side of the transversal is two right angles. Suggestion. JMake use of the preced- ing theorem and give the proof in full. Of what theorem is this the converse ? 100, EXERCISES. 1. State and prove the converse of the theorem in § 92. 2. Prove that if two parallel lines are cut by a transversal the alternate exterior angles are equal. Draw the figure. 3. State and prove the converse of the theorem in Ex. 2. 4. If a straight line is perpendicular to one of two parallel lines, it is perpendicular to the other also. 5. Two straight lines in the same plane parallel to a third line are parallel to each other. Suppose they meet and then use § 96. 6/5 ^ h h D / 6. If ?i II ^2 II Ig and if Zl = 30°, find the other angles in the first figure. 7. If Zj II Zg, how are the bisectors of .^1 and Z3 related? Of Z3 andZ4? OfZlandZ2? Use §102 for the last case. 8. If Ij II Z,, and ^ 0= OB, show that DO=OC. V^ '* State this theorem fully and prove it. a^ '2 9. If ZilU2andZ2 = 5Zl, findZiandZS. '' 10. If two parallel lines are cut by a transversal, the sum of the exterior angles on one side of the transversal is two right angles. 11. State and prove the converse of the preceding theorem. 40 PLANE GEOMETRY. APPLICATIONS OF THEOREMS ON PARALLELS. 101. Problem. Through a given point to construct a line parallel to a given line. Given the line / and the point P outside of it. To construct a line Zj through p || to I. Construction. Through P draw any line making a con- venient angle, as Z 1 with I. Through P draw the line Ip making Z2 = Z1(§47). Then^ilU. Proof : Use the theorem, § 92. Hereafter all constructions should be described fully as above, followed by a proof that the con- struction gives the required figure. 102. Theorem. The siim of the angles of a triangle is equal to two right angles. e Given A ABC with Z 1, Z 2, Z 3. c \ To prove that V~\ \ U g<\5 D Proof: Prolong ^B to some point Z). -i B Through B draw BE II AC. Then Z5-|-Z4 + Z3=2rtJ. (Wliy?) But Z 4 = Z 2 and Z 5 = Z 1. C^^''iy ?) Hence, replacing Z5 and Z-t by their equals. Zl and Z 2, we have Zl + Z2 + Z3 = 2 rt zi. Historical Note. This is one of the famous theorems of geome- try. It was known liy Pytliagoras (500 B.C.). but special oases were known much earlier. The figure used here is the one given bv Aristotle and Euclid. As is aiiparent, the proof depends upon the theorem, § !t7, and thus indirectly upon Axiom VIII. The interdepend- ence of these two propositions has been studied extensively during the last two centuries. BECTILINEAB FIGURES. 41 103. Theorem. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The proof is left to the student. Compare this theorem with that of § 83. 104. Definition. A theorem which follows very easily from another theo- rem is called a corollary of that theorem. U.ff. the theorem in § 103 is a corollary of that in § 102. 105. EXERCISES, 1. Find each angle of an equiangular triangle. 2. If one angle of an equiangular ti-iangle is bisected, find all the angles in the two triangles thus formed. S.'If in.aAABC, AB =^C'and Z^ =ZB + Z C, find each angle. 4. If in the figure AB = AC and Z 4 = 120°, find Zl, Z2, Z3. 5. If one acute angle of a right triangle is 30°, what is the other acute angle? If one is 41° 23'? 6. If in two right triangles an acute angle of one ^ is equal to an acute angle of the other, what can be said of the remaining acute angles. What axiom is involved? 7. If in two right triangles the hypotenuse and an acute angle of one are equal respectively to the hypotenuse and an acute angle of the other, the triangles are congruent. Prove in fuU. 8. Can a triangle have two right angles? Two obtuse angles? Can the sum of two angles of a triangle be two right angles? What is the sum of the acute angles of a right triangle? 9. If two angles of a triangle are given how can the third be found? If the sum of two angles of one triangle is equal to the sum of two angles of another, how do the third angles compare? 10. Prove the theorem of § 102, using each of the figures in the margin. The first of these figures was used by Pythagoras. A 2\l C 42 PLANE GEOMETRY. 106. Definition. An angle viewed from the vertex has a right side and a left side. 107- Theorem. If two angles have their sides respectively parallel, right side to rigid side, and left side to left side, the angles are equal. Given ^ 1 and ^ 2 such that a II a' and & II &' . To prove that Z 1 = Z 2. Proof : Produce a and b' till they meet, forming Z 3. Complete the proof. Wliy do a and V meet when produced ? Make a proof also by producing h and a' till they meet. 108, Theorem. If two angles have their sides re- spectively perpendicular, right side to right .-ude, aiid left side to left side, the angles are equal. Given Z 1 and Z 2 such that aXa' and b ± b'- To prove that Z 1 = Z2. Proof : Produce a and a' till they meet in O. Through O draw a line parallel to b. Now show that Z 2 = Z 3 since each is the complement of Z4. Complete the proof. -,0/^ 109. EXERCISES. If two angles have their sides respectively parallel, or perjiendicular, riglit side to left side, and left side to riylit side, the angles are supple- mentary. In each figure Z 1 = Z:!. (Why?) lIiSToKlCAi. NoTK. The theorems of §§107, 108, 109 are not found in Kuelid's Elements. It b' / .A 3V / b/ A u RECTILINEAR FIGURES. 43 OTHER THEOREMS ON TRIANGLES. 110. Theorem. // in two right triangles the hypote- nuse and one side of one are equal respectively to the hypotenuse and one side of the other, the triangles are congruent. b b' b Given the right A ABC and A'B!C', having AB = A'S and BC=B'C'. To prove that A ABC s AA'b'c'. Proof : Place the triangles so that BC and b'c' coincide, and so that A and A' are on opposite sides of BC. Then AC and CA' lie in a straight line (Why ?), and A aba' isosceles (Why?). Hence, show that Z A = Z a' and complete the proof. 111. Corollary. If from a point in a perpendicular to a straight line equal oblique segments are drawn to the line, these cut off equal distances from the foot of the perpendicular, and make equal angles with the perpendic- ular and with the given line. Give the proof in full, as of an independent theorem. 112, Theorem. If from a point in a perpendicular to a line oblique segments are drawn, cutting off equal distances from the foot of the perpendicular, these seg- ments are equal, and make equal angles with the perpen- dicular and with the given line. Give the proof in full, using A ABA' of § 110, 44 PLANE GEOMETRY. 113. Theorem. The perpendicular is shorter than any oblique segment from a point to a line. Suggestion. Show by § 102 that the angle opposite the oblique segment in the triangle formed is greater, than either of the other angles, and then make use of § 85. 114. The distance from a point to a line means the shortest distance, and hence is measured on the perpen- dicular. 115. Theorem. If from a point in a perpendicular to a line segments are drawn cutting off vnequal dis- tances from the foot of the perpendindar, the segments are unequal, that segment being the greater lohich cuts off the greater distance. Given AO ± BC and OC > OB. To prove that AC > AB. Proof : Let ob' = OB and draw Ab'. Then, Zob' A < rt. Z, and .-.Z AB'O rt. Z. Also Z OCA < rt. Z. .-.aoab'. ■ •- AC > AB since Ab' = AB. Give reasons for each step. 116. Theorem. // frorn a point in a perpendicular to a line unequal segments are drawn, these cut off unequal distances from the foot of the perpendicular, the greater segment cutting off the greater distance. Suggestion. Using the figure of § ll.'i and the liypoth- esis that AC > AB, show that two of the following state- ments are impossible: (1) OC = Oli; (-2) oc OB. See § 8li. RECTILINEAR FIGURES. 45 117. Theorem. // in two triangles two sides of the one are equal respectively to two sides of the other, but the included angle of the first is greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given A ABC and A'BC in which AB=A'&, BC = BC and To prove that AC > A'c'. Proof: Place Aa'b'c' on A ABC so that a'b' coincides with its equal AB and C' is on the same side of AB as C. Let BD bisect Z g'bc, meeting AC in I) . Draw DC'. Then ABDC' ^ABDC. (Why?) .-.DC' = DC and AD + DC' = AC. (Why?) But AD + DC'>AC'. (Ax. Ill, § 61) AOAC'. (Ax. VII, §82) 118, Theorem. // in two triangles two' sides of the one are equal to two sides of the other but the third side of the first is greater than the third side of the second, then the included angle of the first is greater than the included angle of the second. Suggestion. Using the figure of § 117 and the hypothesis that AOA'c', show that two of the three following statements are impossible: (1)ZB = ZB'; (2)ZbZb'. See § 86. 46 PLANE GEOMETRY. 119. Axiom IX. // a, b, c, d are line-segments (or angles) such that a>b and c = d or such that a>b and c>d, then a + ob + d. Also if a>b and c^d, then a — ob — d, provided a>c and b>d. (See §§ 10, 39.) 120. Theorem. The sum of the segments drawn from a point within a triangle to the extremities of one side is less than the sum of the other two sides. c 'B A Given the point O within the A ABC. To prove that AO + OB <:AC+ CB. Proof: AO + OE .•. AO + OE^- OB J, the arc will cut AK only once on the right of A, and hence only one triangle will be found. Repeat this construction, making a separate figure for each case. Make the construction when Z ^ is a right angle. Are all three cases possible then? Make the construction when Z A is obtuse. What cases are possible then ? 48 PLANE GEOMETRY. 123. EXERCISES. 1. A carpenter bisects an angle A as follows: Lay oS. AB = AC. Place a steel square so that BD = CD as shown in the figure. Draw the line A D. Is this method correct ? Give proof. Would this method be correct if the square were not right-angled at Z)? 2. In the triangle ABC, AC = BC. The points D, E, F are so placed that A D = BD a.ndAF = BE. Compare DE and DF. Prove your conclusion. 3. If in the figure Z 2 - Z 1 = 15°, and Z 1 = 120, find each angle of the triangle. .B E 4. If in A.^BC, AC = BC, and ii AC is extended to D so that AC =^ CD, prove that DB is perpendicular to AB. 5. In A ABC, CA = CB, AD = BE. Prove A ADB ^ A ABE. 6. In the triangle KLN, NM is perpendicular to KL, and KM = MN = ML. Prove that KLN is an isosceles right triangle. 7. If in the isosceles /\ ABC a point D lies in the base, and Z 1 = Z 2, determine whether there is any position for D such that DE = DF. 8. If the bisectors AD and BE of the base angles of an isosceles triangle ABC meet in 0, what pairs of equal angles are formed? What pairs of equal segments ? Of congruent triangles? 9. If the middle points of the sides of an equi- lateral triangle are connected iis .shown in the fig- ure, compare the resulting four triangles. BECTILINEAB. FIGURES. 49 10. Triangle ABC is equilateral. AD = BE ■■ Compare the triangles DBE, ECF, FAD. CF. AD B 11. Two railway tracks cross as indicated in the figure. What angles are equal and what pairs of angles are supplementary? State a theorem involved in each case. G 12. In a, A ABC does the bisector oi ZA also bisect the side BC (1) if AC = BC hat AC S. can be answered by measurini; sides instead of angles. Verify the results by this process. 2. \Vhich two of the theorems §§ lof», Itl, 1 1'J are converse? State in detail the hypothesis and conclusion of eaeh. RECTILINEAR FIGURES. 57 144. Theorem. If a quadrilateral has one pair of opposite sides equal and parallel, the figure is a parallel- ogram, p ^ '^^^n 7 2'-,./ A B Given. (State all the items given in the hypothesis.) To prove. (State what needs to be proved in order to show that ABCD is a parallelogram.) Proof : From the data given prove that A ABC ^ A EDO. Use corresponding angles of these congruent triangles to show that the other two opposite sides are parallel. Hence show that the figure is a parallelogram. Write out this demonstration in full. Could the theorem be proved equally well by drawing the other diagonal ? If so, draw it and give the proof. 145. EXERCISES. 1. Prove that if the diagonals of a quadrilateral bisect each other it is a parallelogram. What is the converse of this proposition? 2. Show that if two intersecting line-segments bisect each other, the lines joining their extremities are parallel. 3. The parallel lines ?j and I2 are cut by a trans- versal AB. AC a:nd AD bisect Z2 and /^l respec- tively. Prove that A CBA and DBA are isosceles. Compare the segments CB and BD. 4. If in an isosceles right triangle ABC the bisectors' of the acute angles meet at 0, find how many degrees in the Z 1 thus formed. 58 PLANE GEOMETRY. 146. Theorem. // the two diagonals of a 'parallelo- gram are equal, the figure is a rectangle. D Given O ABCD with AC = BD. To prove that Z 1 = Z 2 = Z 3 = Z 4 = rt. Z. Proof : Show that A ABD ^AABC .-. Zl = Z3. But Z 1 + ^ 3 = 2 rt. A. .-. Z 1 = Z 3 = rt. Z. In like manner prove that Z 2 = Z 4 = rt. Z. Hence the figure is a rectangle (§ 135). (Why ?) (Why ?; (Why?) 147. EXERCISES. 1. State and prove the converse of the theorem, § 146. 2. Show that a diameter of a parallelogram passes through the in- tersection of its diagonals. See § 1:57. . Suggestion. Show that it bisects each diagonal. 3. Prove that the diagonals of a square are perpendicular to each other. 4. Prove that the diagonals of a rhombus are perpendicular to each other. 5. Does the same proof apply to Exs. 3 and 4? 6. Are the diagonals of a square equal ? Is this ti-ue of a rhom- bus? Prove each answer correct. 7. Uo the diagonals of a square bisect each other? Is this true of a rhombus? Of a trapezoid? 8. SIkuv that if two adjacent angles of a parallelogram are equal, the figure is a rectangle. RECTILINEAR FIGURES. 59 148. Theorem. Two parallelograms having an angle and the two adjacent sides of one equal respectively to an angle and the two adjacent sides of the other are con- gruent. ^ ^ ^, ^, A B A' B' . Given [S ABCD and A'^C'D^ such that AB = A'^, AD = A' If a.n&/LA = A'. To prove that O ABCD ^ O a'b'g'd'. Proof: Apply EJ ABCD to O a'b'c'd' so as to make Z.A coincide with ^A\ AB falling on a'b', and AD on A>D'. Then BC takes the direction b'g', since AB = Z.b', being supplements of the equal angles A and A'. (Why ?) G falls on c', since BC = b'c', being segments equal to the equal segments AD and a'd'. (Why ?) Hence CD coincides with g'd'. (Why ?) Therefore O ABCD ^ O a'b'g'd', since they coincide throughout. 149. EXERCISES. 1. Are two parallelograms congruent if they have a side and two adjacent angles of the one equal respectively to a side and two adja- cent angles of the other ? Draw figures to illustrate your answer. 2. Are two parallelograms congruent if they have four sides of one equal to four sides of the other? Show why, and draw figures to illustrate. 3. Compare the theorem of § 148 and Exs. 1 and 2 preceding with the tests for congruence of triangles. 4. Prove that the opposite angles of a parallelogram are equal. 5. State and prove the converse of Ex. 4, 60 PLANE GEOMETRY. OTHER THEOREMS APPLYING PAEALLELS 150. Theorem. // a line bisects one side of a triangle and is parallel to the base, then it bisects the other side and the included segment is equal c to one half the base. Given a line / II AB in A ABC such y^ ^ ^ that AD = DC. To prove that be = EC and jt DE=^AB. Proof : Draw EF through E parallel to CA. Now show that AFED is a parallelogram, and that Adec^Afbe, from which CE= EB, DE= FB, DE= AF, and AB = af+fb = -2de, or I)E=-^AB. State the reasons for each step. * 151. Theorem. A segment connecting the middle points of two sides of a triangle is parallel to the third side and equal to one half of it. Given A ABC and the segment DE such that AD = DC and CE = EB. See figure in § 150. To prove that DE II AB and DE = \ AB. Proof: Since each of the sides AC and BC has but one middle point (§ 65), it follows that there is but one seg- ment DE bisecting both these sides. But by § 150 a certain segment paniUel to the base fulfills this condition. Hence, DE is parallel to the base AB. Then, DE = ^AB as in § 150. RECTILINEAR FIGURES. 61 152. Theorem. // a segment is parallel to the bases of a trapezoid and bisects one of the non-parallel sides, then it bisects the other also and is equal to one half the sum of the bases. d ^c E/ "O A B Given the trapezoid ABCD in which AE=ED and EFWAB. To prove that BF= FC and EF= ^{AB + DC'). Proof : Draw the diagonal AC meeting EF in o. In A 4CZ), 40 = OC and ^0 = 1 DC. (Why?) In A ABC, BF = FC and OF=^A B. (Why ?) Adding, E0+ OF = EF= ^(^AB + DC). (§ 61, V) Prove this theorem also by drawing the diagonal BD. State a similar theorem for a parallelogram and make a proof for that case which is simpler than the above. 153. Theorem. If a segment connects the middle points of the two non-parallel sides of a trapezoid, it is parallel to the bases and equal to one half their sum. Proof : The argument is similar to that in § 151. Give it in full. 154. EXERCISES. 1. Does a theorem similar to that of § 153 hold for a parallelogram ? If so, state it and give a simpler proof in this case. 2. If in the figure DE, FO, HI, etc., are parallel to AC and if FE, HO, JI, etc., are parallel to BC, find the sum of BD, DE, EF, FG, GH, etc. How, if at all, does the length of AB enter into the solution? Suggestion. Produce GF, IH, KJ, to meet BC, forming ZI7. Then show that BD + EF + GH, etc., = BC, and similarly ED + (3F + IH, etc., = AC aTk' 62 PLANE GEOMETRY. 155. Theorem. // a series of 'parallel lines intercept equal segments on one transversal, they intercept equal segments on every transversal. Given the parallel lines /i, i,j /g, l^, cutting the transversal fi sotheitAB = BC=CD. To prove that on the transversal t^ EF = fg = gh. Proof : The figure ACGE is a parallelogram or a trapezoid according as t^ II t^ or not. In either case BF, which bisects AC, also bisects EG (§ 1^2)- .-. EF=FG. Similarly, in the figure DHFB, FG = GH. .-. EF= F0= GH. 156. Theorem. The three altitudes of a triangle vieet in a point. p ,a' Outline of Proof: Through each vertex of the given trianij^le ABC draw a line parallel to the opposite side, forming a triangle A'B'cf. RECTILINEAR FIGURES. 63 Show that ACA'b, and ab'cb are parallelograms, and hence that b'c = ab = ca'. That is, C is the middle point of A'b'. In the same manner show that A and B are the middle points of b'c' and a'c' respectively. Also show that the altitudes of A ABC are the perpendicular bisectors of the sides of A a' b'c', and therefore' meet in a point (§132). 157. Definition. A segment connecting a vertex of a triangle with the middle point of the opposite side is called a median of the triangle. 158. Theorem. The three medians of a triangle meet in a point which is two thirds the distance from each ver- tex to the middle point of its opposite side. o A H B Given A ABC with medians BD and AE meeting in O. To prove that the median from c also passes through o, and that AO = ^AE, -BO = | BD, and CO = | CH. Outline of Proof : Taking F and G, the middle points of OB and OA respectively, use §§ 151 and 144 to show that the figure GFED is a parallelogram, and hence that DO = 0F= FB and EO = 0G= GA. That is, O trisects AE and BD. In the same way we find that AE and CH meet in a point o' which trisects each of them. Hence O and o' are the same point. Therefore the three medians meet in a point which trisects each of them. G4 PLANE GEOMETRY. 159. EXERCISES. 1. If one angle of a parallelogram is 120", how many degrees in each of the other angles ? 2. If the angles adjacent to one base of a trapezoid are equal, then those adjacent to the other base are equal. 3. If the angles adjacent to either base of a trapezoid are equal, then the non-parallel sides are equal and the trapezoid is isosceles. 4. In an isosceles trapezoid the angles adjacent to either base are equal. 5. Divide a segment ^B into three equal parts. ^ Suggestion. From A draw a segment A C, and on it lay off three equal segments, AD, DE, ^■' JSF(§33). Draw FB and construct EO and DH each parallel to FB. Prove AH = HG = GB. 6. To cut braces for a roof, as shown in the figure, a carpenter needs to know the angle DBC when the angle DAB is given, it being given that AB = AD. Show how to find this angle. (See § 123, Ex. U.) 7. If each of the perpendicular bisectors of the sides of a triangle passes through the opposite vertex, what kind of a triangle is it ? If it is given that two of the perpendicular bisectors of sides pass through the opposite vertices, what kind of a triangle is it ? If only one? 8. Find the locus of the middle points of the segments joining a vertex of a triangle to all points on the ojiposite side. 9. Given a line I and a point P not in the line. Find the locus of the middle points of all segments drawn from P to /. 10. The length of the sides of a triangle are 12, 14, 16. Four new triangles are formed by comieoting- middle points of the sides of this triangle. What is the sum of the sides of these four triangles ? 11. Draw any segment PA nioeting a line Mn .1 . p \ ^ Lay off AB on I. AVith P and Ji as centers and with T yv AB and AP as radii respectively, strike arcs meet- / ing in C. Draw PC, and prove it parallel to /. a b^" RECTILINEAR FIGURES. 65 12. If the side of a triangle is bisected by the perpendicular upon it from the opposite vertex, the triangle is isosceles. 13. State and prove the converse of this theorem. 14. If in a right triangle the hypotenuse is twice as long as one side, then one acute angle is 60° and the other 30". Suggestion. Let D be the middle point of AB. Use Ex. 12 and the hypothesis to show that AAC'D is equilateral. Prove the converse by drawing CD so as to make /.BCD = /.B. 15. A stairway leading from a floor to one 12 feet above it is con- structed with steps 8 inches high and 10 inches wide. What is the length of the carpet re- quired to cover the stairway, allowing 10 inches for the last step, which is on a level with the upper floor. 16. A stairway inclined 45° to. the horizon- tal leads to a floor 15 feet above the first. What .is the length of the carpet required to cover it if each step is 10 inches high ? If each is 12 inches? If each is 9 inches? Can this problem be solved without know- ing the height of the steps? Is it necessary to know that the steps are of the same height? 17. If the vertex angle of an isosceles triangle is 60°, show that it is equilateral. 18. By successively constructing angles of 60° divide the perigon about a point into six equal angles. (This is possible because 360 -T- 6 = 60.) With as a center construct a circle cutting the sides of these angles in points A, B, C, D, E, F. Draw the segments AB, BC, etc. Show by Ex. 17 that each of the six triangles thus formed is equilateral. Show also that ABCDEF is equiangular and equi- lateral, that is, AB = BC, etc., and AABC = Z BCD, etc. 66 PLANE GEOMETRY. POLYGONS. 160. Definitions. A polygon is a figure formed by a series of segments, AB, BC, CD, etc., leading back to the starting point A. The segments are the sides of the polygon and the points A, B, C, D, etc., are its vertices. The angles A, B, c, D, etc., are the angles of the polygon. E Convex. Equiangular. Equilateral. Regular. A polygon is convex if no side when produced enters it. Otherwise it is concave. Only convex polygons are here considered. A polygon is equiangular if all its angles are equal and equilateral if all its sides are equal. A polygon is regular if it is both equiangular and equi- lateral. A segment connecting two non-adjacent vertices is a diagonal of the polygon. The perimeter of a polygon is the sum of its sides. 161. Theorem. The sum of the angles of a polygon having n sides is (2 u — 4) right angles. Proof : Connect one vertex with each of the other non-adjacent vertices, thus forming a set of triangles. Evidently the sum of the angles of tliese triangles equals the sum of the angles of the polygon. RECTILINEAR FIGURES. 67 Now show that if the polygon has n sides there are (n — 2) triangles. The sum of the angles of one triangle is 2 rt. A. Hence, the sum of the angles of all the tri- angles, that is, the sum of the angles of the polygon, is 2(m - 2) = (2 jz - 4) rt. A. 162. Theorem. The sum of the exterior angles of a polygon, formed by producing the sides in succession, is four right angles. Outline of Proof : The sum of both exterior and interior angles is 2 «, rt. A. (Why ?) The sum of the interior angles is (2 n — 4) rt. A. (Why ?) Hence, the sum of the exterior angles is 4 rt. A. (Why ?) Write out the proof in detail, using the figure. 163. EXERCISES. 1. What is the sum of the angles of a polygon of 3 sides? of 4 sides? of 5 sides? of 6 sides? of 10 sides? of 18 sides? 2. Find each angle of a regular polygon of 3 sides, 4 sides, 5 sides, 6 sides, 8 sides, 14 sides, n sides. 3. Construct a regular triangle, thus obtaining an angle of 60°. 4. Construct a regular quadrilateral. "What is its common name? 5. Prove that a regular hexagon ABCDEF may be constructed as follows : Let A be any point on a circle with center 0. With A as center and OA as radius describe arcs meeting the circle in B and in 'F. With B as center and the same radius describe an arc meeting the circle in C, and so for points to i3»and E. See § 159, Ex. 18. 68 PLANE GEOMETRY. -A SYMMETRY. 164. Two points A and A' are said to be symmetric points with respect to a line Z if ^ is the per- pendicular bisector of the segment AA' . A figure is symmetric with respect to an axis I if for every point P in the figure there is also a point P' in the figure such that P and P' are symmetric points with respect to I. This is called axial symmetry. Two separate figures may have an axis of symmetry between them. 165. Theorem. Two figures which are symmetric with respect to a line are congruent. i Given two figures F and P symmetric with respect to a line /. To Prove that f^f'. Proof: This is evident, since, by folding figure F over on the line I as an axis, every point in F will fall upon a corresponding point in f' (Why ?). 166. Corollary. If points A and A' and also B and B' are symmetric with re- spect to a line I, then the segments AB b< ^ i^- and A'B' are symmetric with respect to I. 167. EXERCISES 1. How many axes of symmeti'v has a square? A rectangle? A rhombus? An isosceles trapezoid? 2. If a diagonal of a rectangle is an axis of symmetry, what kind of a rectangle is it? BECTILINEAR FIGURES. 69 3. If a triangle has an axis of symmetry, what kind of a triangle is it? Assume that the axis passes through one vertex. 4. If a triangle has two axes of symmetry, what kind of a triangle is it? '' \ 5. How many axes of symmetry has an equilateral triangle? 6. How many axes of symmetry has a regular pentagon (five-sided figure) ? 7. How many axes of symmetry has a regular hexagon ? 8. Show that one figure of § 40 has an axis of symmetry. State this as a theorem. 168. Problem. Given a polygon P and a line I not meeting it, to construct a polygon P' such that P and P' shall be symmetric with respect to I. Solution. Let A, B, C, D, E be the vertices of the given polygon P, and / the given line. Construct A', b', c', 1)', e' symmetric respectively to A, B, C, B, E with respect to the line I. Then the polygon P' formed by joining the points A', b', c', d', e', a' in succession is symmetric to P. Proof : Give the proof in full. Figures having an axis of symmetry are very common in all kinds of decoration and architectural construction. 70 PLANE GEOMETRY. 169. Two points A and A' are symmetric with respect to a point O if the segment AA' is bisected by o. A figure is symmetric with respect to a point if for every point P of the figure there is a point P' also in the figure such that P and p' are symmetric with respect to O. Such a figure is said to have central sym- m metry with respect to the point. The point [\~~,o,''' is called the center of symmetry. A circle has central symmetry. Two separate figures may have a center of symmetry between them. 170. EXERCISES. 1. Prove that if A and A' and also B and K' are symmetric with respect to a point O, tlien the segments AB and A'B' are symmetric with respect to 0. 2. Prove that if the triangles. I BC and A'B'C are symmetiic with respect to a point 0, then they are congruent. 171. Problem. Given a polygon P and a point outside of it, to construct a polygon f symmetric to P with respect to 0. Solution. Construct points symmetric to the vertices of P. Connect these points, forming the polygon p\ and prove this is the poly^fon sought. D' BJSCTILINEAB. FIGURES. 71 172. Theorem. 7/ a figure has two axes of symmetry at right angles to each other, their point of intersection is a center of symmetry of the figure. Outline of Proof : It is to be shown that for every point P in the figure, a point p" also in the figure can be found such that PO = P"o and POP" is a straight line. Draw PP' X I, p'p" ± I', P"p"' ± I and connect the p/_ points P"' and P. Now use the hypothesis that I ± V, and each an axis of symmetry, to show that pp'p"p'" is a rectangle of which dd" and d'd'" are the diameters. Hence o, the inter- section of DD" and d'd"', bisects the diagonal PP", making p" symmetric to P with respect to o, § 147, Ex. 2. Give the proof in full. 173. EXERCISES. 1. Has a square a center of symmetry? has a rectangle? 2. If a parallelogram has a center of symmetry, does it follow that it is a rectangle ? 3. Has a trapezoid a center of symmetry? has an isosceles trapezoid ? 4. If two non-parallel straight lines are symmetric with respect to a line /, show that they meet this line in the same point and make equal angles with it. (Any point on the axis of symmetry is regarded as being symmetric to itself with respect to the axis.) 5. If segments AB undA'B' are symmetric with respect to a point O, they are equal and parallel. 6. Has a regular pentagon a center of symmetry ? See figure of Ex. 6, § 167. 7. Has a regular hexagon a center of symmetry ? Ex. 7, § 167. 8. Has an equilateral triangle a center of symmetry? 72 PLANE GEOMETRY. METHODS OF ATTACK. 174. No general rule can be given for proving theorems or for solving problems. In the case of theorems the following suggestions may be helpful. (1) Distinguish carefully the items of the hypothesis and of the conclusion. It is best to tabulate these as suggested in § 79. ( 2) Construct with care the figure described in the hy- pothesis. The figure should be as general as the terms of the hypothesis per- mit. Thus if a triangle is called for Imt no special triangle is men- tioned, then a scalene triangle should be drawn. Otherwise some particular form or appearance of the figure may lead to unwarranted conclusions. (3) Study the hypothesis with care and determine whether any auxiliary lines may assist in deducing the properties required by the conclusion. Study the theorems previously proved in this respect. A careful review of these proofs will lead to some insight as to liow they were evolved. 175. Direct Proof. The majority of theorem.'? are proved by passing directly from the hj'pothesis to the conclusion by a series of logical steps. 'I'liis is called direct proof. It is often helpful in disivrering a direct proof to trace it backward from tlie conclusion. Thus, we may observe that the oonolusion C follow.-s if st^itement B is true, and that B follows if .V is true. If tlieu we can show thai X is true, it follows that B and C are true and the theorem is proved. Having thus discovered a proof, wc may then start from the beginning and follow it directly throuijh. As an example consider the following theorem: BECTILINEAB FIGURES. 73 The segments connecting the middle points of the opposite sides of any quadrilateral bisect each other. Given segments AB and CD connect- ing the middle points of the quadrilat- eral PQRS. To prove that AB and CD bisect each other. Proof : Draw the diagonals PR and SQ and the segments AD, DB, BC, CA. Now AB and CJ) bisect each other if ADBC is aO, andADBC is aO if AD II CB and ^C II DB. But AD II CB since each is II SQ. See § 151. And ACWDB since each is II PJi. Hence ADBC is a0.and AB and CD bisect each other. Notice that the auxiliary lines PR and SQ divide the figure into triangles and this suggests the use of § 151. 176. Indirect Proof. In case a direct proof is not easily found, it is often possible to make a proof by assuming that the theorem is not true and showing that this leads to a conclusion known to be false. As an example consider the follow- ing theorem : A convex polygon cannot have more than three acute angles. Proof: Assume that such a polygon may have four acute angles, as ^ 1, Z2, Z3, Z4. Extend the sides forming the exterior angles 5, 6, 7, 8. Since Zl + Z5 =2 rt. .J, Z2 + Z.6 = 2 rt. A, etc., and since /^l, Z2, ZS, Zi are all acute by hypothesis, it follows that Z5, Z6, Z7, Z8 are all obtuse, and hence Z5 + Z6 + Z7+ .^8>4rt. A. But this cannot be true since the sum of all the exterior angles Ta exactly 4 rt. ^ by § 162. Hence the assumption that Zl, Z2, ZS, Z4: are all acute is false. That is, a convex polygon cannot have more than three acute angles. 74 PLANE GEOMETRY. The proof by the method of exclusion (§ 86) involves the indirect process in showing that all but one of the pos- sible suppositions is false. 177, The solution of a problem often involves the same kind of analysis as that suggested for the discovery of a direct proof (§ 175). For instance, consider the following problem : To draw a line parallel to the base of a triangle such that the segment included between the sides shall equal the sum of the segments of the sides between the parallel and the base. Given the A^ISC. To find a point P through which to draw DE WAB aothat DP + PE = AD + EB. dX jP.\ )7 Construction. Draw the bisectors of ^ ^ and B A meeting in point P. Then P is the point required. Proof : / 1 = Z 3 since DE II AB, and Z 1 = Z 2 since BP bisects Z£. . . Z 3 = Z 2 and P£ = BE. (Why ?) Likewise in ^ADP, AD =DP. Hence DP+PE = AD+BE. (Ax. Y.) Or DE = AD + BE. This construction is discovered by observing that a point P must be found such that PE = BE and PD = AD. This will be true if Z 3 = Z 2 and this follows if Z 1 = Z 2, while at the same time Z6 = Z5 and Z 4 = Z 5. Hence the bisectors of the base angles will determine the point P. Having thus discovered the process, the construction and proof are made directly. 178. In general the most effective help is a ready knowl- edge of the facts of geometry alrctuhj discovered., and skill in applying these will come with priietice. It is important for this purpose that summaries like the following be made by the student and memorized: SECTILINEAR FIGURES. 75 179. (a) Two triangles are congruent if they have: (1) Two sides and the included angle of the one equal to the corresponding parts of the other. (2) Two angles and the included side of the one equal to the corresponding parts of the other. (3) Three sides of the one equal respectively to three sides of the other. In the case of right triangles : (4) The hypotenuse and one side of one equal to the corre- sponding parts of the other. (5) Two segments are proved equal if: (1) They are homologous sides of congruent triangles. (2) They are legs of an isosceles triangle. (3) They are opposite sides of a parallelogram. (4) They are radii of the same circle. SUMMARY OF CHAPTER I. 1. Make a summary of ways in which two angles may be shown to be equal. 2. Make a summary of ways in which lines are proved parallel. 3. What conditions are sufficient to prove that a quadrilateral is a parallelogram ? 4. Make a list of problems of construction thus far given. 5. Make a list of definitions thus far given. Which of the figures defined are found on page 4 ? 6. Tabulate all theorems on (a) bisectors of angles and segments, (5) perpendicular lines, (c) polygons in general, (d) symmetry. 7. What are some of the more important applications thus far given of the theorems in Chapter I ? 76 PLANE GEOMETRY. >»:« Border, Parquet Flooring. PROBLEMS AND APPLICATIONS. 1. Divide each side of an equilateral triangle into three equal parts (§ 1.59, 5), and draw lines through the division points as shown in the figure. Prove (a) The six small triangles are equilateral and congruent to each other. (First prove them equiangular.) (A) The two large triangles are congruent. (c) The inner figure is a regular hexagon. 2. Let Z, and Iz be parallel lines, with A E per- pendicular to both. Lay off segments AB, BC, CD, . . . and EF, FG, GH, . . . each equal to AE. Connect these points as shown in the figure. Tile Floor Border. Parquet Floor Border. Prove (a) EKA, EFK,AKB,BLC, etc.. are congruent right isosceles triangles. (J) FLBK. etc., are squares. Bisect AB, BC, ... EF, FG ... and join points by lines parallel to EB and ^irrespectively. Xotice that the resulting figure forms the basis for the floor border to the right. 3. Let Zj h be parallel lines, with .1/.V perpendicular to both. Through F, the middle point of MN, draw lines KE and AG, each making an angle of 30° with ^1/.V. riiniiiet Floor Boi-der. Prove FE = AFhy showing A I XF 5; A FE.l/. Also prove KF = FG and that ^ FKA and FEG are equilateral. MECTILINEAB FIGURES. 77 rTVTTT TTTTTl fTTTTT Tile Flooring. Lay off segments AB, BO, ED, DP, KJ, etc., each equal to AK. Connect points as shown in the figure. Prove A SCO ^ A CPD ^AAFK, etc. Prove that ABCDEF and JKFGHl are regiilar hexagons. 4. A network of congruent equilateral tri- angles is constructed in a rectangle ABCD as shown in the figure. (a) How many of these triangles meet in a common vertex? (J) Does this number of equilateral triangles exactly cover the whole plane aboutthe vertices? Why? (c) Do the triangles that meet in one point form a regular hexagon ? (rf) At what angle to the horizontal lines are the oblique lines that form sides of the triangles ? e.g. what is the angle DCK ? (e) Compare the lengths of CK and AK (see § 159, Ex. 14). To construct this network when a side a of the triangles is given, proceed as follows : From the vertex ^ of a right angle lay off segments equal to a along one side AK. With one of these division points K as center and with a radius equal to twice AK strike an arc meeting the side AC nt C. Draw CD parallel to AB and lay oif segments on it equal to a. Connect these points as shown in the figure, and through the intersection points draw the horizontal lines, thus fixing the division points on ^ C and BD. Prove that the resulting triangles are equilateral and congruent to each other. Suggestion. Use in order the converse of § 159, Ex. 14 ; § 144, and the fact that a diagonal divides a parallelogram into two congruent triangles. Notice how this construction is studied. The figure is first sup- posed constructed and its properties tabulated. Some of these (c) and (d) are then used in making the construction. This method is of very general application in problems of construction. Note. This is the method by which a designer would construct a network of congruent regular triangles. 78 PLANE GEOMETRY. 5. Construct a network of triangles as in Ex. 4, using pencil and then ink in parts of the lines, making a set of regular hexagons as shown in the figure. How many such hexagons meet in a point? WiU this number exactly cover the plane about the point? Why? - "M 1 1 m Tile Flooriu'; 6. Point out how the figure constructed uuder Ex. 5 may be made the basis for the two tile floor patterns here given. 7. Given a rhombus whose acute angles are 60 °. Show how to out off two triangles so the resulting figure shall be a regular hexagon. What fraction of the whole rhombus is thus cut off? What fraction of the pattern is black ? 8. In a parallelogram, prove : (a) The diameters divide it into four congruent parallelograms. (V) The segments joining the midpoints of the sides in order form a parallelogram, and the four triangles thus cut off may be pieced together to form another parallelogram congruent to the one first formed. (c) The diameters and diagonals all meet in a point. Suggestion. Prove that a diameter bisects a diagonal. Do these propositions hold for a square ? What part of each of the three large squares in the adjoining figure is white? 9. The octagons (eight-sided polygons) in the figure are equiangular and the small quadriliitcrals arc squares. Find the angles of the irrei;ular hexagons. If the iicl:i)j;ons were regular, would the hexagons be equiangular? equilateral? reg- ular? Tilo I'aUeru. Tile Border. BECTILINEAB FIGUBES. 79 10. In the figure the octagons are equi- angular. (a) Will two such octagons and a square completely cover the plane about a point ? Why ? (6) At what angles do the slanting sides of the octagons meet the horizontal line ? Find all the angles of the two small white ^ „.. _ trapezoids to the right. (c) Could a pattern of this type be constructed by using regular octagons ? 11. Equilateral triangles and regular hexagons have thus far been found to make a complete pavement. What other regular polygons can be used to make a complete pavement? 12. If three regular pentagons (five-sided polygons) meet in a common vertex, will they completely cover the plane about that point? If not, by how great an angle will they fail to do so? 13. If four regular pentagons be placqd about a point, by how great an angle will they overlap? 14. If three regular seven-sided polygons are placed about a point, by how great an angle will they overlap? If two, by how many degrees will they fail to cover the plane ? In Exs. 12, 13, 14 it is understood that, so far as possible, the poly- gons are so placed that no part of one of them lies within another; that is, they are not to overlap. 15. Answer questions like those in Ex. 14 for regular polygons of 8, 9, and 10 sides. Can any regular polygon of more than six sides be used to form a complete pavement ? Note that the larger the number of sides of a regular polygon the larger is each angle. 16. A carpenter divides a board into strips of equal width as fol- lows : Suppose five strips are desired. Place a steel square in two positions, as indicated in the figure, at such angles that the distance in inches diagonally J^^'s^^^s^zizi across the board shall be some multiple of five r^----^^ ■\^ &'.*-■"! A. E Prove that : (i Parquet Flooriug. O N 25. In the figure ABCDEF is a regular hex- jj agon. ABHG, BCLK, etc., are squares. (a) What kind of triangles are BHK, CLM, ^ etc. ? Prove. (J) Is the dodecagon (twelve-sided polygon) HKLMN. . . regular? Prove. (c) How many axes of symmetry has this dodecagon? (d) Has it a center of symmetry ? (e) Are the points 5, F, B, K collinear ? (That is, do they lie in the same straight line ?) 26. If from any two points P and Q in the base of an isosceles triangle parallels to the other sides are drawn, two parallelograms are formed whose perimeters are equal. 27. The middle point of the hypotenuse of a right tri- angle is equidistant from the three vertices. (This is a very important tliporoiii.) 28. State and prove the converse of the theorem in Ex. 27. RECTILINEAR FIGURES. 83 29. The bisectors of the four interior angles formed by a transversal cutting two parallel lines form a rectangle. \/ / '~/~~-i?\ / / / /^- D__F_0 I E B 30. The sum of the perpendiculars from a point in the base of an isosceles triangle to the sides is equal to the altitude from the vertex of either base angle on the side opposite. 31. Find the locus of the middle points of all seg- ments joining the center of a parallelogram to points on the sides. (See § 159, Ex. 8.) 32. In the parallelogram ABCD points E and F are the middle points of ^iJ and CD respectively. Show that AF and CE divide BD into three equal segments. 33. The sum of the perpendiculars to the sides of an equilateral triangle from a point P within is the same for all such points P (i.e., the sum is a constant). Suggestion. Prove the sum equal to an altitude of the triangle. 34. In any triangle the sum of two sides is greater than twice the median on the third side. 3S. ABCD is a square, and EFGH a rec- tangle. Does it follow that A^SHs A FCG and AEBF^AHDGI Prove. 36. If a median of a triangle is equal to half the base, the vertex angle is a right angle. 37. State and prove the converse of the theorem in Ex. 36. CHAPTER II. STRAIGHT LINES AND CIRCLES. 'Outdde Faint 180. A circle (§ 12) divides the plane into two parts such that any point which does not lie on the circle lies within it or outside it. 181. A line-segment joining any two points on a circle is called a chord. A chord which passes through the center is a diameter. 182. If a chord is extended in one or both directions, it cuts the circle and is called a secant. 183. A tangent is a straight line which touches a circle in one point but does not cut it. An indefinite straight line through a point outside a circle is a secant, a tangent, or does not meet the circle. 184. The portion of a circle included between any two of its points is called an arc (§ 12). An arc AB is denoted by the symbol AB. A circle is divided into two arcs by any two of its points. If tlieso arcs are equal, each is :i semicircle. Otlierwise one is called the major arc and the ' iiiciins the minor arc. In case of aiiiliifjuity a tliivd Idtrr may ho used, a> arc AmR. HI STRAIGHT LINES AND CIRCLES. 85 An arc is said to be subtended by the .,»45l^<; chord which joins its end-points. Evi- dently every chord of a circle subtends two arcs. Unless otherwise indicated the arc subtended by a chord means the minor arc. 185. An angle formed by two radii is called a central angle. An angle formed by two chords drawn from the same point on the circle is called an inscribed angle. If the sides of an angle meet a circle the arc or arcs which lie within the angle are called intercepted arcs. If the vertex of the angle is within or on the circle there is only one intercepted arc ; if it is outside the circle there are two inter- cepted arcs, as AB and CD in the figure. 186. If a circle is partly inside and partly outside another circle, then they cut each other. If two circles meet in one and only one point, they are said to be tangent. Arcs of two circles are tangent to each other if the complete circles of which they form a part are tangent to each other. 187. Two circles which can be made to coincide are said to be equal. The word congruent is unnecessary here, since all circles are similar. 188. EXERCISES. 1. Does the word circle as used in this book (§ 12) mean a curved line or the part of the plane inclosed by that line? 2. In how many points can a straight line cut a circle ? 3. In how many points can two circles cut each other ? 86 PLANE GEOMETRY. PRELIMINARY THEOREMS ON THE CIRCLE. 189. Radii or diameters of the same circle or of equal circles are equal. 190. // the radii or diameters of two circles are eqvxil, the circles are equal. 191. A diameter of a circle is double the radium. 192. A point lies within, outside, or on a circle, accord- ing as its distance from the center is less than, greater than, or equal to the radius. 193. If an unlimited straight line contains a point within a circle, then it cuts the circle in two points. 194. // two circles intersect once, they intersect again. See figure, § 186. 195. // a straight line is tangent to each of two circles at the same point, then the circles do not intersect, but are tangent to each other at this point. See § 186. 196. If two arcs of the same circle or equal circles can be so placed that their end-points coincide and also their centers, then the arcs coincide throughout or else form a complete circle. 197' If in two circles an arc of one can be made to coincide with an arc of the other, the circles arc equal. 198. A circle is conveniently referred to by indicating its center and radius. Thus, O OA means the circle ichnse center is O and raditts OA. When no ambiguity arises, the lettir at the center alone may be used to denote the circle. Tlius, O C menus the circle whose center is C. STRAIGHT LINES AND CIRCLES. 87 199. Theorem. In the same circle or in equal circles equal central angles intercept equal arcs. Given the equal circles C and C and /^C= Z. C- To prove that AB = J^B>. Proof : Place O C on O c' so that C falls on c', and Z c coincides with Z Cf. Then A falls on A' and B on b'. (§ 189) Hence AB = Jab'. (§ 196) 200. Theorem. In the same circle or in equal circles equal arcs are intercepted by equal central angles. Given O C = O C, AB = A^. (See figure, § 199.) To prove that Zc = Zc'. Proof : Since AB = A'B', the equal circles can be made to coincide in such manner that the arcs will also coincide. That is, A will fall on A', B on b', and c on c'. Hence Z C coincides with Z c'. 201. EXERCISES. 1. Show that in the same circle or in equal circles equal arcs subtend equal chords, and conversely. 2. Can two intersecting circles have the same center? 3. From a point on a circle construct two equal chords. 4. Show that the bisector of the angle formed by the chords in Ex. 3 passes through the center of the circle. 88 PLANE GEOMETBT. 202. Measurement of Angles. If the perigon at the cen- ter of a circle be divided by radii into 360 equal angles, these radii will divide the circle into 360 equal arcs accord- ing to the theorem, § 199. Hence, vi^e speak of an arc of 1°, 2°, 8°, etc., and similarly for minutes and seconds. For this reason a central angle is said to be measured by the arc which it intercepts, meaning that a given central angle contains a number of unit angles equal to the number of unit arcs in the intercepted arc. 203. Definitions. A quadrant is an arc of 90°. A semicircle is an arc of 180°. A right angle is, there- fore, measured by a quadrant and a straight angle is meas- ured by a semicircle. A sector is a figure formed by two radii and their intercepted arc. Thus, the sector BC^l is formed by the radii CB,CA, a.ndB2. 204. EXERCISES. 1. Show how to bisect an arc, using §§ 48, 199. Divide an arc into four equal parts. Into eight equal parts. 2. Show that in the same circle or in equal circles two sectors hav- ing equal angles are congruent. 3. Show that if two sectors in the same circle or in equal circles have equal arcs, the sectors are congruent. 4. How many degrees in the arc which measiu-es a right angle? a straight angle? half a right angle? three fourths of a straight an- gle? two thirds of a right angle? two thirds of a straight angle? 5. Show that the dianiet<=r is the longest chord of a circle. 6. Show that the two arcs into which the extroniitifs of a diameter di\ idi' a circle are i'i|iial; that is, each is a semicircle. .Sri;(ii>Ti(iN. Fold the figure over on the diameter. 7. Show that by bisecting the angles between two perpendicular diameters, a eiiele is divideil into eight eipial parts. STKAIGHT LINES AND CIRCLES. 89 205. Theorem. A diameter perpendicular to a chord bisects the chord and also its subtended arc. Given the diameter EF ± AB at D. To prove that AD = DB and AF = BF. Proof : Draw the radii CA and CB. If it can be shown that A ACD ^ A BCD, then (1) 4D= BD (Why?), and (2) Zacd^Zbcd (Why?); (3)i>=i> (Why?). 206. Theorem. A line perpendicular to a radius at its extremity is tangent to the circle. Given AB ± CD at D. To prove that AB is tangent to the circle ; that is, does not meet it in any other point than D. Proof : Let E be any point of AB other than D. Draw segment CE. Then CE > CD (Why?). Hence E is outside the circle (§ 192). That is, every point of AB except D is outside the circle, and hence AB is tangent to the circle (§ 183). 90 PLANE GEOMETRY. 207. Theorem. If a line is tangent to a circle, it is perpendicular to the radium drawn to the point of tan- gency. A D E B Given O CD with a line AB tangent to the circle at D. To prove that AB ± CD at D. Proof: If CD is not-L^B, then some other line, as CE, mustbeX^iJ (§ 67), thus making ,CE < CD (Why?). The point E would then lie within the circle (§ 192 ), and the line AB would meet the circle in two points ( § 193). But this contradicts the hypothesis that AB is tani^ent to the circle. Hence no other line than CD can be perpendicular to AB from C, and as one such line exists, it must be CD. 208. EXERCISES. 1. What type of proof is used in the preceding paragraph ? 2. How are the two theorems immediately preceding related to each other. 3. Show that there is only one tangent to a circle at a given point on it, and that the perpendicular from the center upon the tangent meets it at the point of contact. 4. A perpendicular to a tangent at the point of tangency passes through the center of the circle. 5. The periiondicular bisector of any chord passes through the center of the circle. 6. Two taiit;ciits at tlio extivniities of a diameter are parallel. 7. A diaiiK'tcr lii.soi'ts all chords parallel to the tangents at its extremities, and also bisects the central angle subtended by each chord. STRAIGHT LINES AND CIRCLES. 91 8. A diameter bisecting a chord (or its subtended arc) is perpendicular to the chord. 9. The mid-points of parallel chords all lie on a diameter. 10. A tangent to a circle at the mid-point of any arc is parallel to the chord of the arc. 209. Theorem. If two circles meet on the line joining their centers, they are tangent to each other at this point. Given © CB and C'B meeting in a point B on the line CC. To prove that © CB and c'b are tangent to each other at B. Proof : (1) When each circle is outside the other. Let D be ani/ point on Oc'b other than B. Draw CD and do. Then CD + c'd > CB + c'b. (Why ?) But c'b =^ c'd. .-. CD > CB. (Ax. IX, § 119) .•. D is outside of © CB. (2) When O CB is inside of O c'b. Let D be any point on O CB other than B. Draw CD and c'd. Then c'c +CD> c'd. (Why ?) But c'c + CD = c'c + CB = c'b. (Why ?) .-. C'B > C'D. (Ax. VII, § 82) .-. D is within c'b. Therefore O c'b and O CB have only one point in com- mon and hence are tangent to each other (§ 186). 92 PLANE GEOMETRY. PROBLEMS AND APPLICATIONS. 1. If the distance between the centers of two circles is equal to the sum of their radii, how are the circles related? Construct and prove. 2. If the distance between the centers of two circles is equal to the difference of their radii, how are the circles related ? Construct and prove ? 3. If the distance from the center of a circle to a straight line is equal to the radius, how is the line related to the circle ? Consti-uct and prove. 4. Given two circles having the same center, construct a circle tangent to each of them. Can more than one such circle be con- structed? '\^^hat is the locus of the centers of all such circles? 5. Prove the converse of the theorem in § 209. 6. The straight line joining the centers of two intersecting circles bisects their common chord at right angles. 7. A line tangent to each of two equal circles is either parallel to the segment join- ing their centers or else it bisects this seg- ment. 8. In the figure AD = DB. Semicircles are constructed on AD, DB, and AB as di- ameters. "Which semicircles are tangent to each other? 9. In the figure ,1, B, C, D are the vertices of a square. Show how to construct the entire figure. What semicircles are tangent to each other ? This ciiiislriicliim occurs froiiuciitly in (li'.sinns for tilt' flooriiii;'. Si'c aocoinpaiiyiug figure. This is from a Roman inosaic. w I: "^' \-- %J STRAIGHT LINES AND CIRCLES. 93 10. Given two parallel lines BE and AD, to con- struct arcs -which shall be tangent to each other and one of which shall be tangent to BE at B and the other tangent to AD a,t A. Solution. Draw AB and bisect this segment at C; construct ± bisectors oi AC and BC. From A and B draw Js to AD and BE respectively, thus locating the points O and 0'. Prove that O and 0' are the centers of the required arcs. Suggestion. Show that 0, C, and 0' lie in a straight line and use the theorem of § 209. This construction occurs in archi- tectural designs and in many other applications. In the accompanying designs pick out all the arcs that are tangent to each other and also the points of tangency. c„ Scroll Work. 11. On the sides of the equilateral triangle ABC as diameters, semicircles are drawn, as AEFB. Also with A, B, C as centers and AB s,s radius arcs are drawn, as AB, ^. (a) Prove that the arcs AEFB, BDEC, and CFDA meet in pairs at the middle points D, E, F of the sides of the triangle. Sdggestiox. If the mid- dle points of the sides of an equilateral triangle are joined, what kind of triangles are formed ? (h) What arcs in this figure are tangent to each other? (c) Has the figure one or more axes of symmetry ? This figure and the two following occur frequently in church windows and other decorative designs. Fourth Presbyterian Church, Chicago. 94 PLANE GEOMETRY. 12. Construct the design shown in the figure. SuGGKSTiON. Divide the diameter AB into six equal parts and construct the three semicircles. On DC and DC as bases construct equilateral triangles with vertices O and 0'. With radius equal to CB and centers O and 0' construct circles. AC' S C B (a) Prove that O is tangent to each of the three semicircles. Likewise O 0'. (6) Erect a, ± to AB a,t D and prove (D O and 0' tangent to it. (c) Prove circles with centers at and 0' tangent to each other. (d) Has this figure one or more axes of symmetry? 13. In the figure AB, CD and OD are bisected, and O'O" II AB through E. DO' = DO" = I DB. Circles are constructed as shown in the figure. (a) If ^iJ is 4 feet, what is the radius of each circle ? (b) Prove that O O is tangent to O 0' and also to O 0". Suggestion. Show that 00' is the sum of the radii of the two circles. (c) Is O 0' tangent to the arc A CB and also to the line AB ? ((f) Has this figure one or more axes of symmetry? 14. ABCD is a square. Arcs are constructed with A. B, C, D as centers and with radii each equal to one half the side of the square The lines AC, BD, MN, and RS are drawn, and the points E, F, G, H are connected as shown in the figure. The arc SN is extended to P, forming a semicircle. The line LP meets i'.V in K, and BK meets MN in 0'. (a) Prove that EF6^H is a square. ' V ■ '"-. \ (6) Prove that &> KLO' and KPB are mutually equiangular and each isosceles. (c) I'liivc that O O'K is tangent to FG and to N.V. (d) How many axes of symmetry has the figure inside the square? (e) Show that QO'K is tangent to />.V by drawing O'C and fold- ing the figure over on the axis of symmetry MN. STRAIGHT LINES AND CIRCLES. 95 210. Theorem. An angle inscribed in a circle is measured by one half the intercepted arc. Given Z DBA inscribed in O CB. To prove that Z. DBA is measured by J AD. Show that Z 2 = 1 Z 1. Proof : (1) if one side, as BD, is a diameter. Draw the radius CA. But Z 1 is measured by AD (§ 202). Hence Z 2 is measured by ^ AD. (2) j^ the center C lies within the angle. Draw the diameter BE. Now Z DBA = Z 1 + Z 2. Complete the proof. (3) if the center C lies outside the angle. Draw BE and use the equation Z DBA = Z 1 — Z 2. 211, It follows from § 210 that if in equal circles two inscribed angles intercept equal arcs, they are equal ; and conversely, that if equal angles are inscribed in equal circles, they intercept equal arcs. 212. EXERCISES. 1. If the sides of two angles BAD and BA'D pass through the points B and Z) on a circle, and if the vertex A is on the minor are BD and A' is on the major arc BD, find the sum of the two angles. 2. In Ex. 1 if the points B and D remain fixed while the vertex A of the angle is made to move along the minor arc of the circle, what can be said of the angle A 1 What if it moves along the major arc? 96 PLANE GEOMETRY. 213. Theorem. The locus of the vertices of all right triangles on a given hypotenuse is a circle whose diame- ter is the given hypotenuse. Outline of Proof : Let AB be the given hypotenuse. (1) If P is any point on the circle whose diameter is AB, Z APB = rt. Z. (Why ?) (2) If Ap'b is any right triangle with AB as hypote- nuse, then AC=CB = CP'. (See Ex. 27, p. 82. _) State the proof in full. 214. PROBLEMS ON LOCI. Find the following loci : 1. The centers of all circles of fixed radius tangent to a fixed line. 2. The centers of all circles tangent to two parallel lines. 3. The centers of all circles tangent to both sides of an angle. 4. The centers of all circles tangent to a given line at a given point. Is the given point a part of this locus ? 5. The vertices of all triangles \\ hich have a common base and equal altitudes. 6. The middle points of all chords through a fixed point on a circle. Use Ex. 8, § 208, and then § 213. 7. The points of intersection of the diasicjnals of trapezoids formed by the sides of an isosceles triangle and lines parallel to its base. 8. Two vertices of a triangle slide along two parallel lines, ^^'hat is the locus of the third vertex if the triangle is fixc'il in size and shape? 9. .IB( 'D is a parall('li\t;raiii all of whose sides are of fixed lens:th. The side A B is fixed in position. Find the locus of the middle points of the remaining three sides. STRAIGHT LINES AND CIRCLES. 97 10. Prove that in the same circle or in equal circles equal chords are equally distant from the center. Suggestion. MB = ND. Why? Then prove A B3IC = A CND. 11. State and prove the converse of the theorem in the preceding exercise. (What parts of A BMC and CND are now known ?) 12. Find the locus of the middle points of all chords of equal length in the same circle. 13. Find the locus of the middle point of a segment AB of fixed length which moves so that its end-points slide along the sides of a right angle. (Use Ex. 27, p. 82.) 14. Find the locus of the points of contact of two varying circles tangent to each other, and each tangent to a given line at a given point. Suggestion. A and B are the fixed points, \ \\:---'i~.:-f' and P one point of contact of the circles. Draw the common tangent PB. and DB = DP. Prove AD = DP That is, Hence, D is the middle point of AB and DP is constant, the locus is a circle of which AB is a diameter. 15. Find the locus of the centers of all circles tangent to a fixed circle at a fixed point P. Is the fixed point P a part of this locus? Is the center of the fixed circle a part of it? 16. Find the locus of the centers of all circles of the same radius which are tangent to a fixed circle. Under vphat conditions will this locus include the fixed circle itself? The center of this fixed circle ? Will the locus ever contain a circle within the fixed circle? Under what conditions will the locus consist of two circles, each outside the fixed circle ? Under what condition does the locus consist of only one circle? 17. In making core-boxes, pattern makers use a square as indicated in the figure to test whether or not the core is a true semicircle. Is this method correct? Prove. 98 PLANE GEOMETRY. 215. Theorem. The arcs intercepted by two parallel chords or by a tangent and a chord parallel to it are equal. Given AB \\ DE and LK II MN. To prove that AI) = BE and LB = KB. Proof : (1) Draw chord BE. Compare Z 1 and Z 2, and hence show that ^ = BE. (Why ?) (2) Draw the radius CB to the point of tangency. Then CB ± mn and CB -L LK. (Why?) Prove A LCS s A KCS, and hence that LB = BK. (§ 199) 216. .EXERCISES. 1. Prove that a tangent at the vertex of an inscribed angle forms equal angles with the two sides, if these are equal chords. 2. If the vertices of a quadrilateral lie on a circle, any two of its opposite angles are supplementary. 3. If two chords of a circle are perpendicular to each other, find the sum of each pair of opposite arcs into which they divide the circle. 4. If the vertices of a trapezoid lie on a circle, its diagonals are equal. 