PRESENTED TO THJE; CORNELL UNIVERSITY, 1S70, The Hon. William Kelly Of Rhinebeck. Elements of geometry : onn,an? ^^24 031 273 349 The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031273349 /UA.t/ ^t^ f}^^"><>^^ ELEMFy TS GEOMETRY; PRACTICAL APPLICATIONS MENSURATION Br BENJAMIN £REENLEAF, A. M., AUTHOR OP '* THE NATIONAL ARITHUETIC," " TBEATISB ON ALGEBRA." ETC. Permanent Electrotype Edition* BOSTON: PUBLISHED BY ROBERT S. DAVIS & CO. NEW YOKK: OAKLEY AND MASON, AND W. I. POOLEY. PHILADELPHIA: J. B. LIPPINCOTT AND COMPANY. CINCINNATI: u. H. BLANCHARD & CO. CHICAGO: S. C. GRIGGS « CO. 1869. GSEENLEAPS NEW COMPREHENSIVE SERIES. An ENTiBELT NEW MATHEMATICAL CouESE, fuUy adapted to the best methods of Modem Jnstruction. GREENLEAF'S NEW PRIMARY ARITHMETIC. GREENLEAF'S NEW INTELLECTUAL ARITHMETIC. GREENLEAF'S NEW ELEMENTARY AEITHMETIC. GREENLEAF'S NEW PRACTICAL ARITHME«er— — GREENLEAF'S NEW ELEMENTARY AL^EB^ O R ^ F i ^ ^ GREENLEAF'S NEW HIGHER ALGEBHA.y {%| ' V Z R^j ' j V GREENLEAF'S ELEMENTS OF GEOMESRT. \ I >'■''< ^ ^^ ' f GREENLEAF'S ELEMENTS OF TRIGONOME TItfT^ — — GREENLEAF'S GEOMETRY AND TRIGONOMETRY. 1^" Other Books of a Complete Series, in preparation. *^* Keys to the Practical Abithmetic, Algebras, Geometry and Trigonometry, in separate volumes. Entered according to Act of Congress, in the year 1858, by BENJAMtN GrbENLEAP, in tlie Olerk's Office of tlie District Court for tlie District of M:i:^sactiasetta. RIVSnsiDB PRESS : miNTED DT n. 0. UOUQUTON AND COHPAHT. PREFACE. Tbb preparation of this treatise has been undertaken at the earnest solicitation of many teachers, who, having used the au- thor's Aritlunetics and Algebra with satisfaction, have been de- sirous of seeing his series rendered more complete by the addition of the Elements of Geometry. That there are peculiar advantages in a graded series of text- books on the same subject, few, if any, properly qualified to judge, will doubt. The author, therefore, feels justified in intro- ducing this volume to the attention of the public. In common with most compilers of the present day, he has followed, in the main, the simple and elegant order of arrange- ment adopted by Legendre ; but in the methods of demonstra- tion no particular authority has been closely followed, the aim having been to adapt the work fully to the latest and most ap- proved modes of instruction. In this respect, it is believed, there will be ibund incorporated a considerable number of important improvements. More attention than is usual in elementary works of this kind has been given to the converse of propositions. In almost all cases where it was possible, the converse of a proposition has been demonstrated. The demonstration of Proposition XX. of the first book is essentially the one given by M. da Cunha in the Principes Maihe- IV PREFACE. matiques, which has justly been pronounced by the highest mathematical authorities to be a very important improvement in elementary geometry. It has, however, never before been intro- duced into a text-book by an American author. The Application of Geometry to Mensuration, given in the eleventh and twelfth books, are designed to show how the theo- retical principles of the science are connected with manifold practical results. , The Miscellaneous Geometrical Exercises, which follow, are calculated to test the thoroughness of the scholar's geometrical knowledge, besides being especially adapted to develop skill and discrimination in the demonstration of theorems and the solution of problems unaided except by principles. Sufficient Applications of Algebra to Geometry are given to show the relation existing between these two branches of the mathematics. The problems introduced in connection therewith will be found to be, not only of a highly interesting character, but well calculated to secure valuable mental discipline. In the preparation of this work the author has received valua- ble suggestions from many eminent teachers, to whom he would here express his sincere thanks. Especially would he acknowl- edge his great obligations to H. B. Maglathlin, A. M., who for many months has been associated with him in his labors, and to whose experience as a teacher, skill as a mathematician, and abihty as a writer, the value of this treatise is largely due. BENJAMIN GEEENLEAF. Bradford, Mass., June 25, 1858. NOTICE. A Key, comprising the Solutions of the Problems contained in the last four Books of this Geometry, has been published, /or Teachers only ; and the same will be mailed, post-paid, to the address of any Teacher who will forward sixty cents in stamps to the Publishers. CONTENTS. PLANE GEOMETRY. BOOK I, PAGE ELEMENTARY PRINCIPLES . 7 BOOK II. RATIO AND PROPORTION 43 BOOK III. THE CIRCLE, AND THE MEASURE OF ANGLES ... 56 BOOK IV. PROPORTIONS, AREAS, AND SIMILARITY OP FIGURES . . 76 BOOK V. PROBLEMS RELATING TO THE PRECEDING BOOKS . . .118 BOOK VI. BEOULAR POLYqONS, AND THE AREA OF THE CIRCLE . . 142 VI CONTENTS. SOLID GEOMETRY. BOOK VII. PLANES. — DIEDRAL AND POLYBDRAL ANGLES . . .165 BOOK VIII. POLYEDR0N8 184 BOOK IX. THE SPHERE, AND ITS PROPERTIES 214 BOOK X. •THE THREE ROUND BODIES 238 MENSURATION. BOOK XI. APPLICATIONS OP GEOMETRY TO THE MENSURATION OP PLANE FIGURES 263 BOOK XII. APPLICATIONS OF GEOMETRY TO THE MENSURATION OF SOLIDS 2t-l BOOK XIII. MISCELLANEOUS GEOMETRICAL EXERCISES . , . 801 BOOK XIV. APPLICATION OF Al.GBBKA TO OEOMI-.TRY .' . . 811 ELEMENTS OF GEOMETEY. BOOK I. ELEMENTARY PRINCIPLES. DEFINITIONS. 1. Geometky is the science of Position and Extension. The elements of position are direction and distance. Tlie dimensions of extension are length, breadth, and height or thickness. 2. Magnitude, in general, is that which has one or more of the three dimensions of extension. 3. A Point is that which has position, without magni- tude. 4. A Line is that which has length, without either breadth or thickness. 5. A Straight Line, or Right Line, is one which has the same . „ direction in its whole extent; as the line AB. The word line is frequently used alone, to designate a straight line. 6. A Curved Line is one which continually changes its direction ; C '""^ "~-- D as the line C D. The word curve is frequently used to designate a curved line. 8 ELEMENTS OF GEOMETRY. 7. A Broken Line is one which is composed of straight lines, not lying in the same direction ; as the line E P. 8. A Mixed Line is one which is composed 'of straight lines and of ciirved lines. 9. A Surface is that which has length and breadth, without height or thickness. 10. A Plane Suepacb, or simply a Plane, is one in which any two points being taken, the straight line that joins them will lie wholly in the surface. 11. A Curved Surface is one that is not a plane sur- face, nor made up of plane surfaces. 12. A Solid, or Volume, is that which has length, breadth, and thickness. ANGLES AND LINES. 13. A Plane Angle, or simply an Angle, is the difference in the direc- tion of two lines, which meet at a point ; as the angle A. The point of meeting. A, is the vertex of the angle, and the lines A B, A C are the sidex of the angle. An angle may be designated, not D only by the letter at its vertex, as C, but by three letters, particularly when two or more angles have the same vertex; as the angle A CD or " C D C B, the letter at the vertex always occupying the mid- dle place. The quantity of an aaigle does not depend upon the length, but entirely upon the position, of tlie sides ; for the angle remains the same, however the lines containing it be increased or diminished. B BOOK I. 9 D 14. Two straight lines are said to B be perpendicular to each other, when their meeting forms equal adjacent angles ; thus the lines A B and C D are perpendicular to each other. Two adjacent angles, as CAB and BAD, have a com- mon vertex, as A ; and a common side, as AB. 15. A Right Angle is one which is formed by a straight line and a perpendicular to it ; as the angle CAB. B D 16. An Acute Angle is one which is less than a right angle ; as the angle DEP. An Obtuse Angle is one which is greater than a right angle; as the angle EFG. Acute and obtuse angles have their sides oblique to each other, and are sometimes called oblique angles. 17. Parallel Lines are such as, being in the same plane, cannot meet, however far either way both of them may be produced ; as the lines A B, CD. 18. When a straight line, as E P, intersects two parallel lines, as AB, CD, the angles formed by the intersecting or secant line take particular names, thus : — Interior Angles on the same Side are those which lie within the parallels, and on the same D 10 ELEMENTS OF GEOMETRY. side of the secant line ; as the angles BGH, GHD, and also AGH, GHC. Exterior Angles on the same Side are those which lie without the parallels, and on the same side of the secant line ; as the angles BGE, DHP, and also the angles AGE, CHP. Alternate Interior Angles lie within the parallels, and on different sides of the secant line, but are not adja- cent to each other; as the angles BGH, GHC, and also AGH, GHD. Alternate Exterior Angles lie without the parallels, and on dififereut sides of the secant line, but not adjacent to each other ; as the angles E G B, C H F, and also the angles AGE, DHP. Opposite Exterior and Interior Angles lie on the same side of the secant line, the one without and the other within the parallels, but not adjacent to eacli other ; as the angles B GB, GHD, and also EGA, GHC, are, respec- tively, the opposite exterior and interior angles. PLANE FIGURES. 19. A Plane Figure is a plane terminated on all sides by straight lines or curves. The boundary of any figure is called its perimeter. D 20. When the boundary lines are straight, the space they enclose is called a Rectilineal Figure, or Polygon ; as the figure A B C D E. A B 21. A polygon of three sides is called a triangle ; one of four sides, a quadrilateral; one of five, a pentagon; one of six, a hexagon ; one of seven, a heptagon ; ouo BOOK I. 11 of eight, an octagon ; one of nine, a nonagon ; one of ten, a decagon ; one of eleven, an undecagon ; one of twelve, a dodecagon ; and so on. 22. An Equilateral Triangle is one which has its three sides equal ; as the triangle ABC. An Isosceles Triangle is one which has two of its sides equal ; as the triangle D E P. A Scalene Triangle is one which has no two of its sides equal ; as the triangle G H I. 23. A Right-angled Triangle is one which has a right angle ; as the triangle J K L. The side opposite to the right angle is called the hy- pothenuse ; as the side J L. 24. An Acute-angled Triangle is one which has three acute angles ; as the triangles ABC and DEE, Art. 22. An Obtuse-angled J'riangle is one which has an ob- tuse angle ; as the triangle GHI, Art. 22. Acute-angled and obtuse-angled triangles are also called oblique-angled triangles. 25. A Parallelogram is a quadrilateral which has its opposite sides parallel. 26. A Rectangle is any parallel- ogram whose angles are right angles ; as the parallelogram A B C D. ^ ' B 12 ELEMENTS OF GEOMETRY. A Square is a rectangle whose sides are equal ; as tlie rectangle EFGH. 27. A Rhomboid is any parallelo- gram whose angles are not right an- gles ; as the parallelogram IJKL. H E ^K A Rhombus is a rhomboid whose sides are equal ; as the rhomboid MNOP. M^ O 28. A Trapezoid is a quadrilateral which has only two of its sides par- allel ; as the quadrilateral R S T U. R- W A Trapezium is a quadrilateral which has no two of its sides paral- lel ; as the quadrilateral V WXY. 29. A Diagonal is a line joining the vertices of any two angles which are opposite to each other ; as the lines E C and E B in the polygon ABCDB. 30. A Base of a polygon is the side on which the poly- gon is supposed to stand. But in the case of the isosceles triangle, it is usual to consider that side the base which is not equal to either of the other tiides. 31. An equilateral polygon is one wliich has all its sides equal. An equiangular polygon is one which has BOOK I. 13 all its angles equal. A regular polygon is one which is equilateral and equiangular. 32. Two polygons are mutually equilateral, when all the sides of the one equal the corresponding sides of the other, each to each, and are placed in the same order. Two polygons are mutually equiangular, when all the angles of the one equal the corresponding angles of the other, each to each, and are placed in the same order. 33. The corresponding equal sides, or equal angles, of polygons mutually equilateral, or mutually equiangular, are called homologous sides or angles. AXIOMS. 34. An Axiom is a self-evident truth ; such as, — 1. Things which are equal to the same thing, ai-e equal to each other. 2. If equals be added to equals, the sums will be equal. 3. If equals be taken from equals, the remainders will be equal. 4. If equals be added to unequals, the sums will be unequak 5. If equals be taken from unequals, the remainders will be uneqiial. 6. Things which are double of the same thing, or of equal things, are equal to each other. 7. Things which are lialves of the same thing, or of equal things, are equal to each other. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 10. A straight line is the shortest line that can be drawn from one point to another. 11. From one point to another only one straight line can be drawn. 12. Through the same point only one parallel to a straight line can be drawn. 14 ELEMENTS OP GEOMETRY. 13. All right angles are equal to one another. 14. Magnitudes which coincide throughout their whole extent, are equal. POSTULATES. 35. A Postulate is a self-evident problem ; such as, — 1. That a straight line may be drawn from one point to another. 2. That a straight line may be produced to any length. 3. That a straight line may be drawn through a given point parallel to another straight line. 4. That a perpendicular to a given straight line may be drawn from a point either within or without the line. 6. That an angle may be described equal to any given angle. PROPOSITIONS. 36. A Demonstration is a course of reasoning by which a truth becomes evident. 37. A Proposition is something proposed to be demon- strated, or to be performed. A proposition is said to be the converse of another, when the conclusion of the first is used as the supposition in the second. 38. A Theorem is something to be demonstrated. 39. A Problem is something to be performed. 40. A Lemma is a proposition preparatory to the dem- onstration or solution of a succeeding proposition. 41. A Corollary is an obvious consequence deduced from one or more propositions. 42. A Scholium is a remark made upon one or more preceding propositions. 43. An Hypothesis is a supposition, made either in the BOOK I. 15 enunciation of a proposition, or in the course of a demon- stration. Proposition I. — Theorem. 44. The adjacent angles which one straight line makes by meeting another straight line, are together equal to two right angles. Let the straight line D C meet AB, making the adjacent angles A C D, DOB; these angles to- gether will be equal to two right angles. A / g From the point C suppose C E ^ to be drawn perpendicular to A B ; then the angles ACE and E C B will each be a right angle (Art. 15) . But the angle A C D is composed of the right angle ACE and the angle BCD (Art. 34, Ax. 9), and the angles ECD and D C B compose the other right angle, E C B ; hence the angles A C D, D C B together equal two right angles. 45. Cor. 1. If one of the angles A C D, D C B is a right angle, the other must also be a right angle. 46. Cor. 2. All the successive angles, BAC, CAD, DAE, EAP, formed on the same side of a straight line, B P, are equal, when taken together, to two right angles ; for their sum is equal to g. that of the two adjacent angles, ^ BAC, CAP. Proposition II. — Theorem. 47. If one straight line meets two other straight lines at a common point, making adjacent angles, which to- gether are equal to tivo right angles, the two lines form one and the same straight line. 16 ELEMENTS OP GEOMETRY. Let the straight line D C meet the two straight lines AC, C B at the common point C, making the adjacent angles A C D, D C B to- gether equal to two right angles ; then the lines AC and CB will C ~^^Z~'^ form one and the same straight line. If C B is not the straight line A C produced, let C E be that line produced ; then the line ACE being straight, the sum of the angles A C D and D C E will be equal to two right angles (Prop. I.). But by hypothesis the angles A C D and D C B are together equal to two right angles ; therefore the sum of the angles A C D and D C E must be equal to the sum of the angles A C D and D C B (Art. 34, Ax. 2). Take away the common angle A C D from each, and there will remain the angle D C B, equal to the angle D C E, a part to the whole, which is impossible ; therefore C E is not the line A C produced. Hence A C and C B form one and the same straight line. Pkoposition III. — Theorem. 48. Two straight lines, which have ttoo points common, coincide with each other throughout their whole extent, and form one and the same straig'ht line. Let the two points which are F common to two straight lines be A and B. The two lines must coincide between the points A and B, for otherwise there would be two ^ G ^ straight lines between A and B, which is impossible (Art. 34, Ax. 11). Suppose, however, that, on being produced, the lines begin to separate at the point C, the one taking the direc- BOOK I. 17 tioii C D, and the other C E. Prom the point C let the line C F be drawn, making, with C A, the riglit angle A C F. Now, since A C D is a straight line, the angle F C D will be a right angle (Prop. I. Cor. 1) ; and since ACE is a straight line, the angle FOE will also be a right angle ; therefore the angle F C E is equal to the angle F C D (Art. 34, Ax. 13), a part to the whole, which is impossible ; hence two straight lines which have two points common, A and B, cannot separate from each other when produced ; hence they must form one and the same straight line. Proposition IV. — Theorem. 49. When tioo straight lines intersect each other, the opposite or vertical angles which they form are equal. Let the two straight lines AB, ^ CD intersect each other at the point E ; then will the angle AE C be equal to the angle DEB, and the angle CEB to A ED. For the angles AEC, CEB, D which the straight line C E forms by meeting the straight line AB, are together equal to two right angles (Prop. I.) ; and the angles CEB, BED, which the straight line B E forms by meeting the straight line CD, are equal to two right angles ; hence the sum of the angles AEC, CEB is equal to the sum of the angles CEB, BED (Art. 34, Ax. 1). Take away from each of these sxims the common angle CEB, and there will remain the angle AEC, equal to its opposite angle, BED (Art. 34, Ax. 3). In the same manner it may be shown that the angle C E B is equal to its opposite angle, A E D. 50. Cor. 1. The four angles formed by two straight lines intersecting each other, are together equal to four right angles. a* -B 18 ELEMENTS OP GEOMETRY. 51. Cor. 2. All the successive angles, around a com- mon point, formed by any number of straight lines meet- ing at that point, are together equal to four right angles. PaoposiTiON V. — Theorem. 52. If two triangles have two sides and the included angle in the one equal to two sides and the included angle in the other, each to each, the two triangles will be equal. In the two triangles ABC, DEF, let the side A B be equal to the side D E, the side A C to the side D P, and the angle A to the angle D ; then the triangles ABC, D E P will be equal. Conceive the triangle ABC to be applied to the triangle D EP, so that the side AB shall fall upon its equal, D B, the point A upon D, and the point B upon E ; then, since the angle A is equal to the angle D, the side AC wiU take the direction DP. But A C is equal to D P ; therefore the point will fall upon P, and the third side B C will co- incide with the third side E P (Art. 34, Ax. 11). Hence the triangle ABC coincides with the triangle D E P, and they are therefore equal (Art. 34, Ax. 14). 53. Cor. When, in two triangles, these three parts are equal, namely, the side AB equal to DE, the side AC equal to D P, and the angle A equal to D, the other three corresponding parts are also equal, namely, the side B C equal to E P, the angle B equal to E, and the angle C equal to P. Proposition VI. — Theorem. 54. If tivo triangles have two angles and the included side in the one equal to tv)o angles and the included side in the other, each to each, the tico triangles will be equal. BOOK I. 19 In the two triangles ABC, DEP, let the angle B be equal to the angle E, the angle C to the angle F, and the side BC to the side EF; then the triangles ABC, D E F will be equal. Conceive the triangle A B C to be applied to the triangle DEP, so that the side B C shall fall upon its equal, E P, the point B upon E, and the point C upon P. Then, since the angle B is equal to the angle E, the side B A will take the direction E D ; therefore the point A will be ■found somewhere in the line ED. In like manner, since the angle C is equal to the angle P, the line C A will take the direction F D, and the point A will be found some- where in the line F D. Hence the point A, falling at the same time in both of the straight lines E D and F D, must fall at their intersection, D. Hence the two triangles ABC, DEP coincide with each other, and are therefore equal (Art. 34, Ax. 14). 65. Cor. When, in two triangles, these three parts are equal, namely, the angle B equal to the angle E, the angle C equal to the angle F, and the side B C equal to the side EF, the other three corresponding parts are also equal; namely, the side B A equal to E D, the side C A equal to PD, and the angle A equal to the angle D. Peoposition VH. — Theorem. 66. In an isosceles triangle, the angles opposite the equal sides are equal. Let ABC be an isosceles triangle, in which the side AB is equal to the side A C ; then will the angle B be equal to the angle C. Conceive the angle B A C to be bisect- ed, or divided into two eqiial parts, by 20 ELEMENTS OF GEOMETRY. the straight line AD, making the angle BAD equal to D A C. Then the two triangles BAD, CAD have the two sides A B', AD and the included angle in the one equal to the two sides AC, A D and the included angle in the other, each to each ; hence the two triangles are equal, and the angle B is equal to the angle C (Prop. V.). 67. Cor. 1. The line bisecting the vertical angle of an isosceles triangle bisects the base at right angles. 68. Cor. 2. Conversely, the line bisectmg the base of an isosceles triangle at right angles, bisects also the verti- cal angle. 69. Cor. 3. Every equilateral triangle is also equian- gular. Proposition VIII. — Theorem. 60. ^ two angles of a triangle are equal, the opposite sides are also equal, and the triangle is isosceles. Let A B C be a ti-iangle having the an- A gle B equal to the angle C ; then will tlie side AB be equal to the side A C. For, if the two sides are not equal, one of them must be greater than the other. Let A B be the greater ; then take D B equal to AC the less, and draw CD. ^ C Now, in the two triangles D B C, ABC, we have DB equal to A C by constiiiction, the side B C common, and the angle B equal to the angle A C B by hypothesis ; therefore, since two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each, the triangle D B C is equal to the triangle ABO (Prop. V.), a part to the whole, which is impossible (Art. 34, Ax. 8). Hence the sides AB and A C cannot be unequal ; therefore the triangle A B C is isosceles. BOOK I. , • 21 61. Cor. Therefore every equiangular triangle is equi- lateral. Pkoposition IX. — Theorem. 62. Any side of a triangle is less than the sum of the other two. In the triangle ABC, any one side, as AB, is less than the sum of the othei two sides, A C and C B. For the straight line AB is the shortest line that can be drawn from the point A to the point B (Art. 34, Ax. 10) ; hence the side AB is less than the sum of the sides A C and C B. In like manner it may be proved that the side A C is less than the sum of A B and B C, and the side B C less than the sum of B A and A 0. 63. Cor. Since the side AB is less than the sum of AC and C B, if we take away from each of these two unequals the side C B, we shall have the diiference between A B and C B less than A C ; that is, the difference between any tioo sides of a triangle is less than the other side. Proposition X. — Theorem. 64. The greater side of any triangle is opposite the greater angle. In the triangle CAB, let the angle C be greater than B ; then will the side A B, opposite to C, be greater than A C, opposite to B. Draw the straight line C D, making the angle BCD equal to B. Then, in the triangle B D C, we shall have the side BD equal to D C (Prop. VIII.). But the side A C is less than the sum of AD and D C (Prop. IX.), and the 22 • ELEMENTS OP GEOMETRY. Bum of A D and D C is equal to the sum of A D and D B, which is equal to A B ; therefore the side A B is greater than AC. 65. Cor. 1. Therefore the shorter side is opposite to the less angle. 66. Cor. 2. In the right-angled triangle the hypoth&- nuse is the longest side. Proposition XI. — Theorem. 67. The greater anffle of any triangle is opposite the greater side. In the triangle CAB, suppose the a side A B to be greater than A C ; then will the angle C, opposite to A B, be greater than the angle B, opposite to AC. For, if the angle C is not greater than B, it must either be equal to it or less. If the angle C were equal to B, then would the side AB be equal to the side AC (Prop. VIII.), which is contrary to the hypothesis ; and if the angle C were less than B, then would the side AB be less than AC (Prop. X. Cor. 1), wliich is also contrary to the hypothesis. Heuce, the angle C must be greater than B. 68. Cor. It follows, tlierefore, that the less angle is opposite to the shorter side. Proposition XII. — Theorem. 69. If, from any point within a triangle, two straig-ht lines are draion to the extremities of either side, their sum will be less than that of the other two sides of the triangle. BOOK I. 23 Let the two straight lines B 0, C be drawn from the point 0, within the triangle ABC, to the extremities of the side B C ; then will the sum of the two lines B and C be less than the sum of the sides B A and A C. Let the straight line BO be pro- duced till it meets the side A C in the point D ; and be- cause one side of a triangle is less than the sum of the other two sides (Prop. IX.), the side OC in the triangle C D is less than the sum of D and DC. To each of these inequalities add B 0, and %e have the sum of B and C less than the sum of BO, OD, and D C (Art. 34, Ax. 4) ; or the sum of B and C less than the sum of B D and D C. Again, because the side B D is less than the sum of B A and A D, by adding D C to each, we have the sum of B D and D C less than the sum of B A and A C. But it has been just shown that the sum of B and C is less than the sum of B D and D C ; much more, then, is the sum of B and C less than B A and A C. Proposition XIII. — Theorem. 70. From a point without a straight line, only one per- pendicular can be drawn to that line. Let A be the point, and DE the given straight line ; then from the point A only one perpendicular can be drawn to D E. Let it be supposed that we can draw two perpendiculars, AB and A C. Produce one of them, as A B, till B F is equal to A B, and join F C. ^ Then, in the triangles ABC and C B F, the angles C B A and CBP are both right angles (Prop. I. Cor. 1), the side C B is common to both, and the side B F is equal to D ^\ E 24 ELEMENTS OP GEOMETRY. D- — E the side A B ; hence the two triangles are equal, and the angle BCF is equal to the angle BOA (Prop. V.) But the angle BCA is, by hypothesis, a right angle ; therefore BCF must also be a right angle ; and if the two adja- cent angles, BCA and BCF, are to- gether equal to two right angles, the two lines A C and C F must form one and the same straight line (Prop. II.). Whence it follows, that be- tween the same two points, A and F, two straight lines can be drawn, which is impossible (Art. 34, Ax. 1*1) ; hence no mOre than one perpendicular can be drawn from the same point to the same straight line. 71. Cor. At the same point C, in the line AB, it is likewise impossible to erect more than one perpendicular to that line. For, if C D and C E were each perpendicular to A B, the angles BCD, BCE would be right angles; hence the angle BCD would be equal to the angle BCE, a part to the whole, which is impossible. E D / / B Proposition XIV. — Theorem. 72. If, from a point without a straight line, a perpen- dicular be let fall on that line, and oblique lines be drawn to different points in the same line ; — 1st. The perpendicular will be shorter than any oblique line. 2d. Any two oblique lines, which meet the given line at equal distances from the perpendicular, will be equal. 3d. Of any two oblique lines, that trhich meets the given line at the greater distance from the perpendicular will he the longer. BOOK I. 25 Let A be the given point, and D E the given straight line. Draw AB perpendicular to DE, and the oblique lines A E, AC, AD. Produce AB till BE is equal to AB, and join CP, DF. First. The triangle BCP is equal to the triangle B C A, for they have the side C B common, the side AB equal to the side B F, and the angle ABC equal to the angle F B C, both being right angles (Prop. I. Cor. 1) ; hence the third sides, CP and AC, are equal (Prop. V. Cor.). But A B F, being a straight line, is shorter tlian A C F, which is a broken line (Art. 34, Ax. 10); therefore AB, the half of A B P, is shorter than A C, the half of A C F ; hence the perpendicular is shorter than any oblique line. Secondly. If B E is equal to B C, then, since A B is common to the triangles, ABE, ABC, and the angles ABE, ABC are right angles, the two triangles are equal (Prop, v.), and the side AE is equal to the side AC (Prop. V. Cor.). Hence the two oblique lines, meeting the given line at equal distances from the perpendicular, are equal. Thirdly. The point C being in the triangle A DP, the sum of the lines AC, C P is less than the sum of the sides AD, DP (Prop. XII.) But A C has been shown to be equal to C P ; and in like manner it may be shown that A D is equal to D F. Therefore A C, the half of the line A C F, is shorter than A D, the half of the line AD F ; hence the oblique line which meets the given line the greater distance from the perpendicular, is the longer. 73. Cor. 1. The perpendicular measures the shortest distance of any point from a straight line. 74. Cor. 2. Prom the same point to a given straight line only two equal straight lines can be drawn. 26 ELEMENTS OF GEOMETEY. 75. Cor. 3. Of any two straight Hues drawn from a point to a straight line, tliat which is not shorter tlian the other will be longer than any straight line that can be drawn between them, from the same point to the same line. Proposition XV. — Theorem. 76. If from the middle point of a straight line a per- pendicular to this line be drawn, — 1st. Any point in the perpendicular will be equally dis- tant from the extremities of the line. 2d. Any point out of the perpendicular icill be zin- equally distant from those extremities. Let D C be drawn pe^rpendicular to the straight line A B, from its middle point C. First. Let D and E be points, taken at pleasure, in the perpendicular, and join DA, DB, and also AE, EB. Then, since A C is equal to B, the two oblique lines D A, D B meet points which are at the same distance from the perpendicular, and are therefore equal (Prop. XIV.). So, likewise, the two oblique lines E A, E B are equal ; therefore any point in the perpendic- ular is equally distant from the extremities A and B. Secondly. Let P be any point out of the perpendicular, and join F A, P B. Then one of those lines must cut the perpendicular, in some point, as E. Join E B ; then wo have E B equal to E A. But in {\\q triangle P E B, tlie side P B is less than the sum of the sides E P, E B (Prop. IX.), and since the sum of PE, E B is equal to the sum of P E, E A, which is equal to P A, P ]5 is loss than PA. Hence any point out of the perpendicular is at unequal distances from the cxtromifios A and B. 77. Cor. If a straight line have two points, of which each is equally distant from the extremities of another i;ooK I. 27 straiglit line, it will be perpendicular to that line at its middle point. Proposition XVI. — Theorem. 78. If two triangles have two sides of the one equal to two sides of the other, each to each, and the included an- gle of the one greater than the included angle of the other, the third side of that ivhich has the greater angle will be greater than the third side of the other. Let ABC, DEF be two triangles, having the side A B equal to D B, and A C equal to D P, and the angle A greater than D ; ^ then will the side BC be greater than BP. Of the two sides D B, D F, let D P be the side which is not shorter than the other ; make the angle E D G equal to B A C ; and make D G- equal to A G or D P, and join EG, GP. Since D P, or its equal D G, is not shorter than D E, it is longer than D H (Prop. XIV. Cor. 3) ; therefore its extremity, P, must fall below the line E G. The two tri- angles, ABO and D E G, have the two sides AB, AG equal to the two sides D B, D G, each to each, and the ■ eluded angle B A of the one equal to the included angle B D G of the other ; hence the side B is equal to EG (Prop. V. Cor.). In the triangle D P G, since D G is equal to D P, the angle D P G is equal to the angle D G P (Prop. VII.) ; but the angle D G P is greater than the angle E G P ; therefore the angle D P G is greater than B G P, and much more is the angle EPG greater tlian the angle 28 ELEMENTS OP GEOMETEY. E G F. Because the angle E F G- in the triangle E F G is greater than EGF, and because the greater side is opposite the greater angle (Prop. X.), the side E G is greater than E F ; and E G has been shown to be equal to B C ; hence B C is greater than E F. Peoposition XVII. — Theoeem. 