5. Two circles intersect at C and D. Diame- ters CA and CB are drawn. Prove that ^4, Fi. B lie on a straight line. Suggestion. Prove that /L A /" ' = Z. CDB - rt. Z. STRAIGHT LINES AND CIRCLES. 99 217. Theorem. An angle formed by two intersecting chords is measured by one half the sum of the arcs inter- cepted by the angle itself and its vertical angle. Given Z 1 formed by the chords AB and DE. To prove that Zl is measured by J (^AE+ BD}. Proof : Through A draw the chord AF || ED. Compare Z 1 and Z 3. Compare AE and DF, also AE + BD and BD + DF. How is Z 3 measured ? Hence, how is Z 1 measured ? 218. EXERCISES. 1. A chord AB is divided into three equal parts, AC, CD, and DB. OA, OC, OD, and OB are drawn. Compare the angles 4 OC, COD, and DOB. 2. The accompanying table refers to the figure in § 217. Fill out blank spaces. 3. In a circle C with a diameter AB a chord AD is drawn, and a radius CE || A D. Prove that arcs DE and EB are equal. 4. The vertices of a square A BCD all lie on a circle. E is any point on the arc AB. Prove that EC and ED divide the AZigleAEB into three equal parts. Zl AE BD EB A^D 35° 40° 80° 48° 50° 216° 40° 50° 60° 60° 54° 190° 45° 90° 180° 34° 108° 164° 100 PLANE GEOMETEY. 219, Theorem. An angle formed by a tangent and a chord drawn from the point of tangency is measured by one half the intercepted arc. Given Z. 1 formed by tangent BD and chord BA. To prove that Z 1 is measured by J BA. Proof : Draw a chord EF II BH intersecting BA in Q. Compare Z 1 and Z 2, also EB and BF. How is Z 2 measured ? Hence, how is Z 1 measured ? Give the proof in full. 220. Definitions. A segment of a circle, or a circle-seg- ment, is a figure formed by a chord and the arc which it subtends. For each chord there are two y^^^^Sc^ circle-segments corresponding to the two arcs which it subtends. If a chord is a diameter the two circle- segments are equal. An angle is said to be inscribed in an arc if its vert^i-x lies on the arc and its siiU's meet the arc in its end-points. Such an angle is also said to be inscribed in the cirule-segraent formed by the arc and its chord. E.g. Zl is inscribed in tlie aiv APR ov iu the segment .iPB. STRAIGHT LINES AND CIRCLES. 101 221. EXERCISES. 1. Show that an angle inscribed in a semicircle is a right angle. . 2. If the sides of a right angle pass through the extremities of a diameter, show that its vertex lies on the circle. 3. If a triangular ruler il/iVO, right-angled at O, is moved about in the plane so that two fixed points, A and B, lie always on the sides MO and NO respectively, what path does the point O trace? 4. Draw two concentric circles, having different radii, and show that all chords of the outer circle which are tangent to the inner circle are equal. 5. In an equilateral triangle construct three equal circles, each tangent to the two other circles and to two sides of the triangle. Suggestion. Construct the altitudes of the triangle and bisect angles as shown in the figure. Complete the construction and prove that the figure has the required properties. («) Has the figure consisting of the triangle and the three circles one or more axes of symmetry? (i) Has it a center of symmetry? 6. Within a given circle construct three equal circles, each tangent to the other two and to the given circle. Suggestion. Trisect the circle at D, E, c and F by making angles at the center each /:\ equal to 120° Draw tangents at D, E, and F, and prove that A ^5C is equilateral. Construct the altitudes and prove that they meet the sides of the triangle at the points of tangency of the given circle with the sides of ABC, and also that they pass through the center of the given circle. Bisect angles as shown in the figure and prove that the centers of the required circles are thus obtained. (o) Has the figure consisting of the four circles one or more axes of symmetry? (6) Has it a center of symmetry? 102 PLANE GEOMETRY. 222. Theorem. The angle formed by two secants, two tangents, or a tangent and a secant, meeting outside a circle, is measured by one half the difference of the intercepted arcs. Outline of Proof : In each case the given angle is equal to Z 1, and the arc which measures Z 1 is the difference between two arcs, one of which is the larger of the two intercepted arcs and the other is equal to the smaller. E"'or instance, in the first figure, FH=zDF— £h = DF— be. Give the proof in detail for each figure. 223. EXERCISES. 1. If (in left figure, § 222) ZA = 17° and £B = 25", find DF. 2. If ZA = 37° (in middle figure), find the arcs into which the points B and E divide the circle. 3. With a given radius construct a circle passing through a given point. How many such circles can be drawn ? What is the locus of the centers of all such circles ? 4. Draw a circle passing through two given fixed points. How many such circles are there? What is the locus of the centers of all such circles? 5. Construct a circle having a given radius and passing through two given points. How many such circles can be drawn? Is this construction ever impossible? Under what conditions is only one such cii'cle possible? STRAIGHT LINES AND CIBCLES. 103 224. Problem. To construct a circle through three fixed points not all in the same straight line. Given three points A, B, C not in the same straight line. To construct a circle passing through them. Construction. Let the student give the construction and proof in full. (See §132.) 225. Definition. The circle OA in § 224 is said to be circumscribed about the triangle ABG and the triangle is said to be inscribed in the circle. 226. EXERCISES. 1. In the construction of § 224 why do DM and EN meet ? 2. Why cannot a circle be drawn through three points all lying in the same straight line ? Make a figure to illustrate this. 3. Show that an angle inscribed in an arc is greater than or less than a right angle according as the arc in which it is inscribed is less than or greater than a semicircle. 4. Prove that the bisectors of the angles of an equilateral triangle pass through the center of the circumscribed circle. 5. Draw a circle tangent to two fixed lines. How many such circles are there ? What is the locus of their centers '! Is the point of intersection part of this locus? Discuss fully. 6. Show that not more than one circle can be drawn through three given points, and hence that two circles which coincide in three points coincide throughout. 104 PLANE GEOMETRY. 227. Problem. To construct a circle tangent to each of three lines, no two of which are parallel and not all of which pass through the same point. Given the lines h, h, Is- To construct a circle tangent to each of these lines. Construction. Since no two of the lines are II, let 1^ and ^2 meet in A, l^ and Zj in B, and l^ and l^ in D, where A, B, and D are distinct points. Draw the bisectors oi ZA and Z B and let them meet in point C. Then C is the center of the required circle. (See § 131.) Give the proof in full. 228. Definitions. The circle in the construction of § 227 is said to be inscribed in the triangle ABD. Three or more lines which all pass through the same point are called concurrent. Hence the lines /j. 1^. l^ are not con- current. 229. EXERCISES. 1. Why is the construction of § 227 impossible if l^ I:, and /a are concurrent ? 2. If two of the lines are parallel to each other, show that the con- struction is possible. How many tangent circles can be constructed in this case? Draw a iigure and give the construction and proof in full. 3. Is the construction possible wliou all three lines are parallel? Why? STRAIGHT LINES AND CIRCLES. 105 4. If two sides of the triangle are produced, as AB and AD in the figure of § 227, construct a circle tangent to the side BD and to the prolongations of the sides AB and AD. This is called an escribed circle of the triangle. 5. How many circles can be constructed tangent to each of three straight lines if they are not concurrent and no two of them are parallel? 6. Draw a triangle and construct its inscribed and circumscribed circles and its three escribed circles. 230. PiiOBLEM. From a given 'point outside a circle to draw a tangent to the circle. , rn, -^bx. / 'n Given O CA and an outside point P. To consttuct a tangent from P to the circle. Construction. Draw CP. On CP as a diameter con- struct a circle, cutting the given circle in the points A and B. Draw the lines PA and PB. Then PA and PB are both tangents. Give the proof. 231, EXERCISES. 1. If in the figure of § 230 the point P is made to move towards the circle along the line PC until it finally reaches the circle, while PA and PB remain tangent to the circle, describe the motion of the points A and B and also of the lines PA and PB. How does this agree with the fact that through a point on the circle there is only one tangent to the circle ? 106 PLANE GEOMETRY. 2. Can a tangent be drawn to a circle from a point inside the circle ? Why ? 3. Show that the line connecting a point outside a circle with the center bisects the angle formed by the tangents from that point. 4. Why are not more than two tangents possible from a given point to a circle ? 5. The two tangents which can be drawn to a circle from an ex- terior point are equal. 6. In a right triangle the hypotenuse plus the diameter of the inscribed circle is equal to the sum of the two legs of the triangle. 7. If an isosceles triangle inscribed in a circle has each of its base angles double the vertex angle, and if tangents to the circle are drawn through the vertices, find the angles of the resulting triangle. 8. If the angles of a triangle ABC inscribed in a circle are G4°, 72°, and 44'^, find the angles of the triangle formed by the tangents to the circle at the points A , B, and C. SUMMARY OF CHAPTER H. 1. Make a list of all the definitions involving the circle. 2. State the theorems on the measurement of angles by inter- cepted arcs. 3. State the theorems involving equality of chords, central angles, and intercepted arcs. 4. State the theorems on the tangency of straight lines and circles. 5. State the theorems involving the tangency of two circles. 6. Make a list, to supplement that in the summary of Chaptei- I, of ways in which two angles or two line-segments may be proved equal. 7. State the ways in which two arcs of the same or equal circles may be proved equal. 8. State the problems of construction given in Chapter II. 9. Explain what is meant by saying that a central angle is meas- ured liy its intercepted arc. 10. State some of the important applications of Chapter II. (Return to this question aftei- studying those which follow.) STRAIGHT LINES AND CIRCLES. 107 '2 -1 ' PROBLEMS AND APPLICATIONS. 1. Given two roads of different width at right angles to each other, to connect them by a road whose sides are arcs of circles tangent to the sides of the roads. (a) Make the construction shown in the figure and prove that it has the required properties. (6) Is this construction possible when the given roads are not at right angles to each other ? Illustrate. (c) Can the curve be made long or short at will ? (d) Make the construction if the given roads have the same width 2. Two circles C an(\ C" are tangent at the point D. AB is a, segment through D termi- nating in the circles. Prove that the radii CA and C'B are parallel. 3. Through a point on the bisector of an angle to construct a circle tangent to both sides of the angle. Construction. Through the given point P draw EP ±to AP. Lay ofl ED = EP and at D construct DC J- ^7) meeting the line AP in C. Then C is the center of the required circle and CD is its radius. Proof : Draw PD and prove that A DEP is isosceles and hence also APD.C. Is it possible to construct another circle having the properties required? If so, construct it. This construction is used in the accompany- ing design in which the shape is determined by fixing the point P in ad- vance. Such designs are of frequent occurrence in decorative work such as the steel ceiling panel given here. 108 PLANE GEOMETRY. B 4. In an isosceles triangle construct three circles as shown in the figure. Suggestion. First construct the inscribed circle with center C). Let the bisector oi /.A meet a this circle in a point P- Then use Ex. .3. 5. The angles formed by a chord and a tangent are equal respec- tively to the angles inscribed in the arcs into which the end-points of the chord divide the circle. 6. If a triangle whose angles are 48°, 56°, and 7f!' is circumscribed about a circle, find the number of degrees in the arcs into which the points of tangenoy divide the circle. 7. Divide each side of an equilateral triangle into three equal parts (Ex. 5, § 159) and connect points as shown in the figure. Prove that DEFGHK is a regular hexagon. 8. If a circle is inscribed in the triangle of Ex. 7, prove that all sides of the hexagon are tangent to the circle. Suggestion. Show that the perpendicular bisec- tors of the segments HK, KD, DE meet in a point equidistant from these segments. _ 9. "Within a given square construct four equal circles so that each circle is tangent to one side of the square and to two of the circles. Suggestion. First construct the diagonals of the square. 10. In the figure, A BCD is a rectangle D F with AD = i AB. E and F are the middle points of AB and CD respectively. Scniicirrles are con- structed with E and F as centers and J.I/? as a radius, etc. A K B Fan vaulting from (ilouoislor Oatlioiir;il, EngUind. STRAIGHT LINES AND CIRCLES. 109 (a) Prove that these quadrant arcs are tangent to each other in pairs and also to the semicircles. (6) Lines are drawn tangent to the arcs at the points where these are met by the diagonals of the squares AEFD and BCFE. Prove that these lines form squares KLMN and X YZ W. (c) Construct the small circles within each of these squares. The above design occurs in fan-vaulted ceilings. The gothic or pointed arch plaj'S a conspicuous part in modern architecture, and examples of it may be found in al- most any city. Its most common use is in church windows. The figure represents a so-called equilat- eral gothic arch. The arcs AG and BG are drawn from B and A as centers respectively, and with AB as a radius. The segment ^B is called the span of the arch, and the point G its apex. 11. In the figure AD = DB. ABC, ADE, and DBF are equi- lateral gothic arches. (a) Construct the circle with center O tangent to the four arcs as shown. Suggestion. Take X so that DX = XB. With centers A and B and radius AX draw arcs meeting at O. Complete the con- struction and prove that the figure has the required properties. (J) Prove that DE and DF are tangent to each other. Also BF and BC, and AE and A C. (c) What axis of symmetry has this figure ? 12. A triangle ABC whose angles are 45°, 80°, and 55° is inscribed in a circle. Find the angles of the triangle formed by the tangents at A, B, and C. Door, Union Park Church, Chicago. 110 PLANE GEOMETBY. 13. Inscribe a circle in an equilateral gothic arch ABC. Suggestions. Construct CD Xto AB and ex- tend it to P, making DP = AB. From P con- struct a tangent to A C at L. (a) Prove that A BDP s A BLP and hence PL = BD. (b) A OLP = A BDO and hence OD = OL. Then OD is the required circle. See § 209. Notice that this figure is symmetrical with re- spect to the line PD, and hence if the circle is proved tangent to A C, we know at once that it is tangent to BC. 14. In the figure ABC is an equilateral gothic arch with a circle inscribed, as in Ex. 13. (a) Construct the two equilateral arches GHE and HKF, as shown in the figure. Construction. Draw BK and AG±AB. With a radius equal to OD + DB, and with O as center draw arcs meeting BK and AG in K and G respectively. Draw GK, construct the arches and show that each is tangent to the circle. (i) Do the points E and F lie on the circle ? Suggestion. Suppose KF to be drawn, and ^ -- compare Z HKF with / HKO by comparing the sides HK and KF and also GK and KO. 15. Construct an arc passing through a given point B, and tangent to a given line AD at a given point D. 16. In the figure ABC is an equilateral arch. BK is \ of BD. KBF and A HE are equal equi- lateral arches. Arcs KQ and HQ are tangent to arcs KF and HE respectively. (a) Find by construction the center O of the circle tangent to ,1 (', Jl( ', KJ'\ and HE, and give proof. From Lincoln Cathedral, England. STRAIGHT LINES AND CIRCLES. Ill (V) Find by construction the centers of the arcs KQ and HQ. How is this problem related to Ex. 15 ? 17. Two circles are tangent to each other in- ternally. Find the locus of the centers of all circles tangent to both externally. 18. Two circles are tangent to each other ex- ternally. Find the locus of the centers of all circles f tangent to both, but external to one and internal to ' the other. 19. Two equal circles are tangent to each other externally. Find the locus of the centers of all circles tangent to both. 20. AD'B is an angle whose vertex is outside the circle and whose sides meet the circle in the points A and B, while A ADB is an inscribed angle intercepting the arc AB. Prove that Z.ADB~^ Z AD'B, provided each of the segments D'A and D'B cuts the circle at a second point. 21. Through two given points A and B con- struct a circle tangent to a given line which is perpendicular to the line AB. Is this construction possible if the given line passes through either of the points A or Bt If it meets AB between these points? 22. In kicking a goal after a touchdown in the game of football, the ball is brought back into the field at right angles to the line marking the end of the field. The distance between the goal posts being given, and also the point at which the touchdown is made, find by a geo- metrical construction how far back into the field the ball must be brought in order that the goal posts may subtend the greatest pos- sible angle. CHAPTER III. THE MEASUREMENT OP STRAIGHT LINE- SEGMENTS. 232. A straight line-segment is said to be exactly meas- ured when we find how many times it contains a certain other segment which is taken as a unit. The number thus found is called the numerical measure, or the length of the segment. E.g. a line-segment is 9 in. long if a segment 1 in. long can be laid off on it 9 times in succession. Thus, 9 is the numerical measure, or the length of the segment, when 1 in. is taken as a unit. 233. In selecting a unit of measure it may happen that it is not contained an integral number of times in the seg- ment to be measured. Thus, in measuring a line-segment the meter is often a convenient unit. Suppose it has been applied five times to the segment AB and that the last time the end falls on .Ij, A^B being less than one meter. Then, taking a decimeter (one tenth of a meter) as a new unit, suppose this is contained ., 5 5 3" three times in AJi with a remainder A.,B less -^i -^-* than a decimeter. Finally, using as a unit a centimeter (one tenth of a decimeter), sup- pose this is contained exactly six times in ,1„B. Then, the length of AB is fi meters, 3 decimeters, and 6 centi- meters, or ."5.36 meters. The ]irocess of mcusuring- considered here is ideal. Tn practice we cannot say tli;it a given segment is contained cxaclli/ an int^>gral number of times in another segment. Sec § 12;!5. 112 MEA8VBEMENT OF LINE-SEGMENTS. 113 234. It may also happen that, in continuing this ideal process of measuring as just described, no subdivided unit can be found which exactly measures the last interval, that is, such that the final division point falls exactly on B. E.g. it is known that in a square whose sides are each one unit the diagonal is V2, and that this cannot be exactly expressed as an integer or a fraction whose numerator and denomi- nator are both integers. By the ordinary process of extracting square root we find v'2 = 1.4142 •••, each added decimal mak- ing a nearer approximation. But this process never terminates. Hence, in attempting to measure the diagonal of a square whose side is one meter, we find 1 meter, 4 decimeters, 1 centimeter, 4 milli- meters, etc., or 1.414 meters approximately. It should be noticed, however, that 1.415 is greater than the di- agonal and hence the approximation given is correct within one millimeter. 235. Evidently any line-segment can be measured either exactly or to a degree of approximation, depending upon the fineness of the instruments and the skill of the opera- tor. The word measure is commonly used to include both exact and approximate measurement. For practical purposes, a line-segment is measured as soon as the last remainder is smaller than the smallest unit available. It should be noticed that all practical measurements are in reality only approximations, since it is quite impossible to say that a given distance is, for in- stance, exactly 25 ft. It may be a fraction of an inch more or less. E.g. in the above example 1.414 meters gives the length of the diagonal for practical purposes if the millimeter is the smallest unit available. The error in this case is less than one millimeter. 114 PLANE GEOMETRY. 236. Definition. Two straight line-segments are com- mensurable if they have a common unit of measure. Other- wise they are incommensurable. E.g. two line-segments whose lengths are exactly 5.27 and .3.42 meters respectively have one centimeter as a common unit of meas- ure, it being contained .527 times in the first segment and 342 times in the second. But the side and the diagonal of a square have no common unit of measure. In the example of § 234, the millimeter is contained 1000 times in the side and 1414 times in the diagonal, plus a remainder less than one millimeter. A similar statement holds for any unit of measure, however small. 237. For the purposes of practical measurement any two line-segments may he considered as commensurahle. but for theoretical purposes it is necessary to take account of in- commensurable segments also. The theorems in this chapter are here proved for com- mensurable segments only. They are proved for incom- mensurable segments also in Chapter VII. RATIOS OF LINE-SEGMENTS. 238. The ratio of two commensurable line-segments is the quotient of their numerical measures taken %vith respect to the same unit. E.g. if two spgrnonts are respectively •' ft. and 4 ft. in length, the ratio of the first segment to the second is \ and the ratio of the second to the first is \. 239. The ratio of two commensunihle segments is the same., no matter what common unit of mea.suro is used. E.g. two segments whose numerical measures are :> and 4 if one foot is the common unit, have 36 and 18 as their numerical measures if one inch is the common unit. Hut the ratio is the same in both caacs, namely : J8 = J- MEASUREMENT OF LINE- SEGMENTS. 115 240. The approximate ratio of two incommensurable line- segments is the quotient of their approximate numerical measures. It will be seen that this approximate ratio depends upon the length of the smallest measuring unit available, and that the approximation can be made as close as we please by taking the measuring unit small enough. E.g. an approximate ratio of the side of a square to its diagonal is — — = Another and closer approximation is = . In 1.41 141 ^^ 1.414 1414 this case the numerical measure of one of the segments is exact. An approximate ratio of V^ to VS, in which neither has an exact . 1.41 141 . ,, . 1.414 1414 measure, is = — . Another is = . 1.73 173 1.732 1732 241. It should be clearly understood that the numerical measure of a line-segment is a number, as is also the ratio of two such segments. Hence they are subject to the same laws of operation as other arithmetic numbers. For example, the following are axioms pertaining to such numbers : (1) Numbers which are equal to the same number are equal to each other. (2) If equal numbers are added to or subtracted from equal numbers, the results are equal numbers. (3) If equal numbers are multiplied by or divided by equal numbers, the results are equal numbers. It is understood, however, that all the numbers here considered are positive. For a more complete con- sideration of axioms pertaining to numbers, see Chapter I of the Advanced Course of the authors' High School Algebra. 116 PLANE GEOMETRY. 242. A proportion is an equality, each member of which is a ratio. Four numbers, a, J, c, and d, are said to be in proportion, in the order given, if the ratios — and — are a equal. In this case a and c are called the antecedents and b and d tlie consequents. Also a and d are called the extremes and b and c the means. The proportion - = -- is sometimes written a :b = c : d, b a and in either case may be read a is to b as c is to d. If D and E are points on the sides of the triangle ABC, and if m, w, p, and q, the numerical measures respectively of Al), BB, AE, and EC, are such that — = 2, ". .^ then the points D and E are said to divide the sides AB and AC proportionally, that is, in the same ratio. For convenience it is common to let AD, DB, AE, and EC stand for the numerical measures of these segments, and thus to AD AE write the above proportion, — = — or AD : DB = AE:EC. DB EC THEOREMS ON PROPORTIONAL SEGMENTS. 243. Theorem. // a line is parallel to one side of a triangle and cuts the other two sides, then it divides these sides in the same ratio. Given A ABC in which DE II BC. A To prove that ' — = '— ^. DB EC ^ Proof : Choose some common measure of AD and DB, as AK. Suppose it is eon- tained 3 times in .ID and 5 times in DB. B MEASUREMENT OF LINE-SEGMENTS. 117 Then AD DB (1) Through the points of division on AD and DB draw lines parallel to BC, cutting AE and EC. By § 155 these paral- lels divide AE into three equal parts and EC into five equal parts. Hence, ^^ =1- (2) §241 EC from (1) and (2)^ = ^. DB EC For a proof in case AD and DB are incommensurable, see § 410. 244. EXERCISES. 1. If DEWBC in i\ABC, compute the segments left blank from those given in the following table : AD DB aij: EC AB AC 20 24 15 4 56 42 102 12 408 25 18 342 2. B is a point visible from A but inaccessible. Required to com- pute the distance from A to B. Suggestion. Select some accessible point C -? from which A and B are both visible. Through E, a point near A and on the line of sight from A to C, draw ED II CB and meeting the line of sight from ^ to B at D. Now AE, EC, and AD can be measured. Then Z)B and hence AB can be computed by § 243. Note. The advantage of this method over that in § 34, Ex. 5, is that here a small triangle AED is made to do the service, which was there performed by another triangle the same size as ABC. (5) 118 PLANE GEOMETRY. 245. Theorem. If four numbers m, n, p, q are such that — = ^, then it follows that: n q (1)^ = 2. (2)™ = ^. m p VI (3) m + n ^p + q ^^^ m-n _p-q n q n q m + n _ p + q m — n p — q Given - = ^- (a) n g To prove (1) divide the members of 1 = 1 by those of To prove (2) multiply each member of (a) by — To prove (3) add 1 to each member of (a) and reduce each side to a common denominator. To prove (4) subtract 1 from each member of (a) and reduce each side to a common denominator. To prove (5) divide the members of (3) by the members of (4). Write out these proofs in full, giving the reason for each step, and read off the results as applied to the figure. For example, show that (3) gives, when applied to the figure, AB^AC „^ ^ DB EC A B C 246. The results in the above theorem are sometimes named as follows : The proportion {a') is said to be tnken by inversion in (1), by alternation in ('2), by composition in (3), by division in (4;, by composition and division in (^5). MEASUREMENT OF LINE-SEGMENTS. 11& 247. 1. If — =^, prove that n q above figure that EXERCISES. = — " — , aud hence show in the m + n p + q AD^AE AB AC' 2. If in the figure on page 118 DEWBC, compute the segments indicated by blanks in the accompanying table. 3. If 1L = P, n q show that "* + P = P n + q 9 4. If !^ = £, » n q show that '"+■?' = n + q m n ' 5. If ^=S and ™=2, X q y q show that X = y. 6. If : — , show that x = y. AT) AB DB AE AC EC 8 12 6 6 10 16 6 9 7 12 10 10 10 8 18 10 7 14 240 200 380 160 140 20 120 100 50 35 21 14 40 15 30 500 200 400 90 40 70 20 30 8 800 360 300 27 30 48 30 20 50 27 560 48 7. State Exs. 4, 5, and 6 in words. 8. A triangle is formed by a chord and the tangents to the circle at its extremities. Prove that the triangle is isosceles. 9. A triangle with angles A, B, C is circumscribed about a cii-cle. Find the angles of the triangle formed by the chords joining the points of tangency. 120 PLANE GEOMETRY. 248. Theorem. // a line divides two sides of a tri- angle in the same ratio, it is parallel to the third side. Given the points D and E on the sides of the A ABC such .. ^ AD AE that — = — . DB EC To prove that DE II BC. Proof : Suppose DE' is drawn paral- lel to BC. It is proposed to prove that the point e' coincides with E. AB AC Since BE' II BC, we have — = AB e'c AC_ EC But by (3), § 245, ^ : ' ^ ^^ BB Now use the proof of Ex. 5, § 247, to show that e'c = EC, and hence that e' and E coincide, so that BE II BC. Give the proof in full detail. 249. EXERCISES. 1. Show by § 24S that the line joining the middle points of two sides of a triangle is parallel to the third side. Compare § 151. 2. Show by § '243 that the line which bisects one side of a triangle and is parallel to a second side bisects the third side. 3. If in the A ABC a segment f'l/ connects the vertex to any point y of the base, find tlie locus of the point x on this segment such that Cx : Cy is the same for all points y. 4. A BCD is a O whose diagonals meet in 0. If ^ is a point on any side of the O, find the loons of a point X on the si'j;nuMit Oy such that Oy : xy is the sanif for (■\i'iy snch point y. 5. Find the locus of the points of intersection of the nicdiiLiis of all triangles having the same base and e(iual altitudes. (Isc §§ ITiS, 2KS.) MEASUREMENT OF LINE-SEGMENTS. 121 250. Theorem. The bisector of an angle of a triangle divides the opposite side into segments whose ratio is the same as that of the adjacent sides. Given CD bisecting Z C in A ABC. To prove that ^ = fi^. DB BC Proof : Through B draw BE II BC. Prolong ^c to meet BE at E. q,."'y extended. 124 PLANE GEOMETRY. SIMILAR POLYGONS. 255. Two polygons, in which the angles of the one are equal respectively to the angles of the other, taken in order, are said to be mutually equiangular. The angles of the two polygons are thus arranged in pairs of equal angles, which are called corresponding angles. Two sides, one of each polygon, included between cor- responding angles, are called corresponding sides. 256. Two polygons are similar if (1 ) they are mutually equiangular and if (2) their pairs of corresponding sides are proportional. Two pohgons may hare property (1) but not (2). For example, a rectangle and a square. Or they may have property (2) and not (1). For example, a square and a rhombus. Hence any proof that two polygons are similar must show that both (1) and (2) hold concerning them. In the case of triangles it will be proved that either property speci- fied in the definition of similar polygons is sufficient to make them similar. 257. Theorem. If tico triangles arc mutually equi- angular, they are similar. Given A ABC and ABC, in which ZA = /:a,ZB=ZB', and ZC=.:C'. MEASUREMENT OF LINE-SEGMENTS. 125 namely that = ■ — •^ Al al r>//-.' ryl aI To prove that the other property of similarity holds, AB _ BG _ CA A'B' ^ B'C'~ C'A' Proof: Place Aa'b'c' on A ABC with Z. A' upon its equal Z A, and b'c' taking the position DE. Now show that DE II BC and hence — = ^^ that is, AD AE AB AG a'b' a'g' In like manner, placing Z b' upon Z B, AB BC show that ^'s' b'g' Hence, i_^ = .^ = ^. (Why ?) A'B b'c' C'a' ^ ■' Give the full details of this proof. 258. EXERCISES. 1. To measure indirectly the distance from an accessible point A to an inaccessible point B. • Suggestion. Construct j4 2) ± the line of sight ■Ij ,,--'' ^ from A to B, and ED X AD. Let C be the point B-'='-^^ (yn AD which lies in line with E and B. Now show that M\EDC and BAC are mutually equiangular and hence similar. What segments need to be measured in order to com- pute AB ? Give full details of proof. 2. Prove that two right triangles are similar if they have an acute angle of one equal to an acute angle of the other. 3. Two isosceles triangles are similar if they have the vertical angle of one equal to the vertical angle of the other. 4. Two triangles which have the sides of one respectively parallel or perpendicular to the sides of the other are similar. 5. Show by similar triangles that the segment joining the mid- points of two sides of a triangle is equal to one half the third side. 126 PLANE GEOMETRY. 259. Theorem. If two triangles have an angle of one equal to an angle of the other and the pairs of adjacent sides in the same ratio, the triangles are similar. Given A ABC and A'BfC in which Z.A = ZA' and. AB A'B AC A' a' To prove that A4BC~A4'S'C''. Proof: Place AA'b'c' upon A ABC with Z.A' on ZA, b'c' taking the position.!) J?. Then, AB^AC (Why?) AD AE V J y and hence, DE || BC. (Whj- ?) Now show that A ADE and ABC are mutually equian- gular and hence similar. 260. Theorem. If two tri- angles have their pairs of corresponding sides in the same ratio, they are similar. Given A ABC and A'SfC in 7,, . . . AB BC CA t- which — ,— ; = — — ; = , - • A'Bf BC CA' To prove that Aabc ^ A^'ij'c', tliatis, toproveZvl = Z a', Zb = Z b',Z c = Zc'. MEASUREMENT OF LINE-SEGMENTS. 127 Proof : Lay off on AB and AC respectively AD = a'b' and AE = a'c', and draw BE. Now prove, as in § 259, that AADE^^AABC, and hence, that :d^ = i^ . (U DE BG -^ But, ^ = i^. (Why?) (2) B'c' BG Hence, since AD = A'b', it follows from (1) and (2) that DE= b'c', as in Ex. 5, § 247. Now show that AA'b'g'^A ADE, and hence, that AA'b'g' ~ AABG. Make an outline of the steps in this proof and show how each is needed for the one that follows. 261. EXERCISES. 1. Given a triangle whose sides are 2, 3, 4. Construct a triangle having its angles equal respectively to those of the given triangle and having a side 10 corresponding to the given side 2. 2. If each of two triangles is similar to a third triangle, they are similar to each other. 3. If in a right triangle a perpendicular be drawn from the vertex of the right angle to the hypotenuse, show that each of the triangles thus formed is similar to the given tri- angle, and hence that they are similar to each other. 4. In the figure of Ex. 3, make a table showing which angles are equal and which pairs of sides are corresponding in the fol- lowing pairs of triangles: ACD and ACB, CDB and ACB, ACD and CDB. 5. On a given segment as a side show how to construct a triangle similar to a given equilateral triangle. 128 PLANE GEOMETBT. 262. Theorem. The square on the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides. Given A ABC with a right angle at C. Call the lengths of the sides opposite A A. B, C, respectively, a, b, c. To prove that a', then b < V. Proof : Let c be the length of the hypotenuse in each. Then a2 + J2 = c2 and a'2+J'2=c2. (Why?) •. a^ + h^-. ' + J'2. (Why?) and a2 _ a'2 = 6'2 _ J2. (-yVhy ?) Since a>a', the left member of the last equation is positive, and hence the right member is also positive, that is, bA'Bf. To prove that AB is nearer the center than A'b' . Proof: Draw the -Is 6 and V and the radii c and c' . Then a and a' are halves of AB and A' b' respectively. Now complete the proof, using § 264. 266. Theorem. State and prove the converse of the theorem in § 265, using the same figure. 267. Definitions. A continued proportiod is a series of equal ratios connected by signs of equality. a c e Q The perimeter of a polygon is the sum of its sides. MEASUREMENT OF LINE-SEOMENTS. 131 268. Theorem. In a continued proportion, the sum of the antecedents and the sum of the consequents form a ratio equal to any one of the given ratios. n c & Q Given the continued proportion - = - = -=-• b d f h To prove that Proof : Let a + c ■\- e+ g a b + d+f+h^b' a b = '- Then, -^ = r, j = r, |=r. Hence, a = br, c = dr, e =fr, g = hr, and a + c + e + g = br+dr +fr + hr = (b + d +f+ h)r, or g + c+e+g' ^ a b+d+f+h \ Give all reasons in full. 269. Theoeem. The perimeters of two similar poly- gons are in the same ratio as any two corresponding sides. A ■ — ^ f Proof : By definition of similar polygons AB BC CD DE EA a'b' b'c' c'b' d'e' e'a' Complete the proof. 132 PLANE GEOMETRY. 270. Theorem. // the diagonals drawn from one vertex in each of two polygons divide them into the same number of triangles, similar each to each and similarly placed, then the two polygons are similar. Given the diagonals drawn from the vertices A and A' in the polygons P and P', forming the same number of triangles in each, such that A / ~ A /', AII~ £\ir, A in ~ A HI'. To prove that P ~ p' . Outline of Proof : (1) Use the hypothesis to show that Z. A = /- a', Zb = Z b', Z c=Z c', etc. (2) Show that AB A'B' BC '' B'C' CD C'D' , etc. Notice that the proportion — — = follows from b'c' c'd' nc AC A'C CD ' C'D' OVhy?) Give the proof in detail. 271. Theorem. Sidtc the converse of the preceding theorem, and gJve the proof in fill detail. Mako an outline of all t/ie strpx in the proofs of these two tlieoreins. MEASUREMENT OF LINE-SEGMENTS. 133 272, Theorem. If through a fixed point within a circle any number of chords are drawn, the product of the segments of one chord is equal to the product of the seg- ments of any other. Given O C with any two chords AB and DE intersecting in P. To prove that AP pb = EP ■ PD. Proof : Draw EB and DA. Then, Z 1 = Z 2 and Z 3 = Z 4. (Why?) Hence, A EPB ~ A PDA. (Why?) Which are corresponding angles, and which are corre- sponding sides? Show that ^=IR. EP PA Complete the proof. It follows from this theorem that if a chord AB is made to swing around the fixed point P, the product AP ■ PB does not change, that is, it is constant. 273. EXERCISES. 1. Which chord through a point is bisected by the diameter through that point? Why? 2. Through a given point within a circle which chord is the short- est? Why? 3. The product AP ■ PB is the area of the rectangle whose base and altitude are the segments of AB. (See § 307.) Note that this area is constant as the chord swings about the point P as a pivot. 134 PLANE GEOMETRY. 274. Definition. If a secant of a circle is drawn from a point P without it, meeting the circle in the points A and B, then PB is called the whole secant and PA the external segment, provided A lies between B and P. 275. Theorem. If from a fixed point outside a cir- cle any number of secants are drawn, the 'product of one whole secant and its external segment is the same as that of any whole secant and its external segment.

o and /.BAC measured very accurately with an instrument. Sides AC and BG may now be computed by means ^ d of the theorem, § 280, and a table of sines (see page opposite). By measuring Z BAC and Z ACT), CD may be computed. By this process, called triangulating, it is possible to survey over a large territory without directly measuring any line except the first. 284. The saving of labor afPorded by this indirect method of measuring is very great, and especially so in a rough and mountainous country, since measuring the straight line distance from one mountain peak to another by means of a measuring chain is impossible. In practice, tables of logarithms are used and the sines are carried out to a larger number of decimal places, but the general process is that used on the preceding page. 285. EXERCISES. Using the table on the next page, solve the following examples; 1. Given A = 53", B = 65°, a = 11. .'"i. Find ft, c, and /. C. 2. Given B = til , C = 71°, a = 19.8. Find ft, c, and ZA. 3. Given A = 6.5°, a = 14, ft = 1L>. Find ZB.ZC, and c. Solution, ^^i^ = - or sin B = sin .1 n * = .91 x ~ = .78. Sin A a u 14 From the table we find that sin 51' =.78. Hence B = 51°. ZC and c may now be found as before. MEASUREMENT OF LINE-SEGMENTS. 139 4. Given A =71°, a = 19.5, b = 17. Find ^B, ^C, and c. As be- From the fore, sin £ = sin .4 X - = sin 71° x :.95x— = .828. 19.5 a 19.5 table -we find sin 55° = .82 and sin 56° = .83. But sin B is nearer .83 than .82, and hence B = 56° is the nearest approximation using a degree as the smallest unit. Angle Sin Angle Sin An^le Sin Angle Sin Angle Sin Angle Sin 0° 15° .26 30° .50 45° .71 60° .87 75° .97 1° .02 16° .28 31° .52 46° .72 61° .87 76° .97 2° .03 17° .29 32° .53 47° .73 62° .88 77° .97 3° .05 18° .31 33° .54 48° .74 63° .89 78° .98 4° .07 19° .33 34" .56 49° .75 64° .90 79° .98 5° .09 20° .34 35° .57 50° .77 65° .91 80° .98 6° .10 21° .36 36° .59 51° .78 66° .91 81° .99 7° .12 22° .37 37° .60 52° .79 67° .92 82° .99 8° .14 23° .39 38° .62 53° .80 68° .93 83° .99 9° .16 24° .41 39° .63 54° .81 69° .93 84° .99 10° .17 25° .42 40° .64 55° .82 70° .94 85° 11° .19 26° .44 41° .66 56° .83 71° .95 86° 12° .21 27° .45 42° .67 57° .84 72° .95 87° 13° .22 28° .47 43° .68 58° .85 73° .96 88° 14° .24 29° .48 44° .69 59° .86 74° .96 89° 286. Definition. An object AB is said to subtend an angle APB from a point P if the lines PA and PB are sup- posed to be drawn from P to the extremities of the object. 287. EXERCISES. 1. A building known to be 150 feet high is seen to subtend an angle of 20°. How far is the observer Horn the building if he is standing on a level with its base ? 2. Show how to find the distance from an accessible to ah inacces- ible point by means of § 280 and the table of sines. 3. A flagstaff is 125 feet tall. How far from it must one be in order that the flagstaff shall subtend an angle of 25°? 140 PLANE GEOMETRY. PROBLEMS OF CONSTRUCTION. 288. Instruments. In addition to the ruler, compagseg, and protractor described in § 44, tlie parallel ruler is convenient for drawing lines through given points parallel to given lines without each time making the construction of § 101. Description. R and R' are two rulers of equal length and width. AB and CD are arms of equal length pivoted at ^ the points A. B, C, and D, making AC = BD. \ T Why do the rulers, when thus constructed, i -l- •/■ i t;' remain parallel as they are spread ? 289. Segments of equal length can be laid off with great accuracy on a line-segment b}' means of the com- passes. The following construction makes use of the compasses and parallel ruler. 290. Problem. To divide a given line-segment into any number of equal parts. Construction. Proceed as in § I.V.), Ex. 5, using the parallel ruler to draw the parallel lines. Clive the construction and proof in full. Historical Note. The idea of similarity of geometric figures, or "sameness of shape," is one of early origin, as is also the simple theory of proportion. It was probably used by Pytliagoras to prove the famous tlieorem known by liis name. (Si'e § Jli'J.) But the dis- covery by him of Uie iueoinnuMisurable ease (§ 'i^iH) showed that tliis theory was iua(lei]uale for llie rienrous proof of all theorems on simi- lar figures. It remaineil for Kuiloxus, the teadier of Plato, to perfect a rigorous tliemy of ratio and proportion. MEASUREMENT OF LINE-SEGMENTS. 141 Euclid, following his predecessors, deals with ratios of magnitudes in general as well as of numbers. Later writers have frequently in- sisted that ratios in general are not numbers. But nothing is gained by this procedure, since they possess all the properties of numbers. Iq this book a ratio is treated simply as the quotient of two numbers. See §§ 238-242. 291. Problem. Given three line-segments m, n, p, to construct a fourth segment q such that — = - ; that is, to n q find a fourth proportional to m, n, and p. Construction. Draw two indefi- nite straight lines, Ax and A>/ mak- ing a convenient angle. On Ay lay off AB ==m, BC= n. ^^ On Ax lay off AT>=f. Draw BD. ^ P o e Through C draw GE II BD. Then BE=q is the required segment. Give the proof in detail. 292. EXERCISES. 1. Apply the method of § 290 to bisect a given line-segment, and compare this with the method of § 5L 2. Divide a line 7| inches long into 11 equal parts by the method of § 290, and compare with the process of measuring by means of an ordinary ruler giving inches and sixteenths. 3. Using the same segments as in § 291, construct a segment q such that - = -; also such that - = -. p q Pi 4. Using the same segments as in the preceding, construct a seg- ment q such that £- = -; also such that £ = '". m q n q What is q called in each case ? 5. If the given segments m,n,p are respectively 3f inches, 5^^ inches, 45 inches, compute such that — = •£. Also construct j as in § 291 ,, n q and compare results. 142 PLANE GEOMETRY. 293. Definitions. If three numbers a, m, and h are such that a : m = m : b, then m is the mean proportional between a and b, and b is the third proportional to a and m. 294. Pkoblem. To construct a mean proportional be- tween- two given segments m and n. A Construction. On an indefinite line xy lay off AB = m and BC = n. On AC as a diameter construct a semicircle. At B erect a perpendicular to AC meeting the semicircle in D. Then BB =p is the required segment. Proof. Draw AD and CD and prove Complete the proof. 295. EXERCISES. 1. Sliow that in § 277 DP is a mean proportional between PB andP.4. 2. If in tiie above problem m = 3, and h = 5 show that the con- struction gives n/TT). 3. Show how to construct a segment p = V5, also /) = \ 3, ;) = \'2, by means of the above process. 4. Show how to construct a square equal in area to a given rectangle. 5. Show how to construct on a given base a rectangle equal in area to a given square. Suppose AB = m is the given base and BD =p a side of the given , square, the two segments being placed at right angles as in the ligure above. The problem is then to find on the line .IB the center of a circle which passes through tlie points A and 7^. To do this connect .1 and I) and construct a jierpendicular bisector of this segment meeting AC in a point which is tlic cmitcr of the i-equired circle. BC = n is then the required side of the rectangle since m • it =p'^. MEASUREMENT OF LINE-SEGMENTS. 143 296. Pkoblem. To divide a line-segment into three parts proportional to three given segments m, n, p. A D E B Construction. Let AB be the given segment. Con- struct Ax and on it lay off m, n, p as shown in the figure. Complete the figure and give proof in full. 297. EXERCISES. 1. Divide a line-segment 5 inches long into three parts proportional to 2, 3, and 4. 2. Divide a segment 11 inches long into parts proportional to 3, 5, 7, and 9. First compute the lengths of the required segments, then construct them and measure the segments obtained. Compare the results. Which method is more convenient ? Which is more accurate ? 3. Divide a segment 9 inches long into two parts proportional to 1 and V2. Also compute the required segments. Which is more convenient ? More accurate ? 4. Divide a segment whose length is Vll into two parts propor- tional to V2 and VS. First construct the segments whose lengths are y/2, Vs, and vTT. Also compute the required segments. ■ 5. Divide a given line-segment into parts proportional to two given segments m and n, (a) if the division point falls on the segment; (6) if the division point falls on the segment produced. See § 252. 6. A triangle is inscribed in another by joining the middle points of the sides. (a) What is the ratio of the perimeters of the original and the inscribed triangles? (6) Is the inscribed triangle similar to the original? 144 PLANE GEOMETRY. 298. Problem. On a given line-segment as a side construct a triangle similar to a given triangle. Construction. Let ABC be the given triangle and a'b' tlie given segment, and let it correspond to AB of the given triangle. Construct Z 2 = Z 1 and Z 4 = Z -3 and produce the sides till they meet in c'. Then A'b'c' is the required tri- angle (AVhy?). 299. EXERCISES. 1. In tlie problem of § 298 construct on A'B' a triangle similar to ABC such that A'B' and BC are corresponding sides. 2. Show that on a given segment three different triangles may be constructed similar to a given triangle. 3. Solve the problem of § "-'08 by making Z2 = Zl and construct- ing A'C so that — — -= I, ,, • Give the solution and proof in full. A'B' ^l'f_ 4. On a given segment as a side construct a polygon similar to a ' given polygon, liy first dividing the given polygon intcj triangles and then constructing triangles in order similar to these. Apply § 'JTO. 5. On a given segment construct a polygon similar to a given polygon in a manner analogous to (he methoil used in Kx. 3 for con- structing a triangle similar to a given triangle. 6. ABC is a right triangle with the hypotenuse AB. and CDXAB. Prove AC a mean proportional between A FS and AD and likewise ' 'II a mean proportional between AB and DB. 7. Use the thiMirem of § 277 to construct a square equal in area to that of a given rec^lanuli'. 8. The tangents to two intersecting cireles from any point in their common chord picMlneed are eqnab L'se J 277. MEASUREMENT OF LINE-SEGMENTS. 145 9. In the figure ^D and BE are tangents at the extremities of a diameter. H BD and AE meet in a point C on the circle prove that AB is a mean pro- portional between AD and BE. 10. The greatest distance to a chord 8 inches long from a point on its intercepted arc (minor arc) is 2 inches. Find the diameter of the circle. Use § 272. 11. At the extremities of a chord AB tangents are drawn. From a point in the arc AB perpen- diculars PC, PD, PF are drawn to the chord and the tangents. Prove that PC is a mean propor- tional between PD and PF. Suggestion. Draw segments AP and BP and prove A ^PF~ A iJCP and AAPC-^BPD. Then AP^FP_ . AP ^PC PB PC ™ PB PD SUMMARY OF CHAPTER III. 1. Make a list of the definitions in Chapter III. 2. Make a list of the theorems on proportional segments involving triangles. / 3. State the various conditions which make triangles similar. 4. State the conclusions "which can be drawn when it is known that two triangles are similar. 5. State the theorems on proportional segments involving poly- gons. 6. State the theorems on proportional segments involving straight lines and circles. 7. State in the form of theorems the various ways in which a proportion may be taken so as to leave the four terms still in proportion. 8. Under what conditions is a segment a mean proportional be- tween two given segments. For instance, see § 277. 9. Make a list of the problems of construction in Chapter III. 10. State some important applications of the theorems on propor- tion. (Return to this question after studying those which follow.) 146 PLANE GEOMETRY. PROBLEMS AND APPLICATIONS. 1. The middle points of adjacent sides of a square are joined. (a) Prove that the inscribed figure is a square. (J) What is the ratio of the inscribed and the original squares. (c) If a side of the original square is a, find a side of the inscribed square. (d) If a side of the inscribed square is J, find a side of the original square. 2. Two strips intersect at right angles. (a) If the width of each strip is 3 inches, what is the largest square which can be placed on them so that its sides will pass through the corner points as shown in the figure? (The corners will bisect the sides of the square.) (i) What is the side of this square if the width of each strip is n ? (c) Find the side of the square if the width of one strip is 3 and that of the other 5. (rf) Find the side of the square if the width of one strip is a and that of the other 6. 3. In the figure A BCD is a square and EFCH ■■■ is a regular octagon. (a) Show that .1 E, EF, FB are proportional to 1, \/2, 1. (Assume AE = 1 and find XE.) A F 1 J- \ -I \'^ (b) If . I B = f7, show that -ii- = ^ ^ = —= • (c) Find Z FEX. Seo § 163, Ex. 2. Id) Show that A = A F, and hence that the regular octagon may be constructed as shown in the figure. (e) Show that AAOE = 2-2\° by § 21i), and use this to make another proof that EI-'dH ■■■ is a regular ()cla,i;on. 4. Find lh(^ area of a square whose diagonal is rf. Tile Pattern. a a a o -f °'°i°i°! a Tile Pattern. i > L K t y \ ~a MEASUREMENT OF LINE-SEGMENTS. 147 5. If a side of a regular octagon is a, find its area, 6. A floor is tiled withi regular white- octagons and black squares as shown in the design. A\'hat per cent (approxi- mately) of the floor is black? 7. Divide a circle into eight equal parts and join alternate division points. (a) Prove that LMNPQ ■■■ is a regular octagon. Cut Glass Design. (5) If the diameter of the circle is a, show that LE = 2(\/2 — 1), Suggestion. Find HE and then use Ex. 3, {d). 8. Problem. Inscribe a square in a given semicircle. Construction. Let AB be the diameter of the given semicircle. At B construct a perpendicular to AB, making BD = AB. Connect D with the center C, meeting the circle at E. Let fall EF perpendicular to AB. Then EF is a side of the required square. Complete the figure and make the proof by showing that EF=2 CF. 9. Pboblem. Inscribe a square in a given triangle. Construction. Let ABC be the given tri- angle. Draw CD parallel to AB, making CD = CO. Complete the square CDEO. Draw DA meeting CB in F. Draw FG II AB, GH II CO, and FK II CO. Then FGHK'^ CDEO and hence is the required square. 148 PLANE GEOMETRY. Touch down "o'' !»''» 10. The sides of a triangle are 13, 17, 19. Find the lengths of the segmeuts into which the angle bisectors divide the opposite sides. 11. The angles of a triangle are 30"^, W, 90'^- Find the lengths of the segments into which the angle bisectors divide the opposite sides if the hypotenuse is 10. 12. Prove that the perpendicular bisectors of all the sides of a polygon inscribed in a circle meet in a point. 13. An equilateral polygon inscribed in a circle is equiangular. 14. The goal posts on a football field are 1 8J feet apart. If in making a touchdown the ball crosses the goal line 25 ft. from the nearest goal post, how far back should it be carried so that the goal posts shall subtend the gi'eate^t possible angle from the place where the ball is placed? See Ex. 22, page 111, and § 277. 15. Solve the preceding problem if the touchdown is made a feet from the nearest goal post, and thus obtain a formula by means of which the distance may be computed in any case. 16. A triangle has a fixed base and a constant vertex angle. Show that the locus of the vertex .^-'- ^\ consists of two arcs whose end-points are the ex- ,• - ^ y I y'^ \ tremities of the base. See Ex. 2(1. page 111. I ■ /.\\ \ Xote that the locus also includes an arc on the \ x "^ ' tVi *B opposite side of AP, from the one shown in tlie ' figure. Tt follows that if two points .-1 and B subtend the same angle from P and from Q, then a circle may be passed through A,B,P, Q. 17. In the figure, A. B, F lie in the same straight line, as do also />, ( ', F. and /i 1 = Z2. ((0 If /JF = M, (F= !), nn.l DC = 12, find AH. (h) IE A/l = 1(10, AF = 'JM). CF = iW. find IX'- N(i ri-,. l!y holding a small object, say a pencil, at linn's lenj;(,li, and sighting aeross the ends of it, we iiiiiy ileli'nnine ap|iroxiiiiately whether two given (ilijeets subtend the same angle. A MEASUREMENT OF LINE-SEGMENTS. 149 18. Not having any instruments, an engineer proceeds as follows to obtain approximately the distance from an accessible point B to an inaccessible point A . Walking from B along j^ the line AB he takes 50 steps to F. Then he walks in a convenient direction 50 steps to C, and notes that A and B subtend a certain angle. He then proceeds along the same straight line until he reaches a point D at which A and B again subtend the same angle as at C. He then concludes that DC = AB. Is this conclusion correct? Give proof. 19. If the height of a building is known, show how the method of Ex. 18 can be used to determine the height of a flagstaff on it. 20. A building is 1.30 feet high, and a flagstaff on the top of it is 60 ft. high ; 130 feet from the base of the building in a horizontal plane, the flagstaff subtends a certain angle. How far from the build- ing along the same line is there another point at which the staff sub- tends the same angle? At what distance does it subtend the same angle as it does at 300 feet? 21. If the diagonals of an isosceles trapezoid are drawn, what similar triangles are produced? 22. Find the locus of points at a fixed distance from a given tri- angle, always measuring to the nearest point on it. 23. The line bisecting the bases of any trapezoid passes through the point of intersection of its diagonals. Suggestion. Let E bisect DC, and draw EO meeting AB in F. Prove AF = FB. 24. If two triangles have equal bases on one of two parallel lines, and their vertices on the other, then the sides of these triangles intercept equal segments on a line parallel to these and lying between them. 25. A segment bisecting the two bases of a trapezoid bisects every segment joining its other two sides and parallel to the bases. (Prove SF = HG and FK= KH, using Ex. 24.) - ■« - 26. In a triangle lines are drawn parallel to oue side, forming trap- ezoids. Find the locus of the intersection points of their diagonals. eO^ 150 PLANE GEOMETRY. 27. In the figure D, E, F are the middle points of the sides of an equilateral AABi' Arcs are constructed with centers A, B, C as shown in the figure. (a) Prove that these arcs are tangent in pairs at the points D, E, F. (b) Construct a circle tangent to the three arcs. (c) What axes of symmetry has this figure? 28. In the figure ABC is an equilateral triangle. Arcs AB, BC, and CA are constructed with C, A, and B as centers. AF, BE, CD are altitudes of the triangle. Prove O OG tangent to Ah, bT\ and 'T3 . 29. In the equilateral triangle ABC, JB, BC, and CA are constructed as in the preceding ex- ample. DE, EF, FF) are constructed as in Ex. 27. The figure is completed making ADE, F)FB, and EEC similar to A BC. (a) Construct a circle tangent to a7), Si^, and EA as shown in the figure. (h) Construct a circle tangent to ^1), £P, and Pi:. 30. In the figure A CB is a semicircle. Arcs DE and DF are constructed with B and ^-1 as centers. If AD - 4 feet, find the radius of O OC. Solution. Show that the value of r is derived from the equation (4 + r)^ = (4 — r)^ + 4-. 31. Construct the accompanying design. Notice that the points A , B, if, N, F, E divide the circle into six equal arcs. See Ex. .5, From Boynton Cathedral, Englaud. MEASUREMENT OF LINE-SEGMENTS. 151 The § 163. The arcs A 0, OB, etc., and the circle with center 0', are constructed as in Ex. 28. The small circle with center at Q is con- structed as in the preceding example. 32. ABC is an equilateral arch and AEB is a semicircle, (a) If AB (the span of the arch) is 10 feet, _ G find the radius of the small circle tangent to the semicircle and the arcs of the arch. (&) Find the radius OE of the circle if AB = s. (e) If OS = 2, find ^B. 33. The accompanying design consists of three A semicircles and three circles related as shown in the figure. (a) If AB = 12, find OQ by using the triangle OSB. method is similar to that used in Ex. 30. (b) Using the right triangle DRO', find O'x. Notice that in order to have the circles with the centers and 0' tangent to each other the sum of their radii must be equal to RD. (c) If AB = s show that 0Q=^ and 6 O'x = — , and construct the figure. 12 ^ 34. Upon a given segment AB construct the design shown in the figure. Notice that it consists of the two preceding figures put together. Compare the radii of the circles in this design. A E H B Ospedale Maggiore, Milan. 35. Between two parallel lines construct semicircles and circles as shown in the figure. Solution. Suppose the construction made. Let HE = r, and OQ = r'. Using the triangle EHO, show that r' = - and then con- struct the figure. 152 PLANE GEOMETRY. 36. Between two parallel lines construct circles and equilateral arches as shown in the figure. Hint. Hence That is, The Doge's Palace, Venice. 4Q = 2Pff and AH=iPH. HQ' =APH'~'2PH' = 5PH'. PH :HQ = 1: VS. (Why ?) Now divide A C into two segments proportional to 1 and Vb, and so construct the figure. C 37. ABC is an equilateral arch. AD=lil;. AF=EB=^AB. Semicircles are constructed with E and F as centers, and radius FB. (n) Jf AB = s, find OH and construct the figure. Solution. Let OH ■■ F0 = - + r. 3 Then BO -. Fi5=-, Z)F=-, 3 6 From A ODB, OD^ = Olf ~ inf = (n - r)- - i^Y . (\) From A <)I>F, fVT' =?//•'-- /;/.'- = f -I*- + r]" - [i]''. (1>) From (1) and ('J) (s -^ i-y -- (^'LJ = (t+ i^ -(^t^. Hence, sliow that r = J^ s. ( 'onstrnct the liguri^ (/-) If in the precfdinj; 0//= 4 fwt. find .1/.' and .-IF. MEASUREMENT OF LINE-SEGMENTS. 153 38. Between two parallel lines construct cii-cles and semicircles having equal radii as shown in the figure. Prove that the ratio — = — , RLK being HQ RS ^ an equilateral triangle and KS = SL. Divide CA in the ratio of .Si : RS. k's'^'I' 39. Through a fixed point on a circle chords are drawn and each extended to twice its length. Find the locus of the end-points of these segments. Compare Ex. 6, § 2U. 40. If a quadrilateral is circumscribed about a circle, show that the sums of its pairs of opposite sides are equal. 41. On a diameter produced of a given circle, find a point from which the tangents to the circle are of a given length. Solve this problem by construction, and also algebraically. 42. Compare the perimeters of equilateral triangles circumscribed about and inscribed in the same circle. 43. In a given square construct semicircles each tangent to two sides of the square and terminating on the diameters of the square. Construction. Connect E and F two ex- O O tremities oi diameters and on EF as a diame- ter construct a semicircle with center at 0'. Draw O'H perpendicular to ^i? meeting the arc in H. Draw OH meeting AB in K. Draw KO" perpendicular to AB meeting the diagonal AC in 0". Then 0" is the center of the re- quired circle. I FE. ^Ye need to prove that 0"K = O'lL Proof. Draw C'L [\00'E noil A 00"L, and hence ^ = Also A 00' H ' 00" _ OQi ' 0"L ^ 0"K O'E ~ O'H ' But 01 E = O'H, and hence 0"L= 0"K. O'E 0"K O'H' 0"L. (Why?) (1) From (1) and (2) ^ = ; (2) (Why?) This figure occurs in designs for steel ceilings. CHAPTER IV. AEBAS OP POLYGONS. AREAS OF RECTANGLES. 300. Heretofore certain properties of plane figures have been studied, such as congruence and similarity, but no attempt lias been made to measure the extent of surface inclosed by such figures. For this purpose we first con- sider the rectangle. 301. The surface inclosed by a rectangle is said to be exactly measured when we find how many times some unit square is contained in it. E.g. if the base of a rectangle is five units long and its altitude three units, its surface con- tains a square one unit on a side iifteeu times. 302. The number of times which a unit square is con- tained in the surface of a rectangle is called the numerical measure of the surface, or its area. We distinguish three cases. 303. Case 1. If the sides of the gi\en rectangle are in- tegral 'iiiiilli/ilc.'i of the sides of the unit square, then the area of the. ri'claugle is determined by finding into how man}' unit squares it can be divided. Thus if the siili's (if a rectangle are m and n units respectively, then it can be divided inln m rows of unit sniiares, each row containing « squares. Hence the area of such a rectaui;le is m x h unit squares, that is, in this ease, ^^^^ ^ ^^^^ ^ altitude. (1) 164 AREAS OF POLYGONS. 