79. If two triangles have tioo sides of the one equal to two sides of the other, each to each, but the third side of the one greater than the third side of the other, the angle contained by the sides of that which has the greater third side will be greater than the angle contained by the sides of the other. Let ABC, DBF be two triangles, the side AB equal to D B, and AC equal to DF, and the side C B greater than BF, then will the angle b~ A be greater than D. ^^"^ F For, if it be not greater, it must either be equal to it or less. But the angle A cannot be equal to D, for then the side B C would be equal to EP (Prop. V. Cor.), which is contrary to the hypothesis ; neither can it be loss, for then the side B C would be less than E F (Prop. XVI.), which also is contrary to the hypothesis ; there- fore the angle A is not less than the angle D, and it has been shown that is not equal to it ; hence the angle A must be greater than the angle D. Peoposition XVIII. — Theoeem. 80. If tioo triangles have the three sides of the one C(/aal to the three sitlrs of the other, each to each, the triangles themselves will be equal. BOOK I. 29 Let the triangles ABC, D E P have the side A B equal to D E, A C to D F, and B C to E P ; then will the angle A be eqtial to D, B the angle B to the angle E, and the angle C to the angle P, and the two triangles will also be equal. For, if the angle A were greater than the angle D, since the sides A B, A are equal to the sides D E, DP, each to each, the side B would be greater than E P (Prop. XVI.) ; and if the angle A were less than D, it would follow that the side B C would be less than E P. But by hypothesis B C is equal to E P ; hence the angle A can neither be greater nor less than D ; therefore it must be equal to it. In the same manner, it may be shown that the angle B is equal to E, and the angle C to F ; hence the two triangles must be equal. 81. Scholium. In two triangles equal to each other, the equal angles are opposite the equal sides ; tints the equal angles A and D are opposite the equal sides B C and EP. Proposition XIX. — Theorem. 82. If two right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each to each, the triangles are equal. Let the two right-an- a D gled triangles ABC, DEP, have the hypothenuse A C equal to D P, and the side A B equal to D E ; then will the triangle A B C bo equal to the triangle DEP. B G C E F The two triangles are evidently equal, if the sides B C and EP are equal (Prop. XVIII.) . If it be possible, let 3* 30 ELEMENTS OP GEOMETRY. these sides be unequal, and let B C be the greater. Take B G equal to E P, the less side, and join A G. Then, in the two triangles A B G, DBF, the angles B and B are equal, both B G C E J? being right angles, the side A B is equal to D E by hy- pothesis, and the side B G to B P by construction ; hence these triangles are equal (Prop. V.) ; and therefore A G is equal to D P. But by hypothesis D P is equal to A C, and therefore A G is equal to A C. But the oblique line A C cannot be equal to A G, which meets the same straight line nearer the perpendicular A B (Prop. XIV.) ; there- fore B C and B P cannot be unequal, hence they must be equal ; therefore the triangles ABC and DBF are equal. Peoposition XX. — Theorem. 83. If a straight line, intersecting tioo other straight lines, makes the alternate angles equal, the two lines are parallel. Let the straight line E BF intersect the two straight lines A B, CD, making the alternate an- gles B G H, C H G equal ; then the lines A B, C D will be parallel. For, if the lines A B, C D are not parallel, let them meet in some point K, and through 0, the middle point of Gil, draw the straight line IK, luaking 10 equal to K, and join H I. Tlicn the opposite angles K G, I II, formed by the intersection of the two straight lines IK, GH, are eqiial (Prop. IV.) ; and the triangles K G, BOOK I. 31 1 H have the two sides K 0, G and the included angle in the one equal to the two sides LO, OH and tlie includ- ed angle in the other, each to each; hence the angle K G is equal to the angle I H (Prop. V. Cor.). But, by hypothesis, the angle KGO is equal to the angle CHO, therefore the angle 1 H is equal to H 0, so that H I and H must coincide ; that is, the line C D when pro- duced meets I K in two points, I, K, and yet does not form one and the same straight line, which is impossible (Prop. III.) ; therefore the lines A B, C D cannot meet, conse- quently they are parallel (Art. 17). Note. — The demonstration of the proposition is substantially that given by ]\I. da Cunha in the Principes Malhe'matiques. This demon- stration Young pronounces " superior to every other that has been given of the same proposition " ; and Professor Playfair, in the Edinburgh lie- ineio, Vol. XX., calls attention to it, as a most important improvement in elementary Geometry. Proposition XXI. — Theorem. 84. If a straight line, intersecting tivo other straight lines, makes any exterior angle equal to the interior and opposite angle, or makes the interior angles on the same side together equal to tiuo right angles, the tivo lines are parallel. Let the straight line E P inter- sect the two straight lines A B, C D, making the exterior angle E G B equal to the interior and opposite angle, G H D ; then the lines A B, C D are parallel. For the angle A G H is equal to the angle E G B (Prop. IV.) ; F and E G B is equal to G H D, by hypothesis ; therefore the angle A G H is equal to the angle G H D ; and they are alternate angles ; hence the lines A B, C D are parallel (Prop. XX.). 32 ELEMENTS OP GEOMETKT. D Again, let the interior angles E on the same side, B G H, G H D, \ be together equal to two right a A^^ — angles ; then the lines A B, C D \ are parallel. \ For the sum of the angles ^ lO BGH, GHD is equal to two \ right angles, by hypothesis ; and F the sum of AG H, BGH is also eqiial to two right an- gles (Prop. I.) ; take away BGH, which is common to both, and there remains the angle GHD, equal to the angle A G H ; and these are alternate angles ; hence the lines AB, CD are parallel. E 85. Cor. If two straight lines are perpendicular to another, they are parallel ; thus AB, CD, per- pendicular to E P, are parallel. -D Proposition XXH. — Theorem. 86. If a straight line intersects two parallel lines, it makes the alternate angles equal; also any exterior angle equal to the interior and opposite angle ; and the two in- terior angles upon the same side together equal to two right angles. Let the straight line E P inter- sect the parallel lines A B, C D ; the alternate angles AGH, GHD are equal ; the exterior angle E G B is equal to the interior and opposite angle GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles. BOOK I. 33 For if the angle A G H is not equal to G H D, draw the straight Hue K L tlirough the point G, making the angle K G H equal to G H D ; then, since the alternate angles G H D, K G H are equal, K L is parallel to C D (Prop. XX.) ; but by hypothesis A B is also parallel to CD, so that through the same point, G, two straight lines are drawn parallel to G D, which is impossible (Art. 84, Ax. 12). Hence the angles AGH, GHD are not unequal ; that is, they are equal. Now, the angle E G B is equal to the angle AGH (Prop. IV.), and AGH has been shown to be equal to GHD; hence B G B is also equal to G H D. Again, add to each of these equals the angle B G H ; then the sum of the angles E G B, B G H is equal to the sum of the angles B G H, G H D. But E G B, B G H are equal to two right angles (Prop. I.) ; hence B G H, GHD are also equal to two right angles. 87. Cor. If a line is perpendicular to one of two parallel lines, it is perpendicular to the other ; thus EP (Art. 85), perpendicular to AB, is perpendicular to CD. Proposition XXIII. — Theorem. 88. If tv30 straight lines intersect a third line, and make the two interior angles on the same side together less than tioo right angles, the two lines will meet on being produced. Let the two lines KL, CD make j, with E F the angles K G H, G H C, togetlier less than two right angles ; then KL and CD will meet on being produced. For if they do not meet, tliey are parallel (Art. 17). But they are not parallel ; for then the sum 34 ELEMENTS OP GEOMETRY. of the interior angles K G H, G II C would be equal to two right angles (Prop. XXII.) ; but by hypotliesis it is less ;■ therefore the lines K L, C D will meet on being produced. 89. Scholium. The two lines K L, CD, on being pro- duced, must meet on the side of E F, on which are the two interior angles whosQ sum is less than two right angles. Proposition XXIV, - Theorem. E- n -D tt 90. Straight lines which are parallel to the same line are parallel to each other. Let the straight lines A B, C D be each parallel to tlie line E P ; then are they parallel to each other. Draw GHI perpendicular to EP. Then, since AB is parallel to EF, GI will be perpendicular to AB (Prop. XXII. Cor.) ; and since CD is pai-allel to EF, GI will for a like reason be perpendicular to CD. Conse- quently AB and CD are perpendicular to the same straight line ; hence they are parallel (Prop. XXI. Cor.). Proposition XXV. — Theorem. 91. Two parallel straight lines are evert/where equully distant from each other. Let AB, CD be two parallel straight lines. Through any two points in A B, as E and F, draw the straight lines EG, FH, per- pendicular to AB. Tliese lines will be equal to each other. For, if G F bo joined, the angles G P E, F G H, consid- ered in reference to the parallels AB, CD, will be alter- H G D BOOK I. 35 nate interior angles, and tlicrcfore equal to each other (Prop. XXII.). Also, since the straight lines E G, F H are perpendicular to the same straight line A B, and con- sequently parallel (Prop. XXI. Cor.), the angles EG-P, GPH, considered in reference to the parallels EG, PH, will be alternate interior angles, and therefore equal. Hence, the two triangles EPG, PGH, have a side and the two adjacent angles of the one equal to a side and the two adjacent angles of the other, each to each ; therefore these triangles are equal (Prop. YL) ; hence the side E G, which measures the distance of the parallels AB, CD, at "the point E, is equal to the side P H, which measures the distance of the same parallels at the point P. Hence two parallels are everywhere equally distant. Proposition XXVI. — Theorem. 92. If hoo angles have their sides parallel, each to each, and lying' in the same direction, the two angles are equal. Let A B C, D E P be two angles, ^ -p which have the side AB parallel / / to DE, and BC parallel to EP; / ^/ ^ then these angles are equal. / 7 Por produce DE, if necessary, ^ '/' C till it meets B C in the point G. h ^ F Then, since B P is parallel to G C, the angle D B P is equal to D G C (Prop. XXII.) ; and since D G is parallel to A B, the angle D G C is equal to ABC; hence the angle DEP is equal to ABC. 93, Scholium. This proposition is restricted to the case wliere the side E P lies in the same direction with B C, since if P E were produced toward H, the angles D E H, ABC would only be equal when they are right angles. 3G ELEMENTS OP GEOMETRY. PropositioK XXVII. — Theorem. 94. If any side of a iriang-Ie be prodvced, the exterior angle is equal lo the sum of the tivo interior and opposite angles. Let ABC be a triangle, and let one of its sides, B C be pro- diiced towards D ; then the ex- terior angle A C D is equal to the two interior and opposite g^ angles, CAB, ABC. For, draw E C parallel to the side A B ; then, since A C meets the two parallels A B, EC, the alternate angles BAC, ACE are equal (Prop. XXII.). Again, since B D meets the two parallels A B, EC, the exterior angle E C D is equal to the interior and opposite angle ABC. But the angle A C E is equal to BAC; therefore, the whole exterior angle A C D is equal to the two interior and opposite angles CAB, ABC (Art. 34, Ax. 2). Proposition XXVIII. — Theorem. 95. In every triangle the svni of the three anglrs is eqval to two right angles. Let A B C be any triangle ; tlien will the sura of the angles ABC, BCA, CAB be equal to two right angles. For, let the side B C be pro- duced towards D, making the exterior angle A C D ; then the CAB and ABC (Prop. XXVII.) D angle A C D is equal to T(i onch of these equals add the angle A C B, and we shall have the sum of BOOK I. 37 A C B and A C D, eqtial to the sum of A B C, BOA, and CAB. But the sum of A C B and A C D is equal to two right angles (Prop. I.) ; hence the sum of the three an- gles ABC, B C A, and C AB is equal to two right angles (Art. 34, Ax. 2). 96. Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles. 97. Cor. 2. If two angles in one triangle be respective- ly equal to two angles in another, their third angles will also be equal. 98. Cor. 8. A triangle cannot have more than one an- gle as great as a right angle. 99. Cor. 4. And, therefore, every triangle must have at least two acvite angles. 100. Cor. 6. In a right-angled triangle the right angle is equal to the sum of the other two angles. 101. Cor. 6. Since every equilateral triangle is also equiangular (Prop. VII. Cor. 3), each of its angles will be equal to two thirds of one right angle. Proposition XXIX. — Theokem. 102. The sum of all the interior angles of any polygon is equal to twice as many right angles, less four, as the figure has sides. Let A B C D E be any polygon ; then the sum of all its interior angles, A, B, C, D, E, is equal to twice as many right angles as the figure has sides, less four right angles. For, from any point P within the pol- A B ygon, draw the straight lines PA, PB, PC, P D, P B, to the vertices of all the angles, and the polygon will be 4 38 ELEMENTS OP GEOMETRY. divided into as many triangles as it has sides. Now, the sum of the three angles in each of these triangles is equal to two right angles (Prop. XXVIII.) ; therefore the sum of the angles of all these triangles is equal to twice as many right angles as there are triangles, or sides, to the polygon. But the sum of all the angles about the point P is equal to four right angles (Prop. IV. Cor. 2), which sum forms no part of the interior angles of the polygon ; therefore, deducting the sum of the angles about the point, there remain the angles of the polygon equal to twice as many right angles as the figure has sides, less four right angles. 103. Cor. 1. The sum of the angles in a quadrilateral is equal to four right angles ; hence, if all the angles of a quadrilateral are equal, each of them is a right angle ; also, if three of the angles are right angles, the fourth is likewise a right angle. 104. Cor. 2. The sum of the angles in a pentag-on is equal to six right angles ; in a hexagon, the sum is equal to eight right angles, &c. 105. Cor. 3. In every equiamgular figure of more than four sides, each angle is greater than a right angle ; thus, in a regular pentagon, each angle is equal to one and one fifth right angles ; in a regular hexagon, to one and one third right angles, &c. 106. Scholium. In applying this prop- osition to polygons which have re-en- trant angles, or angles whose vertices are directed inward, as B P 0, each of these angles must be considered greater than two right angles. But, in order to avoid ambiguity, we shall hereafter ^ limit our reasoning to polygons with snlicnt angles, or with angles directed outwards, and which may be called convex polygons. Every convex polygon is such that a BOOK I. 39 straight line, ho-n^ever drawn, cannot meet the perimeter of tlie jjolygdii iu more than two points. Peoposition XXX. — Theorem. 107. The sum of all the exterior angles of any polygon, formed by producing each side in the same direction, is equal to four right angles. Let each side of the polygon ABODE be produced in the same direction ; then the sum of the exterior angles A, B, C, D, E, will be equal to four right angles. For each interior angle, together with its adjacent exterior angle, is equal to two right angles (Prop. 1.) ; hence the sum of all the angles, both interior and exterior, is eqiial to twice as many right angles as there are sides to the polygon. But the sum of the interior angles alone, less four right angles, is equal to the same sum (Prop. XXIX.) ; therefore the sum of the exterior angles is equal to four right angles. Proposition XXXI. — Theorem. 108. The opposite sides and angles of every parallelo- gram are equal to each other. Let A B C D be a parallelogram ; then the opposite sides and angles are equal to each other. Draw the diagonal B D, then, since the opposite sides A B, DC are paral- A B lei, and BD meets them, the alternate angles ABD, BDC are equal (Prop. XXII.) ; and since AD, BC are parallel, and B D meets them, the alternate angles A D B, D B C are likewise eqiial. Hence, the two triangles AD B, DBG have two angles, ABD, AD B, in the one, equal to two angles, BDC, DBC, in the otlier, each to eacli; and since 40 ELEMENTS OP GEOMETRY. the side BD included between these equal angles is common to the two tri- angles, they are equal (Prop. VI.) ; hence the side AB opposite tlie angle A D B is equal to the side D C opposite A B the angle D B C (Prop. VI. Cor.) ; and, in like manner, the side A D is equal to the side B C ; hence the opposite sides of a parallelogram are equal. Again, since the triangles are equal, the angle A is equal to the angle C (Prop. VI. Cor.) ; and since the two angles D B C, A B D are respectively equal to the two angles A D B, B D C, the angle ABC is equal to the angle ADC. 109. Cor. 1. The diagonal divides a parallelogram into two equal triangles. 110. Cor. 2. The two parallels AD, B C, included be- tween two other parallels, AB, CD, are equal. Proposition XXXII. — Theorem. 111. J^ the opposite sides of a quadrilateral are equal, each to each, the equal sides are parallel, and the ^figure is a parallelogram. Let A B C D be a quadrilateral having its opposite sides equal ; then will the equal sides be parallel, and the figure be a parallelogram. For, having drawn the diagonal A B B D, the triangles A B D, B D C have all the sides of the one equal to the corresponding sides of the other ; there- fore they are eqxial, and the angle A D B opposite tlie side A B is equal to D B C opposite C D (Prop. XVIII. Sch.) ; hence the side AD is parallel to B C (Prop. XX.). For a like reason, A B is parallel to C D ; therefore the quad- rilateral A B C D is a parallelogram. BOOK I. 41 Pkoposition XXXIII. — Theokem. 112. If two opposite sides of a quadrilateral are equal and parallel, the other sides are also equal and parallel, and the figure is a parallelogram. Let A B C D be a quadrilateral, having the sides A B, CD equal and parallel ; then will the other sides also be equal and parallel. Draw the diagonal B D ; then, since -^ ^ A B is parallel to C D, and B D meets them, the alternate angles A B D, B D C are equal (Prop. XXII.) ; moreover, in the two triangles A B D, D B C, the side B D is com- mon ; therefore, two sides and the included angle in the one are eqiial to two sides and the included angle in the other, each to each ; hence these triangles are equal (Prop, v.), and the side AD is equal to B C. Hence the angle A D B is equal to D B C, and consequently A D is parallel to B C (Prop. XX.) ; therefore the figure A B C D is a parallelogram. Proposition XXXIV. — Theobem. 113. The diagonals of every parallelogram bisect each other. Let A B C D be a parallelogram, D C and A C, D B its diagonals, intersect- ing at E ; then will A E equal E C, and B E equal E D. For, since A B, C D are parallel, ^ B and B D meets them, the alternate angles ODE, ABE are equal (Prop. XXII.) ; and since A C meets the same parallels, the alternate angles BAE, ECD are also equal; and the sides AB, CD are equal (Prop. XXXI.). Hence the triangles ABE, ODE liave two angles and the in- 4# 42 ELEMENTS OP GEOMETRY. eluded side in the one equal to two 'angles and the includ- ed side iu the other, each to each ; hence the two triangles are equal (Prop. VI.) ; therefore the side A E opposite the angle A B E is equal to C E opposite C D E ; hence, also, the sides B E, D E opposite the other equal angles are equal. 114. Scholium. In the case of a rhom- biis, the sides A B, B C being equal, the triangles .A E B, B B C have all the sides of the one equal to the corresponding sides of the other, and are, therefore, B equal ; whence it follows that the angles A E B, BBC are equal. Therefore the diagonals of a rhombus bisect each other at right angles. Proposition XXXV. — Theorem. 115. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Let A B C D be a quadrilateral, and D C AC, D B its diagonals intersecting at B ; then will the figure be a parallelogram. For, in the two triangles ABE, CDE, the two sides A E, E B and the included -^ B angle in the one are equal to the two sides C E, E D and the included angle in the other ; hence the triangles ai-e equal, and the side AB is equal to the side CD (Prop. V. Cor.). For a like reason, A D is equal to C B ; therefore the quadrilateral is a parallelogram (Prop. XXXII.). . BOOK II. RATIO AND PROPORTION. DEFINITIONS. 116. Ratio is the relation, in respect to quantity, ■wliicli one magnitude bears to another of the same kind ; and is the quotient arising from dividing the first by the second. A ratio may be written in tlie form of a fraction, or witli the sign : . Thus the ratio of A to B may be expressed cither by -, or by A : B. 117. The two magnitudes necessary to form a ratio are called the terms of the ratio. The first term is called the ANTECEDENT, and tlic last, the consequent. 118. Ratios of magnitudes may be expressed by num- bers, either exactly, or approximately. This may be illustrated by the operation of finding the }iumerical ratio of two straight lines, AB, CD. Prom the greater line AB cut off a part equal I' to the less C D, as many times as possible ; for ex- E G ample, twice, with the remainder B E. From the line C D cut off a part equal to tlie remainder B B as many times as possible ; once, for example, with tlie remainder D P. From the first remainder B E, cut off a part equal to the second D F, as many times as possible ; once, for ex- ample, with the remainder B G. 44 ELEMENTS OP GEOMETRY. Prom the second re- ^ P mainder DP, cut off a 3? •part equal to B G, the r ^ . A' ' ' ' — 'R third, as many times as EG possible. Proceed thus till a remainder arises, -which is exactly contained a certain number of times in the preceding one. Then this last remainder will be the common measure of the proposed lines ; and, regarding it as unity, we shall easily find the values of the preceding remainders ; and, at last, those of the two proposed lines, and hence their ratio in numbers. Suppose, for instance, we find G B to be contained ex- actly twice in P D ; B G will be the common measure of the two proposed lines. Let B G equal 1 ; then will P D equal 2. But E B contains P D once, plus G B ; there- fore we have E B equal to 3. CD contains E B once, plus P D ; therefore we have C D equal to 5. A B con- tains C D twice, plus E B ; therefore we have A B equal to 13. Hence the ratio of the two lines is that of 13 to 5. If the line C D were taken for unity, the line A B would be -'s* ; if A B were taken for unity, C D would be ^. It is possible that, however far the operation be con- tinued, no remainder may be found which shall be con- tained an exact number of times in the preceding one. In that case there can be obtained only an approximate ratio, expressed in numbers, more or less exact, according as the operation is moi'e or less extended. 119. When the greater of two magnitudes contains .the less a certain number of times without having a remain- der, it is called a multiple of the less ; and the less is then called a submultiple, or measure of the greater. Thus, 6 is a- multiple of 2 ; 2 and 3 are siibmultiplos, or measures, of 6. 120. Equimultiples, or like multiples, are those which contain their respective submultiples the same number of BOOK II. 45 times ; and equisubmultiples, or like submultiples, are those contained in tlieir respective multiples the same number of times. Thus 4 and 5 are like submultiples of 8 and 10 ; 8 and 10 are like multiples of 4 and 5. 121. Commensurable magnitudes are magnitudes of the same kind, which have a common measure, and whose ratio therefore may be exactly expressed in numbers. 122. Incommensurable magnitudes arc magnitudes of the same kind, whicli have no common measure, and whose ratio, therefore, cannot be exactly expressed in numbers. 120. A DIRECT ratio is the quotient of the antecedent by the consequent ; an inverse ratio, or reciprocal ratio, is the quotient of the consequent by the antecedent, or the reciprocal of the direct ratio. Thus the direct ratio of a line 6 feet long to a line 2 feet long is I or 3 ; and the inverse ratio of a line 6 feet long to a line 2 feet long is | or \, which is the same as the reciprocal of 3, the direct ratio of 6 to 2. The word ratio when used alone means the direct ratio. 124. A compound ratio is the product of two or more ratios. Tlius the ratio compounded of A : B and : D is A AJ^C B ^ D'^"" BXD' 125. A proportion is an equality of ratios. Four magnitudes are in proportion, when the ratio of the first to the second is the same as that of the third to the fourth.- Thus, the ratios of A : B and X : Y, being equal to A X each other, when written A : B = X : Y, or — = =, form a proportion. 46 ELEMENTS OP GEOMETRY. 126. Proportion is written not only with the sign =, but, more often, with the sign : : between the ratios. Thus, A : B : : X : Y, expresses a proportion, and is read. The ratio of A to B is equal to the ratio of X to Y ; or, A is to B as X is to Y. 127. TliQ first and third terms of a proportion are called the ANTECEDENTS; the second and fourth, the consequents. The Jlrst and fou/rth are also called the extremes, and the second and third the means. Thus, in the proportion A : B : : C : D, A and C arc the antecedents ; B and D are the consequents ; A and D are the extremes ; and B and C are the means. The antecedents are called homologous or like terms, and so also are the consequents. 128. All the terms of a proportion are called propor- tionals ; and the last term is called a fourth propor- tional to the other three taken in their order. Thus, in the proportion A : B : : C : D, D is the fourth proportional to A, B, and C. 129. When both the means are the same magnitude, either of them is called a mean proportional between the extremes ; and if, in a series of proportional magnitudes, each consequent is the same as the next antecedent, those magnitudes are said to be in continued proportion. Thus, if we have A : B : : B : C : : C : D : : D : E, B is a mean proportional between A and C, C between B and D, D between and E ; and the magnitudes A, B, C, D, E are said to be in continued proportion. 130. "When a continued proportion consists of but three terms, the middle term is said to be a mean peoportional between the other two ; and the last term is said to be the third proportional to the first and second. Thus, when A, B, and C arc in jirdportiou, A : B : : B : C ; in which case B is called a mean proportional between A and C ; and C is called the third proportional to A and B. BOOK II. 47 131. Magnitudes are in proportion by inversion, or INVERSELY, when each antecedent takes the place of its consequent, and each consequent the place of its antece- dent. Thus, let A : B : : C : D ; then, by inversion, B : A : : D : C. 132. Magnitudes are in proportion by alternation, or ALTERNATELY, when antecedent is compared with antece- dent, and consequent with consequent. Thus, let A : B : : D : C ; then, by alternation, A : D : : B : 0. 133. Magnitudes are in proportion by composition, when the sum of the first antecedent and consequent is to the first antecedent, or consequent, as the sum of the second antecedent and consequent is to the second ante- cedent, or consequent. Thus, let A : B : : : D ; then, by composition, AH-B:A::C-(-D:C, or A + B:B::C + D:D. 134. Magnitudes are in proportion by division, when the difference of the first antecedent and consequent is to the first antecedejit, or consequent, as the difference of the second antecedent and consequent is to the second ante- cedent, or consequent. Thus, let A : B : : C : D ; then, by division, A — B:A::C — D:C, or A — B : B : : C — D : D. Proposition I. — Theorem. 135. If four magnitudes are in proportion, the product of the tivo extremes is equal to the product of the tivo means. Let A : B : : C : D ; then will A X D = B X C. For, since the magnitudes are in proportion, A B = D' 48 ELEMENTS OF GEOMETRY. and reducing the fractions of this equation to a common denominator, we have AX i) ^ B X C B X D B X D' or, the common denominator being omitted, AX D = BX 0, Peoposition II. — Theorem. 136. If the product of two magnitudes is equal to the product of two others, these four magnitvdes form a proportion. Let A X D = B X C ; then will A : B : : C : D. For, dividing each member of the given equation by B X D, we have AX D _ BX C B X D B X D' which, reduced to the lowest terms, gives A _ B D' Whence A : B : : C : D. Proposition III. — Theorem. 137. If three magnitvdes are in proportion , the product of the two extremes is equal to the square of the viean. Let A : B : : B : C ; then will A X C == B\ For, since the magnitudes ai-e in proportion, A ^ B B C' and, by Prop. I,, A X C = B X B, or A X C = B". BOOK II. 49 Proposition IV . — Theorem . 138. If the product of any two quantities is equal to the square of a third, the third is a mean proportional between the other two. Let Ax C ^ B'' ; then B is a mean proportional be- tween A and C. For, dividing eacli member of the given equation by B X C, we have A _ B B ~ C' whence A : B : : B : C. Proposition V. — Theorem. 139. If four magnitudes are in proportion, they will be in proportion when taken inversely. Let A : B : : C : D ; then will B : A : : D : C. For, from the given proportion, by Prop. I., we have AXD = BXC, or BxC = AxD. Hence, by Prop. II., B : A : : D : C. Proposition VI. — Theorem. 140. If four magnitudes are in proportion, they will be in proportion when taken alternately. Let A : B : : C : D ; then will A : C : : B : D. For, since the magnitudes are in proportion, A C B = D' T> and multiplying each member of this equation by p, we have Ax B _ C X B B X C — D X C 5 60 ELEMENTS OP GEOMETRY. 4 which, reduced to the lowest terms, gives A _ B C ~ D' whence A : C : : B : D. Proposition VII. — Theorem. 141. If four magnitudes are in proportion, they will be in proportion by composition. Let A : B : : C : D ; then will A + B : A : : C + D : C. For, from the given proportion, by Prop. I., we have B X C = A X D. Adding A X C to each side of this equation, we have AXC + BXC = AXC + AXD, and resolving each, member into its factors, (A + B) X C = (C + D) X A. Heuce, by Prop. II., A+B:A::C + D:C. Proposition VIII. — Theorem. 142. If fomr magnitudes are in proportion, they will be in proportion by dlvisioi. Let A : B : : C : D ; then will A — B : A : : C — D : C. For, from the given proportion, by Prop. I., we have B X C = A X D. Subtracting each side of this eqiiation from A X C, we have AXC — BXC = AXC — AXD, and resolving each member into its factors, (A — B) X C = (C — D) X A. Hence, by Prop. II., A — B : A : : C — D : C. BOOK II. 51 Proposition IX. — Theorem. 143. Equimultiples of two magnitudes have the same ratio as the magnitudes themselves . Let A and B be two magnitudes, and my, A. and wi X B their equimultiples, then will m X A : m X B : : A : B. For . A X B = B X A ; Multiplying each side of this equation by any number, m, we have TOXAxB = mXBxA; therefore (m X A) X B = (m X B) X A. Hence, by Prop. II., mXA:mXB::A:B. Proposition X. — Theorem. 144. Magnitudes which are proportional to the same proportionals, will be proportional to each other. Let A : B : : E : F, and C : D : : E : F ; then will A: B : : C:D. For, by the given proportions, we have A E , C B - = ^, and j^ = -. Therefore, it is evident (Art. 34, Ax. 1), A _ C B ~ D' Hence A : B : : C : D. 145. Cor. 1. If two proportions have an antecedent and its consequent the same in both, the remaining terms will be in pi-oportion. 146. Cor. 2. Therefore, by alternation (Prop. VL), if two proportions have the two antecedents or the two con- 52 ELEMENTS OF GEOMETRY. sequents the same in both, the remaining terms will be in proportion. Peoposition XI. — Theorem. 147. If any number of magnitudes are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let A : B : : C : D : : B : P ; then will A:B::A+C + E:B + D + F. For, from the given proportion, we hare A X D = B X C, and A X F = B X E. By adding A X B to the sum of the corresponding sides of these equations, we have AXB+AXD + AXF = AXB + BXC + BXE. Therefore, A X (B + D + F) = B X (A + C + E). Hence, by Prop. II., A:B::A+C + E:B + D+F. Proposition XII. — Theorem. 148. If four magnitudes are in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. Let A : B : : C : D ; then will A + B:A— B::C + D:C — D. For, from the given proportion, by Prop. VII., we have A + B: A: : C + D : C; and from the given proportion, by Prop. Till., we have A — B : A : : C — D : C. Hence, from these two proportions, by Prop. X. Cor. 2, we have A + B:A — B::C + D:C — D. BOOK II. 53 Peoposition XIII. — Theorem. 149. If there be two sets of proportional magnitudes, the products of the corresponding terms unll be propor- tionals. Let A : B : : : D, and E : F : : G : H ; then will AxB:BxF::CxG:DxH. For, from the first of the given proportions, by Prop. I., we have A X D = B X C; and from the second of the given proportions, by Prop. I., we have B X H = F X 0. Multiplying together the corresponding members of these equations, we have AXDXEXH = BXCXPXG. Hence, by Prop. II., AXE:BXF::CXG:DXH. Proposition XIV. — Theorem. 150. If three magnitudes are proportionals, the first vnll be to the third as the square of the first is to the square of the second. Let A : B : : B : C ; then will A : C : : A^ : Bl For, from the given proportion, by Prop. III., we have A X C = B^ Multiplying each side of this equation by A gives A^ X C = A X B^ Hence, by Prop. II., A : C : : A"" : B^ 5* 54 ELEMENTS OF GEOMETRY. Proposition XV. — Theorem. 151. If four magnitudes are proportionals, their like ■powers and roots will also be proportional. Let A : B : : C : D ; then will A" : B" : : C" : D", and A» : Bi : : C* : D". For, from the given proportion, we have A _ C B ~ D' Raising both members of this equation to the «th power, we have and extracting the «th root of each member, we have A* _ Ci B* " Di" Hence, by Prop. II., the last two equations give A" : B" : : C : D", and A* : Bi : : Ci : D*. BOOK III. THE CIRCLE, AND THE MEASURE OF ANGLES. DEFINITIONS. 152. A CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure A D B E. 153. The ciHCUMFBRENCE or PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of which are equally distant from a point within called the centre. 