155 o ::::xi;: + ::::::;:±::^ffr; J::: 84 cm. 304, Case 2. If the sides of the rectangle are not inte- gral multiples of the side of the chosen unit square, but if the side of this square can be divided into equal parts such that the sides of the rectangle are integral multiples of one of these parts, then the area of the rectangle may be expressed integrally in terms of this smaller unit square, and fractionally in terms of the original unit. For example, if the base is 3.4 deci- meters and the altitude 2.6 decimeters, then the rectangle cannot be exactly di- vided into square decimeters, but it can be exactly divided into square centi- meters. Each row contains 3i centi- meters and there are 26 such rows. Hence, the area is 31 x 26 = 884 small squares or 8.84 square decimeters. But 3.4 X 2.6 = 8.84. Hence, in this case also, area = base x altitude. (2) 305. Case 3. If a rectangle is such that there exists no common measure whatever of its base and altitude, then there is no surface unit in terms of which its area can be exactly expressed. But by choosing a unit sufficiently small we may determine the area of a rectangle which differs as little as we please from the given rectangle. E.g. if the base is 5 inches and the altitude is v'5 inches, then the rectangle cannot be exactly divided into equal squares, however small. But since v'5 = 2.2361. ••, if we take as a unit of area a square whose side is one one-thousandth of an inch, then the rectangle whose base is 5 inches and whose altitude is 2.236 inches can be exactly meas- ured as in cases 1 and 2, and its area is 5 x 2.236 = 11.18 square inches. The small strip by which this rectangle differs from the given rec- tangle is less than .0002 of an inch in width, and its area is less than 5 X .0002 = .001 of a square inch. By expressing Vb to further places of decimals and thus using smaller units of area, successive rectangles may be found which differ less and less from the given rectangle. 166 PLANE OEOMETRY. 306. An area thus obtained is called an approximate area of the rectangle. For practical purposes the surface of a rectangle is measured as soon as the width of the remaining strip is less than the width of the smallest unit square available. From the foregoing considerations we are led to the following preliminary theorem : 307. Theorem. The area of a rectangle is equal to the product of its base and altitude. 308. The argument used above shows that the theorem (§ 307) holds for all rectangles used in the process of approximation, and hence it applies to all practical meas- urements of the areas of rectangles. AREAS OF POLYGONS. 309. From the formula for rectangles, area = base x altitude, we deduce the areas of other rectilinear figures by means of the principle : Two rectilinear figures are equivalent (that is, have the same area) if they are congruent, or if they can be divided into parts which are congruent in pairs. E.g. the two figures here shown are equiv- / iv / alent siiiip AT and Til are congruent respec- "^ "I/' /ii' / tively to II and IV. ~ ' L.—J It can be shown that for any two given rectilinear figures, eillier it is ]i(>ssible to divide them into parts which ;ire congruent in pairs, or inic of the figures incloses a greater ;ire;i than the other. Hence the lesl s|>iTified is suffioient for all rectilineal' figures. The s\ inbol = joining two polygons luoaiis I'tjiiiralent ov f(jU(il in iirra. AREAS OF POLYGONS. 157 310. Theorem. The area of a parallelogram is equal to the product of its base and altitude. D E G F in Given the parallelogram ABCD whose base is AB and whose altitude is AE. To prove that area ABCD = AB x AE. Proof : Draw BF ± to DC produced, forming the rec- tangle ABFE, whose base is AB and altitude AE. Then area ABFE= AB x AE (Why ?). If now we prove A I ^ A II, then the parallelogram is composed of parts I and III which are congruent respec- tively to parts II and III of the rectangle, and hence the parallelogram and rectangle are equal in area. Give this proof in full. 311. EXERCISES. 1. In the figure CF is perpendicular to AB and AE to CB. Prove that AB x CF = BC y. AE. Suggestion. (Show that A.4££ ~AC£F.) It follows that the same result is obtained if either of two adjacent sides of a parallelogram is taken as the base. ^ ^ -^ 2. Construct the figure described in Ex. 1 in such manner that the point E falls on BC extended and prove the theorem in that case. In the following Zj and l^ are parallel lines. 3. Two parallelograms have equal bases lying on Zj and l^. Show that they are equivalent. 4. One base of a parallelogram is fixed on Zj and the other moves along l.^. Does the area change ? 158 PLANE GEOMETRY. 312. Theorem. The area of a triangle is equal to one half the product of its base and altitude. Given the A ABC whose altitude upon the side AB is CD. To prove that the area of AABC= I {AB X CD). Proof : Let be the middle point of AC. Complete the parallelogram ABFG. Prove that A CGH ^ A BFH, and hence show that A ABC and O ABFG have equal areas. Hence the area of A ABC = ED x AB. Hence the required area is i (^AB x CD). But ED = ^ CD. 313. EXERCISES. 1. Prove the theorem of § 312 us- ing the accompanying figure. 2. Prove this theorem also liy draw- ing a figure in which the given triangle is one half of a parallelogram. 3. If the middle points of two A B D adjacent sides of a parallelogram are joined, a triangle is formed whose area is equal to one eighth of the area of the parallelogram. In the following exercises /, and l^ are parallel lines. 4. Show that two triaiit;les are equivalent if their vertices lie in /j and their equal bases in /j. 5. A triangle has a fixed base in L,. If its vertex moves along /j, what can you say of its area? 6. If the base of a trianult' is tixcd and if its vertex moves so as to preserve the area constant, wliat is the locus of the vertex? 7. A line is drawn from a vertex of a triangle to the middle point of the opposite side. Cumpare the areas of the triangles thus formed. If a scgiiicnt i.s drawn from tlii> \i>rtt>x to the point P in the base AB, show that the areas of the triangles are in the ratio AP:PB. AREAS OF POLYGONS. 159 314. Theorem. The area of a trapezoid is equal to one half the sum of its bases multiplied by its altitude. Proof : Draw a diagonal, thus forming two triangles having a common altitude. Form the ex- pression for the sum of the areas of these triangles. ^ ^ Give the proof in detail. 315. Theorem. The square erected on the sum of two line-segments as a side is equal to the sum of the squares erected on the two segments separately, plus twice the rec- tangle whose base is one segment and whose altitude is the other segment. Proof : Let a and h represent the numerical measures of the two line-segments. Then, the square erected upon the segment a+ h may be subdivided as shown in the figure, giving Give the construction and jDroof in full. 316, EXERCISES. 1. Show by a figure that (a + b + cy = a^ + b^ + c^ + 2 ab + 2 ac + 2 be. 2. If ABC is any triangle and AD, BE, and CF are its altitudes, show that ^ = ^, and M = -i^. BE BC CF AC Hence show that AD X CB = BE X CA = CFx AB. 3. The side AB of O ABCD is fixed. What is the locus of the points C and D if CD moves so as to leave the area of the parallelogram fixed ? 160 PLANE GEOMETHr. 317. Theorem. The square erected on the difference of two line-segments as a side is equal to the sum of the squares erected on the two segments separately, minus twice the rectangle whose base is one of the segments and whose altitude is the other. Proof : Use the figure to show that (a _ J)2 = a2 + j2 _ 2 ab. Give the construction and proof in detail. . &*[ /,A ,1 1 a-b 1 . 1 ab (o-bY j a-b j 6 b 318. EXERCISES. 1. Show by a figure that (a — 6) (a — c) = a^ + Ac — aJ — ac. 2. Likewise, show that (a -\- b)(a — b) = a^ — b^. 3. Show by counting squares in the accompany- ing figure that in case of an isosceles right triangle the square constructed on the hypotenuse is equal in area to the sum of the squares on the two legs. 4. In case the triangle is scalene, show by counting squares and congruent parts of squares that the square constructed on the hypotenuse is equal in area to the sura of the squares on the other two sides. 5. Using the third figure, show that the same theorem liolds for any right triangle. II INT. Call the legs of the given triangle n and b. Describe the construction of the auxiliary lines and give the proof in full. This theorem is proved again in the next paragraph iind was also ])roved in § ■_*(i"J. The proof just given is diw to lUiiiskara (born 111! A.D.), who constructed the " Behold 1 " V 1 t : 1 '/Tx ' ! I i / 1^ / q : Q \ 1 ; bv. oJT,-, y; -16- X 8- r^ \ N 1 ■ ' 1 fi iU, ' ■ ' ' ' ' [ 1 1 1 1 |,--^^ ' _^>^l .61 |\> ^ 1 ' ■ <\ \| ifi^fiV 3 ft 1 \^ -—16, ,b-V \ i 1 \ l\ 6' ^^-"^ i \ ^^^^ ' 1 i 1 simply said AREAS OF POLYGONS. 161 319. Theorem. The area of the square described on the hypotenuse of a right triangle is equal to the sum of the areas of the squares described on the other two sides. Given the right triangle ABC, and the squares I, II, and III con- structed as shown in the figure. To prove that D I = D II + D III. Proof : Complete the rectangle ABCE. Construct AGFH^AACE. Draw EH and produce AE to meet the line FH at K. Prove that (a) AAEC^Ahek. (5) AEHG is a parallelogram whose base AE and alti- tude HK are each equal to a side of D III. (e) ECFH a parallelogram whose base EC and altitude EK are each equal to a side of DII. But these two parallelograms together are equivalent to DI (Why?). The student should give the proof iu detail, making an outline of the steps and showing how each step is needed for the next. For ex- ample, why is it necessary to prove AHEK^AAEC. Compare the various proofs of this theorem that are given in this book. The method of counting squares shown on the opposite page is applicable to all cases where the two sides are commensurable. 162 PLANE GEOMETRY. Historical Note. The theorem of § 319 is one of the most impor- tant in all mathematics. It is now fairly certain that the general theorem was first stated and proved by Pythagoras, though the story that he sacrificed 100 oxen to the gods on the occasion may be ques- tioned. Special cases of the theorem were known to the Egyptians as early as 2000 B.C., e.g. that a triangle whose sides are 3, 4, 5 is right-angled. In this connection the Pythagoreans also discovered the irrational number, that is, that there are numbers such as V2 which cannot be expressed exactly as integers or as ordinary fractions. 320. EXERCISES. 1. The bases of a trapezoid are 8 and 12 inches, and its altitude is 8 inches. Find its area. 2. The bases of a trapezoid are 16 and 20 inches, respectively, and the area of the smaller of its component triangles is •'iO square inches. Find the area of the trapezoid. See the figure in § \r,2. 3. State in words the geometric theorem indicated in each of the following, and draw a figure to illustrate each case : (a) h{a +h) = ah + hh. (h) (a + by + (a-by = 2 (a^ + 52). (c) (a + by~(a~by = 4ab. 4. Show that the rectangle whose base is a -f 5 and whose altitude is a — b has the same perimeter as the square whose side is a. By means of Ex. 2, § 318, compare their areas. 5. If two triangles have the same base but their vertices are on opposite sides of it, and if the segment joining their vertices is bi- sected by the common base, extended if necessary, then the two triangles are equivalent. 6. State and prove the converse of the theorem in the preceding exercise. 7. The bases of a parallelogram lie on the parallel lines /, and /,. A triangle whose base is equal to that of the parallelogram has its vertex in /, and its base in I.,. Compare their areas. 8. Prove that all triangles having the same vertex and equal bases lying in the same straight line are equal in area. AREAS OF POLYGONS. 163 321. Theorem. The areas of similar triangles are in the same ratio as the squares of any two corresponding sides, or as the squares of any two corresponding altitudes. A ~C^D A' C'~D'' Given A ABC and A'ffC, with altitudes BD and B'D'. To prove that areaA^BC ~" " " " AB BC CA BD a.Tea.AA'B'c' A'b'^ b'c' c'a'" b'd'' Proof: But Hence AC BD area A ^^C _ \(AC y. BD) area Aa'b'c' ~ ^(a'c' x b'd') a'c' " b'd' AC BD area A ABC B'D' Ic" area AA'B'C' A'C' BD Why? Why? Give the proof for each of the other pairs of corre- sponding sides. 322. EXERCISE. Verify the preceding theorem by counting triangles in the accom- panying figure. 164 PLANE GEOMETRY. 323. Theorem. The areas of two similar polygons are in the same ratio as the squares of any two corre- sponding sides or any two corresponding diagonals. E E' Outline of Proof: Let I, II, • ■-, I', 11', ••• stand for the areas of pairs of similar triangles. I ^11 _III ^IV 1' 11' 111' IV' ' I +11 4.111 + IV ^ I 1' -2 Show (1) (2) (3) a'b'c'd'e'f' Give the proof in detail. 1' + 11' + 111' + IV' ABCDEF _ I ~1' (§ ^"1-) (By §321.) (By § 2t;8.) AB BC B^C' 324. EXERCISES. 1. Verify the above theorem by counting the number of equal hexagons into which these two similar hexagons are divided, and also taking the s( J I Hire of the length of one side in e:irh, using a sidt' of a small hexagon as a unit. 2. A certain triangular field containing 2 aiTcn is 10 ruils Idiif;' 1111 one .side. Find the area of a similar triangular Held whose corresponding side is 50 rods. AREAS OF POLYGONS. 165 3. The areas of two similar triangular flower beds are 24 square feet and 36 square feet respectively. If a side of one bed is S feet, find the corresponding side of the other. 4. If similar polygons are constructed on the three sides of a right triangle, show that the one described on the hypotenuse is equivalent to the sum of the other two. Suggestions. Let a, h, c represent the two legs and hypotenuse of the triangle and A,B,C the areas of the corresponding polygons. Then A a" , B h^ — = — and — = — . C c^ C c^ Hence - — ^ — = — — — = — = 1 or A + B = C. Give all reasons in C c2 c2 full. 5. Find a line-segment such that the equilateral triangle described upon it has four times the area of the equilateral triangle whose side is 3 inches long. 6. Show that the square on the altitude of an equilateral triangle is three fourths the square on a side. 7. If in a right triangle a perpendicular is let fall from the vertex of the right angle to the hypotenuse, show that the areas of the two triangles thus formed are in the same ratio as the adjacent segments of the hypotenuse, and also as the squares of the adjacent sides of the triangle. 8. Draw a line from a vertex of a triangle to a point in the oppo- site side which shall divide the triangle into two triangles whose ratio is 2 : 5. Also 2 : 1. 9. Divide a parallelogram into three equivalent parts by lines drawn from one vertex. Use the last construction in Ex. 8. 10. The sides of two equilateral triangles are 8 and 6 respectively. Find the side of an equilateral triangle whose area shall be equivalent to their sum. Use the result in Ex. 4. 11. State and solve a problem like the preceding for the difference of the areas. 12. Corresponding sides of two similar triangles are a and b. Find the side of a third triangle similar to these whose area is equal to the sum of their areas. 13. Likewise for any two similar polygons. 166 PLANE GEOMETRY. CONSTRUCTIONS. The theorems of this chapter lead to numerous construc- tions of practical importance. 325. Problem. To construct a square equivalent to the sum of two or more given squares. Construction. In the case of two given squares, con- struct a right triangle whose legs are sides of the two given squares. How is the desired square obtained ? In the case of three given squares, use the square result- ing from the iirst construction together with the third square, and so on. Give the construction and proof in full. 326. Problem. To construct a square equivalent to a given rectangle. Construction. If the base and altitude of the rectangle are h and a respectively, we seek the side s of a square such that s^— ah ; that is, we seek a mean proportional between a and h. See Ex. 4, § 295. Give the construction and proof in full here. 327. Problem. To construct a square equivalent to a given triangle. Construction. Show how to modify the preceding con- struction to suit this case. 328, EXERCISES. Show how to construct each of the following : 1. A square equivalent to the difference of two given squares. 2. A square equivalent to the sum of two given rectangles. 3. A square equivalent to tlie sum of two given triangles. 4. A square equivalent to the difference of two given rectangles, also to the difference of two given triangles. AREAS OF POLYGONS. 167 329. Problem. To construct a triangle equivalent to a given polygon. c PA EG Given the polygon ABODE. To construct the triangle PCO equivalent to ABODE. Construction. Cut off A ABC by the segment AC. Through B draw BP II AC to meet EA extended at P. Draw CP. In a similar manner draw CO. Then PCO is the required triangle. Proof : PCDE= ABODE since AAPG=AABC (Why?). Further, PCDE has one less vertex tjian ABCDE. But A PCO = PCDE since A ECD = A ECO (Why?). Hence, A PCO is equivalent to the given polygon. 330. Problem. To construct a rectangle on a given base and equivalent to a given parallelogram. Construction. Let b and h be the base j,' and altitude of the given parallelogram, b' the base of the required rectangle, and x the unknown altitude. Then we are to determine x so that h'x = bh, that is, b' : b = h : x. Hence, x is the fourth proportional to b', h, and h. (See § 291.) Construct this fourth proportional, showing the complete solution. This construction is attributed to Pythagoras. It represents a much higher achievement than the discovery of the Pythagorean proposition itself. 168 PLANE GEOMETRY. 331. EXERCISES. 1. Show how to modify the last construction in case the given figure is a triangle. Give the construction. 2. Construct a rectangle on a given base equivalent to a given irregular quadrilateral. 3. Construct a rectangle on a given base equivalent to an irregu- lar hexagon. 4. On a side of a regular hexagon as a base construct a rectangle equivalent to the hexagon. 5. Construct a parallelogram on a given base equivalent to a given triangle. Is there more than one solution? 6. Construct a square whose area shall be three times that of a given square ; five times ; one half the area ; one fifth. 7. Construct an isosceles triangle, with a given altitude h, equiva- lent to a given triangle. 8. Draw a line parallel to the base of a triangle and cutting two of its sides. How will the resulting triangle and trapezoid compare in area, (a) If each of the two sides of the triangle is bisected? (6) If each of the two sides of the triangle is three times the length of the corresponding side of the trapezoid ? 9. Construct a triangle whose base and altitude are equal and whose area is e(|ual to that of a given triangle. 10. In a parallelogram A BCD, any point E on the diagonal BD is joined to .4 and ('. Prove that A JUCA and BEt' are equivalent, and also that &> /^A'. I and !)/•:( ' are equivalent. 11. The sides of a triangle are 0, S. 9. A line parallel to the longest side divides the triangle into a tra]iezoid and a triangle of equal areas. Find tlie ratio in which the line divides the two sides. 12. Draw a line ]iarallel to the base of any triangle, and cutting two of its sides. How do tlie altitudes of (he resulting triangle and trape- zoid comjiMie, (ii) If they are ecpial in area? (i) ir the area of tlie triangle is Wiree times that of the trapezoid? ABEA8 OF POLYGONS. 169 13. Through a point on a side of a triangle draw a line dividing it into two equivalent parts. Solution. Let P be the given point. Draw the median BD. Draw BE II PD and draw PE. This is the required line. Prove. Suggestion. Is^tice that ADPE = ADPB by- Ex. 4, § 313, and hence that AEOD = ABOP. 14. Through a given point on a triangle draw a line which divides it into two figures whose areas are in the ratio ^. 15. Inscribe a circle in a triangle, touching its sides in the points D, E, F. With the vertices as centers, construct circles passing through these points in pairs. Show that each of these latter cii'cles is tangent to the other two. SUMMARY OP CHAPTER IV. 1. State what is meant by the area of a rectangle. Give the formula. 2. Give formulas for areas of parallelograms and triangles. 3. How is the formula for the area of a trapezoid obtained ? 4. What theorems of this chapter can be stated algebraically, as (a + 6)2 = a2 + 2o6 + b^ 5. State the theorem on the ratio of the areas of two similar tri- angles; two similar polygons. Give examples. 6. Tabulate the problems of construction given in this chapter. 7. If two rectangles have the same base, how does the ratio of their areas compare with the ratio of their altitudes? 8. If two triangles have equal altitudes, how does the ratio of their areas compare with the ratio of their bases? 9. State all theorems of this chapter proved by means of the Pythagorean proposition. 10. State some of the more important applications of the theorems in this chapter. Return to this question after studying the succeeding list of problems. 170 PLANE GEOMETRY. PROBLEMS AND APPLICATIONS. 1. Find the area of a square whose diagonal is 6 inches. 2. Find the area of a square whose diagonal is d inches. 3. ABCD is a square placed at the crossing of two strips of equal width, as shown in the figure. The small black square has two vertices on the sides of the horizontal strip and two on the sides of the vertical strip. (a) Find the area of each square when the width of the strips is 4 inches. (V) Compare the area of the black square and the white border surrounding it. (c) Can squares be placed as in the figure in case the strips are of unequal width? In the two following questions let the small square be drawn with two vertices on the sides of the horizontal strip and one diagonal parallel to these sides. (d) If the horizontal strip is 4 inches wide, what must be the width of the vertical strip in order that the large square may have twice the area of the small one ? Hint. The diagonal of the small square is 4 inches. (e) If the horizontal strip is a inches wide, what must be the width of the vertical strip in order that the area of the black square shall be - the area of the larger square ? 4. Prove that the area of a rhombus is one half the product of its diagonals. 5. Prove that the area of an isosceles right triangle is equal to the square on the altitude let fall upon the hypotenuse. 6. If the diagonals of a quadrilateral intersect at right angles, prove that the sum of the squares on one pair of opposite sides is equal to the sum of the squares on the other two sides. 7. Inscribe a square in a semicircle and in a quadrant of the same circle. Compare their areas. See Ex. 8, pai;e 14". 8. In the triangle ABC, CD is au altitude. E is any point on CI). If /'/•> is one half CD. compare the area of the triangle AllB and tlie sum of the areas of the triangles .1 /if and BEC. Also compare these areas if DE is one nih of DC. AREAS OF POLYGONS. 171 M M m W 9. ABCD is a square, and E, F, G, H are the middle points of its sides. On EF, points iV and Q are taken so that PN = PQ. Similarly KL = KM, also KL = NP, and so on. (a) Prove that AMOL is a rhombus. (b) It AB = 6 inches and if KM = ME, find the sum of the areas of the four rhombuses AMOL, BQON, etc. (c) If HE = 8 inches, what is the area of the whole square? (d) What part of KE must KM be in order that the sum of the rhombuses shall be \ the area of the square? (e) If AB = a, find KM so that the sum of the rhombuses shall be _ the area of the square. — „, n Parquet Flooring. (/) Prove that L, 0, R lie in the same straight line. 10. ABCD is a square and E, F, G, Ha.re the middle points of its sides. SN = PK= QL = RM. (a) li AB^Q inches and if Pi? = 1 inch, find the sum of the areas of the triangles EHN, EFK, FGL, and GHM. (6) If AB = 6, find PK so that the sum of the four triangles EFK, etc., shall be one fifth of the whole square. (c) li AB = % and if the points E, K, Q lie in a straight line, find the sum of the areas of these triangles. (d) If AB = a and if PK = one rath of PO, find the sum of the areas of the triangles. (e) li AB = a, find PK so that the sum of the triangles shall be one mth of the whole square. What will be the length of PK in case the tri- angles occupy one half of the whole square? 11. The sides of two equilateral triangles are a and h respectively. Find the side of an equilateral triangle whose area is equal to the sum of their areas. Parquet Flooring. 172 PLANE GEOMETRY. D \ M \ 1 /' 1 / F 'P H \ t) / A/\ L V -E.V 12. Construct a triangle similar to a given triangle and having 16 times the area. 13. The middle points of the sides of a quadrilateral are con- nected. Shovf that the area of the parallelogram so formed is half the area of the quadrilateral. 14. A BCD is & square. Each side is divided into four equal parts and the construction com- pleted as shown in the figure. (a) Prove that QNOH, KLOE, etc., are par- allelograms. (6) AN'liat part of the area of the square is oc- cupied by these four parallelograms? (c) What part of the area of the square is oc- cupied by the four triangles NEO, LGO, PFO, andil/fiO? (rf) U AB = %, find the lengths of KO and QP- (e) Find the ratio of the segments KO and QP. Does this ratio depend upon the length of ABl (/) In the parquet floor design what fraction is made of the dark wood? Does this depend upon the size of the original square? 15. On a given line-segment .I7J as a hypotenuse construct a right triangle such that the altitude upon the hypotenuse shall meet it at a given point D. 16. If ABC is a right triangle and CDA.AB, /'' ~^^C ■Hz = -AH . i^^-^-""'^^ 'X 5-2 ' •N// ^// "^M Parquet Flooring. prove that CB' DB 17. By means of Exs. I.t and 16 construct two segments HK and LM such that the ratio of the squares on these segments shall equal a given ratio. 18. Divide a given segment into two segments such that tlie areas of the s(iiiares constructed upon tliem shall be in a given ratio. ■> 19. On a given segment .17? find a point D such that = 2. Air 20. Construct a line parallel to the base of a triangle such that the resulting trianj^le and trapezoid shall be equivalent. AREAS OF POLYGONS. 17a 21. Construct two lines parallel to the base of a triangle so that the resulting two trapezoids and the triangle shall have equal areas. J5-J^— _.i__c 22. In each of the accompanying de- signs for tile flooring find what fraction of the space between the lines AB and CD is occupied by tUes of each color. Study each design with care to see that the character of the figure determines the relative sizes of the various pieces of tile. 23. ABCD is a square. Points E, F, O, •■■ are so taken that AE = AF= GB = BH=-. . {a) li AB=Qs.ndiAF=l, find the sum of the areas of the four triangles EFO, GHO, KLO, MNO. (h) Find the sum of these areas if AB = a and AF=h. (c) \i AB = Q and if the sum of the areas of the triangles is 9 square inches, find AF. (d) li AB = a and if the sum of the areas of the triangles is — , find n AF. Interpret the two results. 24. Show how to construct a square whose area is n times the area of a given square. 25. Construct a triangle similar to a given triangle and equivalent to n times its area. 26. Construct a hexagon similar to a given hexagon and equivalent to n times its area. 27. Show how to construct a polygon similar to a given polygon and equivalent to n times its area. Q B 174 PLANE GEOMETRY. 28. The alternate middle points of the sides of a regular hexagon are joined as shown in the figure. (a) Are the triangles thus formed equilateral? Prove. (h) Is the star regular (i.e. are its six acute angles equal and its sides equal) ? (c) Compare the three segments into 'which each triangle divides the sides of the other. (rf) Is the inner hexagon regular? Prove. See Ex. 1, p. 76. Tile Pattern. (e) If AB = 8 in., find the area of the large hexagon, the star, and the small hexagon. 29. A border is to be constructed about a given square with an area equal to one half that of the square. (a) By geometrical construction find the outer side of the border if the side of the square is given'. (6) If an outer side of the border is 21 in., find a side of the square, its area being two thirds that of the border. 30. A border is constructed about a given regular octagon, sueli that its area is equal to that of the octagon. ('/) If a side of the given octagon is a given segment All, find by geometrical construction a si'unient equ;il to an outer side of the border. (i) If a side of the given octagon is 1(1 ill., find an outer side of tlie Ij'Jidei. Ceiling Pattern. Ceiling Pattern. AREAS OF POLYGONS. 175 Find to the 31. The accompanying design is based on a set of squares such as ABCD. The small triangles are equal isosceles right triangles constructed as shown. (a) Are the vertical and the horizontal sides of the octagons equal? (6) Are the octagons regular? (c) If a side of one of the squares is 6, find the area of one of the octagons. (d) What fraction of the whole tile design is occupied by the light-colored, tiles? Does this depend upon the size of the original squares ? 32. Given two lines at right angles to each other. Find the locus of all points such that the sum of the squares of the distances from the lines is 25. 33. Given two concentric circles whose radii are r and r' the length of a chord of the greater which is tangent smaller. 34. If two equal circles of radius r intersect so that each passes through the center of the other, find the length of the common chord. 35. The square on the hypotenuse of a right triangle is four times the square on the altitude upon the hypotenuse. Prove it isosceles. 36. In a right triangle the hypotenuse is 10 feet and the difference between the other sides is 2 feet. Find the sides. , 37. Two equal circles are tangent to each other and each circle is tangent to one of two lines perpendicular to each other. Find the locus of the points of tangency of the two circles. Suggestion. Note that the point of tangency bisects their line of centers and that the centers move along lines at right angles to each other. 38. The square on a diagonal of a rectangle is equal to half the sum of the squares on the diagonals of the squares constructed on two adjacent sides of the rectangle. 39. Show that the diagonals of a trapezoid form with the non- parallel sides two triangles having equal areas. CHAPTER V. EBGULAR POLYGONS AND CIRCLES. REGULAR POLYGONS. 332. A regular polygon is one which is both equilateral and equiangular. According to this definition, determine whether each of the following polygons is regular or not and state why : An equilateral triangle, an equiangular triangle, a rectangle, a square, a rhombus. Draw a figure to illustrate each. Make a triangle which fulfills neither condition of the definition, also a quadrilateral. 333. The general problem of constructing a regular polygon depends upon the division of a circle into as many equal parts as the polygon has sides. The problem of dividing the circle into equal parts can be solved in some cases by the methods of elementary geometry, and some of these methods will be considered in this chapter. In most cases this problem cannot be solved by elementary methods. E.g. the circle may be divided into 2, •'^. 4, 5, 6, S. 10, 12, l.i, equal parts, but not into 7, 9, 11, 13, 11, equal parts. If a circle has already been divided into a certain num- ber of equal jiarts, it may then be divided into twice, four times, eight times, etc., that number of parts by repeated bisection of the arcs. (See § 204, Ex. 1.) The division of the circle into equal parts depends upon the theorem (§!'.•'.>) that c(iual central angles intercept 170 REGULAR POLYGONS AND CIRCLES. 177 equal arcs on the circle, and hence it involves the subdi- vision of angles into equal parts. The following exercises depend upon principles already familiar. With each solution give the reasons in full for each step. All constructions are to be made with ruler and com- passes only. 334. EXERCISES. 1. Divide a given angle or arc into four equal parts. 2. Divide a given angle or arc into eight equal parts. 3. If an angle or arc is already divided into a certain number of equal parts, show how to divide it into twice that number of equal parts. 4. Divide a circle into four equal parts. 5. Divide a circle into eight equal parts. 6. If a circle is already divided into a certain number of equal parts, show how to divide it into twice that number of equal parts. 7. Divide a circle into six equal parts. Suggestion. Construct at the center an angle of 60°. 8. Draw the chords connecting in order the four division points in Ex. 4, and show that the figure is a regular quadrilateral. 9. Draw the tangents at the four division points in Ex. 4, and show that a regular quadrilateral is formed. 10. Draw the chords connecting in order the division points in Ex. 7, and show that a regular hexagon is formed. Prove that the side of the hexagon is equal to the radius of the circle. 11. Draw chords connecting alternate division points in Ex. 7, and show that a regular triangle is formed. 12. Construct tangents to the circle at alternate division points in Ex. 7, and show that a regular triangle is formed. 13. Draw tangents to the circle at the division points in Ex. 7, and show that a regular hexagon is formed. NoTB. See also the construction of regular polygons of 3, 4, and 6 sides, § 163, Exs. 3, 4, 5. 178 PLANE GEOMETRY. 335. Theorem. If a circle is divided into any nvm- ber of equal arcs, the chords joining the division points, taken in order, form a regular polygon. Proof : Show (1) that the chords are equal ; (2) that the angles are inscribed in equal arcs and hence are equal. 336. Theorem. If a circle is divided into any num- her of equal parts, the tangents at the pjoints of divi- sion, take?! in order, form a regular polygon. A N B Given the O OA divided into equal arcs by the points M, N, P, etc., with tangents drawn at these points forming the polygon ABCDEF. To prove that ABCDEF is a regular polygon. Analysis of Proof : (1) To prove that ^B = BC= CD, etc. We know that BP = BN (Why ?). Hence, if we can show that AN = yB and BP = PC. then it will follow that AB = BC. To prove AN = NB, it must be shown tliatZl = Z 2, and this is done by showing that Z 1 and Z 2 are halves of the equal angles NOM and NOP. This necessitates proving A Aox ^ A .lo.v. Now state the proof in full. (2) To prove Zabc = Z iscD = Z cm:, etc. From the triangles proved cougnient under (1") show REGULAR POLYGONS AND CIRCLES. 179 that ^MAO = Z OAN = Z NBO = Z OBP; and hence, Z Jlf^O+ Z O^IV^ = Z.NBO + Z O^P, or Z i^^B = Z ^Ba. In like manner it is proved that Z. ABC = Z BCD, etc. Hence the polygon is equilateral and also equiangular, and hence regular. 337. Theorem. If a polygon is regular, a circle may he circumscribed about it. A B Given a regular polygon ABCDE. To prove that a'point can be found such that OA = OB = OC = 0D= OE. Outline of Proof : Bisect A A and B and let the bisectors meet in some point o. Then o is the center sought. For we have (1) OA = OB, Why ? (2) AaoB^ BOC. .-. 0A= OC. (3) AboC^ ACOjD. .-. ob=od. (4) Now prove OC = OE. .-. OA = OB = OC = OD = OE. Prove each step, showing how it depends upon the pre- ceding. 338. EXERCISES. 1. Find the locus of the vertices of all regular polygons of the same number of sides which can be circumscribed about the same circle. 180 PLANE GEOMETBY. 2. Find the locus of the middle jioints of the sides of all regular poly- gonsof the same number of sides which are in,sciibed in the same circle. 3. Is an equilateral circumscribed polygon regular? Prove. 4. Is an equiangular circumscribed polygon regular? Prove. 339, Theorem. // a polygon is regular, a circle may he inscribed In it. Given the regular polygon ABODE. To prove that a point O ni;iy be found such that the per- pendiculars OM, ON, OP, OQ, OR, are equal. Outline of Proof : Determine a point O as in the proof of the pre- ceding theorem. Then A AOB, BOO, etc., are equal isosceles trian- gles, and their altitudes are equal. Hence O is the required point. Give proof in full. 340. Theorem. An equilateral ^polygon inscribed in a circle is regular. Suggestion for Proof : Use the fact that equal chords subtend equal arcs, and apply § 335. 341. EXERCISE. Show that the inscribed and circumscribed circles of a regular polygon have the same center. 342. Definitions. The center of a regular polygon is the common center of its inscrihod and circumscribed circles. The radius of a regular polygon is the distance from the center to one of its vertices. Tlie apothem of a regular polygon is the perpendicular distance from its center to a sitU'. REGULAR POLTGONS AND CIRCLES. 181 343. Theorem. The area of a regular polygon is equal to half the product of its apothem and perimeter. Suggestion for Proof : A regular polygon is divided by its radii into a series of congruent triangles, the area of each of which is one half the product of the apothem and a side of the polygon. Complete the proof. 344. Theorem. The area of any polygon circum- scribed about a circle is equal to one half the product of the perimeter of the polygon and the radius of the circle. The proof is left to the student. 345. EXERCISES. 1. Is every equiangular polygon inscribed in a circle regular? Prove. 2. Show that the radius of a regular polygon bisects the angle at the vertex to which it is drawn. 3. Show that the perimeter of a regular polygon of a given num- ber of sides is less than that of one having twice the number of sides, both being inscribed in the same circle. 4. Show that the perimeter of a regular polygon is greater than that of one having twice the number of sides, both being circum- scribed about the same circle. 5. Compare the areas of the two polygons in Ex. 3. 6. Compare the areas of the two polygons in Ex. 4. 7. Sliow that the area of a square inscribed in a circle of radius r is 2 r'^. How does this compare with the area of the circum- scribed square. 8. Compute the apothem and area of a regular inscribed hexagon if the radius of the circle is r ; also of the regular circumscribed hexagon. 9. A regular triahgle is inscribed in a circle of radius 10. Find the apothem and a side of the triangle. 182 PLANE GEOMETRY. 346. Theorem. Two regular polygons of the same number of sides are similar. Outline of Proof: (1) Show that all pairs of corre- sponding angles are equal. (2) Show that the ratios of pairs of corresponding sides are equal. Hence the polygons are similar (Why?). 347. Theorem. The perimeters of two regular poly- gons having the same number of sides are in the same ratio as their radii or their apothems. Outline of Proof : Show (1) that each triangle formed by a side and two radii in one polygon is similar to the corresponding triangle in the other polygon. A'Q 7* ft C2') That — r-, = -7 = -;, where AB and A'B' are the two ^ A'B' r' a' sides, r and r' the corresponding radii and a and a' the corresponding apothems. And so for the remaining pairs of triangles. That ^^ + BO+c^-f - ^AB_ r a^, A'b' + b'c' + CD' + •■■ a'b' r' a' Draw the figure and give the proof in full. 348. Theorem. The areas of two regular polygons of the same number of sidc'^ are in the same ratio as the squares of the corresponding radii or apothems. Outline of Proof : Divide the polygons into pairs of corresponding triangles as in the preceding proof. CI) Show that — - = -r;, = —5, and so for each pair of triangles. (2) Hence A T + A 11 + A 111+ ... _,-_.' AI' + AII'+A iir+ - Draw figure and give the pruof in full. r '2 REGULAR POLYGONS AND CIRCLES. -183 PROBLEMS AND APPLICATIONS. 1. Find the ratio of the perimeters of squares inscribed in and circumscribed about the same circle. 2. Find the ratio of the perimeters of regular hexagons inscribed in and circumscribed about the same circle. 3. Find the ratio between the perimeters of regular triangles inscribed in and circumscribed about the same circle. 4. Find the ratio of the areas of regular triangles in scribed in and circumscribed about the same circle. Also find the ratio of the areas of such squares and of such hexagons. 5. The perimeter of a regular hexagon inscribed in a circle is 24 inches. Find the perimeter of a regular hexagon circumscribed about a circle of twice the diameter. 6. The area of a regular triangle circumscribed about a circle is 64 square inches. What is the area of a regular triangle inscribed in a circle of one third the radius? 7. The area of a regular hexagon inscribed in a circle is 48 square inches. What is the area of a regular hexagon circumscribed about a circle whose diameter is If times that of the first? 8. A chord AB bisects the radius perpendicular to it. Find the central angle subtended by the chord. State the result as a theorem. 9. State and prove the converse of the theorem in Ex. 8. 10. Find the area of a regular triangle inscribed in a circle whose radius is 6 inches. 11. Find the area of a regular triangle inscribed in a circle whose radius is r inches. 12. One of the acute angles of a right triangle is 60° and the side adjacent to this angle is r inches long. Find the remaining sides of the triangle. 13. A regular triangle is circumscribed about a circle of radius r. Find its area. Suggestion. First find DC in the figure. 14. A regular triangle of area 36 square inches is inscribed in a circle. Find the radius of the circle. 184. PLANE GEOMETHY. 15. Find the radius of a circle if the area, of its regular inscribed triangle is a. 16. Find the radius of the circle if the area of the regular circum- scribed triangle is a. 17. Find the radius of a circle if the difference between the perimeters of the regular inscribed and circumscribed triangles is 12 iuches. 18. Find the radius of a circle if the difference of the perimeters of the reguUir inscribed and circumscribed hexagons is 10 inches. 19. If the area of a circumscribed square is L'.j square inches greater than that of an inscribed square, what is the diameter of the circle? 20. Find the radius of a circle if the difference between the areas of the inscribed and circumscribed regular triangles is l'.j square inches. 21. Find the radius of a circle if the difference between the areas of the regular inscribed and circumscribed hexagons is 25 square inches. 22. The difference between the areas of the squares circumscribed about two circles is 50 S(iuare inches and the differ- ence of their diameters is 4 inches. Find each diameter. 23. If the inscribed and escribed circles O and 0' of an equilateral triangle are constructed as shown in the figure, find the ratio of their radii. Does this ratio depend upon the size of the triangle ? 24. (iiyeii a triangle ABC and a segment a, show how to construct a segment 1>E II AB and equal to the segment a, such that the points D and 7? shall lie on the siil(^s ( 'A and CB respectively or on these sides extended. 25. A triangular plot of ground ABC is to be laid out as a triangular flower bed witli a walk of uniform widl-li extending around it. (a) I'rove that the flower bed is similar to the original triauLjIe. {h) Show that tlic corners of the flower bed lie on the bisectors of the angles of the original triangle. /: ± REGULAR POLYGONS AND CIRCLES. 185 AB^ (f) Fiud a segment S so that — ~ = 2 and construct A'B' equal to S and parallel to AB, the points A' and B' lying on the bisectors of the angles A and B respectively. See Exs. 16-19, page 172. {d) Draw A'C" II AC and £'C'' II BC. Prove that the area of the flower bed A'B'C, as thus constructed, is equal to the area of the walk. (e) Construct the figure for the flower bed so that its area is five times that of the walk. 26. Given a rectangular plot of ground. Is it always possible to lay off on it a walk of uniform width running around it so that the plot inside the walk shall be similar to the original figure? Prove. 27. Is the construction proposed in Ex. 26 always possible in the case of a square ? of a rhombus ? Prove. 28. If a side of a regular hexagon circumscribed about a circle is a, find tlie radius of the circle. 29. On a regular hexagonal plot of ground whose side is 12 feet a walk of uniform width is to be laid off around it. Find by algebraic computation the width of the walk if it is to occupy one half the whole plot. 30. Find by geometric construction the width of the walk in Ex. 29. 31. Show that the figure inside the walk in Ex. 29 is a regular hexagon; 32. Given a segment AB, find three points C, D, E on it so that, AC'' ^1 Air ^ 1 ^^^^ AM. = ^ AB^ 4' Jb'' 2' Xfc'' 4 33. A regular hexagon ABCDEF is to he divided into four pieces of equal area by segments drawn parallel to its sides forming hexagons as shown in the figure. If AB = 24: feet, find a side of each of the other hexa^ gons and also the apothem of each. 34. "Without computing algebraically the apo- them or sides of the inner hexagons in Ex. 33, show how to construct the figure geometrically. See Ex. 6, § 331. Also Ex. 18, page 172. Use § 348. 186 PLANE GEOMETRY. 35. In a given hexagonal polygon whose side is a find in terms of (I and n the width of a walk around it which wiU occupy one nth of the area of the whole polygon. 36. By means of hexagons similar to those in Ex. 33, divide a given regular hexagon into three parts such that the outside part is J the whole area and the next | of the whole area. 37. Compute the sides and the apothein of the two hexagons con- structed in Ex. 36 if the side of the given hexagon is 12 inches. 38. In the adjoining pattern find two regular hexagons whose areas are in the ratio 1 : 4 and show that this agrees with theorem, § 348. 39. If a polygon is circumscribed about a circle, show that the bisectors of all its angles meet in a point. 40. Given any polygon circumscribed about a circle, di-aw segments parallel to each of its sides and at the same distance from each side. Show that these segments form a polygon similar to the first. 41. If the bisectors of all the angles of a polygon meet in a point prove that a circle may be inscribed in it (tangent to all its sides). What is the relation of the theorems in Exs. 39 and 41? 42. Given a polygonal plot of gi-ound (boundarj^ is a polygon) such that a path of equal width on it around the border leaves a polygonal plot similar to the first. Prove that a circle may be in- scribed in the polygon wh^f h forms the boundary of either plot. 43. In the figure, A BCF) is a square and EFGHKLMXis a regular octagon. (.() If KF = I inches, find AB. d (b) If AB = 12 inches, find EF. (c) If KF = a, find AB. (d) If ^B = s, find FF. (/') If AB = s, find the apothem. (/) M AB = s, find (he radius of the octagon. BEGULAE POLYGONS AND CIRCLES. 187 44. Find the ratio between the areas of regular octagons inscribed in and circumscribed about the same circle. 45. (a) The apothem of a regular octagon is 10 feet. Find the width of a uniform strip laid off around it which occupies \ its area. (6) Show how to construct this strip geometrically without first computing its width. Prove that the inside figure is a regular octagon. 46. A regular octagon ABCDEFGH is to be divided into four parts of equal area by means of octagons as shown in the figure. (a) If AB = 12 inches find a side of each octagon. (h) Show how to construct the figure without first computing the sides. (c) Show how to construct such a figure if the four parts I, II, III, IV, are to be in any required ratios. (d) Measure the sides of the two inner parts of the ceiling pattern and hence find the ratio of their areas. A ^ B 47. The middle points, A,B, C, D, of alternate sides of a regular octagon are joined as shown in the figure. AH is, perpendicular ia AE and equal to ili. BG, CK, and DL are con- structed in the same manner. (a) Prove that ABCD and HGKL are squares. (b) Find the areas of ABCD and HGKL, if EF= 10 inches. (c) What fraction of the whole octagon is occupied by the square HGKL. See the accompanying tile border. 188 PLANE GEOMETRY. MEASUREMENT OF THE CIRCLE. 349. If in a circle a regular polygon is inscribed, its perimeter may be measured. For example, the perimeter of a regular inscribed hexa- gon is 6 r if r is the radius of the circle. St-o § o'ii, Ex. 10. If the number of sides of the inscribed polygon be doubled, the resulting perimeter may be measured or completed in terms of r. See § 357. If again tlie number of sides be doubled, the resulting perimeter may be computed in terms of r, and so on. In a similar manner a regular polygon, say a hexagon, may be circumscribed about a circle and its perimeter- expressed in terms of r. If the number of sides of the circumscribed polygon be doubled, its perimeter may again be com- puted, and so on as often as desired. 350. By continuing either of these processes it is evident that the inscribed or the circumscribed polygon may be made to lie as close to the circle as we please. 351. The word Ie?igth has thus far been used in connec- tion with straight line-segments only. Thus, the perime- ter of any polygon is the snm of the lengths of its sides. 352. We now assume tliat a circle has a definite length and that tliis can be approximated as nearly as we please by taking the perimeters of the successive inscribed or cir- cumscM'iliod polj'gons. The lengtli of a circle is often called its perimeter or circumference. REGULAR POLYGONS AND CIRCLES. 189 It is evident that approximate nfeasurement is the only kind possible in the case of the circle, since no straight line unit of measure, how- ever small, can be made to coincide with an arc of a circle. 353. Comparison of the Lengths of Two Circles. In each of two circles, O and o', let regular polygons of 6, 12, 24, 48, 96, 192, etc., sides be inscribed. Call the perimeters of the polygons in O 0, Pg, P^2i -^24' ^^c., and those in © o', p'g, p'j2, P'241 etc. Let the radii of the circles be -B and r' respectively. Then by the theorem of § 347, we have iL = 5} = Zi2 = £24= etc., 7?' p' p' p' -" -^^S ■'12 -^24 however great the number of sides of the inscribed poly- gons. Show that the same relations hold if polygons are circumscribed about the circle. From these considerations we are led to the following : 354. . Preliminary Theorem. The lengths of two circles are in the same ratio as their radii. 355. Hence, if C and c' are the circumferences of two circles, R and r' their radii, and D and d' their diameters, we have £ = ^ = ^ ; and, also ^ = ^ • (See § 245) c' r' d' d d' Hence, the ratio of the circumference to the diameter in one circle is the same as this ratio in any other circle. 356. This constant ratio is denoted by the Greek letter tt, pronounced pi. The argument used above shows that a theorem like that of § 354 holds for every pair of poU^gons used in the approximation process, and hence it is established for all purposes of practical measurement or computation. 190 PLANE GEOMETRY. 357. Problem. To compute the approximate value Solution. Suppose a regular polygon of n sides is in- scribed in a circle whose radius is r and let one of the sides AB be called s„. We first obtain in terms of S„ the length AC, or S^n, of a side of a regular polygon of "2 n sides inscribed in the same circle. Let AB be a side of the first polygon. Bisect AB at C and draw AC and BC. Then AC is a side of a regular inscribed polygon of 2w sides (Why ■'). Then in the figure ( S,„y = IP = (1 ASy +P = (i S„y + W. ( Why (1) But h-KDE = h{1r—h') = ADy.DB = I^, (Why/) or AD^ = ils;f=h('lr-h) Solving equation (2) for h, we hiive h = •lr±xAr'- Taking only the negative sign, since // < r, and squaring, we have, h^ = "Ir^ — r ^' 4 r" — n,,'-* — \ ^■„'-. or, ¥ + Q ^S'= •- '" - rv 4 *•- - .s,;-. Hence, from (1), S^^ = 2 r^ — r \ i r- ~ s^. REGULAR POLYGONS AND CIRCLES. 191 and s^^ = V2r2-rV4r2-s„2. Since, by § 355, the value of "ir is the same for all circles, we take a circle whose radius is 1. In this case s^„ = V2 _ V4^^^. (3) If the first polygon is a regular hexagon, then -Sg = 1. Hence, 8^^ = ^2-VT^ = 0.51763809. Denoting the perimeter of a regular inscribed polygon of n sides by P„, we have, P12 = 12(0.51768809)= 6.21165708. In the formula (3) let n = 12. Then s^^ = V2 _ V4 - (0.51768809)2 = 0.26105238. Hence, P^^ = 24(0.26108238) = 6.26525722. Computing S^g, P^g, etc., in a similar manner, we have. 5,2 = V2 - V4 - 1 = .51763809. .-. P^ = 6.21165708. 5^2^ = V2 - VJ^:^ (.51763809)2 ^ .26105238. .-. P^^ = 6.26525722. S^ = V2 - V4: - (.26105238)2 = .13080626. .-. P^ = 6.27870041. 5<,g = V2 - V4 - (.13080626)2 = .06543817. . . Pgj = 6.28206396. S,g2 = V2 - Vd - (.06543817)2 = .03272346. .-. Pjgj = 6.28290510. S384 = V2 - V4 - (.03272346)2 = .01636228. .-. P^g^ = 6.28311.544. Sjus = V2 - v'4 - (.01636228)2 = .00818126. .-. P,^ = 6.28316941. 358. The Length of the Circle. By continuing this process it is found that the first five figures in the decimal remain unchanged. Hence 6.28317 is an approximation to the circumference of a circle whose radius is 1. 192 PLANE GEOMETRY. 359. By a process similar to the preceding, if circum- scribed polygons of ■!, 8, 16, etc., sides are used, the follow- ing results are obtained: NUMHEE OF SlIiFH Lf.vgtii of Each Side Peri.metee 4 2.000000 8.00000(1 6.6274T8 8 O.Sl'sIl'h 0.:;!)7s24 16 0.30.1196 32 lL'8 0.1tH:!)s4 I3.8034.jO f;.28s2:J0 0.0r).S2.->4 (1.(110078 (i.28444S LTiO 0.024.544 (i.2s:!.)00 5lL' 102 1 0.012272 6.2882(34 0.0061 :!0 6.28:;i'0,) 2048 0.00:50(iS (j.288100 360. Since the diameter is 2, the ratio of the circumfer- ence to the diameter is r..i!S317 gmro 6.28319 o -, n -., = 3.141.:)9, or = 3.l4l.3Vi according as the inscribed or circumscribed polygons are used. That is, these approximations of tt agree to five decimal places. If ,so great accuracy is not required, we mav use a smaller number of decimal places; such as, -3.141(3, 3.14. or ^. .Sincr — =7r, it follows that ('=7rD= 27ri?. n That 19, ///(' rircumfcroicc is '2 it times the radius. REGULAR POLYGONS AND CIRCLES. 193 361. The area of the circle. In § 359 the area of each circumscribed polygon is half the product of the perimeter and the apothem, which in this case is the radius of the circle. The area inclosed by the circle is called the area of the circle. That is, for every such polygon, A=lP- R, or area equals one half perimeter times radius. We now assume that a circle has a definite area which can be approximated as closely as we please by taking the areas of the successive circumscribed polygons. 362. Since the perimeters of the circumscribed polygons can be made to approximate the length of the circle as nearly as we please, and since c = 2 ttt* we have, area of circle = | • 2 irr . r = ur^- The degree of accuracy to which this formula leads depends entirely upon the accuracy with which tt is determined. The old problem of squaring the circle, that is, finding the side of a square whose area equals that of a given circle, involves therefore determining the value of ir. Much time and labor have been expended upon this in the hope that this value could be exactly constructed by means of the ruler and compasses, but it is now known that this is impossible. 363. Since the area of a circle is tt t^, if we have two given circles whose radii are r and r' and whose diameters are d and d', then the ratio of their areas A and A' is A irr^ r^ d? +t, 4. ■ Theorem. The areas of two circles are in the same ratio as the squares of their radii, or of their diameters. 194 PLANE GEOMETRY. 364. The area of a sector bears the same ratio to the area of the circle as the angle of the sector does to the perigon. E.g. the area of the sector whose arc is a quadrant is one fourth of the area of the circle, that is, ^ — . The area of a semicircle is - — . i 2 Find the area of a sector whose angle is 60^ ; 30° ; 45' ; 72'. 365. The area of a segment of a circle is known if that of the sector having the same arc, and of the central triangle on the same chord, can be determined. Show that the areas of segments of a circle whose arcs are respec- tively 90° and 60° are !r«f_^= = ^\,_2) and E^^ - ^ V3 = ^^2. - 3^3). 424^^ 64 12^ ^ The accompanying figure shows a circle cut into sectors by a 5erip=: of radii. Each sector ap- proximates the shape of /i (\ / 'i ,'\ / \ i\ i\ j\ /, /\ ,' , ,", ; ■, a triangle, whose altitude / \' \/ \/ w' \/ \( \! \! v "" \'' \'' > is the radius of the circle, and whose base is an arc of the circle. Since the sum of the areas of these triangles is the product of the altitude and the sum of the bases, we obtain a verification of the theorem that the area of a circle is one half the product of the circumference and radius. SUMMARY OF CHAPTER V. 1. ]\Iake a list of the definitions pertaining to regular polygons. 2. Statfl the theorems concerning regular polygons inscribed in or circumscribed about a circle. 3. State the theorems involving similar regidar polygons. 4. Give an outline of the discussion concerning tlio value of - and its use in approximating the length and area of the oirclo. 5. 'What are some of llu> move important applications of the theo- rems in ('lia]itiT V ? (Kctuvn to this question after studying the A\>- plications which follow.) J "kJ ^yK. )§>-©G )g^^( ^r^r^r REGULAR POLYGONS AND CIRCLES. 195 PROBLEMS AND APPLICATIONS ON CHAPTER V. Note. In the following problems, unless otherwise specified, use the value ir = 3f, which differs from 3.1416 by about t'iy of one per cent. 1. The diameter of a circle is 8 inches. Find the circumference and ai-ea. 2. The circumference of a circle is 10 feet. Find its radius and area. 3. The area of a circle is 24 square inches. Find its radius and circumference. 4. Centers of circles are arranged at equal distances on a network of lines at right angles to each other as shown in the figure. If r = i AB, what part of the whole area is inclosed by the circles? Suggestion. Consider what part of the square ABCD lies within the circles. 5. If in Ex. 4 the radius of each circle is 3 inches, how far apart must the centers be in order that one half the area shall lie within the circles ? 6. If the radius of each circle is r, how far apart must the centers be located in order that - of the whole area may lie within the circles ? 7. If the circles occupy - of the area, and if the centers are d inches apart, find the radii. " 8. A square is inscribed in a circle. Find the ratio between the areas of the square and the circle. 9. A square is circumscribed about a circle. Find the ratio be- tween their areas. Do the ratios required in Exs. 8 and 9 depend upon the radius of the circle? 10. Find the ratio between the area of a circle and its regular in- scribed hexagon. il. Find the ratio between the area of a circle and its regular cir- cumscribed hexagon. Do the ratios required in Exs. 10 and 11 depend upon the radius of the circle ? 196 I'LANE GEOMETRY. 12. Find the ratio between the area of a square inscribed in a circle and another circumscribed about a circle having B Cr C 3 times the radius. 13. Find the ratio of the area of a regular hexagon inscribed in a circle to that of another circumscribed about a circle having a radius \ as great. 14. AB< ' I> is a square whose side is a and the -^ ^ points .1, B, C, D, are the centers of the arcs HE, EF, etc. (a) Find the area of the figure formed by the arcs FLE, EKH, HNG, GMF. (b) Find the area of the figure formed by the arcs ELF and EQF. SrG(;ESTiijN. Find the difference between the areas of a circle and its inscribed .square. (c) Find the area of the figure formed by the arcs EK,1<.L, and LE. 15. If the sides of the rectangle A BCD are 8 and 12 inches, resi tively, find the radii of the circles.. (a) What fraction of the area of the rectangle lies within only one circle? (&) Prove that at each vertex of a square two circles are tangent to a diagonal of the square. 16. Two concentric circles are such that one divides the area of the other into two equal parts. (a) Find the ratio of the radii of the circles. (J) (liven the outer circle, construct the inner one. 17. Construct three circles concentric with a circle of radius r, which shall divide its area into four equal parts. 18. Prove that six circles of e(|Lial radii can be constructed each tan;;ent to two of the others and to a yiven circle. (a) Show that :i circle can be constructed around the six circles tanj^ciil to each of lliem. (b) What fraction of the area of tlie last circle is occii|iieil by llie sevi'n ciri'les within it? 19. .1 O/; is acentral auLjli' oljl(t\ Find the area bounded by the cluu-d ,1/.' and ill if the radius of the circle is 3, REGULAR POLYGONS AND CIRCLES. 197 20. ABC is an equilateral gothic arch. (See page 109.) area inclosed by the segment AB and the arcs AC and BC, iiAB^S feet. Suggestion. Find the area of the sector with center A, arc BC, and radii AB and AC, and add to this the area of the circle-segment whose chord is A C. 21. ABC is an equilateral triangle and A, B, and C are the centers of the arcs. Show that the area of the figure formed by the arcs is three times the area of one of the sectors minus twice the area of the A ABC. 22. In the figure ABC is an equilateral triangle and D, E, and F are the mid- dle points of its sides. Arcs are constructed as shown. (a) If ^^=6 feet, find the area inclosed by CE, El , Ft'. {b) Find the area inclosed by i£, E(:, and f^. (c) Find the area inclosed by i5>, f^, and EB. 23. In the figure ABC is an equi- lateral arch and O is constructed on AB as a diameter. AH and BK are perpendicular to AB. (a) Construct the equilateral arches HED and DFK tangent to the circle as shown in the figure. Suggestion. OH = OA + DH. (h) Prove that the vertices E and F lie on the circle. Suggestion. What kind of a triangle is HOKt In the following let AB — 8 feet. (c) Find the area bounded by the arcs DF, FE, and ED. (d) Find the area of the rectangle ABKH. (e) Find the area bounded by AH and the arcs HE and EA. (f) Find the area bounded by the upper semicircle AB and the arcs AC and BC. (g) Find all the areas required in (c) ••• (/) if AB — a. From the Union Park Church, Chicago. C H D K From the First Presbyterian Church, Chicago. 198 PLANE GEOMETRY. 24. In the figure, yliS C-DEF is a regular hexagon. jB is the center of the arc ALC, D the center of ( HE, and F the center of EKA. li AB = 16 inches, find H (a) The circumference and area of the circle, (6) The area bounded by the arc ALC and the segments AB and BC, (c) The area bounded by Ky\, AL and LK, „v (rf) The area bounded by JTc, CUE and EK'a . (c) Find the areas required in (o)-(rf) if AB = a inches. 25. ABC is a regular octagon. Arcs are con- structed with the vertices as centers as in the figure. If AB = 10 inches, find the area inclosed by the whole figure. Also ii AB = a. 26. ABCDEFGH is a regular octagon. Semi- circles are constructed with the sides as diameters. (a) If AB = 10 inches, find the area of the whole figure. Also hi AB = a. (J) Complete the drawing in the outline figure to make the steel ceiling pattern here shown. 27. In the figure A CB, AFD, DEB, and ECF are semicircles. EF is tangent to two semicu-cles. (a) Prove that the semicircles AFD, FCE, and DEB are equal, D being given the middle point of AB. If ^B = 48 inches, find: (ft) The area bounded by AC, FT and FA, (c) The area bounded by FD and DE, and the , line-segment FE, ^ jj"^ — r (d) The area bounded by .4 F, FCE, EB, and the line-segment AB. (e) Find the areas required iu (h)-{rl) if .1 R = a. 28. (a) If a side of the regular hexagon in the figure is a, find the area inclosed by the arcs (including the area of the hexagon). (J) Show that a circle may be cii- cumscribed about the \vlioli> figuro. IMP (c) Find the area inside this circle and out.side the figure in (o). REGULAR POLYGONS AND CIRCLES. 