154. A RADIUS of a circle is any straight line drawn from the centre to the circumference ; as the line C A, C D, or C B. 155. A DIAMETER of a circlc is any straight line drawn through the centre, and terminating in both directions in the circumference ; as the line AB. All the radii of a circle are eqiial ; all the diameters are also equal, and each is double the radius. 156. An ARC of a circle is any part of the circumference ; as the part AD, AB, or EGF. 157. The CHORD of an arc is the straight line joining its extremities ; thus B P is the chord of the arc EGP. 66 ELEMENTS OP GEOMETRY. 158. The SEGMENT of a circle is the part of a circle included be- tween an arc and its chord ; as the surface included between the arc -^' EGP and the chord EP. E 159. The SECTOR of a circle is the part of a circle included between an arc, and the two radii drawn to the extremities of the arc ; as the surface included between the arc A D, and the two radii CA, CD. 160. A SECANT to a circle is a straight line which meets the cir- cumference in two points, and lies partly within and partly without the circle ; as the line A B. ^^^__^_^ "IT D 161. A TANGENT to a circle is a straight line which, how far so ever produced, meets the circumference in but one point; as the line CD. The point of meeting is called the point op contact ; as the point M. 162. Two circumferences touch each other, when they have a point of contact without cutting one an- other ; thus two circumferences touch each other at the point A, and two at the point B. 163. A straight lixe is ix- SCRIBED in a circle wlicn its ex- tremities are in the circumference; as the lino A B, or B C. 1()4. An INSCRIBED ANfiLE is ono wliich has its vortex in the circumference, and is formed by two cliords ; as the angle ABO. BOOK III. 57 165. An INSCRIBED POLYGON is 0116 ■which has the vertices of all its angles in the circuiufercnce of the circle ; as the triangle ABO. 166. The circle is then said to be ciecumscribed about the polygon. D E 167. A POLYGON is CIRCUMSCRIBED about a circle when all its sides are tangents to the circumference ; as the polygon ABO DBF. 168. The circle is then said to be inscribed in the polygon. Proposition I. — Theorem. 169. Every diameter divides the circle and its circum- ference each into two equal parts. Let A E B F be a circle, and A B r a diameter ; then the two parts AEB, APB are equal. For, if the figure A E B be applied to AFB, their common base AB re- taining its position, the curve line ABB, must fall exactly on the curve Ihic AFB; otherwise there would be points in the one or the other unequally distant from the centre, which is contrary to the definition of the circle (Art. 152). Hence a diameter divides the circle and its circumference into two equal parts. 170. Cor. 1. Conversely, a straight line dividing the circle into two equal parts is a diameter. 58 ELEMENTS OP GEOMETRY. Tor, let the line AB divide the circle ABBCP into two equal parts; then, if the centre is not in A B, let A C be drawn through 'it, which is therefore a diameter, and conse- quently divides the circle into two equal parts ; hence the surface AFC is equal to the surface A F C B, a part to the whole, which is impossible. 171. Cor. 2. The arc of a circle, whose chord is a diameter, is a semi-circumference, and the included seg- ment is a semicircle. Proposition II. — Theorem. 172. A straight line cannot meet the circumference of a circle in more than two points. For, if a straight line could meet the circumference ABD, in three points. A, B, D, join each of these points with the centre, C ; then, since the straight lines C A, C B, C D are radii, they are equal (Art. 155) ; hence, three equal straight lines can be drawn from the same point to the same straight line, which is impossible (Prop. XIV. Cor. '2, Bk. 1.). Proposition III. — Theorem. 173. In the same circle, or in equal circles, equal arcs are subtended by equal chords; and, concerseli/, equal chords svbtend equal arcs. Let A D B and E G F bo two equal circles, and let the arc AD be equal to EG; thou will the chord AD be equal to the chord E G. BOOK III. 59 For, since the diameters A B, EP' are equal, the semicircle ^f 7 ^B E A D B may be applied to the semicircle E G F ; and the curve line ADB will coincide with the curve line BGF (Prop. I.). But, by hypothesis, the arc AD is equal to the arc E G ; hence the point D will fall on G ; hence the chord A D is equal to the chord E G (Art. 34, Ax. 11). Conversely, if the chord A D is equal to the chord E G, the arcs A D, B G will be equal. For, if the radii CD, G are drawn, the triangles A C D, E G, having the tlu'ee sides of the one equal to the three sides of the other, each to each, are themselves equal (Prop. XVIII. Bk. I.) ; therefore the angle A C D is equal to the angle E G (Prop. XVIII. Sch., Bk. I.). If now the semicircle A D B be applied to its equal BGF, with the radius A on its equal B 0, since the angles A C D, E G are equal, the radius D will fall on G, and the point D on G. Therefore the arcs A D and B G coincide with each other ; hence they must be equal (Art. 34, Ax. 14). Proposition IV. — Theorem. 174. In the same circle, or in equal circles, a greater arc is subtended by a greater chord; and, conversely, the greater chord subtends the greater arc. In the circle of which C is the centre, let the arc A B be greater than the arc A D ; then will tlie chord A B be greater than tlie chord AD. Draw the radii CA, CD, and C B. The two sides AC, GO ELEMENTS OP GEOMETRY. C B in the triangle A C B are equal to the two A C, C D in the triangle A CD, and the angle AC 13 is greater than the angle A C D ; therefore the tliird side AB is greater than the third side A D (Prop. XVI. Bk. I.) ; hence tlie chord which subtends the greater arc is the greater. Converselij, if the chord A B be greater than the chord AD, the arc A B will be greater than the arc AD. For the triangles ACB, ACD have two sides, AC, CB, in the one, equal to two sides, AC, CD, in tlie other, while the side A B is greater than the side A D ; therefore the angle ACB is greater than the angle ACD (Prop. XVII. Bk. I.) ; hence the arc A B is greater than the arc A D. 175. Scholium. The arcs here treated of are each less than the semi-circumference. If they were greater, the contrary would be true ; in which case, as the arcs in- creased, the chords would diminish, and conversely. Proposition V. — Theorem. 176. In the same circle, or in equal circles, radii ichich make equal angles at the centre intercept equal arcs on the circumference ; and, conversely, if the intercepted arcs are equal, the angles made by the radii are also equal. Let ACB and D C E be equal angles made by nidii at the centre of equal cir- cles ; thou will the intercepted arcs A B and DE be also equal. F First. Shico tlio angles A C B, D C E are equal, the ono may ba applied to the other; and since their sides, BOOK III. 61 being radii of equal circles, are eqiial, the point A will coincide with D, and the point B witli B. Therefore the arc A B must also coincide witli the arc I) E, or there would be points in the one or the other unequally distant from the centre, which is impossible ; hence the arc A B is equal to the arc D E. Second. If the arcs A B and D E are equal, the angles A C B and D C E will be equal. For, if these angles are not equal, let ACB be the greater, and let A C P be taken equal to D C E. From what has been shown, we shall have the arc AF equal to the arc D B. But, by hypothesis, A B is equal to D E ; hence A F must be equal to A B, the part to the whole, wliich is impossible ; hence the angle A C B is equal to the angle DOE. Proposition VI. — Theorem. 177. The radius which is perpendicular to a chord bi- sects the chord, and also the arc subtended by the chord. Let the radius C E be perpendicu- lar to the chord A B ; then will C B bisect the chord at D, and the arc AB at E. Draw the radii C A and C B. Then C A and C B, witli respect to the perpendicular C B, are equal oblique lines drawn to the chord AB ; tlierefore their extremities are at equal distances from the perpendicular (Prop. XIY. Bk. I.) ; hence A D and D B are equal. Again, since the triangle ACB has the sides A C and C B equal, it is isosceles ; and the line C E bisects tlie base A B at right angles ; therefore C E bisects also the angle ACB (Prop. VII. Cor. 2, Bk. I.). Pince the an- gles A D, D C B are equal, the arcs A E, B B are equal 6 62 ELEMENTS OP GEOMETRY. (Prop. V.) ; hence the radius C E, which is perpendicular to the chord A B, bisects the arc A B subtended by the chord. 178. Cor. 1. Any straight line which joins the centime of the circle and the middle of the chord, or the middle of the arc, must be perpendicular to the chord. For the perpendicular from the centre C passes through the middle, D, of the chord, and the middle, E, of the arc subtended by the chord. Now, any two of these three points in the straight line C E are sufficient to determine its position. 179. Cor. 2. A perpendicular at the middle of a chord passes through the centre of the circle, and through the middle of the arc subtended by the chord, bisecting at the centre the angle which the arc subtends. Proposition VII. — Theorem. 180. Tlirough three given points, not in the same straight line, one circumference can be made to pass, and but one. Let A, B, and C be any three points not in the same straight line ; one circumference can be made to pass tlirough them, and but one. Join A B and B C ; and bisect these straight lines by the perpendic- ulars D B and P E. Join D P ; then, tlie angles B D E, B P E, being each a right angle, are together . equal to two right angles; therefore the angles E D P, E P D are together less than two right angles ; hence D E, P.E, produced, must meet in some point E (Prop. XXIII. Bk. I.). Now, since the point E lies in the perpendicular D E, it is equally distant from the two points A and B (Prop. XV. Bk. I.) ; and since the same point E lies in the per- BOOK III. 63 pendicular PE, it is also equally distant from the two points B and C ; therefore the three distances, E A, E B, E C, are equal ; hence a circumference can be described from the centre E passing tlu-ough the three points A, B, C. Again, the centre, lying in the perpendicular DE bi- secting the chord AB, and at the same time in the per- pendicular F E bisecting the chord B C (Prop. VI. Cor. 2), must be at the point of their meeting, E. There- fore, since there can be but one centre, but one circiim- ference can be made to pass through three given points. 181. Cor. Two circumferences can intersect in only two points ; for, if they have three points in common, they must have the same centre, and must coincide. Proposition VIII. — Theoeem. 182. Equal chords are equally distant from the centre ; and, conversely, chords which are equally distant from the centre are equal. Let AB and D E b3 equal chords, and the centre of the circle ; and draw OF perpendicular to AB, and C G perpendicular to D E ; then these perpendiculars, which measure the distance of the chords from the centre, are equal. Join C A and CD. Then, in the righlrangled triangle C A F, C D G, the hypothenuses C A, C D are equal ; and the side A F, the half of A B, is equal to the side D G, the half of D E ; therefore the triangles are equal, and C P is equal to C G (Prop. XIX. Bk. I.) ; hence the two equal chords A B, D E are equally distant from the centre. Conversely, if the distances C F and C G are equal, the chords A B and D E are equal. For, in the right-angled triangles A C F, D C G, the hypothenuses C A, C D are equal ; and the side C F is 64 ELEMENTS OP GEOMETRY. equal to the side C G ; therefore the triangles are equal, and AF is equal to D G ; hence AB, the double of AP, is equal to D E, the double of D G (Art. 34, Ax. 6). Peoposition IX. — Theokem. 183. Of two wnequal chords, the less is the farther from the centre. Of the two chords D E and A H, let A H be the greater ; then will D E be the farther from the cen- tre C. Since the chord AH is greater than the chord D E, the arc A H is greater than the arc D E (Prop. IV.). Cut oif from the arc A H a part, A B, equal D E ; draw C P perpendicular to this chord, C I perpendicular to A H, and C G perpendicular to D E. C P is greater than C (Art. 34, Ax. 8), and C O than C I (Prop. XIV. Bk. I.) ; therefore C P is greater tlian C I. But C F is equal to C G, since tlie chords A B, D E are equal (Prop. VIII.) ; therefore, C G is greater than C I ; hence, of two unequal chords, the less is the farther from the ceutre. Pboposition X. — Theorem. 184. A straight line perpendicular to a radius at its termination in the circumference, is a tangent to the circle. Let the straight line B D be per- pendicular to the radius A at its termination A ; then will it be a tangent to the circle. • Draw from the centre C to B D any other straight lino, as C E. Tlien, since C A is perpendicular to B D, it is shorter than the oblique BOOK in. 65 line C E (Prop. XIV. Bk. I.) ; hence the point E is with- out the circle. The same may be shown of any other point in the line B D, except the point A ; therefore B D meets the circumfei-ence at A, and, being produced, does not cat it ; hence B D is a tangent (Art. 161). Proposition XI. — Theorem. 185. If a line is a tangent to a circumference, the ra- dius drawn to the point of contact with it is perpendicular to the tangent. Let B D be a tangent to the cir- cumference, at the point A; then will the radius C A be perpendicu- lar to BD. For every point in B D, except A, being without the circumference (Prop. X.), any line CE drawn from the centre C to B D, at any point other tlian A, must terminate at E, without tlie cir- cumference ; therefore the radius C A is tlie shortest line that can be drawn from the centre to B D ; hence C A is perpendicular to the tangent B D (Prop. XIV. Cor. 1, Bk. I.). 186. Cor. Only one tangent can be drawn throiigh the same point in a circumference ; for two lines cannot both be perpendicular to a radius at the same point. Proposition XII. — Theorem. 187. Two parallel straight lines intercept equal arcs of the circumference. First. "When the two parallels are secants, as AB, DE. Draw the radius C H perpendicular to A B ; and it will also be perpendicular to D E (Prop. XXII. Cor., Bk. I.) ; 6* 66 ELEMENTS OP GEOMETRY. therefore the point H will be at the same time the middle of the arc AH B and of the arc D H E (Prop. VI.) ; therefore, the arc AH is equal to the arc H B, and the arc D H is equal to the arc H B ; hence A H diminished by D H is equal to H B diminished by H E ; that is, the in- tercepted arcs AD, BE are equal. Second. When of the two parallels, one, as A B, is a secant, and the other, as D E, is a tangent. Draw the radius C H to the point of contact H. This radius will be D ■ perpendicular to the tangent D E ^ ■ (Prop. X.), and also to its parallel AB (Prop. XXn. Cor., Bk. I.). But, since C H is perpendicular to the chord AB, the point H is the middle of the arc A H B ; hence the arcs AH, HB, included between the parallels AB, D E, are equal. Third. When the two parallels are tangents, as D E, IL. Draw the secant AB parallel to either of the tangents, and it will be parallel to the other (Prop. XXIV. Bk. I.) ; then, from what has been just shown, the arc AH is equal to the arc H B, and also the arc A G is equal to the arc G B ; hence the whole arc HAG is equal to the whole arc H B G. It is further evident, since the two arcs IT AG, 11 P. are equal, and together make up the whole circumroroneo, that each of thorn is a semi-circumference. 188. Cor. Two parallel (angeiits meet Ihc circumfe:- enco at tho extremities of tho same diameter. BOOK III. 67 Proposition XIII. — Theorem. 189. If tivo circumferences touch each other externally or internally, their centres and the point of contact are in the same straight line. Let the two circumferences, whose centres are C and D, touch each other externally in the point A ; the points C, D, and A will be all in the same straight line. Draw from the point of con- tact A the common tangent AB. Then the radius C A of tlie one circle, and the radius D A of the other, are each perpendicular to A B (Prop. XI.) ; but there can be but one straight line drawn through the point A perpendicular to A B (Prop. XIII. Bk. I.) ; therefore the points C, D, and A are in one perpendicular ; hence they are in one and the same straight line. Also, let the two circumferences touch each other internally in A ; then their centres, C and D, and the point of contact. A, will be in the same straight line. Draw the common tangent AB. Tlien a straight line perpendicular to A B, at the point A, on being suf- ficiently produced, must pass through the two centres C and D (Prop. XI.) ; but from the same point there can be but one perpendicular ; therefore tlie points C, D, and A are in that perpendicular ; hence tl\ey are in the same straight line. 190. Cor. 1. When two circiimferences touch each other externally, the distance between their centres is equal to the sum of their radii. 68 ELEMENTS OP GEOMETRY. 191. Cor. 2. And when two circumferences touch each other internally, the distance between their centres is equal to the difference of their radii. Proposition XIV. — Theorem. 192. If two circumferences cut each other, the straight line passing through their centres will bisect at right an- gles the chord which joins the points of intersection. Let two circumferences cut each other at the points A and B ; then the straight line passing through the centres C and D will bisect at right angles the chord A B common to the two circles. For, if a perpendicular be erected at the middle of this chord, it will pass through each of the two centres C and D (Prop. VI. Cor. 1). But no more than one straight line can be drawn through two points ; hence the straight line C D, passing through the centres, must bisect at right angles the common chord A B. 193. Cor. The straight line joining the points of intei-- section of two circumferences is perpendicular to the straight line which passes through their centres. Proposition XV. — Theorem. 194. If hi)o circiiiiifrrciices nit each other, the distance betiveen their centres ivill be less than the sum of their radii, and greater than their difference. BOOK III. 69 Let two circumferences whose centres are and D cut each other in the point A, and draw the radii C A and D A. Then, in order tliat the intersec- tion may take place, tlie triangle CAD must be possible. And in this triangle the side CD must be less than the siim of AC and AD (Prop. IX. Bk. I.) ; also CD must be greater than the diiference between D A and C A (Prop. IX. Cor., Bk. I.). Proposition XVI. — Theorem. 195. In the same circle, or in equal circles, if two an- gles at the centre are to each other as tivo whole numbers, the intercepted arcs will be to each other as the same numbers. Let us suppose, for example, that the angles ACB, D C E, at the centre of equal circles, are to eacli other as 7 to 4 ; or, which amounts to the same thing, that the angle M, whicli will serve as a common measure, is con- tained scTcn times in the angle ACB, and four times in the angle D C B. The seven partial angles A C w, m C w, n Cjo, &c. into which A C B is divided, being each equal to any of the four partial angles into which D C E is divided, each of the partial arcs Aw, mn, np, &c. will 70 ELEMENTS OP GEOMETRY. be also equal to each of the partial arcs Dx, xy, >iam A B C E, which has the same base B C, and the same altitude A D (Prop. II.) ; but the area of the parallelogram is equal to B C X A D (Prop. V.) ; hence the area of the triangle must bo i B C X A D, or B C X i A D. BOOK IV. 83 228. Cor. Triangles of equal altitudes are to each other as their bases, and triangles of equal bases are to each other as their altitudes ; aud, in general, triangles are to each other as the products of their bases and alti- tudes. Proposition VII. — Theorem. 229. The area of any trapezoid is equal to the product of its altitude by half the sum of its parallel sides. Let ABC D be a trapezoid, EP D E_ its altitude, and A B, C D its par- allel sides ; then its area will be equal to the product of B F by half the sum of A B and C D. Through I, the middle point of A F LB the side B C, draw K L parallel to AD ; and produce DO till it meet K L. In the ti-iangles I B L, I C K, we have the sides I B, I C equal, by construction ; the vertical an- gles LIB, C I K are equal (Prop. IV. Bk. I.) ; and, since CK and BL are parallel, the alternate angles IBL, ICK are also equal (Prop. XXII. Bk. I.) ; therefore the trian- gles I B L, 1 K are equal (Prop. VI. Bk. I.) ; hence the trapezoid A B C D is equivalent to the parallelogram A D K L, and is measured by the product of E F by A L (Prop. v.). But we have AL equal DK; and since the triangles IBL and K C I are equal, the sides B L and C K are equal ; therefore the sum of A B and C D is equal to tlifi sum of A L and D K, or twice A L. Hence A L is half the sum of the bases A B, CD; hence the area of the trapezoid A B, C D is equal to the product of the altitude ]^ P by half the sum of the parallel sides AB, CD. Cor. If through I, the middle point of B C, tlie line IH be drawn parallel to the base A B, the point H will also be the middle point of AD. For, since the figure AH IL 8-4 ELEMENTS OF GEOMETRY. is a parallelogram, as is likewise D H I K, their opposite sides being parallel, we have A H equal to I L, and D H equal to I K. But since the triangles B I L, C I K are equal, we have I L equal to I K ; hence A H is equal to D H. Now, the line H I is equal to A L, which has been shown to be equal to half the sum of A B and C D ; there-' fore the area of the trapezoid is equal to tlie product of E F by H I. Hence, the area of a trapezoid is equal to the product of its altitude by the line connectuig tlic mid- dle points of the sides which are uot parallel. Proposition VIII. — Theorem. 230. If a straight line be divided into two parts, the stjvare described on the whole line is equivalent to the sum of the squares described on the parts, together with twice the rectangle contained by the parts. Let A C be a straight line, divided £ h D into two parts, AB, B C, at the point B ; l 1 1 then the square described on A C is F G equivalent to the sum of the squares described on the parts AB, B C, togeth- er with twice the rectangle contained . byAB,BC; that is, ^ ^ ^ AC- = AB^ + B~C^ + 2 AB X B C. On A C describe the square A C D E ; take AP equal to A B ; draw F G parallel to A C, and B H pai-allel to A E. The square A C D E is divided into four parts: tlie first, ABIF, is the square described on AB, since A F was taken equal to A B. The second, I O D H, is the square described upon B C ; for, since A C is equal to A E, and A B is equal to AF, AC minus AB is equal- to AE miiuK A F, which fiivos B C equal to K F. But I G is equal to B C, and D G to E F, since the lines are parallels ; thei-e- fore IGDII is equal to tin square described on BC. I BOOK IV. 85 These two parts being taken from the whole square, there remain two rectangles B C G I, E T I H, each of which is measured by A B X B C ; hence the square on the whole line AC is equivalent to the squares on the parts AB, BC, together with twice the rectangle of the parts. E H D 231. Coi\ The square described on tlie \\'hole line A C is equivalent to four times the square described on the half AB. ^ A B C 232. Scholium. This proposition is equivalent to the algebraical formula, I Proposition IX. — Theorem. 233. The square described on the difference of two straight lines is equivalent to the sum of the squares de- scribed on the two lines, diminished by twice the rectangle contained by the lines. Let A B and BC be two lines, and LP G I A C their difference ; then will the square described on A C be eqiiiva- lent to the sum of the squares de- scribed on AB, B C, diminished by twice tlie rectangle AB, B C ; that is. K E D H (AB — B Cy or A C- = A B- + B C^ K C B 2 A B X B C. On A B describe the square A B I F ; take A E equal to A C ; draw C G- parallel to B I, H K parallel to AB, and complete the square E P L K. Since AP is equal to AB, and AE to AC, EP is equal to B C, and LP to GI ; therefore L G is equal to PI; hence the two rectangles CBIG, GLKD are each 8 86 ELEMENTS OP GEOMETRY. measured by A B X B C. Take these rectangles from the whole figure ABILKE, which is equivalent to A B^ -j- B C^, and there will evidently remain the square A C D B ; hence the square on A C is equivalent to the sum of the squares on A B, B C, diminished by twice the rectangle contained by AB, B C. 234. Scholium. This proposition is equivalent to the algebraical formula, {a — by=a' — 2ab + bK Proposition X. — Theorem. 235. The rectangle contained by the sum and difference of two straight lines is equivalent to the difference of the squares of these lines. F G I Let A B, B C be two lines ; then E will the rectangle contained by the sum and difference of A B, B C, be equivalent to the difference of the squares of A B, B C ; that is, A C B~ (A B + B C) X (A B — B C) = AB^ — BCl On A B describe the square A B I F, and on A C the square A C D E ; produce CD to G ; and produce A B until B K is equal to B C, and complete the rectangle AKLE. The base A K of the rectangle is the sum of the two lines A B, B C ; and its altitude A E is the difference of the same lines; therefore the rectangle AKLE is that contained by the sum and the differoneo of the linos A B, B C. But this rectangle is composed of the two parts A B H E and B H L K ; and the part B H L K is equal to the rectangle EDGE, since B H is eq\ial to D E, and BK to EP. Hence the rectangle AKLE is equivalent to A B II E plus EDGE, which is equivalent to the dif- H B BOOK IV. 87 ference between the square ABIF described on AB, and D H I G described on B ; hence (AB + BC) X (AB — BC) = AB^ — BCl 236. Scholium. This proposition is equivalent to the algebraical formula, (a + V) X (a — b') = aP- — b\ PeOPOSITION XI. — THEOREiVI. 237. The square described on the ht/potheAuse of a right-angled triangle is equivalent la the sum of the squares described on the other tivo sides. Let ABC be a right-angled L triangle, having the right angle at A ; then the square described H^ on the hypothenuse B C will be equivalent to the sum of the squares on the sides BA, AC. On B C describe the square BCGF, andon AB, AC the squares ABHL, ACIK; and through A draw AE parallel to BF or CG, and join AF, HC. The angle A B F is composed of the angle ABC, to- gether with the right angle C B F ; the angle C B H is composed of the same angle ABC together witli the right angle A B H ; therefore the angle A B F is equal to the angle H B C. But we have A B equal to B H, being sides of the same square ; and B F equal to B C, for the same reason ; therefore the triangles A B F, H B C have two sides and the included angle of the one eqiTal to two sides and the included angle of the other ; hence they are themselves equal (Prop. V. Bk. I.). But the triangle A B F is equivalent to half the rectan- gle B D E F, since they have the same base B F, and the E G 88 ELEMENTS OP GEOMETRY. E G same altitude BD (Prop. 11. Cor. 1). The triangle HBC is, in like manner, equivalent H to half the square A B H L ; for the angles BAG, BAL .being both right, AC and AL form one and the same straight line parallel to HB (Prop. II. Bk. I.) ; and consequently the tri- angle and the square have the same altitude A B (Prop. XXV. Bk. I.) ; and they also have the same base B H ; hence the triangle is equivalent to half the square (Prop. II.). The triangle ABP has already been proved equal to the triangle HBC; hence the rectangle BDEF, wliich is double the triangle ABP, must be equivalent to the square A B H L, which is double the triangle HBC. In the same manner it may be proved that the rectangle C D E Gr is equivalent to the square A C I K. But the two rectangles B D B P, C D E G, taken together, compose the square B C G P ; therefore the squai-e B C G F, de- scribed on the hypothenuse, is equivalent to the sum of the squares A B H L, A C I K, described on the two other sides ; that is, B C is equivalent to A B + A C • 238. Cor. 1. The square of either of the sides lohich form the right angle of a right-angled triangle is equiva- lent to the square of the hypothenuse diminished by the square of the other side ; thus, AB^ is equivalent to B'C'' — AC^. 239. Cor. 2. The square of the hypothenuse is to the square of either of the other sides, as the hypothenuse is to the part of the hypothenuse eut off, adjacent to that side. BOOK IV. 89 by the perpendicular let fall from the vertex of the right angle. For, on account of the common altitude B P, the square BOGF is to the rectangle BDEP as the base BC is to the base BD (Prop. III.) ; now, the square ABHL has been proved to be equivalent to the rectangle BDEP; therefore we have, BC^ : AB^ : : B C : B D. In like manner, we have, BG^ : AC^ : : B C : C D. 240. Cor. 2>. If a perpendicular be draion from the vertex of the right angle to the hypothenuse, the squares of the sides about the right angle will be to each other as the adjacent segments of the hypothenuse. For the rec- tangles BDEP, DOGE, having the same altitude, are to each other as their bases, B D, C D (Prop. III.). But these rectangles are equivalent to the squares ABHL, A C I K ; therefore we have, A^ : AC^ : : B D : D C. 241. Cor. 4. Tlie square described on the diagonal of a square is equivalent to double the square described on a side. For let A B C D be a square, and A C its diagonal ; the triangle ABC being right-angled and isosceles, we have, -*■ ^ AC' = AB^ + BC^ = 2 A B^ = 2 x AB CD. 242. Cor. 5. Since AC" is equal to 2 AB^, we have AC' : AB' : : 2 : 1 ; and, extracting the square root, we have AC : AB : : ^2 : 1; hence, the diagonal of a square is incominensurable with a side. 8* 90 ELEMENTS OP GEOMETRY. -The proposition may also be demonstrated M 243. Note, as follows : — Let ABC be a right-angled triangle, having the right angle at A ; then the square described on the hypothenuse B C will be equivalent to the sum of the squares on the sides BA, AC. On B C describe the square BCGF, and on AB, AC the squares ABHL, ACIK; pro- duce F B to N, HL and IK to M ; and through A draw EDA parallel to FBN, and meeting the prolongation of H L in M. Then, since the angles H B A, NBC are both right an- gles, if the common angle NBA be taken from each of these equals, there will remain the equal angles H B N, ABC; and, consequently, since the triangles H B N, A B C are both right-angled, and have also the sides B H, B A equal, their hypothenuses B X, B C are equal (Prop. VI. Cor., Bk. I.). But B C is equal to B F ; therefore B N is equal to B F ; hence the parallelograms B A M N, B D E F, of which the common altitude is B D, have equal bases ; therefore the two parallelograms are equivalent (Prop. I.). But the parallelogram B A M X is equivalent to the square ABHL, since they have the same base B A, and the same altitude A L ; hence the parallelogram B D E P is also equivalent to tlio square ABHL. In like manner it may bo shown that the rectangle D C G E is equivalent to tlio square ACIK; hence the two rectan- gles together, that is, the square BCGF, arc equivalent to the sum of the squares ABHL, ACIK. BOOK IV. 91 Proposition XII. — Theorem. 244. In any triangle, the square of the side opposite an acute angle is less than the sum of the squares of the base and the other side, by twice the rectangle contained by the base and the distance from the vertex of the acute angle to the perpendicular let fall from the vertex of the opposite angle on the base, or on the base produced. Let AB C be any triangle, C one of its acute angles, and A D the perpendicular let fall on the base B C, or on B C produced ; then, in either case, will the square of A B be less than tlie sum of the squares of AC, B C, by twice the rectangle B C X CD. First. When the perpendicular falls within the triangle ABC, we have BD = BC — CD; and consequently, BD^ = BO^ + OD^ — 2 B C X D (Prop. IX.). By adding A D to each of these equals, we have BD^+ AD^= BC'+ CD' + AD' — 2BC X CD. But the two right-angled triangles AD B, AD C give AB' = BD' + AD^ and AO' = CD' +. AD' (Prop. XI.) ; therefore, AB' = BC' + AC' — 2 B C X C D. Secondly. When the perpendicular A D falls without the triangle ABC, we have BD=CD — BC; and conse- quently, B D' = CD' + BC' — 2 C D X B C. By add- ing A D" to each of these equals, we find, as before, AB' = BC' + AC' — 2 B X C D. 92 ELEMENTS OP GEOMETRY. Proposition XIII. — Theorem. 245. In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equivalent to the sum of the squares of the two other sides plus twice the rectangle contained by the one of those sides into the distance from the vertex of the obtuse angle to the perpendicular let fall from the vertex of the opposite angle to that side produced. Let A C B be an obtuse-angled triangle, A having the obtuse angle at C, and let A D be perpendicular to the base B C produced ; then the square of A B is greater than the sum of the squares of B C, AC, by twice tlie rectangle BC X CD. Since BD is the sum of the lines B C + C D, we have B D^ = Bl3^ + C^ + 2 B C X C D (Prop. YIII.). By adding AD to each of these equals, we have BD^ -f AD^ = FC^ + CD^ + AD^ + 2 B C X C D. But the two right-angled triangles A D B, A D C give AB* = B D^ + AD^, and AC^ = C D^ + AD' (Prop. XI.) ; therefore, AB^ = B C"" + AC^ + 2 B C X C D. 246. Sclwliuni. The right-angled triangle is the only one iu which the sum of the squares of two sides is equiv- alent to the square of the third ; for if the angle contained by the two sides is acute, the sum of their squares will be greater than the sqiiaro of the opposite side ; if obtuse, it will be less. Proposition XIV. — Theorem. 247. //* any triangle, if a straight line be drawn from the vertex to the middle point of the base, the sum of the BOOK IV. 93 squares of the other two sides is equivalent to tivice the square of the bisecting line, together with twice the square of half the base. In any triangle ABC, draw the line A A E from the vertex A to the middle of the base B C ; then the sum of the squares of the two sides, AB, AC, is equivalent to twice the square of A B together with twice the sqiiai'e of B E. /_ On B C let fall the perpendicular AD ; ^ ^ ^ ^ then, in the triangle ABB, AB^ = AE^ + EB' +'^ E B X E D (Prop. XIII.), and, in triangle ABC, AC^ = Al' + EC' — 2 E C X E D (Prop. XII.). Hence, by adding the corresponding sides together, observing that since E B and E C are equal, WW is equal to¥C^ and E B X E D to E C X B D, we have AB^ _)_ AC' = 2 AE^ + 2 EB^- Peoposition XV. — Theorem. 248. In any parallelogram the sum of the squares of the four sides is equivalent to the sum of the squares of the two diagonals. Let A B C D be any parallelogram, the diagonals of which are AC, BD; then the sum of the squares of A B, B C, CD, D A is equivalent to the sum of the squares of A C, B D. For the diagonals A C, B D bisect each other (Prop. XXXIV. Bk. I.) ; hence, in the trian- gle ABC, XB^ + BO^ = 2 AE^ + 2 BE® (Prop. XIV.) ; also, in the triangle A D C, ad' + DC' = 2 A E® + 2 D E^ 94 ELEMENTS OP GEOMETRY. Hence, by adding the corresponding sides together, and observing that, since B E and D E are equal, B E and D E^ must also be equal, we shall have, AB' + BC'' + ad' + DC' = 4 AB' + 4 D"e'. But 4 AE' is the square of 2 A E, or of A C, and 4 DE' is the square of 2 D E, or of B D (Prop. VIII. Cor.) ; hence, BA' + BC' + CD' + ad' = A^' + BD'. Pkoposition XVI. — Theoeem. 249. In any quadrilateral the sum of the squares of the sides is equivalent to the sum of the squares of the diag- onals, plus four times the square of the straight line that joins the middle points of the diagonals. Let A B C D be any quadrilateral, the C diagonals of -whicli are AC, D B, and E P a straight line joining their mid- dle points, E, P ; then the sum of the squares of A B, B C, C D, A D is equiv- alent to AC' -f BD' + 4 EF', a Join E B and E D ; then in the triangle ABC, AB' -j- BC' = 2 AE' -f 2 BE^ (Prop. XIV.), and in the triangle ADC, AD' + CD' = 2AE'+ 2DE'. Hence, by adding the corresponding sides, ve have AB' -f BC' + AD' -f CD' = 4 AE- + 2 BE^+ 2 DE^. feut 4 AE' is equivalent to A C" (Prop. A'lll. Cor.) and 2 be' -t- 2 'dI:' is equivalent to 4 Fp'* -f 4 EP' (Prop XIV.); hence, A li' -I- BC' + ad' -I- CD' = Ajf + BD^ -f 4 EP'. BOOK IV. 95 250. Cor. If the quadrilateral is a parallelogram, the points E and F will coincide ; then the proposition -will be the same as Prop. XV. 251. Scholium. Proposition XV. is only a particular case of this proposition. Proposition XVII. — Theorem. 252. If a straight line be dravm in a triangle parallel to one of the sides, it will divide the other two sides pro- portionally. Let A B C be a triangle, and D E a ^ straight line drawn within it parallel to the side B C ; then will AD : DB:: AE:EC. Join B E and D C ; then the two trian- gles BDB, DEO have the same base, D E ; they have also the same altitude, since the vertices B and lie in a line parallel to the base ; therefore the ti'iangles are equivalent (Prop. II. Cor. 2). The triangles A D E, B D E, having their bases in the same line A B, and having the common vertex E, have the same altitude, and therefore are to each other as their bases (Prop. VI. Cor.) ; hence ADE:BDE::AD:DB. The triangles A D E, D E C, whose common vertex is D, have also the same altitude, and therefore are to each other as their bases ; hence ADE: DEC : : AB : EC. But the triangles B D E, DEC have been shown to be equivalent ; therefore, on account of the common ratio in the two proportions (Prop. X. Bk. II.), AD:DB::AE:EC. 253. Cor. 1. Hence, by composition (Prop. VII. Bk. 96 ELEMENTS OF GEOMETRY. II.), wc have AD + DB : AD : : AB + EC : AE, or AB:AD::AC:AB; also, AB:BD::AC:EC. 254. Cor. 2. If two or more straight lines be drawn in a triangle parallel to one of the sides, they will divide the other two sides proportionally. For, in the triangle ABC, since D E A is parallel to B C, by the theorem, A D : D B : : A E : E C ; and, in the triangle A D E, since F G is parallel to D E, by the preceding corollaiy, AD : FD : : A E : G E. Hence, since the antece- dents are the same in the two propor- tions (Prop. X. Cor. 2, Bk. II.), P D : D B : : G E : E C. B Pkoposition XVIII. — Theorem. 255. If a straight line divides two sides of a triang-le • proportionally, the line is parallel to the other side of the triangle. Let ABC be a triangle, and D E a j^ straight line drawn in it dividing the sides AB, AC, so that AD : D B : : AE: E C ; then will the line D E be parallel to the side B C. Join B E and D ; then the triangles A D E, B D E, having their bases in the same straight line AB, and having a common vertex, E, are to each other as their bases AD, DB (Prop. YI. Cor.) ; that is, ADE:BDE::AD:DB. Also, the triangles A D E, DEC, liaving the common vertex D, and their bases in the same line, are to each other as these bases, \ E, E ; that is, ADE:DEC::AE:EC. BOOK IV. 97 But, by hypothesis, AD:DB::AE:EC; hence (Prop. X. Bk. II.), ADE : BDE: : ADE: DEC; that is, BDE, DE C have the same ratio to ADE ; there- fore the triangles BDE, DEC have the same area, and consequently are equivalent (Art. 211). Since these tri- angles have the same base, D B, their altitudes are equal (Prop. VI. Cor.) ; hence the line B C, in which their ver- tices are, must be parallel to D E. Proposition XIX. — Theorem. 256. The straight line bisecting any angle of a triangle divides the opposite side into parts, which are proportional to the adjacent sides. In any triangle, ABO, let the an- E gle B A C be bisected by the straight line A D ; then will B D : D C : : A B : A C. Through the point C draw C E parallel to AD, meeting BA pro- duced in E. Then, since the two parallels AD, EC are met by the straight line A C, the alternate angles D A C, ACE are equal (Prop. XXII. Bk. I.) ; and the same parallels being met by the straight line B B, the op- posite exterior and interior angles BAD, AE C are also equal. (Prop. XXII. Bk. I.). But, by hypothesis, the an- gles D AC, BAD are equal; consequently the angle ACE is equal to the angle A E C ; hence the triangle A C E is isosceles, and the side AE is equal to the side A C (Prop. VIII. Bk. I.). Again, since AD, in the triangle EEC, is parallel to E 0, we have BD:DC::AB:AE (Prop. XVII.), and, substituting A C in place of its equal A E, BD:DC::AB:AC. 98 ELEMENTS OF GEOMETUY. Proposition XX. — Theorem. 257. If a straight line drawn from the vertex of any ang-le of a triangle divides the opposite side into parts which are proportional to the adjacent sides, the line bi- sects the an^le. Let the straight line AD, drawn E from tlie vertex of the angle BAG, in the triangle ABC, divide the op- posite side B 0, so that B D : D C : : AB:AC; then will the line AD bisect the angle BAG. Through the point C draw CE parallel to AD, meeting B A produced in E. Then, by hypothesis, B D : D C : : A B : A G ; and since AD is parallel to E C, B D : D G : : AB : AE (Prop. XVII.); then AB : AG : : AB : AE (Prop. X. Bk. II.) ; consequently A C is equal to A E ; hence the angle A E C is equal to the angle AGE (Prop. VII. Bk. I.). But, since G E and AD are parallels, the angle A E G is equal to the opposite exterior angle BAD, and the angle A G E is equal to the alternate angle DAG (Prop. XXII. Bk. I.); hence the angles BAD, DAG are equal, and consequently the straight line A D bisects the angle BAG. Proposition XXI. — Theorem. 258. If the exterior angle formed by producing one of the sides of any triangle be bisected by a straight line which meets the base produced, the distances from the ex- tremities of the base to the point xchcre the bisecting line meets the Ixisr produced, will l>c to each other as the other two sides of the triangle. Lot the exterior angle C A E, formed by producing the side B A of the triaiglo A B 0. bo bisectcil liy the straight BOOK IV. 99 AD , ■which meets the side D, tlieu will 111 line B C produced BD : DC : : AB: AC. Through C draw C F parallel to A D ; then the angle A C P is equal to the alternate angle CAD, and the exterior angle D A E is equal to the interior and opposite angle CFA (Prop. XXII. Bk. I.). But, by hypothesis, the angles CAD, DAB are equal ; consequently the angle A C F is equal to the angle CFA; hence the triangle A C F is isosceles, and the side A C is equal to the side A F (Prop. VIII. Bk. I.). Again, since A D is parallel to F C, BD : D C : : B A : AF (Prop. XVII. Cor. 1), and substituting AC in the place of its equal A F, we have BD:DC::BA:AC. Peoposition XXII. — Theorem. 259. Equiangular triangles have their homologous sides proportional, and are similar. Let the two triangles AB C, D C E ^ be equiangular ; the angle B A C / ' \. being equal to the angle C D E, the angle A B C to the angle D CE, and the angle A C B to the angle DEC, then the homologous sides will be proportional, and we shall have BC:CE::AB:CD::AC:DE. For, let the two triangles be placed so that two homol- ogous sides, B C, C E, may join each other, and be in the same straight line ; and produce the sides B A, E D till they meet in F. Since B C E is a straight line, and the angle B C A is equal to the angle CED, AC is parallel to FE (Prop. XXI. Bk. I.) ; also, since the angle A B C is equal to the 100 ELEMENTS OP GEOMETKY. angle D C B, the line B F is parallel to the line CD. Hence the figure A C D P is a parallelogram ; and, consequently, A F is equal to C D, and AC to FD (Prop. XXXI. Bk. I.). B C E In the triangle B B F, since the line A C is parallel to the side F B, we have BC : CB : : BA : AP (Prop. XVII.) ; or, substituting CD for its equal, AP, BC:CB::BA:CD. Again, C D is parallel to B F ; therefore, B C : C E : : F D : D E ; or, substituting A C for its equal F D, BC:CB::AC:DE. And, since both these proportions contain the same ratio B C : C B, we have (Prop. X. Bk. II.) AC:DB::BA:CD. Hence, the equiangular triangles B A C, C D B have their homologous sides proportional ; and ' consequently the two triangles are similar (Art. 210). 260. Cor. Two triangles having two angles of the one equal to two angles of the other, each to each, are similar; since the third angles will also be equal, and the two tri- angles be equiangular. 261. Scholium. In similar triangles, the homologous sides are opposite to the equal angles ; thus the angle A C B being equal to DEC, the side AB is homologous to D C ; in like manner, A C and D E are homologous. Proposition XXIII. — Theorem. 262. Triangles which have their homologous sides pro- portional, are ci/inani'-itlar and simi/ar. Let the two triangles A B C, D E P liave their sides pro- portional, so that we have BC:EF::AB:DE::AC:DF; BOOK IV. 101 then will the triangles have their angles equal ; namely, the angle A equal to the angle D, the angle B to the angle E, and the angle C to tlie angle P. At the point E, in the B "c G straight line EP, make the angle PEG equal to the angle B, and at the point P, the angle E P G equal the angle C ; the third angle G will be equal to the third angle A (Prop. XXVIII. Cor. 2, Bk. I.) ; and the two triangles ABC, E P G will be equiangular. Therefore, by the last theorem, we have BC:i]P::AB:EG; but, by hypothesis, we have BC :EP: : AB:DE; hence, E G is equal to D E. By the same theorem, we also have BC : EP: : AC :PG; and, by hypothesis, BC :EP : : AC :DF; hence P G is equal to D P. Hence, the triangles E G P, D E P, having their three sides equal, each to each, are themselves equal (Prop. XVIII. Bk. I.). But, by con- struction, the triangle E G P is equiangular with the tri- angle ABC; hence the triangles D B P, ABC are also equiangular and similar. 263. Scholium. The two preceding propositions, together with that relating to the square of the hypothenuse (Art. 237), are the most important and fertile in results of any in Geometry. They are almost sufficient of themselves for all applications to subsequent reasoning, and for the 9* 102 ELEMENTS OP GEOMETRY. solution of all problems ; since the general properties of triangles include, by implication, those of all figures. Peoposition XXIV . — Theorem. 264. Two triangles, which have an angle of the one equal to an angle of the other, and the sides containing these angles proportional, are similar. Let the two triangles ABC, ^ D E F have the angle A equal to the angle D, and the sides contain- ing these angles proportional, so that AB:DE::AC:DF; then the triangles are similar. Take A G equal D E, and draw " w ^ - GH parallel to BC. The angle AGH will bo equal to the angle ABO (Prop. XXII. Bk. I.) ; and the triangles AGH, ABC will be equiangular ; hence we shall have A B : A G : : A C : A H. But, by hypothesis, AB:DE : : AC :DP; and, by construction, A G is equal to D E ; hence AH is equal to D P. Therefore the two triangles A G H, D E P, having two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, are themselves equal (Prop. Y. Bk. I.). But the triangle A G H is similar to ABC; therefore D E F is also similar to ABC. Proposition XXY. — Theorem. 265. Two triangles, irliich have their sides, taken tico and tiro, either parallel or perpcndicnlar to each other, are similar. Lot the two triangles ABC, DEF have the side AB parallel to the side D E, B C parallel to 10 F, and A C BOOK IV. 103 parallel to D P ; these triangles will then be similar. For, since tlie side AB is parallel to the side D E, and B C to EF, the angle ABC is equal to the angle D E F (Prop. XXVI. Bk. I.). Also, since AC is parallel to DP, the angle AOB is equal to the angle D FE, and the angle BAG to EDP; therefore the triangles ABC, DEP are equiangular; hence they are similar (Prop. XXII.). Again, let the two triangles ABC, DEP have the side D E perpendicular to the side AB, DP perpendicular to AC, and E F perpendicular to B C ; these triangles are similar. Produce F D till it meets A C at G ; then the angles D G A, D E A of the quadrilateral A E D G are two right angles ; and since all the four an- gles are together equal to four right angles (Prop. XXIX. Cor. 1, Bk. I.), the remaining two angles, E D G, E A G, are together equal to two right angles. But the two angles E D G, B D F are also together equal to two riglit angles (Prop. I. Bk. I.) ; hence the angle B D P is equal to E AG or B AC. The two angles, GP C, GCP, in the right-angled trian- gle F G C, are together equal to a right angle (Prop. XXVIII. Cor. 5, Bk. I.), and the two angles GPC, GPE are together equal to the right angle E F C (Art. 34, Ax. 9) ; therefore G F E is equal to G C P, or D F B to B C A. Therefore the triangles ABC, DEP have two angles of the one equal to two angles of the other, each to each ; hence they are similar (Prop. XXII. Cor.). 266. ScJioliiim. "When the two triangles have their sides parallel, the parallel sides are homologous ; and when they have them perpendicular, the perpendicular sides are 104 ELEMENTS OF GEOMETRY. homologous. Thus, D E is homologous with A B, D F with A , aud B F with B C. Proposition XXVI. — Theorem. 267. In any triangle, if a line be drawn parallel to the base, all lines drawn from the vertex will divide the par- allel and the base proportionally. In the triangle BAG, let DB he drawn parallel to the base B C ; then will the lines AF, A G, AH, drawn from the vertex, divide the parallel D B, and the base B C, so that DI:BF::IK:FG::KL:GH. For, since DI is parallel to B F, the triangles AD I and A B F are equiangular ; and we have (Prop. XXII.), DI: BF: : AI: AF; and since I K is parallel to F G, we have in like manner, AI: AF: :IK:FG; and, since these two propositions contain the same ratio, AI : AF, we shall have (Prop. X. Cor. 1, Bk. II.), D I : B F : : I K : F G. In the same manner, it may be shown that IK:FG: :KL: GH: : L E : H C. Therefore the line D E is divided at the points I, K, L, as the base B C is, at the points F, G, H. 268. Cor. If B were divided into equal parts at the points F, G, H, the parnllol D E would also be divided into equal parts at the points I, K, L, Proposition XXVII. — Theorem. 269. In a righf-aiii^'Icd triiitilo 10 FG. For the two iircs, B C and E, huve C(]ual radii and equal chords ; thoroforo thoy are equal (^I'rop. III. Bk. COOK V. 121 III.) ; hence the angles C AB, E F G-, measured by tliese arcs, are also equal (Prop. V. Bk. III.). Problem VI. 300. To bisect a given arc, or a given angle. First. Let A D B be the given arc C which it is required to bisect. /\\ Draw the chord A B ; from the cen- / i \ tre C draw the line C D perpendicular / 1 \ to AB (Prob. III.) ; it will bisect the a^t — ! z^n arc A D B in the point D. ^^^"1^''''^ For C D being a radius perpendicu- lar to a chord A B, must bisect the arc A D B which is subtended by that chord (Prop. VI. Bk. III.). Secondly. Let A C B be the angle which it is required to bisect. From C as a centre, with any radius, describe the arc A D B ; bisect this arc, as in the first case, by drawing the line C D ; and this line will also bisect the angle A B. For the angles A C D, B C D are equal, being measured by the equal arcs AD, D B (Prop. V. Bk. III.). 301. Scholium. By the same construction, we may bi- sect each of the halves AD, D B ; and thus, by successive subdivisions, a given angle or a given arc may be divided into four equal parts, into eight, into sixteen, Ac. Peoblem VII. 302. Through a given point, to draw a straight line parallel to a given straight line. Let A be the given point, and A "=^::::;; B C D the given straight line. From A draw a straight line, ■■■..., AE, to any point, E, in CD. E Then draw A B, making the angle E A B equal to the u 122 ELEMENTS OP GEOMETRY. angle A E C (Prob. V.) ; and A B is parallel to CD. For the alternate angles E A B, D A E C, made by the straight line E A E meeting the two straight lines A B, CD, being equal, the lines AB and CD mnst be parallel (Prop. XX. Bk. I.). Problem VIII. 303. Tvjo ang-les of a triangle being given, to find the third angle. Draw the indefinite straight line ABE. At any point, B, make the angle ABC equal to one of the given angles (Prob. V.), and the _ angle C B D equal to the other given ABE angle ; then the angle D B E will be the third angle re- quired. For these three angles are together equal to two right angles (Prop. I. Cor. 2, Bk. I.), as are also the three an- gles of every triangle (Prop. XXVni. Bk. I.) ; and two of the angles at B having been made equal to two angles of the triangle, the remaining angle D B E must be equal to the third angle. Problem IX. 304. Two sides of a triangle and the included angle being given, to construct the triangle. Draw the straight line A B equal to one of the two given sides. At the point A make an angle, C A B, oqiiiil to the given angle (Prob. Y.^ ; and tnko A C equal to the otlior given side. Join A B B ; and the triangle ABC will be the one ro(\uired (Prop. V. Bk. 1.). BOOK V. 123 Problem X. 305. One side and two angles of a triangle being given, to construct the triangle. The two given angles -will either be . ^ both adjacent to the given side, or one adjacent and tlie other opposite. In the latter case, find the third angle (Prob. VIII.) ; and the two angles ad- ^ jacent to the given side will then be known. .In the former case, draw the straiglit line AB equal to the given side ; at the point A, make an angle, BAG, equal to one of the adjacent angles, and at B an angle, ABC, equal to the other. Then the two sides AC, BC will meet, and form with A B the triangle required (Prop. VI. Bk. I.) Problesi XI. 306. Two sides of a triangle and an angle opposite one of them being given, to construct the triangle. Draw the indefinite straight C line AB. At the point A make an angle BAC equal to the given angle, and make A equal to that side which is adjacent to the given angle. Then from C, as a centre, with a radius equal to the other side,- describe an arc, which must either touch the line AB in D, or cut it in the points E and F, otherwise a tri- angle could not be formed. When the arc touches A B, a straight line drawn from C to the point of contact, D, will be perpendicular to AB (Prop. XI. Bk. III.), and the right-angled triangle CAD will be the triangle required. When tlie arc cuts AB in two points, E and P, lying 124; ELEMENTS OF GEOMETRY. on the same side of the pouit A, draw the straight lines C E, C P ; and each of the two triangles CAB, CAP will satisfy the conditions of the problem. If, however, the two points E and P should lie on different sides of the point A, only one of the triangles, as CAP, will satisfy all the conditions; hence that will be the triangle required. 307. Scholium. The problem would be impossible, if the side opposite the given angle were less than the per- pendicular let fall from the point C on the straight line AB. Problem XII. 308. The three sides of a triangle being- given, to corir- struct the triangle. Draw the straight line AB equal to C one of the given sides; from the point A as a centre, with a radius equal to either of the other two sides, describe an arc; from the point B, with a radius equal to the third side, describe another ^ arc cutting the former in the point C ; draw the straight lines AC, BC ; and the triangle ABC will be the one required (Prop. XVni. Bk. I.). 309. Scholium. The problem would be impossible, if one of the given sides were equal to or greater than the sum of the other two. Problem Xin. 310. Two adjacent sides of a parallelogram and the in- cluded angle being given, to constnirl the parallelogram. Draw the straight line A B equal to one of the given sides. At the point A make an angle, B A D, cqunl to tiio given angle, and take A D equal to tho other given side. From BOOK V. 125 the point D, -with a radius equal to AB, describe an arc ; and from the point B as a centre, with a radius equal to AD, describe anotlier arc cutting the former in the point 0. Draw the straight lines CD, C B ; and the parallel- ogram A B C D will be the one required. For the opposite sides are equal, by construction ; hence the figure is a parallelogram (Prop. XXXll. Bk. 1.) ; and it is formed with the given sides and the given angle. 311. Cor. If the given aiigle is a right angle, the figuie will be a rectangle ; and if the adjacent sides are ako equal, the figure will be a square. Pkoblkm XIV. 312. A circumference, or an arc, being given, to find the centre of the circle. Take any three points, A, B, C, ^ ^^^ on the given circiimference, or arc. /^ \. Draw the chords AB, BC, and / \ bisect them by the perpendiculars ^( ,.-,^ J DE and PE (Prob. I.) ; the point Tx / X /j E, in. which these perpendiculars \ ^\ /y meet, is the centre required. ^-Oi^^-'^^ For the perpendiculars DE, FB must both pass through the centre (Prop. VI. Cor. 2, Bk. III.), and E being the only pouit througli which they both pass, E miist be the centre. 313. Scholium. By the same construction, a circumfer- ence may be made to pass through three given points, A, B, C, not in the same straight line ; and also a cir- cumference described in which a given triangle, ABC, shall be inscribed. Problem XV. 314. Tlirovgh a given point to draw a tangent to a p-iven circle. " 11* 126 ELEMENTS OP GEOMETRY. First. Let the given point A be in the circumference. Find tlie centre of the circle, C (Prob. XIV.) ; draw the radius C A ; through the point A draw AB perpendicular to C A (Prob. IV.) ; and A B will be tlie tangent required (Prop. X. Bk. III.). Secondly. Let the given point B be without the circum- ference. Join the point B and the centre C by the straight line B C ; bisect B C in D ; and from D as a centre, with a radius equal to CD or D B, describe a circumference intersecting tlie given circumference in the points A and E. Draw AB and EB, and each will be a tangent as required. For, drawing C A, the angle CAB, being inscribed in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.) ; tlierefore A B is perpendicular to the radius C A at its ex- tremity, A, and consequently is a tangent (Prop. X. Bk. III.). In like manner it may be shown that EB is a tangent. Problem XVI. 315. On a gwen straight line to construct a segment of a circle that shall contain an angle equal to a given angle. Let AB be the given straight line. Through the point B draw the straight lino BD, making the angle A B D equal to the given angle ; draw B E perpendicular to 1> D ; bisect A B, and from F erect the perpendicular F E. FrouT the point E, whore thcso G l)(!i'poiidiculiirs meet, as a centre, with the distance EB BOOK V. 127 or E jV, describe a circumference, and A C B will be the segment required. For, since B D is a perpendicular at the extremity of the radius E B, it is a tangent (Prop. X. Bk. III.) ; and the angle A B D is measured by half the arc A G B (Prop. XX. Bk. III.). Also, the angle ACB, being an inscribed angle, is measured by half the arc A G B ; there- fore the angle ACB is equal to the angle ABD. But, by construction, the angle ABD is equal to the given angle ; hence the segment ACB contains an angle equal to the given angle. 316. Scholium. If the given angle ■were acute, the centre must lie within the segment (Pi'op. XVIII. Cor. 3, Bk. III.) ; and if it were right, the centre m\ist be in the middle of the line A B, and the required segment would be a semicircle. Problem XVII. 317. To inscribe a circle in any given triangle. Bisect any two of the angles, as A and B, by the straight lines A B and B E, meeting in the point E (Prob. VI.). From the point E let fall the perpen- diculars ED, EP, EG (Prob. 11.) on the three sides of the triangle ; these perpendiculars will all be equal. For, by construction, we have the angle DAE equal to the angle E A G, and the right angle A D E equal to the right angle AGE; hence the third angle A E D is equal to the third angle AEG. Moreover, A B is common to the two triangles ABD, AEG; hence the triangles them- selves are equal, and E D is equal to E G. In the same manner it may be shown that the two triangles BEP, 128 ELKMENTS OF GEOMETRY. BEG are equal ; thei-efore E F is equal to E G ; heuce the three perpendiculars ED, E F, E G are all equal, and if, from the point E as a centre, with the radius ED, a circle be described, it must pass tlirougli tlie points F and G. 318. Scholium. The three lines which bisect the angles of a triangle all meet in the centre of tlie inscribed cii-cle. Pkoblem XVIII. 319. To inscribe a circle in a given square. Draw the diagonals A C, D B, and d from the point E, where the diagonals mutually bisect each other (Prop. XXXIV. Bk. I.), draw the straight line EF perpendicular to a side of the sqtiare. From E as a centre, with a radius equal to EP, describe a circle, and it will touch each side of the square. For the square is divided by the diagonals into four equal isosceles triangles ; hence, the perpendicular, from the vertex E to the base, is the same in each triangle ; therefore the circumference described from the centre E, with the radius E F, passes through the extremities of each perpendicular ; consequently, the sides of the square are tangents to the circle (Prop. X. Bk. III.). Problem XIX. 320. To find the aide of a square v/iich shall be equii alent to llic sum of tiro si-imi S(/ii(nrs. Draw the two straight lines A B, B C perpondictilar to each other, taking A B equal to a side <>( one of the given squares, and B C equal to a side of llic other. Joi;i AC; this will bo the side of the equaro required. C BOOK V. 129 For, the triangle ABC being I'iglit-angled, the square that can be described upon the hypothenuse A is equiv- alent to the sum of the squares that can be described upon the sides A B and B C (Prop. XL Bk. IV.). 321. Scholium. A square may thus be found equivalent to the sum. of any number of squares ; for the construc- tion which reduces two of them to one, will reduce three of them to two, and these two to one. Peoblem XX. 822. To find the side of a square v^hich shall be equiv- alent to the difference of tioo given squares. Draw tlie two straight lines AB, AC C perpendicular to each other, making A C equal to the side of the less square. Then from as a centre, with a radius equal to the side of tlie otlier square, describe an arc intersecting A B in the point B, and AB will be the side of the required square. For, join B C, and the square that can be described upon AB is equivalent to the difference of the squares tliat can be described on B C and A (Prop. XI. Cor. 1, Bk. IV.). Problem XXI. 323. To construct a rectangle that shall be equivalent to a given triangle. Let ABC be the given triangle. Draw the indefinite straight line C D parallel to the base A B ; bi- sect A B by the perpendicular EF, and make E G equal to F B. Then, by drawing Gr B, the rec- tangle B F B G is eq\ial to the tri- angle ABC. ICO ELEMENTS OP GEOMETRY. For the rectangle EFBG has the same altitude, EP, as the triangle A B C, and half its base (Prop. II. Cor. 1, Bk. IV.). Pboblem XXII. 324. To construct a triangle that shall be equivalent to a given polygon. Let A B C D E be the given poly- C gon. ^^^<^^^Vv Draw the diagonal CE, cutting off ^ K / i \ \\ ,^ the triangle ODE; through the i A / ; V^\ point D draw D P parallel to C E, 1/ v ■/ \ and meeting A E produced in P. Cl^ A E F Draw C P ; and tlie polygon ABODE will be equivalent to the polygon A B C P, whicli has one side less than the given polygon. Per the triangles D E, C P E have the base C E com- mon ; they have also the same altitude, since their ver- tices, D, P, are situated in a line, DP, parallel to the base ; these triangles are therefore equivalent (Prop. II. Cor. 2, Bk. IV.). Add to each of them the figure ABCE, and the polygon A B C D E will be equivalent to the polygon ABC P. In like manner, the triangle C G A may be substituted for the equivalent triangle ABC, and tluis the pentagon ABODE will be changed into an equivalent triangle GOP. The same process may be applied to every other polygon ; for, by successively diminishing the number of its sides, one at eacli step of the process, the equivalent triangle will at last bo found. Problem XXTII. 325. To divide a g-ivcii straight line into any number of equal parts. BOOK V. 131 Let A B be the given straight line proposed to be divided into any number of equal parts ; 'for example, six. Tlirough the extremity A draw the indefinite straight line A E, making any angle with A B. Take A C of any convenient length, and apply it six times upon A E. Join the last point of division, E, and the extremity B by tlie straight line E B ; and througli tlie point C draw C D par- allel to E B ; then A D will be the sixth part of the line A B, and, being applied six times to AB, divides it into six equal parts. For, since C D is parallel to E B, in the triangle ABE, we have the proportion (Prop. XVII. Bk. IV.), AD:AB::AC:AB. But A C is the sixth part of A B ; hence A D is the sixth part of A B. Problem XXIV. 326. To divide a given straight line into parts that shall be proportional to other given lines. Let A B be the given straight line proposed to be divided into parts proportional to the given lines A G, C D, D B. Through the point A draw the indefinite straight line AB, mak- ing any angle with A B. On A E lay off A C, C D, and D E. Join the points E and B by the straight line E B, and through the points and D draw C G- and D H par- allel to E B ; and the line A B will be divided into parts proportional to the given lines. For, since C G and D H are each parallel to E B, we have the proportion (Prop. XVII. Cor. 2, Bk. IV.), AC:AG::CD:GH::DB:HB. 132 ELEMENTS OP GEOMETRY. Problem XXV. 327. To find a fourth proportional to three given, straight lines. Draw the two indefinite straight lines AB, AE, forming any angle with each other. On AB make AD equal to the first of the proposed lines, and A B equal to the second ; and on A E A D B make A E equal to the third. Join B E ; and through the point D draw D C parallel to B E, and A C will he the fourth proportional required. For, since D C is parallel to B E, we have the propor- tion (Prop. XVII. Cor. 1, Bk. IV.), AB: AD : : AE : AC. 328. Cor. A third proportional to two given lines, A and B, may be found in the same manner, for it will be tlie same as a fourth proportional to the three lines, A, B, and B. Problem XXVI. 329. To find a mean proportional beticeen tiro given straight lines. Draw the indefinite straight line A B. On A B take A C equal to the first of the given lines, and C B equal to the second. On A B, as a diameter, describe a semicircle, and at the point C draw the perpendic- ular CD, meeting the somi-circnniforciico in D ; CD will be the moan proportional required. For the perpendicular C D, drawn from a point in the circumforonco to a point in the diameter, is a moan pro- / \ ■••? / Q..-^ '\ ,.^' •1 D>- /■ ^y / BOOK V. 133 portional between the two segments of the diameter A C, C B (Prop. XXXII. Cor., Bk. IV.) ; and these segments are equal to the given lines. Problem XXVII. 330. To divide a given straight line into two such parts, that the greater part shall be a mean proportional between the whole line and the other part. Let A B be the given straight line; At the extremity, B, of the line AB, erect the perpendicular BC, equal to the half of A B. From the point C as a centre, with the radius C B, describe a circle. ^ E B Draw A C cutting the circumference in D ; and take AE equal to AD. The line AB will be divided at the point E in the mamier required ; that is, AB: AE : : AE:BB. For A B, being perpendicular to the radius at its ex- tremity, is a tangent (Prop. X. Bk. III.) ; and if A C be produced till it again meets the circumference, in F, we shall have (Prop. XXXV. Bk. IV.), AF: AB: : AB: AD; hence, by division (Prop. VIII. Bk. II.), AF — AB:AB::AB — AD:AD. But, since the radius is the half of A B, the diameter D F is equal to A B, and consequently A F — A B is equal to AD, which is equal to AE ; also, since A E is equal to A D, we have A B — AD equal to E B ; hence, AE : AB : : EB : AD, or AE; and, by inversion (Prop. V. Bk. II.), AB: AE: : AE: EB. 12 1S4 ELEMENTS OP GEOMETRY. 331. Scholium. This sort of division of the line A B is called division in extreme and mean ratio. Problem XXVIH. 