199 Kaeaa^ MISCELLANEOUS PROBLEMS AND APPLICATIONS. 1. Given a rhombus, two of whose angles are 60°, to divide it into a regular hexagon and two equilateral tri- angles. See Ex. 7, page 78. 2. M'hat fraction of the accompanying design for tile flooring is made up of the black tiles? Show how to construct this design by marking of£ points along the border and drawing parallel lines. 3. In the accompanying design for a parquet border two strips of wood appear to be intertwined. (a) If the border is 8 inches wide and 3 feet 4 inches long, find the area of one of these strips, including the part which ap- pears to be obscured by the other strip. (b) If the figure consists of squares, find the angle at which the strips meet the sides. Use the table on page 139. (c) If the width of the border is a and its length b, find the com- bined area of these strips. Compare the total area of the obscured part of these strips with the sum of the areas of the small triangles along the edge of the border. 4. A solid board fence 5 feet in vertical height running due north and south is to be built across a valley, 8 rods connecting two points of the same elevation. Find the number of square feet in the fence if the horizontal distance is 80 rods. 5. Are the data given in the preceding problem sufficient to solve it if the fence required is to be an ordinary four- board fence,' each board 6 inches wide ? 6. Two circles are tangent internally at a point A. Chords AB and AC oi the larger circle are drawn meeting the smaller circle in D and E respectively. Prove that BC and DE are parallel. 7. Two circles, radii r and r', are tangent internally. Find the length of a chord of the larger circle tangent to the smaller if : (a) The chord is parallel to the line of centers, (b) The chord is perpendicular to the line of centers, (c) Meets the larger circle at the same point as the line of centers. 'JOO PLANE GEOMETRY. 8. Givi'ii a straight line and two points A and B on the same side of it. Find a point < ' on the line such that the sum of the seg- ments A I ' and BC shall be the least possible. Solution. In the figure let B' be symmetric to B with respect to the line. Draw AB' meeting the line in C. Then C is the required point. For let C be any other point on the line. Then A f' + (•'B>AC + CB. The proof depends upon § 128 and Ax. Til, § 61. Give I , ', it in full detail. I '' 9. If in the figure preceding, A L> is perpendicular to the line, prove that A ADC '^ A (BE a,nd hence - — = . ^ BE CE 10. If in Ex. 8, DE = a, AD = h, BE = c, find CD and CE. 11. Two towns, A and B, ai-e 10 and 6 miles respectively from a river and A is 12 miles farther up the rivei- thau B. A pumping station is to be built which shall serve both towns. AVhere must it be located so that the total length of water main to the two towns shall be the least possible ? 12. 'I'wij factories are situated on the same side of a railway at different distances from it. A spur is to be built to each factory and these are to join the railway at the same point. State just what measurements umst be made and how to locate the point where these spurs should join the main line in order to permit the shortest length of road to be built. 13. Two equal circles of radius ?• intersect su that their common chord is equal to ;■. Find the area of the figure which lies witliin both circles. 14. In the accompanying design for i>ak and mahogany parquet flooring the large squares ;ue inches and the small black ones '2\ inches un a side. 'What fraction of the whole is the ma- hogany (the black squares) ? 15. Construct oireles on the three sides of a right triangle as diameters. Conqiare tlie area ii! the circle eonstrucli'd on the hypotenuse with the sum of the areas of the other two. I'mve. !■ ■ ■ ■ ■ B BXJ REGULAR POLYGONS AND CIRCLES. 201 16. In the accompanying design for grill work : (a) Find the angles ABC, BAD, and ^GC. (5) If the radius of each circle is r, find the distance AF. (c) What fraction of the area of the parallelogram GBCE lies within one circle only? (rf) If the radius of each circle is r, find the distance between two horizontal lines. (e) Construct the whole figure. Suggestions. (1) Find AF and lay off points on AB. (2) Find ZBAD and construct it. (3) Through the points of division on AB draw lines parallel to AD. (4) From A along AD lay off segments equal to AF and through these division points construct lines parallel to AB. (5) Along DC lay off segments equal to AF. Connect points as shown in the figure. (/) From the construction of the figure does it foUow that ZDAB = ZEGC1 17. Prove that the sum of three altitudes of a triangle is less than its perimeter. 18. In the accompanying design for grill work, the arcs are con- structed from the vertices of tlie equilateral triangles as centers. (a) Prove that two arcs are tangent to each other at each vertex of a triangle. (b) Find the area bounded by the arcs AB,BC, CD, DA. (c) Find the ratio between the area in (6) and the area of the triangle DBC. 202 PLANE GEOMETRY. 19. The character of the accompanying design for a window is obvious from the figure. Denote the radius of the large circle by R, of the semicircles by /J', and of the small circles by r. (a) If iJ = 8 feet find R' and , . (i) Find R' and r in terms of it. (c) What fraction of the area of the large circle lies within the four small circles? (d) What fraction of the area of the large circle lies outside the four semicircles ? (e) If i? = 10, find the area inclosed within the four small circles. 20. In the accompanying design for a stained glass window : (a) What part of the square A'B'C'D' lies ■within ABCD'I (5) If A'B' = 4 feet, find the sum of the areas of the semicircles. (c) Find the area inclosed by the line-seg- ments FB', B'E and the arcs FB and BE, if EB' is 1^ feet. ((/) Find the areas required in (6) and (c) if A'B' = a. 21. The accompanying design for tile flooring consists of regular octagons and squares. The design can be constructed by drawing parallel lines as shown in the figure. (o) If a side of the octagon AB is given, find BC, DE, and EF by construction. Find the ratio of any two of these seg- ments. (A) li AB = a, findBC. (c) li AB = a find the area of the square XJ/zw. (rf) At what angles do the oblique lines meet the horizontal? (e) Ciiiisl.iucl. the figure by layiiii;' off llu' required ]i(iintsoii the sides, drawini;- parallel lines in pencil, inking in the sides of the nclagoiis and I'rasiug llie remainder of the lines. REGULAR POLYGONS AND CIRCLES. 203 D T_T_T_T_-i X_X_X_^. T T T X X X X X ^ ^ XIX ^ X^r XI3 vH 3 22. This design for tile flooring is constructed by first making a network of squares and then drawing horizontal lines cutting of£ equal tri- angles from the squares. (a) At what angle to the base of the design are the oblique lines ? (b) If each of the small squares is 6 inches on a side, find EF and HL. (c) Find EF and HL if the side of a small square is a. (rf) What fraction of the whole area is occupied by the black triangles? 23. Five parallel lines are drawn at uni- form distances apart, as shown in the figure. (a) If these lines are 4 inches apart, find the width of the strip from which the squares are made, so that their outer vertices shall just touch ?j and Zz, and the corresponding inner vertices shall touch Zg and l^. (b) What part of the area between l^ and Zg would be occupied by a series of such squares arranged as shown in the figure ? 24. ABC is an equilateral triangle. AO and BO bisect its base angles. OD and OE are drawn parallel to CA and CB, respectively. Show that AD = DE =EB. 25. If one base of a trapezoid is twice the other, then each diagonal divides the other into two segments which are in the ratio 1 : 2. 26. If one base of a trapezoid is n times the other, show that each diagonal divides the other into two segments which are in the ratio l:n. 27. Prove that if an angle of a parallelogram is bisected, and the bi- sector extended to meet an opposite side, an isosceles triangle is formed. Is there any exception to this proposition ? Are two isosceles triangles formed in any case ? 28. Prove that two circles cannot bisect each other. 29. Find the locus of all points from which a, given line-segment subtends a constant angle. 204 PLANE GEOMETRY. 30. In the figure, equilateral arches are constructed on the base AB, and on its subdivisions into halves, fourths, and eighths. C ... ^.. ^ From Lincoln Cathedral, England. (a) Show how to construct the circle O tangent to the arcs as shown in the figure. Suggestion. The point is determined by drawing arcs from A and B as centers with BF as a radius (Why?). (i) Show how to complete the construction of the figure. (c) If AB = 12 feet, find the radii of the circles O, 0', 0". (d) Find these radii if AB = s (span of the arch). (e) What part of the area of the arch ABC is occupied by the arch ADE? by the arch AFS"} by ALK'l (f) The sum of the areas of the seven circles is what part of the area of the whole arch? (g) The sum of the areas of the two equal circles O* and O" is what part of the area of the circle O? 31. The accompanying church window design consists of the equilateral arch ABC and the six smaller equal equilateral arches. (a) If . I B = 8 feet, find the area bounded by tlie arcs AfCl. GE, ED, DM. (J) If . I B = 8 feet, find tlie area bounded bv the arcs AC. CE, ICD, 7). I. (c) Find the areas required under (a) and (6) if AB = . / / ■: ''1 -> / <^A - -\ ,- P y / 1 ': - / / I 1 .4 - II 1 i / J 1 (^ L ) AlU tud Ax 9 VARIABLE GEOMETRIC MAGNITUDES. 209 If we suppose that while the base of the rectangle re- mains fixed, the altitude varies continuously through all values from \ = 10 to h^= 1Q = 2 y. 10, then it must take among other values the value 10 V2. Using 10 a/2 as an abscissa, the question is, whether CD is the area ordinate corresponding to it. This area ordinate is not less than CD and commen- surable with the base, for in that case the altitude would be less than 10 V2, and for the same reason it is not greater than CD and commensurable with the base. But we can find line-segments less than 10 V2 or greater than 10 V2 and as near to 10 v'2 as we please. Hence we conclude that the area ordinate for the rectangle whose altitude is 10 V2 is CD, that is, the point C lies in the line OP^- In like manner, the point determined by any other ab- scissa incommensurable with the base is shown to lie on the line OP^- Since the abscissa and ordinate of any point on OP are equal, we have for any altitude, A_^'h_^ ^1 h 369. The preceding theorem may also be stated : The area of a rectangle with a fixed base varies directly as its altitude. This means that if ^ and h are the varying area and altitude re- spectively, and if Ai and hi are the area and altitude at any given instant, then — =: — or ^ = — 1 • h OT A = Jch, where k is the Jixed ratio — -. A\ hi hx hi The graph representing the relations of two variables when one varies directly as the other is always a straight line. 210 PLANE GEOMETRY. 370. EXERCISES. 1. Make a graph to show that the ari'ii of two rectangles having equal altitudes are in the same ratio as their bases. 2. .Sl]GS, which question in Ex. -3 applies to the altitude and area of a parallelogram having a fixed base and varying altitude ? Which applies to a triangle having a fixed base and varying altitude? 371. TnEOREir. The area of a rectangle is equal to the product of its base and altitude. Proof : Using the graphic representations of §§ 368, 370, Ex. 1, we have for all cases, A^ a T Ao b — 2 = - and —i—-. Hence — ^ = ^-2 x ^ ''•2 3 _ A.2 ah , Rut ^j is the unit of urcn. Hence ah represents the numerical measure of Ar. liy the area unit. ^, 1 A^ 1 -4, b That is, Ari'it = husr x attitude. 372, Problem. Make a graphic representation of the theorem: Thr jiiri))irt(rs (f similar j)oIygons are in tJic sajne nitio as (tny tiro correspondimj sides. SoLirioN. First (•(iiisIiUm- tlio ."Special ease of equilat- eral triaiiLi'les. On (lie hurizuntal axis lay off the lengths of one side of several siuih triangles, and on the vertical VARIABLE GMOMETRIC MAGNITUDES. 211 axis lay off the lengths of the cor- responding perimeters. Show that the points so obtained lie in a straight line. 373. EXERCISES. 1. In the manner above graph the rela- tion between the perimeters and sides of squares. Of regular pentagons. Of regu- lar hexagons. Of rhombuses. 2. If a side of a given regular polygon is a and its perimeter p, graph the relation of the perimeters and corresponding sides of polygons similar to the given polygon. 7 / / / / ^ / in / -t .-, / 1 1 s / / fll ' p- / V a -i 1 f ai / « / < ' 1 1 ' 1 / ,_ _J _ Axis ioi Sides 374. Theorem. If a line is parallel to one side of a triangle and cuts the other two sides, then it divides these two sides in the same ratio. c Graphic representation : Lay off CA along the hori- zontal axis and CB along the vertical axis, thus locating the -A point Pj. In like manner find Pg with the coordinates CD and CE. If CD and CA are commensurable, we know that CD _CE CA Cb' Hence O, P^ and Pj are collinear. If cd' and CA are incommensur- able, show, as in § 368, that the corresponding point P lies on the line 0?^, and hence, as in that case, also / c ^ i_ — _E ^ = - 3 ~x - — / / e if / CI / / r r> /- - ^ — Li _ _ >CB OA CD' CA ce' cb' 212 PLANE GEOMETBT. 375. Problem. To represent graphically the rela- tion hetiueen the area and side of a square as the side variety continiiondy. ^\ ■ ■ ■ T^ 1 I 7 -^,7 i 1 ■ ' ' 1 1 j 1 \ 1 1 1 1 1 / 1 / ! ^c / \ , i/ / 1 i !_U J § ■/ 1 *J W / 1 1 I <^ Iw r t 1 1 ; / 1 1 1 O«o 1 1 ^/ 1 •§ / / 1 •3 V ■^ vy 1 n ^ / , > / 1 i / I 1 / i f ) / 1 -ip- B, / 1 I -m- '/ 1 1 / 1 1 1 / 4 ~ _ / 1 1 1 1 1 1 ■ TE '2 ^^ '.^ 1 ' ' ti, _L - ±. ±. ■•!:!! 12 3 4 6 7 8 Aiis for Sides Solution. On the horizontal axis lay off the segments equal to various values of the side «, and on the vertical axis lay off secfnients equal to the corresponding areas A. (1) If one hori/imtal space represents one unit of length of side, and one vertical space one unit of area, then the points Pj, Pj, Pj, etc., are found to lie on the %teep curve. VABIABLE GEOMETRIC MAGNITUDES. 213 (2) If five horizontal spaces are taken for one unit of length of side and one vertical space for one unit of area, then the points Pj', P^', P3', etc., are found and the less steep curve is the result. The student should locate many more points between those here shown and see that a smooth curve can be drawn through them all in each case. The graph of the relation between two variables, one of which varies as the square of the other, is always similar to the one here given. 376. The area of a square is said to vary as the square of one of its sides, that is, A = s^- For example, the theorem : The areas of two similar polygons are in the same ratio as the squares of any two corresponding sides, means that if a given side of a polygon is made to vary continuously while the polygon remains similar to itself, the area of the polygon varies con- tinuously as the square of the side. 377. EXERCISES. 1. From the last graph find approximately the areas of squares whose sides are 3.4 ; 5.25 ; 6.35. 2. Find approximately from the graph the side of a square whose area is 28 square units ; 21 square units ; 41.5 square units. 3. Construct a graph showing the relation between the areas and sides of equilateral triangles. 4. Given a polygon with area A and a side a. Construct a graph showing the relation between the areas and the sides corresponding to a in polygons similar to the one given. 5. From the graph constructed in Ex. 3, fi.nd the area of an equi- lateral triangle whose sides are 4. Also of one whose sides are 6. Compute these areas and compare results. 6. Construct a graph showing the relation between a side and the area of a regular hexagon. By means of it find the area of a regular hexagon whose side? we 6. Compare with the computed area. 214 PLANE GEOMETRY. 378. Pkoblem. To construct a graph showing the rcldtion hctwecn the rai^ius and the circumference of a circle, (ind between the radius and area of a circle, as the radius varies continumisl/j. SoLUTi(_)X. Taking ten horizontal spaces to represent one unit of length of radius, and one vertical space for one unit of circumference in one graph and one unit of area in the other, we find the results as shown in the figure. 50 45 -§30 U 3 12S i 1 ' ' ' l_L^ ; II / ■ -P 1 / ' / 1 1 1 1 1 P~' f 1 1 1 ' 1 1 x: 1 1 ' i 1 ' 1 MM I -^ f_ 1 r : i: 1 / ' ! i r\ 1 TZJl- ^ ' jZ. 7 in -L/L it ' 1 1 3 1 ■ 1 1 / 7 QI it y^ ± y^ VL it ^ : 1 r~ z ! 1 ^^it i ! I t r^ 1 1 J y^ 1 ^4 1 / ^iL 1 / ^ 1 (^ ^' \ 1 ^' ^" ' /''' ' I II c>^ / 1 ^^'^ _,7 -I ^ ^ y^ ^^ / ^'^ X " £—«-'''' i 1 A.xisforlladU VARIABLE GEOMETRIC MAGNITUDES. 215 379. EXERCISES. 1. From the graph find approximately the circumference of a circle whose radius is 2.7, also of one whose radius is 3.4. 2. Find the radius of a circle whose circumference is 17, also of one whose circumference is 23. 3. Find the areas of circles whose radii are 1.9, 2.8, 3.6. 4. Find the radii of circles whose areas are 13.5, 25.5, 37, 45. 5. How does the circumference of a circle vary with respect to the radius? 6. How does the area of a circle vary with respect to the radius ? 7. Find the radius of that circle whose area in square units equals its circumference in linear units. DEPENDENCE OP VARIABLES. 380. In the preceding pages we have considered certain areas or perimeters of polygons as varying through a series of values. For example, if a rectangle has a fixed base and varying altitude, then the area also varies de- pending on the altitudes. The fixed base is called a con- stant, while the altitude and area are called variables. The altitude Avhich we think of as varying at our pleas- ure is called the independent variable, while the area, being dejDendent upon the altitude, is called the dependent variable. 381. The dependent variable is sometimes called a func- tion of the independent variable, meaning that the two are connected by a definite relation such that for any definite value of the independent variable, the dependent variable also has a definite value. Thus, in A = s'^ (§ 376), A is a function of s, since giving s any definite value also assigns a definite value to A . Similarly, C is a function of r in C = 2 ttt, and ^ is a function of r m A = Trr^. 216 PLANE GEOMETRY. 382, EXERCISES. Justify each of the following statements, remembering that a variable y varies directly as another variable x ii y = kx, and directly as the square oi x ii y — kx^, where k is some constant. Find k in each case. 1. The area of a rectangle with constant base varies directly as its altitude. Find the value of k. Suggestion. By § 368, — = — or A = — >■ . h. Hence A = kh. Ai Aj A| In this case k = — ^ — b, the constant base. See also § 369. 2. The area of a square varies directly as the square of its side. Suggestion. Since A = s^ and A, = s.'-, we have ■^— = — or A=^-s^. That is, ,4 = ts^ where /t = ^ = 1. See § 370. 3. An angle inscribed in a circle varies directly as the intercepted arc. Show that in this case A; = |. 4. A central angle in a circle varies as the intercepted arc. Show that in this case k = 1. 5. In the figure of § 37i, show that DE varies directly as CD if DE moves, remaining parallel to AB. 6. An angle formed by two chords intersecting within a circle varies directly as the sum of the two arcs intercepted by the angle and its vertical angle. 7. An angle formed by two secants intersecting outside a circle varies directly as the difference of the two intercepted arcs. 8. If a polygon varies so as to remain similar to a fixed polygon, then its perimeter varies directly as any one of its sides. Show that p k = —1, where P, and «, are the perimeter and side of the fixed «i jKilyH'im. 9. Ill the preceding', the area of the polygon varies directly as the square of any one of its sides. 10. The circumference of a eirele varies diivotly as the radius, and its area varies directly as the square of its radius. VARIABLE GEOMETRIC MAGNITUDES. 217 ai. Plot y = kxiork = 1, -J, 3, 4. 12. Plot y = kx^ for k = 1, i, 3, 4. Notice that the graphs in Ex. 11 are all straight lines, while those in Ex. 12 are curves which rise more and more rapidly as the inde- pendent variable increases. See also § 378. LIMIT OF A VARIABLE. 383. If a regular polygon (§ 357) is inscribed in a circle of fixed radius, and if the number of sides of the polygon be continually increased, for instance by repeatedly doubling the number, then the apothem, perimeter, and area are all variables depending upon the number of sides. That is, each of these is a function of the number of sides. Now the greater the number of sides the more nearly does the apothem equal the radius in length. Indeed, it is evident that the difference between the apothem and the radius will ultimately become less than any fixed number, however small. Hence we say that the apothem approaches the radius as a limit as the number of sides increases indefinitely. 384. Similarly by § 352 the perimeters of the polygons considered in the preceding paragraph may be made as nearly equal to the circumference as we please by making the number of sides sufficiently great. Hence we define the circumference of a circle as the limit of the perimeter of a regular inscribed polygon as the number of sides increases indefinitely. It also follows from § 353 that the circumference of a circle may be defined as the limit of the perimeter of a circumscribed polygon as the number of sides is increased. Likewise we may define the area of a circle as the limit of the area of the inscribed or the circumscribed polygon as the number of sides increases indefinitely. See § 361. 218 PLANE GEOMETRY. 385. The notion of a limit may be used to define the length of a line-segment which is incommensurable with a given unit segment. Thus, the diagonal d of a square whose side is unity \s d = y/2. Hence d may be defined as the limit of the variable line-segment whose successive lengths are 1, 1.4, l.il, 1.414r---. See §§ iyi, '_'4U. In like manner, the length of any line-segment, whether commensurable or incommensurable with the unit seg- ment, may be defined in terms of a limit. Thus, if a variable segment is increased by successively adding to it one half the length previously added, tlien the segment will approach a limit. If the initial length is 1, then the successive additions are \, Ji i> A' -3^) 6tc., and the successive lengths are 1, 1\, 1|, 1|, \\l, \\\, etc. Evidently this segment approaches the limit 2. Hence 2 may be defined as the limit of the variable segment, whose successive lengths are 1, IJ, 1|, 1 J, etc., as the number of suc- cessive additions is increased indefinitely. 386. A useful definition of a tangent to a circle, or to any other smooth curve, may be given in terms of a limit. Let a secant cut the curve in a fixed point A and a variable point B, and let the point B move along the curve and approach coincidence with A, thus making the secant continually vary its direction. Then tlie tangent is defined to be the limiting position of the secant as B approaches A indefinitely. This definition of a tangent is used in all higher mathematical work. It includes the definition given in the case of the circle in § 183. 387. Till! funcdonal relation holwofn variables and the idea of a limit as illustrated above are two of the most im- portant e(ine(']its in all matliemat ies. The whole subject is much too ilifiiciilt for rigorous consideration in this course. VARIABLE GEOMETRIC MAGNITUDES. 219 388. EXERCISES. 1. Find the limit of a variable line-segment whose initial length is 6 inches, and which varies by successive additions each equal to one half the preceding. 2. Find the limit of a variable line-segment whose initial length is 1 and whose successive additions are .3, .03, .003, etc. 3. Construct a right triangle whose sides are 1 and 2. By approxi- mating a square root, find five successive lengths of a segment which approaches the length of the hypotenuse as ^ limit. 4. If one tangent to a circle is fixed and another is made to m6ve so that their intersection jioint approaches the circle, what is the limiting position of the moving tangent? What is the limit of the measure of the angle formed by the tangents ? 5. The arc AB of 74° is the greater of the two ares intercepted be- tween two secants meeting at C outside the circle. The points A and B remain fixed while C moves up to the circle. AVhat is the limit of the angle formed by the secants ? The limit of the measure of their angle ? 6. If in the preceding the secants meet within the circle, what is the limit of their angle and also of the measure of this angle ? 7. If in Ex. 5 one secant and the intersection point remain fixed, while the other secant approaches the limiting position of a tangent at the point A, find the limit of the measure of the included angle. 8. If in Ex. 6 one secant and the point of intersection remain fixed while the other secant swings so as to make the included angle approach a straight angle, find the limit of the measure of the angle. 9. If in Exs. 5 and 6 the moving point crosses the circle, state the theorem on the measurement of the angle in question so as to apply equally well whether the point is inside or outside the circle. 10. A fixed segment AB is divided into equal parts, and equilateral triangles are constructed on each part as a base, as ABEC, CGD, DHA. Then each base is divided into equal parts and equilateral tri- angles are constructed on these parts as bases. What is the limit of the sum of the perimeters of these triangles as the number of them is increased indefinitely? What is the limit of the sum of their areas? CHAPTER VII. REVIEW AND FURTHER APPLICATIONS. ON DEFINITIONS AND PROOFS. 389. A definition is a statement that a certain word or phrase is to be used in place of a more complicated expression. Thus the word "triangle "is used instead of "thp figure formed by three segments oonueotiug three non-coUinear points." In geometry we may distinguish two classes of words: (a) Technical words representing geometric concepts such as line, plane, polygon, circle, etc. (6) Words of ordinary speech not included in the iirst class. The meaning of words in the second class is taken for granted without any definition whatever. It is not possible to define t'l'ery word of the first class, for every definition brings in new technical terms which in turn require definition. Thus, one of the many definitions of "point" is "that which separates one part of a line from the adjoining part." In this definition the technical terms "separate," "line," "adjoin- ing" are used. If we Iry to di'tine these, still other therms are brought in which need to be defined, and so on. The only escape is defining in a circle, which is not permitted in a logical science. Since it is thus t'\ idont that some terms must be used willuuit being defined, it is best to state which ones are left undefined. 220 REVIEW AND FURTHER APPLICATIONS. 221 390. In this book point, straight line, or simply line, plane, size and shape of figures are not defined. Nor do we define the expression, a point is between two other points, or a point lies on a segment. Descriptions of some of these terms which do not, how- ever, constitute definitions in the logical sense are given in §§ 1-5. Other technical terms are defined by means of these simple undefined words with which we start. Thus, " a segment is that part of a line which lies between two of its points " is a definition of segment in terms of the three undefined words, point, line, and between. 391. A geometric proposition consists of an affirmation that a geometric figure has some property not explicitly specified in its definition. A proposition is said to be proved if it is shown to fol- low from other propositions which are admitted to be true. Hence every geometric proposition demands for its proof certain other propositions. 392. It is obvious, therefore, that certain propositions must be admitted without proof. Such unproved propo- sitions are called axioms. In order that a set of axioms for geometry shall be complete it must be possible to prove tkat every theorem of geometry follows from them. The set of axioms used in this text is not complete. For instance, it is assumed without formal statement that if A, B, C, are three points on a segment we cannot have at the same time B between A and C and C between A and B ; that the diagonals of a convex quadrilateral intersect each other ; that a ray drawn from the vertex of an angle and included between its sides intersects every segment determined by two points, one on each side of the angle. In like manner many other tacit assumptions are made. 222 PLANE GEOMETRY. 393. In a complete logical treatment every uudefined term must occur in one or more axioms, since all knowl- edge of this term in a logical sense comes from the axioms in which it is found. In this text not every undefined term occurs in an axiom, for instance, the word between. The axioms are, of course, based on our apace intuition, or on our experience with the space in which we live. It is interesting to notice, however, that the axioms tran- scend that experience both as to exactness and extent. For instance, we have had no experience with endless lines, and hence we cannot know directly from experience whether or not there are complete lines which have no point in common. See §§ 89, 96. 394. Proofs are of two kinds, direct and indirect. A direct proof starts with the hypothesis and leads step by step to the conclusion. An indirect proof starts with the hypothesis and with the assumption tliat the conclusion does not hold, and shows that this leads to a contradiction with some known proposition. Or it starts simply with the assumption that the conclusion does not hold and shows that this leads to a contradiction with the hypothesis. This kind of proof is based upon the logical assumption that a proposition must.either be true or not true. The proof consists in showing- that if the proposition were not true, impossible consequences would follow. Hence the only remaining possibility is that it must be true. 395. Every proposition in geometry refers to some figure. See § 12. The essential characteristic of a figure is its description in words and not the drawing that repre- sents it. I'^ach drawing represents just one figure from a ehiss of figures lU'lined by the description. Thus we say REVIEW AND FURTHER APPLICATIONS. 223 "let ABCB be a convex quadrilateral," and we construct a particular quadrilateral. We must then take care that all we say about it applies to any figure whatever so long as it is a convex quadrilateral. The logic of the proof must be entirely independent of the appearance of the constructed figure. The description of the figure must contain all the con- ditions given by the hypothesis. A good way to show that the description of the figure is what really enters into the proof, is to let one pupil describe the figure in words and each of the others draw a figure of his own to correspond to that description. The proof must then be such as to apply to every one of these figures though there may not be two of them'exactly alike. 396, EXERCISES. 1. Every word in the language is defined in the dictionary. How is this possible in view of what has been said about the impossibility of defining every word ? 2. Can we determine experimentally whether or not the space in which we live satisfies the parallel line axiom (§ 96) ? 3. Can we determine experimentally whether or not there can be more than one straight line through two given points? 4. Which theorems of Chapter I are found by direct proof and which by indirect proof ? 5. If two triangles have two angles of the one equal to two angles of the other, and also any pair of corresponding sides equal, the triangles are congruent. 6. If two triangles have two sides of the one equal to two sides of the other, and also any pair of corresponding angles equal, the triangles are congruent in all cases except one. Discuss the various cases according as the given equal angles are greater than, equal to, or less than a right angle, and are, or are not, included between the equal sides, and thus discover the exceptional case. 7. State a theorem on the congruence of right triangles which is included in the preceding theorem. 224 PLANE GEOMETRY. 8. What theorems of Chapter I on parallel lines can be proved without the parallel line axiom? 9. Wliat theorems of Chajiter I on parallelograms can be proved without the parallel line axiom ? 10. >\'hat regular figures of the same kind can be used to exactly cover the plane about a point used as a vertex ? 11. A\'hat combinations of the same or different regular figures can be used to exactly cover the plane about a point used as a vertex ? 12. Suppose it has been proved that the base angles of an isosceles triangle are equal but that the converse has not been proved. On this basis can it be decided whether or not the base angles are equal by simply measuring the sides ? Can it be decided on the same basis whether or not the sides are equal by simply measuring the base angles? Discuss fully. 13. The sum of the three medians of a triangle is less than the sum of the sides. See Ex. 34, p. 83. 14. The sum of the three altitudes of a triangle is less than the sum of the sides. 15. A triangle is isosceles (1) if an altitude and an angle-bisector coincide, (2) if an altitude and a median coincide, (3) if a median and an angle-bisector coincide. 16. If two sides of a triangle are unequal, the medians upon these sides are unequal and also the altitudes. 17. An isosceles triangle has two equal altitudes, two equal medians, and two equal angle-bisectors. 18. ABCD is a square and the points E, F, G, H, are so taken that AE = AH = CF = CO. Prove that EFGH is a rectangle of constant perimeter, whatever the length of AE. 19. Tlie liispctors of the exterior angles of a parallekijjrara form a rectangle the sum of the diagonals of which is the same as the sum of the sides of the parallelogram. 20. I'ldvi' that the perpendicular bisectors of the sides of ■■! iinlyumi iiiscrilied in a circle meet in a jioiiit. Use this tlieorein to show that the statcnient is true of any triangle. BEVIEW AND FURTHER APPLICATIONS. 225 21. The bisectors of the exterior angles of any quadrilateral form a quadrilateral whose- opposite angles are supplementary. Definition. In a polygon of n sides there are n angles and hence 2 n parts. Parts a, Z ^, J, Z B, etc., are said to be consecutive if a lies on a side oi /.A,h lies on the other side of Z^, and also on a side of Zb, etc. 22. Is the following proposition true? If in two polygons each of n sides 2n — 3 consecutive parts of- one are equal respectively to 2 n — 3 consecutive parts of the other, the polygons are congruent. Suggestion. Try to prove this proposition for » = 3, then for re = 4, and finally for the general polygon. 23. What theorems on the congruence of triangles are included in the preceding proposition ? A proposition may be proved not true by giving one ex- ample in which it does not hold. 24. Is the following proposition true? If in two polygons each of n sides 2 n — 3 parts of one are equal respectively to 2 re — 3 correspond- ing parts of the other, the polygons are congruent provided at least one of the equal parts is a side. 25. Given three parallel lines, to construct an equilateral triangle whose vertices shall lie on these lines. Solution. I^et P be any point on the middle line Zj. Draw PC and PA, making an angle of 60° with l^. Through the points A, P, C construct a circle meeting l^ in B. Then ABC is the required triangle. Suggestion for Proof. Compai'e A BPA and EC A also A BPC and BAC. 26. Show how to modify the construction of the preceding example so as to make ABC similar to any given triangle. 27. If tangents are drawn to a circle at the extremities of a diameter and if another line tangent to the circle at P meets these two tangents in ^ and -B respectively, show that AP PB = r^, where r is the radius of the circle. 226 PLANE GEOMETRY. LOCI CONSIDERATIONS. 397. Two methods are available to show that a certain geometric figure is the locus of points satisfying a given condition. First method : Prove (a) Every point satisfying the condition lies on Ihefgure. (h) ]'^ eery point on the figure satisfies the condition. Second method : Prove (a') every point not on the figure fails to saiisfy the condition. (b') Every point on the figure satisfies the condition. The first of the methods is more direct and usuall}' more simple. See § 127. The second is likely to lead to proofs that are not general. PROBLEMS AND APPLICATIOHS. 1. AB is n fixed segment connecting two parallel lines and per- pendicular to each of them. Find the locus of the ^ vertices of all isoscclr.s triaiiffles whose common base is AB. Is the middle point of AB a part of --^^ — this locus V - _:6._c A A 2. If in Ex. 1 AB is allowed to move always remaining perpendicular to the given lines, and if .4BC is ony tri- angle remaining fixed in shape, find the locus of the , point C. -T-t — - — 3. Find the locus of the centers of all parallelo- grams which have the same base and equal altitudes. 4. Find the locus of the centers of parallelograms obtained by cutting two parallel lines by parallel seeant line<. 5. Find the locus of the vertices of all triangles which have the same l)as(' and equal areas. 6. Find thi' locus of a point whose distances from two intersect- ing lines are in a fixed ratio. Note tliat the whole figure is symmetrical with respect to the bi.seelur of each pair of veilieal angles formed liy the two given lines. REVIEW AND FURTHER APPLICATIONS. 227 7. Find the locus of a point such that the difference of the squares of its distances from two fixed points is a constant. Note that the locus must be symmetrical with respect to the perpendicular bisector of the segment connecting the two given points A and B. Show that in the figure AP - PB = AC - CB . A'^ -^ ^B 8. The lines l^ and l^ meet at right angles in a point A, O is any fixed point on Zj. Through O draw a line meet- ing /j in B. P is a varying point on this p y /\ p' A ^^ , 5- -T'o D B / line such that OB • OP is fixed. Find the locus of P as the line swings about as a pivot. Suggestion. Draw PD J. to BP. Show that BO ■ OP = AO ■ OD. Hence we obtain a set of right triangles whose common hypotenuse is OD. Find the locus of the vertex P. 9. Find the locus of all points the sum of whose distances from two intersecting lines is equal to a fixed segment a. Suggestion. If AB and CB are the given lines, investigate points on the base AC oi the isosceles triangle ABC in case the perpendicular from C on AB ia equal to a. 10. The figure ABCDEF is a regular hexagon. AG, BH, etc., are equal segments bisecting the angles A, B, etc. (a) Prove GHWAB. (b) Prove that the inner figure is a regular hexagon. Suggestion. Show ABHG congruent to the other similar parts by superposition. (c) Find how many degrees in ZAGH and /.BHG, as needed by a carpenter in making a hexagonal frame. 11. The figure ABCD is a parallelogram. The points E, F, G, H are taken so that AE = CG and AH = CF. Prove A FCG s A AEH and A EOH ^ A FOG. This figure is found in an old Roman pavement in Sussex, England. x> g c A E B 228 PLANE GEOMETRY. 12. The iigure A BCD is a square. A E = BF = CG = DH, and Ey, Fz, Gw, Hx are so drawn that z:l = Z2 = Z3 = Z4. (a) Find AEi/F, Fz i.s a square. J'Q and A'.s' an' iliaiiu'lirs. The jioints /,'. /■', C. IT, ••. are so taUen that A E ~ BF - l:(7 = lie = .... Also O.V = OK = 0L= 0.\L BEVIEW AND FURTHER APPLICATIONS. 229 ^ ^ ^ ^ ^ ^ ^ «^ ^ (a) Prove that A ENF = A GKH. (b) Prove that TMONEA = FNOKGB. (c) How many axes of symmetry has the figure TAENFBGKH ...1 (d) Has this figure a center of symmetry ? See the four squares in the accompanying design. 15. A cross-over track is constructed as shown in the figure. The rails of the curved track are tangent to those of the main line at A, B, F and G. The curves are tan- gent to each other at D and E. The arcs GE and BD have equal radii as have the arcs ^S and DF. Cj is the center of the arcs GE and FD and C2 is the center of the arcs BD and AE. (a) Prove that C^HC^ is a right triangle. (b) If the distance between the tracks is b feet and the distance be- tween the rails in each track (the gauge) is a feet, show that (2x + 3a + 2by = y^+ (2x + 2a + by. Since a and b are given, this equation may be solved for x in terms of y or for y in terms of x. Hence if the distance AK is known, we may use this equation to compute the radii of the arcs used in constructing the figure. On the other hand, if the radii of the arcs are known, we may compute the distance AK. This is a very common problem in railway construction. The construction is also used in laying out a curved street to connect two parallel streets. 16. If in the preceding problem a = 4 feet 8} inches, b = 9 feet, and AK = 200 feet (an actual case), find the radii of the arcs. Using the same values for a and 6 as in the preceding example, &ndAKii C^E = 300 feet. 230 PLANE GEOMETRY. FURTHER DATA CONCERNING CIRCLES. 398. In Chapter II numerous theorems were proved concerning the equality of arcs, angles, chords, etc. The three following theorems involve inequalities of these elements. The student should construct figures in each case and give the proof in full. 399. Theorem. Li the same circle or in equal circles, of two unequal angles at the center the greater is sub- tended by the greater arc ; and of two unequal arcs the greater subtends the greater angle at the center. The proof is made by superposition. 400. Theorem. In the same circle or in equal circles, of two unequal minor arcs the greater is subtended by the greater chord; and of ttoo unequal chords the greater subtends the greater minor arc. The proof depends upon the theorem of § 117. 401. THEOREjr. In the same circle or in equal circles, of tioo unequal angles at the center, both less than straight angles, the greater is subtended by the greater chord; and of two unequal chords the greater subtends the greater angle at the center. 402. Problem. On a given line-segment as a chord to construct an arc of a circle in ivhich a ginen angle may be in- scribed. \C Construction. Let K be the given angle and AB the given segment. REVIEW AND FURTHER APPLICATIONS. 231 It is required to construct a circle in which AB shall be a chord and such that an angle inscribed in the arc AMB shall be equal to the given angle K. At A construct an angle BAG equal to Z K. If now a circle were passed through A and B so as to be tangent to ^c at A, then one half the arc AB would be the measure of the angle BAG. See § 219. Then any angle inscribed in the arc AMB as Z.APB would be equal to Zbag=Zk. Hence the problem is to find the center of a circle tan- gent to AG at A and passing through A and B. Let the student complete the construction and proof. 403. EXERCISES. 1. Prove that the problem of § 402 may be solved as follows: With A, any point on one side of the angle K, as center and AB as radius strike an arc meeting the other side of the angle at B. Cir- cumscribe a circle about the triangle ABK. 2. Show that from any point within or outside a circle two equal line-segments can be drawn to meet the. circle and that these make equal angles with the line joining the given point to the center. 3. Show that if two opposite angles of any quadrilateral are sup- plementary, it can be inscribed in a circle. Suggestion. Let A BCD be the quadri- p lateral in which /LA +ZC = 2 rt. d. y — JPxX Pass a circle through B, C, and D. To prove / /ji ^s^ri that A lies also on the circle, and not at some / /7/ \ inside or outside point as A' or A". V?- '/ '^l' 1/ (1) Show that Z .4 ' > Z .1 and /.A" 1xt.A if A' is /.-'''-^ within the circle and Z .4 " -)- Z C < 2 rt. ^ if ^ " A'' is outside the circle, both of which are contrary to the hypothesis. Hence the fourth vertex must lie on the circle. Show that the condition that the polygon is convex follows from the hypothesis. 232 PLANE GEOMETRY. 404. Problem. 2o draw a common tangent to two circles which lie wholly outside of each other. Construction. Let the given circles be and C of which the radius of the first is the greater. Required to draw a tangent common to both circles. Draw an auxiliaiy circle with center o and radius equal to the difference of the two given radii in one figure and equal to the sum of these radii in the other figure. In each case draw a tangent to this auxiliary circle from o', thus fixing the point D. See § 230. Draw the radius OB, thus fixing the point P. Draw o'p' II DP, thus fixing the point P'. Then PP' is the required tangent. Proof: Show that, in each case, PP'o'd is a rectangle, thus making PP' perpendicular to the radii OP and o'p', that is, tangent to eacli circle. Definition. A common tangent to two circles is called direct if it does not cross the segment connecting the centers, and transverse if it does cross it. 405. EXERCISES. 1. Describe the relative positions of two circles if they have tvco direct common tangents. Also if they have only one. 2. Describe the positions of two circles if they have two transverse common tangents ; one ; none. BEVIEW AND FURTHER APPLICATIONS. 233 PROBLEMS AND APPLICATIONS. 1. Find the locus of the middle points of all chords of a circle drawn through a fixed point within it. _ 2. The sides AD and BC of the square ABCD are each divided into four equal parts and a circle inscribed in the square. Lines are drawn as shown in the figure. Show that EFGHKL is a regular hexagon. This is the construction by means of ^ which a regular hexagonal tile is cut from a square tile. 3. In the preceding example what fraction of the area of the square is covered by the hexagon. 4. A circle of constant radius passes through a fixed point A. A line tangent to it at the point P remains parallel to a fixed line BD. Find the locus of P. Suggestion. In the figure prove ./I CPC a parallelogram. Show that the locus consists of two circles. 5. The same as the preceding except that the circle of constant radius remains tangent to a fixed circle instead of passing through a fixed point. Suggestion. On that diameter of the fixed circle which is perpendicular to the fixed line lay onAC = C'P. Prove A C'PC a parallelo- gram. Show that the locus consists of two circles. 6. In the square ABCD, AM = OB -. BU = VC, etc. The point E is the intei'section of the segments MV and TU, F is the inter- section of OR and TU, G is the intersection of O W and US, etc. "When these segments are partially erased, we have the figure. (a) Prove that TMEN, OUGF, etc., are squares. (J) What kind of a figure is MOFE ? Prove, (c) How many axes of symmetry has the figure NEFGH ••. ? Has it a center of symmetry? w a 234 PLANE GEOMETRY. -E F \/HK 7. ABCD is a square the mid-points of whose sides are joined. On its diagonals points E, K, P, T are taken so that AE = BK = CP = DT. The construction is then completed as shown in the figure. (F6' and MN lie in the same straight line.) (a) Prove that E, K, P, T are the vertices of a square. (J) If AB = 6 find AE so that the area of the square EKPT shall be half that of the square ABCD. (c) If AB = 6, and if CP = PS, find the area of the figure EFGHKI.M ■■■ . Also find the area of the trapezoid BKLR. (d) If CP = I CS, find the area of the inside figure and also of BKLR. (e) What fraction of CS must CP be in order that the inside figure shall be half the square ? (/) If the inside figure is J of the square and if AB = 8 inches, find CP. 8. In the figure ABCD is a square. Each of its sides is divided into three equal parts by the points E, F, G, H, ■■■. The points E, X, Y, H; F, A', W, M; etc., lie in straight lines. (a) If AB = 6 inches, find the area of that part of the figure which lies outside the shaded band. ( h) If AB = 6 inches, and if the width of the shaded band is } inch, find the area of the band. (c) If ,1 /} = S inches and XP = ' inch, find the area of that part of the figure which lies inside the band. (d) If ,l/; = 8 inches, what must be the width cif the liaiul in order that it shall oeenpy 10% of I lie area cil' the whole design? (c) If A I! = a inches find the width of the baud if it occupies ti |ici- cent ol' the area of the whole ilesien. "^ J 1_ Si^. ^"^ Parquet Flooring. BEVIEW AND FURTHER APPLICATIONS. 235 9. By means of the accompanying figure find the area of a regular dodecagon whose sides are 6 inches. Notice that the dodecagon con- sists of the regular hexagon in the center, the six equilateral triangles, and the six squares. Note how this figure enters into the accom- panying tile design. 10. Find an expression for the area of a regu- lar dodecagon whose sides are a. 11. Find the apothera of a regular dodecagon by dividing its area by half the perimeter. Also find it by finding the apothem of the hexagon and then adding a to this. Compare results. 12. Find the radius of a circle circumscribed dodecagon whose side is a. 13. If the accompanying design for par- quet border is 10 inches wide, find the width of the strips from which the small squares are made. Also find this result if the width of the border is a. 14. In the figure of the preceding example find the dimensions of the small triangles along the two edges if the width of the border is a inches. 15. Show that the figure given below may be used as the basis for the design shown beside it. (See Ex. 7, p. 147.) 16. If an outside of each of the two squares shown in the figure is 6 inches, find the width of the strips of which they are made in order that each shall fit closely into the corner of the other? Suggestion. Find the altitude on EM in A UEM above if 45 = 6 inches. This is a common Arabic ornament. about a regular 236 PLANE GEOMETRY. 17. In the figure ABCfJ is a square. On each side a triangle, two of whose side.s are parallel to the diagonals of the square, is so constructed that the points E, F, and G lie in the same straight line. (a) Prove that the triangles are right isos- celes triangles. (6) What part of the square lies within the triangles ? (c) If AB = a, find AE so that the tri- angles shall occupy — of the area of the square, n See the accompanying tile pattern. 18. ABCD is a square and the small fig- ures in the corners are squares. (a) Show that if the lines intersect as shown in the figure A E must be J of A B. (6) \i AB = % inches find the areas of the squares A'B'C'D' and XYZW. 19. In the tile design show that the figure within the square is the same as that of Ex. 18. What part of the large square is occupied by each shaded part ? 20. ABCD is a square. AE = FB = BK = LC= rp = QD=DR = .S.l. (a) li AB = n and ii AE = b, find the sum of the areas of the four shaded rectangles. (h) If .IS = 8 inches, find ,12? so that the sum of these rectangles shall be ^ of the whole square. Interpret the two solutions. (c) If AB = a inches, find AE so that the sum of tlie rectangles shall be - of the square. n 21. ((j) Show that in the accompanying de- sign for tile llooriiig the sizo of any ono piece ilrtfrmines the size of every piece in (lie figure. (/;) What fraction of the figure is occupied by eacli colurV m. M D 'd' . / 'C' \ Z\ / 'V Y X' \ A', ' \ B- D Q P r* L S H G A EEVIEW AND FURTHER APPLICATIONS. 237 22. This tile floor design is based on the plane figure and this in turn is based on the figure of Ex. 17 where AB 6 ■ What part of the whole design AE = - 0^/ '.^- A E is occupied by tiles of the various colors ? 23. ABCD is a square. Lines are drawn parallel to the sides and intersecting as shown in the figure. (a) Show how the sides of the square must be divided in order to make the lines intersect as they do. (6) If AB = 6 inches, find the areas of the squares SXZQ and KTYP. (c) Find these areas i£ AB = a. (d) Show that the outline figure is the basis of the tile design here given. 24. ABCD and A'B'C'D' are equal squares, rounded by strips of equal width. (a) If AB = a, find AF, FE, and EB. (b) Find the width of the strips if their outer edge passes through the points A,B, C, D, when AB = S inches. Also when AB = a. (e) If AB = a, find A'H and hence KL. (d) If the width of the strip is 1 inch, find KL. (e) If the width of the strip is J, find KL. (/) If KL = c, find the width of the strip. When the width of the parquet floor border shown in the figure is given, (/) is the problem one needs to solve to know how wide a dark strip to use. Compare Ex. 16, page 235. A'B'C'D' is sur- 238 PLANE GEOMETRY. THE INCOMMENSURABLE CASES. 406. We have seen, in § 2o4, that there are segments which are incommensurable ; that is, which have no com- mon unit of measure, — for instance, the side and the diagonal of a square. For practical purposes the lengths of such segments are approximated to any desired degree of accuracy, and their ratios are understood to be the ratios of these approximate numerical measures. See §§ 23b-240. All theorems involving the ratios of incommensurable segments, and the lengths and areas of circles, have thus far been proved only for such approximations, and these are quite sufficient for any refinements of measurement which it is possible to make. But for theoretical purposes it is important to consider these incommensurable cases further, just as in algebra we not only approximate such roots as Vi. V3, V5, etc., but we also deal with these surds as exact numbers. For instance, in such an operation as (VS + V2)( V3 - V;j)= 3 - '2 = 1. While the length of the diagonal of a unit square cannot be expressed as an integer or as a rational fraction, that is, the quotient of two integers, we nevertheless think of such a segment as having a di-Jinite length, or what is the same thing, a definite ratio with the unit segment forming the side of the sqnare. E.g. If d is the diagonal of the square whose side is 1, then (/2 = 1 + 1 or (/ = V2. Xow suppose %'2 = ", a fraction in it,s lowest fcrms. Then 2=—, a fraction also in its lowest terms. But a fiacttion in its lowest terms cannot be equal to 2. Hence \ 2 is ueitliei- an inteyfr nor the quotient of two integers. REVIEW AND FURTHER APPLICATIONS. 239 407. The following axiom and that of § 409 ure funda- mental in the consideration of incommensurable line-seg- ments. Axiom X. Every line-segment ab has a definite length, which is greater than ac if a lies between A and B. The length of a line-segment is in every case a number, which is rational (an integer or the quotient of two inte- gers) in case the segment is commensurable with the unit segment, but which otherwise is irrational. Under the operations of arithmetic these irrational numbers obey the same laws as the rational numbers. The length of a line-segment is often called its numeri- cal measure. 408. The exact ratio, or simply the ratio, of two line- segments is the quotient of their numerical measures, whether these are rational or irrational. That is, every such ratio is a number. It is obvious that a segment may be constructed whose length is any given rational number. We have- also seen how to construct with ruler and compasses segments whose lengths are certain irrational numbers, such as V2, V3, V5, etc. See Ex. 3, § 295. 409. We now assume the following Axiom XI. For any giveii number K there exists a line segment whose length is K. This does not imply that it is possible by means of the ruler and compasses to construct a segment whose length is any given irrational number. For instance, we cannot thus construct a segment whose length is V2. We now prove the fundamental theorem on the propor- tionality of sides of triangles. 240 PLANE GEOMETRY. 410. Theorem. A line parallel to the base of a tri- angle, and meeting the other two sides, divides these ^i/les propiirtionalhj. G Given A ABC with DE 1 1 AB- /\ rr CD CE / V To prove — = — v/ \g CA cB jy- — Y Proof : Whether cd and CA are / ^ commensurable or not, we know Ijy SS 407, 408 that — and — are definite numbers. We prove ^'' CA CB that these numbers cannot be different. ■p. , CD ^ CE h irst, suppose — < CA CB CTi f^W Take F between c and E so that — = • — Ax. XI (1) CA CB Divide CB into equal parts, each less than EF. Then at least one of the division points, as G, lies between E and F. Draw GH II AB. Since CG and CB are commensurable, we have, by § 243, CH^CG o^ CA Cb' Dividing- (1) by (2"), we have — = - — -. But this can- en CG not be true, since CD > rfl" and CF < CG. — • cannot be less tlian — - CA CB Secondlv, prnve in the same manner that — cannot be " CE . '^•' greater tliau — ^. Hence, since the one is neither less than nor i^n-ealer than tlie otlier, tliese ratios must be equal. Tlio following treatment of im-ommensurable ares and angles is exaelly similar to the above. Hence — • cannot be less than — -. Whv ? BEVIEW AND FURTHER APPLICATIONS. 241 411. Axiom XII. Any given arc ab has a definite ratio ivith a unit arc, which is greater than that of an arc AC, if c lies on the arc ab. 412. Axiom XIII. A7iy given angle abc has a defi- nite ratio ivith a unit angle, which is greater than that of ABD if BD lies within the angle abc 413. Theorem. In the same or equal circles the ratio of two central angles is the same as the ratio of their intercepted arcs. E Outline of Proof : We show that in the figure j— arc AB "^ can neither be less than nor greater than . arc CD Z AOB arc AB Suppose ;— < . Z CO'd arc CD 1 hen take ^ so that p- = . (1) Z CO'd arc CD Divide arc CD into equal parts each less than arc EB. Lay off this unit arc successively on AB reaching a point F between E and B. Then arcs AF and CD are commen- surable and by a proof exactly similar to that of § 243, making use of § 199, we can show that Zy40-F __ arc AF ^n-^ /.CO'd arc CD' Complete the proof as in § 410. 242 PLANE GEOMETRY. 414. Axiom XIV. Any given rectangle with base b and altitude a has a definite area which is greater than that of another rectangle luith base V and altitude a' if a>a' and b>b' or if a> a' and b > V. 415, Theorem. The area of a rectangle is the j^/'od- uct of its base and altitude. A. B Proof : Denote the base and altitude by b and a, respec- tively, and the area by A. Suppose A < ab, and let a' be a number such that A = a'b. Lay off BE= a'. Consider first the case where b is commensurable with the unit segment and a is not. Divide the unit suyment into equal parts each less than CE and lay off one of these parts successively on BC reach- ing a point F between E and ('. Denote the length of BF by a", and draw FF' II AB. Tlien by § 307 the area of ABFF' i.s a"b. By hypothesis A = a'b, but a'b < a"b since a' < a" . Hence A < <('7). (1) But by Ax. XIV ,1 > a"b. (2^ Hence the assumption that A rt^ cannot hold. Tli(^ prodf in cast! both sides are inoomnu'nsurable with tlu^ unit se,t,nnont is now oxarlly like the above and is left to the student. REVIEW AND FURTHER APPLICATIONS. 243 416. Axiom XV. A circle has a definite length and incloses a definite area which are greater than those of any inscribed polygon and less than those of any cir- cumscribed polygon. 417. Theoeeji. For a given circle and for any number K, hoivever small, it is possible to inscribe and to circumscribe similar polygons such that their perimeters or their areas shall differ by less than K. Proof : First, Let p and p' be the perimeters of two similar polygons, the first circumscribed and the second in- scribed, and let a and a' be their apothems. Then P=^, py P-P' p' a' p' a' Hence p -p' =pi .'LZ^. §245 Now a— a' can be taken as small as we like. Hence p • — j-~ can be made as small as we please; that is, p —p' can be made smaller than any given number K. Second, letting P and p' be the areas of the circum- scribed and inscribed polygons respectively, we have P a2 — = — ^, and the proof proceeds exactly as before. 418. Since the length c of the circle is greater than p' and less than p, it follows that C is thus made to differ from either p or p' by less than K. And since the area A of the circle is greater than p' and less than P, it follows that A is made to differ from either P or p' by less than K. 244 PLANE GEOMETRY. 419. Theorem. The lengths of two circles are in the same ratio as their radii. Proof: Let c and t/ be the lengths of two circles whose centers are O and & and whose radii are r and r'. C T We shall prove that — = — by showing that— is neither less than nor greater than — . r First, suppose that — < — or c<(^ ■—:. i: r r T And let c differ from c' • -r by some number K. r' ■' Now circumscribe a regular polygon P about O O with perimeter^ such 'that pc' and hence ^>c' •—. (2) C T Hence the supposition that -7 < — leads to the contra- diction expressed in (1) and (2) and is untenable. Now prove in same way that -y> — is untenable. 420. Theorem. The arco^ of tiro cirrh s are in the savic mill) ax the squares of their radii. I'sing §§ :ltS and 418, the proof is exactly similar to that (if the jireceding tlieiuein. REVIEW AND FURTHER APPLICATIONS. 245 PROBLEMS AND APPLICATIONS. 1. A billiard ball is placed at a point P on a billiard table. In ■what direction must it be shot in order to return to the same point after hitting all four sides ? (The angle at which the ball is reflected from a side is equal to the angle at which it meets the side, that is, Zl = /.-I, and Z3 = Z4.) Suggestion, (a) Show that the opposite sides of the quadrilateral aloug which the ball travels are parallel. Qi) If the ball is started parallel to a diagonal of the table, show that it will return to the starting point. 2. Show that in the preceding problem the length of the path traveled by the ball is equal to the sum of the diagonals of the table. 3. Find the direction in which a billiard ball must be shot from a given point on the table so as to strike another ball at a given point after first striking one side of the table. Suggestion. Construct BS _L to that side of the table which the ball is to strike and make ED — BE. 4. The same as the preceding problem except that the cue ball is to strike two sides of the table before striking the other ball. Suggestion. B'E' = E'D', D'H = HF. 5. Solve Ex. 4, if the cue ball is to strike three sides before strik- ing the other ball, — also if it is to strike all four sides. 6. In the figure ABCDEF is a regular hexagon. Prove that: (a) AD, BE, and CF meet in a point. (6) ABCO i&a. rhombus, (c) The inner circle with center at and the arcs with centers &\, A,B, C, etc., have equal radii. (rf) The straight line connecting A and C is tangent to the inner circle and to the arc with center at B. (e) The centers of two of the small circles lie on the line connecting A and C. r-^F E.^ ■■ i'---— -'B (y) Find by construction the centers of the small circles. 246 PLANE GEOMETRY. 