332. Tlirough a given point in a given angle., to draw a straight line, tvhich shall have the parts included be- tiveen that point and the sides of the angle equal to each other. Let E be the given point, and ABC the given angle. Through the point E draw E F paral- lel to BC, make AF equal to BF. Through the points A and E draw tlie straight line A E C, and it will be the line required. For, E F being parallel to B C, we have (Prop. XVII. Bk. IV.), AF:FB: : AE: EC; but A F is equal to F B ; therefore A E is equal to E C. Problem :^XIX. 333. On a given straight line to construct a rectangle that shall be equivalent to a given rectangle. Let AB be the given straight line, and C D E F the given rectangle. Find a fourth proportional to the throe straight lines AB, CD,DE (Prob.XXY.); ^^ B C B and let B G- be that fourth proportional. The rectangle constructed on A B and B G will be equivalent to the rec- tangle C D E F. For, since A B. : C D : : D E : B G, it follows (Prop. I. Bk. IL) that AB X BG = CD X DE; H BOOK V. 135 hence, the rectangle A B G H, which is constructed ou the line A B, is equivalent to the rectangle C D E P. A E Problem XXX. 334. To construct a square that shall be equivalent to a given parallelogram, or to a given triangle. First. Let ABCD be the given D C parallelogram, A B its base, and I) E its altitude. Pind a mean proportional between A B and D E (Prob. XXVI.) ; and tlie square constructed on that proportional will be equiv- alent to the parallelogram ABCD. For, denoting the mean proportional by xy, we have, by construction, A B : a; ?/ : : a; «/ : D E ; therefore, ^^ = A B X D B ; but A B X D E is the measiire of the parallelogram, and X y that of the square ; hence they are eqiiivalent. Secondly. Let A B C be the given tri- angle, BO its base, and AD its altitude. Find a mean proportional between BC and the half of AD, and let xy denote that proportional ; the square constructed on xy will be equivalent to the triangle ABC. For since, by construction, B C : xy : : xy : 2- A D, it follows that a;/ = BC X ^ AD; hence the square constructed on xy is equivalent to the triangle ABC. 136 ELEMENTS OP GEOMETRY. Problem XXXI. 335. To construct a rectangU equivalent to a given square, and having the sum of its adjacent^ sides equal to a given line. Let the straight line AB be equal ^ '~ ~^r to the sum of the adjacent sides of the required rectangle. Upon A B as a diameter describe L _____^ _ \ a semicircle ; at the point A, draw A E B A D perpendicular to A B, making A D equal to the side of the given square ; then draw the line D C parallel to tlie diameter A B. From the point C, where the parallel meets the circumference, draw C E perpendicular to the diameter ; A E and E B will be the sides of the rectangle required. For their sum is equal to A B ; and their rectangle A E X E B is equivalent to the square of C E, or to the square of A D (Prop. XXXII. Cor., Bk. IT.) ; hence, this rectangle is equivalent to the given squai-e. 336. Scholium. The problem is impossible, when the distance A D is greater than the half the given line A B, for then the line D C will not meet the circumference. Problem XXXII. 337. To construct a rectangle that shall be equivalent to a given square, and the differciu-e of whose adjacent sides shall be equal to a given line. Let the straight line AB be equal to the difference of the adjacent sides of the required rectangle. Upon XB as a diameter, describe a circle. At the ex- tremity of the diameter, draw the tangent AD, making it equal to the side of the given square. BOOK V. 137 Through the point D and the centre C draw the secant D C F, intersecting the circumt'erence in E ; tlieu D E and D P will be the adjacent sides of tlie rectangle required. For the difference of these lines is equal to the diameter E P or A B ; audjthe rectangle D E x D P is equal to ad' (Prop. XXXV. Cor., Bk. IV.) ; hence it is equiv- alent to the given square. Problem XXXIII. 338. To construct a square that shall be to a given square as one given line is to another given tine. Draw the indefinite line AB, on which take AC equal to one of the given lines, and C B equal to the other. Upon A B as a di- ameter, describe a semicircle, and at the point C draw the perpendicular C D, meeting the circumference in D. Through the points A and B draw- the straight lines D E, DP, making the former equal to the side of the given square ; and through the point E draw E P parallel to A B ; D F will be the side of tlie square required. For, since E F is parallel to A B, DB : DP : : D A: DB; consequently (Prop. XV. Bk. II.), DB' : DP' : : DA' : D B'. But in the right-angled triangle A D B the square of A D is to tlie square of D B as the segment A C is to the seg- ment C B (Prop. XI. Cor. 3, Bk. TV.) ; hence, DE' : DF' : : a C : C B. 12* 138 ELEMENTS OP GEOMETRY. But, by construction, D E is equal to the side of the given s,quare ; also, A is equal to one of the given lines, and C B to tlie other ; hence, tlie given square is to that con- structed on D P as the one given line is to the other. Problem XXXIV. 339. Upon a given base to construct an isosceles tri- angle, having each of the angles at the base double the vertical angle. Let A B be the given base. Produce A B to some point C till the rectangle A C X B C shall be equiva- lent to the square of A B (Prob. XXXn.) ; then, with the base A B and sides each equal to A 0, construct the isosceles triangle DAB, and the angle A will double the angle D. For, make DB equal to AB, or make AE equal to BC, and join E B. Then, by construction, AD : AB: : AB: AE; for AE is equal to BC ; consequently the triangles DAB, B A E have a common angle. A, contained by proportional sides ; hence they are similar (Prop. XXIV. Bk. IV.) ; therefore these triangles are both isosceles, for D A B is isosceles by construction, so that A B is equal to E B ; but A B is equal to D E ; consequently D E is equal to E B, and therefore the angle D is equal to the angle E B D ; hence the exterior angle A E B is equal to double the an- gle D, but the angle A is equal to the augle AEB ; thei-e- fore the angle A is double the angle D. Problem XXXV. 340. Upon a given straight line to construct a polygon similar to a given poli/gon. BOOK V. 1S9 Lot ABODE he the given poly- gon, and P G the given straight line. Draw the diag- onals AC, AD. At the point P in the straight line F G, make the angle G P H equal to the angle BAG; and at the point G make the angle F G H equal to the angle ABC. The lines PH, GH will cut each other in H, and P G H will be a triangle similar to ABC. In the same manner, upon P H, homologous to A C, construct the triangle P I H similar to A D C ; and upon PI, homologous to AD, construct the triangle PIK similar to A D E. The polygon P G li I K will be similar to A B C D E, as required. For these two polygons are composed of the same num- ber of triangles, similar each to each, and similarly situ- ated (Prop. XXX. Cor., Bk. IV.). Problem XXXVI. 341. Two similar polygons being- given, to construct a similar polygon, which shall be equivalent to their sum or their difference. Let A and B be two homologous sides of the given poly- gons. Find a square equal to the sum or to the difference of the squares described up- on A and B ; let a; be the side of that square ; then will x in the polygon required be the side which is homologous to the sides A and B in tlie given polygons. The polygon itself may then be constructed on x, by the last problem. 140 ELEMENTS OP GEOMETRY. For similar figixres are to each other as the squares of their liomologous sides ; but tlie square of the side x is equal to the sum or the differeuce of the squares described upon the homologous sides A and B ; therefore the figure described upon the side x is equivalent to the sum or to the difference of- the similar figures described upon the sides A and B. Problem XXXVII. 342. To construct a polygon similar to a given polygon, and which shall have to it a given ratio. . Let A be a side of the given polygon. Find the side B of a square, which is to the square on A in the given ratio of the polygons (Prob. XXXIII.). Upon B construct a polygon similar to the given polygon (Prob. XXXV.), and B will be the polygon required. -^ For the similar polygons constructed upon A and B have the same ratio to each other as the squares con- structed upon A and B (Prop. XXXI. Bk. IV.). Problem XXXVIII. 343. To construct a polygon similar to a given polygon, P, and which shall be equivalent to another polygon, Q. Find M, the side of a square, equivalent to the polygon P, and N, the side of a square equivalent to the polygon Q. Let a; be a fourth propor- tional to the three given lines M, N, A B ; upon the side r, homologous to A B, describe a polygon similar to the polygon P (Pi'ob. XXXV.) ; it will also be equivalent to the polygon Q. BOOK V. 14]^ For, representing the polygon described upon the side X by y^ we have P : y : : AB^ : x" ; but, by construction, AB:a:::M:N, or XB^ : a? : : W : W ; hence, V:y::W:W. But, by construction also, M^ is equivalent to P, and W is equivalent to Q ; therefore, P:y::P:Q; consequently y is equal to Q ; hence the polygon y . is similar to- the polygon P, and equivalent to the poly- gon Q. BOOK YI. EEGULAE POLYGONS, AND THE AREA OF THE CIRCLE. DEFINITIONS. 344. A Regular Polygon is one which is both equi- lateral and equiangular. 345. Regular polygons may have any number of sides : the equilateral triangle is one of three sides ; the square is one of four. Proposition I. — Theorem. 346. Regular polygons of the same nuviber of sides are similar figures. LetABCDEP, E D GHIKLM, be two regular poly- gons of the same number of sides ; then these poly- gons are similar. A B G H For, since the two polygons have the same number of sides, they have the same number of angles ; and the sum of all the angles is the same in the one as in the other (Prop. XXIX. Blc. I.). Also, since the polygons are cquianguliu', oacli of tlic iiiiglos A, T?, C, t^-c. is equal to each of tlie luiglL-s (i, II, I, &.c. ; lionco tho two polygons are mutually e(iuiiuigular. BOOK VI. 143 Again ; the polygons being regular, the sides A B, B C, CD, &c. are equal to each other; so likewise are the sides GH, HI, IK, &c. Hence, AB:GH::BC:HI::CD:IK, &c. Therefore the two polygons have their angles equal, and their homologous sides proportional ; hence they are simi- lar (Art. 210). 347. Cor. The perimeters of two regular polygons of the same number of sides, are to eacli other as their lio- mologous sides, and their areas are to each other as the squares of those sides (Prop. XXXI. Bk. IV.). 348. Scholivm. The angle of a reg-ular polygon is de- termined by the number of its sides (Prop. XXIX. Bk. I.). Proposition II. — ■ Theorem. 349. A circle may be circumscribed about, and another inscribed in, any regular polygon. Let A B C D E P G H be any reg- ular polygon ; then a circle may be circumscribed about, and another inscribed in it. Describe a circle whose circum- ference shall pass through the three points A, B, C, the centre being ; let fall the perpendicular P from to the middle point of tlie side B ; and draw the straight lines A, OB, C, OD. Now, if the quadrilateral P C D be placed upon the quadrilateral OPB A, they will coincide ; for the side OP is common, and the angle P C is equal to the angle OPB, each being a right angle ; conseqiiently the side P will fall upon its equal, P B, and the point C on B. Moreover, from tlie nature of the polygon, the angle PCD is equal to the angle P B A ; therefore C D will take the 144 ELEMENTS OF GEOMETRY. direction B A, and C D being equal to B A, the point D will fall upon A, and the two quadrilaterals will coincide throughout. Therefore OD is equal to AO, and the circum- ference wliicli passes through the tliree points A, B, C, will also pass through the point D. By the same mode of reasoning, it may bo sliown that the circle which passes tlirougli tlie three vertices B, C, D, will also pass througli the vertex E, and so on. Hence, the circumfer- ence wliich passes througli the tliree points A, B, C, passes through the vertices of all the angles of the polygon, and is circumscribed about the polygon (Art. 166). Again, with respect to this circumference, all the sides, AB, B C, CD, &c., of the polygon are equal chords ; con- sequently they are equally distant from the centre (Prop. VIII. Bk. III.). Hence, if from the point 0, as a centre, and with the radius P, a circle be described, the circum- ference will touch the side B C, and all the other sides of the polygon, each at its middle point, and the circle will be inscribed in the polygon (Art. 168). 350. Scholium 1. The point 0, the common centre of the circumsGribed and inscribed circles, may also be re- garded as the centre of the polygon. The angle formed at the centre by two radii drawn to the extremities of tlie same side is called the ariffle af the centre : and the per- pendicular from the centre to a side is called the apothegm of the polygon. Since all the chords AB, B C, CD, }>ular lioxagon, is equal to AO, the radius of the clrjlu (Prop. VIII. Cor. Bk. 1.). BOOK VI. 147 356. Cor. 1. To inscribe a regular hexagon in a given circle, apply the ra- dius, A 0, of the circle six times, as a chord to the circumference. Hence, beginning at any point A, and applying A six times as a chord to the circum- ference, we are brought round to the point of beginning, and the inscribed figure A B C D E F, thus formed, is a regular hex- agon. 357. Cor. 2. By joining the alternate angles of the in- scribed regular hexagon by the straight lines A C, C E, E A, the figure ACE, thus inscribed in the circle, 'vvill be an equilateral triangle, since its sides subtend equal arcs, ABC, CDE, EFxi, on the circumfoi-ence (Prop. III. Bk. III.). 358. Cor. 3. Join A, OC, and the figure ABCO is a rhombus, for each side is equal to the I'adius. Hence, the sum of the sqiiares of the diagonals A C, B is equiv- alent to the sum of the squares of the sides (Prop. XV. Bk. IV.) ; or to four times the square of the radius B ; that is, A C^ + OB^ is equivalent to 4 AB^, or 4 B^ ; and taking away B from both, there remains A C^ equivalent to 3 B ; hence AC^0B^•:3:1, or AC:0B::VI:1; hence, the side of the inscribed equilateral triangle is to the radius as the square root of 3 is to unity. Proposition VI. — Problem. 359. To inscribe a regular decagon in a given circle. Divide the radius, A, of the given circle, in extreme and mean ratio, at the point M (Prob. XXVII. Bk. V.). 148 ELEMENTS OP GEOMETRY. Take the chord AB equal to OM, and A B will be the side of a regular decagon iuscribed in the circle. For we have by construction, A : M : : M : A M ; or, since A B is equal to M, A : A B : : A B : A M. Draw MB and BO; and the triangles ABO, AMB have a common angle. A, included between proportional sides ; hence the two triangles are similar (Prop. XXR''. Bk. IV.). Now, the triangle A B being isosceles, AMB must also be isosceles, and A B is equal to B il ; but A B is equal to M, consequently M B is equal to M ; hence the triangle M B is isosceles. Again, the angle A M B, being exterior to the isosceles triangle BMO, is double the interior angle O (Prop. XXVII. Bk. I.). But the angle AMB is equal to the angle M A B ; hence the triangle A B is such, that each of the angles at the base, OAB, OB A, is double the angle 0, at its vertex. Hence the three angles of ithe triangle are together equal to five times the angle 0, which conse- quently is a fifth part of two right angles, or the tenth part of four right angles ; tlierefore the arc A B is the tenth part of the circumfei'ence, and the chord A B is the side of an inscribed regular decagon. 360. Cor. 1. By joining the vertices of the alternate angles A, C, kc. of the regular decagon, a regular penta- gon may be inscribed. Hence, the chord A C is the side of an inscribed regular pentagon. 3G1. Cor. 2. A B being the side of the inscribed regu- lar decagon, let A L bo the side of an inscribed regular liexagon ( Prop. V. Cm-. 1). Join BL : then B L will bo the side of an inscribed ronular ]i(Mi(i'docagon, or regular polygon of fifteen sides. For A B cuts off an arc equal to a lentil part of the circumference ; and A L subtends an BOOK VI. 149 arc equal to a sixth of the circumference ; therefore B L, the difference of tliese arcs, is a fifteentli part of tlio cir- cumference ; and since equal arcs are subtended by equal cliords, it follows that the chord B L may be applied ex- actly fifteen times around the circumference, thus forming a regular pentedecagon. 362. Scholium. If the arcs subtended by the sides of any inscribed regular polygon be severally bisected, the chords of those semi-arcs will form another inscribed polygon of double the number of sides. Thus, from having an inscribed square, there may be inscribed in suc- cession polygons of 8, IG, 32, 64, &c. sides ; from the hexagon may be formed polygons of 12, 24, 48, 96, &c. sides ; from the decagon, polygons of 20, 40, 80, i-o. ; conse- quently, the sides of the ciirmnsoribod polygon are all equal ; hence this polygon is regular, and similar to the inscribed one. BOOK VI. 1;j1 364. Cor. 1. Conversely, if the circumscribed polygon G H I K, &c. is given, and it is required, by means of it, to construct a similar inscribed polygon, draw the straight lines O G, H, &c. from the vertices of the angles G, H, I, &c. of the given polygon to the centre ; the lines will meet the circumference in the points A, B, C,, &c. Join these points by the chords A B, B C, &c., which will form the inscribed polygon. Or simply join the points of con- tact, M, N, P, &c., by chords, MN, NP, &c., wliich like- wise would form an inscribed polygon similar to the cir- cumscribed one. 365. Cor. 2. Hence, we may circumscribe about a cir- cle any regular polygon similar to an inscribed one, and conversely. 366. Cor. 8. It has been shown that N H and H M are equal ; therefore the sum of N H and H M, wliich is equal to the sum of H M and M G, is equal to H G, one of the equal sides of the polygon. 367. Scholium. From having a circumscribed regular polygon, another having double the number of sides may be readily constructed, by drawing tangents to the points of bisection of the arcs, intercepted by the sides of the pro- posed polygon, limiting tliese tangents by those sides. In like manner other circumscribed polygons may be formed ; but it is plain that each of the polygons so formed will be less than the preceding polygon, being entirely compre- hended in it. Proposition YIII. — Theorem. 368. T^ie area of a regular polyg-on is equivalent to the product of its perimeter hij half of the radius of the in- scribed circle. Lot ABCPEP be a regular polygon, and the centre of the inscribed circle. From let the straight lines A, OB, &c. bo drawn to i:2 ELEMENTS OF GEOMETRY. the vortices of all the angles of the polygon, and the polygon will bo divided into as many equal triangles as it has sides ; and let tiie radii M, N, &c. of the in- scribed circle be drawn to the centres of the sides of the polygon, or to the points of tangency M, N, &c., and these radii are perpendicular to the sides re- spectively (Prop. XI. Bk. III.) ; therefore the radius of the circle is equal to the altitude of the several triangles. Now, the triangle A B is measured by the product of AB by half of M (Prop. VI. Bk. IV.) ; the triangle B C by the product of B C by half of X. But M is equal to N ; hence the two triangles taken together are measured by the sum of A B and B C by half of M. In like manner tlie measure of the other triangles may be found ; hence, the sum of all the triangles, or the whole polygon, is equal to the s\im of the bases AB, BC, itc, or the perimeter of the polygon, multiplied by half of OM, or half the radius of the inscribed circle. Proposition IX. — Theorem. 369. The perimeters of tiro regtilar polygons, hnvvig the same number of sides, are to each other as tht radii of the circumscribed circles, and, also, as the radii of the inscribed circles ; and their areas are to each othi r as the squares of those radii. Let A B be a side of one polygon, the centre, and consequently A the radius of the circuniscribod cir- cle, and M, perpendicular to A B, the radius of the inscribed circle. Lot G H be a side of the other poly- gon, C the centre, OG and ON the A B G radii of the circumscribed and the inscribed circles. BOOK YI. 153 The perimeters of the two polygons arc to each other as the sides AB and GH (Prop. XXXI. Bk. 1\^), but tlie angles A and G are equal, being each half of the angle of the polygon ; so also are the angles B and H ; hence, drawing OB and C H, the isosceles triangles ABO, GHC are similar, as are likewise the right-angled triangles A M 0, G N C ; hence AB:GH::A0:GC::M0:NC. Hence, the perimeters of the polygons are to each other as the radii A 0, G C of the circumscribed circles, and, also, as tlie radii M 0, N of the inscribed circles. The surfaces of these polygons are to each other as the squares of the homologous sides A B, G H (Prop. XXXI. Bk. IV.) ; they are therefore to each other as the squares of AO, GO, the radii of the circumscribed circles, or as the squares of M, N, the radii of the inscribed circles. Proposition X. — Pboblem. 370. The surface of a regular inscribed poiygon, and that of a similar circumscribed polygon, beifig given; to find the surfaces of regular inscribed and circumscribed polygons having double the number of sides. Let AB be a side of the given inscribed polygon ; E P, parallel to AB, a side of the circumscribed polygon, and C the centre of the circle. Draw the chord AM, and the tangents A P, B Q ; then A M will be a side of the inscribed poly- '■■l' gon, having twice the number of sides ; and P Q, the double of P M, will be a side of the similar circumscribed polygon. Let A, then, be the surface of the inscribed polygon whose side is AB, B that of the similar cii-cumscribcJ polygon ; A' the surface of the polygon whose side is A il. 154 ELEMENTS OP GEOMETRY. B' that of the similar circumscribed polygon : A ajid B are given ; we have to find A' and B'. First. The triangles A C D, A C M, whose common vertex is A, are to ■each other as their bases C D, C M (Prop. VI. Cor., Bk. IV.) ; they are likewise as the polygons A and A' ; hence A : A' : : C D : C M. Again, the triangles C A M, C M E, whose common vertex is M, are to each other as their bases C A, C E ; they are likewise to each other as the polygons A' and B ; hence A' : B : : C A : C E. But, since A D and M E are parallel, we have, CD : CM: : CA: CE; hence A : A' : : A' : B ; hence, the polygon A' is a mean proportional between the two given polygons. Secondly. The altitude C M being common, the tri- angle C P M is to the triangle CPEasPZHistoPE; but since C P bisects the angle MCE, we have (Prop. XIX. Bk. IV.), P M : P E : : C M : C E : : C D : C A : : A : A' ; hence C P M : C P E : : A : A' ; and, consequently, C P M : C P M + C P E or C ^[ E : : A : A -f A'. But C M P A or 2 C M P and C il E ai-e to each other as the polygons B' and B ; hence B':B: : 2 A : A + A'; which gives r>, _ 2 A X B A + A' ' BOOK VI. l~>r> or, the polygon B' is equal to the qvotient of tiuice the product of the g-irru polygons dirided by the sum of the inscribed polygons. Thus, by means of the polygons A and B, it is easy to find tlie polygons A' and B', which have double the num- ber of sides. Proposition XI. — Theorem. 371. A circle being given, two similar polygons can always be formed, the one circumscribed about the circle, the other inscribed in, it, which shall differ from- each other by less than any assignable surface. Let Q be the side of a square I less than the eiven surface. H.^-^;^^C^^55^ -c- Bisect A C, a fourth part of \iff^^'-^-~^^ -^ \\ the circumference, and then hi- /iT'-'-Ox ""--.^ \V sect the half of this fourth, and ^ \tK~ % -..-j^-.J)L so proceed until an arc is found \ Q // whose chord AB is less than ^^^T Z-i^ Q. As this arc must be an ex- ' ^^^^^"^^ act part of the circumference, if we apply the chords A B, B C, &c., each equal to AB, the last will terminate at A, and there will be inscribed in the circle a regular polygon, ABODE, &c. Next describe about the circle a similar polygon, G H I K L, &c. (Prop. VII.) ; and the difference of these two polygons will be less than the square of Q. Find the centre, ; from the points G and H draw the straight lines GO, HO, and they will pass through the points A and B (Prop. VII.). Draw also OM to the point of tangency, M ; and it will bisect A B hi N, and be per- pendicular to it (Prop. VI. Cor. 1, Bk. III.). Produce A to B, and draw B E. Let P represent the circumscribed polygon, and p the inscribed polygon. Tlien, since these polygons are simi- lar, they are as the squares of the homologous sides G H, 1E6 ELEMENTS OF GEOMETRY. A B (Prop. XXXI. Bk. lY.) ; but the triangles G H, A B are similar (Prop. XXIV. Bk. IV.) ; hence they are to each other as the squares of the homologous sides G and A (Prop. XXIX. Bk. IV.) ; therefore F:p:: OG^ : OA^ or OM^. Again, the triangles G M, E A B, having their sides respectively parallel, are similar ; therefore F:p:: OW : OM^ : : A E' : JTe'; and, hy division, P:P—p:: A~E^ : AE^ — EB^ or AB^. But P is less than the square described on the diameter A E ; therefore P — p is less than the square described on AB, that is, less than the given square Q. Hence, the difference between the circumscribed and inscribed polygons may always be made less than any given bur- face. 372. Cor. Since the circle is obviously greater than any inscribed polygon, and less than any circumscribed one, it follows that a polygon may be inscribed or c ire u inscribed., which will differ from the circle by less than any asstgrir able magnitude. Proposition XII. — Problem. 373. To find the approximate area of a circle whose radius is unity. Let the radius of the circle be 1, and lot tlie first in- scribed and circumscribed polyaoas bo squares ; the side of the inscribed sqnare will be V- (Prop. IV. Cor.), and that of tlie circumscribed square will bo oiiiial to the di- ameter 2. Hence the surface of the inscribed square is 2, and tiiat of the circumscribed square is 4. Lot, tlicre- foro A = 2, and B = 4. Now it has been proved, in Proposition X., that the surl'aco of the inscribed octagon, or, as it has boon rojjroscnli'd, A', is a moan i)roportional BOOK VI. 157 between the two squares A and B, so that A' = a/F= 2.8284271 ; and it has also been proved, in the same prop- osition, that the circumscribed octagon, represented by B', = ~A-r-A^; so that B' = -^ — = 3.3137085. The A + A' 2 + v''8 inscribed and the circumscribed octagons being thus determined, we can easily, by means of them, determine the polygons having twice the number of sides. We have only in this case to put A = 2.8284271, B = 3.3137085 ; and we shall find A = V A X B = 3.0614674, and B' = ^A>criinctron, which signifies circumference. BOOK VI. 163 386. Scholium 2. The Quadrature op the Circle is the problem which requires tlie finding of a square equiv- alent in area to a circle haviug a given radius. Now, it has just been proved that a circle is equivalent to the rec- tangle contained by its circumference and half its radius ; and this rectangle may be changed into a square, by find- ing a mean proportional between its length and its breadth (Prob. XXVI. Bk. V.). To square the circle, therefore. is to find the circumference when the radius is given; and for effecting this, it is enough to know the ratio of the cir- cumference to its radius, or its diameter. But this ratio has never been determined except approx- imately ; but the approximation has been carried so far, that a knowledge of the exact ratio would afford no real advantage whatever beyond tliat of the approximate ratio. Professor Rutherford extended the approximation to 208 places of decimals, and Dr. Clausen to 250 places. The value of n, as developed to 208 places of decimals, is C . 14159265358979323846264388327950288419716939937 5105820974944592307816406286208998628034825342717 0679821480865132823066470938446095505822317253594 0812848473781392038633830215747399600825931259129 40183280651744. Such an approximation is evidently equivalent to per- fect correctness ; the root of an imperfect power is in no case more accurately known. Proposition XVI. — Problem. 387. To divide a circle into any number of equal parts by means of concentric circles. Let it be proposed to divide the circle, whose centre is 0, into a certain number of equal parts, — three for in- stance, — by means of concentric circles. Draw the radius AO; divide AO into three equal parts, A B, B C, C 0. Upon A describe a semi-circumference, 164 ELEMENTS OF GEOMETRY. and draw tlie perpendiculars, B E, C D, meeting tliat semi-circuinfer- ence in the points E, D. Join E, D, and with these lines as radii from the centre, 0, describe circles ; these circles will divide the given circle into the required number of equal parts. For join A E, AD; then the angle ADO, being in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.) ; hence the triangles D A 0, D C are similar, and consequently are to each other as the squares of their homologous sides ; that is. but hence DAO:DCO::OA^:OD^ DAO: DCO: :0A: OC; OA': OD^: : OA: OC; consequently, since circles are to each other as the squares of their radii (Prop. XIII.), it follows that the circle whose radius is OA, is to that whose radius is OD, as OA to OC ; that is to say, the latter is one third of the former. In the same manner, by means of the right-angled tri- angles E A 0, E B 0, it may be proved that the circle ■whose radius is E, is two thirds that whose radius is A. Hence, the smaller circle and the two surroiuiding annular spaces are all equal. Note. — This useful problem ivas first solved by Dr. Hutton, tlio justly distinguished English mathematician. BOOK VII. PLANES. — DIEDEAL AND POLYEDEAL ANGLES. DEFINITIONS. 888. A STRAIGHT line is per- pendicular to a plane, when it is perpendicular to every straight line which it meets in that plane. Conversely, the plane, in the same case, is perpendicular to the line. The foot of the perpendicular is the point in which it meets the plane. Thtis the straight line A B is perpendicular to the plane M N ; the plane M N is perpendicular to the straight line A B ; and B is the foot of the perpendicular AB. 389. A line is parallel to a plane when it cannot meet the plane, however far both of them may be produced. Conversely, the plane, in the same case, is parallel to the line. 390. Two planes are parallel to each other, when they cannot meet, however far both of them may be produced. 391. A DiEDEAL Angle is an angle formed by the intersec- tion of two planes, and is meas- ured by the inclination of two straight lines drawn from any point in the line of intersection, perpendicular to that line, one being di-awn in each plane. 166 ELEMENTS OF GEOMETRY. The line of common section is called the edge, and the two planes are called the faces, of the diedral angle. Tims the two planes ABM, A B N, whose line of intersec- tion is AB, form a diedral angle, of which the line AB is the edge, and the planes ABM, ABN are faces. 392. A diedral angle may be actite, right, or obtuse. If the two faces are perpendicular to each other, the angle is right. 393. A POLYEDRAL AnGLE Is an angle formed by the meeting at one point of more than two plane angles, which are uot in the same- plane. The common point of meeting of the planes is called the vertex, eacli of the plane angles a face, and the line of common section of any two of the planes an edg-e of the polyedral angle. Thus the three plane angles A S B, B S C, C S A form a polyedral angle, whose vertex is S, whose faces are the plane angles, and whose edges are the sides, AS, BS, CS, of the same angles. 394. A polyedral angle formed by three faces is called a triedral angle ; by four faces, a tetraedral; by five faces, a pentaedral, /s 180 ELEMENTS OP GEOMETKY. hence, since the ratio A G : G D is common to both pro- portions, we have AE:EB: : CF : FD. Proposition XIX. — Theorem. 428. The sum of any two of the plane angles which form a triedral angle is greater than the third. The proposition requires dem- onstration only when the plane angle, which is compared to the sum of the other two, is greater than either of them. Let the triedral angle whose vertex is S be formed by the three plane angles ASB, ASC, BSC; and suppose the angle A S B to be greater than either of the other two ; then the angle A S B is less than the sum of the angles ASC, BSC. In the plane ASB make the angle BSD equal to BSC; draw tlie straight line A D B at pleasure ; make S C equal S D, and draw A C, B C. The two sides B S, S D are equal to the two sides B S, S C, and the angle BSD is equal to tlie angle BSC; therefore the triangles BSD. BSC are equal (Prop. V. Bk. I.); hence the side BD is equal to the s-idc B C. But A B is less than the sum of A C and B C : taking B D from the one side, and from tlie other its equal. B C, there remains A D less than A 0. The two sides A S, S D of the triangle A S D, are equal to the two sides A S, S C, of the triangle iV S C, and the third side A D is less than the third side A C ; hence the angle A S D is less than the angle ASC (Prop. XVII. Bk.'l.^ one, and its equal, BSC, t(i the other, wi sum of ASD, BSD, or A S B, loss than (ho sum of A S C, BSC. Adding B S D to shall have the BOOK VII. 181 Proposition XX. — Theorem. 429. The sum of the plane angles which form any poly- edral angle is less than four right angles. Let the polyedral angles whose a. vertex is S be formed by any number /f \ of plane angles, A S B, B S C, C S D, /f \ \ &c. ; the sum of all these plane angles / ,7 \r\ is less than four right angles. / '■ \ .V 'X Let the planes forming the poly- Air ■■--yr;--\ — Ajj edral angle be cut by any plane, \// \\/ ABCDEF. From any point, 0, B C in this plane, draw the straight lines A 0, BO, CO, DO, E 0, F 0. The sum of the angles of the triangles A SB, BSC,'&c. formed about the vertex S, is equal to the sum of the angles of an equal luimber of triangles A OB, BOG, :]. A Regular Polyedron is one whose faces are all equal and regular polygons, and whoso polycdral angles are all equal to each other. BOOK VIII. 187 Proposition I. — Theorem. ihi. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Let ABCDE-K be a right prism; then will its convex surface be equal to the perimeter of its base, AB + BC + CD + I)E + EA, multiplied by its altitude A T. For, the convex svirface of the prism is equal to the sum of the parallelo- grams AG, BH, CI, DK,'EF (Art. 436). Now, the area of each of those parallelograms is equal to its base, AB, B C, CD, &c., multiplied by its altitude, AF, BG, CH, &c. (Prop. V. Bk. IV.). But the altitudes AF, BG, OH, &c. are each equal to A F, the altitude of the prism. Hence, the area of these paral- lelograms, or the convex surface of the prism, is equal to (AB + BC+CD + DE + EA) XAF; or the product of the perimeter of the prism by its alti- tude. 455. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases. Proposition II. — -Theorem. 