7. In the figure EG and FH are diameters of the square ABCD. On the diagonals, points K, U, V, W are laid off so that AK = BU= CV = DW. Also LM = NP = RS = etc. EN and SF are in the same straight line, and so on around the figure. Prove that : ((() A' t/TI-F is a square. (6) AKME and ENUB are equal trapezoids. (c) L, O, 2' lie in a straight line. (rf) The four heavy six-sided figures are congruent. D ( ? r ■w / \ T / / / ^ H z /^ \. F (e) liAB = a, AK , A0 6 and ML = KL, find the areas of the figures KUVW, AKME, LOPXEM. (/) Find tlie areas required under («) if AK = — and LM = m ■ LK. {d) IflA- AO , what is the length of ML if the four heavy figures occupy half the square ABCDl 8. Find a side of a regular octagon of radius r. See Ex. 3, p. 146. 9. Find the area of a regular octagon of radius r. 10. In the figure ABCD-- is a regular C - — - P octagon inscribed in a circle of radius i. Find the area of the triangle ABC. SrcGESTiON. Find first the altitude on .1 ( ' in the A .1 0< ' and thus the altitude of A.l/jr 11. In the same fig- ure the liiii-s < 7/ and .1 /' arc l \/ \o A B E F 14. The design opposite consists of white figures constructed like the inner figure preced- ing, together with the remaining black, figures. What part of the figure is white? 15. Show that the altitude of an equilateral triangle with side s is - VS. 16. If the angles of a triangle are 30°, 60°, 90°, and if the side opposite the 60° angle is r, show that .MKMMMM .hkkM» Lmm^ Design from the Alhambra. the other sides are o v 3 and ir Vs. See §159, Ex. 14. 17. Three equal circles of radius 2 are inscribed in a circle as shown in the figure. Find the radius of the large circle. Suggestion. The center of the large circle is at the intersection of the altitudes of the equi- lateral triangle O' O" 0'" (Why ?) Hence O'OD is a triangle with angles 30°, 60°,- 90° as in Ex. 16. Solve this problem for any radius r of the 2 r /— small circles. Ans. iJ = r-f- — v3. o s/^ \ / T \ /fr 248 PLANE GEOMETRY. 18. A', B', C, are the middle points of the sides of the equilateral triangle ABC. The sides of the triangle A'B'C are trisected and segments drawn as shown in the figure. From C'luireb of Or San iiiehele. If AB = a. (a) Pi-ove A 'B"C" an equilateral triangle. (6) Find the area of A"B"C" and of the dotted hexagon, (c) Find the area of the triangle GFC and of the trapezoid DEB"A". A P From Westminster Abbey. 19. In the figure ARC is an equilateral triangle. On the equal bases P.\f, MD, DE equilateral arches are constructed, two of them tangent to thi» sides qf the triangle at L and ,V respectively. Circles 0' and 0" an' eacli tanj;ent to a side of the triangle and to two of the arches. Circle is tangent to circles (>' and 0" and to both sides of the triannle. (a) \i AB = II, find /)/•:. .SiKjGESTioN. In I he right triangle DEL one acut«> angle is 60''. REVIEW AND FURTHER APPLICATIONS. 249 (b) Find the ratio r:r',r being the base DE of the arch and H the radius of the circle 0'. Suggest ion. GD = 2 DL, GO' = 2 r', and O'D = V(r+r'y-r^= y/2 n-i + r'K (e) By what fraction of a will the circles 0' and 0" fail to touch each other? 20. In the figure CD ^ DA = AG = GB, and DE = EA = AF = FG. Semicircles are constructed on the diameters CB, CG, CD, DG, DB, GB. HG = KD = CE. Arcs GL and DM have centers H and K respectively. ^ fi S — p — i — ^ — i ' Jr ■ From Church of Or San K L D E A F C B H Michele. Circles are constructed tangent to the various arcs as shown in the figure. Thus O 0" is tangent to semicircles on the diameters CB, CG, and DB. O 0' is tangent to the semicircles on the diameters CG and DB and to the arc DM. Let CB = a. (a) Find the areas of each of the six semicircles. (6) Find the radius r". SuGGESTiox. EA, EP and AN are known. (c) Find the radius r'. Suggestion. Enumerate the known parts In A EDO' and FDO'. (rf) Find r and r^n (e) Having determined the radii of the various circles, show how to construct the whole figure. (/) What fraction of the area of the whole figure is occupied by the six circles ? 21. Prove that two segments drawn from vertices of a triangle to points on the opposite sides cannot bisect each other. 250 PLANE GEOMETRY. FURTHER APPLICATIONS OF PROPORTION. 421. Definition. A segment is said to be divided in extreme and mean ratio by a point on it, provided the ratio of the whole segment to the larger part ^ OB equals the ratio of the larger part to the smaller. E.g. the segment AB is divided in extreme and mean ratio by the point C 11 = — — . ^ A a CB 422. Problem. To divide a given line-segment in extreine and iiwan ratio. Solution. At one extremity B ^--^ erect BO\.AB and make BO = \ ^ ,--'' ; •^ 71-' AB. Draw OA. ,'-' I On OA take OB = OB and with ,--'' I ^ as a center and AB as a radius a c B draw the arc DC. Then C is the required point of division. Proof : Let a be the length of AB. Then OB = - . Hence ao^ = Jff' + ^2 = a2 + £ = 2 ^2, or ^0 = ^ ^ 57 Then AC = AD = Ao - BO = -^ V5 - ^ =0 (^ ^' - l)' and BC = AB — AG = a — 'l^ T Vo — 1 j = -^ Ts — ^ ^j. I R \0 Substituting these values in ' — and ' — , and simplifying, '^ AC CB ^ ■' ° WO have _ V5 - 1 3 - % ,V By rutidiializing donominators these fractions are shown to be equal, and licme — = - — AC CB REVIEW AND FURTHER APPLICATIONS. 251 423, Problem. To construct with ruler and com' passes angles of 36° and 72°. Solution. On a given segment AB, determine a point such that ^ = :^. (See 5 422.) AC RC -^ Using BB = AC zs a base and AB as one leg, construct the isosceles triangle ABB. Then Zl = 72° and Z4 = 36°. Proof : Draw BG. Then A ABB ~ A BBC, (§ 259) AB BD J / 1 . since — = — andZ 1 is common. _ BB BG A C Hence A ABB and BBC are both isosceles (Why ?), and Z 1 = Z 2, Z 3 = Z 4. Also Z2 = Z3 + Z4 = 2Z4=Z1. (Why?) Thus in A ABB each base angle is double Z4, making Z4 = | of 2 rt. Z = 36°, (Why?) and Z 1 = 2 Z 4 = 72°. 424, EXERCISES. 1. Inscribe a regular decagon in a circle. /^ ^V^ Suggestion. Construct the central angle j n'-^ i-^ A OB = 36°, thus determining the sjde AB of the V J decagon. N^ y' 2. Inscribe a regular pentagon in a circle. Suggestion. Join alternate vertices of a decagon, or construct a central angle equal to 72°- 3. Inscribe a regular polygon of fifteen sides in a circle. Suggestion. Since } — jV = -h> i* follows that the arc subtended by one side of a hexagon minus that subtended by one side of a deca- gon is the one which subtends one side of a polygon of fifteen sides. 252 PLANE GEOMETRY. 4. The radius of an inscribed regular polygon is a mean propor- tional between its apothem and the radius of a similar circumscribed polygon. 5. A circular pond is surrounded by a gravel walk, such that the area of the walk is equal to the area of the pond. What is the ratio of the radius of the pond to the width of the walk? 6. If a, b, c are three line-segments such as - = - show that - equals he c the ratio of the area of any triangle described on a as a base to the similar triangle described on J as a base. 425. Definitions. A line-segment AB is said to be divided internally in a given ratio — by a point G lying on the segment, if — = — • See § 252. A line-segment is said to be divided externally in a given ratio - by a point c' lying c b r" ACf T """^""^ on AB produced, if = -• C'b 8 A line-segment AB is said to be divided harmonically if the points O and c', lying respectively on AB and on AB A r* A r^' produced, are such that — = CB c'b 426. EXERCISES. 1. Show that if AB is divided harmonically by C and C, then CC is divided harmonically by A and B. 2. Show that the base of any triangle is cut harmonically by the bisectors of the internal and external vertex angles. 3. Show how to divide a line-segment externally in extreme and mean ratio, that is, in the figure below, so that ^^-^ = -^-i . C'A C'B Suggestion. In the figure of § 422 produee BA to a ix>int C such that C'A = vl -(- OU. .. Then use the method there A B used, to show that -'— and ■— ^ each reduces to ^-^J- C. 1 CB 2 REVIEW AND FURTHER APPLICATIONS. 253 427. Theorem. If an angle of one triangle is equal to an angle of another, their areas are in the same ratio as the products of the sides including the equal angles. A B A' B' A Given A ABC and A'^C in which /:LA=AA'. To prove that ^^^^ ^^'^^ Proof : Then Hence, aa'b'g' a'b'-a'c' Place aa'b'g' so that Z.A' coincides with Z.A. AABC=^h-AB and AA'B'c' = ^h''A'B'. AABC _ Ih AB _ h AB Now show that aa'b'g' h ih'- AC a'b' h' a'b' h' A'G' and hence that AABC __ AC ^ AB _ AB- AC aa'b'g' ~ a' c' a'b'~~ a'b' -a'c'' 428. Theokem. The square on the bisector of an angle of a triangle is equal to the product of the tioo adjacent sides minus the product of the segments of the opposite side. Outline of proof: Produce the bisec- tor of the given angle to meet the cir- cumscribed circle. Since Abdc~Aaeg AC • bc = gd • GE. (1) But GD- CE= CDQOB + BE') = CD^ + CD • DE (2) and CD-I)E = AD-DB. (Why?) (3) Using (1), (2), and (3), complete the proof. 254 PLANE GEOMETRY. 429. THEOitEjr. The product of tivo sides of a tri- angle is equal to the product of the altitude from the vertex in luhich these sides meet and the diameter of the circmnscrihed circle. Outline of Proof : Using the figure, show from the similar triangles ACD and EBC that AC-BC= CE- CD. 430, EXERCISES. 1. The areas of two parallelograms having aa angle of the one equal to an angle of the other are in the same ratio as the product of the sides including the equal angles. 2. Three semicircles of equal diameter are arranged as shown in the figure. (a) li AD = a, find the area bounded by the arcs AB, BEC, and CA. (h) If the area just found is 2 square feet, find.lZ). 3. Prove that the bisectors of the angles of any quadrilateral form a quadrilateral whose opposite angles are supiilementary. SuddESTioN. Show that Z3 + Zi + Z5 + ^Q = 2 rt. A, and hence that Z 1 + Z 2 = 2 rt. ^. 4. On each of two sides of a given triangle ABC as chords con- struct arcs in which an angle of 120° may be inscribed. If these arcs meet in a point inside the triangle, show that the three sides of the triangle subtend the same angle from the point 0. 5. If Z 1 + Z 2 + Z :? = t rt. A, sliow liow to find a point within a given triangle AIW so that ZAOB = Zl, ZBOC = Z-2, and ZCOA = z:i. REVIEW AND FURTHER APPLICATIONS. 255 6. Given any two segments AB and CD and any two angles a and 6, find a point suoli that ZAOB — Aa and Z COD = Ah. Discuss the various possible cases and the number of points in each case. 7. If Z A OB is a central angle of a circle and if Z CDE is inscribed in an arc of the same circle and if Z 4 OB = Z CDE -. the chords AB and CE are equal. 431. Definition. A set of lines which all pass through a common point is called a pencil of lines, and the point is called the center of the pencil. 4 rt. A, then 432. EXERCISES. CA m". /A" A % \ ■ <4' B' to' 1. The lines of a pencil intercept proportional segments on parallel transversals. Given three lines meeting in O cut by three parallel transversals. To prove that the corresponding segments are proportional, that is, to show that AB ^ A'B' ^ A"B" BC B'C B"C" ' 2. If two polygons are symmetrical with respect to a point, they are congruent. (See § 170, Ex. 2.) 3. Any two figures symmetrical with respect to a point are congruent. Suggestion. About the point O as a pivot swing one of the figures through a, straight angle. Then any point P' of the right-side figure will fall on its sym- metrical point P of the left-side figure. 4. By means of the theorem in the preceding example show how to make an accurate copy of a map. Suggestion. Fasten the map to be copied on a drawing board. A long graduated ruler is made to swing freely about a fixed point 0, and by means of it construct a figure symmetrical to the map with respect to the point 0. How are the distances from measured off ? 256 PLANE GEOMETRY. 433. Theorem. If corresponding vertices of tioo polygons lie on the same lines of a pencil and if they cut off proportional dis- tances on these lines from the center, then the poly- gons are similar. Prove the theorem first for triangles. OB OC OA J OA OB OC Given. — : = — 7 = — 7 and — - ni3' r\n' n a' n a' i OA' OA" OB" OC" To prove that A ABC, A'b'c', a"b"c" are similar. Prove the theorem for polygons of any number of sides. 434. Definition. Any two figures are said to have a center of similitude O, if for any two points P^ and P^ the lines P^O and P^O meet the other figure in points P'j and p'o such that p\o P'^O Then P\ and P'j are said to correspond to the points Pj and Pj. Thus in the figure of § 433 o is called the center of similitude of the two polygons. Any two figures which have a center of similitude are similar. This affords a ready means of constructing a figure similar to a given figure and having some other required property that is sufficient to determine it. Definition. The ratio of any two corresponding sides of similar polygons is called their ratio of similitude. REVIEW AND FURTHER APPLICATIONS. 257 435. EXERCISES. 1. Construct a polygon similar to a given polygon such that they shall have a given ratio of similitude. Suggestion. Let ABCDEF be the given polygon and — the n given ratio of similitude. Select any convenient point O, such that OA = m. Draw lines OA, OB, etc. On OA lay off OA', n units. Lay oS points B', C, etc., so that OA ^ OB ^ PC ^ OP ^ OE ^ OF OA' OB' OC OD' OB' OP' Prove that A'B'C'D'E'F' is the required polygon. 2. Show how the preceding may be used to enlarge or reduce a map to any required size. Suggestion. Arrange apparatus as under Ex. 4, §432. 3. In the figure ABCD is a parallelogram whose sides are of con- stant length. The point A is fixed, while the remainder of the figure is free to move. Show that the points P and P' trace out similar figures and that their ratio of simili- tude is • AD+ CP This shows the essential parts of an in- strument called the pantograph, which is much used by engravers to transfer figures and to increase or decrease their size. The point P is made to trace out the figure which is to be copied. Hence P' traces a figure similar to it. The scale or ratio of similitude is regulated by adjusting the length of CP. 4. Construct a triangle having given two angles and the median a from one specified angle. Solution. Construct any triangle ABC having the required angles and construct a median AD. Prolong DA to D', making AD' =a. Extend BA and CA and through D' draw B'C'WBC, making A AB'C. Prove that this is the required triangle. (Notice that ^ is the center of similitude.) y\ 5. Liscribe a square in a given triangle using the figure given here. Compare this method with that given on page 147. 258 PLANE GEOMETRY. 6. Construct a circle through a given point tangent to two given- straight lines. Solution. Let a and h be the given lines and P the given point. Construct any circle tangent to a and 0. Draw AP meeting the circle O in C and D. Draw CO and OD and through P draw lines parallel to these meeting the bisector of the angle formed by a and b in O' and 0". Prove that and 0" are centers of the re- quired circles. Observe that A is a center of similitude. This method is used to construct a railway curve through a fixed point connecting two straight stretches of road. 7. On a line find a point which is equidistant from a given point and a given line. The following is another instance of the use of this very important device in constructing figures that resist other methods of attack. It consists essentially in first constructing a figure similar to the one required and then constructing one similar to this and of the proper size. From Westminster Abbey. 8. Given a circle with radius OE. Construct within it the design shown in the figure. That is, the inner semicircles have as diameters the sides of a res'ular sixteen-sided polygon. Each of the small circles is tangent to two semicircles. The outer nres have their centers on the given circle and each is tangent to two small circles. All these arcs and circles have equal radii. BEVIEW AND FURTHER APPLICATIONS. 259 Solution. Construct an angle equal to one sixteenth of a peri- gon. Through any point D' in the bisector of this angle draw a seg- ment EF perpendicular to the bisector and terminated by the sides of the angle. On EF as a diameter construct a semi- circle. Make B'D' = EF = B'A' = D'C = C'A' and construct the small figure. Now draw radii of the given circle dividing it into sixteen equal parts and bisect one of the central angles by a radius OA. Construct Z BAB = Z D'A'B'. Then ^ JS is twice the radius of the required arcs and circles. The whole figure may now be constructed. 9. Given any three non-collinear points A', B', C, to construct an equilateral triangle such that A', B', C shall lie on the sides of the triangle, one point on each side. Suggestion. Through one of the points as A' draw a line such that B' and C lie on the same side of it. 10. Given an equilateral triangle j4i?C, to construct a triangle sim- ilar to a given triangle A'B'C with its vertices on the sides of ABC. Suggestion. Construct an equilateral triangle such that A',B', C lie on its sides. Then construct a figure similar to this and of the required size. 11. lip andp' are similar polygons inscribed in and circumscribed about the same circle, and if 2 s is a side of the circumscribed polygon p', show that the difference of the areas of p and p' is equal to the area of a polygon similar to these and having a radius s. Suggestion. Let the areas of the polygons whose radii are r, r', and s be A, A', A". Prove that A< and A' r' Complete the proof. 12. Two circles are tangent at A. A secant through A meets the circles at B and C respectively. Prove that the tangents at B and C are parallel to each other. 13. Prove that the segments joining one vertex of a regular poly- gon of n sides to the remaining vertices divide the angle at that vertex into n-2 equal parts. 260 PLANE OEOMETBT. FURTHER PROPERTIES OF TRIANGLES. 436. Definition. A segment AB is said to be projected upon a line I if perpendiculars from A and B are drawn to I. If these meet I in points C and D, then CD is the pro- jection of AB upon I. 437. Theorem. The square of a side opposite an acute a7igle of a triangle is equal to the sum of the squares of the other two sides minus twice the product of one of these sides and the projection of the other upon it. c c c m Outline of Proof : In either figure let Z 5 be the given acute angle, and in each case BH is the projection of BC upon AB. Call this projection m. We are to prove that J^ = a^ + e^ — 2 cm. In the left figure, J2 = A^ + (e - mf. (1) In the right figure, V^ = h^. + (m — c)^. (2) In either case }? = c? — m^. (3) Substitute (3) in (1) or in (2), and complete the proof. Modify each figure so as to draw the projection of AB upon BC and call this n. Then give the proof to show that Ifi = a^ + c"^ — 2 an. 438. EXERCISE. 1. The area of a polygon may be found by drawing its longest diagonal and letting fall perpendiculars upon this diagonal from each of the remaining vertices. Draw a figure like the one in the margin, only on a much larger scale, measiiro the necessary lines, compute the areas of the various parts (see § 314), and thus find its total ai'ea. BEVIEW AND FUBTHEB APPLICATIONS. 261 439. Theorem. 27ie square of the side opposite an ohtuse angle of a triangle is equal to the sum of the squares of the other two sides plies twice the product of otie of these sides and the projection of the other upon it. A C B D Outline of Proof : Let Z B be the given obtuse angle and BD the projection of BC upon AB. Call this projection m. As in the preceding theorem show that Also modify the figure so as to show the projection of AB on BC and call this n. Then show that 62 = a2 + c2 + 2 an. 440. Theorem. The sum of the squares of two sides of any triangle is equal to tioice the square ^ of half the third side plus twice the square of the median drawn to that side, a d b Suggestion. Make use of the two preceding theorems. 441. EXERCISES. 1. Compute the medians of a triangle in terms of the sides. 2. Show that the difference of the squares of two sides of a tri- angle is equal to twice the product of the third side and the projection of the median upon that side. 3. The base of a triangle is 40 feet and the altitude is 30 feet. Find the area of the triangle cut off by a line parallel to the base and 10 feet from the vertex. 262 PLANE GEOMETRY. 442. Problem. To express the area of a triangle in terms of its three sides. a or Solution. The area of A ABC =\AB y. CD = I ho. It is first necessary to express h in terms of a, b, c. We have b^= a^+ c^-2cm, (^Vhy ' m = ■ Also h?= a? — m^. (Why ■') Hence A2= ,2_/^«^+'^^-^7^ 4 aV- f.^ + .^ - /. ,^ V lie y 4c2 ^ (2 ac + a^ + c^ - ^-)(2 ac - a^ - r^ + ^2) _ f(a + cf - h'^'] \b'i -(a- f)2] 4c-2 ^ (ffl + c+ ^)(^? + c - b)(b + a ~ c){b - a + c) 4 (-2 Now call a + b + c= 2s, or a + c — b = '2 s — '2 b. Then a + c- i= :2(s - J) b + a— (•= 2{s-c) b + c— a = 2{s— (i). Hence h^ = ^^1^( ^^l^ir^^lil:^. or ^= ~ Va(« — (iX* — ^Xs — <•)■ BEVIEW AND FURTHER APPLICATIONS. 263 \ 12 Then -hc = -c-- ^s{s - a)(s - J)(s - c). Hence area of A ABC = Vs{s — a^a — &)(« — c). 443. Problem. To express the area of a triangle in terms of the three sides and the radius of the circum- scribed circle. ^ -^^ c In the figure CE= 2r, where r is the radius of the cir- cumscribed circle. Then But ab=2r-h. (§429) A = ^Vs(s-a)(s-6)(8-c). (§442) e A. 1. Hence, al = ^^Vs{8- a) (s - 6) (s - e), c .7 . "^"-VKs a)(s_6)(8-c). 4r or But area of A ABC = ^s(is - a)(« - J)(8 - c)- (§ 442) Hence, area of A ABC = -— ■ 4r 444. EXERCISES. 1. Compute the areas of the triangles whose sides are (1) 7, 9, 12; (2) 11,9,7; (3) 3,4,5. 2. Express the radius of the circumscribed circle of a triangle in terms of the three sides. 3. Find the area of an equilateral triangle whose side is a by § 443, and also without this theorem, and compare results. 264 PLANE GEOMETRY. PROBLEMS AND APPLICATIONS. 1. Given a regular dodecagon (twelve-sided polygon) with radius / . Connect alternate vertices. (a) Prove that the resulting figure is a regular hexagon. (6) Find the apothem of the hexagon. (c) Find the area oi each of the triangles formed by joining the alternate vertices of the dodecagon. 2. Find the area of a regular dodecagon of radius r. 3. Find the side of a regular dodecagon of radius r. 4. Find the apothem of a regular dodecagon of radius r, (a) by means of Exs. 2 and 3 and § 343. (6) by means of Ex. 3 and § 319. 5. Given the side 8 of a regular dodecagon. Find the apothem. Also find the apothem if the side is s. See Exs. 9-12, page 2'i'j. 6. Given a circle of radius 6 : (a) Find the area of a regular hexagon circumscribed about it. (b) Find the area of a regular octagon inscribed in it, also of one circumscribed about it. 7. Using the formula obtained in Ex. 5, find the side of a regu- lar dodecagon whose apothem is b. 8. Find the radius of a circle circumscribed about a dodecagon whose apothem is b. 9. Find the difference between the radii of the regular dodecagons inscribed in and circumscribed about a circle of radius 10 inches. 10. Solve Ex. 9 if the radius of the circle is r. 11. What is the radius of a circle if the difference between the areas of the inscribed and circumscribed regular dodecagons is 12 square inches ? 12. AVhat is the radius of a circle if the difference between the areas of the inscribed and circumscribed regular hexagons is 8 square inches? See Ex. 11, page 259. 13. A regular hexagon and a retjular dodecagon have equal sides. Find these sides if the area of the dodecagon is six square inches more than twice the area of the hexagon. 14. Solve Ex. 13 if tlie area of the dodecagon exceeds twice the area of the hexagon by b square inches. REVIEW AND FURTHER APPLICATIONS. 265 15. Is there any common length of side for which the area of a regular dodecagon is twice that of a regular hexagon 1 Three times ? Four times? Prove your answer. 16. Solve a problem similar to that of Ex. 15 if the side of the dodecagon is twice that of the hexagon. 17. The nineteen small circles in the accompanying figure are of the same size. The centers of the twelve outer circles lie on a circle with center at O, as do the six circles between these and the inner- most one. The centers of these small circles divide the large circles on which they lie into equal arcs. If r is the radius of the small circles, then OB = 2J r, BD = 2^ r, mADG=\\r. (a) Find AB and CD in terms of r. (A) FindZJB. (c) What part of the circle OG is contained within the nineteen small circles ? (d) If the radius of the large circle is 36 inches, find the radii of the small circles so that they ^hall occupy half its area. (e) If the radius of the large circle is r inches, find the radii of the small circles so they shall occupy one half the area of the large circle. (/) Under (rf) how far apart will the outer twelve centers be? The inner six ? (^) Under (e) how far apart will the outer twelve centers be? The inner six ? In Exs. (/) and {g) it is understood that the distances are meas- ured along straight lines. 266 PLANE GEOMETRY. From First Congregational Church, Chicago. 18. In the figure ABC is an equilateral triangle. A, B, C are the centers of the arcs BC. CA,&nd. AB. Semicircles are constructed on the diameters AB, BC, CA. Let .15 = a. (See Ex. 11, p. 93.) Circles are constructed tangent to the various arcs as in the figure. (a) Find the radius r". Suggestion. Find in order BM, ^rN, BN, HO", BO" (h) Find the radius )■'. Suggestion. BX is known from the solution of (a). BN = B0' + r' and BO' = 2 KO'. Hence KO' = ^^"^ ~ '"' • KB = iV3.B0' = i V3 (£iV - r') and HO' = "-r'. But HO''-^HK^+KO'\ (^^^y + KO'-' (1) Substituting for HO', KB and KO' in (1) we may solve for r'. (c) Find the radius r. Suggestion. Use A BLO and HLO and proceed as under (b). (d) Using the radii thus found, sho\v how to construct the figure. (e) What fraction of the whole area is contained in the circles? 19. yl BCZ) is a parallelogram with fixed liase and altitude. Find the locus of the intersection points of the bisectors of its interior base angles. p 20. Find the locus of a point P such that the sum of the sqiuires of its distances from two fixed points is constant. SiKiciESTioN. By § I m, . I /'" + /'//' :^ 2 A C" + 2 ~CP'. Solve for ( 7'. BEVIEW AND FURTHER APPLICATIONS. 267 21. Find the locus of a point P such that the ratio of its distances from two fixed points is equal to the constant ratio m : n. Suggestions; Let A- and B be the fixed points. On the line AB there are two points P' and P" on the locus, i.e. AP'i : P"B = AP' -.P'B^m-.n. Let P be any other point on the locus. Then AP:BP = AP': P'B = AP" : P"B. Show by the converses of §§ 250, 253 that PP' bisects /.APB and PP'i bisects Z BPK, and hence P'P and P"P form a right angle. 22. Find the locus of a point P from which two circles subtend the same angle. Suggestion. C and C are the centers of the given circles. Prove APDC-^APD'C and PC ^ r' PC r ' 23. The points ABCD are collinear. Find the locus of a point P from which the segments AB and CD subtend the same angle. Suggestions. The &.ABP, ACP, BDP, axiA ^ CDP have a common altitude, and hence their areas are to each other as their bases. Also A j4 BP and CDP have equal vertex angles, whence by § 427 their areas are to each other as the products of the sides forming these angles. Similarly for A A CP and BDP. A P-PB _AB . AP-PC _AC cd^'^'^bT^pd-'bd hence that B C Hence CP.PD CD " BP-PD From these equations show that AP : PD is a constant ratio. 24. ABC is a fixed isosceles triangle. With P center C and radius less than AC, construct a circle, and from A and B draw tangents to it meet- ing in P. Find the locus of P. A\/\)\P Suggestions, (a) Show that part of the locus is the straight line PP'- (ft) Show that Z 1 =Z 2 and hence that Z AP"B is constant. 268 Plane geometry. maxima and minima. 445. Definitions. Of all geometric figures fulfilling certain conditions it often happens that some one is greater than any other, in which case it is called a maxi- mum. Or it may happen that some one is less than any other, in which case it is called a minimum. E.g. of all chords of a circle the diameter is the maximum, and of all segments drawn to a line from a point outside it the perpendicular is the minimum. In the following theorems and exercises the terms maximum and minimum are used as above defined. However, a geometric figure is often thought of as con- tinuously varying in size, in which case it is said to have a maximum at any position where it may cease to increase and begin to decrease, whether or not this is the greatest of all its possible values. Likewise it is said to have a minimum at any position where it may cease to decrease and begin to increase. E.g. if in the figure a perpen- dicular from a point in the curve to the sta-aight line be moved continu- ously parallel to itself, the length of this perpendicular vfill have maxima at ^4, C, and E, and minima at B, D, and F. Certain simple cases of maxima and minima problems have already been given. Some of these will be recalled in the following exercises. 446. EXERCISES. 1. If from a point within a circle, not the center, a line-segment be drawn to meet the firclo, show that this soginent is a maximum ■when it passes thmugh tlu> ceutor and a miuinium wlien, if produced in the opposite direction, it would pass tln-ougli the center. REVIEW AND FURTHER APPLICATIONS. 269 2. Show that of all chords through a given point within a circle, not the center, the diameter is a maximum and the chord perpendicu- lar to the diameter is a minimum. 3. Of all line-segments which may be drawn from a point outside a circle to meet the circle, that is a maximum which meets it after passing through the center, and that is a minimum which, if pro- duced, would pass through the center, 4. Show that if a square and a rectangle have equal perimeters, the square has the greater area. Suggestion. If s is the side of the square and a and 6 are the altitude and base of the rectangle respectively, then 2J + 2a = 4sor ■ _ & + a * r- Hence ,2^ ''^ + a^ +2ah _ (h'^ -2ah + b^) + i^ah _ {h - aY , ^^_ 4 4 4 That is, s^ is greater than ab by ' — — ^ • 5. Use the preceding exercise to find that point in a given line- segment which divides it into two such parts that their product is a maximum. 6. Find the point in a given line-segment such that the sum of •the squares on the two parts into which it divides the segment is a minimum. Suggestion. If a and 5 are the two parts and h the length of the segment, then „ + j = i- ^nd a^ + 2 a6 4- 5^ = F, or, a-^ + h^ = k^-2 ah. Hence, o^ +6^ is least when 2 ah is greatest. Now apply Ex. 5. 7. In the preceding exercise show that a^ + b^ increases as 6 grows smaller, that is, as the point of division approaches one end. When is this sum a maximum ? 8. Two points A and B, on opposite sides of a straight stream of uniform width, are to be con- nected by a road and a bridge crossing the stream at right angles. Find by construction the location of the bridge so as to make the total path from A to -B a minimum. 270 PLANE GEOMETRY. 447, Theorem. Of all triwigh's having equal perim- eters and ihe ^a;nie base, the isosceles triangle has the maximum area. A E B Given A ABC isosceles and having the same perimeter as A ABD. To prove that area A ABO area A ABD. Outline of proof: Draw ce A. AB. Construct AAFB having its altitude FE the same as that of A ABD. Pro- long AF making FG = AF. Draw GB and GD. The ob- ject is to prove that EF, the altitude of A ABD, is less than EC, the altitude of A ABC. (1) Show that A .1 FB, FBG, and GBD are all isosceles, for which purpose it must be shown tliat GB J.AB and FD II AB. Then AD + DB = AD+ 1)G>AG. Or AF+FB area A AltD. Give all the steps and i-easons in full. 448. CoROLTiAifV. 0/ 1(11 tri(in)jl(s having iJic same arr/i, and sf((ii(lln(/ (»i the saiiir Ixise. that ic/iieh is isosceles has the least j)ei'i»ieter. BEVIEW AND FURTHER APPLICATIONS. 271 449. Theorem. Of all polygons having the same perimeter and the same number of sides, the one loith maxim,um area is equilatetal. E Given ABCDEF the maximum of polygons having a given perim- eter and the same number of sides. To prove that AB = BC = CD = DE = EF = FA. Suggestion. Show that every A, such as FDE, is isos- celes by use of the preceding tlieorem. 450. Theorem. Of all polygons having the same area and the same number of sides, the regidar polygon has the viinimum pterhneter. Given polygons A and B with same number of sides .and equal area, A being regular and B not. To prove tliat the perimeter of A is less than that of B. Outline of Proof : Construct a regular polygon c having the same number of sides and same perimeter as B. Then, area of B < area of C, or area of ^ < area of C. But A and C are both regular and have the same num- ber of sides. Hence the perimeter of A is less than that pf C. That is, perimeter of ^ < perimeter of B. 272 PLANE GEOMETRY. 451. Theorem. The polygon with maximum area which can be formed by a series of line-segments of given lengths, starting and ending on a gixen line, is that one whose vertices all lie on a semicircle con- structed on the intercepted part of the given line as a diameter. Given the line-segments AB, BC, CD, DE and EF meeting the line RS at A and F so as to form the polygon ABCDEF with maxi- mum area. To prove that B, C, Z), and E lie on the semicircle whose diameter is AE. Proof : Suppose any vertex as Z) does not lie on this semicircle. Join D to F and to A. Then Z ADF is not a right angle, else it would be inscribable in a semicircle (§ 213). . If now the extremities A and F be moved in or out on the line US, keeping AD and DF unchanged in length, till Z ADF becomes a right angle, then A ADF will be increasid in area (Ex. 3, § 456), while the rest of the polj-gon is unchanged in area. This would increase the total area of the polygon ABCDEF, which is contrary to tlie hypothesis that this is the polygon with maximum area. Hence the vertex D must lie on the semirirrle. In the same manner it can lie jiroved tliat eiich vertex lies on the semicircle. REVIEW AND FUBTHEE APPLICATIONS. 273 452. Theorem. Of all polygons with the same num- her of sides equal in pairs and taken in the same order, the one with maximum area is that one which can be inscribed in a circle. Given a polygon ABCDE inscribed in a circle and A'BC'D'Ef not inscribable but with sides respectively equal to those of ABCDE. To prove that ABCI>e>a'b'c'd'e'. Proof: In the first figure draw the diameter DK and draw AK and BK. On a'b' construct Aa'b'k^ Aabk and draw d'k'. The circle whose diameter is d'k' cannot pass through all the points a', b', c', and E', else the polygon would be inscribed contrary to hypothesis. If either A' or E' is not on this circle, then AKBE>A'k'b'e'. (Why?) (1) Likewise if either b' or c' is not on the circle, then KBCB>k'b'c'd'. (Why?) (2) In any case, adding (1) and (2), AKBCDE> A'k'b'c'b'e'. ' (3) By construction, A AEB ^ A a'k'b'. (4) From (3) and (4), ABCDE > a'b'c'd'e'. 274 PLANE GEOMETRY. 453. Theorem. Of all regular polygons having (■(jnal perimeters, that which has the greatest number of sides is the maximum. F Given T a regular triangle and S a square having the same perimeter. To prove that s> T. Proof : Take any point E in AB. Draw CE and on CE construct A CEF ^ A AEC. Then polygon BCFE has a perimeter equal to that of T and the same area, and has the same number of sides as >. Hence, s > BCFE. (§ 449) That is, s > T. In like manner it can be shown that a regular polygon of five sides is greater than a square of equal perimeter, and so on for any number of sides. 454, EXERCISE Of the three medians of a sialeiie triaiii;le that one is the shortest which is chawu to the longest side. Sii(i(3ESTi(i.N. If vlC' , I /<;. .1 D In AA'ix ■ anil BDC. A /).= DB and DC is common. But AC< Ii( '. Ilf niM. Z 1 < Z •_' ( Why '). Now usi! &iAI)0 unci UDO to sliow that /l'(>>.^0 and hence that BF>AE. REVIEW AND FURTHER APPLICATIONS. 275 455. Theorem. Of all regular polygons having a given area, that one which has the greatest number of sides has the least perimeter. Given regular polygons P and Q such that P = Q while Q has the greater number of sides. To prove that perimeter of Q < perimeter of P. Proof : Construct a regular polygon R having the same number of sides as P and the same perimeter as Q. Then r figure. Tlien linr / J. plane PAP'. (Why?) Now sliiiw tliat (1) 7M=/''.l, (-J) A vlP« ^ A.4P'B, (;i) A )iOl' ^- A BOP'. Hence, PO±.i/. (Why?) p' LINES AND PLANES IN SPACE. 287 475. EXERCISES. 1. Only one perpendicular to a plane can be drawn from a point outside the plane. Suggestion. Show that the hypothesis of two perpendiculars from an outside point to a plane leads to a contradiction. 2. A perpendicular is the shortest distance from a point to a plane. 3. A line cannot be perpendicular to each of two intersecting planes. Suggestions. (1) If a line / is perpendicular to the planes M and N at the points A and B, and C is a point in their intersection, then h. ABC would contain ^<~~ two right angles. (2) If I is oV^^ perpendicular to M and TV at \ ■ a point P in their intersec- tion, pass a plane through I, meeting M and A'^ in /, and l^. 4. A line which is perpendicular to each of two lines at their point of intersection is perpendicular to their plane. See § 470. 5. All perpendiculars to a line at a point in it lie in the same plane, namely, the plane determined by any two of them. The following are theorems proved in the preceding con- structions and exercises. 476. Theorem. Through any given ])oint there is one and only one 'plane perjyendicidar to a given line. 477. Theorem. Through any given point there is one and only one line perpendicular to a given plane. 478. Theorem. All lines perpendicular to a line at a point lie in one and the same plane. 479. Theorem. A line perpendicular to each of two intersecting lines is perpendicular to their plane. 480. Theorem. The shortest distance from a point to a plane is the perpendicular to the plane. 288 SOLID OEOMETEY. 481, EXERCISES. 1. Show how a carpenter could use the theorem of § 479 to stand a post perpendicular to the floor, by use of two ordinary steel squares. 2. If a steel square be made to rotate about one of its edges as an axis, what kind of surface does the other edge describe? 3. In making long timbers perpendicular to each other, a car- penter may use simply a measuring rod, such as a ten-foot pole. Show how to make a timber perpendicular to the floor by this process. 4. Show how a back-stop on a ball field can be adjusted perpen- dicular to the line through sec- ond base and the home plate. | -^ s^ bok What theorems of solid geometry .^.f^i'lf ,''^ ^\ are used? ■; ly* ; / \^ 5. If a plane is perpendicular y ^ Wo'ie'X x'ioi to a line-segment PF' at its middle ' X ,,-'' point, prove : (1) Every point in l"Baee the plane is equally distant from P and J" ; (2) every point equally distant from P and P' lies in this plane. What is the locus of all points in space equidistant from P and P' ? See § 127. 6. Given the points ^1 and B not in a plane ^f■ Find the locus of all points in Af equidistant from .4 and B. Suggestion. All such points must lie in the plane M and also in the plane which is the perpendicular bisector of the segment .-15. Is there any position of the points A and B for which there is no such locus? For which the locus contains the whole plane .1/? 7. On a line find a point equidistant from two given points not on the line. Is there any position of the points for wliich there is no such point on the line? For which all points on the line satisfy tliis condition? 8. Find the locus of all points in space equidistant from the three vei-tices of a triangle. 9. Find the locus of all points in a given plane equidistant from the verl iees of a triangle not in the plane. Is there, any position of the triangle for which this loons contains more than one point? No point? LINES AND PLANES IN SPACE. 289 10. Ask and answer questions similar to the two preceding, using the vertices of a square instead of a triangle ; the vertices of a rec- tangle ; of any pol3'gon which can be inscribed in a circle. 11. Find the locus of all points in apace equidistant from the points of a circle and also of all points in a plane equidistant from the points of a circle not in that plane. Discuss as in Ex. 9. 12. Find the locus of all points equidistant from two given points A and B, and also equidistant from two points C and D. Discuss. 13. In geometry of two dimensions, how many lines may be per- pendicular to a given line at a given point? In geometry of three dimensions? 14. In how many points can a straight line cut a plane ? In how many points may it cut a curved surface, such as a stovepipe? In how many points can a straight line cut a circle? In how many points can a plane cut a circle, the plane being distinct from the. plane of the circle? 15. State and prove a theorem of solid geome- try corresponding to the theorem in the Plane Geometry, comparing the lengths of oblique seg- ments cutting off equal distances from the foot of a perpendicular. Also state and prove the converse. State and prove a theorem comparing the lengths of segments cutting off unequal distances. See §§ 112, 115, and 116. 16. Find the locus of all points in a plane which are equally dis- tant from a given point outside the plane. If a perpendicular be drawn to the plane from this point, how is its foot related to this locus? 17. If in the figure PZ)± plane M, and DC ±AB any line of the plane, prove that PC±AB. Suggestion. Lay off CA = CB, and compare triangles. 18. If in the same figure PD ± M, and PC X AB any line of the plane, prove that DCXAB. 290 SOLID GEOMETRY. THEOREMS ON PARALLELS. 482. Theorem. If two linc>s arc perpendicular to the same plane, they .are parallel. CoNVKKSELY. If One of two parallel linen ii pe/pen- dkidar to a plane, the other is also. Given: (i) AB and CD each J. to the plane M. To prove that AB II CB. Proof : Draw BD and make DE±DB. Take points A and E so that BA = DE, and draw AB, AE, and BE. Now prove (1) A ABB ^ A BBE. (:2 ) A ABE ^ A ABE. Hence, Z ABE is a right angle, and BC, DA, and BB all lie in the same plane. (\V]i\ ') Therefore CB and AB are in the same plane and perpen- dicular to the same line BD. • (Wliy'') Henee, AB || CD. Given : (2) AB II CD and CD ± M. To prove th;it AB ± M. Proof: If AB is not -L M. suppose A' B -L .v. Then A'b II CI). But through B there is only one line II CB. Henee, a'b and AB coincide. I!ut A'B was taken ± .V. Hence. AB ± M. Ilis loiMc M. NdiK. The aliove pi'oof for the dii-eot case is the one nivi'ii liy ICiK'liil. iMaiiy proofs have been given, but this seems the most elegant. 483, ('()H(iT,i,,\KY. Iiro lines in sp, s. and 1(1 inches re- speelively, and a shiulinj; rod iuterseeting them has a 7-inch segment betwi'en the first two shelves, find the other two segments of the rod. LIXES AND PLANES IN SPACE. 295 3. If -we attempt to stand up a ten-foot pole in a room eight feet high, find the locus of the foot of the pole on the floor when the top is kept at a fixed point on the ceiling. 4. Show that, if three line-segments not in the same plane are equal and parallel, the triangles formed by joining their extremities, as in the figure of § 489, are congruent, and their planes are parallel. 5. Given a plane M and a point P not in M. Find the locus of the middle points of all segments connecting P with points in M. 6. Given a plane 31 and a point P not in M. Find the locus of a point which divides in a given ratio each segment connecting P with a point in M : (a) if the segments are divided internally; (b) if they are divided externally. 7. The locus required in Ex. 6 consists of two planes, each parallel to the given plane M. Are these two planes equally distant from M? 8. Show that a plane containing one only of two parallel lines is parallel to the other. 9. If in two intersecting planes a line of one is parallel to a line of the other, then each of these lines is parallel to the line of intersec- tion of the planes. 10. Show that three lines which are not concurrent must all lie in the same plane, if each intersects the other two. 11. Show that three planes, each of which intersects the other two, have a point in common unless their three lines of intersection are parallel. Suggestion. Suppose two of the intersection lines are not parallel, and meet in some point 0. Then show that the other line of inter- section passes through 0, and hence that O is the point common to all three planes. 12. Given two intersecting planes M and N. Find the locus of all points in M at a given distance from N. 13. Given two non-intersecting lines Zj and l^. Find the locus of all lines meeting /j and parallel to l^. 14. Prove that the middle points of the sides of any quadrilateral in space are the vertices of a parallelogram. Suggestion. Use the fact that a line bisecting two sides of a tri- angle is parallel to the third side. Note that the four vertices of a quadrilateral in space do not necessarily all lie in the same plane. 296 SOLID GEOMETRY. DIHEDRAL ANGLES. 495. Definitions. The part of a plane on one side of a line in it is called a half-plane. The line is called the edge of the half-plane. Two half-planes meeting in a common edge form a dihedral angle. The common edge is the edge of the angle and the half-planes are its faces. Lines in the faces of a dihedral angle perpendicular to its edge at a common point form a plane a.^. angle, which is called the plane angle of the dihedral angle. Thus, in the figure, the half-planes il/ and N have the common edge AB, and form the dihedral angle M-AB-N, read by naming the two faces and the edge. The Z ('DE, whose sides are CD X AB \\i N and ED ±AB in M is the plane angle of the dihedral angle M-AB-N. 496. By § 489 all plane angles of a dihe- dral angle are equal to each other. A dihedral angle may be thought of as generated by the rotation of a half-plane about its edge. The magnitude of the angle depends solely upon the amount of rotation. 497. Two dihedral angles are equal when they can be so placed that their faces coincide. 498. Theorem. Ttoo diliedral angles are equal if tliclr plane angles ((re equal, a. Given: the dihedral angles M- AB-N and M'-A'B'-N' in which the i). plane A CDE and CD'E' are equal. To prove tlial M-AIi N = m'-A'h'-n'. Proof : Place the I'qual . CDE and c'd'e' in coincidence. ^^>.v,_^ .V "->.v^^^ .V M M' __c. d' _^v ^/; b' ^A'' ^\^ LINES AND PLANES IN SPACE. 297 Then the edges AB and A'b' must also coincide, since they are both perpendicular to the plane CDE at the point D (§ 477). Face Mthen coincides with m', since its determining lines, AB and DE, coincide respectively with those of m', namely, a'b' and d'e'. Likewise, face N coincides with n'. Hence, M-AB-n = m'-a'b'-c' since their faces have been made to coincide (§ 497). 499. Theorem. State and prove the converse of the preceding theorem. 500. Definitions. It follows from §§ 498 and 499 that the plane angle of a dihedral angle may be regarded as its measure. A dihedral angle is right, acute, or obtuse according as its plane fcgle is right, acute, or obtuse. Two dihedral angles are adjacent, vertical, supplementary, or comple- mentary, according as their corre- sponding plan e angles, with a common vertex, are adjacent, vertical, supple- mentary, or complementary. In the figure pick out all the dihedral angles and describe them and their relations, (1) if Z CDE is acute, (2) if Z. CDE is a right angle. 501. EXERCISES. 1. State theorems on dihedral angles corresponding to those on plane angles in §§ 64, 68, 69, 72-76. 2. State theorems concerning two planes cut by a transversal plane corresponding to those on lines in §§ 90, 92, 93, 97, 99. Note that in Exs. 1 and 2 the proofs are exactly analogous to those in the Plane Geometry. 298 SOLID GEOMETRY. 502. Definition. Two planes are perpendicular to each other if their dihedral angle is a right angle. 503. Theoreji. If a line is perpendicular to a plane, every plane containing this line is perpendicular to the plane. h/ Given : a line / X plane N at P, To prove that a plane M containing I is A_ N. Proof : Let ?j be the intersection of N and M. In .v draw ?2 -L Zj at P. Then Hence, I ± ?i, /^ ± ?i, and ?2 -L ?• (Why '') M±y. CWhv?) 504. EXERCISES. 1. Show that the above theorem is equivalent to the following: If a plane is perpendicular to a line lying in another plane, then the first plane is perpendicular to the second. 2. Name all the dihedral angles in the accom- panying figure. 3. Find the locus of all points equidistant from two given ])arallel planes and also equi- distant friim two given points. 4. Find thi' locus of all points at a given dis- tance from a given jilane and equidistant from two given points. Diacu.ss Exs. 3 and 1 for the various cnsos possible. 5. IIow can the theorem of § .'>(>:! bo used to erect a plane perpen- dicular t(i ,1 given plane? LINES AND PLANES IN SPACE. 299 505. Theorem. If two planes m and n are mutually perpendicular, and if p is any point in m, then a line through p perpendicular to n lies wholly in m and is perpendicular to the line of intersection of m and n. /m p h/o Given : plane M J. plane N, and P any point in M. Let /i be the intersection of M and iVand let / be a line through P1.N. To prove that I lies wholly in M and that I J_ Zj. Proof : (1) Let P be any point in M outside of /j. Suppose I does not lie wholly in M. Then through P draw a line V in the plane M perpen- dicular to Zj at O. Through O draw line l^\a N 1. ly Then Zg and V form a right angle (§ 502). Hence ?' -L iV (§ 479). But from P there can be but one perpendicular to N. Hence, I coincides with V , and we have I lying wholly in M and 1 1. \. (2) Let the point P be in the intersection Zj. The proof is similar to case (1). Give it in full. 506. EXERCISES. 1. Show that if in the figure of § 505, M XN and Z is a line in M such that I X h, then I X N. 2. Through a given point, on or outside of a given plane, how many planes can be constructed perpendicular to the given plane ? 300 SOLID GEOMETRY. 507. Theorem. If a plane is perpendicular to each of two planes, it is perpendicular to their line of inter- section. Given : Q 1 M and Q ^N and / the intersection of M and N, To prove that Q ± I. Proof: From a point P common to M and N draw a line Then V lies wholly in both M and N. (Why ?) Hence, V and I are the same line. That is, 11. (), ov Q A. I. 508. EXERCISES. 1. Any plane ± to the edge of a dihedral angle is J. to each of its faces. 2. If three lines are X to each other at a common point, tlien each is A- to the plane of the other tw o. 3. Through a given line in space pass a plane -L to a given plane. IIow many such planes can be constructed? .SuiKiESTioN. From some point in the given line draw a line ± to tlie given plane. Then use § ."lU:). 4. What is the answer to the question in Ex. 3 in ease the given line is perpendicular to the given plane? 5. Throiii;]i :i given point in spaee pass a plane J. to eaeh of two given planes. How many ,sueh can bo passed? Sr(i(iKs 1 iiiN. Consider tlie relation of the required plane to the inter.sectidn of the two given planes. 6. What is the answer to the qnestion in Kx. ."> in case the given [loint is on the line of interseetion of the given planes? LINES AND PLANES IN SPACE. 301 509. Definition. A plane through the edge of a dihedral angle bisects the angle if it forms equal angles with the faces of that angle. 510. Theorem. The locus of all points in space equally distant from the faces of a dihedral angle is the half-plane bisecting the angle. A Given : the half -plane P bisecting the dihedral Z M-AB-N. To prove : (1) that any point £ in P is equally dis- tant from M and N, and (2) that any point which is equally distant from M and N lies in P. Outline of proof : (1) Draw EG A.M and ED ± N. Then the plane CED cuts ii/, N, and P in OG, OD, and OE re- spectively and is J- to both M and N. (Why ?) Then Z EOD is the measure of P-AB-N and Z EOC is the measure of P-AB-M. (Why?) Now use A EOG and EOD to show that EC = ED. (2) Take E any point such that EG= ED, and let p' be the plane through AB containing E. Now argue as in § 125, to show that Z EOD= Z EOC, and hence that p' is the same plane as the bisector plane P. Give the proof in full. 511. Corollary. If from any point within a dihe- dral angle perpendiculars are drnivn to the faces, the angle between the perpendiculars is the supplement of the dihedral angle. 302 SOLID GEOMETRY. 512, Definition. The projection of a point on a plane is the foot of the perpendicular from the point to the plane. The projection of any figure on a plane is the locus of the projections of all points of the figure on the plane. 513. Theorem. Tlie projection of any straight line on a plane is a straight line in that plane. ^ B A D Outline of proof: Pass a plane through the given line andJ_ to the given plane. Is this always possible ? (Why?) Now show that the intersection of these two planes con- tains the projection of every point of the given line upon the given plane. 514. Theoreji. The acute angle formed hy a straight line ivith its oivn projeetion on a plane />■ the least angle lohich it makes with any line in that plane. Outline of proof: Let BC l>e tlie projection of AB, and let BB be any other line in 3/ through B. Lay off BD = BC and draw AD. In the A ABD and ABC show that /^ ABD > Z ABC. 515. Definition. The angle between a plane and a line oblique to it is understood to mean tlie ariite any;le furmed by the line and its projci^lion upon the plane. LINES AND PLANES IN SPACE. 303 516, Problem. To construct a common perpendicular to two non-parallel lines in space. _ C 7 Given : 1^ and 4 two non-parallel lines, and also non-intersecting. To construct a line BC perpendicular to each of them. Construction : Through A, any point in l^, draw l^ II ly Let M be the plane determined by Zg and l^. Through Zj pass a plane P ± Jf and meeting M in l^. At B the intersection of l^ and l^, in the plane P, erect BO ± M. Then BC is the required perpendicular. Proof : Give the proof in full. 517. EXERCISES. 1. If a line is J- to a plane, its projection on that plane is a point. 2. If a line-segment is II to a plane, show that its projection is a segment equal to the given segment. 3. If a line-segment is oblique to a plane, show that ^its projection is less than the given segment. 4. If a line-segment 6 in. long makes an angle of 30° with a plane, find the length of its projection on the plane. If it makes an angle of 45° ; an angle of 60°. 5. If two parallel lines meet a plane, they make equal angles with it. (Why?) Is the converse true? 6. If a line cuts two parallel planes, it makes equal angles with them. (Why?) Is the converse true? 7. If two parallel line-segments are oblique to a plane, their pro- jections on the plane are in the same ratio as the given segments. 304 SOLID GEOMETRY. POLYHEDRAL ANGLES. 518. Definitions. Given a convex polj'gon and a point P not in its plane. If a half-line I witli end point fixed at P moves so that it always touches the poly^-^iin and is made to traverse it completely, it is said to generate a convex polyhedral angle. The fixed point is the vertex of the polyhedral angle, and the- rays through the ver- tices of the polygon are the edges of the polyhedral angle. Any two consecutive edges de- termine a plane, and the portion of such a plane included between these edges is called a face of the polyhedral angle. The plane angles at the vertex are called the face angles of the polyhedral angle. A poly- hedral angle having three faces is called a trihedral angle. 519. Theorem. If two trihedral anijlcs hace the three face angles of the one equal rcspectiiuhj to the three face angles of the other, the dihedral angles eipposite the equal face angles are equed. Given : the trihedral angles P and P'. in which Z a ■. LINES AND PLANES IN SPACE. 305 To prove lliat the pairs of dihedral angles are equal whose edges are PA and p'a', PB and p'b', PC and p'c'. Proof : Let p and p' be cut by the planes ABC and a'b'c', making PA = PB = PC = p' A' = p'b' = P'c'. On the edges PA and p'a' lay off AF= a'f', and through F and f' pass planes J. to PA and p'a' respectively. Let the first plane cut ^C in some point E, and AB in some point Z), and likewise the second plane cut A'c' and a'b' in corresponding points E' and n'. Why must there be such intersection points? Must they be between A and C, A and B, A' and c', J.' and b' respectively ? Then ^ DFE and d'f'e' are the measures of the dihedral angles whose edges are AP and A'p', and these are to be proved equal to each other. Now use the pairs of face triangles APB, a'p'b', etc., to show that A ABC ^ A a'b'c'. Then show in order (1) aadf^aa'd'f'. (2) aaef^aa'e'f'. (3) A AVE ^ A a'b' E'. (4) ADFE ^Ad'f'e'. In like manner the other pairs of dihedral angles may be proved equal to each other. 520. Two polyhedral angles are congruent if they can be so placed that their vertices and edges coincide. A polyhedral angle is read by naming the vertex and one letter in each edge, as P-ABCDE, or by the vertex alone where no ambiguity would arise. 521. Corollary. If tioo trihedral angles have their face angles equal each to each and arranged in the same order, they are congruent. For by the theorem their dihedral angles are equal each to each, and if the equal faces are arranged in the same order, the trihedral angles may be made to coincide throughout. 306 SOLID GEOMETRY. SUMMARY OF CHAPTER VIU. 1. Describe the various ways of determining a plane. 2. State the axioms used in this chapter. 3. State the definitions and theorems on perpendicular lines and planes. 4. State the definitions and theorems on parallel lines and planes. 5. Name some applications in connection with perpendiculars and parallels which have impressed you. 6. State some facts in regard to perpendiculars and parallels in the plane which do not hold in space. 7. Give the definitiong and theorems on dihedral angles. 8. What theorems on perpendicular planes are proved in connec- tion with dihedral angles? 9. Give the definitions and theorems on projections used in this chapter. 10. Give the definitions and theorems on polyhedral angles thii< far used. 11. Make a list of all the loci problems in this chapter. 12. Study the following collection of problems and applications, and then make a collection of those which impiess you as most inter- esting or useful. Include in this list any applications in the chapter. PROBLEMS AND APPLICATIONS. 1. A Christmas tree is made to stand cm a cross-shaped ba^f. If the tree is perpendicular to each piece of the cross is it j^rpeiidicular (o tlie floor? 2. If the projections on a plane of a number of points outside the plane lie in the same straii^ht line, do the points tlieniselves neces- sarily lie in a straight line? 3. If A , /.', and C do not lie in the same line, and if their projections on a plane .1/ do lie in a straight line, what is the relation of the planes .I/and AliCl 4. Is it possililo to project, a circle upon a plane so that the projec- t ion shall l>e a, straight line-segment? If so, how must the circle and tlie plane be related ? LINES AND PLANES IN SPACE. 307 5. How many planes may be made to pass through a given point parallel to a given line ? Discuss the mutual relations of all such planes. 6. Through a point P construct a line meeting each of two lines l^ and Zj. Suggestion. Let M and N be the planes determined by P and /j and by P and l^. Is this construction always possible ? 7. Given a plane M and lines /,, l^- Construct a line perpendicular to M and meeting both Z, and l^. Suggestion. Project l^ and Z2 on M. 8. A line ? is parallel to a plane M, and lines Zj and I2 in i1/ are not parallel to I. Show that the shortest distance between I and Z, is equal to the shortest distance between I and l^. 9. Prove that the planes bisecting the dihedral angles of a trihedral angle meet in a line. 10. Find the locus of all points equidistant from the planes deter- mined by the faces of a trihedral angle. 11. Find the locus of all points equi- distant from the edges of a trihedral angle. Suggestion. On the edges lay off PA = PB = PC. Let O be equidistant from A, B, and C. Then any point Q in PO is equidistant from the edges. 12. Planes determined by the edges of a trihedral angle and the bisectors of the opposite face angles meet in a line. Suggestion. If, in the figure of Ex. 9, PA = PB = PC, and if PE, PF, PG bisect the face angles, then E, F, G are the middle points of the sides of the triangle A BC. Now apply the theorem that the medians of a triangle meet in a point. 13. Planes perpendicular to the faces of a trihedral angle and bisecting its face angles meet in a line. 308 SOLID GEOMETRY. IP'B 14. Show how to locate a point which is at a distance of 2 feet from each face of a trihedral angle. Suggestion. Puss a plane parallel to each face of the trihedral angle and at a distance of 2 feet from it. 15. Show how to locate a point which is 2 feet from one face of a trihedral angle, 3 feet from the second, and 4 feet from the third. 16. Find the locu.s of a point in space such that the difference of the squares of its distances from t two fixed points is constant. Suggestion. First solve the problem in the plane, obtaining as the required locus the line PP'. Then rotate the figure about the line AB as an axis. See page 227, Ex. 7. 17. Find the locus of all points in a, plane at which lines from a fixed point P not in the plane meet the plane at equal angles. 18. Find the locus of all points which are at the same fixed distance from each of two inter- secting planes. Son'TioN. Pass a plane perpendicular to the intersection of the two given planes, making the plane cross section shown in the figure. Draw the bisector of Z AOB and a line at the re- quired distance a from OA and parallel p^ to it. Then P is at the distance a from each of the lines OA and OB. In the same manner three other such points, P^tP^- Pr,^ are constructed. Now let this figure move through space iiarallil to itself, the point O moving along the intersection of the given planrs. The points P, /',, /'.„ P, will tlius move along titiaight lines whii-h constitute the required locus. 19. Find the locus of points 3 feet from one of two intersecting plain's and (i feet from tlie other. 20. It is reijiiii-ed that a series of eleetrie lights sliall be 7 feet above tlic floor of a romn and 3 (eil frcun the walls. Find the locus of all points at which such lights may be placed. LINES AND PLANES IN SPACE. 309 21. Find the locus of all points in space equally distant from each of two intersecting straight lines. 22. Find the locus of all points in space equally distant from two parallel lines. 23. Given two points A and B on the same side of a plane M. Determine a point P in M such that AP + PB shall be a minimum. Suggestion. Pass a plane through A and B perpendicular to M, and proceed as in Ex. 8, page 200. 24. Show that if the edge of a dihedral angle is cut by two parallel planes, the sections which they make with the faces form equal angles. 25. Show that if all edges of a trihedral angle are out by each of a series of parallel planes, the intersections form a series of similar tri- angles. 26. Find the locus of the intersection points of the medians of the ti'iangles obtained in Ex. 25. Also of the altitudes. 27. If three planes are so related that the segments intercepted on any transversal line are in the same ratio as the segments intercepted on any other transversal / ^ M I then the planes are parallel. Suggestion. Let M, N, Q be the three / ^':^Ag^c/ nJ K I^ planes. From A any point in M draw three lines, not in the same plane, meeting N in A', ■ ^^ — ^c" QJ B', C, and Q in A", B", C"- Use the hypothe- sis to show that A'B' II A"B" and C'B' II C'B". Hence Q II N. Similarly show that M II N. 28. A segment AB of fixed length is free to move so that its end- points lie in two fixed parallel planes. Find the locus of a point C on ^B if ^ C is of fixed length. 29. If the projections of a set of points on each of two planes not parallel to each other lie in straight lines, show that the points them- selves lie in a straight line. 30. If l\ and Z2 are not parallel and non-intersecting, show that there is only one plane through h parallel to l^. 31. Show that if two lines are not parallel and do not lie in the same plane, they have only one common perpendicular, and that the shortest distance between the lines is measured along this perpen- dicular. CHAPTER IX. PRISMS AND CYLINDERS. 522. Definitions. Any portion of a plane entirely bounded by line-segments or curves is called a plane-segment. E.g. the portion of the plane M inclosed by the __ , triangle ABC is the plane-segment ABC. / /^\ „/ If the boundary is composed entirely of ^ ' straight line-segments, the inclosed portion is called a polygonal plane-segment. 523. A polyhedron is a three-dimensional figure formed by polygonal plane-segments which entirely inclose a portion of space. The line-segments which are common to two polygons are tlie edges of the polyliedron. The plane-segments inclosed by the edges are the faces, and the intersections of the edges are the vertices of the polyliedron. .V pol}-- hedron is convex if every section of it made by a plane is a convex jiolygon. Note. The word polyhedron, as here defined, means the surface inclosing a portion of space and not tliat portion of space itself. However, it is sometimes eoiiveuient to use the v, old pol i/hed ron when referring to the inclosed space. Thus, we speak of dividing a poly- hedron into oilier iiolyheclnms when, strictly, we mean that smaller polylieilroiis are constructed which divide into parts the space inclosed in the given polyhedron. In the same manner the word poli/i/mi is sometimes used to indicate the plane-segment lioinuled by it. and the word aiii/le to indicat*? the part of the plaue within il. 310 PHISMS AND CrLIXDERS. 311 Thus, a face of a polyhedron is sometimes called a polygon, and the face of a polyhedral angle is sometimes called an angle. In all cases the context will clearly indicate what is meant. 524, Definitions. Given a convex polygon and a straight line not in its plane. If the straight line move so as to remain parallel to itself while it always touches the / polygon and is made to traverse it '/ completely, the line is said to generate a closed prismatic surface. The moving line is the generator of the surface, and the guiding polygon the directrix. Tlie generator in any one of its positions is an element of the surface. A cross section of a prismatic surface is made b}- any plane cutting all its elements. \ 525. A prism is that ' part of a closed prismatic surface included between two parallel cross sections, together with the inter- ^ _cepted plane-segments. " The parallel plane-segments are the bases of the prism, and the portion of the prismatic surface between the bases is called the lateral surface of the prism. The lateral surface is composed of par- allelograms (why ?), and these are called the lateral faces of the prism. The sides of these parallelograms, not common to the bases, are the lateral edges of the prism. A right section of a prism is made by a plane cutting each of its lateral edges, extended if neces- sary, and perpendicular to them. 312 SOLID GEOMETRY. 526. Prisms are classified according to the form of their right sections, as triangular, quadrangular, pentagonal, hex- agonal, etc. A regular prism is one whose right section is a regular polygon. A prism is a right prism if its lateral edges are perpen- dicular to its bases ; otherwise it is oblique. The altitude of a prism is the perpendicular distance between its bases. The altitude of a right prism is equal to its edge. The lateral area of a prism is the sum of the areas of its lateral faces. The total area is the sum of its lateral area and the area of its bases. THEOREMS ON PRISMS. 527. Theorem. The cross sections of aprimn made hy parallel planes are congruent poly gorn. Given a prism cut by two parallel planes forming the polygons ABCDE and A'B'CUfE'. To prove that abcde ^ a' b' c' d' Ef . Outline of proof: (1) Show that ab = a'b\ BC=B'd, etc., by proving that ABB' a' , BCC'b', etc., are ^. (2) Show that Zaiu'= Za'b'c', Z. BCD= Z b'c'b\ etc. (3) Hence, show that ABODE and a'b'c'd'e' can be made to coincide. PRISMS AND CYLINDERS. 313 528. Theorem. JTie lateral area of a prism is'equal to the product of a lateral edge and the perimeter of a right section. Suggestion. Show that the lateral edges are mutually- equal and that the area of each face is the product of a lateral edge and one side of the right-section polygon. Complete the proof. Note. The form of statement in this theorem is the usual abbreviation for the more precise form : The lateral area of a prism is equal to the product of the numerical measures of a lateral edge and the perimeter of a right section. Similar abbreviations are used throughout this text. 529. Corollary. The lateral area of a right prism is equal to the product of its altitude and the per- iineter of its base. 530. Definitions. A polyhedron which is a part of a prism cut off by a plane meeting all the lateral edges, but not parallel to the base, is called a trun- cated prism. Two polyhedrons are said to be added when they are placed so that a face of one coincides with a face of the other, but otherwise each lies outside the other. 314 SOLID GEOMETRY. 531. Theorem. Tioo iirisms are congruent if three faces having a rowvian vertex in flu; one eire eemgruent respectireh/ to three faces haeing a eovivcon vertex in the other, and sinularlg placed. l' F, K' B G B' C Given the three faces meeting at B in prism P congruent re- spectively to the three faces meeting at B in prism P, and similarly placed. To prove tliat P can be made to coincide with P' . Outline of proof: Trihedral angles B and b' are con- gruent. (\Miy?) Now apply the two prisms, making B coincide with b', and thus show in detail that : (1) The lower bases coincide. (2) The lateral faces at B and b' coincide. (3 ) The upper bases coincide. (4) All the lateral faces coincide. Hence P and P' cnincide throughout. 532. Corollary. Tiro right prisms arc congruent if they have eongrnent bases and cepial attitude. 533, ( "oi;oLi,Ai!Y. Two truncated prisms are con- gruent if the faces forming a trihedral angle of (oie are et/n(d resjn'ctiri'Ii/ to the eorresj)onding faces of the ot/ier. PBISMS AND CYLINDERS. 315 534. Definitions. A parallelopipecl is a prism whose bases, as well as lateral faces, are parallelograms. A rectangular parallelopiped has all its faces rectangles. A cube is a parallelopiped whose faces are all squares. 535. Theorem. Any two opjjosite faces of a parallel ojxiped are piarallel and con- gruent. r Suggestions. Consider the opposite faces ABFE and DCGH. (1) Show that the sides of the angles ABF and BCG are parallel, and hence the planes determined by them are parallel. (2) Show that these faces are congruent. In like manner argue about any other pair of opposite faces. 536. EXERCISES. 1. Can a parallelopiped be a i-ight prism without being a, rectan- gular parallelopiped ? Illustrate. 2. Show that in a rectangular parallelopiped each edge is perpen- dicular to all the other edges meeting it. 3. Is it possible to construct a prism which has no right section according to the definition of § 525? 316 SOLID GEOMETRY. 4. Show that any section of a parallelo- B R piped made by a plane cutting four parallel edges is a parallelogram. 5. A section made by a plane passed through two diagonally opposite edges of a parallelopiped is a parallelogram. E.g. the section through DH and BF in the figure of Ex. 4. 6. Sliow that any two of the four diagonals of a parallelopiped bisect each other. E.g. AG and CE in the figure. Make use of the preceding example. 7. Show that the diagonals of a rec- tangular parallelopiped are all equal to each other. 8. Show that the sljuare on the diag- onal of a rectangular parallelopiped is equal to the sum of the squares on the sides meeting at a vertex from which it is drawn. E.g. in the figure of Ex. 6, . W' = AC^ + CG' = AB^ + BC^ + CG^- 9. Find the ratio of the diagonal to one edge of a cube. 10. Find the edge of a cube whose diagonal is 14 inches. Find the diagonal of a cube whose edge is 16 inches. 11. Find a diagonal of a rectangular parallelopiped whose edges are 6, 8, and 10 inches respectively. 12. Are the diagonals of a cube perpendicular to each other? 13. Tf two diagonals of a rectangular parallelopiped meet at right angles, and if two of its faces are squares, find the ratio of the sides of the remaining faces. 14. If two .congruent right prisms whose bases are equilateral tri- angles are placed logctlier so as to form one prism whose base is a parallclunraMi, comjiare the lateral area of the prism so formed with the sum of the lateral areas of the original prisms. 15. A riglit prism whose bases are regular hexagons is divided into six prisms whose liases are equilateral triangles. Compare the lateral area of the original prism with the sum of the lateral areas of the resulting prisms. PBI8MS AND CYLINDERS. 317 16. If the perimeter of a right section of a prism is 24 inches and its altitude 6 inches, what is the smallest possible lateral area? What can be said about the largest possible area of such a prism ? 17. Any section of a prism made by a plane parallel to a lateral edge is a parallelogram. 18. Show that for every prism there is at least one set of parallel planes each of which cuts the prism in a rectangular section. See suggestions Ex. 7, p. 327. VOLUMES OF RECTANGULAR PARALLELOPIPEDS. 537. Thus far certain properties of prisms have been studied, but no attempt has been made to measure the space inclosed by such a solid. For this purpose we con- sider first a rectangular parallelopiped. 538. Definition. In case each edge of a rectangular par- allelopiped is commensurable with a unit segment, the number of times which a unit cube is contained in it is the numerical measure or the volume of the parallelopiped. 539. In the case just described in the definition, the volume is easily com- puted. E.g. if in the figure one edge ^C is 4 units, and an adjoining edge AB is 3 units, then a cube as AK, whose edge is one unit, may be laid ofE 4 times along AC and a tier of 3 • 4 = 12 such cubes will adjoin the face AD, while 5 such tiers will exactly fill the space within the solid. That is, 5 • 3 • 4= 60 is the number of cubic units in the solid. Again, if the given dimensions are 3.4,. 2.6, 4.5 decimeters re- spectively, then unit cubes, with edge one decimeter, cannot be made to fill exactly the space inclosed by the figure, but cubes with edge each one centimeter will do so, giving 34, 26, and 45 respectively along the three edges of the solid. Hence the volume is 34 . 26 ■ 45 = 39,780 cubic centimeters, or 39,78 cubic decimeters. 318 SOLID GEOMETRY. 540. Axiom XIX. Anij rrrtangular parallelopiped in- closes a flcfiiiitc volume which is greater than that of another, provided no dimension of the first is less fh.rni the rori'espond/nr/ dimension of the second, and at least one dimension is greater. 541. Theorem. The volume of any rectangidar par- allelopiped is equal to the product of the numerical measures of its linear dimensions. Proof: Case 1. If each, dimension is commensurable with the unit segment. This is tiie case treated in § 539. A a B T'ase 2. If tiro dimensions are commensurable with the unit sei/ment and the third is not. \>y Axiom XIX the parallelopiped has a definite volume r. Suppose tills is not equal to abc and that r< abc. Let c' lie a number such that V = abc' . Then c' abo". (2) Hence, the assumption that V < abo cannot hold. In the same manner show that V > abo cannot hold. Hence, we have V = abo. Case 3. If one dimension is commensurable with the unit segment and two are not. Case 4. If all three dimensions are incommensurable with the unit segment. The proofs in these cases are similar to .he above. Thus, in Case 3, when b and o are both incommensu- rable with the unit segment, we obtain tlie parallelopiped AL, two of whose dimensions a and c" are commensu- rable with the unit. Hence, by Case 2, its volume is abo". Then the argument is identical with that given before. Hence, in all cases V= abc. 542. Corollary 1. The volume of a rectangular parallelopiped is equal to the product of the numerical measures of its base and altitude. 643. Corollary 2. If tioo rectangular parallelopipeds have tioo dimensions respectively equal to each other, their volumes are in the same ratio as their third di- mensions ; and if they have one dimension the same in each, their volumes are in the same ratio as the products of their other two difmensions. For if V and V are the volumes, and a, b, c and a', V, c' the dimen- , , F n -b ■ c a -c , , , ■ , F o • 6 . , , sions, then —, = ——- — - = —-db—b andc = c'; or — ,= , if c = c'. I' a • b' ■ c' a' V a'-b' 320 SOLID GEOMETRY. VOLUMES OF PRISMS IN GENERAL. 544. From the formula for rectangular parallelepipeds, Volume = length x width x altitude, or V= abc we deduce the volumes of prisms in general by means of the principle : Two polyhedrons are equal (that is, have the same volume) if they are congruent, or if they can be divided into parts which are congruent in pairs. The sign = between two polyhedrons means that they are equal in volume. The word equivalent is sometimes used to mean equal in volume. 545. Theorem. The volume of an oblique prism is equal to that of a right prism having for its base a right section of the oblique prism and for its altitude a lateral edge of the oblique prism. Given the oblique prism AD', with FGHJK a right section, and F'G'S'fK' a right section of the prism extended 80 that the edge AA' = KK'. To prove that the oblique prism ad' has the same volume as the right prism KH'. Outline of proof: Show that (1) a^cde^a'b'c'd'e' ; (2) ABFK ^ A'b'f'k' ; (3) KAKJ ^K'a'k'j'. Ilcneo, liy § 5;5'5 AH and a'ii' are cungruent Now the f^ivcn prism AD' = AB + Kn' and the right prism KII' = A'u' + KD'. Honce, AD' = KU' . ( W hy ? ) (§ o30) PBISMS AND CYLINDESa. 321 546. Theorem. TTie volume of any parallelopiped is equal to the product of its base and altitude. Given the oblique parallelopiped AK, with base b and alt. h. To prove that its volume is equal to 5 • A. Proof : Considering face AL as the base of prism AK, pro- duce the four edges parallel to AB, and lay off a'b' = AB. Through A' and b' erect planes -L to Ab', thus cutting off the right prism A'k' with a'l' as one base. Now considering c'l' as the base of prism a'k', produce the four edges U to c'b' and lay off c/'b" = c'b'. At c" and b" erect planes X to c'b", cutting off the right prism a"k", which is a rectangular parallelopiped. Then show that (1) h = h' = h"; (2) b = b' = h" ; (3) prisms AE, a'k', A"k" are equal (§ 545). But prism a"k" = b" • h". (Why?) Hence, prism AK=b- h. Write out this proof in full. 322 SOLID r; KOMETli Y. E H kL y A / f- \ 1 ^ ^~^ D\ ,-iM 547. Theorem. 77ie rolarne of a triangular prism is equal to the j^rodact of its base and altitude. Given the triangular prism whose base is ABC. To prove that the volume of this prism is equal to the area of A ABC multiplied by the altitude h, that is, by the perpendicular distance between ABC and EFG. Proof: Complete the HJ ABCD and EFGH and draw DH. Now show that CDHG and ADHE are /I7, and hence that BH is a parallelepiped. Let KLMO be a right section of BH, and let s:^T be the line in which the plane ACGE cuts tlie plane KLMO. Now (1) ELMO is a O. Hence, A KLM ^ A KMO. (2) Volume of prism ABC-F = area A KLM times BF. (3) Volume of prism BH = area O KLMO times BF. (4) Hence, prism ABC-F = l prism BH. (5) But prism BH = h x area of O ABCB. (0) Hence, prism ABC-F = 7t x | area of O ABCB = A X area of A ABU. Therefore prism ABC-F is equal to the product of its base and altitude, (iive reasons for each step. e' 648. Corollary 1. "Jlie ruhiDw of aiii/ jjrism is (-(jiial tn the product of its base and altitude. For any prism can In- Jividt'd into triangu- lar jirisins by jilaues iiassini,' tliroin;h one edge and ciu'li of the oilier iion-adjaeeiit edges. PRISMS AND CTLINDEES. 323 549. CoEOLLARV 2. Any tivo prisms of equal alti- tudes have the same volumes if their bases are equal. 550. CoKOLLARY 3. I%e volumes of tioo prisms have the same ratio as their altitudes if their bases are equal; and the same ratio as the areas of their bases if their altitudes are equal. 551. EXERCISES. 1. The proposition that the volume of any prism is equal to the product of its base and altitude is of great importance. "What theo- rems of this chapter were used directly or indirectly in proving it ? 2. If the base of a prism is 36 square inches and its altitude 12 inches, what is its volume? 3. If the perimeter of the base of a prism and the length of a lateral edge are knovfn, is the lateral area thereby determined ? 4. If the area of the base and the length of an edge are known, can the volume be found ? 5. What dimensions of a prism must be known in order to deter- mine its lateral area by means of the theorems of this chapter? What dimensions must be known to determine its volume ? 6. Parallel sections of a closed prismatic surface are congruent polygons. Prove. 7. Find the edge of a cube if its total area is equal numerically to its volume, an inch being used as a unit. 8. A side of the base of a regular right hexagonal prism is 3 inches. Find its altitude if its volume is 54'v/3 cubic inches. What is the total area of this prism? 9. A side of one base of a regular right triangular prism is equal to the altitude of the prism. Find the length of the side if the total surface is numerically equal to the volume. 10. Solve the preceding problem in case the prism is a regular right hexagonal prism. 11. The volume of a regular right prism is equal to the lateral area multiplied by half the apothem of the base. Prove. 324 SOLID OEOMETBT. of which is ^3- CYLINDERS. 552. Definitions. A surface no segment plane is called a curved surface. E.g. the surface of an eggshell or of a stovepipe is a curved surface. A closed plane curve is one which can be traced continu- ously by a point moving in a plane so as to return to its original posi- tion without crossing its path. A convex closed curve is one which can be cut by a straight line in only two points. 553. If a straight line moves so as to remain parallel to itself, while it always touches a closed convex curve and is made to traverse it completely, the line is said to generate a closed convex cylin- drical surface. The moving line is the generator, and the guiding curve the di- rectrix. The generator in any one of its positions is an element of the surface. A cross section of a cylindrical sur- face is made by a plane cutting all its elements. 554. A cylinder is that part of a closed cylindrical sur- face included between two parallel cross sections, together with the plane-segments thus intercepted. The plane-segments are the bases of the t'ylinder, and the part of the cylindrical surface between the bases is tho lateral surface. The portion of the generating line in any position which is included between the bases is an element of tho cylinder. The altitude of a cylinder is the perpen- dicular distance between its bases. PRISMS AND CYLINDERS. 325 A right section of a cylinder is made by a plane cutting each of its elements at right angles. A circular cylinder is one whose right section is a circle. The radius of a circular cylinder is the radius of its right section. A cylinder whose bases and right sections are circles is called a right circular cylinder. A right circular cylinder is called a cylinder of revolutions, since it may be generated by revolving a rectangle about one of its sides as an axis. The side opposite the axis generates the lateral sur- face, and the sides adjacent to the axis generate the bases. 555. Theorem. If a plane contains an element of a cylinder and meets it in one other point, then it contains another element also, and the section is a parallelogram,. Given the cylinder AC and a plane containing the element AD and one other point as B. To prove that it contains another element BC and that ABCD is a parallelogram. Proof : Through B pass a line BC II AD. Then BC is an element of the cylinder. (Why ?) Also BC lies in the plane ABD. Now show that ABCD is a parallelogram. What is the section ABCD if AC is a right cylinder? 326 SOLID GEOMETRY. 556. Definition. If a plane contains an element of a cylinder and no other point of it, the plane is said to be tangent to the cylinder, and the element is called the ele- ment of contact. 557. Theorem. The bases of a cylinder are con- gruent plane-segments. Given a cylinder with the bases b and 6'. To prove that b ^b'. Proof : Take any three points A, B, C in the rim of the base b and draw the elements at these points, meeting the base b' in C, E, F. Show that A ABC^ A DEF. Now, while the elements AD and BE remain fixed, con- ceive CF to generate the cylinder. Evidently A ABC ^ A DEF in every position of CF. Hence, if base b' is applied to base b with these triangles coinciding in one position, they will coincide in every position corresponding to the moving generator. That is, b ^ b'. 558, EXERCISES. 1. Sliijw t]ial right sections cif any cylinder are congruent, and any section iiarallel to the base is congruent to the base. PRISMS AND CYLINDERS. 2>21 2. If two intersecting planes are each tangent to a cylinder, show that their line of intersection is parallel to an element of the cylinder and also parallel to the plane containing the two elements of contact. 3. What is the locus of all points at a perpendicular distance of 2 feet from a given line ? 4. If the section of a cylinder made by every plane parallel to an element of it is a rectangle, what kind of cylinder is it ? 5. If the radius r of a right circular cylinder Is equal to its alti- tude, find the distance from the center of the base to a plane whose intersection with the cylinder is a square. 6. Roll a sheet of paper so as to form a circular cylinder, that is, one whose right section is a circle. Now determine by inspection the shape of the base if the paper is cut so as to let the cylinder stand in an oblique position. Also deform the cylinder so as to make the oblique base circular, and ,then determine the shape of the right section. 7. Show that for every cylinder there is at least one set of parallel planes which cut the cylinder in rectangular sections. Suggestion. Project an element on the plane of the base, and draw lines in the base at right angles to this projection. Through these lines pass planes parallel to an element. 8. Is a polygon circumscribed about a convex closed curve neces- sarily a convex polygon ? See § 160. 9. Is a polygon inscribed in a convex closed curve necessarily convex? (Note that a broken line AB, BC, CD, DE, EA which cuts itself ag shown in the figure is not here regarded as a polygon.) The answers to Exs. 8 and 9 limit the polygons to be used under Ax. XX to convex polygons. Hence it is not necessary to state in that axiom that these polygons must be convex. 10. If polygons other than convex are permitted, is it possible to construct one within a convex closed curve whose perimeter is greater than the length of the curve ? Can such polygons be inscribed in the curve? 328 80LID GEOMETRY. MEASUREMENT OF THE SURFACE AND VOLUME OF A CYLINDER. 559, Thus far in geometry the word area has been used in connection with plane-segments only. In some cases the computation of an area has been possible by an approximation process only, as in the case of some rec- tangles and of tlie circle. In the case of any curved surface it is evident that ap- proximate measurement is the only kind possible in terms of a plane area unit, since no such unit, however small, can be made to coincide with a segment of a curved sur- face. Theorems concerning the surface and the volume of a cylinder are based upon the following definitions and as- sumptions : 560, Axiom XX. Any convex closed curve has a defi- nite length and incloses a definite area, ichich are less rcHpcrfively than the jjcri meter and area of any circumscribed polygon and greater than tliose of any inscribed polygon. Also the perimeter and area of either the inscribed or the circumscribed i^olygon may be made to difer from those of the curve by as little as ice please by taking all the sides sufficiently small. 561, Definitions. A prism is said to be inscribed in a cylinder if its lateral edges are elements of the cylin- der, and their bases lie in the same planes. A prism is said to be circumscribed about a cylinder if its lateral f^ucs are all tangent to the lyliiuk'r, and their bases lie in the same planes. PRISMS AND CYLINDERS. 329 562 Axiom XXI. The lateral surface of a convex cylinder has a definite area, and the cylinder incloses a definite volume, lohich are less respectively than those of any circumscribed prism, and greater than those of any inscribed prism. 563. Theorem. The lateral area of a convex cylin- der is equal to the product of an element and the perimeter of a right section. Given a convex cylinder of which L is the lateral area, e an ele- ment, and p tjie perimeter of a right section. To prove that L = ep. Proof : It will be shown that i can be neither greater than nor less than ep. First, suppose L < ep. Let L=eK. Then^ eK. (2) Hence, ep' is greater than the supposed value eK of L. But this is impossible, since ep' is the area of an in- scribed prism (§ 562). Hence, the supposition L < ep leads to a contradiction. Secondly, the supposition that L > ep may be shown to be impossible by using a circumscribed prism. Hence, we have L = ep. 564. Corollary. If r is the radius of any circidar cylinder, and e is an element, then i = 2 nre. In the case of a right circular cylinder e = h, the altitude, and we have i = 2 Trrh. 330 SOLID GEOMETRY. 565. Definition. Two right circular cylinders are simi- lar if they are generated by similar rectangles revolving about corresponding sides. 566. Theorem. The lateral areas or the entire areas of two similar right circular cylinders are in the same ratio as the squares of their altitudes or as the squares of the radii of their bases. Suggestion. If h and h' are the altitudes, r and r' the radii, L and i' the lateral areas, and A and A' the total areas, we are to show that -^ = i- =^ = ^ L' A' r'2 h"^' Make use of the following, giving each in detail. (2) i = 2 TrrA, (2) ^ = 2 7rr(r + K), h . HTld I ^ I ■ ^ — ^ h' (3) - = 77, and (4) -yf-r,-- r h' r + w r h' 567. Theorem. The volume of an;/ convex cylinder is equal to the product of its altitude and the area of its base, or to the product of an clement and the area of its right section. Suggestions. (1) If h is tho altitude and b the base, sliow hy an argument similar to that of § 563, that T', the volume, can be neither greater nor less than bh. PRISMS AND CYLINDERS. 331 (2) If e is an element and c the area of a right section, give a similar argument to show that F = eo, using § 545. Give all the steps in full. 568, Corollary. If a cylinder has a right circu- lar section whose radius is r^ and an element e, then V=->rrU If a cylinder has a circular base of radius r^, and an altitude h, then V = ■^rlh. In the case of a right circular cylinder, n = r-i and h =e. Hence, the two formulas become identical. 569. Theorem. The volumes of two similar right circidar cylinders are in the same ratio as the cubes of the radii of their bases or as the cubes of their altitudes. Give the proof in full, using suggestions similar to those given in § 566. PROBLEMS AND APPLICATIONS. 1. What dimensions of a cylinder must be known in order that its lateral area may be computed? State fully for different kinds of cylinders. 2. What dimensions of a cylinder must be known in order that its volume may be computed ? 3. If the base of a cylinder is a circle with radius 5 inches, find its volume if the altitude is 8 inches. 4. If the lateral surface of a cylinder and the length of an element are known, can the perimeter of a right section be found? If the lateral area is 400 tt, and an element 15, find the perimeter of a right section. 5. The diameter of a right circular cylinder is 8, and the diagonal of the largest rectangle which can be cut from it is 16. Find its altitude. 6. The volume of a right circular cylinder is 128 7r cubic inches and its altitude is equal to its diameter. Find the altitude and the diameter. 832 SOLID GEOMETBT. 7. If the diameter of a right circular cylinder is equal to its alti- tude, deterinitie the diameter so that the total area of the cylinder is equal numerically to its volume. 8. Find the diameter of a right circular cylinder if the total area of the inscribed regular triangular prism is equal numerically to the volume of the cylinder, the diameter of the cylinder being equal to its altitude. 9. In the preceding find the diameter of the cylinder if the volume of the prism is equal numerically to the total area of the cylinder. 10. Solve a problem like Ex. 8 if a regular hexagonal prism is used instead of a triangular prism. 11. Solve a problem like Ex. 9 if a regular hexagonal prism is used instead of a triangular prism. 12. A rectangle whose sides are a and b is turned about the side a as an axis and then about the side b. Find the ratio of the volumes of the tv^o cylinders thus developed. 13. Compare the total surfaces of the two figures developed in Ex. 12. 14. Find the diameter of a right circular cylinder if its lateral area is equal numerically to its volume. Does the result depend upon the altitude of the cylinder? 15. If the altitude of a right circular cylinder is equal to its diameter, find the ratio of the numerical values of its total area and its volume. Does this depend on the radius? 16. A regular octagonal prism is inscribed in a right circular cylinder whose altitude is equal to the diameter. Find the difference liutween the volumes of the cylinder and the prism, if a side of the octagon is i inches. 17. A cylindrical tank S feet in diameter, partly filled with water, is lying on its side. If the greatest di'p(h of the water is 6 feet, what fraction of the volume of the tank is filled with water? 18. In the pri'ci'dini;- problem find the fraction of the volume nccnineil l)y water if the width of the toi> il.£_-/^--^_^^ making an angle of SO'' with the / -/-,, / ^^~~"~>.,^ right section p, then the section c' \Zy~ e'~Z^ Lj_::s:«. thus made is an ellipse whose axes are 6 in. and 6 -^ cos 30= = 6 -^ .866 = 6.93 in. nearly. Hence, the area of this elliptical section is ■TTob = 3.14 X 6.93 x 6 = 130.56 sq. in. SUMMARY OF CHAPTER IX. 1. Make a list of the definitions on prisms and also a list of those on cylinders and compare them. 2. Make a list of the theorems on prisms and also of those on cylinders and compare them. 3. State the axioms of this chapter and note that they all refer to areas and volumes which involve curved surfaces. Compare these with the axioms on the circle in Plane Geometry. 4. Make a list of all the formulas given by the theorems and corollaries of this chapter. 5. Make an outline of the definitions and theorems concerning pro- jections of lines and surfaces. 6. Explain how the area of an ellipse is obtained either by project- ing it into a circle or by projecting a circle into it. 7. Make a list of the applications in this chapter which have appealed to you as interesting or practical or both. Return to this question, after studying the problems and applications which follow. ;j44 SOLID GEOMETRY. PROBLEMS AND APPLICATIONS. 1. Given a right circular cylinder the radius of whose base is 6 inches. Find the area of an oblique cross section inclined at an angle of 45° to the plane of the base. 2. Given an oblique circular cylinder the radius of whose right section is 10 inches. Find the area of the base if it is inclined at an angle of 60° to the right section. 3. If an oblique circular cylinder has an altitude h, an element e, radius of right section r, and ^A the inclination of the base to the right section, express the volume in two ways and show that these are equivalent. 4. A six-inch stovepipe has a 45° elbow, that is, it turns at right angles. (The angle CAB is called the elbow angle.) Find the area of the cross section B at AB. Likewise if it has a 60° elbow angle. 5. At what angle must the damper in a circular stovepipe be turned in order to obstruct just half the right cross sectional area of the pipe? SuGGESTiox. The damper must be turned through an angle ^ueh that the area of the projection of the damper upon a right cross section is equal to half that cross section. 6. The comparatively low temperature of the earth's surface near the pole, even in summer, when the sun does not set for months, is due largely to the obliqueness with which tlie sun's rays strike the earth. Tluit is, a given amount of sunlight is spread over a larger area than in lower latitudes. Thus, if in the figure D'C is a horizontal line, and D'D the direc- linn of the sun's rays, then a beam of light whose right cross section is A BCD is spread over the rectangle A' BCD'. In other words, a patch of ground A' BCD' receives only as much sunlight as a patch the size of ABCD receives when tlie sun's ravs strike it vertically. Aiv.i AliCD = a.rv:i AliCD'cosZl, or ■area A'BCD'-- arcn.lBC/^x — i cos Z 1 PRISMS AND CYLINDERS. 345 Hence, each unit of area in A'BCD' receives cos Z 1 times as much light as a unit in A BCD. Hence, to^ compare the heat-producing power of sunlight in any latitude with that at the place where the sun's rays fall vertically, we need to know how the projection angle, Z 1, is related to the difference in latitude of the two places. 7. If Z 1 = 30°, compare the amount of heat received by a unit of area in ABCD and A'BCD'. 8. What must Z 1 be in order that a unit of ar^a in A'BCD' shall receive only J as much light as a unit in ABCDl 9. The figure represents a cross section of the earth with an indi- cation of the direction of the rays of light as they strike it at the summer solstice "when they are vertical at A, the ti'opic of Cancer. B represents the latitude of Chicago, Cthe polar circle, and P the north pole. The angles PDE, CGF, BKH represent the projection angle, Z. 1, for the various latitudes. Prove that Z PDE = ZPOK, ZCGF = ZCOK, ZBKH = ZBOK. That is, Z 1 for each place is the latitude of that place minus the latitude of the place where the sun's rays are vertical. 10. Find the relative amount of sunlight received by a unit of area at the tropic of Cancer and at the north pole at the time of the summer solstice. Suggestion. The required ratio is = = . cos Z 1 cos .661° .399 11. Find the ratio between the amount of light received by a unit of the earth's area at Chicago and at the tropic of Cancer at the time of the summer solstice. 12. Find the same ratio for the polar circle and the tropic of Cancer at the spring equinox, when the sun is vertical over the equator. 13. Find the same ratio for the equator and Chicago at the winter solstice when the sun is vertical at latitude 23J° south. CHAPTER X. PYRAMIDS AND CONES. 585. Definitions. Given a convex polygon and a fixed point not in its plane. If a line through the fixed point moves so as always to touch the polygon and is made to traverse it completely, the line is said to generate a convex pyramidal surface. The moving line is the generator of the sur- face, and in any of its positions it is an element of the surface. The guiding polygon is the directrix, and the fixed point the vertex. A pyramidal surface has two parts, called nappes, on op- posite sides of the fixed' point. Compare definitions in § 603. A polyhedral angle is a pyramidal surface of oue nappe. See § 518. 586. That part of a pyramidal surface included between the fixed point and a plane cutting all its elements, together with tlie intuioepleil segment of the piano, is I'allod a pyramid. The iiiti'icepted plaiio-segment is the base of tlie pyra- mid, and the part of the pyramidal surface between the base and the voilex is the lateral surface. 346 PYRAMIDS AND CONES. 347 The lateral surface is composed of triangles having a common vertex at the vertex of the pyramid, and having as bases the sides / of the polygonal base. The sides common / , to two such triangles are the edges of the \ / pyramid. * ^ Pyramids are classified, according to the shape of the base, as triangular, quadrangular, pentagonal, etc. A pyramid having a triangular base has, in all, four faces, and is called a tetrahedron. In this case every face is a triangle, and any one may be taken as the base. The altitude of a pyramid is the perpendicular distance from the vertex to the base. A regular right pyramid, or simply a regular pyramid, is one whose base is a regular polygon such that the perpen- dicular from the vertex upon it meets it at the center. EXERCISE. Show that the faces of a regular right pyramid are congruent isosceles triangles, and hence have equal altitudes. The slant height of a regular right pyramid is the alti- tude of any one of its triangular faces. p 587. Theorem. Hie lateral area of /// '\\ a regular right pyramid is equal to /J '\A \ one half the product of its slant height .L-f't \~\n and the perimeter of the base. k\1__\/ B C Suggestion. Calling L the lateral area, s = KP the slant height, and p = AB + BC + CD + DE + EA, the perimeter, show that Z = J sp. 348 SOLID GKOMETRY. 588. Theorem. If a pyramid is cut by a plane parallel to the bane r (1) The edges and the altitude are divided in the same ratio. (2) The polygonal section is similar to the hai^e. (3) The areas of this section and of the base are in the same ratio as the squares of their perpendicular distances from the vertex. Outline of proof : Given ABODE W a'b'c'd'e'. PM' PA' PB' (1) To prove that — ^ — = — ^— = , etc.. pass another PM PA PB plane through P parallel to the base and then use § 492. (2) To prove that ABCDE ^ a'b'c' d'e'. we show that Z A = Z a', Z b = Z b', etc., and also ii-L = i^-!-, etc. AB BC (3) C'iiUing the area of the cross section b' and that of PM the hasc 6, we are to prove that — = ^ PM' ,1 and for this we need to show tliat Ali 1>A PM (ii\e all the steps in detail. PYRAMIDS AND CONES. 349 589. Corollary. If two pyramids have equal alti- tudes and bases of equal areas lying in the same plane, the sections made by a plane parallel to the plane of the bases have equal areas. Suggestion. If t and t' are the areas of the section's and b and b' those of the bases, show by the theorem that t t' 7 = — , and hence that t = t' ilb = b'. b b 590. EXERCISES. 1. What is the slant height of a regular right pyramid if its lateral area is 160 sq. in. and the perimeter of its base is 20 in.? 2. What is the perimeter of the base of a regular right pyramid if its lateral area is 250 sq. in. and its slant height is 17 in.? 3. How could you find the lateral area of a regular right pyramid? Of any irregular pyramid? What measurements would be necessary in each case ? Why are fewer measurements needed in the case of a regular right pyramid? 4. The base of a regular right pyramid is a regular hexagon whose side is 8 ft. Find the lateral area if the altitude of the pyra- mid is 6 ft. 5. The lateral area of a regular right hexagonal pyramid is 48 sq. ft. and the slant height is 12 ft. Find the altitude of the pyramid. 6. The base of a regular right pyramid is a square whose side is 16 ft., and the altitude of the pyramid is 6 ft. B^ind the lateral area. 7. A pyramid with altitude 8 and a base whose area is 36 is cut by a plane parallel to the base so that the area of the section is 18 sq. in. Find the distance from the base to the cutting plane. 8. If the altitude of a pyramid is h, how far from the base must a plane parallel to it be drawn so that the area of its cross section shall be half that of the base of the pyramid ? 9. In a regular rig^t pyramid a plane parallel to the base cuts it so as to make a section whose area is one half that of the base. Find the ratio between the lateral area of the pyramid and that of the small pyramid cut off by the plane. 350 SOLID GEOMETRY. 591. Definition. A triangular pyramid is cut by a series of planes parallel to the base, including one through the vertex and also the one in which the base lies. Through the lines of intersection of these planes with one of the faces, planes are constructed par- a allel to the opposite edge, thus forming a set of prisms all lying within the pyramid, as a', h\ e', in pyramid p', or a set lying partly outside the pyramid, as a, 6, c, d in pyramid P. The inner prisms thus constructed are called a set of inscribed prisms, and the other a set of circumscribed prisms. 592. Axiom XXII. A pi/ramid has a definite volume which is less than the comhined volume of any set of circumscribed prisms and greater than that of any set of inscribed prisms. 593. EXERCISES. 1. In the figure above prove that prisms d and c' are equal in volume. Also that c = V, etc. 2. If the area of the base of pyramid P is V2 sq. in., vrhat is the altitude of prism a if its volume is 1 cu. in.? If its volume is -j^ cu. iu.? 3. If the altitude of the pyramid in the preceding exercise is 16 in., into at least how many equal parts must it be divided if the volume of prism a is to he less than 1 cu. in.? less fflian .01 ou. in.? Is it possible to divide the altitude of the pyramid int r'. Now a can be made less than K by taking the divisions on PM small enough. Hence, P — P'EF and E-CFD are only different notations for tlie same pyramid. Hence, E-ABC = C-DFF= E-ACD. 'r\\',i[ is, E-Aiu^ is one third of the prism, ]}iit the volume of prism = bh. ( Wliv?) Hence, T' = volume of pyramid = ,^ hh. State in detail tlie reasons for eaeli step. PYRAMIDS AND CONES. 353 596. Theorem. The volmne of any jjyramid is one third of the product pf its base and altitude. A B Given the pyramid P-ABCDE. Let V, b, and h be the numeri- cal measures respectively of the volume, base, and altitude. To prove that r=lbh. Proof: By means of the diagonal planes PAC and PAD, divide the given pyramid into three triangular pyramids. Complete the proof. 597. Corollaries. 1. The volumes of any two pyramids having the ' same or equal altitudes are in the same ratio as the areas of their bases. 2. The volumes of any tioo ^jyraviids having the same or equal bases are in the sam,e ratio as their altitudes. 5.98. EXERCISES. 1. The altitude of a certain pyramid is 14 in. and its volume is 380 ou. in. Find the area of its base. 2. The area of the base of a pyramid is 48 sq. ft. and its volume 260 cu. ft. Find its altitude. 3. Find the locus of the vertices of pyramids having the same base and equal volumes. 4. A diagonal of the square base of a regular right pyramid is TV'S in. and its volume 147 cu. in. Find its altitude and lateral area. 354 SOLID GEOMETRY. 5. A flower bed is in the form of a regular right pyramid, with a square base 5 ft. on a side. The altitude is 2 ft. Find the number of cubic feet of soil in its construction. 6. A tent is to be made in the form of a right pyramid, with a regular hexagonal base. If the altitude is fixed at 15 ft., what must be the side of the base in order that the tent may inclose o'A) cu. ft. of space ? 7. Two marble ornaments of equal altitudes are pyramidal in form. One has a square base 2 in. on a side and the other a regular hexagonal base 1 in. on a side. Compare their volumes. 8. Two monuments having bases of equal areas are pyramidal in shape, one being 15 ft. high and the other 18 ft. Compare their volumes. 9. If the base and the volume of a pyramid are known, is it possi- ble to determine its lateral area ? 10. Given a pyramid with rectangular base. By how much is its volume multiplied if the length and width of the base and also the altitude are each multiplied by 2; by 3; by any number n? 11. Given a pyramid with altitude 10 and a regular hexagonal base, each of whose sides is 5. By how much is its volume multiplied if each side of the base and also the altitude is multiplied by 2 ; by 3 ; by 4 ; by any number n? For a general statement of the law exemplified in these exercises, see § 650. / 599. Definitions. The figure formed b}- the base of a pyramid, any cross section, and tlae portion of the lateral faces included be- / tween these planes, is called a truncated pyramid. If ihe cross seeliun is parallel to the base, the figure is called a frustum of a pyramid, and this section is the upper base. The altitude of a frustum is the perpendicular distance between its bases. The slant height of the frustum of a regular right pyramid is the common altitude of its trapezoidal faces. PYRAMIDS AND CONES. 355 600. Theorem. The volume of a frustum of a pyramid is equal to the coinhined volume of three pyramids ivhose common altitude is the same as that of the frustum, and whose bases are the upper and lower bases of the frustum and the mean proportional between these bases. p \ A B Given the frustum AC with lower base b, upper base b', and altitude h. Let h' be the altitude PM of the completed pyramid P-ABCDE. Then V&6' is the mean proportional between b and V. To prove that the volume of AC' is V=lh\h + h' + ^W-\. Proof: The altitude of the pyramid P-A'b'c'd'e' is h' -h. Hence, ,^ = ^.' .(§^88) from which h'= — = —• (1) V6-V6' Now V is the difference between the pyramids whose altitudes are A' and h' — h. Hence, F = f 6A' - J J' (A' - A), or, rearranging, V = ^b'h + ^ h'(b — *')• (2) Substituting (1) in (2), r=^h[h + b' + y/W}. Show all the details. 356 SOLID GEOMETRY. 601. Theorem. Tlie volumes of two tetrahedrons, having a trilicilral aiujle of the one congruent to a trihedral angle of the oilier, are in the same ratio as the products of the edges ivhich meet in the vertices of these angles. o Given the tetrahedrons P-ABC and P'-A'&C whose volumes are V and V and in which Tri. Z i'= Tri. ZP'. To prove that PA ■ PB ■ PC p'a' ■ p'b' ■ p'c' Proof : Place p'-A'b'c' so that Tri. Z p' coincides with Tri. Z pi. Let CM and c'm' be the altitudes of P-ABC and P-A'b'c' from the vertices C and c' upon the plane PA It. Let AN and ^'iV^'be the altitudes of the A PAB and PA'b'. „, V _ ^CM ■ area PAB O f PB A X r' ~ I C' m' ■ avea, pa' b' c'm' PB' A'y' „ CM PC , ^.Y PA Now prove — — -, = — ; and —r- = -.• ^ c'm' Pc' A'y' PA' Hence, substituting in (1), (1) we have PC- PB ■ PA v' p'd ■ p'b' ■ p'a' Give all the steps and reasons in detail. PYRAMIDS AND CONES. 357 602. EXERCISES. 1. Show that the lateral faces of a frustum of a regular pyramid are congruent isosceles trapezoids. Hence find the area of its lateral surface in terms of the slant height and the perimeters of the bases. 2. Show that sections of a pyramid made by two planes parallel to the base are similar polygons whose areas are in the same ratio as the squares of the distances from the vertex. 3. Show that any two pyramids standing on the same base, or on equal bases in the same plane, have the same volume if their vertices coincide or lie in a plane parallel to the base. 4. Show that the volumes of two pyramids have the same ratio as the areas of their bases if they have equal altitudes, and the same ratio as their altitudes if they have equal bases. 5. A frustum of a pyramid is cut from a pyramid the perimeter of whose base is 60 inches and whose altitude is 15 inches. What is the altitude of the frustum, if the perimeter of its upper base is 40 inches ? Does the result depend upon the number of sides of the pyramid ? 6. Solve the preceding problem if the perimeter of the upper base of the frustum is one nth that of the lower base. Does this result depend upon the number of sides of the pyramid ? 7. The area of the base of a pyramid is 180 square inches and its altitude is 20 inches. Cut from it a frustum, the area of whose upper base is 45 square inches ; also one the area of whose upper base is one nth of 180 square inches. Do these results depend upon the number of sides of the pyramid? 8. Two triangular pyramids have equal trihedral angles at the vertex. The lateral edges of one pyramid are 14, 16, and 18, and those of the other 7, 8, and 9. Find the ratio between their volumes. Are the data given sufficient to find the volume of each pyramid ? 9. If two triangular pyramids have equal trihedral angles at the vertex and if the lateral edges of one are a, b, and c, and two lateral edges of the others are ",' and h', find the third lateral edge of the second pyramid so that their volumes shall be equal. 10. The slant height of a frustum of a regular pyramid is 10 inches and the apothems of its bases 8 and 6 inches respectively. Find its altitude. 358 SOLID GEOMETRY. CONES. 603, Definition. Given a closed convex curve and a fixed point not in its plane. If C" a line through the fixed point moves so as always to touch the curve and is made to traverse it completely, it is said to generate a convex conical surface. The moving line is called the generator of the surface, and in any particular posi- tion it is an element of the surface. The fixed curve is called the directrix, and the fixed point the vertex. A conical surface consists of two parts, called nappes, on opposite sides of the fixed point. 604, That part of a convex conical sur- face included between its vertex and a plane cutting all its elements, together with the intercepted portion of the plane, is called a cone. The intercepted part of the plane is the base of the cone, and the curved surface is its lateral surface. The altitude of a cone is the perpen- dicular distance from the vertex to the plane of the base. A circular cone is one which has a cir- cular cro.ss section such that the pei-pen- . (liruhir upon it from the vertex meets it at the center. If the base is such a rii-cli>. the •- cone is tlien called a right circular cone. Otherwise, it is an oblique circular cone. PTBAMIDS AND CONES. 359 A right circular cone may be generated by rotating a right triangle PMB about one of its legs, PM, as an axis. The hypot- enuse PB generates the con- ical surface, and the other ^( leg, MB, generates the base. jT The generator of a right circular cone in any position is called the slant height. 605. Theorem. If a plane contains an element of a cone and meets it in one other point, then it contains another element also, and the section is a triangle. Let a plane contain the element PD of the cone P-ABC, and also one other point B. To prove that this plane contains another element PB, and that the section is a triangle PBD. Suggestion. Connect P and B. This segment lies in the conical surface. (Why ?) Complete the proof by showing that BB lies in the base of the cone. Compare this proof with that of § 555. 606. Definition. If a plane contains an element of a cone and no other point of the cone, the plane is tangent to the cone, and the element is called the element of contact. 360 SOLID GEOMETRY. 607, Theorem. If the base of a cone is circular, every plane section parallel to the base is also circular. B- c Given a cone with a circular base AD. To prove that the || section EH is also circular. Proof : Draw the straight line from P to the center 31 of the base, and let it meet the section EB in the point O. Let F and G be any two points on the perimeter of the section EH. Pass planes containing PM through the points F and G, and let them cut the base in MB and MC respectively. Now in the A PMB and PMC prove tliat 0F= OG. Hence^ as F and G, ani/ two points on the perimeter of this section, are equally distant from O, this shows tiiat EH is a circle whose center is 0. 608. Corollary. Tf a cane has a circular base, the areas of two paiydlrl cro./ramid may be inscribed whose slant height differs from the slant height of the cone by less than any given fixed number. Proof : Let d be the given fixed number. From plane geometry we know that a regular polygon may be inscribed in the base of the cone whose apothem differs from the radius of the base by less than d. Let tliis regular poly- gon be the base of an inscribed pyramid. Then the slant height of this pyi-amid differs from that of the cone by less than d, since the difference of two sides of a triangle is less than the third side. 612. Axiom XXIII. The latend surface of a convcv cone has a dcjiiiite area, aiid the eo)ie incloses a definite vohntie, ir/iich are less resj)ectireh/ than those of any eireiDuserilied j)i/rainid and greater than those of any inscribed j)i/rainiil. PYRAMIDS AND CONES. 363 613. Theorem. The area of the lateral surface of a right circular cone is equal to one half the product of its slant height and the circumference of its base. Given a right circular cone of which s is the slant height, c is the circumference of the base, and L the lateral area. To prove that L = |- sc. Proof : Suppose that L > |- sc. Then L = \ sK (1) where K>c. Circumscribe about the cone a pyramid the perimeter of whose base is p, such that plsK. (4) But (4) contradicts (3) because of § 612. Since therefore L is neither less than nor greater than ^ sc, it must be equal to ^ sc. 614, Corollary. If r is the radius of the base of a right circular cone and s the slant height, then L = 1 • 2 irrS = irrS. 364 SOLID GEOMETRT. 615, Definition. Two right circular cones are similar if they are generated by two similar right triangles revolving about corresponding sides. 616. Theorem. The lateral areas or the entire areas of tioo similar right circular cones are in the same ratio as the squares of their altitudes, their slant heights, or the radii of their bases. Give all the Given the cones P and P', with altitudes h, h', slant heights s, s', radii of bases r, r*, lateral areas L, L', and entire areas A, A'. To prove that A = ^ = Z = i^ = Zi A' L' A'2 8'2 /2' Proof: See the suggestions under § 566. steps. 617. EXERCISES. 1. The lateral area of a cone is ;Ui si]uare inclios. M'hat is the lateral area of a similar cone whcisc altitude is | that of the given cone? 2. The total .area of one of two similar cones is three times that of the other. Compare their altitmles ami also their radii. 3. The sum of tlie total areas of two similar cones is 144 square inches. Find the area of each cone if one is 1 J times as high .is the other. PYRAMIDS AND CONES. 365 4. Prove that if in two tetrahedrons three faces of one are congru- ent respectively to three faces of the other and similarly placed about a vertex, the tetrahedrons are congruent. 5. Prove that if in two tetrahedrons two faces and the included dihedral angle are congruent and similarly placed, the tetrahedrons are congruent. 6. A pedestal for a monument is in the shape of a frustum of a regular hexagonal pyi-amid, the radius of the upper base being 4 ft., that of the lower base 6 ft., and the altitude of the frustum 8 ft. Find its volume, slant height, and lateral surface. 7. Find the volume of a frustum of a pyramid the areas of whose bases are 25 sq. in. and 18 sq. in. and whose altitude is 6 in. 8. The area of the lower base of a frustum is 42 sq. ft., its alti- tude 8 ft., and volume 200 cu. ft. Find the area of the upper base. 9. The area of the base of a pyramid is 480 sq. ft. and its altitude 30 ft. Find the volume of the frustum remaining after a pyramid with altitude 10 ft. has been cut off by a plane parallel to the base. 10. The area of the base of a pyramid is 250 sq. in. If a plane section of the pyramid parallel to the base and at a distance of 5 in. from it has an area of 175 sq. in., find the altitude of the pyramid. 11. The sides of the base of a triangular pyramid are 6 ft., 8 ft., 10 ft., and its volume 96 cu. ft. Find its altitude. 12. What part of the volume of a cube is a frustum of a pyramid cut from a pyramid whose base is one face of the cube and whose vertex lies in the opposite face, if the altitude of the frustum is one half the edge of the cube ? 13. Find the dihedral angle at the base of a regular pyramid if the altitude is one half the slant height. p 14. A right triangular prism is cut by a plane not parallel to the base, but such that its inter- section DE is parallel to the base segment AB. Show that the volume of the part thus cut ofE is one third the product of the sum of the three vertical edges and the area of the base. Suggestion. Draw plane DEK II ABC. 15. Find the volume of a truncated triangular prism, the area of whose base is 25 square inches and whose lateral edges are 8, 7, 8. 366 SOLID GEOMETRY. 618. Theorem. The lateral area of a frustum of a right circular cone is equal to one half of the sum of the circumferences of the bases multiplied by the slant height. p Given the frustum ABCD, with slant height s and radii r and Let L represent its lateral area. To prove that L = 1(2 ttt + 2 7rr')« = 7rs(r + r'). Proof : Complete the cone, and let PC = s'. Then i = ^ [2 Trr (s + «') - 2 Trr's'] = irrs + TTs' (r — r'). But 8 + 8' , from which s' = - r 8' Substituting s' from ( 2 ) in (1), L ==irrs -j- 7r/8 = 7rs(r + r'). (1) 619. Corollary. The lateral area of a frustum of a right circular cone /.s equal to the circmnference if n si'cfion midivaji between the bases multiplied by the slant height. Suggestion. From the theorem L = 7r8(r +'•') = 2 7r^+-^ S. r -i~ r' Now show that -^ — is the radius of the section mid- way between the two bases. PYRAMIDS AND CONES. 367 620. EXERCISES. 1. The lateral surface of a right circular cone is 75 sq. ft. Find the altitude if the radius of the base is 4 ft. 2. A circular chimney 100 ft. high is in the form of a frustum of a right cone whose lower base is 10 ft. in diameter and upper base 8 ft. Find the lateral surface. 3. The lateral area of a frustum of a right circular cone is 60 ir sq. in., the radii of the two bases are 6 in. and 4 in. Find the slant height of the frustum. 4. Find the altitude of the frustum in the preceding example, and also the altitude of the cone from whieh it is cut. 5. A frustum of a right circular cone has an altitude one half that of the cone from which it is cut. If its slant height is 8 ft. and lateral area 64 ir sq. ft., find the diameters of its bases. 6. Find the altitude of the frustum of the cone in the preceding example ; also the lateral area of the cone from which it was cut. 621. Theorem. The volume of any convex cone is equal to one third the product of its hose and altitude. Suggestion. Let h be the altitude, h the area of the base, and V the volume. Show that V cannot be different from ^ bh by an argu- ment similar to that of § 563, making use of § 596. 622, CoEOLLART. If a cone has a circular base of radius r and altitude h, then v=^'7Tr^h. 368 SOLID GEOMETRY. 623. Theorem. The volume of the frustum of a convex cone is equal to the combined volumes of three cones whose common altitude is the cdtitude of the frustum and tvhose bases are the upper and loioer bases of the frustum and a mean proportional between these bases. Suggestion. The proof is ex- actly like that of § 600. Note. Observe that the two preceding theorems apply to any con- vex coue because the altitude h is constant. The actual computation is practicable only when the areas of the bases can be found, as in the case of the circle or ellipse. SUMMARY OF CHAPTER X. 1. Make a list of definitions on pyramids and also one on cones and compare them. 2. l\Iake a list of theorems on p^Tamids and also one on cones and compare them. 3. What axioms have been used in this chapter ? Compare these with the axioms in Chapter IX. 4. Make a list of all the formulas given by the theorems of this chapter and compare them with the correspondiug formulas in Chap- ter IX. 5. ^^'hat theorems on cylinders have no corresponding theorems for cones? 6. Show that a frustum of a cone becomes more and more nearly identical with a cylinder if the yertox of the cone is removed farther and farther from the base. 7. Make a list of the applications in this chapter which have im- pressed you afi inteivsling or practical or both. Return to this ques- tion ayain after studying the problems aiul applications which follow. PYBAMIDS A^^D CONES. 369 PROBLEMS AND APPLICATIONS. 1. Show that the volume of a right triangular prism is equal to one half the product of the area of one face and the distance from the opposite edge to that face. 2. A mound of earth in the shape shown in the figure has a rec- tangular base 16 yards long and 8 yards wide. Its perpendicular height is 5 yards, and the length on top is 8 yards. Find the number of cubic yards of earth in the mound. Suggestion. If from each end a pyramid i 8 i with a base 8 yd. by 4 yd. is removed, the remaining part is a tri- angular prism. 3. Given a figure in geoeral shape the same as the preceding, with a rectangular base of length 24 ft. and width 6 ft. Find its volume and lateral area if the dihedral angles around the base are each 45°. 4. The accompanying figure represents a solid whose base is a rectangle 50 units long and 40 units wide. Its height is 12 units and its top a rectangle 20 units by 10 units. Find its volume. Suggestion. Divide the solid as indicated in the figure. ISTotice that this is not a frus- tum of a pyramid. 5. In a figure like the foregoing, how can we determine whether or not it represents the frustum of a pyramid? 6. Show how to pass a plane through a tetrahedron so that the section shall be a parallelogram. Suggestion. Pass a plane parallel to each of two opposite edges. See Ex. 6, page 293. 370 SOLID GEOMETRY. 7. If the middle points of four edges of a tetrahedron, no three of which meet at the same vertex, are joined, a parallelogram is formed. Prove. G A^ 8. If through any point P in a diagonal of a parallelopiped planes KN and RM are drawn parallel to two faces, show that the parallele- pipeds DQ and LN thus formed have equal volumes. 9. Find the volume and area of a figure formed by revolving an equilateral triangle about an altitude, the sides of the triangle being s. 10. Find the area and volume of the figure developed by an equilateral triangle with sides i if it is revolved about one of its sides. 11. Find the area and volume of the figure developed by revolving a square whose side is s about one of its diagonals. 12. Through one vertex of an equilateral tri- angle with sides s draw a line / perpendicular to the altitude upon the opposite side. Find the vol- ume and area of the figure developed by revohing the triangle about the line /. 13. Through a vertex of a square with sides s draw a line I perpendicular to the diagonal through that vertex. Find the ai-ea and volume of the figure developed by turning the square around the line I. 14. In a regular hexagon with sides . half that of the base. Find the distance from the vertex to the plane. 