456. In every prism, the sections formed by parallel planes are equal polygons. Let the prism ABCDE-K be intersected by ' the parallel planes NP, S V; then are the sections NOPQR, S T V X Y equal polygons. For the sides S T, N are parallel, being the intersec- tions of two parallel planes with a third plane A B G F 188 ELEMENTS OF GEOMETRY. (Prop. XIII. Bk. VII.) ; these same sides ST, NO, are included between tlie parallels N S, OT, which arc sides of the prism ; hence N is equal to ST. For like reasons, the sides OP, P Q, Q R, &c. of the section N P Q R, are respectively equal to the sides TV, VX, XY, &c. of the section S T V X Y ; and Since the equal sides are at the same time parallel, it fol- lows that the angles NOP, OP Q, &c. of the first section are respectively equal to the angles S T V, T V X of the second (Prop. XVI. Bk. VII.). Hence, the two sections N P Q R, S T V X Y, are equal polygons. 457. Cor. Every section made in a prism parallel to its base, is equal to that base. Proposition III. — Theorem. 458. Two prisms are equal, when the three faces which form a triedral ang-le in the one are equal to those which form a triedral ang-le in the other, each to each, and are similarly situated. , Let the two prisms ABCDE-K and LMOPQ-Y have the faces which form the triedral angle B equal to the faces which form the tri- edral angle M ; that is, the base ABODE B C MO equal to the base L M N P Q, the parallelogram A B .G F (•(|uiil to the parallelogram LMSR, and the parallelogram B H G equal to MOTS; then the two prisms arc equal. BOOK VIII. 189 For, apply the base ABODE to the equal base LMOPQ ; then, the triedral an- gles B and M, being equal, will coincide, since the plane an- gles which form these triedral angles are equal each to each, and similarly situated (Prop. XXI. Sch. 2, Bk. VII.) ; hence the edge B G will fall on its equal M S, and the face B H will coincide with its equal MT, and the face BF with its equfil MR. But the iipper bases are equal to their corresponding lower bases (Art. 436) ; therefore the bases P G II I K, R S T V Y are equal ; hence they coincide with each other. Therefore H I coin- cides with T V, I K with V Y, and K P with Y R ; and consequently the lateral faces coincide. Hence the two prisms coincide throughout, and are equal. 459. Cor. Two right prisms, which have equal bases and equal altitudes, are eqiial. For, since the side AB is equal to LM, and tlie altitude B G to M S, the rectangle A B G P is equal to the rectan- gle L M S R ; so, also, the rectangle B G H C is equal to M S T ; and thus the three faces which form the triedral angle B, are equal to the three faces which form the trie- dral angle M. Hence the two prisms are equal. Peoposition IV. — Theoeem. 460. In every parallelopipedon the opposite faces are equal and parallel. Let ABCD-H be a parallelopipedon ; then its oppo- site faces are equal and parallel. The bases ABCD, EPGH are equal and parallel (Art. 436), and it remains only to be shown that the same is 190 ELEMENTS OF GEOMETRY. H -.J> G C true of any two opposite lateral faces, as B C G P, A D II E. Now, since the base A B C D is a parallelogram, the side A D is equal and parallel to B C. For a similar reason, AE is equal and par- allel to B F ; hence the angle D A E is equal to the angle C B P (Prop. XVI. Bk. VII.), and the planes DAE, CBP are parallel ; hence, also, the parallelogram B C G P is equal to the parallelogram AD HE. In the same w;iy, it may be shown that the opposite faces A B P E, D C G II are equal and parallel. 461. Cor. Any two "opposite faces of a parallelopipe- don may be assumed as its bases, since any face and the one opposite to it are equal and parallel. Proposition V. — Theorem. 462. The diagonals of every parallelopipedon bisect each other. Let ABCD-Hbe a parallelo- e H pipedon ; then its diagonals, as B H, DP, will bisect each other. Por, since B P is equal and par- allel to D H, the figure B P H D is a parallelogram ; hence the diago- nals B H, D P bisect each other at B C the point (Prop. XXXIV. Bk. I.). In the same man- ner it may be shown that the two diagonals A G and C E bisect each other at the point ; hence tlie several diag- onals bisect each other. 463. Srholiinii. Tlio point at which tlic diagonals mu- tually bisect each other may be regarded as tlio centre of the parallelopipedon. \ ..■-"'/ ,\ A --"■ \ .Q M .,-'■ BOOK VIII. 191 Proposition VI. — Theorem. 464. Any paraUelopipedon may be divided into Iwo equivalent triang-ular prisms by a plane passing through its opposite diagonal edges. Let any paraUelopipedon , A B C D-H , G be divided into two prisms, A B C - G, AD 0-G, by a plane, A C G E, passing through opposite diagonal edges; then will the two prisms be equivalent. Through the vertices A and E, draw the planes A K L M, E N P, perpen- dicular to the edge AE, and meeting BF, C G, DH, the three other edges of the paraUelopipedon, iu the points K, L, M, and in N, 0, P. The sections A K L Jl, E N P are equal, since they are formed by planes perpendicular to the same straight lines, aud hence parallel (Prop. II.). They are parallelograms, since the two opposite sides of the same section, AK, LM, are the intersections of two parallel planes, A B P E, D C G H, by the same plane, A K L M (Prop. XIII. Bk. VII.). For a like reason, the figure A M P E is a parallelo- gram ; so, also, are A K N E, K L N, L M P 0, the other lateral faces of the solid A K L M - P ; consequently, this solid is a prism (Art. 436) ; and this prism is right, since the edge AE is perpendicular to the plane of its base. This right prism is divided by the plane ALOE into the two right prisms AKL-0, AML-0, which, having equal bases, A K L, A M L, and the same altitude, A E, are equal (Prop. III. Cor.). Now, since AEHD, AEPM are parallelograms, the sides D H, M P, being each-equal to A E, are equal to each other ; and taking away the common part, D P, there re- mains DM equal to HP. In the same manner it may be shown that C L is equal to GO. 192 ELEMENTS OF GEOMETRY. Conceive now E P 0, the base of the solid E P - Gr, to be applied to its equal A M L, the point P fallhig upon M, and the point upon L ; the edges GO, HP will coincide with their equals CL, DM, since they are all perpen- dicular to tlie same plane, A K L M. Hence the two solids coincide through- out, and are therefore equal. To each of these equals add the solid AD C-P, and the right prism AML-0 is equivalent to the prism ADC-G. In the same manner, it may be proved that the riglit prism AKL-0 is equivalent to the prism ABC-G. The two right prisms AKL-0, A]ML-0 being equal, it fol- lows that two triangular prisms, ABC-G, ADC-G, arc equivalent to eacli other. 4Go. Cor. Every triangular prism is lialf of a parallelo- pipedon having the same triedral angle, with the same edges. Proposition VII. — Theorem. 466. Two parallelopipedons, having" a common lower base, and their upper bases in the same plane and betwei ii the same parallels, are equivalent to each otfier. Let the two parallelo- pipedons A G, A L have the common base ABCD, and their upper bases, EFGH, IKLM, in the same plane, and be- tween the same paral- lels, EK, III; ; then the parallelopipedons will bo 0(]uiv!ilont. BOOK VIII. 193 There may be three cases, according as E I is greater or loss tliau, or equal to, E F ; but the demonstration is the same for each. Since A E is parallel to BF, and HE to G-P, the plane angle A E I is equal to B F K, HEIto GPK, andHEA to Gr F B. Of these six plane angles, the three first form the polyedral angle E, tlie tlu-ee last the polyedral angle F ; consequently, since these plane angles are equal each to each, and similarly situated, the polyedral angles, B, F, must be equal. Now conceive the prism A E I - M to be applied to the prism B F K - L ; the base A E I, being placed upon the base B F K, will coincide with it, since they are equal ; and, since the polyedral angle E is equal to the polyedral angle F, the side E H will fall upou its equal, F G. But the base A E I and its edge E H deter- mine the prism A E I - M, as the base B F K and its edge FG determine the prism BFK-L (Prop. III.); hence the two prisms coincide throughout, and therefore are equal to each other. Take away, now, from the whole solid AELC, the prism AEI-M, and there will remain the parallelopipedon AL; and take away from the same solid A L the prism BFK-L, and there will remain the parallelopipedon A G ; hence the two parallelopipedons AL, AG are equivalent. Proposition VIII. — Theorem. 467. Two parallelopipedons having the same base and the same altitude are equivalent. Let the two parallelopipedons A G, A L have the com- mon base ABCD, and the same altitude ; then will the two parallelopipedons be equivalent. For, the upper bases EFGH, IKLM being in the same plane, produce the edges E F, HG, L K, IM, till by their intersections they form the parallelogram N P Q ; this parallelogram is equal to either of the bases I L, E G, and 17 194 ELEMENTS OP GEOMETKY. is between the same par- allels ; hence N P Q is equal to the common base A B C D, and is parallel to it. Now, if a third paral- lelopipedon be conceived, which, with the same lower base A B C D, has for its upper base NOPQ, this third parallelopipe- don will be equivalent to the parallelopipedon A G, since the lower base is the same, and the upper bases lie in the same plane and between the same parallels, GQ, FN (Prop. VII.). Tor the same reason, this third parallelopipedon will also be equivalent to the parallelopipedon A L ; hence the two parallelopipedons A G, A L, which have the same base and the same altitude, are equivalent. Pboposition IX. — Theorem. 468. Any oblique parallelopipedon is equivalent to a reciaTigvlar parallelopipedon having the same altitude and an equivalent base. Let A G be any paral- H G lelopipedon ; then A G wiU be equivalent to a rectangular parallelopip- edon having the same altitude and an equiv- alent base. From the points A, B, C, D,draw AI, BK, C L, D M, perpendicular to the lower base, and equal in altitude to AG; there w ill thus be formed the BOOK VIII. 195 parallelopipedon AL, equivalent to AG (Prop. VIII.), and having its lateral faces, AK, B L, &c., rectangular. Now, if the base ABCD is a rectangle, AL will be a rec- tangular parallelopipedon equivalent to A G. But if A B C D is not a rectangle, draw A 0, B N, each perpendicular to C D ; also Q, N P, each perpendicular i\_ -AK to the base ; then we shall have a rec- tangular parallelopipedon A B N - Q. For, by construction, the bases ABNO, I K P Q are rectangles ; so, also, are the lateral faces, the edges A I, Q, &c. being perpendicular to the plane of the base ; therefore the solid A P is a rectangular parallelopipedon. But the two parallelopip- MQ ] \ D f) A I C ■N :F edons A P, A L may be considered as having the same base, ABKI, and the same altitude, AO ; hence they are equivalent. Hence the parallelopipedon A G, which was shown to be equivalent to the parallelopipedon AL, is also equivalent to the rectangular parallelopipedon A P, having the same altitude, A I, and a base, ABNO, equivalent to the base ABCD. G Proposition X. — Theorem. 469. Two rectangular parallelopipedons , ivlnch have the same base, are to each other as their altitudes. Let the two parallelopipedons A G, ^ -^ A L have the same base, ABCD ; then they are to each other as their altitudes, AE, AI. First. Suppose the altitudes AB, AI are to each other as two whole numbers ; for example, as 15 is to 8. Divide A B into 15 eqiial parts, of which A I will contain 8. Through x, «/, z, &o., the points of division, coiiceive planes to g \ \ \ \ m I \ P M z y K A I \ 196 ELEMENTS OP GEOMETRY. E I I \ K m T p Kf \ * . K. A \ E S. G B pass parallel to the common base. These planes will divide the solid A G into 15 small parallelopipedons, all equal to each other, having equal bases and equal altitudes ; equal bases, since every section, as 1 K L M, parallel to the base A B C D, is equal to that base (Prop. II.), and equal altitudes, since the alti- tudes are the equal divisions A x, x y, y z, &c. But of those 15 equal parallel- opipedons, 8 are contained in A L ; hence tlie parallelo- pipedon A G is to the parallelopipedou A L as 15 is to 8, or, in general, as the altitude A E is to the altitude A I. Secondly. If the ratio of A E to A I cannot be exactly expressed by numbers, we shall still have tl^p proportion, Solid A G : Solid A L : : A E : A I. For, if this proportion is not correct, suppose we have Solid A G : Solid AL : : AE : AO greater than A I. Divide A E into equal parts, each of which shall be less than I ; there will be at least one point of division, m, between I and 0. Let P represent the parallelopipedon, whose base is A B C D, and altitude A m. ; since the alti- tudes A E, A TO are to each other as two whole numbers, we shall have Solid A.G: 7: : AE : Awi. But, by hypothesis, we have Solid A G : Solid A L : : A E : A ; hence (Prop. X. Cor. 2, Bk. II.), Solid Ah : P : : AO : Ato. But AO is greater than Aw; ; hence, if the proportion is correct, the parallelopipedon A L must be greater than P. On tiic contrary, however, it is less ; consequently the solid A G cannot be to the solid A L as the line A E is to a lino greutor than A I. BOOK VIII. 197 By the same mode of reasoning, it may be shown that the fourth term of the proportion cannot be less than A I ; therefore it must be equal to A I. Hence rectangular parallelopipedons, having the same base, are to each other as their altitudes. Peoposition XI. — Theorem. 470. Tioo rectang-ular parallelopipedons, having- the same altitude, are to each other as their bases. Let the two rectan- gular- parallelopipedons AG, AK have the same altitude, A E ; then they are to each other as their bases. Place the two solids so that their faces, B E, E, may have the com- mon angle B A E ; pro- duce the plane ONKL till it meets the plane D C G H in P Q ; we shall thus have a third parallelopipedon, A Q, wliich may be compared with eacli of the parallelopipedons A G, A K. The two solids, A G, A Q, having the same base, A B H D, are to each other as their altitudes A B, A (Prop. X.) ; in like manner, the two solids A Q, A K, having the same base, A L E, arc to each other as their altitudes have the two proportions. AD, AM. Hence we Solid A G : Solid A Q : : A B : A 0, Solid A Q : Solid A K : : A D : A M. Multiplying together the corresponding terms of these 17* 198 ELEMENTS OP GEOMETRY. proportions, and omitting, in the result, the common factor Solid A Q, we shall have, Solid A Q : Solid AK::ABxAD:AOxAM. But A B X A D measures tlie base A B C D (Prop. IV. Sell., Bk. I^.}; and AO X AM measures tlie base A M N ; hence two rectangular parallclopipedous of the same altitude are to each other as their bases. Phoposition Xn. — Theorem. 471. Any two rectangular parallelopipedons are to each other as the product of their bases by their altitudes. Let A G, A Z be two E H M rectangular parallelo- pipedons ; then they are to each other as the product of their bases, ABCD, AMNO, by their altitiides, AE, AX. Place the two solids so that their faces, B B, X, may have tlie com- mon angle B A E ; pro- duce tlie planes neces- sary for completing the third parallelopipedon, A K, having the same altitude with the parallelopipedon A G. By the last proposition, we shall have Solid A G : Solid A K : : A B C D : A ]\[ X 0. But the two parallelopipedons A K, A Z, having the same base, AMNO, are to each nthei- as their altitudes, AE, A X (Prop. X.) ; hence wo luivo Solid A Iv : Solid A Z : : A E : A X. Multiplying together the corresponding terms of these V %. V \ \ x; '■ 1 A M N o\ \ \ B BOOK VIII. 199 proportions, and omitting, in the result, the common fac- tor Solid A K, we shall have Solid A.Gr: &/tcZAZ::ABCDx AE: AMNOx AX. -Hence, any two rectangular parallelopipedons are to each other as the products of their bases by their altitudes. 472. Scholium 1. We are consequently authorized to assume, as the measure of a rectangular parallelopipedon, the product of its base by its altitude ; in other words, tk* product of its three dimensions. But by the product of two or more lines is always meant the product of the num- bers which represent them ; those numbers themselves being determined by the particular linear unit, which may be assumed as the standard. It is necessary, therefore, in comparing magnitudes, that the measuring unit be the same for each of the magnitudes compared. 473. Scholium 2. The measured magnitude of a solid, or volume, is called its volume, solidity, or solid contents. We assume as the unit of volume, or solidity, the cube, each of whose edges is the linear unit, and each of whose faces is the unit of surface. Peoposition XIII. — Theorem. 474. Tlie solid contents of a parallelopipedon, and of any other prism, are equal to the product of its base by its altitude. First. Any parallelopipedon is equivalent to a rectan- gular parallelopipedon having tlie same altitude and an equivalent base (Prop. IX.). But the solid contents of a rectangular parallelopipedon are equal to the product of its base by its altitude ; therefore the solid contents of any parallelopipedon are equal to the product of its base by its altitude. Second. Any triangular prism is half of a parallelopip- edon, so constructed as to have the same altitude, and a 200 ELEMENTS OP GEOMETRY. base twice as great (Prop. VI.). But the solid contents of the parallelopipedou are equal to the product of its base by its altitude ; hence, that of the triangular prism is also equal to the product of its base, or half that of tiie paral- lelopipedou, by its altitude. Third. Any prism may be divided into as many trian- gular prisms of the same altitude, as tliere are triangles in the polygon taken for a base. But the solid contents of each triangular prism are equal to the product of its base by its altitude; and, since the altitude is tlie same in each, it follows that the sura of all these prisms is equal to the sum of all the triangles taken as bases multiplied by the common altitude. Hence the solid contents of any prism are equal to the product of its base by its altitude. 475. Cor. When any two prisms liave the same altitude, the products of the bases by the altitudes will be as tlie bases (Prop. IX. Bk. II.) ; hence, prisms of the savte altitude are to each other as their bases. For a like reason, prisms of the same base are to each other as their altitudes. Proposition XIY. — Theorem. 476. Similar prisms are to each other as the cubes of their homologous edges. Let ABC-E, FHI-M be two similar prisms; these prisms are to each other as the cubes of their homologous edges, A B and FII. For, from D and K, ho- mologous angles of tlio two prisms, draw tlic perpendiculars J) \, KO, to the bases A B C, F H I. Talce A K' cquiil to F K, and join A N. BOOK Tin. 201 Draw K' 0' perpendicular to A N in the plane AND, and K' will be perpendicular to the plane ABC, and equal to K 0, the altitude of the prism F H I - M. For, con- ceive the triedral angles A and F to be applied the one to the other ; the planes containing them, and therefore the perpendiculars K'O', KO, will coincide. Now, since the bases A B C, F H I are similar, we have (Prop. XXIX. Bk. IV.), Base ABC: Base F H I : : AB^ : PH^ ; and, because of the similar triangles DAN, KFO, and of the similar parallelograms D B, K H, we have DN:KO::DA:KF::AB:FH. Hence, multiplying together the corresponding terms of these proportions, we have Base A B X D N : Base F HI X K : AB' : FH'. But the product of the base by the altitude is equal to the solidity of a prism (Prop. XIII.) ; hence Prism ABC-B: PnsmFHI-M: : AB^ FHI Proposition XV. — Theoeem. 477. The convex surface of a right pyramid is to the perimeter of its base, multiplied by half the height. Let A B C D E - S be a right pyra- mid, and S M its slant height ; then the convex surface is equal to the perimeter AB + BC + CD + DE + E A mul- tiplied by i S M. The triangles SAB, SBC, S C D, &c. are all equal ; for the sides A B, BC, CD, &c. are equal (Art. 445), and the sides S A, S B, S C, &c., being ob- lique lines meeting the base at equal equal slant 202 ELEMENTS OP GEOMETRY. distances from a perpendicular let fall from the vertex S to the centre of the base, are also equal (Prop. V. Bk. VII.). Hence, these triangles are all equkl (Prop. XVIII. Bk. I.) ; and the altitude of each is equal to the slant height S M. But the area of a triangle is equal to the product of its base mul- tiplied by half its altitude (Prop. VI. Bk. IV.). Hence, the areas of the tri- angles SAB, SBC, SCD, &c. are equal to the sum of the bases A B, B C, CD, &c. multiplied by half the common altitude, S M ; that is, the convex surface of the pyramid is equal to the perimeter of the base multiplied by half the slant height. 478. Cor. The lateral faces of a right pyramid are equal isosceles triangles, having for their bases the sides of the base of the pyramid. Pkoposition XVI. — Theorem. 479. If a pyramid be cut by a plane parallel to its base, — 1st. The edffes and the altitude will be divided propor- tionally. 2d. The section will be a polygon similar to the base. Let the pyramid A B C D E - S, whose g altitude is SO, be cut by a plane, G H I K L, parallel to its base ; then will the edges S A, S B, S C, &c., with the altitude SO, be divided proportion- ally ; and the section G H I K L will be similar to the base A B C D E. First. Since the planes ABC, GHI are parallel, their intersections A B, GH, by the third plane SAB, arc parallel (Prop. XIII. Bk. VII.) ; hence BOOK vm. 203 the triangles SAB, SGH are similar (Prop. XXV. Bk. IV.), and we have SA: SG: : SB: SH. For the same reason, we have SB:SH::SC:SI; and so on. Hence all the edges, S A, S B, S C, &c., are cut proportionally in G, H, I, &c. The altitude S is likewise cut in the same proportion, at the point P ; for B and H P are parallel ; therefore we have SO:SP::SB:SH. Secondly. Since G H is parallel to AB, H I to B C, I K to CD, &c. the angle G H 1 is equal to A B C, the angle H I K to B C D, and so on (Prop. XVI. Bk. VII.). Also, bj reason of the similar triangles SAB, S G H, we have AB: GH: : SB : SH; and by reason of the similar triangles SBC, SHI, we have SB: SH: : BC: HI; hence, on account of the common ratio SB: S H, AB:GH::BC:HI. For a like reason, we have B C : H I : : C D : I K, and so on. Hence the polygons ABCDE, GHIKL have their angles equal, each to each, and their homolo- gous sides proportional ; hence they are similar. 480. Cor. 1. If two pyrandds have the same altitude, and their bases in the same plane, their sections made by a plane parallel to the plane of their bases are to each other as their bases. Let ABCDE- S, MNO-S be two pyramids, having the same altitude, and their bases in the same plane ; and let G n I K L, P Q R be sections made by a plane parallel 204 ELEMENTS OP GEOMETRY. S A^ : SG^ to the plane of their bases ; then these sections are to each other as the bases ABODE, MNO. For, the two polygons ABCDE,GHIKL be- ing similar, their surfaces are as tlie squares of the homologous sides AB, GH (Prop. XXXI. Bk. IV.). But AB: GH Hence, ABODE: GHIKL For the same reason, M N : P Q R : : SM^ : SP^- But since GHIKL and P Q R are in the same plane, we have also (Prop. XVIII. Bk. VII.), SA: SG: : SM: SP; ll6nC6 ' ABODE: GHIKL: :MNO:PQR; therefore the sections GHIKL, PQR are to each other as the bases ABODE, MNO. 481. Cor. 2. If the bases A B D E, M X are equiv- alent, any sections, GHIKL, PQR, made at equal dis- tances from those bases, are likewise equivalent. Proposition X \'I I. — Theorem. 482, Tlie coiive.c'surfacc of a frustum of a rig-ht pyra- mid is equal to half the sum of the perimeters of its two bases, vtultiplied by its slant height. Let ABODE-L be the frustum of a right pyramid, ftnd MN its giant height; then the convex surface is equal to the sum of the perimeters of tlie two bases A B D E, GHIKL, multiplied by lialf of JI N. BOOK VIII. 205 For the upper base G H I K L is similar to the base ABODE (Prop. XVI.), and ABODE is a regular polygon (Art. 445) ; hence the sides G H, HI, IK, K L, and L G are all equal to each other. The angles GAB, A B H, HBO, &c. are equal (Prop. XV. Oor.), and the edges AG, B H, 01, &c. are also equal (Prop. XVI.) ; therefore the faces A H, B I, OK, &c. are all equal trapezoids (Art. 28), having a common altitude, M N, the slant height of the frustum. But the area of either trapezoid, as A PI, is equal to i-(AB+GH)xMN (Prop. VII. Bk. IV.) ; hence the areas of all the trapezoids, or the convex surface of frustum, are equal to half the sum of the perimeters of the two bases multiplied by the slant height. Proposition XVIII. — Theorem. 483. Triangular pyramids, having equivalent bases and the same altitude, are equivalent. T S S' B B' Let A B - S, A' B' 0' - S' be two triangular pyramids, having equivalent bases, ABC, A' B' 0', situated in the same piano ; and let them have the same altitude, A T ; then these pyramids are equivalent. For, if the two pyramids are not equivalent, let ^/ J5' c — S' be the smaller, and suppose AX to be the 18 206 ELEMENTS OP GEOMETRY. altitude of a prism, which, having ABC for its base, is equal to their difference. B B' Divide the altitude A T into equal parts, each less than AX ; through each point of division pass a plane parallel to the plane of the base, thus forming corresponding sections in the two pyramids, equivalent each to each, namely, DBF to D' E' P', G H I to G' H' I', &c. Upon the triangles ABC, D E P, G H I, &c., taken as bases, construct exterior prisms, having for edges the parts AD, D G, GK, &c. of the edge S A ; in like manner, on the bases D' E' P', G' H' I', &c. in the second pyramid, construct interior prisms, having for edges the correspond- ing parts of S' A'. It is plain that the sum of all the ex- terior prisms of the pyramid A B C-S is greater than this pyramid ; and also that the sum of all the interior prisms of the pyramid A'B'C'-S' is less than this pvramid. Hence, the difference between the sum of all the exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids themselves. Now, beginning with the bases ABC, A' B' C', the second exterior prism, D E P - G, is equivalent to the first interior prism, D'E'P'-A', since they have equal alti- tudes, and their bases, DEP, D'E'P', are equivalent. Por a like reason, the third exterior prism, G HI - K, and the second interior prism, G' H' 1' - D', are equivalent ; and so BOOK vm. 207 on to the last in each series. Hence, all the exterior prisms of tlie pyramid A B C - S, excepting the first prism, ABC-D, have equivalent correspondhig ones in tlie in- terior prisms of the pyramid A B C'-S'. Therefore the prism A B C - D is the difference between the sum of all the exterior prisms of the pyramid ABC-S, and the sum of tlie interior prisms of the pyramid A'B'C'-.S'. But the difference between these two sets of prisms has been proved to be greater than that of the two pyramids, which latter difference we supposed to be equal to tlie prism A B C - X. Hence, the prism ABC-D must be greater than tlie prism A B C -X, which is impossible, since tliey have the same base, ABO, and the altitude of the first is less than AX, the altitude of the second. Hence, the supposed inequality between the two pyramids cannot exist ; therefore the two pyramids ABC-S, A'B' C'-S', having the same altitude and equivalent bases, are them- selves equivalent. Proposition XIX. — Theorem. 484. Every triangular pyramid is a third part of a tri- angular prism having the same base and the same altitude. Let A B C - F be a triangu- lar pyramid, and A B C - D E P a triangular prism of the same base and the same altitude ; then the pyramid is one third of the prism. Cut off the pyramid AB C-P from the prism, by the plane F A C ; there will remain the solid A C D B - P, which may be considered as a quadrangu- lar pyramid, whose vertex is P, and whose bas^e is the parallelogram A C D B. Draw the 208 ELEMENTS OP GEOMETRY. E D diagonal C E, and pass the plane F C E, which will cut the quad- rangular pyramid into two tri- angular ones, ACE-F, EDC-P. These two triangular pyramids have for their common altitude the perpendicular let fall from F on the plane A C D E ; they have equal bases, since the tri- angles A C E, C D B are halves of the same parallelogram ; hence the two pyramids A C E - P, C D E-F are equivalent (Prop. XVIII.). But the pyra- mid CDE-F and the pyramid ABC-F have equal bases, ABC, D E F ; they have also the same altitude, namely, the distance between the parallel planes ABC, D E F ; hence the two pyramids are equivaleut. Now, the pyramid CDE-F has been proved equivalent to ACE-F; hence the three pyramids ABC-F, CDE-F, A C E-F, which compose the whole prism AB C-D EF, are all equivalent; therefore, either pyramid, as ABC-F, is the third part of the prism, which has the same base and the same altitude. 485. Cor. 1. Every ti-iangular prism may be divided into three equivalent triangular pyramids. 486. Cor. 2. The solidity of a triangular pyramid is equal to a third part of the product of its base by its altitude. Proposition XX. — Theorem. 48T. Tlie solidity of every pyramid is equal to the pro- duct of its base by one third of its altitude. Let A B C D E - S be any pyramid, whose base is A B C D E, and altitude S ; then its solidity is equal toABCDEXiSO. BOOK viri. 209 Draw the diagonals AC, AD, and pass the planes SAC, SAD through these diagonals and the vertex S ; the polygonal pyramid ABCDE-S will be divided into several triangular pyra- mids, all having the same altitude, S 0. But each of these pyramids is measured by the product of its base, B A C, C AD, D A E, by a third part of its altitude, S (Prop. XIX. Cor. 2) ; hence, the sum of these triangular pyramids, or the polygonal pyra- mid ABCDE-S, will be measured by the sum of the triangles B A C, CAD, D A E, or the polygon A B C D B, multiplied by one third of S ; hence, every pyramid is measured by the product of its base by one third of its altitude. 488. Cor. 1. Every pyramid is the third part of the prism which has the same base and the same altitude. 489. Cor. 2. Pyramids having the same altitude are to each other as their bases. 490. Cor. 3. Pyramids having the same base, or equiv- alent bases, are to each other as their altitudes. 491. Cor. 4. Pyramids are to each other as the pro- ducts of their bases by their altitudes. 492. Scholium. The solidity of any polyedron may be found by dividing it into pyramids, by passing planes through its vertices. Proposition XXI. — Theorem. 493. A frushim of a pyramid is equivalent to the sum of three pyramids, having for their common altitude the altitude of the frustum.., and whose bases are the two bases of the frustum and a mean proportional betioeen them. First. Let A B C-D B F be the frustum of a pyramid, whose base is a triangle. Pass a plane through the points 18* 210 ELEMENTS OF GEOMETRY. A, B, C ; it cuts oif the triangular pyramid A B C - E, whose altitude is that of the frustum, and whose base, ABC, is the lower base of the frustum. Pass another plane through tlie points D, E, C ; it cuts off the triangular pyramid D E P- C, whose altitude is that of the frus- tum, and whose base, D E F, is the upper base of the frustum. There now remains of the frus- tum the pyramid A C D - E. Draw E G parallel to AD ; join C G and D G. Then, since E G is parallel to AD, it is parallel to the plane A C D (Prop. XI. Bk. VII.) ; and the pyramid A C D - E is equivalent to the pyramid A C D - G, since they have the same base, A C D, and tlieir vertices, E and G, lie in the same straight line par- allel to the common base. But the pyramid A C D - G is the same as the pyramid A G C-D, whose altitude is that of the frustum, and whose base, AGO, as will be proved, is a mean proportional between the bases ABC and DEP. The two triangles A G C, DEE have the angles A and D equal to each other (Prop. XVI. Bk. VII.) ; hence we have (Prop. XXVIII. Bk. IV.), AGC:DEF::AGXAC:DEXDF; but since A G is equal to D E, AGC : DEF: : AC : D F. We have, also (Prop. VI. Cor., Bk. IV.), A B C : A G C : : A B : A G or D E. But the similar triangles ABC, DEF give AB:DE::AC:DF; henco (Prop. X. Bk. 11.), ABC:AGC::AGC:DEF; BOOK VIII. 211 that is, the base A G C is a mean proportional between the bases A B C, D E P of the frustum. Secondly. Let G H I K L - M N P Q be the frustum of a pyramid, whose base is any polygon. Let ABC- S be ^ a triangular pyramid having the same alti- tude, and an equiva- lent base, witli any polygonal pyramid, G H I K L - T ; these pyramids are equiva- lent (Prop. XX. Cor. 3.) The bases of the two pyramids may be regarded as situ- ated in the same plane, in which case the plane MNOPQ produced will form in the triangular pyramid a section, D B F, at the same distance above the common plane of the bases ; and therefore the section D E P will be to the section MNOPQ as the base ABC is to the base GHIKL (Prop. XVI. Cor. 1) ; and since the bases are equivalent, the sections will be so likewise. Hence, the pyramids MNOPQ- T, DEP-S, having the same al- titude and equivalent bases, are equivalent. Por the same reason, the entire pyramids G H I K L - T, A B C — S are equivalent ; conseqiiently, the frustums GHIKL- MNOPQ, ABC-DEP, are equivalent. But the frustum A B C - D E F has been shown to be equivalent to the sum of three pyramids having for their common altitude the altitude of the frustum, and whose bases are the two bases of the frustum, and a mean proportional between thcni. Hence the proposition is true of the frustum of any pyramid. Proposition XXII. — Theorem. 494. Similar pyramids are to each other as the cubes of their Jiomolo^ous edges. 212 ELEMENTS OP GEOMETRY. Let ABC-S and DEF-S be two sim- ilar pyramids ; these pyramids are to each other as the cubes of their homologous edges AB and DE, or BC and EF, &c. For, the two pyra- mids being similar, the homologous polyedral angles at the vertices are equal (Art. 452) ; hence the smaller pyramid may be so applied to the larger, that the polyedral angle S shall be common to both. In that case, the bases ABC, D E F will be parallel ; for, since the homologous faces are similar, the angle S D E is equal to S A B, and S E F to SBC; hence the plane A B C is parallel to the plane D E P (Prop. XVI. Bk. VII.). Then let SO be drawn from the vertex S perpendicular to the plane ABC, and let P be the point where this perpendicular meets the plane D E F. From what has already been shown (Prop. XVI.), we shall have SO: SP: : SA: SD : : AB:DE; and consequently, ^SO: iSP: : AB: DE. But the bases ABC, D E P are similar ; hence (Prop. XXIX. Bk. IV.), A B C : D E F : : AB^ : DE*. Multiplying together the corresponding terms of these two proportions, we have ABCXiS0:DEFx4SP:: AB' : Dl'. Now, A B C X ^ S represents the solidity of the pyra- mid ABC-S, and D E F X i S P that of the pyramid DEP-S (Prop. XX.) ; hence two similar pyramids are to each other as the cubes of their homologous edges. BOOK VIII. 213 Proposition XXIII. — Theorem. 