27. Show that the lateral area of the small cone cut off in the preceding example is one half the lateral area of the original cone. 28. Cut a pyramid of altitude h hy n planes, each parallel to the j4 2 a B a base, so that the areas of the sections shall be -, -, , ..., n + ln + ln + 1 372 SOLID GEOMETRY. (n - \)A ~~j — , A being the area of the base. Show that the distances from the vertex to the planes are /< \i , A-v — = — , h\ — '- — , •■■. '» + l ^(1 + 1 ^n + 1 29. Cut a cone with altitude h by a plane parallel to the base .so that the volume of the frustum formed shall equal half that of the colli'. Find the distance from the vertex of the cone to the plane. 30. Cut a cone of altitude h by n planes, each parallel to the base, so that the frustums formed and the one pyramid cut off at the top shall all have equal volumes. 31. An equilateral triangle ABC is swung around the line DE as an axis, D and E being middle points of the sides of the triangle. Find the volume of the figure thus developed by the trapezoid ABED if AB = a. 32. Find the total surface of the figure in the preceding example. 33. In a right circular cone, with altitude h, and r the radius of its base, a cylinder is inscribed as shown in the figure. Find the radius OF of the cylinder if the area of the ring bounded by the circles OF and OA is equal to the lateral area of the small cone cut off by the upper base of the cylinder. 34. The same as the preceding, except that the lateral area of the small cone is to equal the lateral area of the cylinder. 35. Find the dihedral angles of a regular tetrahedron. CHAPTER XI. REGULAR AND SIMILAR POLYHEDRONS. REGULAR POLYHEDRONS. 624. Definitions. A polyhedron is said to be regular if its polyhedral angles are all congruent and its faces are congruent regular polygons. Certain regu- 625. Construction of regular polyhedrons lar polyhedrons are very simple of construction, as indicated below. (1) The regular tetrahedron. At the center E of an equilateral tri- angle ABC erect a perpendicular to the plane of the triangle. On this take a point 2) so that AD = AC. Now prove that the four triangles, ABC, ACD, ABB, BCD, are regular and congruent, and that the four trihedral angles are congruent. Suggestion. AE= BE= CE. (2) The regular hexahedron or cube. At the vertices of a given square erect perpendiculars to its plane equal in length to the sides, and join their up- per extremities as shown in the figure. Show that six equal and congruent squares are formed and also eight congruent trihedral angles. 373 B_ a U 374 SOLID GEOMETRY. F /^j> ^w / (3) The regular octahedron. Through the center o of a square ABCD draw a perpendicular to the plane of the square. On this take points E and F such that AF= AE = AB. Join ^ and J'' to each of the four vertices. A, B, c, D. Now prove that the eight faces are congruent regular triangles, and that the six polyhedral angles are con- gruent. There are two other regular polyhe- drons, namely : (4) The regular dodecahedron, having for faces twelve regular pentagons, and E TETRAHEDRON HEXAHEDRON OCTAHEDRON DODECAHEDRON ICOSAHEDRON REGULAR AND SIMILAR POLYHEDRONS. 375 (5) The regular icosahedron, having for faces twenty equilateral triangles. The geometric constructions for (4) and (5) are not so simple as for the others. However, cardboard models of all five may be made by cutting out the figures, as shown herewith, and folding them along the dotted lines. They may be held in shape by means of gum paper stuck over the joints. 626. The number of regular polyhedrons. It will now be shown that there are not more than these five regular polyhedrons. There must be at least three faces meeting at each vertex. If these are regular triangles, there may be three, as in the tetrahedron, or four, as in the octahedron, or Jive, as in the icosahedron ; hut there cannot be six, for in that case the sum of the angles about a vertex would be 360°, and it is readily seen that this sum cannot be as great as 360°. For a proof of this fact, see § 726. If the faces are squares, there may be three about a vertex, as in the cube, but there cannot be four, for in that case the sum of the face angles at a vertex would be 360°. If the faces are regular pentagons, there may be three about a vertex, making the sum of the face angles 3 ■ 108° = 324°, b2d there cannot be four, for then the sum would be greater than 360°. See § 726. Regular polygons of more than five sides cannot form the faces of a regular polyhedron, for the sum of the face angles at a vertex would in any such case be more than 360°. Show why this is so. Hence, there cannot be more regular convex polyhedrons than those exhibited above. 376 SOLID GEOMETRY. 627, Theorem. If v l.i the number of vertices of a convex j^ohjluidron, e its number of edges, and f its number of faces, then e + 2 = v +f. This is called Euler's Theorem. The proof is too dif- ficult for an elementary textbook such as this. The proofs given in the current texts are not conclusive. 628. EXERCISES. ' 1. Verify the above theorem bj' counting the number of edges, faces, and vertices in each of the regular figures given in § 625. 2. The following is a form of proof of this theorem which is often given : Denote the number of vertices, edges, and faces of a polyhedron by V, E, and F, respectively. To prove E + 2 = V + F. Proof: Taking the single face ABCD, the number of edges equals the number of vertices, oi E = V. If another face be annexed, three new edges and two new vertices are added. Hence the number of edges gains one on the number of vertices, as E= V + \. If still another face be added, two new edges and one new vertex are added. Hence E = V + 2. With each new face that is annexed the number of edges gains one on the number of vertices, till but one face is lacking. The last face increases neither the number of edges nor vertices. Hence, etc. Show by putting together the faces of a cube in a certain order that the statement in italics need not be true, and hence that the proof is not conclusive. 629. Theorem. The sttm of the face angles of any eu)ii'i'.f poli/hcdron Is e/jiial to four times as mani/ right (i/igles, less eicjlit, as tJie poh/hedron has rertiees. The proof de[)en(l.s on the preceding theorem and is not given licre. 630. EXERCISE. Verify this theorem by uu exaniiiuition of the regular polyhedrons. REGULAR AND SIMILAR POLYHEDRONS. 377 nfSCRIPTION OF REGULAR POLYHEDRONS. 631. Problem. To find a point equally distant from the four vertices of any tetrahedron. Given the tetrahedron P-ABC. To find a point o equidistant from P, A, B, c. Construction. At D, the middle point of BC, construct a plane perpendicular to BC. The plane will contain the point E, the center of the circle circumscribed about A PBC, and also the similar point ii' in A ABC. (Why '?) In the plane DEF draw EG ± ED and FS ± FD. Then EG and FH cannot be parallel (why ?), and hence meet in some point 0. Also EG -L to plane PBC and FH -L plane ABC. (Why ?) Proof : Now show that is equidistant from P, A, B, C. 632. Definitions. The locus of all points in space equi- distant from a given fixed point is a surface called a sphere. The fixed point is the center of the sphere, and a line-segment from the center to the surface is called a radius. A polyhedron is inscribed in a sphere if all its vertices lie in the sphere. The sphere is also said to be circumscribed about the polyhedron. 378 SOLID GEOMETRY. 633. EXERCISES. 1. In the construction of § 631 show that is the only point equi- distant from P, A, B, and C. 2. Show that the planes perpendicular to each of the six edges of the tetrahedron at their middle points meet in the point 0. 3. Does the construction of § 931 depend upon the tetrahedron being regular ? Cau a sphere be circumscribed about any tetrahedron ? 4. Is there any limitation on the relative position of four points in order that a sphere may be passed through them? 634. Four points not all lying in the same plane are said to determine a sphere. Any other point, taken at random, will not, in general, lie on a sphere determined by four given points. Hence, while any tetrahedron may be inscribed in a sphere, a polyhedron, in general, cannot. However, any regular polyhedron may be inscribed in a sphere. 635. Problem. To inscribe a cube in a sphere. H a E K "•-- \i \ /D LK x\ Suggestion. Show that all the diagonals meet in a com- mon point O which is equally distant from all the vertices. < )r show that a perpendicular to one face at its center L meets the ojiposite face at its center K and is perpendic- ular to this face also, and that the middle point O of KL is the point required. REGULAR AND SIMILAR POLYHEDRONS. 379 636. Problem. To inscribe an octahedron in a sphere. Suggestion. Making use of the con- struction, § 625 (3), show that o is the point equidistant from A, B, C, D, E, F. Give all the steps in full. Note. The dodecahedron and icosahedron may each be inscribed in ,a sphere, but the proof in these cases is much more complicated. 637. EXERCISES. 1. If in the figure of § 631 the tetrahedron is regular, show that OE = OF. 2. If in Ex. 1 a sphere is described with center and radiu.s OE, would it touch the face PBC at any other point than E'l (Why?) Would any part of the surface lie on the opposite side of ABC from ? (Why ?) Is the same true of each of the other faces 'I Such a sphere is said to be inscribed in the tetrahedron and the faces are said to be tangent to the sphere. See § 669. 3. In the figure of § 635, show that the point is equidistant from the six faces of the cube and hence that a sphere may be inscribed. 4. In the figure of § 636 show that the point is equidistant from the eight faces and hence that a sphere may be inscribed. The preceding exercises show that a sphere may be inscribed in three of the regular polyhedrons, and the center in each case is the same as that of the circumscribed sphere. This is true also of the other two regular poly- hedrons, but the proof is not so simple as in these cases. In the case of the tetrahedron a sphere maj"- be inscribed whether it is regular or not; but if it is not regular, the center is not the same as that of the circumscribed sphere. 380 SOLID GEOMETRY. SIMILAR POLYHEDRONS. 638. Definitions. Two polyhedrons are similar if they have the same number of faces similar each to each and similarly placed, and have their corresponding polyhedral angles congruent. Any two parts which are similarly placed are called corresponding parts, as corresponding faces, edges, vertices. 639, Theorem. Two tetrahedrons are similar if three faces of one are similar resjyectively to three faces of the other, and are similarly placed. ^ B' Given the tetrahedrons P-ABC and P-A'ffC having A APB ~ A A'PE, A APC ~ A A'P'C, and A BPC ~ A SP'C. To prove p-abc~p'-a'b'c' . Proof : (1) Show that A ^sr- ~ A .1 'b' c'. (li) Show that trihedral A P and p' are congruent. Likewise Z A ^ Z A', Z B ^ Z b\ Z C ^ Z c'- Hence, by definition, the polyhedrons are similar. 640. EXERCISES. 1. If the Iwii pri.sms in the fig- ure iii-p similar, iiiinie tlic pairs of corresponding jiiirls. I^iliow iso for two similar ])vranuils. BEGULAB AND SIMILAR POLYHEDRONS. 381 2. Show that a plane parallel to the base of a pj'ramid cuts ofi a pyramid similar to the given pyramid. Suggestion. Use the principles of similar triangles and § 521 to show that all the requirements of the definition (§ 638) are fulfilled. 3. Does a plane parallel to the base of the prism cut ofi a prism similar to the given prism ? Prove. 4. Show that two tetrahedrons are similar if they have a dihedral angle in one equal to a dihedral angle in the other and the including faces similar each to each and similarly placed. 5. Show that the total areas of two similar tetrahedrons are in the same ratio as the squares of any two corresponding edges. 6. Show that if each of two polyhedrons is similar to a third they are similar to each other. 641, Theorem. The volumes of tioo similar tetra- hedrons are in the same ratio as the cubes of their corresponding edges. Given P-ABC'^P'-A'B'C', with volumes 7 and V. To prove that r PA^ p'a 73 Proof: We have ^= f^'^^]^^, , ■ (§601) f' p' a! ■ p'b' ■ p'c' Now use the properties of similar triangles to complete the proof. Use the figure of § 639. 642. EXERCISES. 1. Two similar tetrahedral mounds have a pair of corresponding dimensions 3 ft. and 4 ft. If one mound contains 40 cu. ft. of earth, how much does the other contain ? 2. The edges of a tetrahedron are 3, 4, 5, 6, 7, and 10. Find the edges of a similar tetrahedron containing 64 times the volume. 3. Find what fraction of the altitude of a tetrahedron must be cut off by a plane parallel to the base, measuring from the vertex, in order that the new pyramid thus detached may have one third of the original volume. 382 SOLID GEOMETRY. 643. Definitions. Two figures are said to have a center of similitude 0, if for any two points A and B of the one the lines AO and BO meet the other in two points, A' and b' , called corresponding points, such that AO ^BO a'o b'o See figures under §§ 180-194. 644. Theorem. Amj tioo figures lohich have a center of similitude are similar. Proof : (1) Two triangles. 0.1 _ OB _ DC OA' ~ OB' " OC' Given Let the student prove that A abc~Aa'b'c'. In case the triangles do not lie in the same plane, use § 489 to show that the corresponding A are equal. (2) Two polygons. Given OA OA' OB or OB' DC'' e' etc. • .1 Ciive tho proof both for polygons in the same plane and not in the same plane. MEGULAR AND SIMILAR POLYHEDRONS. 383 (3) Two tetrahedrons. With the same hypothesis as before, we must prove A PAB~p'A'b', a pbc~A p'b'c', etc., and then use §639. (4) Any two polyhedrons. (a) Prove corresponding polygonal faces similar to each other. (6) Prove corresponding polyhedral angles equal to each other. The last step requires not only equal face angles about the vertex, as in the case of the tetrahedron, but also equal dihedral angles. Note that two dihedral angles are equal if their faces are parallel right face to right face and left face to left face. (Why?) (5) Consider any two figures whatsoever having a center of similitude. (a) Take any three points A, B, C in one figure and the three corresponding points A', b', c' in the other. Then AB and A'b', AC and A'c', etc., are called corre- sponding linear dimensions, and the triangles ABC and A'b'c' are corresponding triangles. (J) It is clear that any two corresponding linear dimen- sions have the same ratio as any other two, and that any two corresponding triangles are similar. In this sense the two figures are said to be similar. 384 SOLID GEOMETRY. 645. Definition. The ratio of similitude of two similar figures is the common ratio of their corresponding linear dimensions. This ratio is the same as the distance ratio of corresponding points from the center of similitude. 646. Theorem. Two similar triangles may he so placed as to have a center of similitude. t')b' A Given the similar triangles T and T, in which A'B' ^A'c ^sa AB AC BC ' To prove that they may be placed so as to have a center of similitude. Proof : From any point O draw OA, OB, OC. On these rays take ^j, Bj, c^ so that OA^ _ OBj _ OCi _ a'b' OA OB OC AB Now show the following : (1) A Tj ~ A T, and hence A Tj ~ A r'. C-^) ATj^Ar'. For this show that ,ljBj = a' B' by means of the .• A,B, O.I, , a'b' OA, equations — i— i = i and = - -1- AB OA AB OA 0M~ OA' REGULAR AND SIMILAR POLYHEDRONS. 385 Likewise A^C-^ = a'g' and b^C.^=b'c'. (3) Finally, where M and M^ are any two corresponding points what- ever. Hence O is the required center of similitude. 647. Theorem. Two similar tetrahedrons may he so placed as to have a center of similitude. Given the similar tetrahedrons T and T'. To prove that they can be placed so as to have a center of similitude. Proof : With o as a center of similitude, construct Tj, making oA ^ OB ^ ^^^ ^ AB OA^ OB-i ' A'b'' Now show as in § 646 that Tj ^ t', and hence that t' can be placed in the position T^ so as to have with T the center of similitude O. Give all the steps in detail. 648. Theorem. Any two similar polyhedrons m,ay be placed so as to have a center of similitude. Suggestion. The argument is precisely similar to that of § 647. Give it in full. 386 aOLtJ) GEOMETRY. 649. In the figures for the preceding theorems the center of similitude has been taken between the two figures or beyond them both. Tlie center may be taken equally well within them, as in the following illustrations : * In the case of similar convex polyhedrons with the center of similitude thus placed, the faces are the bases of pyramids whose vertices are all at the center of similitude. If, further, the polygonal faces be divided into triangles by drax^'ing their diagonals, these triangles become the bases of tetrahedrons, all of whose ver- tices are at the center of similitude. Moreover, each inner tetrahedron is similar to its corresponding outer tetrahedron. (Why ') The volumes of the two similar polyhedrons are thus composed of the sums of sets of similar tetrahedrons. 650. Theorem. Tlte volumes of any t)co similai' ■pohilicdronx hare the name ratio as the cubes of their eorrcspundiiKj e(/(/es. Proof : Place the polyhedrons wliose volumes are ]' and V' so as U> ha\ (■ tlirir crntors of similitude within them as in the figures of § VA'.K REGULAR AND SIMILAR POLYHEDRONS. 387 Call the volumes of the similar tetrahedrons Tj, T^, T^, •■•, and Tj', T^', Tg', ••-, and let AB and a'b' be two corre- sponding edges. Then we have =^0=^ = ^=^=-. (Why?) And ^1 + ^2+^3+ - =J1 = j^. (Why?) But ^1 + ^2 + ^3 ■■■ = T^and r/ + t^' + T^' ■■■ = r'. Hence, — - = . 651. CoKOLLAKT. The volumes of any two similar solids are in the same ratio as the cubes of any two corresponding linear dimensions. This proposition may be rendered evident by noticing that any two similar three-dimensional figures may be built up to any degree of approximation by means of pairs of similar tetrahedrons similarly placed. The proposition then holds of any two corresponding figures used in the approximations. Note that the ratio of similitude of two similar figures may be obtained from the ratio of any pair of corresponding linear dimensions. 552, EXERCISES. 1. If two coal bins are of the same shape and one is twice as long as the other, what is the ratio of their cubical contents ? 2. What is the ratio of the lengths of the two bins in the preced- ing example if one holds twice as much coal as the other ? 3. Two water tanks are of the same shape. Find the ratio of their capacities if their ratio of similitude is |. 4. In the preceding what must be the ratio of similitude in order that the ratio of their capacities shall be |^ ? 388 SOLID GEOMETRY. APPLICATIONS OF SIMILARITY. 653. The theorem that any two figures which have a center of similitude are similar is the geometric basis of many mechanical contrivances for enlarging or reducing both plane and solid figures ; that is, for constructing figures similar to given figures and having a given ratio of similitude vpith them. The essential property of all such contrivances is that one point O is kept fixed, while two points A and B are allowed to move so that 0, A, and B always remain in a straight B' line, and so that the ratio — remains the same. See OB §§ 432-435. In the first figure O is a fixed point. Segments OD, CB, and the sides of the parallelogram ACED are of fixed length. Prove that if B is once so taken on the line EC as to be in the line OA, the points O, A, and B will always remain collinear, and that — remains r OB 2^ a fixed ratio. \S. /^^^\ In the second figure is shown ^^C j< an ordinary pantograph used X/^ \\ y^ <\ for copying and at the same ^- f%-=(^ --- /^^a time for reducing or enlarg- "^n^ ^ r / ing maps, designs, etc. The ^^^ \ f lengths of the various segments are adjustable, as shown, thus obtaining any desired scale. Tlie same contrivance may bo used for copying fignies in Sface and at the same time reducing or enlarging them. REGULAR AND SIMILAR POLYHEDRONS. 389 654. Now consider any two similar figures whatever so placed as to have a center of similitude o. We have seen that if points AB and A'b' are corresponding points of the two figures, then the ratio of the corresponding linear dimensions AB and a'b' is equal to the ratio of similitude — of the two figures. n Also if A, B, c, D and A', b', c', d' are corresponding points, then A ABC and A'b'c', and the tetrahedrons ABCD and a'b'c'd' are similar, and we have Area ABC nfi ■, vol. ABCD m^ Aiea. a'b'c' w2 yoLa'b'c'd' n^ The points A,B, cand^', b' , c' determine two planes, each of which intercepts a certain plane figure in the solid figure to which the points belong. These two plane figures we call corresponding cross sections. We assume without full argument that : — Theorem. (1) Tlie ratio of the areas of any pair of corresponding cross sections or any pair of corre- sponding surfaces of similar figures is equal to the square of their ratio of similitude, and (2) The ratio of the volumes of any two similar fig- ures is equal to the cube of their ratio of similitude. " Thus the ratio of the radii or of the diameters of two spheres is their ratio of similitude ; likewise the ratio of the lengths or of the diametei-s of two shells used in gunnery, or the ratio of the heights of two men of similar build. The fact that the ratio of the areas of corresponding surfaces of similar solids is equal to the square of their ratio of similitude, while the ratio of their volumes equals the cube of this ratio is one of the most important and far-reaching conclusions of geometry. 390 SOLID GEOMETRY. SUMMARY OF CHAPTER XI. 1. Describe the five regular solids according to the form and number of their faces. Why can there not be more than these five ? 2. Compare the relation of the regular solids to the sphere with that of regular polygons in the plane to the circle. 3. Review the process of construction of the tetrahedron, cube, and octahedron. 4. Form all five regular polyhedrons by cardboard models as indicated in § 625. 5. Make a list of the definitions concerning similar polyhedrons. 6. Make a list of the theorems concerning similar polyhedrons. 7. Explain the relation of two figures which have a center of similitude. 8. What theorem of this chapter is referred to as of unusual im- portance in the problems and applications? 9. State the applications of this chapter which appeal to you as especially interesting or useful. Return to this question, after study- ing those which follow. PROBLEMS AND APPLICATIONS. 1. If it is known that a steel wire of radius r will carry a certain weight w, how great a weight will a wire of the same material carry if its radius is 2 r ? Suggestion. The tensile strengths of wires are in the same ratio as their cross-section areas. 2. Find the ratio of the diameters of two wires of the same mate- rial if one is capable of carrying twice the load of the other; three times the load. 3. In a laboratory experiment a heavy iron ball is suspended by a steel wire. In suspending another ball of twice the diameter a wire of twice the radius of the first one is used. Is this perfootly safe if it is known that the first wire will just safely cany the ball suspended from it? Discuss fvilly. 4. In two schoolrooms of the same sliape (similar figures) but of different sizn, thi' sanii' proportion of tlie fioor space is occupied by desks. Which contains the larger amount of air for each pupil? REGULAR AND SIMILAR POLYHEDRONS. 391 5. It Ls decided to erect a school building exactly like another already built, except that every linear dimension is 'to be increased by ten per cent ; that is, each room is to be ten per cent longer, wider, and higher, and so for all parts of the building. If the air in the ventilating flues flows with the same velocity in the two build- ings, in which will the air in a room be entirely renewed the more ■quickly? Suggestion. Note that the ratio of the amount of air discharged by two flues under such conditions is equal to the ratio of their cross- section areas. 6. If the shells used in guns are similar in shape, find the ratio of the total surface areas of an eight-inch and a twelve-inch shell. 7. Find the ratio of the weights of the shells in the preceding problem, weights being in the same ratio as the volumes. 8. If a man 5 ft. 9 in. tall weighs 165 lb., what should be the weight of a man 6 ft. 1 in. tall, supposing them to be similar in shape ? 9. What is the diameter of a gun which fires a shell weighing twice as much as a shell fired from an eight-inch gun, supposing the shells to be similar bodies? 10. The ocean liner Mauretania is 790 feet in length. What must be the length of a ship having twice her tonnage, supposing the boats to be similar in shape ? 11. The steamship Lusitania is 790 feet long, with a tonnage of 32,500, and the Olympic is 882 feet long, with a tonnage of 45,000. Are these vessels similar in shape? If not, which has the greater capacity in proportion to its length ? • 12. Supposing two trees to be similar in shape, what is the diame- ter of a tree whose volume is three times that of one whose diameter is 2 feet? What is the diameter if the volume is 5 times that of the given tree? What if it is n times that of the given tree? 13. Two balloons of similar shape are so related that the total surface area of one is 5 times that of the other. Find the ratio of their volumes. See page 444 for further applications. CHAPTER XII. THE SPHERE. PLANE SECTIONS OF THE SPHERE. 655. Definitions. A sphere consists of all points in space which are equally distant from a fixed point, and of these points only. The fixed point is called the center of the sphere. (See § 632.) A sphere divides space into two parts such that any point which does not lie on the sphere lies within it or outside it. The sphere may be developed by revolving a circle about a diameter as a fixed axis. A line-segment joining any two points on a sphere and passing through its center is a diameter. A segment joining the center to any point on the sphere is a radius. If the distance from a point to the center of the sphere is less than the radius, the point is within the sphere, and if greater than the radius it is out- side the sphere. A sphere may be designated by a single let- ter at its center, or more explicitly by naming its center and radius. Thus the sphere C means the sphere who.^e center is C, and the sphere CA is the sphere whose center is C and whose radius is CA . Two spheres are said to be equal if they have equal radii. 892 THE SPHERE. 393 656. Theorem. All radii of the same sphere or of equal spheres are equal. All diameters of the same sphere or of equal spheres are equal. These statements follow directly from the definitions. 657. EXERCISES. 1. How does the definition of a sphere differ from that of a circle ? State each in terms of a locus. 2. If two spheres have the same center, show that they are either equal or one lies entirely inside the other. 3. In how many points can a straight line meet a sphere V 4. Does every line through an interior point of a sphere meet it? In how many points ? 658. Theorem. A section of a sjjhere made by a plane is a circle. Given a sphere with center C cut by the plane M. To prove that the points common to the sphere and the plane form a circle. Proof : From the center C draw CA perpendicular to the plane M. Let B and D be any two points common to the plane and the sphere. Complete the figure, and prove AB = AD. Hence, any two points common to the plane and the sphere are equidistant from A. How must this proof be modified in case the plane M passes through the center of the sphere ? 394 SOLID GEOMETRY. 659. Definitions. A circle is said to be on a sphere if all its points lie on the sphere. The line perpendicular to the plane of a circle at its center is called the axis of the circle. The points in which the axis of a circle on a sphere meets the sphere are called the poles of the circle. If the plane of a circle on a sphere passes through the center of the sphere, it is called a great circle of the sphere, and if not, it is called a small circle. 660. Theorem. (1) Hie axis of nny circle on a sphere passes through the renter of the sphere. (2) The center of a great circle is the center of the sphere. (3) All rjrerit circles are eq^ial and bisect each other. (4) Three paints on a sphere determine a circle on the sphere. (5) TJirnni/h tiro giren j)oints on a sphere there is one anel onhj one great circle nnless tJiese paints are at opposite ends of a dianuti'r. (6) Ererji great circle bisects a sphere. The proofs of these statcuu'iits fallow easily from the definitions. Let the student gi\o the proofs in detail. THE SPHERE. 395 661, Definition. The distance between two points on a sphere is tlie distance measured between these points along the minor arc of the great circle through thera. 662. Theorem. All points of a circle on a sphere are equidistant from either pole of the circle. Given P a pole of the circle whose center is A, and let B and D be any two points on this circle. To prove that the great circle arcs PB and PB are equal. Suggestion. Let G be the center of the sphere. Prove that Z.ACB = Z ACB. Hence, chord PB = chord PD and PB = PD. Extend the radius PC to meet the sphere in P' . Prove that P' is also equidistant from any two points of the given circle. 663. Definition. The common distance from the pole of a circle to all points on it is called the polar distance of tlie circle. One fourth of a great circle is a quadrant. 664. CoEOLLART 1. The polar distance of a great circle is a quadrant. 665. Corollary 2. If a point p is at a quadrant's distance from each of tivo points not at the extremities of the same diameter, it is the pole of the great circle through these points. 396 SOLID GEOMETRY. 666. It follows from the preceding theorem that, if a spherical blackboard y^ is at hand, circles may be constructed If~i on it by means of crayon and string the same as on a plane blackboard. Like- wise, curve-legged compasses may be used. 657_ EXERCISES. 1. How many small circles can be passed through two points on a sphere ? How many great circles ? Show why. 2. If two points are at the extremities of the same diameter of a sphere, how many great circles can be passed through these points ? 3. AVhat great circles on the earth's surface pass through both poles? If a great circle passes through one pole, must it pass through the other ? 4. If P is at a quadrant's distance from each of two points A and B, and if these points are at opposite ends of the same diameter, is P the pole of any circle through .1 and Bl Of how many circles? 5. If .4 and B are at opposite ends of a diameter, can a small circle be passed through them ? 6. If two circles on a sphere have the same poles, prove that their planes are parallel. 7. What is the locus of all points on a sphere at a quadrant's dis- tance from a given point? 8. What is the locus of all points on a sphere at any fixed distance from a given point on the sphere? What is the greatest such dis- tance possible? Discuss fully. 9. If two planes cutting a sphere are parallel, compare the posi- tions of the poles of the circles thus formed. 10. Find the locus of the centers of a set of circles on a sphere formed by a set of parallel planes eulting it. 11. AB is a fixed diameter of a sphere. .\ plane containing; AR is made to revolve about it as an axis. Find the locus of the poles of the great circles on the sjihere made by this revolving plane. How are the points .1 ami B related to this locus? THE SPHERE. 397 668. Theorem. ^ two planes cutting a sphere are equidistant from the center, the circles thus formed are equal; and conversely, If two planes cut a sphere in equal circles, the planes are equidistant from the center. Suggestion. In the figure show that '(1) if GA = CA', then AB =a'b', and (2) if AB = a'b', then CA = CA' . 669. Definitions. A plane which meets a sphere in only one point is tangent to the sphere. Two spheres are tangent to each other if they have only one point in common. A line is tangent to a sphere if it contains one and only one point of the sphere. 670. EXERCISES. 1. If a plane has more than one point in common with a sphere, must it have a circle in common with the sphere ? 2. If a plane is tangent to a sphere, how many lines in the plane are tangent to the sphere ? 3. Can two spheres be tangent to each other and still one be in- aide the other 1 398 SOLID GEOMETRY. 671. Theorem. A plajie tangent to a sphere is perpendicular to the radius from the point oftarirjency ; and conversely, A plane perpendicular to a radius at its extremity h tangent to the spliere. Given a sphere C with plane M tangent to it at A. To prove that CA is perpendicular to the plane il. Proof : (1) It is only necessary to prove that CA is per- pendicular to every line in M through A. (Why?) Draw any such line AB. The plane BAG cuts the sphere in a circle. Prove that AB is tangent to this circle, and hence perpendicular to AC. (2) To prove the converse, note that c.i is the shortest distance from C to the plane 3/. (Why'?") And hence that every point of M except A is exterior to the sphere. Hence, JIf is a tangent plane. (Why?) 672. Definition. A sphere is saitl to be inscribed in a polyhedron if every face of tlie polyhedron is tangent to the sphere. The polyhedron is also said to be circumscribed about the sphere. 673. TiiEORKM. ^4 sphere may he inscribed in any ietrdhcdron. The proof is left to the student. See § G37, THE SPHERE. 399 674. EXERCISES. ■ 1. How many planes may be tangent to a sphere at one point on the sphere ? How many lines ? 2. Through a given point exterior to a sphere construct a line tangent to the sphere. Suggestion. Let be the center of the sphere and P the given exterior point. Pass any plane M through P and O. In the plane J/ consti'uct a line through P tangent to the great circle in which M cuts the sphere. 3. How many lines tangent to a sphere can be constructed from a point outside the sphere? 4. Through a given point exterior to a sphere construct a plane tangent to the sphere. Sdggestion. As in Ex. 2 construct a line through the given point tangent to the sphere. Through the point of tangency of this line pass a plane tangent to the sphere. 5. How many planes can be passed through a given exterior point tangent to the sphere ? 6. How many planes tangent to a sphere can be passed through two given points A and B outside a sphere ? Discuss fully if the line AB (1) meets the sphere in two points ; (2) is tangent to the sphere ; (3) does not meet the sphere. 675. Theorem. The intersection of tivo spheres is a circle. Proof: The two intersecting spheres may be developed by rotating about a fixed axis CC' two intersecting circles with centers C and c'. Let A and B be the two points common to both circles. Then BA±CC'. (Why?) As the figure rotates about the line CC', AB remains fixed in length and perpendicular to CC'. Hence, B traces out a circle. See § 478, 400 676. Problem. material sphere. SOLID GEOMETRY. To find the diameter of a given With any point P of the sphere as a pole, construct any circle, and on this circle select any three points A, B, C. Using a pair of compasses, measure the straight line- segments AB, BC, CA, and construct the triangle a'b'c' congruent to ABC. Let b'd' be the radius of the circle circumscribed about A'B'C'. If PP' is the axis of the circle ABC on the sphere and, BD the radius of this circle, then BB = b'd'. Measure PB by means of the compasses. Then PBP' is a right triangle, with BD perpendicular to its hypotenuse PP'. PB and BD being known, we may now compute PD from the right triangle PBD and then compute PP' from the similar triangles PBZ» and PP'T), finding PD: PB = PB : PP' or PD X PP' = PB^. See P2x. 3, § 1)6. The segment PP' may also be found by ijeometric con- struction ; namely, draw a triangle congruent to p'bp. Show how to do this when BP aud BD are known. THE SPHERE. 401 677, EXERCISES. 1. How many points are necessary to determine a sphere ? See §632. 2. If the center of a sphere is given, how many points on the sphere are required to determine it? 3. If a plane M is tangent to a sphere at a point A, show that the plane of every great circle of the sphere through A is perpendicular to Af. 4. Show that the line of centers of two intersecting spheres meets the spheres in the poles of their common circle. 5. Find the locus of the centers of all spheres tangent to a given plane at a given point. 6. Find the locus of the centers of all spheres tangent to a given line at a given point. 7. Find the locus of the centers of all spheres of given radius tangent to a fixed plane. 8. Find the locus of the centers of all spheres of given radius tangent to a fixed line. 9. Find the locus of the centers of all spheres tangent to two given intersecting planes. 10. Find the locus of the centers of all spheres tangent to all faces of a trihedral angle. 11. Show that two spheres are tangent if they meet on their line of centers. Distinguish two cases. Compare § 209. 12. State and prove the converse of the preceding proposition. 13. In plane geometry how many cii'cles can be drawn through a given point tangent to a given line at a given point ? 14. If a given sphere is tangent to a given plane iW" at a given point A, how many points on the sphere are required to determine it? Suggestion. Suppose one point P given. Pass a plane through P JL to plane M a,t A. Is there only one such plane? Discuss fully. 15. Describe the set of all lines in space whose distances from the center of a sphere are all equal to the radius of the sphere. 16. Describe the set of all planes whose distances from the center of a sphere are all equal to the radius of the sphere. 402 SOLID GEOMETRY. TRIHEDRAL ANGLES AND SPHERICAL TRIANGLES. 678. Definitions. When two curves meet, they are said to form an angle ; namely, the angle made by the tangents to the curves at their common point. Any two planes through the center of a sphere cut out two great circles which intersect in two points and form four spherical angles about each of these points. Two of these angles with a common vertex are either adjacent or vertical in the same sense as the angles formed by two in- tersecting straight lines. A spherical angle is acute, right, or obtuse according to the form of the angle between the tangents to its sides (arcs) at their common point. Any two circles on a sphere which meet, whether great circles or not, form angles according to the above defini- tion. Only angles formed by great circles are considered in this book and the expression spherical angle will be under- stood to refer to such angles. (Jnly angles greater than zero and not greater than two right angles are considered. A spherical angle may be denoted by a single letter or by three letters, as in the case of a plane angle. 679. TFiEOREjr. A sjjhcrical anr/Ic is iiiensurcd hi/ dii (trc of the ijrcat circle lohose pole Is the vertex of the intercepted (UiijJe find tvliieh is the. sides of the d/ii/Ie. % Suggestion. Show that AB meas- ures the dihedral angle formed by the planes PAC PliC and that Z«cM = ZrPS. md THE SPHERE. 403 680. Definitions. The section of a sphere made by a convex trihedral angle, whose vertex is at the center of the sphere, is called a spherical triangle. -^ The face angles of the trihedral angle /\ \ are measured by the sides (arcs) of the / \ "\^ spherical triangle, and its dihedral / \c_ \^ angles are equal to the angles of the I y^^-"^'"^ spherical triangle. ^""^^ Since each face angle of a trihedral angle is less than two right angles, it follows that each side of a spherical triangle is less than a semicircle. 681. EXERCISES. 1. Show that a spherical angle is equal to the plane angle of the dihedral angle formed by the planes of the great circles whose arcs are the sides of the spherical angle. See figure under § 679. 2. Prove that vertical spherical angles are equal. 3. Prove that the sum of the spherical angles about a point is four right angles. 4. At what angle does a meridian on the earth's surface intersect the equator? 5. Denote by P the North Pole on the earth's surface. Consider any two meridians forming an angle of one degree at P and meeting the equator in points A and B, respectively. What is the sum of the angles of the spherical triangle PAB ? Compare with the sum of the angles of a plane triangle. 6. Is it possible to construct a spherical triangle each of whose angles is a right angle? Suggestion. Consider two meridians forming a right angle at P. Such a triangle is called a trirectangular triangle. 7. In a trirectangular spherical triangle what is the length of each side in terms of degrees ? The question as to whether or not the sum of the angles of a spherical triangle is ever equal to two right angles is answered in §712. 404 SOLID GEOMETRY. 682. Theorem. TTie sum of two face angles of a trihedral angle is greater than the third face angle. c Proof : Connect points A and D on two sides. Suppose not all three face angles are equal and that Z ACD > Z ACB. Construct CE in the face J. CD, making Z ACE= Z ACB. Lay off CB = CE and draw AB and BD. Now show that (1) AB = AH, (2 ) AD < AB + BD, (8 ) ED < BD, (4) Z ECD < Z BCD. Hence, Z ACD < Z ACB + Z BCD. 683. Corollary 1. The sum of tioos id e^ of n spheri- cal triangle is grrnter thmi the third side. 684. Corollary 2. The sum (f the three sides of a spherical tria)ujlr is less than a gnat circle. Proof: aca' +aisa'= a great circle, But AC < aca' and .IB < ABA', and BC < Ba' + ca' liy Corollary 1. Hence, AC + CH + BA < a yruat cirole. 685. CoKoi.LAiiv ?K Sliife and prove the fhenn ids on triliedrtd angles corresponding to Corollaries 1 and 2. THE SPHERE. 405 686. Theorem. Tlie shortest distance on a sphere be- tiveen two of its points is measured along the minor arc of a great circle passing through these points. Proof: Let A and B be any two points on a sphere, AB the minor arc of a great circle through them, and ABCB any other curve on the sphere connecting A and B with points c and B on it in the order ADCB. Neither C nor B is on the arc AB. Draw tlie great circle arcs AB, AC, BC, and CB. Then by § 683 AC+ CB > ab and AB + DC> AC. Hence, AB -i- BC + CB > AB. Continuing in this manner, we obtain a succession of paths, each longer than the preceding. But by this process we get closer and closer to the length of the curve ACB. Hence, it must be greater than that of AB. 687. Definitions. Two trihedral angles are symmetrical one to the other if the face angles and the dihedral angles of one are equal respectively to the face angles and the dihedral angles of the other, but arranged in the opposite order. Similarly, two spherical trian- gles are sym- metrical one to the other if the sides and the angles of one are equal respectively to the sides and the angles of the other, but arranged in the opposite order. 406 SOLID GEOMETRY. 688. Theorem. If the radii drawn from the ver- tices of a sjjherieal trlavfjle are extended, theij meet the sphere in the vertices of a triangle synarietrlcal to the given triangle. The proof is left to the student. 689. Corollary. State and prove the e(>ri't>nd- ing tJteorent uri trihedral angles. 690. Theorem. Tivo trlJiedral angles having their vertices at the center of tJie same or of rijiial spjtrrrs intercept congruent sjjherlcal triangles If the frlhedrcd angles ((re congruent, and sgrnnietrlc(d spJterlccd tri- angles if they are si/nunetrlcrd. o- ■^ 1 Tliis is an iunnediatc^ consequence of §§ t'iSO, iIST. 691, (.^.(>i;ollai;y. //'//( twa sj)herlcal trlaiKjlcs three sides (f one are e(/U(d resjxctlreh/ fo three sides ide of the cii-cle through B' ( ". StinoKsTiDN. Can the two extremities of a diameter lie in the same hemisphei-e? 3. If .1 is a pole of the great circle through B'C and if A' is on the same side of this circle as .1, show that .1 and .1' are less than one quadrant's distance apart.. THE SPHEEM. 411 706. Theorem. If a'b'c' is the polar triangle of ABC, then ABC is the polar triangle of a'b'c'. ^_JProof : It is required to prove (1) that A' is the pole of BC, b' the pole of AC, and c' the pole of AB, and also (2) that A and A' lie on tiie same side of BC, B and B' on the same side of AC, and C and c' on the same side of AB. (1) To prove that a' is the pole of BC we need only to show that a' is at a quadrant's distance from two points in BC. Wliy ? Now a' is at a quadrant's distance from B because B is the pole of a'c. a' is also at a quadrant's distance from C because C is the pole of a'b'. Hence, A' is the pole of BC. Similarly, b' is the pole of AC and c' the pole of AB. (2) To show that A and A' lie on the same side of the circle BC, we note that since A is the pole of the circle b'c' and A lies on the same side of this circle with A', then A and a' are at less than a quadrant's distance. Herice, it follows that if A' is at a quadrant's distance from BC, A and a' must be on the same side of BC. In like manner we show that B and b' lie on the same side of AC, and C and c' on the same side of AB. 707. Definition. If ABC and a'b'c' are polar triangles, and if ^ is a pole of B^, then Z A and b'c' are said to be corresponding parts. 708. EXERCISES. 1. In the two polar triangles ABC and A'B'C name all pairs of corresponding parts. 2. Is there any spherical triangle such that its polar triangle is identical with the given triangle ? 3. If one side of a spherical triangle is greater than a quadrant show that it is cut by two circles in the construction of its polar triangle. 412 SOLID GEOMETRY. 709. Theorem. The sum of the measures of an angle of a spherical triangle and the corresponding arc of its polar triangle is 180°. Given the polar triangles ABC and A'EIC Denote the meas- ures in degrees of the angles by A, B, C, ■■■ and of the correspond- ing sides by a', b', c', •••. To prove that ^ + a' = 180° B + b' = lSO° c+c' =180° = 180° B' + b =180° c' + c = 180° Proof: Extend (if necessary) arcs A' b' and A'c' till they meet the great circle BC in points D and E, respectively. Then arc BE is the measure of Z A'. Also BE =20°, and DC=20°. (Why') But BE+ BC= BC + ED= a + A'. Hence, a' + a = 180°. Complete the proof for the other cases. 710, Corollary. If two spJieriad triangles are con- gruent or si/minetric(tl, their polar trianghs arc con- gruent or symmetrical. 711, EXERCISES. I 1. Does the above proof apply to the secoijd figure ? 2. If two angles of a splioiioal triangle are equal, it is isosceles. SuiiOKSTiON. I'si' § 7lti), § (i'.l5, and .igain § "Oil. ■A THE SPHERE. 413 3. State and prove the theorem on trihedral angles corresponding to the preceding. c 4. If two angles of a spherical triangle are unequal, the sides opposite them are unequal, the greater side being opposite the greater angle. B Suggestion. In the triangle ABC let ZB>ZA. Draw BD, making Z ABD = ZA. Then, AB = BD, and BD + DC > BC. Hence, show that AC > BC. 5. State and prove a theorem on trihedral angles corresponding to the preceding. 712, Theorem. The sum of the angles of a spheri- cal triangle is less than six right angles and greater than two right angles. a a Given the spherical triangle ABC. To prove that (1) ZA + ZB + Zc<6vt. angles. (2) Z A + Z B + Zc> 2 vt. angles. Proof: Construct the polar triangle a'b'c', with sides a', h', c'. (1) By § 709 Z 4 + Z B + Z c + a' + 6' + c'= 6 rt. angles. Since a' + h' + c' is greater than zero, it follows that ZA+ZB + Zc<&vt. angles. (2) Using § 684, show that ZA-\-Zb + Zc> 2rt. A. 713. Corollary. State and prove the theorem on trihedral angles which corresponds to the preceding . 414 SOLID GEOMETRY. 714. Problem. On a given sphere to construct a spherical triangle ivhen its sides are given. Solution. Let o be the given sphere, and a, b, c the arcs of the required triangle, and let .l.l' be any diameter of the sphere. With ^ as a pole, construct circles DBE and FCG whose polar distances from A are c and b respectively. With B as a pole, construct a circle HCK, whose polar distance from B is a. Tiien construct the three great circle arcs, AB, BC, CA. ABC h the required triangle. 715. BXERCISES. 1. What restrictions if any is it necessary to impose upon the three given sides of the triangle in § 714? (§§ lis:!. (iS|.) 2. In plane gednietry two congruent triangles may l^e constructed upon the same hase and on the same side of it. Is a corresponding construction possible on the sphere? 3. If in the aliove construction each of two sides of the required triangle is very great, that is, nearly a scnucircle. show fron\ the con- struction that tlie third side must be very small. 4. If line side of the proposed triaiinle in §714 were equ.al to or greatej' than 1st)', wh^- wouhl that nuike the construction impossible? THE SPHERE. 415 716. Problem. To construct a spherical triangle when its three angles are given. Solution. Let the three given angles be A, B, C, and let a',b',c' be arcs such that a' +Za = 180°, J' + Zb = 180°, c' + Z c = 180°. Then the triangle whose arcs are a', b', c' will be the polar triangle of the required triangle. This latter triangle A'b'c' may be constructed by the method of § 714. Then construct the polar triangle of A'b'c', which will be the required triangle. Give reasons in full for each step. 717. Problem. To construct a trihedral angle tchen its face angles are given. Solution. Construct the corresponding spherical triangle by the method of §714. Give the construction in full and prove each step. 718. Problem. To condiuct a trihedral angle when its dihedral angles are given. Solution. Construct the corresponding spherical tri- angle by the method of § 716. Give reasons in full for each step. 719. EXERCISES. 1. If two spherical triangles having angles respectively equal are constructed on the same sphere, how are these triangles related? Prove. 2. If two trihedral angles with face angles respectively equal are constructed as in § 717, how are they related? Prove. 3. If two trihedral angles each with the same dihedral angles are constructed as in §718, how are the trihedral angles related? Prove. 4. What restrictions if any must be placed upon the given angles A, 5, C in § 716? Compare Ex. 1, §715. 5. What restrictions if any are needed in Exs. 2 and 3 ? 416 SOLID GEOMETRY. 720. Theorem. Two spherical triangles having tivo angles and the included side of one equal reHpectively to two angles and the included side of the other are con- gruent, if the given equal parts arc arranged in the same order, and symmetricaJ , if arranged in the oppo- site order. Proof: By §§ 709 and 699 the polar triangles of the given triangles are congruent or symmetrical. Hence, by § 710, the given triangles themselves are congruent or symmet- rical. 721. Corollary. >Stnte and prove the theorem on trihedral angles which corresponds to the preceding. 722. Theorem. Two spherical triangles having the angles of one equal respecti velii to the angles of the other are congruent, if the equcd angles are arranged in the same order, and sj/mmctrical, if they are arranged in the opposite order. Proof: By §§709, 691, and 693 the polar triangles of the given triangles are equal or symmetrical. Hence, by § 710, the given triangles themselves are equal or sym- metrical. 723. Corollary. jState and prove the theorem on trihedral angles ivhich (■orrcsj)(})ids to the preceding. 724. Is there a tlieorem on piano triangles correspond- ing t(i that of § 7'22/ THE SPHERE. 417 725. EXERCISES. 1. If the sides of a spherical triangle are 60°, 80°, 120°, find the angles of its polar triangle. 2. If the angles of a spherical triangle are 72°, 104°, 88°, find the sides of the polar triangle. 3. If a triangle is isosceles, prove that its polar triangle is isosceles. 4. If each side of a spherical triangle is a quadrant, describe its polar triangle. 5. In case each side of spherical triangle ABCis a quadrant, show that the eight triangles formed by drawing the great circles whose poles are A, B, and C are all congruent. 6. Show that for any triangle the construction of the polar triangle gives four pairs of symmetrical triangles. 7. If a triangle ABC is isosceles, show that of the eight triangles of Ex. 5 there are four pairs of congruent triangles. 8. If the triangle ABC is not isosceles, show that of the pairs of triangles proved congruent in Ex. 7 none are now congruent. 9. If the angles of a spherical triangle are 70°, 80°, and 110° respectively, find the sides of each of the eight triangles formed by the polar construction. 10. Is it possible to construct a spherical triangle whose angles are50°, 60°, 120°? 11. Is it possible to construct a spherical triangle whose angles are 60°, 120°, 150° ? Suggestion. Consider the polar triangle of such triangle. 12. Consider the questions on trihedi-al angles corresponding to the two preceding. 13. If the sides of a spherical triangle are 75°, 95°, and 115° re- spectively, find the angles of each triangle formed by the polar construction. 14. How can the theorem of § 712 be used to prove that a side of a spherical triangle cannot be as great as a semicii-cle? _ 15. If it is given that a spherical triangle is equilateral, can we infer from the theorems thus far proved that its polar triangle is equilateral ? 418 SOLID GEOMETRY. POLYHEDRAL ANGLES AND SPHERICAL POLYGONS. 726, Theorem. The suvi of the face angles of any convex polyhedral angle is less than four right angles. p Proof : Let ABODE be a polygonal section of the given polyhedral angle. The number of triangles thus formed having P for a vertex is equal to the number of face angles of the polyhedral angle. Let o be any point in the base, and draw OA, OB, oc, etc. Then Z pbA + Z PBC > ZABC, and Z PCB + Z PCD > Z BCD, and so on. (Why?) But the sum of the A of the A OAB, OBC, etc., is equal to the sum of the A of the A PAB, PBC, etc. Hence, Z APB + Z BPC + ■■■ < Z A0B + Zb0C-\ . But the sum of tlie A about O is four right angles. Therefore, the sum of the face angles of the polyhedral angle is less than four right angles. 727. Definition. The section of a sphere made by a convex polyhedral angle whose \ ertex is the center of the sphere is called a spherical polygon. Since a plane may be passed through the vertex of a polyliedral angle sueh that the polyhedral angle lies entirely on one side of it, it follows that a spherical polygon lies within one hemi- sphere. THE SPHERE. 419 728. Corollary. State and prove the theorem on spherical polygons ivhich corresponds to that of § 726. 729. Theorem. The sum of the angles of a spherical polygon of n sides is greater than 2{n—2) right angles and less than 2 n right angles. Proof : Divide the polygon into n — 2 triangles. Hence, by § 712 the sum of the angles is greater than 2(w — 2) right angles. Since the polygon has n angles, and since each angle is less than two right angles, it follows that their sum is less than 2 n right angles. 730. Corollary. State and prove the theorem on polyhedral angles ivhich corresponds to the p)Teceding. AREAS OF SPHERICAL POLYGONS. 731. Definitions. A spherical polygon divides a sphere into two parts, an exterior and an interior, so that every path on the sphere passing from one to the other must cross the polygon. We have seen (§ 727) that any spherical polygon lies within one hemisphere. The interior is that one of the two parts which lies entirely within this hemisphere. Two spherical polygons are said to inclose equal areas or to be equal if they are congruent, or if they can be divided into polygonal surfaces which are congruent in pairs. The terms spherical triangle, spherical polygon, are sometimes used to refer to the part of the sphere inclosed by these figures. The context will always indicate clearly in which sense they are used. 420 SOLID GEOMETRY. 732. Theorem. Tuxj symmetrical spherical tri- angles are equal. Proof : Let ABC and a'b'c' be the given triangles. Extend the radii OA, OB, or to meet the sphere in A^, Bj, Oj, thus forming a triangle symmetrical to A ABC (§ 688), and hence congruent to A a'b'c' (§ 698). Let P be a pole of the circle through a. b, c. Extend PO to meet the sphere in Py Draw pa, pb, PC, and Pj^j. PjBj, and PiCy Suppose that P lies within Aabc. Now prove that A PAB ^ A P^A^B^ A PAC ^ A P^i^Cy A PBC ^ PjBjCi. Note that these triangles are isosceles. Hence, show that AABf = A A-^B^C^ and therefore A ABC = A a'b'c'. 733. Definitions. A lune is a figure formed b}- two great semicircles having the same end-points. The angle between these semicircles is the angle of the lune. A birectangular spherical triangle is one having two right angles. If one angle of a birectangular triangle is 1°, the triangle incloses one of 720 equal parts of the sphere. The area THS SPHERE. 421 inclosed by such a triangle is called a spherical degree and is used as a unit of measure of areas inclosed by spherical polygons. In a similar manner we define a spherical minute and a spherical second. 734. Theorem. Tlie area inclosed by a lune in terms of spherical degrees is twice the angle of the lune. 735. Definition. The number of spherical degrees by which the sum of the angles of a spherical triangle exceeds 180° is called the spherical excess of the triangle. 736. Theorem. The area of a spherical triangle in terms of spherical degrees is equal to its spherical excess. Proof : We are to show that area AAIiC = ZA + AB+ZC-lSO°. Consider the lunes ACDB, CAEB, BCFA. We have A ABC + AbCD = ACBB = 2 Z ^. A ABC + ABAE = CAEB = 2 Z C. A ABC + A GFA = BCEA = 2 Z B. Hence, adding, 3 A ABC + A BCD, BAE, CFA = 2(^ZA+Zb + Zc-). Now A BCD and AEF are symmetrical and have equal areas. Hence, 2 A ABC + A ABC, AEF, BAF, CFA = 2(Z A + Z B +Z C). But A ABC, AEF, BAF, CFA together constitute a hemi- sphere or 360 spherical degrees. Hence, 2 A ABC + 360° = 2 (Z a + Z B + Z c). Solving, A ABC = ZA + /LB+ZC- 180°. 422 SOLID (lEOMETRY. 737. Definition. The spherical excess of a spherical polygon is the sum of its angles less (n — 2)180°, where n is the number of sides of the polygon. 738. Theorem. The area of a .yjherical polygon in terms of -yjJierical de/jrees la equal to its spherical excess. Proof : Join one vertex of the polygon to each non- adjacent vertex, thus forming w — 2 spherical triangles. Now prove that the sum of the spherical excesses of these triangles is the spherical excess of the polygon and thus complete the proof. 739. EXERCISES. 1. AVIiat is the area in spherical degrees of a birectangular tri- angle one of whose angles is .54"? If one angle is 7!'' oO' ; loij- : 14' ; :3lJ" V 2. What is the area inclosed hy a lune whose angle is 4.") ? Note that the lune may lie divided into two birectangular triangles. What is the third angle of each ? 3. Between what limits is the sum of the angle-i of a spherical polygon of eight side.s? 4. Tf the sum of the angles of a spherical polygon is 11 right angles, what is known about the number of its sidi's? 5. If the sum of the angles of a spherical ]>olyg'on is 14 right ani;li's, what is known about the munber of its sides? 6. The sides of a spherical polygon are 8.')', Il.'i , lilt . I'iud the area of each of the eight triani;les formed by the polar construction fnim this triani;l('. 7. 'I'lic aira of a sjiherical triangle is 71 spherical degrees. One angle is 1()."i'- ()!' the other two angles one is twice the other. Find all the angles of the triangle. THE SPHERE. 423 AREA AND VOLUME OF THE SPHERE. 740. In §§ 731-738 we discussed the areas inclosed by spherical polygons iu terms of a unit directly applicable to the sphere, namely, the spherical degree. We now consider the area of the sphere in terms of a plane Unit of measure. It is clear that such measurement can be approximate only, since no plane segment however small will coincide with the spherical surface. Similarly, if the cube is used as the unit of volume, the measurement of the volume inclosed by a sphere must be approximate, since no set of cubes however small can be made exactl)' to coincide with a sphere. The two following theorems are needed: 741. Theorem. The lateral area of a frustum of a right circular cone /'*■ ecjual to the altitude of tlie frustum multij)Ued Jjy the length of a circle ivhose radius is the perpendicular distance froon a jjoint in the axis of the frustum to the middle point of an element. Given AA' an element of the frustum whose middle point is £', CC the axis of the frustum, BD± CC and EB1.AA'. To prove that the lateral area is equal to 2 tt • EB • CC'. Proof: Bj- § 619, the lateral area is 2 7r • BD • AA'. Hence, we must show that EB ■ CC' = BD ■ AA'. To do tills, draw A'f± .IC and show that A AFA' ~ A edb. 742. CoEOLLART. The lateral area of a cone is equal to its altitude times the length of a circle whose radius is the perpendlcida?' from a point in the axis to the middle point of an element. 424 SOLIB GEOMETRY. 743. Theorem. Given a fixed line through the ver- tex of a triangle, but not crossing it. The volume swept out by the triangle as it rotates about the fixed line as on axla is equal to the numerical measure of the area generated by the side opposite the fixed vertex multiplied by one third the altitude upon that side. i>v Fig Given A ABC with vertex A in the fixed line /. Case 1. When one side as AB lies in line I. (Fig. 1.) Draw AD ± BC, or BC produced, and CE±AB. Then Vol. ABC= Vol. AEC+ Vol. BEc = '^^vi-r-AB. (Why;') But CB^ ■ AB=CE ■ CE- AB = CK ■ liV AD. ( Wliy ') And by § (11:5, tr ■ CE ■ BC is the area swopt out by BC. Henee, \'ol. ABC = ^ AD • (area generated by lie). Case 2. Wfien neither AB nor AC liett in line I. (Fig. 2.) Produce CB to meet I in F and thaw .iD and CE as before. THE SPHERE. 425 N Then Vol. ABC = Vol. AFC — Vol. AFB. By Case 1 Vol. AFC = ^ AD ■ (area generated by FC), and Vol. 4FB = J ^D • (area generated by FSy. Hence, Vol. ABC = ^ AD ■ (area generated by FC — FB'), or Vol. ABC = ^ AD ■ (area generated by BC}. Let the student give the proof when the triangle is of other forms, e.g. in Case 1 when AD falls loithin the triangle, and in Case 2 when BC is parallel to I. 744. Area of the sphere. About a circle circumscribe a polygon as follows : Construct two diameters AB and CD at right angles to each other and divide each quadrant into an even number of parts by points, as D, E, F. At each alternate division point, beginning with the first point D, draw a tangent. There results a regular poly- gon with vertices on the diam- eters AB and CD. If now we construct another polygon in the same manner by dividing each quadrant into twice as many arcs, it will like- wise have two vertices on each of the diameters AB and CD, and this may be repeated at pleasure. Now inscribe a polygon similar to the one just circum- scribed by joining the points C and E, E and A, and so on. If now the whole figure is made to revolve about AB as an axis, the circle generates a sphere, and the circumscribed and inscribed polygons generate sets of circumscribed and inscribed cones and frustums of cones. We assume the following : "^1%^ \1d Jxf ^.J^^^___^^^ ^^^^ M 4:^(1 ,S0LZO GEOMETRY. 745. Axiom XXIV. A sphere hds a dfp'nite area and wrloscs a definite volume loluch are less respectively titan the volume and the surface of amj eircumserihed jhjnrc and (jreatcr than those of an if inscribed convex fiijnrc. 746. Theorem. Tlie area of a sphere is 4 Trr. Proof : Denote liy r the radiu.s of the circle which gener- ates the spliere. That is, in the figure, r = 0F= 0D= OG. X ^^^ ■^^^F i 1\ 1 Mil \ '\ '-' / JG ^^ v-.^ / ^-^^i? /"^^ Then by § 741 the lateral areas of the frustums whose axes are Oh anil OK are 'Iirr ■ OL and 2 Trr • OK, and by § 742 the lateral areas of the cones whose axes are l._Yand KM are respectively 2 tt;- LX and 2 Trr • KM. Hence, the total surface of the ■whole eireuniseribed figure 2 Trr • MK + 2 Trr KO + 2 irr ■ OL + 2 vrr LX. or 2 Trr (MK + KG + OL + LX ) = - irr • MX. If now a iiolygon of twice the number of sides is con- structed, in a manner similar to the above, we obtain another eireuniseribed ligurc whose area is 2 Trr • m' X' whore M' and X' arc llie two vertices on the line AB. THE SPHERE. 427 As this process is repeated indefinitely, the vertices which lie on the line AB may be made to approach as near as we please to A and B, and hence the total surface gen- erated approaches as near as we please to 4 = 2 Trr • AB = 4 TTJ^. We now prove that A cannot differ from 4 irr^. (a) Suppose ^ > 4 tt?^, and let ^ = 4 '7rr^-\-d. (1) Let k be the difference between m'n' and AB. Then the area of the circumscribed figure is 2 Trr (2 r + A;) = 4 -n-r^ + 2 irrh. Since k can be made as small as we please, 2 irrk can be made smaller than d. Hence, by Axiom XXIV, the area > 4 7rr^ + d, which contradicts the hypothesis (1). Therefore the area cannot be greater than 4 ttj^. (b~) In the inscribed figure let OF' be the apothem. Then by § 741 and an argument like that used above, we find that the area of the inscribed figure is 217 ■ AB ■ of' = 4:7rr ■ OF'. By continuing to double the number of sides, OF' may be made to approach as nearly as we please to OF = r. Suppose that A <\ irr^ and let ^ = 4 irr^ — d'. (2) Let k' be the difference between r and OF' . Then the area of the inscribed figure is 4 irr ■ of" = 4 7rr(r — k') = 4 ttv^ — 4 irrk'. Since k' can be made as small as we please, ^Trrk' can be made less than d'. Hence there is some inscribed fig- ure whose area is greater than iirr^ — d'. Hence by Axiom XXIV, ^ > 4 ttt^— d', which contradicts (2). Therefore the area cannot be less than 4 irr^. Since the area is neither less nor greater than 4 Trr\ it is therefore equal to 4 ttj^. 428 SOLID GEOMETRY. lil. Theorem. Jlie volume of a sphere is f tt?^. Proof : From § 743 tlie volume of the figures formed by- rotating tlie triangles oia and OHN about AB as an axis IS ~ (area generated by IH and BN). N ^^^ L / ''' i\o l>'-^""\ \d V\/ ^^^B J^ M That is, the volume of the whole circumscribed figure is equal to the surface of the figure multiplied by -- o Now suppose the volume of the sphere not equal to I Trr^, but equal to ^ ttt^ + d. Since by the argument of § 746 we can obtain a circum- scribed figure whose surface is as near as we please to 4 7rr^, it follows that one can be found whose volume is less than ^ irr^ + d. Hence, the volume cannot be greater than ^ trr^. In a similar manner, using the inscribed figures, show that the volume cannot be less tlian | -wi-^. 748. The volume of a sphere by inspection. The fact that the volume of a sphere is ecjual tn its area multiplied by one third its radius may be inferred directly from the accompanying figure. THE SPHERE. 429 The sphere is covered with a network of small spher- ical quadrilaterals. If these are taken small enough, they may be regarded as approximately plane surfaces. On this supposition we have a set of pyramids with a common altitude r and the sum of their bases approximately equal to the area of the sphere. Hence, their combined volume is Jr x (area of sphere) or |- r • 4 7rr^. That is, the volume is | irr^. It is clear that, by making these quadrilaterals suffi- ciently small, this result may be approximated as nearly as we please. 749. EXERCISES. 1. The surface of a polyhedron circumscribed about a sphere of radius 4 inches is 420 square inches. Find its vohime. 2. The volume of a polyhedron circumscribed about a sphere of radius 3.5 inches is 450 cubic inches. Find its surface. 3. Given a sphere of radius 6 inches, is there any upper limit to the volume of its circumscribed polyhedrons ? That is, can polyhe- drons be circumscribed having a volume as large as we please? 4. On the same sphere is there any lower limit to the volume of its circumscribed polyhedrons ? 5. Show that the areas of two spheres are in the same ratio as the squares of their Tadii or of their diameters. 6. Show that the volumes of two spheres are in the same ratio as the cubes of their radii or of their diameters. 7. Find the area of a sphere whose radius is 8 inches. 8. Find the volume of a sphere whose radius is 10 feet. 9. If the area of a sphere is 227 sq. ft., find its radius. 10. If the volume of a sphere is 335 cu. in., find its radius. 11. If the volumes of two spheres are 27 cu. in. and 729 cu. in., compare their radii. 430 SOLID GEOMETRY. 750. Definition. That part of a sphere included between two parallel planes cutting it is called a zone. The per- pendicular distance between the planes is the altitude of the zone. The figure formed by a zone, together with the circular plane-segments cut out by the sphere, is called a spheri- cal segment, and the circular plane-segments are its bases. If one of the cutting planes is tangent to the sphere, then the spherical segment and the corresponding zone are said to have but one base. The altitude is the perpen- dicular distance from the base to the tangent plane. If one nappe of a convex conical surface has its vertex at the center of a sphere, the portion cut off by the sphere, together with the intercepted part of the spherical surface, is called a spherical cone. If two spherical cones have the same axis, one lying within the other, the figure formed by their two lateral surfaces, to- gether with the part of the sphere inter- cepted between tliem, is called a spherical sector. If the two cones are right circular cones, they intercept circles on the sphere, and tlie zone thus included is called the base of tlie spherical sector. If the accompanying!; figure he revolved aliout i^Uas an axis, then any arc, as (ID or .1//), generates a zone, thr former with two bases, the latter with one. The figure MDF or FDGE generates a spherical segment, the former with one liasr, the latter with two. The figure CI L or CBL generates a splierical cone, CL being tlie common axis. 'J'lie figure ('/>'. I generates a spherical sector. (■ y. F M ^' -' XR — 1 ^n _^ THE SPHERE. 431 751. Area of a zone. An argument precisely like that of § 746 shows that the area of a zone is A = 2 irrft where h is the altitude of the zone. That is, instead of AB, the diameter in case of the sphere, we should have the sum of the altitudes of the frustums circumscribed about the zone equal to A, the alti- tude of the zone. 752. Volume of a spherical cone. An argument pre- cisely like that of § 747 shows that the volume of a spherical cone is v—'' A where A is the area of the zone cut out of the sphere by the cone. Hence, if h is the altitude of this zone, we have 3 3 In like manner the volume of a spherical sector is 753_ EXERCISES. 1. The radius of a sphere is 6 and the altitude of a zone is 5. Find the area of the sphere and of the zone. 2. The area of a zone is 36 tt and its altitude i. Find the radius of the sphere. 3. If the radius of a sphere is r, find the perpendicular distance from a point A to the plane of a given great circle if the distance on the sphere from A to ,_ "^'"XA the great circle is 35°. r l^j_-^r--*?0 gg,„ 4. Solve the preceding problem if the dis- {::'_'. .il^ ."1~J) tance on the sphere from A to the given great circle is 23^°. Also if the distance is 66|°. (Use the tables, page 335.) 432 SOLID GEOMETBY. 754, Problem. To find the volume of a spherical segment. Solution. Let r be the radius of the sphere, and r^ and r,^ the radii of the bases of the segment, a the altitude of the zone, and let the segment be generated by revolving the figure ACDB about ^O as an axis. ZTT We have Vol. generated by ODB = "^r^a. §752 Vol. generated by OAB =—r^(a-\-d~). (Why?) o Hence, From w^e obtain Vol. generated by OCB = "^r^d. = I [2 r-'a + r^a + d{r^ - r^")] . 'fi=r^ -\- 6^ and r^ = r^^ + (a + (f)^ d: fn — ^1 — a (Why?) (1) (2) Substituting this value of d in r~ = r^ + d^, we gret r^ = -^ — ■ — ^ — ■ 1 — ' — ' — ■ '- 4 a- (3) Substituting (2) and (3) in (1) and reducing, we have ira (n^ + r.=) + ^ 0^ = ? (irrr+ ^r.?) + ^ irf^ 2 -' 6 2 Hence, we have the result : TriEoiiEM. The voluDie of a spherical seijine)it is nnmerieallti equal to the sum of the areas of its bases multiplied hij half its altitude, plus the volume of a sphere mheisc diameter is equal to the altitude of the seipment. THE SPHERE. 433 SUMMARY OP CHAPTER XTI. 1. Collect the theorems involving plane sections of the sphere. 2. Collect the definitions involving plane sections of the sphere. 3. State some of the principal exercises and problems connected with plane sections of the sphere. 4. Collect the definitions on trihedral angles and spherical tri- angles. 5. Arrange in parallel columns the pairs of theorems on trihedral angles and spherical triangles, which are proved without the use of polar triangles. 6. Collect the definitions on polar triangles. 7. Collect the theorems on polar triangles. 8. Continue the lists started in Ex. 5, adding the theorems proved by means of polar triangles. 9. Make a list of the definitions involving polyhedral angles and spherical polygons. 10. Collect the theorems involving polyhedral angles and spherical polygons. 11. Collect the theorems on the areas of spherical triangles and polygons. 12. Give the definitions and axioms pertaining to the area and volume of the sphere. 13. State all the theorems pertaining to the area and volume of the sphere. 14. Give the definitions and theorems pertaining to spherical figures, such as zones, cones, sectors, segments. 15. Collect all the mensuration formulas in this chapter. 16. Collect all the mensuration formulas of Solid Geometry. 17. Which of these formulas are illustrations of the general theorem that the surfaces of similar solids are in the same ratio as the squares of their corresponding linear dimensions? Which ones are special instances of the theorem that the volumes of sinnilar solids are in the same ratio as the cubes of corresponding linear dimensions? 18. Describe some of the most important applications in this chapter. Return to this question after studying the following list. 4i54 SOLID GEOMETRY. PROBLEMS AND APPLICATIONS. 1. What part of the earth's surface lies in the torrid zone? What part in the teinju'ratc- zones? A\'hat part in the frigid zones? The parallels '±1Y north and south of the equator are the boundaries of the torrid zone, and the parallels 07 ^' north and south are the boundaries of the frigid zones. 2. Fuid to four places of decimals the area of a sphere circum- scribed about a cube wliosp edge is (J. Xi> square root is to be approximated in the process, and the value of tt is taken 3.1-116. 3. Can the volume of the sphere in the preceding exercise be approximated without finding a square root? Find tlie volume. 4. Find the area of a sphere circumscribed about a rectangular paraUelopiped whose sides are a, b, and c. 5. Find the volume of the sphere in the preceding example. 6. ^V fixed sphere with center lias its center on another sphere with center 0'. Show that the area of the part of 0' which lies within is equal to the area of a great circle of the sphere O, provided the radius of the sphere is not greater than the diameter of 0' . Su(:(iESTioN". Let the figure represent a cross section through the centers of the two spheres. Connect with .4 and B. Then OA"^ =0B X OD. But OD is the altitude of the zone of W. which lies within 0, and OB is the diameter of the sphere 0'. Hence, the area of the zone is tt BO x OD = ir OA'-. 7. (iiviMi a solid sphere of radius 12 inches. A cylindrical hole is liiired through it so that tli(> axis of the cylinder passes througli the center of the sphere. What area of the sphere is removed if the diameter of the hole is 1 inches? 8. Find the volume removed from tlie sphere by the process described in the precedini;' exercise. 9. Through a sphere of i-adiiis r a liole of radius ;■, iS bored so that the axis of the eylhnler cut cuit passes through the center of the sphoe. Find the volume o! the splieiv which remains. THE SPHERE. 435 10. A cylindrical ppst 6 in. in diameter is surmounted by a part of a sphere 10 in. in diameter, as shown in the figure. Find the surface and the volume of the part of the sphere used. 11. A cylindrical post 5 ft. long and 4 in. in diameter is turned so as to leave on it a part of a sphere 7 in. in diameter and having its center in the axis of the post. Find the volume of the whole post. 12. Find the volume of a spherical shell 1 inch thick if its outer diameter is 8 inches. 13. What is the diameter of a spherical shell an inch thick whose volume is half that inclosed within a sphere of the same diameter V 14. Compare the volumes and areas of a sphere and the circum- scribed cylinder. 15. In a sphere of radius r inscribe a cylinder whose altitude is equal to its diameter. Compare its volume and area with those of the sphere. 16. In a sphere inscribe a cylinder whose altitude is n times its diameter. Compare its area and volume with those of the sphere. 17. Compute the length of the diagonal of a cube in terms of its side, and also the length of the side in terms of half the diagonal. 18. Express the volume of a cube inscribed in a sphere in terms of the radius of the sphere. 19. Three spheres each of radius r are placed on a plane so that each is tangent to the other two. A fourth sphere of radius r is placed on top of them. Find the distance from the plane to the top of the upper sphere. 20. Find the vertical distance from the floor to the top of a triangular pile of spher- ical cannon balls, each of radius 5 inches, if there are 3 layers in the pile. 21. Solve a problem like the preceding if there are 16 layers in the pile, each shot of radius r. 22. Solve a problem like the preceding for a square pile of shot of 12 layers. 23. Solve Exercises 21 and 22 if there are n layers in each pile. CHAPTER XIII. VARIABLE GEOMETRIC MAGNITUDES. GRAPHIC REPRESENTATION. 755. It is often useful to think of a geometric figure as continuously varying in size, or in shape, or both. E.g. if a parallelepiped has a fixed base, say 24 square inches, but an altitude which varies continuously from 3 inches to .5 inches, then the volume varies continuously from .3 • 24 =72 to 5 ■ 24 = 120 cubic inches. We may even think of the altitude as starting at zero inches and increasing continuously, in which case the volume also starts at zero. From this point of view many theorems may be repre- sented graphically. The graph has the advantage of ex- hibiting the theorem at once for all its particular oases. For a description of graphic representation, see Chapter V of the authors' High School Algebra, Elementary Course. 756. Theokem. The volumes of two jyarallelopipeds having equal bases are in the same ratio as their altitudes. Graphic Representation. By § 541 we have Volume = base X altitude. Consider parallelopipeds each with a base whose area is ,4, and with altitudes h.\, h^, hs, and corresponding volumes T'l. T'.., I'j. Then J:i=iL^ = *i, H^-Ah^h^etc. (1) Consider the case where .1 =10. Let one horizontal space repre- sent one unit of altilude, and one verlieal spaee ten units of volume. Thus, the point P, lias the ordinate 1' = 10 vortieal units (repre- senting 100 unitsof \ohiino), ami the abscissa ^2= 10 horizontal unit.s. 430 VABIABLE GEOMETRIC MAGNITUDES. 437 Similarly, locate the points Pi and Pg. Using equations (1), show that 0, Pi, P^, Ps lie in a straight line. A. i a ^J / / m I ./ M ^ ■ r^ / 1 r ~ ~~ / ~ a V / p ^ > / ]■ in 1(1 / ; ^ — 1 1 ■ — -^ ^^ 1 — L— V / / > / ■V - - / / 7i, / 1 V r- ^ -^ , |i / / 1 / 'l / ) II cA E Ait tud i — Ax IS >n If we suppose that while the base of the parallelopiped remains fixed, the altitude varies continuously through all values from ^j = 10 to Ag = 20 = 2 x 10, then the volume varies continuously from 10 x 10 = 100 to 10 x 20 = 200. Using any altitude as an abscissa and the corresponding volume as an ordinate, show, as in § 368, that the point so determined lies on the line OP^P^. 757. The preceding theorem may also be stated: Tlie volume of a parallelopiped with a fixed base varies directly as its altitude. This means that if T' and h are the varying volume and altitude, and F'j and Ji^ the volume and altitude at any given instant, then i_ = _ or F = — h.OT V = kh, where k is the fixed mtio —^■ Fi hi hi ^1 The graph representing the relation of two variables when one varies directly as the other is a straight line. 438 SOLID GEOMETRY. 758. Problem. To represeid fjrapliicnl.hj the rela- tion hefineen the area and the edge of a cube a-s the edge varies continuoushj. — - :-- I i /--' 4- 1 : . ! i ! 1 ' ' I i 1 1 1 i 1 1 ' ^ Ml ^ ' 1 1 "5 / 1 ( / / t/ 1 ■^1 ' ^ J s -f ^-^ f-r/- 1 ■ i 1 ; 1 ^i: ' ! ■ 1 i 1 1 / -[J- — ' i> / I 1 i / ___.J_ 1 ( P-HX ' ■ 1 — / < 1 / --, >/- ^ i^\/- M ' 30 -. L tof O-iD/l n-^y 1 1 -h/- ' 1 ' V -^ i ' i \ I 1 ■ ,/l V^ "~7 i ■ 1 ■ ^1 — ^ ,., ^^ -/ M- V ---^ '-r ! /L i_ / ; ii II.' 'ill! y ' 1 ' / "-/^^■-■~ 1. ±b^^ 1 „ "Hjq " , / V - /__._ y , - --^ -— ] ~- "~ '. " ' 1 ' ' __^±T 1 M ■ 1 _^Li 1 1 — /— r^ -rf^/T n^ 'i-^-i- zjiiziT :^;_~ j±:a" ^:7~: " -I. 7 J^ - - - - ■ '- 1 J /y ' U I -J :l I 6 I. r ¥ A.i-is fur sides Solution. Oil till' liori/.iiiital axis lay off tii.'L;nients equal Id the \ ai'idiis values of tlu' edge (', ami on the veilieal axis lav olT si'Ljnieiits e([iial to the eorrespoiuling areas .1. VARIABLE GEOMETRIC ilAGNITUDES. 439 If one vertical space represents one unit of area, then the points Pj, P^, P3, etc., lie on the steep curve. The student should locate many more points between those here shown, and see that a smooth curve can be drawn through them all. 759. The graph of the relation between two variables, one of which varies as the square of the other, is always similar to the one just given. The lowest curve represents the relation between the side of a square and its area, and the third curve rep- resents the relation between the edge of a regular tetra- hedron and its total surface. In each of these cases the area is said to vary as the square of the side of the figure. These are special cases of the theorem, § 654, that the surfaces of similar figures are in the same ratio as the squares of their corresponding near dimensions. This theorem may also be stated : The areas of similar figures vary as the squares of their linear dimensions. 760. EXERCISES. 1. From the gi-aph find approximately the area of a regular tetra^ hedron whose side is 2.5; one whose side is 3.7; 4.3. 2. Find approximately from the graph the edges of cubes whose total areas are 25 square units ; 40 square units ; 56 square units. 3. Find approximately from the graph the edges of regular tetra- hedrons whose total areas are 42 square inches; 17 square inches; 55 square inches. 4. Construct a graph showing the relation between the edge aud the total area of a regular octahedron. 440 SOLID GEOMETRY. 761. Problem. To construet a graph shoiving the relation hetivern the edge of a cube and its volume. 1 [ I 1 ^„_ . n — ,- )\(\ ' 1 1 " 1" M ■ 1 — — „_ — 7 —.11 ~cl 11 ji^- '-:-'---^/ -' ' ■ :' ..T . , — ^ ; i II _y— ^ "^ tnnz -^Ittr: -: :-; — -_vr-:v /=r 7Z-- i^-rr ■l4i!--t-H-" ■ / -^ ^^7 s^JL' ~L " ~r ■-TIL "".."::: ;;:i:/^W/!^ ITzE ''\L'-^- ■ ; i; ^ 1 ^ - 4 y i -i] ! 1 : ' 1^'''^ -■■i V^^'i- viii^: '^-" "ft .l.iiN/or .s-ic/t's VABIABLE GEOMETRIC MAGNITUDES. 441 Solution. Taking ten horizontal spaces to represent one unit of length of side of the cube and one vertical unit to represent one unit of volume, construct the middle graph shown on the page opposite. 762. The lower graph represents the relation between the length e of the edge of a regular tetrahedron and its volume V; and the iipper graph represents the relation between the radius and the volume of a sphere. In each of these cases the volume is said to vary as the cube of the given linear dimension. These are special cases of the theorem, § 654, that the volumes of similar solids are in the same ratio as the cube of their ratio of similitude. This theorem may now be stated: The volumes of similar solids vary as the cubes of their linear dimensions. 763. EXERCISES. 1. From the graph read off appi-oximately the cubes of 1.3; 2.4; 3.7. 2. From the same graph read off approximately the cube roots of 17; 46; 54; 86. 3. Find approximately by means of the graph the volume of a sphere whose radius is 1.5 ; 2.3 ; 2.7. 4. What is the radius of a sphere whose volume is 26; 52; 71; 80? 5. Construct a graph giving the relation .between the volume and the length of a side of regular octahedrons. 6. By means of the graph just constructed find the volume of a regular octahedron whose side is 1.3; 2.7; 3.6. 7. From the graph constructed under Ex. 5 read off approximately the lengths of an edge of a regular octahedron whose volume is 18 ; also of one whose volume is 46. 442 80LID GEOMETRY. 220 :nT- 2]0i*: V'4 ^ 1 • -\'V -1- T — 1" -.J — _ -M — 1- — m — is ,1 — n +±1^ ::4^ TT j— I^Z 1 1 '• — ^ 1 — --^ ^ ^=h — 160^ 150-4+ ,40^ — — / ^ ^= 130 r^t^ ^^ ISO^TX lio^rr 1-;-— L- -j ^ ^ , r r'^r * ~^7 1^ — ^ — 100^ 90-^- . 1 1 . 1 1 1 =i 1 70^ 60=^ so=n4t 40 ^iJ: "J H / 41 ^1 ^ — - — — ir "^1^ / 'A / -^- - 1 20 — rhf — ^ r— ~ 10 ~^— TTf TT| 1 1 • ^ /\ y / 1 II ^-^ ^ ■• '• i 1 1 % 8 4 5 6 Aivisfor sides VARIABLE GEOMETRIC MAGNITUDES. 443 764. The graph on the opposite page exliibits the varia- tion of the area and the volume of the cube. We notice that for small values of e the numerical value of the area is greater than that of the volume. For e = 6 these values are equal, and for e greater than 6, the numerical value of the volume is greater than that of the area. This is an instance of the general fact that if a solid figure increases in size but remains similar to its first form, then after a certain point its volume increases more rapidly than its area or the area of any cross section. 765. The fact that one variable y varies as the square or the cube of another variable x is expressed algebraically by the equations y = kaP' and y = Ici?, respectively, where A is a constant number to be determined in any particular case. If the value of Tc is known, the relations represented by these equations may be shown by a graph like those on the preceding pages. The equation y = kx^ may be plotted by multiplying the ordinate of each point of the graph V= e^ (see page opposite) by the value of k. SUMMARY OF CHAPTER XHI. 1. Give a brief summary of the graphic representation process in general and in particular, as applied to geometrical figures. 2. Make a list of the theorems which are represented graphically in this chapter. 3. Show in what respects such a representation has advantages over the algebraic representation. 4. State as formulas the important laws of variation for geometric magnitudes discussed in this chapter. 5. Make a collection of important applications given in this chap- ter, including those which occur in the following set. 444 SOLID GEOMETRY. PROBLEMS AND APPLICATIONS. 1. Assuming that the weights of schoolboys vary as the cubes of their heights, construct a graph representing the relation between their heights and weights, if a boy 5 feet 8 inches tall weighs 130 pounds. Suggestion. If w represents the number of pounds in weight and h the number of feet in height, ir — kh^. From w = 130, when h = 5|, we have /c = .71i. For the purpose of the graph, k = .7 is accurate enough. Hence, we obtain the required graph by multiply- ing each ordinate of the graph of V = e^ by .7. 2. From the graph constructed in the preceding example find the weight of a boy 5 feet tall ; one 5 feet 4 inches ; one 5 feet 6 inches. Compare with the weights of boys in your class. 3. If a man 6 feet tall weighs 18.') pounds, construct a graph rep- resenting the weights of men of similar build and of various differ- ent heights. 4. If steamships are of the same shape, their tonnages vary as the cubes of their lengths. The Mauretania is 790 feet long, with a net tonnage of 32,500. Construct a graph representing the tonnage of ships of the same shape, and of various different lengths. Other ships which at one time or another have held ocean records are: the Deutschland, length GSii ft. and tonnage 16,.")00: the Kaiser Wilhelm der (7 ;■».<,«', length 64,s ft. and tonnage 14,:)iiii: the Lucania, length OJ.) ft. and tonnage 13,000 (nearly) ; and the Etruria, length iVJO ft. and tonnage 8000. By means of this graph decide whether or not these boats have greater or less tonnage than the ^^aurel(lnill as compared with their lengths. XoTE. The following more general problems further ilUistiate the wide range of aii]ilicali()n of the fundanieiital theorem that the sur- faces and the volumes of similar solids vary respectively as the squares and the eubes of tlieii- ooirespcmding linear dimensions. These prob- li'ms need not be solved by means of graphs. 5. Raindro]iH as tliey start to fall are extremely small. In the course of their deseiMit a great many unilo to form larger and larger drops. If 1000 such drops unite inio one, what is tlie ratio of the sur- face of the largo drop to the sum of the surfaces of the small drops? VARIABLE GEOMETRIC MAGNITUDES. 445 6. Consider two machines simil&,r in shape and of heights Aj and ^2- Since the tensile strength of corresponding parts varies as the areas of their cross sections, it follows that the tensile strengths of corresponding parts are in the ratio h^ : h^. But the total volumes of the machines vary as the cubes of the heights ; that is, the total weights are in the ratio li-^ : h^. Does this fact offer any hindrance to the building of machines indefinitely large? 7. The strength of a muscle varies as its cross-section area, which in turn varies as the square of the height or length of an animal, while the weight of the animal varies as the cube of its height or length. Use these facts to explain the greater agility of small ani- mals. For example, compare the rabbit and the elephant. 8. Assuming the velocities the same, the amounts of water flow- ing through pipes vary directly as their cross-section areas. How many pipes, each 4 in. in diameter, will carry as much water as one pipe 72 in. in diameter ? 9. What must be the diameter of a cylindrical conduit which will carry enough water to supply ten circular intakes each eight feet in diameter? 10. A water reservoir, including its feed pipes, is replaced by another, each of whose linear dimensions is twice the corresponding dimension of the first. If the velocity of the water in the feed pipes of the new system is the same as that in the old, wiU it take more or less time to fill the new reservoir than it did the old? What is the ratio of the new time to the old ? 11. If two engine plants are exactly similar in shape, but each linear dimension in one is three times the corresponding dimension of the other, and if the steam in the feed pipes flows with the same velocity in both, compare the speeds of the engines. 12. If two men, one 5 ft. 6 in. and the other 6 ft. 2 in. in height, are similar in structure in every respect, how much faster must the blood flow in the larger person in order that the body tissues of both shall be supplied equally well ? Suggestion. Xote that the amount of tissue to be supplied varies as the cube of the height, while the cross-section area of the arteries varies as the square of the height. THAPTER XIV. THEORY OF LIMITS, GENERAL PRINCIPLES. 766. In the Plane Geometry it was found that there are segments which have no common unit of measure, that is. which are incommensurable, and that the ratro of the lengths of two .such segments could be expressed only approximately by means of integers and ordinary frac- tions. Other incommensurables occurred in dealing with the length of the circle and the area inclosed by it. In Chapters III, IV, and V we confined ourselves to an informal first treatment of these incommensurable ratios, tacitly assuming their existence and computing them ap- proximately. In Chapter \'II their existence was explicitly assumed, and certain theorems proved rigorously on the basis of these assumptions. In the Solid Geometry the areas and volumes of the cjdinder, cone, and sphere have been treated accordinq- to this latter method. 767, In returning to this subject once more we fix our attention on the incommensurable ratios themselves, and the method of determining them, rather than on the process of approach and the practical computation liased on it. We have already used symbols such as \- to represent the ratio of the lengths of incommensurable segments. In general, the ratio between any two incommensurable giM)metric niaqiutudes of the same kind may be represented by wlial is called an irrational number: that is. a number which is ticithcr dii iiitci/cr nor it ijiiotii'iit of tivo iiiti-gerit. I III THEORY OF LIMITS. 447 768. The following method for determining irrational numbers is, for simplicity, applied first to the integer 1. Throughout this discussion the words " point on a line " and " number " will be used interchangeably. In a straight line mark a certain point (zero), and one unit to the right of it mark another point 1. Also lay off points such that their ^- ^ % l distances from are ^, f, |, \\, ■■■. If this sequence of points is carried ever so far, it -will never reach the point 1. If, however, we select a point k to the left of 1, no matter how^ near it, we may always go far enough along this sequence to reach points between k and 1. 769. The point 1 has two definite relations to this sequence. (a) Every point of the sequence is to the left of 1. (h) For any fixed point k to the left of 1 there are points of the sequence between k and 1. 770. We now note that 1 is the only point on the whole line such that both (a) and (6) are true of it. For every point to the right of 1 (a) is true, but (6) is not. For every point to the left of 1 (6) is true, but (a) is not. It follows therefore that, while the points of the sequence merely approach the point 1, the sequence, taken as a whole, serves to determine that point just as definitely as if the numeral 1 itself were used to indicate the point. 771. The sequence 3, 2\, 2^, 2^, ■•■ is a decreasing sequence, and the number 2 sustains relations to it similar to those described above. That is, {a') Every point of the sequence is to the right of 2. (&') For every point k to the right of 2 there are points of the sequence hetween K and 2. 448 SOLID GEOMETRY. 772. Definitions. An endless sequence of the sort just described is called an infinite sequence. Not every infinite sequence serves to single out a defi- nite point in the manner shown above. Thus the sequence 1, 2, 3,4, ... fails to do so, because its terms grow large beyond all bound. Such sequences are said to be un- bounded, while the sequence |, |, |, ... is bounded. Again, the sequence 1, 2, 1, 2, 1, ... fails to single out a definite point. This sequence is said to be oscillating, since its terms increase, then decrease, then increase, etc., M'hile ^, |, ^, ... is non-oscillating. 773. The number 1 is said to be the least upper bound of the sequence |, f, |, That is, 1 is the smallegt num- ber beyond which the sequence does not go. 1 is also said to be the limit of the sequence. Similarly, 2 is the greatest lower bound or the limit of the sequence 2|, 2|, 2^, ... ; that is, - is the greategt number such that the sequence contains no number less than it. 774. Axiom XXV. Every hounded increasing sequence has a least uj)pcr hound, and every hounded decreasing sequence has a grcnkst lower hound. This axiom may also be stated : Every hounded increat>ing or decreasing sequence has a limit. This axiom simply means that every such sequence singles out a definite number in the manner stated. Thus, if we attempt to approximate the square root of 2, we obtain a sequence 1,1.4, l.tl, 1.111, 1.4142..., having for its limit a dejinile number rejiresented by \"J, which corresponds to the length of tlie diagonal of a square whose side is unity. THEORY OF LIMITS. 449 775. If two sequences ctj, a^, ag, ... and ij, b^, ig, ... are equal term by term, that is, if aj = Sj, a^ = ij' *$ = ^3) • • • then they are one and the same sequence and hence by Ax. XXV they determine the same number. The same number may be defined by two different sequences. Thus each of the sequences, J, J, |, ... and |, f , f, ..., has 1 as its limit. We notice, however, that no matter what definite num- ber we select in either of these sequences, there is a number in the other greater than it. 776. Theorem. Two increasing bounded sequences define the same number as their limit if neither sequence contains a number greater than every number of the other. Let aj, a^. a^i ■■■ ^^^ ^v ^2' ^3' ••• denote two infinite sequences, with limits A and B, such that there is no a gi'eater than every b, and no b greater than every a. To prove that A = B. Proof : Suppose that A is not equal to B and that A is less than B. Then there must be numbers of the sequence 5 J, ^2, bg, ••• greater than A, since by § 769 there are num- bers of bp §2' ^31 ■•• greater than any fixed number whatever which is less than B, which contradicts the hypothesis of the theorem. , In like manner we show that B is not less than A. Hence A = B. 777. Theorem. Tioo decreasing sequences define the same number as their limit if neither sequence contains a number less than every number of the other. 450 SOLID GEOMETRY. APPLICATION OP LIMITS TO GEOMETRY. 778. Problem. On a given segment ab to lay off a seqninice of points B^, B^, B^, of loliich B is the limit, such that each of the segments AB^, AB.^, ab.^, ■■■ is commensurable with a given segment CD. c rm I 1 1 — A B • ' 1 tr^ Solution. Using Wj, an exact divisor of CD, as a unit of measure, lay off on AB a segment AB-^, such that the remainder B^B is less than my Then CD and AB^ are com- mensurable. Using a unit m^, likewise a divisor of CD, and less than BjB, lay off AB,^ such that B^B is less than m.^ Then AB^ > ABy and CD and AB^ are commensurable. Continuing in this manner, using as units of measure segments Wg, m^, •••, each an exact divisor of CD and each less than Bj B, B^ B, ■■■ respectively, we obtain a sequence of segments AB^ AB^, AB^, •••. each greater than the pre- ceding and each commensurable with CD. If tlie units Wj, m^, wig, ••■ are so selected that they approach zero as a limit, it follows that B is the limit of the sequence B^ B^, Bg, ••■. If a different sequence of divisors of CD, as »ii', m-i. m^' ■••, is used, •we obtain a sequence Bi', BJ , B/ •.., likewise satisfying the conditions of the problem. We note in regard tu any two such sec|uenoes B,, 5-., Bs, .--and •Bi', B2', B3', •■• determined as aliove, tliat they are both increasii)g and that each is sueh that no jioint of it is to the riglit of erery point of the other. llenee, by § 770 any two such seiiueiiees have the same limit B. THEORY OF LIMITS. 451 779. If two arcs, AB and CD, of the same circle or of equal circles are given, then, in the same manner as above, points JBj, Bj' ^31 ••■ ™3'y ^6 constructed on arc AB, forming a sequence whose limit is B, such that the arc CD is com- mensurable with every arc of the sequence AB^, AB^, AB^ ■■•. B..B, p AG 780. Definitions. If B^, B^, Bg, •■■ is a sequence of points on the segment AB having the limit B, then the segment AB is said to be the limit of the segments AB.^, AB^, AB^, ■■■. 781- The ratio of two commensurable segments has been defined as the quotient of their numerical lengths. The ratio of two incommensurable segments has not been explicitly defined, but it is now possible to do so in terms of the limit of a sequence. Consider two incommensurable segments AB and CD. Let ftp a^, ttg, ■■■ be the lengths of the segments each com- mensurable with CD, forming a sequence whose limit is the segment AB, and let b be the length of the segment CD. Then ^, ^, ^, ••• is an increasing bounded sequence b b b having a limit which we call R. If aj', a^', ag', ■•• are the lengths of another sequence of a ' a ' a ' segments whose limit is AB, the sequence -J-, -f-, -f-, ■■■ b is another increasing bounded sequence with limit r'. By § 776 we now know that B = r'. This number R is defined as the ratio of the segments AB and CD. 452 SOLID GEOMETRY. 782. The application of the theory of limits to geometry consists chiefly in showing that two numbers are equal because they are the limit of the same sequence, or of sequences having the property stated in the theorem of § 776. In the following paragraphs the applications are made to the chief cases both in plane and in solid geome- try. 783. Theokem. In the same circle or in equal circles the ratio of tioo central angles is the same as the ratio of their intercepted arcs. B2B1 Proof : In case the arcs are commensurable the proof is obvious. See § 413. If the arcs AB and CD are not commensurable, let ABy AB^y ABg, •■■ be a sequence of arcs whose limit is AB, each arc being commensurable with the arc CD. AB^ AB^ AB, Cd' Cd' CD by definition, is the ratio of the ares AB and CD. Then the sequence ■ ■ has a limit R which. Similarly the sequence 1, Z COD Z AOB, Z AOB, COD Z COD Z COD has a limit R' which, by definition, is the ratio of Z AOB and Z COD. AB^ _ ZAOB., AB^ Z AOBo CD Z COD ' CD Since Z COD ' these two sequences are identical and lionce defint same limit. Therefore it follows that r=r'. the THEORY OF LIMITS. 453 784. Theorem. A line parallel to the base of a tri- angle, and meeting the other two sides, divides them in the same ratio. Given the A ABC with DE II BC and a cutting AB and AC. To prove that ^ = ^. ■^B AC T. , s Proof: Consider the case when d^ AD and AB are incommensurable. ^/ Let z»j, iJj, Dg • • • be a sequence on AB whose limit is D. Through _-„ these points draw parallels to BC, meeting AC in ^j, E^., E^, ■■■. Then E is the limit of the sequence E^, E^, E^, ■■•. For suppose it is not, and that there is a point K on AE such that there is no point of E-^, E^, E^, ■■■ between K and E. Then draw a line parallel to BC through K, meeting AB in H. But there are points of the sequence D^, n^, Kg, between H and D, and hence points of the sequence Ej, i'g, E^, •■• be- tween K and E, which shows that E is the limit of E-y, E^, £g, •••. N AD^_AEj^ Ali^_AE^ AD^ _ AE^ AB~ Ac' AB~ AC' AB ~ Ac' Hence, the two sequences, AD, AD„ ADg ■, AE, AE, AE^ AB AB AB AC AC AC are identical and define the same limit ; ^^- , , AB AE that IS, — = AB AC 454 SOLID GEOMETRY. 785. Definitions. If a■^, a^, ag, •■• is a sequence with limit a, then ka^, ka^, ka^, ■■■ is a sequence with limit ka. If aj, ag, ttg, ■■■ and ij, b^, b^, ■■■ are two sequences -with limits a and b respectively, then ab is the limit of the sequence a^Sj, a^^, agig, •••. Similarly if aj, a^, ag, •■•, by b^, ig, ••• and Cj, Cg, Cg, ••• are sequences with limits a, J, c, then aSe is the limit of the sequence a^b^Cy «2^2'^2' ^s^a'^s' ■"• For a complete treatment it would be necessary to show that these definitions of multiplication of irrational numbers are consistent with the rest of arithmetic and also that these new sequences are such as to determine definite limits. That, however, is beyond the scope of this book. If the sides of a rectangle are incommensurable with the unit segment, we define its area as follows : Let flj, a^, ag, ••• be a sequence of rational numbers whose limit is the altitude a, and let Jj, b^, Jg, ••• be a sequence whose limit is the base b. Then the area of the rectangle is the limit of the sequence OjJj, a^Jg, a^bg, ■■■ b, l\b. But by definition the limit of o^b^ a.^b^, a^b^, ■■• is the product ab. Hence wc lia\eT- 786. Theorem. The ana of a rectangle is the product of its base and altitude. 787. Definition. In a circle inscribe a sequence Pj, Pj, Pg, ••• of regulai- polygons, each having twice the num- ber of sides of the one preceding it. THEORY OF LIMITS. 455 Let the perimeters of these polygons be p-^, p^, p^, ■■■ and their areas A-^, A^, A^, ••■. Then the length o of the circle is the limit of the sequence p^ p^, p^, ■■■ and its area A is the limit of Ap A^, jij, •••. The sequence of polygons thus inscribed is called an approximating sequence of polygons. That these sequences are increasing and bounded is obvious from the figure. 788. Theorem. TJie lengths of two circles are in the same ratio as their radii, and their areas are in the same ratio as the squares of their radii. Proof : Let the radii of the circles whose centers are o and o' be r and r'. Denote — by k. Then r' = kr. Inscribe in O an approximating sequence of polygons with perimeters Pi, p^^ ^3, ••• and areas ^^ A^, A^, ■■■. In 0' inscribe a sequence of similar polygons. By §§ 347, 348, the perimeters of the latter are kp-^, kp^, kp^, ••• and their areas are k^A^, k^A^, k^A^, •••. By § 785, if the limit of jOj, p^, p^, ••■ is c and the limit of ^j, A^, Ag, ■■■ is A, then the limits of kpj^, kp^, kp^, ■■■, and k^A-^, k^A^, k^A^, ••• are ko and Pa, respectively. That is, the ratio of the lengths of the circles is — = k = —, and the ratio of their areas is =. ^2 _ — c r A r'^ If regular circumscribed polygons are used, we obtain decreasing sequences both for the perimeters and the areas. That the limits of these sequences are identical with those obtained above is a direct consequence. See § 417. 456 SOLID GMOMETBY. 789. Definition. If the three concurrent edges a, h, c of a rectangular parallelepiped are incommensurable with the unit segment, the volume in- closed is defined as follows : Let «!, a^, ag, ••• be a sequence of rational numbers whose limit is the side a. Let 6^,521^31 ••• and Cj, CgiCg, ■■•, be similar sequences whose limits are respectively the dimensions b and c. Then the volume is the limit of the sequences ajijCi, a,'}> m, the set of n — 1 prisms will be greater than the set of m — 1 prisms. 793. Theorem. Tioo pyramids loith the same alti- tudes and equal bases inclose equal volumes. Proof: Divide the two equal altitudes into the same number of equal parts and construct inscribed prisms as in § 591. Since corresponding sets of prisms have equal volumes, it follows that the volumes of the pyramids are the limits of the same sequence, and hence identical. 794. Convex curves. Without specifically defining a convex closed curve (see § 552), we assume that in such a curve it is possible to inscribe a sequence of polygons Pj, Pg, P3, •••, having perimeters j»j, p^, p^, ••• and areas A^, A^, Ag, ■■■ with limits p and A respectively, and to cir- cumscribe a sequence of polygons P/, Pg', P3', •••, having perimeters p-^', p^^p^i ••■ ^nd areas a{^ a^, A^ , ■•■ with limits p' and A' respectively, such that p = p' and A = A'. These limits p and A we now define as the perimeter and the area respectively of the curve. 458 SOLID OEOMETBY. 795. Definition. Given any cylinder with a convex right cross section and an element e. In this cross section inscribe a sequence Pj, Pg, P3, •••of polygons, as in § 787, with perimeters Jh^ Pt Ps^ '" ^^^ areas A^, A^, A^, ••-, thus defining the perimeter p and the area A of the cross section. Consider a set of prisms inscribed in this cylinder, of which Pj, Pg, Pg, •■• are right cross sections. Then the areas and the volumes of these prisms are respectively jOj«, ^2^' /"s^i ■■■ ^^'^ ^i^> -^2^' ^3*' "■■• The lateral area and the volume of the cylinder are now defined as the limits of these sequences. But these limits are by § 785 equal to pe and Ae respectively. Hence, we have — 796. Theorem. The lateral area of a cylinder is the product of an element, and tJie perimeter of a right section and its volume w the product of an element and the area of a right section. 797. EXERCISES. 1. Give examples other than those given in the text of infinite sequences which do not determine definite numbers. 2. Give two distinct sequences wliieh determine the number '2. Show that the theorem of § 776 applies and proves that these se- quences determine the same number. 3. (live two decreasing sequences each of which determines tlie number :!. Apply § 777 to show that these sequences determine the same number. 4. State fully the rclaliou between abounded increasing sequence and the number determined by it. .Slate also the relations between a bounded decreasing sequenoo and tlio number determined by it. 5. State fully wliat is meant by "a limit of a sequence" both for increasing and decreasing sequences. THEOBT OF LIMITS. 459 6. Given two incommensurable segments ^£ and CZ>. Lay off on the line AB a decreasing sequence of segments, eacli of which is com- mensurable with CD, such that the limit of the sequence is the seg- ment AB. 7. If ai, 02, as, ••■ is an increasing sequence defining the number 4, prove that 3 ai, 3 aa, 3 as, ••• defines the number 3 x 4 = 12. Note that in case the sequence oi, ai,, as, ••• defines an irrational number, we should not be able to prove the corresponding proposition without first defining what is meant by the product of a rational and an irrational number. But such definition would in that case be the proposition itself. See § 785. 8. If a\, 02, as, ■■■ and bi, hi, Js, •■• are increasing sequences defin- ing the numbers 3 and 5, show that the sequence ai6i, 0262, aebs, ■■• de- fines the number 15. In case these sequences defined irrational numbers, could a cor- responding proposition be proved? 9. In the same manner as in § 781 define the ratio of two incom- mensurable arcs. 10. Show as above that the ratio so obtained is independent of the sequence of units of measurement used, so long as the limit of this sequence is zero. 11. Treat the ratio of two incommensurable angles in a manner similar to the treatment of arcs in the two preceding exercises. 12. Define the lateral area of a cylinder by means of circumscribed prisms, and show that this definition leads to the same result as that given in § 795. 13. Prove as above that the volume of any convex cylinder is equal to the product of its altitude and area of its base. 14. Prove that the lateral area of a right circular cone is equal to half the product of the slant height and the perimeter of its base. 15. Prove that the volume of any convex cone is equal to one third the prodnct of its altitude and the area of its base. Suggestion. The treatment required in the last three exercises is a very close paraphrase of the definitions and proof given in § 789. Observe that we cannot begin to make a proof until we have defined the subject matter of the theorem. That is, we must first define the areas and volumes in question. 460 SOLID GEOMETBT. VOLUME OF THE SPHERE. 798. Through two points P and Q of a sphere pass a great circle forming' a hemisphere with center O. A .V r""^ -'' ~~^ <£^-'-" ' -[b- ' - -J3^^ - - - s-~_ __^<^ <^^!^v:::^VJSS^£::> / '-''- ■ --^*~-\^l --' \.-—".'--'>-f,^r."-.^ '•• k ^^■^^i^:~r--'*''^--—:i^^^\ ""-ll^-V^ii-^s'fc/^Ji^^^i;:.^ -A --_: ,Q^^—^^^^^-^ <^^^__'-'"':'^:^^---_;j^ Divide OA, the radius perpendicular to the plane of the great circle, into the equal parts OC, CB, and BA. Through C and B pass planes parallel to the plane of POQ, meeting the circle in points D and E, respectively. Construct right circular cylinders with axes OC and CB and radii CD and BE. Denote by Fj the sum of the volumes of these cylinders. Now divide the radius 0^4 iuto six equal parts and con- struct five cylinders in the same manner as above. Let the sum of these volumes be V^- ('ontinuing in this manner, each time dividing OA into twice as many equal parts as in the preceding, we obtain a sequence of sets of cylinders and a corresponding sequence Fj, T'ji Fg, ••• of volume. ^Ve now define the volume inclosed by the hemisphere as the limit of the sequence ]\, F,, I'g, ■••. 799. Construct a right circular cylinder with its base in the plane of POQ and altitude o'a' = OA. Denote by F the figure formed by the lo\\er base of the cylinder, its lateral surface and the lateral surface of the cone wliiise base is tlie ujiper base of the cylinder and wbosc vortex is at o'. THEORY OF LIMITS. 461 Draw segments o'm and o'n. Let the planes through C and B cut a' A' in c' and b' and o'Jtf in H and jk:. Now form the cylinder o'g'b whose axis is o'o' and whose radius is C'h. Likewise form c'b'k. Let Vi denote the sum of the volumes of o'c'd' and c'b'e' minus the sum of the volumes O'c's and c'b'k. In a similar manner, using the planes which divide OA and hence o'a' into six equal parts, we form another set of five cylinders, the sum of whose volumes minus that of the smaller inside cylinders we denote by V^'. Continuing in this manner, we obtain a sequence of vol- umes Fj', V^', Fg', ••• whose limit v' we define as the volume of the given figure F. We now prove that F^ = Fj', V^ = V^', ••■■ Denote OA by r, and note that o'b' = b'k. (1) Yo\. jfB^E' - Vol. C'B'K = ttC'b' (Ve'"^ - F?) = ■jrC'B'(r^-¥B'^). _ _ (2) Vol. CBE^-ttCB ■ BE^ = -n-CB (r^ — OE^), since BE^ = r2 _ ^^ But OB = o'b' and CB = c'b'. Hence, Vol. CBE= Vol. c'b'e' -Vol. c'b'k. Similarly we show that Vol. OCD = Vol. o'c'd' —Vol. o'c'n. Hence, Fj = Fj'. In like manner v^ = V^', Fg = Fg', ••-. Hence, F = v', since they are the limits of the same sequences. But the volume of the cylinder o'a'm is tt?^ and of the cone whose volume was subtracted, — r^- That is, the vol- 9 2 TT ume of F is— ^ r^, and hence that of the hemisphere is -— r^. 3 o Hence, we have — 800. Theorem. ITie volume of the sphere is f ttt^. 462 aOLID GEOMETRY. 801. -Note that the above proof consists essentially in showing that the area of the circle BE is equal to that of the ring between the circles b'e' and b'k, and that the area of the circle CD is equal to that of the ring between c'd' and c'h and so on. Indeed this theorem and also that of § 594 are special cases of what is known as Cavalieri's Theorem. If two solid figures are re- garded as resting on the same plane b, and if in every plane parallel to b the sections of the tioo figures have equal areas, the figures have equal volumes. The proof of this more general theorem is more difficult than any thus far given, inasmuch as it involves sequences which oscillate : that is, which are neither constantly increasing nor constantly decreasing. THE AREA OF THE SPHERE. 802. About a sphere of radius r construct a sequence of circumscribed polyhedrons such that the largest face in each polyhedron becomes as small as we please when we proceed along the sequence. Let «j, s^, 83, ■•• be the total surfaces of these polyhedrons. This forms a decreasing sequence with limit S which we define as the surface of the sphere. The volumes of these polyhedrons will be -Sj, -s.,. -s^, ■■•. 000 Then the volume V of the sphere is defined as the limit of this sequence of volumes. Hence, by § 785, r = ^ s. But by § 700, r = f 77?^. Then ■ s = - ■ ~ tt?^ = i -n-r^. r 'S Hence, we have — 803. Theorem. The area of the sphere is 4 7rr*. THEORY OF LIMITS. 463 This argument is incomplete in that two distinct defi- nitions have been given of the volume of the sphere, namely, as the limit of an increasing sequence of volumes of inscribed cylinders and also as the limit of a decreasing sequence of volume of circumscribed polyhedrons. But it has not been proved that these two definitions are equiva- lent ; that is, that they lead to the same formula for the volume of the sphere. The treatment of the area and volume of the sphere in most of the current textbooks on geometry is open to a similar objection. Usually two distinct definitions are used for the area of the sphere, namely, as a limit of the areas of circumscribed frustums of cones and also as the limit of the surfaces of circumscribed polyhedrons. The argument used in §§ 740-748 is not subject to any such objection. SUMMARY OF CHAPTER XIV. 1. Explain the separate stages of treatment of incommensurables and of limits as given in this book, including both the Plane and Solid Geometry. 2. Give an outline of the introduction to this final treatment by means of sequences. 3. Make a list of the definitions, principles, and theorems upon which the treatment of sequences is based. 4. Explain fully the way in which the principles of sequences are applied to geometric theorems. Illustrate by line-segments and arcs. 5. Make a list of the theorems which are proved here by the theory of limits, using the sequence process. 6. State in some detail how this process is used in case of the area and volume of the sphere. 7. Give all the mensuration formulas proved in this chapter. 8. Review the entire list of mensuration formulas for both Plane and Solid Geometry. INDEX. (KeferenceB are to sections unless otherwise stated. ) Abbreviations 80 Addition, of angles . 39 of line-segments .... 10 Adjacent, parts of a polygon 134 parts of a triangle 21 Alternate, exterior angles . . 88 interior angles ... 88 Alternation ... . . 246 Altitude, of a cone . . . (j04 of a cylinder .... 354 of a frustum . . . 599 of a parallelogram . ... 137 of a prism 526 of a pyramid . . .... 386 of a spherical segment 750 of a trapezoid .... . 137 of a triangle ... . . 24 of a zone .... 730 Analysis, of proof . ... 175 of problems . . . 177 Angle, acute, oblique, obtuse, reflex, right ... 19 between two curves . . 678 dihedral .... 495 exterior of a triangle . . 81 face . 518 of projection . 571, 580 polyhedral 518 reentrant . . . 134 spherical . . . . 678 straight, zero, perigon . 16, 17 trihedral ... . 518 Angles, adjacent .39 complementary 70 multiplication and division of . supplementary vertical Antecedents . . Apothem of a regular poly- gon Approach Approximate area 39 70 70 242 342 767 306 1 Approximate measurement 235, 305 Approximate ratio 240 Arc of a circle .... 12, 184 Arc of a spherical triangle . 680 Area, approximate of a circle of a cone . of a cylinder of a polygon of a prism . of a pyramid of a rectangle . 306 . . . 361, 416 ... 612 562, 563, 795 . 309 . 528 587 300-307, 414, 785 of a sector of a circle . . 364 of a segment of a circle . . 365 of a sphere . 740, 745, 746, 802, 803 of a spherical polygon . . . 738 of a spherical triangle 736 of a surface . . . 302 of a zone . 751 lateral 526 total . ... 526 Aristotle . . . page 31 Axial symmetry .... 164 Axioms . 60, 61, 82, 96, 119, 392, 407, 409, 411, 412, 416, 460, 462, 464, 560, 562, 592, 612, 745, 774 on numbers ... ... 241 Axis, in a graph . 368, Chapter XIV of a circle ... . , 659 of symmetry ... . . 164 Base, of a cone . . . . 604 of a cylinder . . . 554 of a parallelogram . . 135 of a prism 525 of a pyramid 586 of a spherical sector . 750 of a spherical segment 750 of a trapezoid .... 136 of a triangle . 24 Bhaskara ... 318 465 466 INDEX. (References are to sections unless otherwise stated.) Birectang'ular spherical tri- angle . . . . 733 Bisector, of an .-ingle .... 19 of dihedral angle . . . 509 of a line-segment . . . . 50 Bound, greatest lower . . . 773 least upper . 773 Broken line .... 11 Cavalieri's theorem . . 801 Center, of a circle . 12 of a pencil .... . . 431 of a regular polygon . . . 342 of similitude 434, 643 of a sphere 655 169 185 169 181 416 225 12 :2!), Ex.4 659 352, of symmetry Central angle . symmetry . Chord of a circle Circle, circumference of circumscribed definition of . . escribed great inscribed . . . 228 inside of, outside of ■ ■ 180 length of, perimeter of . . 3.52, 416 poles of 659 secant of . .... . 182 segment of . 220 small . ... .659 Circles, equal 187 intersecting . . 186 tangent ... .... 186 Circular, cone . . 604 cylindor . 554 Circumscribed, cone 610 cylinder 561 polyhedron 637 prism .... . . 561, 591 pyramid . . 610 sphere . . . (i:!2. I>72 Coincidence . 19, 27 angles, arcs .... . 7S:i Commensurable segments . 23t; ratios . 2:I8, 7S1 Kogments . . 781 Compasses . . . 3;? Complementary angles 70 Composition . . . 246 Composition, and diyision . . Concave polygon Conclusion Concurrent lines Cone, definition of oblique ... . . right circular Congruence Congruent, polyhedral angles . spherical polygons Conical surface Consequents .... Constants Construction, of figures . of regular polygons ■ Ifi-l Continued proportion . Converse theorems Convex polygon Corollary Corresponding, angles . cross-sections linear dimensions parts of polar triangles parts of polygons parts of similar polyhedrons parts of triangles Cosine of an angle . . Cube . .... Curved line Curves, convex . . Cylinder, definition of . . of revolution . . Cylindrical surface, definition of ... . Decagon Definition, technjcal Degree of an angle .... Demonstration 59, 7:', SO. typo of Dependence of variables . . Diagonal of a polygon AM, Diameter, of a circle of a parallelogram . of a I rapo7t>id Dihedral angles . adjacent bisector of coiiiplcuicntary .... 246 160 78 228 604 604 604 27 .520 7:il 242 ;:s(i 4i 3.33 207 98 160 104 8^ I ',.-,4 654 707 29 534 11 .V.4 .".."4 553 424 17 .3 160 ISl i:;7 i:;7 49,5 500 .509 .500 INDEX, (References are to sections unless otherwise stated.) 467 Dihedral angles, measure of 500 supplementary 500 vertical 500 Direct proof 175, 399 Direct tangent to two circles 403 Discussion of a construction 122 Distance, between two points in a plane 61 between two points on a sphere 661 from a point to a line . . . 114 from a point to a plane . . 475 Dividers 33 Division, of line-segments . . 10 external, internal 425 in extreme and mean ratio . . 421 of augles 39 of the circle 333 Dodecagon .... pages 82, 235 Dodecohedron 625, 636 Edge, lateral 525 of a dihedral angle 495 of a polyhedral angle .... 518 of a polyhedron 523 Element, of a cone 603 of a cylinder . . . . 554 of a prismatic surface . . . 524 Ellipse, area of 583 definition of ...... . 582 End-point, of a segment . . . 8, 13 of a ray ... 9 Equal, equivalent 27 solids . 541 Equiangular figures . . .22, 160 mutually • . ■ ... 255 Equilateral figures . . .22, 160 Equivalent solids . . . . 544 Euclid .... pages 3, 31, 40, 141 Eudoxus page 140 Euler's theorem . ... 627 Exterior, angle of a triangle . 81 angles of parallel lines ... 88 External, division . . 252, 425 segment of a secant . . . 274 Extreme and mean ratio . 421 Extremes ... ... 242 Faces, of a dihedral angle . . 495 of a polyhedral angle . . . 518 of a polyhedron . . ... 523 525 Paces, of a prism .... of a pyramid 586 Figure in space 453 Figures, geometric 12 Fourth proportional ... 291 Frustum 599 Generator, of a conical surface 603 of a cylindrical surface . 553 of a pyramidal surface . . . 585 of a spherical surface . . . 744 Geometric proposition . . 59 Geometric solid . . . .3 Geometry 1 Gothic arch, pages 109, 151, 197, 204 Graphic representation, Chapters VI, XIII Great circle, axis of .... 659 pole of . . - . .... 663 Half-plane 495 Harmonic' division. . . 425 Hexagon, regular, pages 65, 67, 7(., 78, 82, 108, 174, 177, etc. Historical notes, pages 3, 31, 38, 40. 42, 128, 140, 162, 167, 193, 290 Hypotenuse ... .23 Hypothesis 78 Incommensurahles, 236, 240 305, 36S ,541 766 767 Icosahedron . . . .625 636 Indirect proof . 91 176 394 Inscribed, angle in an arc 220 angle in a circle . 185 angle in a segment . 220 circle in a triangle . 228 cone 610 "ifil prism . . .561 592 pyramid 610 sphere . . . 632 ,033 634 672 triangle in a circle . 225 Intercepted arc 185 Interior angles 88 Internal division . 252 425 Inversion 246 Irrational, number 762 ratio , , 762 468 INDEX. (References are to sections unless otherwise stated.) Isosceles, spherical triangle trapezoid plane triaiiglr Land surveying Lateral, edges .... l;iC(.'S . surfaces . . j54, 561, 604 Legs of a right triangle . Length 691 22 283, 284 . 525 (il7 23 3 ?,-,o 232, 407 383 of a circle of a segment Limit, of Hpothem of area 384 of perimeter . . 384 of rotating secant .... 386 of segments .... 780 of a sequence . . . 773, 774 Line 1,3,4,5,7,11 half-line . . 9 of centers . . 403 -segment . . 8 Lines, concurrent . . 228 oblique, obtuse, perpendicular 19 Loci, determination of, pages .50-52, 22i>-l."J7 problems, pages 52, 64, 83, 92, 96, 97, 102, 103, 111, 120, 148, 149, 158, 173, 179, 226, 228, 233, 283, 28S, 28il, 293, 2il4, 295, 298, 307, .ms, 309, 396, 401 Lobachevsky . . page 38 Lune, angle of . . 733 Major and minor area Maxima and minima . Mean proportional Means . . Measurement, approximate, ex;ict of augle.s of litic-s('j;nu'iits of tliiM'ircle . . of snrfari's of v(tliiiiics r»;'.7, M4, ,301 17 Median of a triangle Methods of attack Minute of angle . . , 610, (lill, 174 184 443 293 242 :iixi :'>ii4, 202 :;49 740 Multiplication, of angles . . 39 of line-segments ... 10 Nappes, of a conical surface . 'J03 of a pyramidal surface . -''85 Number .... 241, 407 Numerical measure 232, 302, 407 Octahedron . . . 62.'), 6;)6 Oblique lines, angles 19 Obtuse triangles 23 Octagons . pages 7n, 7i), etc. regular pages 146, 147, IS7, 198, 202, 235, etc. Opposite parts of figures . 21, 1:4 Origin of ray 9 OutUne of proof .52 Pantograph .... im.i Parallel, lines . . . 89 axiom ... '.»l line to a plane .... 484 plane to a plane . . . 4s4 rulers 2.^8 Parallelogram . . i:a Parallelepiped, rectangular . 54 \olurae inuhised by 5.17-541 Pencil of lines . . 4^1 Pentagon . . 424 Perigon ... 17 Perimeter, of a circle . .'Cc' of a polygon lui. 2'i7 Perpendicular . 19 bisector of a segment . . 50 line to a plane . . . .5iis plane to a line . . .'x.^ plane to a plane . . . .592 n (pi), computation of . . :!,51 Plane, determiuation of 4<;.i proji-'ctions uinui 5.'^') -segment 522 surf;ifi' . ... 1. 2 Plane angle of :i dihe.">!) Solids, congruent ..KO, 731 equal . HH equivalent .... . . 544 geometrical ... . 458 Solution of problems . . 177, 17H Sphere, area of . 744-741), mi determination of . . . CM diameter of ... . 655 great circle of 659 inscribed KM, 672 points within, without . 655 radius of 6.'i2, 655 small circle of ... 6.';!i Spherical, angle 678 degree, minute, second . 73:i excess 7.15, 737 731 . 7:15, . 727, .75U, polygons sector ... tri;ingle 6H0 Square . ... . 135 Subtended, angle .... 286 arc . . 184 Subtraction, of angles .39 of line-segments 10 Sum of angles 39 Superposition .... page 14 Surface, conical . 603 curved 552, 659 cylindrical .... . 553 definition of 2 lateral 554 prismatic 524 spherical i'*^^^^ Symbols of abbreviation . . 80 Symmetrical, spherical triangles 687 triliedral angles . .... 687 Syrometry 164-173 problems involving, pages 81, 82, 93, 94, 101, 109, 150, 2l)."i. 229, 233, etc. with respect to a point lU:! Tangent, of an iicntc ans;li' .572 circles . . . 186 to a circle . 183 to two circles . . . 403, 404 to u cone 606 Tangent, to a cylinder . . 5.56 to a sphere 6(j9 Technical words 3S9 Tests for congruence . 32, 35, 40 Tetrahedron, circumscribed . 637 inscribed 586, 631, 6:12 regular .... . . 625 Thales pages 3, 17 Theorem, definition of . 59 Theory of limits . . Chapter Xl\' Third proportional . . 2!i3 Transversal ... . . . 8S Transverse tangent . . . 403 Trapezoid . . 136 Triangle . 21-24, 41, 157. 225, 228, etc. 5:10 599 . 389 60, 391 Trtincated, pyramid prism Undefined words . Unproved propositions Variables . . , Chapters VI dependence of limit of . ... Variation, continuous . . Vertex, of a cciiie . . . of a polyhedral angle . . of a pyramid . . of a spluTiral angle Vertex angle of a trians' Vertical angles . . . Vertices, of a polyhedron of a polygon of a quadrilateral of a triangle . Volume, as a limit of asequeno of a cylinder . of a prism . . of a sphere of spherical i-ono of spherical sector . of spherical segment Whole secant Zone, altitude of . bases of