495. There can be no more than five regular poly edrons. For, since regular polyedrons have equal regular poly- gons for their faces, and all their polyedral angles equal, there can be but few regular polyedrons. First. If the faces are equilateral triangles, polyedrons may be formed of them, having each polyedral angle con- tained by three of these triangles, forming a solid bounded by fjur equal equilateral triangles ; or by four, forming a solid bounded by eight equal equilateral triangles ; or by fi'i^e, forming a solid bounded by twenty equal equilateral triangles. No others can be formed witli equilateral triangles. For six of these angles are equal to four right angles, and caiuiot form a polyedral angle (Prop. XX. Bli. VII.). Secondly. If tlie faces are squares, their angles may be arranged by threes, forming a solid bounded by six equal squares. Pour angles of a square are equal to four right angles, and cannot form a polyedral angle. Thirdly. If the faces are regular pentagons, their angles may be arranged by threes, forming a solid bounded by twelve equal and regular pentagons. We can proceed no farther. Three angles of a regular hexagon ai"e equal to four right angles ; three of a hepta- gon are greater. Hence, there can be formed no more than five regular polyedrons, — three with equilateral tri- angles, one with squares, and one with pentagous. 496. Scholium. The regular polyedron bounded by four equilateral triangles is called a tetraedron ; the one bounded by eight is called an octaedron ; the one bound- ed by twenty is called an icosaedron. The regular polye- dron bounded by six equal squares is called a hexaedron, or CUBE ; and the one bounded by twelve equal and regu- lar pentagons is called a dodecaedron. BOOK IX. THE SPHERP:, and its PROPERTn<:S. DEFINITIONS. 497. A Sphere is a solid, or volunie, bounded by a curved surface, all points of which are equally di&ta:it from a point within, called the centre. The sphere may be con- ceived to be formed by the revolution of a semicircle, DAE, about its diameter, D E, which remains fixed. 498. The Eadius of a sphere is a straight line drawn from the centre to any point in surface, as the line C B. " The Diameter, or Axis, of a sphere is a line passing through the centre, and terminated both ways by the sur- face, as the line D E. Hence, all the radii of a sphere are equal ; and all the diameters are equal, and each is double the radius. 499. A Circle, it will be shown, is a section of a sphere. A Grkat Circle of the sphere is a scctiou made by a piano passing through the centre, and having the centre of the sphere for its centre ; as the section A B, whose centre is C. 500. A Small Circle of the sphere is any section made by a plane not passing tlirough Iho f(Mitro. 501. The Pole of a circle of the splicro is a point in the BOOK IX. 215 surface equally distant from every point in the circumfer- ence of the circle. 602. It will be shown (Prop. V.) that every circle, great or small, has two poles. 503. A Plane is tangent to a sphere, when it meets the sphere in but one point, however far it may be produced. 604. A Spherical Angle is the difference in the direc- tion of two arcs of groat cir- cles of the sphere ; as AE D, formed by the arcs E A, D B. It is the same as the angle resulting from passing two planes through those arcs ; as the angle formed on the edge EF, by tlie planes EAP, BDP. 505. A Spherical Triangle is a portion of the surface of a sphere bounded by three arcs of great circles, each arc being less than a semi-circumference ; as A E D. These arcs are named the sides of the triangle ; and the angles which their planes form with each other are the ang-les of the triangle. 506. A spherical triangle takes the nahie of right-angled, isosceles, equilateral, in the same cases as a plane triangle. 507. A Spherical Polygon is a portion of the surface of a sphere bounded by several arcs of great circles. 608. A Lune is a portion of the surface of a sphere com- prehended between semi-cir- cumferences of two great cir- cles; as AIGBDF. 509. A Spherical Wedge, or Ungula, is that portion of a sphere comprehended between 216 ELEMENTS OF GEOMETRY. two great semicircles having a common diameter. 610. A Zone is a portion of tlie surface of a sphere cut off by a plane, or comprehended between two parallel planes ; as EIFK-A, or CGDH- EIFK. 511. A Spherical Segment is a portion of the sphere cut off by a plane, or compre- hended between two parallel planes. 512. The Altitude of a Zone or of a Spherical Seg- ment is the perpendicular distance between the two parallel planes which comprehend the zone or segment. In case the zone or segment is a portion of the sphere cut off, one of the planes is a tangent to the sphere. 513. A Spherical Sector is a solid described by the revolution of a circular sector, in the same manner as the semicircle of which it is a part, by revolving round its diameter, describes a sphere. 514. A Spherical Pyramid is a portion of the sphere comprehended between the planes of a polyedral angle whose vertex is the centre. The base of the pyramid is the spherical polygon inter- cepted by the same planes. Proposition I. — Theorem. 515. Every section of a sphere made by a plane is a circle. Let ABE be a section made by a plane in the sphere whose centre is C. From the centre, C, draw C D per- pendicular to the plane A 15 E ; and draw the lines C A, C 1?, C E, to different points of the curve ABE, which bounds tlio section. BOOK IX. 217 The oblique lines C A, C B, C E arc equal, being radii of the sphere; therefore they are equally distant from the perpendicular, C D (Prop. V. Cor., Bk. VII.). Hence, tlio lines DA, D B, DE, and, in like manner, all the lines drawn from D to tlie boundary of the section, are equal ; and therefore the section whose centre is D. ABE is a circle 516. Cor. 1. If the section passes through the centre of the sphere, its radius will be the radius of the sphere ; hence all great circles are equal. 517. Cor. 2. Two great circles always bisect each other. Eor, since the two circles have the same centre, their com- mon intersection, passing through the centre, must be a common diameter bisecting both circles. 518. Cor. 3. Every great circle divides the sphere and its surface into two equal parts. For if the two hemi- spheres were separated, and afterwards placed on the com- mon base, witli their convexities turned the same way, the two surfaces would exactly coincide. 519. Cor. 4. The centre of a small circle, and that of tlie sphei'e, are in a straiglit line perpendicular to the plane of the small circle. 520. Cor. 5. Small circles are less according to their distance from the centre ; for, the greater the distance CD, the smaller the chord AB, the diameter of the small circle ABE. 521. Cor. 6. The arc of a great circle may be made to pass through any two points on the surface of a spliere ; for the two given points and the centre of the sphere deter- mine the position of a plane. If, however, the two given points be the extremities of a diameter, these two points 19 218 ELEMENTS OP GEOMETRY. and the centre would be in a straight line, and any jium- ber of great circles may be made to pass through the two given points. Pkoposition II. — Theorem. 522. Any one side of a spherical Iricmg-le is less than the sum of the other two. Let AB C be any spherical triangle ; then any side, as A B, is less than the sum of the other two sides, A C, B C. For, draw the radii A, B, C, and the plane angles A OB, AOC, COB will form a triedral angle, 0. The angles AOB, AOC, COB will be measured by A B, AC, B C, the side of the spherical triangle. But each of the three plane angles forming a triedral angle is less than the sum of tlie other two (Prop. XIX. Bk. VII.). Hence, any side of a spherical triangle is less than the sum of the other two. Proposition III. — Theorem. 523. The shortest path from one point to another, on the surface of a sphere, is the arc of the great circle irhich joins the tioo given points. Let ABD be the arc of the great circle which joins the points A and D ; then the line A B D is the shortest path from A to D on tlie surface of the sphere. For, if possible, lot the shortest path on the surface from A to D pass through the point C, out of the arc of the great circle ABD. Draw AC, D C, arcs of great circlos, and take D B equal to D G. Then in the spherical triangle ABDC the side ABD is less than the sum of the sides A C, D C (^Prop. II.) ; and BOOK IX. 219 subtracting the equal D B and D C, there will remain AB less than A C. Now, the shortest path, on the surface, from D to C, whether it is the arc D C, or any other line, is equal to the shortest path from D to B ; for, revolving D C about the diameter which passes through D, the point C may be brouglit into the position of the point B, and the shortest path from D to C be made to coincide with the shortest path from D to B. But, by hypothesis, the shortest path from A to D passes through C ; consequently, the shortest path on the surface from A to C cannot be greater than that from A to B. Now, since AB has been proved to be less than AC, the shortest path from A to C must be greater than that from A to B ; but this has just been shown to be impos- sible. Hence, no point of the shortest path from A to D can lie out of the arc A B D ; consequently, this arc of a great circle is itself the shortest path between its extrem- ities. 524. Cor. The distance between any two points of sur- face, on the surface of a sphere, is measured by the arc of a great circle joining the two points. Peoposition IV. — Theorem. 525. The sum of all the sides of any spherical polygon is less than the circumference of a great circle. Let A B C D E be a spherical polygon ; then the sum of the sides A B, B C, CD, &c. is less than the circumference of a great circle. For, from 0, the centre of the sphere, draw the radii OA, OB, C, &c., and the plane angles AOB, BOC, COB, &c. will form a polyedral angle at 0. Now, the sum of the plane angles which 220 ELEMENTS OF GEOMETRY. form a polyedral angle is less than four right angles (Prop. XX. Bk. VII.). Hence, the sura of the arcs AB, B C, CD, &c., which measure these angles, and bound the spherical polygon, is less than the circumference of a great circle. 626. Cor. The sum of the three sides of a spherical tri- angle is less than the circumference of a great circle, since a triangle is a polygon of three sides. Peoposition V. — Theoeem. 527. The extremities of a diameter of a sphere are the poles of all circles of the sphere whose planes are perperir dicular to that diameter. Let D E be a diameter per- pendicular to AHB, a great circle of a sphere, and also to tlie small circle FIG; then D and E, the extremities of this diameter, are the poles of these two circles. For, since D E is perpendic- ular to the plane AHB, it is perpendicular to all the straight lines, AC, H C, B C, \\ 1/ 246 ELEMENTS OP GEOMETRY. frustum of the cone the frustum of a regular pyramid. The solidity of the frustum of this pyramid is equivalent to the sum of three pyramids, having for their common altitude the altitude of the frustum, and whose bases are the two bases of the frustum, and a mean proportional between them. Conceive now the number of the sides of the polygons to be indefinitely increased ; and tlie bases of the frustum of the pyramid will equal the bases of the frustum of the cone ; and the two frustums will coincide. Hence the frustum of a cone is equivalent to the sum of three cones, having for their common altitude the altitude of the frus- tiun, and whose bases are the two bases of the fiaistum, and a mean proportional between them. Peoposition VII. — Theorem. 587. If any regular svmi-polygon he revolved abovt a line passing through the centre and the vertices of oppo- site angles, the surface described will be equal to the pro- duct of its axis by the circumference of its inscribed circle. Let the regular semi-polygon A B C D E F ^ be revolved about A P as an axis ; then the surface described by the sides A B, EC, CD, &c. will equal the product of A P by the inscribed circle. For, from the vertices B, C, D, E of the semi-polygon, draw BG, CH, DM, EN, perpendicular to the axis A F ; and from the centre, 0, draw 01 perpendicular to one of the sides; also draw IK perpendicular to AF, and B L perpendicular to C II. Now 1 is the radius of the inscribed circle (Prop. II. Bk. A'^I.) ; and the surface described by the revolution of a side, B C, of a regular polygon, is equal to B C multiplied by the circumference, IK (Prop. IV. Cor.). BOOK X. 247 The two triangles OIK, B C L, having their sides per- pendicular to each other, are similar (Prop. XXV. Bk. IV.) ; therefore, BO:BLorGH::OI:IK:: Circ. 01 : Circ. IK, Hence (Prop. I. Bk. II.), BOX Circ. IK = GH X Circ. 01 ; that is, the surface described by B C is equal to the pro- duct of the altitude G H by the circumference of the in- scribed circle. The same may be shown of each of the other sides ; hence, the surface desci-ibed by all the sides taken together is equal to tlie product of the sum of the altitudes AG, GH, HM, MN, NF, by the circ. 01, or to tlie product of tlie axis A P by the circ. 1. Pboposition VIII. — Theorem. 588. The surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Let ABCDEP be a semicircle in A which is inscribed any regular semi-poly- ]3^««=^ gon ; from the centre, 0, draw 1 per- I^ pendicular to one of the sides. C r — .:.~ If now the semicircle and the semi- polyf^on be revolved about the axis AP, Dv- the surface described by the semicircle \ will be the siirface of a sphere (Art. 497), - and that described by the semi-polygon will be equal to the product of its axis, AP, by the cir- cumference, O I (Prop. VII.) ; and the same is true, whatever be the number of sides of the polygon. Conceive the number of sides of the semi-polygon to be made, by continual bisections, indefinitely great ; then its perimeter will coincide with the semi-circumference ABCDEP, and the perpendicular 01 will be equal to the radius A ; hence, the surface of the sphere is equal 2-i'8 ELEMENTS O? GEOMETRY. to the product of the diameter by the circumference of a great circle. 589. Cor. 1. The surface of a sphere is equal to the area of four of its great circles. For the area of a circle is equal to the product of the circumference by lialf the radius, or one fourth of the diameter (Prop. XV. Bk. VI.). 590. Cor. 2. The surface of a zone or segment is equal to the product of its altitude by the circumference of a great circle. For the surface described by the sides B C, C D of the inscribed polygon is equal to the product of the altitude G M by the circumference of the inscribed circle 1. If, now, the number of the sides of an inscribed polygon be indefinitely increased, its perimeter will equal the circle, and B C, C D will coincide with the arc BCD; conse- quently, the surface of the zone described by the revolution of B C D is equal to the product of its altitude by the cir- cumference of a great circle. In like manner, the same may be proved true of a segment, or a zone having but one base. 591. Cor. 3. The surfaces of two zones, or segments upon the same sphere, are to each other as their altitudes ; and any zone or segment is to the surface of the sphere as the altitude of that zone or segment is to the diameter. 592. Cor. 4. If the radius of a sphere is represented by R, and its diameter by D,its surface will bo represented by 4 71 X R', or n X D'. 593. Cor. 5. Hence, the surfaces of spheres are to eacli other as the sqiiares of their radii or diameters. 594. Cor. 6. If the altitude of a zone or seuiuent is BOOK X. 249 represented by H, tlie surface of a zoue or segment will be represented by 2 ji X R X H, or jr X D X H. Proposition IX. — Theorem. 595. The solidity of a sphere is equal to the product of its surface by one third of its radius. For a sphere may be regarded as composed of an indefi- nite number of pyramids, each having for its base a part of the surface of the sphere, and for its vertex the centre of the sphere ; consequently, all these pyramids have the radius of the sphere as their common altitude. Now, the solidity of every pyramid is equal to the pro- duct of its base by one third of its altitude (Prop. XX. Bk. VIII.); hence, the sum of the solidities of these pyra- mids is equal to the product of the sum of their bases by one third of their common altitude. But the sum of their bases is the surface of the sphere, and their common altitude its radius ; consequently, the solidity of the sphere is equal to the product of its surface by one third of its radius. ^ 596. Cor. 1. The solidity of a spherical pyramid or sector is equal to the product of the polygon or zone ivhich forms its base, by one third of the radius. For the polygon or zone forming the base of the spheri- cal pyramid or sector may be regarded as composed of an indefinite number of planes, each serving as a base to a pyramid, having for its vertex the centre of the sphere. 597. Cor. 2. Spherical pyramids, or sectors of the same sphere or of equal spheres, are to each other as their bases. 598. Cor. 3. A spherical pyramid or sector is to the sphere of which it is a part, as its base is to the surface of the sphere. 599. Cor. 4. Hence, spherical sectors upon the same 2r;0 ELEMENTS OP GEOMETRY. sphere are to each other as the altitudes of the zones form- ing their bases (Prop. VIII. Cor. 3) ; and any spherical sector is to the sphere as the altitude of the zone forming its base is to the diameter of the sphere. 600. Cor. 5. If the radius of a sphere is represented by R, its diameter by D, and its surface by S, its solidity will be represented by Sx4R = 4^XR'X^E.= ^"Xll'ori;iXD'. 601. Cor 6. Hence, the solidities of spheres are to each other as the cubes of their radii. 602. Cor. 1. If the altitude of the zone which forms the base of a sector be represented by H, the solidity of the sector will be represented by 2«xB.XHXiR=|"XR'XH. 603. Scholium. The solidity of the spher- ical segment less than a hemisphere, and of one base, formed by the revolution of a por- tion, AB C, of a semicircle about the radius OA, is equivalent to the solidity of the spherical sector formed by A OB, less the solidity of the cone formed by B C. The solidity of the spherical segment greater than a hemisphere, and of one base, formed by tlie revolution of A D E, is equivalent to the solidity of tlie spherical sector formed by AOD, plus the solidity of the cone formed by D E. Tlie solidity of tlie spherical segment of two bases formed by the revolution of C B D E about the axis A F, is equivalent to the solidity of the segment formed by ADE, less the solidity of the segment formed by ABC. Proposition X. — Theorem. 604. The surface of a sphere is equivalent to the convex surface of the circumscribed cylinder, and is two thirds BOOK X. £.31 of the ivhole surface of the cylinder ; also, the solidity of the sphere is two thirds of that of the circumscribed cyl- inder. Let ABPI be a great circle of the sphere ; D E G H the circum- scribed square ; then, if the semi- circle A B P and tlie semi-sqxiare ADEP be revolved about the di- ameter AP, the semicircle will describe a sphere, and the semi- square a cylinder circumscribing the sphere. The convex surface of the cylinder is equal to the cir- cumference of its base multiplied by its altitude (Prop. I.). But the base of the cylinder is equal to the gi-eat circle of the sphere, its diameter E G being equal to the diameter B I, and the altitude D E is equal to the diam- eter A F ; hence, the convex surface of the cylinder is equal to the circumference of the great circle multiplied by its diameter. This measure is the same as that of the surface of the sphere (Prop. VIII.) ; hence, the surface of the sphere is equal to the convex surface of the circum- scribed cylinder. But the surface of the sphere is equal to four great cir- cles of the sphere (Prop. VIII. Cor. 1) ; hence, the convex surface of the cylinder is also equal to four great circles ; and adding the two bases, each equal to a great circle, the whole surface of the circumscribed cylinder is equal to six gi-eat circles of the sphere ; hence, the surface of the sphere is f or f of the whole surface of the circum- scribed sphere. In the next place, since the base of the circumscribed cylinder is equal to a great circle of the sphere, and its altitude to the diameter, the solidity of the cylinder is equal to a great circle multiplied by its diameter (Prop. II.). But the solidity of the sphere is equal to its sur- 252 ELEMENTS OP GEOMETRY. face, or four great circles, multiplied by one third of its radius (Prop. IX.), which is the same as one great circle multiplied by ^ of the radius, or by § of the diameter ; hence, the solidity of the sphere is equal to | of that of the circumscribed cylinder. 605. Cor. 1. Hence the sphere is to the circumscribed cylinder as 2 to 3 ; and their solidities are to each other as their surfaces. 606. Cor. 2. Since a cone is one third of a cylinder of the same base and altitude (Prop. V. Cor. 1), if a cone has the diameter of its base and its altitude each equal to the diameter of a given sphere, the solidities of the cone and sphere are to each other as 1 to 2 ; and the solidities of the cone, sphere, and circumscribing cylinder are to each other, respectively, as 1, 2, and 3. BOOK XI. APPLICATIONS OF GEOMETRY TO THE MENSU- RATION OF PLANE FIGURES. DEFINITIONS. 607. Mensuration op Plane Figures is the process of determining the areas of plane surfaces. 608. Tlie AREA of a figure, or its quantity of surface, is determined by the number of times the given surface con- tains some otlier area, assumed as the unit of measure. 609. The measuring unit assumed for a given surface is called the superficial unit, and is usually a square, tak- ing its name from the linear unit forming its side ; as a square whose side is 1 inch, 1 foot, 1 yard, &c. Some superficial units, however, have no corresponding linear unit ; as the rood, acre, &c. 610. Table of Linear Measures. 12 Indies make 1 Foot. 3 Feet u 1 Yard. 5.^ Yards a 1 Rod or Pole. 40 Rods ii. 1 Furlong. 8 Furlong- S " 1 Mile. Also, TtV^ Inches a 1 Link. 25 Links u 1 Rod or Pole. 100 Links u 1 Chain. 10 Chains a 1 Furlong. 8 Furlong: S " 1 Mile. Note. — For other linear measures, see National Aritlimetic, Art. 133, 134, 136. 22 254 elements op geometry. 611. Table op Surpacb Measures. 144 Square Inches make 1 Square Foot. 9 Square Feet 1 Square Yard. 30^ Square Yards 1 Square Rod or Pole 40 Square Kods 1 Rood. 4 Eoods 1 Acre. 640 Acres 1 Square Mile. 625 Square Links 1 Square Rod. 16 Square Rods 1 Square Chain. 10 Square Chains 1 Acre. Also, 612. Since an acre is equal to 10 chains, or 100,000 links, square chains may be readily reduced to acres by pointingoff one decimal place from the right, and square links by pointing off five decimal places from the right. Problem I. 613. To find the area of a parallelogram. Multiply the base by the altittide, and tlie product will be the area (Prop. V. Bk. IV.). Examples. 1. What is the area of a square, A B C D, whose side is 25 feet ? 25 X 25 = 625 feet, Ans. 2. What is the area of a square field whose side is 35.25 chains ? Ans. 124 A. 1 R. 1 P. 3. How many square feet of boards are required to lay a floor 21 ft. 6 in. square ? 4. Required the area of a squai-o farm, whose side is 3,525 links. 6. What is the area of the rectangle D C A B C D, whose length, A B, is 56 feet, and whose width, AD, is 37 feet ? 56 X 37 = 2,072 feet, Ans. ,V.- XV,V.,, X..11.. ^ g BOOK XI. 255 6. How many square feet in a plank, of a rectangular form, which is 18 feet long and 1 foot 6 inches wide ? 7. How many acres in a rectangular garden, whose sides are 326 and 153 feet ? Ans. 1 A. 23 P. 6^ yd. 8. A rectangular court 68 ft. 3 in. long, by 56 ft. 8 in. broad, is to be paved with stones of a rectangular form, each 2 ft. 3 in. by 10 in. ; how many stones wiU be re- quired ? Ans. 2,062| stones. 9. Required the area of the rhom- boid A B C D, of which the side A B ^ E c is 354 feet, and the perpendicular dis- / tance, E F, between A B and the oppo- / site side C D, is 192 feet. / A F B 354 X 192 = 67,968 feet, Ans. 10. How many square feet in a flower-plat, in the form of a rhombus, whose side is 12 feet, and the perpendicular distance between two opposite sides of wliich is 8 feet ? 11. How many acres in a rhomboidal field, of which the sides are 1,234 and 762 links, and the perpendicular dis- tance between the longer sides of which is 658 links ? Ans. 8 A. 19 P. 4 yd. 6^ ft. Problem II. 614. The area of a square being given, to find the side. Extract the square root of the area. Scholium. This and the two following problems are the converse of Prob. I. Examples. 1. What is the side of a square containing 625 square feet? <\/625 = 25 feet, the side required. 2. The area of a square farm is 124 A. 1 R. 1 P. ; how many links in length is its side ? 3. A certain corn-field in the form of a square contains 256 ELEMENTS OF GEOMETRY. 15 A. 2 R. 20 P. If the corn is planted on the margin, 4 hills to a rod in length, how many hills are there on the margin of the field ? Ans, 800 hills. Problem III. 615. The area of a rectangle and either of its sides being given, to find the other side. Divide the area by the given side, and the quotient will be the other side. Examples. 1. The area of a rectangle is 2,072 feet, and the length of one of the sides is 56 feet ; what is the length of the other side ? 2072 H- 56 = 37 feet, the side required. 2. How long must a rectangular board be, which is 15 inches in width, to contain 11 square feet ? 3. A rectangular piece of land containing 6 acres is 120 rods long ; what is its width ? Ans. 8 rods. 4. The area of a rectangular farm is 266 A. 3 R. 8 P., and the breadth 46 chains ; what is the length ? Ans. 58 chains. Problem IV. 616. The area of a rhomboid or rhombus and the length of the base being given, to find the altitude ; or the area and the altitude being given, to find the base. Divide the area by the length of the base, and the qvo- tient will be the altitude ; or divide the area by the alti- tude, and the quotient will be the length of the base. Examples. 1. The area of a rhomboid is 67,9i)8 square feet, and the length of the side taken as its base 354 feet ; what is the altitude ? 07,968 -r 354 = 192 feet, the altitude required. 2. The area of a piece of land in the form of a rhombus BOOK XI. 257 is 69,452 square feet, and the perpendicular distance be- tween two of its opposite sides is 194 feet ; required the length of one of the equal sides. Ans. 368 ft. 3. On a base 12 feet in length it is required to find the altitude of a rhomboid containing 968 square feet. 4. The area of a rhomboidal-shaped park is lA. 8R. 34 P. 5^ yd. ; and the perpendicular distance between the two shorter sides is 96 yards ; required the length of each of these sides ? Ans. 18 rods. Problem V. 617. The diagonal of a square being given, to find the area. Divide the square of the diagonal by 2, and the quotient will be the area. (Prop. XI. Cor. 4, Bk. IV.) Examples. D ■ C 1. The diagonal, A C, of the square A B C D, is 30 feet ; what is the area ? SO'' = 900 ; 900 -^ 2 = 450 square feet, [the area required. 2. The diagonal of a square field is 45 chains ; how many acres does it contain ? 3. The distance across a public square diagonally is 27 rods ; what is the area of the square ? Problem VI. 618. The area of a square being given, to find the diagonal. Extract the square root of double the area. Scholium. This problem is the converse of the last. Examples. 1. The area of a square is 450 square feet ; what is its diagonal ? 450 X 2 = 900 ; VMO = 30 feet, the diagonal required. 22* 258 ELEMENTS OF GEOMETRY. 2. The area of a public square is 4 A. 2 R. 9 P. ; vhat is the distance across it diagonally ? 3, The area of a squai-e farm is 57.8 acres ; what is the diagonal in chains ? Aus..34 chains. Problem Vn. 619. The sides of a rectangle being given, to cut off a given area by a line parallel to either side. Divide the given area by the side which is to retain its length or width, and the quotient will be the length or width of the part to be cut off. (Prop. IV. Sch., Bk. IT.) Examples. 1. If the sides of a rectangle, ABCD, are 25 and 14 feet, how wide an area, E B C F, to contain 154 square feet, can be cut off by a line parallel to the side AD? 154 -7- 14 = 11 feet, the width required. 2. A farmer has a field 16 rods square, and wishes to cut off from one side a rectangular lot containing exactly one acre ; what must be the width of the lot ? 3. A carpenter sawed off, from the end of a rectangular plank, in a line parallel to its width, 5 square feet. From the remainder he then sawed off, in a line parallel to the length, 8 square feet. Required the dimensions of the part still remaining, provided the original dimensions of the plank were 20 feet by 15 inches. Ans. 16 feet by 9 inches. 4. The length of a certain rectangular lot is 64 rods, and its width 50 rods ; how far from the longer side must a parallel line be drawn to cut off an area of 4 acres, and how far from the shorter side of the remaining portion to cut off 5 acres and 2 roods ? Ilow many acres will re- main after the two portions arc cut off? BOOK XI. 259 Problem VIII. 620. To find the area of a triangle, the base and alti- tude being given. Multiply the base by half the altitude (Prop. ^"1. Bk. IV.). 621. Scholium. The same result can be obtained by multiplying the altitude by half the base, or by multiply- ing together the base and altitude and taking half the product. Examples. 1. Required the area of tlie triangle ^ ABC, whose base, B C, is 210, and alti- tude, AD, is 190 feet. 190 210 X -g- = 19,950 square feet, the [area required. 2. A piece of land is iu the form of a right-angled tri- angle, having the sides about the right angle, the one 254 and the other 136 yards ; required the area in acres. Ans. 3A. 2R. 10 P. 29jyd. 3. Required the number of square feet in a triangular board whose base is 27 inches and altitude 27 feet. 4. What is the area of a triangle whose base is 15.75 chains, and the altitude 10.22 chains ? 5. What is the area of a triangular field whose base is 97 rods, and the perpendicular distance from the base to the opposite angle 40 rods ? Ans. 12 A. 20 P. Problem IX. 622. To find the area of a triangle, the three sides being given. From half the sum of the three sides subtract each 260 ELEMENTS OP GEOMETRY. side ; multiply the half sum and the three remainders to- gether, and the square root of the product will be the area required. For, let A B C be a triangle whose three c sides, AB, BC, AC, are given, but not the altitude C D, and let the side B C be represented by a, AC by b, and A B by c. Now, since A is an acute angle of the triangle ABC, we have (Prop. XII. Bk. IV.), fls = &» :f- c= _ 2 c X AD, or AD = '^ + f ~°^ Hence, in the right-angled triangle A D C, we have (Prop. XI. Cor. 1, Bk. IV.), CD' = ft2 — C^' + g^-"')^ ^ 4 6aca-(&a-t-c3-aa)3 4 c2 4 e8 » and, by extracting the square root. Q-Q ^ j^ 4 tji c^ — (bfi -\- c" — a^ 2c But the area of the triangle A B C is equivalent to the product of c by half of C D (Prob. VIII.) ; hence A B C = i V 4 6« C-i — (68 -I- cS — 0^)3. The expression 4 6" c» — (b^-\- c» — a')', being the difference of two squares, can be decomposed into (2 Z> c + 6" + c» — aO X (2 & c — 6» — c" + a")- Now, the first of these factors may be transformed to Q) + cy — a", and consequently may be resolved into (Jb -\- c -\- a) X (b -\- c — a) ; and the second is the same thing as a' — (b — r)^ which is equal to (a-\-b — c) X (o — 6 + r). We have then A b^ c'' — (b^ -\- c^ — ay = (a + 6-l-c) X (6 + c — fl) X (a + c — ft) X (a + 6 — c). BOOK XI. 261 Let S represent half the sum of the three sides of the triangle ; tlien a + i»+c = 2S; Z/ + c— a = 2(S — «); a+c — 6 = 2 (S — i); « + 6 — c = 2 (S — c); hence A B C = J- V 16 S (S — a) X (S — J) X (S — e), which, ))cing reduced, gives as the area of the triangle, as given above, V S (S — a) X "(S — i) X (S — c). Examples. C 1. What is the area of a triangle, AB C, ■whose sides, AB, B C, CA, are 40, 30, and 50 feet? A B 30 4- 40 + 50 -^ 2 = 60, half the sum of the three sides. 60 — 30 = 30, first remainder. 60 — 40 = 20, second remainder. 60 — 50 = 10, third remainder. 60 X 30 X 20 X 10 = 180,000; Vl807)00 = 424.26 square feet, the area required. 2. How many square feet in a triangular floor, whose sides are 15, 16, and 21 feet ? 3. Required the area of a triangular field whose sides are 834, 658, and 423 links. Ans. lA. IR. 20 P. 4 yd. 1.6 ft. 4. Required the area of an equilateral triangle, of which each side is 15 yards. 5. What is the area of a garden in the form of a paral- lelogram, whose sides are 432 and 263 feet, and a diagonal 342 feet ? Ans. 2 A. 10 P. 11.46 yd. 6. Required the area of an isosceles triangle, whose base is 25 and each of its equal sides 40 rods. 7. What is tlie area of a rhomboidal field, whose sides are 57 and 83 rods, and the diagonal 127 rods ? Ans. 22 A. 3R. 21 P. 26 yd. 5 ft. 262 ELEMENTS OP GEOMETET. Problem X. 623. Any two sides of a eight-angled triangle being given, to find the third side. To the square of the base add the square of the perperv- dicular ; and the square root of the sum will give the hy- pothenuse (Prop. XI. Bk. IV.). From the square of the hypothenuse subtract the square of the given side, and the sqziare root of the difference will be the side required (Prop. XI. Cor. 1, Bk. IV.). Examples. 1. The base, AB, of the triangle ABC is 48 feet, and the perpendicular, B C, 36 feet ; what is the hypothenuse ? 48^ -f. 36» = 3600 ; V 3600 = 60 feet, [the hypothenuse required. /- i 2. The hypothenuse of a triangle is 53 feet, and the per- pendicular 28 feet ; what is the base ? 8. Two ships sail from the same port, one due west 50 miles, and the other due south 120 miles ; how far are they apart ? Ans. 130 miles. 4. A rectangular common is 25 rods long and 20 rods wide ; what is the distance across it diagonally ? 5. If a house is 40 feet long and 25 feet wide, with a pyramidal-shaped roof 10 feet in height, how long is a rafter which reaches from the vertex of the roof to a cor- ner of the building ? 6. There is a park in the form of a square containing 10 acres ; how many rods less is the distance from the centre to each corner, than tlie length of the side of the square ? Ans. 11.716 rods. Problem XI. 024. The sum of the hypothenuse and perpendicular BOOK XI. 263 and the base of a right-angled triangle being given, to find the hypothenuse and the perpendicular. To the square of the sum add the square of the base, and divide the amount by twice the sum of the hypothe- nvse and perpendicular, and the quotient icill be the hy- pothenuse. From the sum of the hypothenuse and perpendicular subtract the hypothenuse, and the remainder will be the perpendicular. 625. Scholium. This problem may be regarded as equiv- alent to the Slim of two mimbers and the difference of their squares being given, to find the numbers (National Arithmetic, Art. 553). Note. — The learner should be required to give a geometrical demon- stration of the problem, as an exercise in the application of principles. Examples. 1. The sum of the hypothenuse and the perpendicular of a right-angled triangle is 160 feet, and the base 80 feet ; required the hypothenuse and the perpendicular. Ans. Hypothenuse, 100 ft. ; perpendicular, 60 ft. 1602 ^ SO'' = 32,000 ; 32,000 -f- (160 X 2) = 100 ; 160 — 100 = 60. 2. Two ships leave the same anchorage ; the one, sailing due north, enters a port 50 miles from the place of depart- xire, and the other, sailing due east, also enters a port, but by sailing thence in a direct course enters the port of the first ; now, allowing that the second passed over, in all, 90 miles, how far apart are the two ports ? 8. A tree 100 feet high, standing perpendicularly on a horizontal plane, was broken by the wind, so that, as it fell, while the pai-t broken off remained in contact with the upright portion, the top reached the ground 40 feet from the foot of the tree ; what is the length of each part ? Ans. The part broken off, 58 ft. ; the upright, 42 ft. 264 ELEMENTS OF GEOMETRY. Problem XII. 626. The area and the base of a triangle being given, to find the altitude ; or the area and altitude being given, to find tlie base. Divide double the area by the base, and the quotient will be the altitude ; or divide double the area by the alti- tude, and the quotient will be the base. 627. Scholium. This problem is the converse of Prob. VIII. Examples. 1. The area of a triangle is 1300 square feet, and the base 65 feet ; what is tiie altitude ? 1300 X 2 = 2600 ; 2600 4- 65 = 40 ft., altitude required. 2. The area of a right-angled triangle is 17,272 yards, of which one of the sides about the right angle is 136 yards ; required the other perpendicular side. 3. Tlie area of a triangle is 46.25 chains, and the alti- tude 5.2 chains ; what is the base ? 4. A triangular field contains 30 A. 3 R. 27 P. ; one of its sides is 97 rods ; required the perpendicular distsuice from the opposite angle to that side. Ans. 102 rods. Problem XIII. 628. To find the area of a trapezoid. Multiply half the sum of its parallel sides by its altitude (Prop. VII. Bk. IV.). Examples. DEC 1. What is the area of the trapezoid / A B C D, whose parallel sides, A B, / D C, are 32 and 24 feet, and the alti- / tude, E F, 20 foot ? A F B ;12 -f 24 = 5G ; 66 -^ 2 = 28 ; 28 X 20 = 500 sq. ft., [tlie area required. BOOK XI. 2G5 2. How many square foot in a board in the form of a trapezoid, wliosc width at one end is 2 feet 3 inches, and at tlie other 1 foot 6 inches, the length being 16 feet ? 3. Required tlie area of a garden in the form of a trape- zoid, wliose parallel sides are 786 and 473 links, and the perpendicular distance between them 986 links. Ans. 6 A. 33 P. 3 yd. 4. How many acres in a quadrilateral field, having two parallel sides 83 and 101 rods in length, and which are distant from each other 60 rods ? Problem XIV. 629. To find the area of a regular polygon, the pe- rimeter and apothegm being given. f Multiply the perimeter by half the apothegm, and the product will be the area (Prop. VIII. Bk. VI.). 630. Scholium. This is in effect resolving tlie polygon into as many equal triangles as it has sides, by drawing lines from tlie centre to all the angles, then finding their areas, and taking their sum. Examples. 1. Required the area of a regu- lar hexagon, A B C D E F, whose sides, AB, B C, &c. are each 15 yards, and the apothegm, M, 13 yards. 15 X 6 = 90 ; 90 4- J/-= 585 yd., [the area i-equired. 2. What is the area of a regular pentagon, whose sides are each 25 feet, and the perpendicular from the centre to a side 17.205 feet ? 3. A park is laid out in tlie form of a regular heptagon, whose sides are each 19.263 cliains ; and the perpendicular 23 2G6 ELEMENTS OF GEOMETRY. distance from the centre to each of the sides is 20 chains. How many acres does it contain ? Ans. 134A. 3R. 14P. Problem XV. 631. To find the area of a regular polygon, its side or perimeter being given. Multiply the square of the side of the polygon by the area of a similar polygon ivhose side is unity or 1 (Prop. XXXI. Bk. IV.). 632. A Table op Regular Polygons whose Side is 1. NAMES. AREAS. KAMES. AREAS. Triangle, Square, Pentagon, Hexagon, Heptagon, 0.4330127 1.0000000 1.7204774 2.5980762 3.6339124 Octagon, Nonagon, Decagon, Undecagon, Dodecagon, 4.8284271 6.1818242 7.6942088 9.3656399 11.1961524 The apothegm of any regular polygon wliose side is 1 being ascertained, its area is computed readily, by Prob. XIV. Examples. 1. Required the area of an equilateral triangle, whose side is 100 feet. 100' = 10,000 ; 10,000 X 0.43301-27 = 4330.127 square [feet, the area required. 2. What is the area of a regular pentagon, whose side is 37 yards ? H. How many acres in a field in the form of a regular undecagon, whoso side is 27 yards ? Alls. lA. 1 R. 2") P. 21 vd. 2.7 ft. BOOK XI. 267 4. What is the area of an octagonal floor, whose side is 15 ft. 6 in. ? 6. How many acres in a regular nonagon, whose perim- eter is 228(3 feet ? Ans. 9 A. 24 P. 28 yd. Problem XVI. 633. To find the side of any regular polygon, its area being given. Divide the given area by the area of a similar polygon whose side is 1, and the square root of the quotient vnll be the side required. 634. Scholium. This problem is the converse of Prob. XY. ExA:\rPLKS. 1 . The area of an equilateral triangle is 4330.127 square feet ; what is its side ? 4830.127 H- .4330127 = 10,000 ; V 10,000 = 100 feet, [the side required. 2. The area of a regular hexagon is 1039.23 feet ; what is its side ? 3. The area of a regular decagon is 7 P. 18 yd. 5 ft. 128.55 in. ; what is its side ? Ans. 10 ft. 5 in. Problem XVII. 635. To find the area of an irregular polygon. Divide the polygon into triangles, or triangles and trapezoids, and find the areas of each of them separately ; the. sum of these areas will be the area required. 636. Scholium. When the irregular polygon is a quad- rilateral, the area may be found by multiplying together the diagonal and half the sum of the perpendiculars drawn from it to the opposite angles. 2G8 ELEittKNTS OP GEOMETRY. Examples. 1. Required the area of the irregular pentagon A B C D E, of which the diag- onal A is 20 feet, and A D 36 feet ; and the perpendicular distance from the angle B to A is 8 feet, from C to AD 12 feet, and from E to AD 6 feet. 20 X I = 80 ; 36 X ¥ = 216 ; 36 X | = 108 ; 80 + 216 + 108 = 504 sq. ft., the area required. 2. What is the area of a trapezium, whose diagonal is 42 feet, and the two perpendiculars from the diagonal to the opposite angles are 16 and 18 feet ? 3. In an irregular hexagon, A B C D E F, are given the sides A B 636, B C 498, D 620, D E 580, EP 308, and AF 492 links, and the diagonals AC 918, CE 1048, and A E 652 links ; i-equired the area. Ans. 6A. 2R. 9P. 23 yd. 8.4 ft. 4. In measuring along one side, AB, of a quadrangular field, A B C D, tliat side and the perpendiculars let fall on it from two opposite corners measured as follows : A B 1110, AE 110, AF 745, DE 352, CF 595 links. Wliat is the area of the field ? Aus. 4 A. 1 R. 5 P. 24 yd. 5. In a four-sided rectilineal field, A B C D, on account of obstructions, there could be taken only the following measures : the two sides B C 2Go and AD 220 yards, tlic diagonal A C 378, and the two distances of the perpendic- \ilars from the ends of the diagonal, namely, AE 100, and C F 70 yards. Required the area in acres. Problem XVIII. 637. To find the circumference of a circle, when the diameter is given, or the diameter wlien the circumfcrcuce is given. MuUijiIi/ the diamc/rr by 3.141(), and the product trill be the circumference ; or, divide the circumference by BOOK XI. 269 3.1416, and the quotient ivill be the diameter (Prop. XY. Cor. 3, Bk. VI.). 638. Scholium. The diameter may also be found by multiplying the circumference by .31831, the reciprocal of 3.1416. E Examples. 1. The diameter, AB, of the cir- cle A E B P is 100 feet ; what is its circumference ? 100 X 3.1416 = 314.1(3 feet, the [circumference required. F 2. Requii-ed the circumference of a circle whose diam- eter is G2S links. Ans. Ifur. 38 rd. 5 yd. 1.56 in. 3. If the diameter of the earth is 7912 miles, what is its circumference ? 4. Kequired the diameter of a circular pond whose cir- cumference is 928 rods. Ans. 7 fur. 15 rd. 2 yd. 5.55 in. 5. The circumference of a circular garden is 1043 feet ; what is its radius ? Ans. 10 rd. 1 ft. Problem XIX. 639. To find the length of an arc of a circle containing any number of degrees, the radius or diameter being given. Multiply the number of degrees in the given arc bij 0.01745, and the product by the radius of the circle. For, when the diameter of a circle is 1, the circumfer- ence is 3.1416 (Prop. XV. Sch. 1, Bk. VI.) ; hence, when the radius is 1, the circumference is 6.2832 ; which, divided by 360, the number of degrees into which every circle is supposed to be divided, gives 0.01745, the length of the arc of 1 degree, when the radius is 1. 640. Scholium. Each of the 360 degrees of a circle, 23* 270 ELEMENTS OP GEOMETRY. marked thus, 360% is divided into 60 minutes, marked thus, 60', and each minute into 60 seconds, marked thus, 60" (National Arithmetic, Art. 143). Examples. 1. What is the length of an arc, AD, containing 60° 30' on the cir- cumference of a circle whose radius, A C, is 100 feet ? 60° 30' = 60.5° ; 60.5 X 0.01745 = 1.055725 ; 1.055725 x 100 = 105.5725 ft., arc required. iLN-~___^i 2. Required the length of an arc of 31° 1.?, the radius being 12 yards. 3. Required the length of an arc of 12° 10', the diam- eter being 20 feet. Ans. 2.12C1 feet. 4. Wliat is the length of an arc of 57° 17' 44j", the radius being 25 feet ? Ans. 25 feet. Problem XX. 641. To find the area of a circle. Multiply the circumference by half the radius (Prop. XV. Bk. VI.) ; or, multiply the square of the radius by 3.1416 (Prop. XV. Cor. 2, Bk VI.). 642. Scholium. Multiplying the circumference by half the radius is the same as multiplying the circumference and diameter together, and taking one fourth of the pro- duct. Now, denoting the circumference by c, and the diameter by d, since c = 3.1416 X d (Prob. XVIII.), we have (d X 3.1416 X rf) -r 4 = c?" X 0.7854 = the area of a circle. Again, since d=c-T- 3.1416 (Prob. XVIII.), wo have c -!- 3.1416 X c -=- 4 = c" -=- 12.5664, which is, by taking the reciprocal of 12.5664, equal to c" X 0.07958 = the area of the circle. Honce the area of the circle may also bo found by multiplying the square of the diam- BOOK XI. 271 eter by 0.7854 ; or by multiplying the square of the cir- cumference by 0.07958. Examples. 1. The circumference of a circle is 314.16 feet, and its radius 60 feet ; what is its area? 314.16 X V- = 7854 feet, the area required. 2. If tlie circumference of a circle is 355 feet, and its diameter 113 feet, what is the area ? 3. Wiiat is the area of a circular garden, whose radius is 281^ links ? Ans. 2 A. 1 R. 38 rd. 9 yd. 5 ft. 4. A horse is tethered in a meadow by a cord 39.25075 yards long ; over how much ground can he graze ? 5. Requii'cd the area of a semicircle, the diameter of the whole circle being 751 feet. Ans. 5 A. 13 P. 16 yd. Pboblem XXI. 643. To find the diametek or ciecumfeeence, the area being given. Divide the area by 0.7854, and the square root of the quotient will be the diameter; or, divide the area by 0.07958, and the square root of the quotient will be the circumference . 644. Scholium. This problem is the converse of Prob. XX. EXAMPLKS. 1. The area of a circle is 314.16 feet ; what is the diameter ? 814.16 -^ 0.7854 = 400 ; VIOO = 20 feet, the diameter [required. 2. What must be the length of a cord to be used as a radiiis in describing a circle which shall contain exactly 1 acre ? 3. The area of a circular pond is 6 A. 1 R. 27 P. 18.2 yd. ; what is the circumference ? Ans. 625 yd. 272 ELEMENTS OF GEOMETRY. 4. The area of a cii'cle is 7856 feet ; what is the cir- cumference ? 5. The length of a rectangular garden is 32, and its width 18 rods ; required the diameter of a circular garden having the same area. Ans. 27 rd. 1 ft. 4 in. Problem XXII. 645. To find the area of a sector of a circle. Multiply the arc of the sector by half of its radius (Prop. XV. Cor. 1, Bk. VI.) ; or. As 360° are to the degrees in the arc of the sector, so is the area of the circle to the area of the sector. Examples. 1. Required the area of a sector, D E, whose arc is 80 feet, and its radiiis, E, 70 feet. 80 X ^i- = 2800 square feet, the area [required. 2. Required the area of a sector, of which the arc is 90 and the radius 112 yards. 3. Required the area of a sector, of which the angle is 137° 20', and the radius 456 links. Ans. 2 A. IR. 38 P. 21.92 yd. Problem XXIII. 646. To find the area of a segment of a circle. Find the area of the sector having- the same arc tcith the segment, and also the area of the triangle formed by the chord of the segment and the radii of the sector. Then, if the segment is less than a semicircle, take the difference of these areas; but if greater, take their sum. 647. Scholium. When the height of tlio seginout and BOOK XI. 273 Hie diameter of the circle are given, the area may be readily found by means of a table of segments, hij divid- ing the height by the diameter, and looking in the table for the quotient in the column of heights, and taking out, in the next column on the right hand, the corresponding area ; lohich, multiplied by the square of the diameter, ivill give the area required. When the quotient cannot be exactly found in the table, proportions may be instituted so as to find the area be- tween the next higher and the next lower, in the same ratio that the given height varies from the next higher and lower heights. 648. Table op Segments. 1 1 .01 Area. i a K Seg. Area. 5 Seg. Area. .31 Seg. Area. aj .41 Seg. Area. .00133 .11 .04701 .21 .11990 .20738 .30319 .02 .00375 .12 .05339 .22 .12811 .32 .21667 .42 .31304 .03 .00687 .13 .06000 .23 .13646 .33 .22603 .43 .32293 .04 .01054 .14 .06683 .24 .14494 .34 .23547 .44 .33284 .05 .01468 .15 .07387 .25 .15354 .35 .24498 .45 .34278 .06 .01924 .16 .08111 .26 .16226 .36 .25455 .46 .35274 .07 .02417 .17 .08853 .27 .17109 .37 .26418 .47 .36272 .08 .02944 .18 .09613 .28 .18002 .38 .27386 .48 .37270 .09 .03502 .19 .10390 .29 .18905 .39 .28359 .49 .38270 .10 .04088 .20 .11182 .30 .19817 .40 .29337 1 .50 .39270 The segments in the table are those of a circle whose diameter is 1, and the first column contains the corre- sponding heights divided by the diameter. The method of calculating the areas of segments from the elements in the table depends upon the principle that similar plane figures are to each other as the squares of their like linear dimensions. Examples. 1. What is the area of the segment ABE, its arc AEB being 73.74°, its chord AB being 12' feet, and 274 ELEMENTS OP GEOMETRY. the radius, CB, of tlio circle 10 feet? 0.7854 X 20^ = 314.1G, area of circle; then SCO" : 73.74° : : 814.1G : G4.C.:04, area of sector A E B C ; and, by Prob- lem IX., 48 is the area of the trian- gle ABC; 64.3504 — 48 = 10.3504 feet, the area required. 2. Required the area of a segment whose lieiglit is 18, and the diameter of the circle 50 feet. 18 -H 50 = .36 ; to which the corresponding area in the ta- ble is .25455 ; .25455 X 50= = 636.375, area required. 3. Required the area of a segment whose arc is 100°, chord 153.208 feet, and the diameter of the circle 200 feet. 4. What is the area of a segment whose height is 4 feet, and the radius 51 feet ? Ans. 106 feet. 5. Required the area of a segment, the arc being 160°, chord 196.9616 feet,and the radius of the circle 100 feet. Problem XXIV. 649. To find the area of a circular zone, or the space included between two parallel chords and their intercepted arcs. From the area of the vjholc circle subtract the areas of the segments on the sides of the zone. Examples. 1. What is the area of a zone whoso chords arc each 12 feet, subtending each an arc of 73.74°, when the radius of the circle is 10 feet ? Area of the whole circle by Prob, XX. = 314.10 ; ai-ca of each segment by Prob. XXllI. = 10.3504; 10.3504 X 2 = 3-2.7008 =: area of both segments ; 314.16 — 32.7003-?= 281.4592, the aroa required. BOOK XI. Zib 2. What is the area of a circular zone whose longer cliord is 20 yards, subtending an arc of 60°, and the shorter chord 14.66 yards, subtending an arc of 43°, the diameter of the circle being 40 yards ? 3. A circle whose diameter is 20 feet is divided into three parts by two parallel chords ; one of the segments cut off is 8 feet in height, and the other 6 feet; what is llie area of the circular zone ? Ans. 117.544 ft. Problem XXV. 650. To find the area of a crescent. Find the difference of the areas of the two segments formed by the arcs of the crescent and its chord. Examples. 1. The arcs A C B, A E B, of circles having tlie same radius, 50 rods, intersecting, form the crescent A G B E ; the height, D C, of the segment A B is 60 rods, and the height, D E, of the segment ABE is 40 rods ; what is the area of the crescent ? The area of the segment A C B, by Prob. XXIII., is 4920.3 rods, and that of the segment A B E is 2933.7 rods ; 4920.3 — 2933.7 = 1986.6 rods, the area of the crescent. 2. If the arc of a circle whose diameter is 24 yards in- tersects a circle whose diameter is 20 yards, forming a crescent, so that the height of the segment of the first cir- cle is 6.072 yards, and that of the segment of the second circle is 8 yards, what is the area of the crescent ? Problem XXYI. 051. To find the area of a circular ring, or the space included between two concenti-ic circles. Find the areas of the two circles separately (Prob. XX.), and take the difference of these areas; or sub- 276 ELEMENTS OF GEOMETKY. tract the square of the less diameter from the square of the greater, and viultiply their difference by 0.7854 (Prob. XX. Sch.). Examples. 1. Required tlie area of the ring formed by two circles ■wliose diameters are 30 and 60 feet. 50^ — 30* = 1400 ; 1400 X 0.7854 = 1099.56 sq. feet, [tlie area of the ring. 2. Wliat is the area of a ring formed by two cii-clcs wliqse radii are 36 and 24 feet ? 3. A circular park, 256 yards in diameter, has a car- riage-way running around it 29 feet wide ; what is tlie area of tlie carriage-way ? Ans. lA. 2R. 26P. 21.5 yd. Problem XXVH. 652. The diameter or circumference of a circle being given, to find the side of an equivalent square. Multiply the diameter by 0.8862, or the circumference by 0.2821 ; the product in either case toill be the side of an equivalent square. For, since 0.7854 is the area of a circle whose diameter is 1 (Prob. XX. Sch.), the square root of 0.7854, which is 0.8862, is the side of a square which is equivalent to a circle whose diameter is 1. Now when the circiimfercnce is 1, the side of an equivalent square must liavo the saiiio ratio to 0.8862 as the diameter 1 has to its circuuifcroiicc 3.1416 (Prop. XV. Cor. 4, Bk. VI.) ; and 0.88G2 ^ 3.1416 gives 0.2821 as the side of the equivalent square when the circumference is 1. Examples. 1. The diameter of a circle is 120 feet ; what is the side of an equivalent square ? 120 X 0.88G2 = 106.844 feet, the side required. BOOK XI. 277 2. The circumference of a circle is 100 yards ; what is the side of an equivalent square ? Ans. 28.21 yd. 3. There is a circular floor 30 feet in diameter ; what is the side of a square floor containing the same area ? 4. If 500 feet is the circumference of a circular island, what is the side of a square of eqiial area ? Ans. 141.05 ft. Problem XXYIII. 653. The diameter or circumference of a circle being given, to find the side of the inscribed square. Multiply the diameter hy 0.7071, or the circumference by 0.2251 ; the product in cither case will be the side of the inscribed square. For 0.7071 is the side of the inscribed square when the diameter of the circumscribed circle is 1, since the side of the inscribed square is to the radius of the circle as the square root of 2 to 1 (Prop. IV. Cor., Bk. VI.) ; conse- quently, the side is to the diameter, or twice the radius, as half the square root of 2 is to 1, and half the sqiiare root of 2 is 0.7071, approximately. Now, the ratio of the diameter of a circle to the side of its inscribed square being as 1 to 0.7071, and the ratio of the circumference of a circle to its diameter as 3.1416 to 1, the ratio of the inscribed square is to the circumference of the circle as 0.7071 to 3.1416 ; and 0.7071 -^ 3.1416 gives 0.2251 as the side of the inscribed square when the circumference is 1. Examples. 1. The diameter, AC, of a circle is 110 feet ; what is the side, A B, of the inscribed square ? 110 X 0.7071 = 77.781 feet, the side [required. 24 278 ELEMENTS OP GEOMETRY. 2. The circumference of a circle is 300 feet ; wliat is tlie side of the inscribed square ? Aiis. 67.63 ft. 8. A log is 36 inches in diameter ; of how many inches square caii a stick be hewn from it ? 4. There is a circular field 1000 rods in circuit ; what is the side of the largest square that can be described in it ? Ans. 225.10 rods. Problem XXIX. 654. The diameter or circumference of a circle being given, to find the side of an inscribed equilateral tri- angle. Multiply the diameter by 0.8660, or the circumference hy 0.2757 ; the product in either case will be the side of the inscribed equilateral triangle. For 0.8660 is the side of the inscribed equilateral trian- gle when the diameter of the circumscribed circle is 1, since the side of the inscribed equilateral triangle is to the radius of the circle as the square root of 3 is to 1 (Prop. V. Cor. 3, Bk. VI.) ; consequently, the side is to the diam- eter, or twice the radius, as half the square root of 3 is to 1, and half the square root of 3 is 0.8660, approximately. Also, since the ratio of the circumference of a circle to its diameter is as 3.1416 to 1, tlie side of the inscribed equilateral triangle, wlien the circiimference is 1, equals 0.8660 H- 3.1416, or 0.2757. Examples. 1. Required the side of an equilateral triangle that may be inscribed in a circle 101 feet in diameter. 101 X 0.8660 = 87.4660 feet, the side rcqiiircd. 2. Required the side of an equilateral triangle that may be inscribed in a circle 80 rods in circumference. Ans. 22.05 rods. 3. Required the side of the lai'gest C(]uilatoral triangular beam that can be liown from a piece of round timber 36 ijiches in diameter. BOOK XI. 279 4. Required the side of an equilateral triangle that can be inscribed in a circle 251.33 feet in circumference. 6. How much less is the ai'ca of an equilateral triangle that can be inscribed in a circle 100 feet in diameter, than the area of the circle itself ? Ans. 4606.4 sq. ft. The Ellipse. 655. An Ellipse is a plane figure bounded by a curve, from any point of which the sum of the distances to two fixed points is equal to a straight line drawn through those two points, and terminated both ways by the curve. Thus A D B C is an ellipse. The two fixed points G and H are called ^ — i — -.^ the/oci. The longest diameter, A B, /^q. ■ jj n, of the ellipse is called its major or r j^ transverse axis, and its shortest di- \^_^__^^^ ameter, CD, is called its minor or ^ conjugate axis. 656. The aeea of aa ellipse is a mean proportional between the areas of two circles whose diameters are the two axes of the ellipse. This, however, can only be well demonstrated by means of Analytical Geometry, a branch of the mathematics with which the learner here is not supposed to be acquainted. Problem XXX. 657. To find the area of an ellipse, the major and minor axes being given. Multiply the axes tog-ether, and their product by 0.7854, and the result will be the area. For A B^ X 0.7854 expresses the area of a circle whose diameter is A B, and C D^ X 0.7854 expresses the area of a circle whose diameter is C D ; and the product of these two areas is equal to A B^ X C D^ X 0.7854% which is 280 ELEMENTS OP GEOMETRY. equal to the square of A B X C D X 0.7854 ; hence, A B X C D X 0.7854 is a mean proportional between the areas of the two circles whose diameters are A B and C D (Prop. IV. Bk. II.) ; consequently it measures the area of an ellipse whose axes are AB and CD (Art. 656). Examples. 1. Required the area of an ellipse, of which the major axis is 60 feet, and the minor axis 40 feet. 60 X 40 X 0.7854 = 1884.96 sq. ft., the area required. 2. What is the area of an ellipse whose axes are 75 and 35 feet ? 3. Required the area of an ellipse whose axes are 526 and 354 inches. Ans. 112 yd. 7 ft. 84.62 in. 4. How many acres in an elliptical pond whose semi- axes are 436 and 254 feet ? Ans. 7A. 3R. 37 P. 27 yd. 7 ft. BOOK XII. APPLICATIONS OF GEOMETRY TO THE MENSU- RATION OF SOLIDS. DEFINITIONS. 658. Mensuration of Solids, or Volumes, is the pro- cess of determining their contents. The SUPERFICIAL contents of a body is its quantity of surface. The SOLID contents of a body is its measured magni- tude, volume, or solidity. 659. Tlie UNIT op volume, or solidity, is a cube, whose faces are each a superficial unit of the surface of the body, and whose edges are each a linear unit of its linear di- mensions. 660. Table of Solid Measures. 1728 Cubic Inches make 1 Cubic Foot 27 a Feet u 1 " Yard. 4492i a Feet « 1 " Rod. 32,768,000 li Rods t( 1 " Mile. Also, 2C1 a Inches (t 1 Liqiiid Gallon. 2G8t a Inches a 1 Dry Gallon. 2150x^^0 . (( Inches a 1 Bushel. 128 a Feet « 1 Cord. Problem I. 6G1. To find the surface of a right prism. BluUiphj the perimeter of the base by the altitude, and the product will be the convex surface (Prop. I. Bk. 24* 282 ELEMENTS OP GEOMETRY. VIII.). To this add the areas of the two bases, and the result will be the entire surface. Examples. 1. Required the entire surface of a pentangular prism, having each side of its base, ABODE, equal to 2 feet, and its altitude, A P, equal to 5 feet. 2 X 5 = 10 ; 10 X 6 = 50 square feet, [the surface required. 2. The altitude of a hexangular prism is 12 feet, two of its faces are each 2 feet wide, three are each 2^ feet wide, and the remaining face is 9 inches wide ; what is the convex surface of the prism ? 3. Required the entire surface of a cube, the length of each edge being 25 feet. 4. Required, in square yards, the wall siirface of a rec- tangular room, whose height is 20 feet, width 30 feet, and length 50 feet. Ans. 355f sq. jd. Problem II. 662. To find the solidity of a prism. Multiply the area of its base by its altitude, and the product will be its solidity (Prop. XIII. Bk. A'^III.). Examples. 1. Required the solidity of a pentangular prism, having each side of its base equal to 2 feet, and its altitude equal to 5 feet. 2' X 1.72048 = 6.88192 ; 6.88102 X 5 = 34.40960 cubic [feet, the solidity required. 2. Required the solidity of a triangular prism, whose length is 10 feet, and the throe sides of whose base are 3, 4, and 5 feet. Ans. 60. 3. A slab of marble is 8 foot long, 3 feet wide, and 6 inches thick ; roquh-od its solidity. BOOK xii. 283 4. There is a cistern in the form of a cube, whose edge is 10 feet ; what is its capacity in liquid gallons ? Ans. 7480.519 gallons. 5. Required the solid contents of a quadrilateral prism, the length being 19 foot, the sides of the base 43, 54, 02, and 38, and the diagonal between the first and second sides, 70 inches. Ans. 306.047 cu. ft. 6. How many cords in a range of wood cut 4 feet long, the range being 4 feet 6 inches high and 160 feet long ? Problem III. 663. To find the surface of a right pyeamid. MuUiply the perimeter of the base by half its slant hrii^-Iit, and the product ivill be the convex surface (Prop. XV. Bk. YIII.). To this add the area of the base, and the result will be the entire surface. 664. Scholium. The surface of an oblique pyramid is found by taking the sum of the areas of its several faces. S Examples. 1. Required the convex surface of a pentangular pyramid, A B D E - S, each side of whose base, A B C D E, is 5 feet, and whose slant height, S M, is 20 feet. 5 X 5 = 25 ; 25 X ^^''- = 250 square [feet, the surface required. 2. What is the entire surface of a triangular pyramid, of which the slant height is 18 feet, and each side of the base 42 inches ? Ans. 99.804 sq. ft. 3. Required the convex surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet. 4. What is the entire surface of a quadrangular pyra- mid, the sides of the base being 40 and 30 inches, and the slant height upon the greater side 20.04, and upon the less side 20.07 feet ? Ans. 125.308 ft. 284 ELEMENTS OP GEOMETBY. Problem IV. 665. To find the surface of a frustum op a eight PYRAMID. Multiply half the sum of the perimeters of its two bases by its slant height, and the product will be the convex surface (Prop. XVII. Bk. VIII.) ; to this add the areas of the two bases, and the result will be the entire surface. Examples. 1. What is the entire surface of a rectangular frustum whose slant height is 12 feet, and the sides of whose bases are 5 and 2 feet ? 5x4=20; 2X4=8; 20+8 = 28; ^x 12= 168; 52 _^ 2= = 29 ; 168 + 29 = 197 sq. ft., area required. 2. Required the convex surface of a regular hexangular frustum, whose slant height is 16 feet, and the sides of whose bases are 2 feet 8 inches and 3 feet 4 inches. 3. What is the entire surface of a regular pentangular frustum, whose slant height is 11 feet, and the sides of whose bases are 18 and 34 inches ? Ans. 136.849 sq. ft. Problem V. 666. To find the solidity of a pyramid. Multiply the area of its base by one third of its altitude (Prop. XX. Bk. VIII.). Examples. 1. Required the solidity of a pen- tangular pyramid, A B C D E - S, each side of whose base, ABODE, is 5 feet, and whose altitude, SO, is 15 feet. 5^ X 1.7205 = 43.0125 ; 43.0125 X •V- = 215.0575 cu. ft., the solidity required. BOOK XII. 285 2. What is the solidity of a hexangular pyramid, the altitude of which is 9 feet, and each side of the base 29 inches ? 3. What is the solidity of a square pyramid, each side of whose base is 30 feet, and whose perpendicular height is 25 feet ? Ans. 7500. 4. Required the solid contents of a triangular pyramid, the perpendicular height of which is 24 feet, and the sides of the base 34, 42, and 50 inches. Ans. 39.2354 cu. ft. Problem VI. 667. To find the solidity of a frustum op a pyramid. Add together the areas of the two bases and a viean proportional between them, and niuUiply that sum by one third of the altitude of the frustum (Prop. XXI. Bk. VIII.). ExAMn.KS. 1. Required the solidity of the frustum of a quadran- gular pyramid, the sides of whose bases are 3 feet and 2 feet, and whose altitude is 15 feet. 3X3=9; 2x2 = 4; ^"9x4=6 (Prop. IV. Bk. II.) ; (9 + 4 + 6) X J]^- = 95 cu. ft., solidity required. 2. How many cubic feet in a stick of timber in the form of a quadrangular frustum, the sides of whose bases are 15 inches and 6 inches, and whose altitude is 20 feet ? 3. Required the solid contents of a pentangular frus- tum, whose altitude is 5 feet, each side of whose lower base is 18 inches, and each side of whose upper base is 6 inches. Ans. 9.319 cu. ft. 4. Rcqiiired the solidity of the frustum of a triangular pyramid, the altitude of which is 14 feet, the sides of the lower base 21, 15, and 12, and those of the upper base 14, 10, and 8 feet. Ans. 868.752 cu. ft. 286 ELEMENTS OF GEOMBTUT. The Wedge. 6G8. A Wedge is a polyedron bounded by a rectangle, called the base of the wedge ; by two trapezoids, called the sides, which meet in an edge parallel to the base ; and by two triangles, called the ends of the wedge. Thus ABCD-GH is a wedge, of which ABCD is the rectangular base ; ABHG,DCHG, the tra- pezoidal sides, which meet in the edge GH ; and ADG, B C H, the triangular ends. The altitvde of a wedge is the perpendicular distance from its edge to the plane of its base ; as G P. Problem VII. 669. To find the solidity of a wedge. Add the length of the edg-e to twice the length of the base ; multipli/ the sum by one sixth of the product of the altitude of the wedge and the breadth of the base. For, let L equal AB, the length of the base ; I equal GH, the length of the edge ; b equal B C, the breadth of the base ; and h equal PG, the height of the wedge. Then L — / = A B — G ri = A IL Now, if the length of tlic base and the edge be equal, the polyedron is equal to half a parallelopipodon having the same base and altitude (Prop. \1. Bk. ^'11I.\ and its isolidity will be equal io \b1 h (Prop. XIII. Bk. VIIL). If the length of the base is greater than that of the edge, let a section, M N G, bo made parallel to B C II. BOOK XII. 287 This section will divide the whole wedge into the quad- rangular pj'ramid A M X D - G, and the triangular prism BCH-G. The solidity of A :M N D - G is equal to ^ & A X (L — (Prob. Y.) ; and the solidity of BCH-G is equal to ^ bill ; hence the solidity of the whole wedge is equal to i- Z; A Z + ^ & A X (L — = i i* A 3 Z + i 6 A 2 L ~ ibh2l=ibh X (2L + 0. But, if the length of the base is less than that of the edge, the solidity of the wedge will be equal to the prism less the pyramid ; or to ibhl — ibhx (l—'L')=lbh 2,l — ^bh2l + ibh2L = ibhX i-2L-{-l). Examples. 1. Required the solidity of a wedge, the edge of which is 10 inches, the sides of the base 12 inches and 6 inches, and the altitude 14 inches. 10 + (12 X 2) = 34 ; 34 X ~^ = 476 cu. in., the [solidity required. 2. What is the solidity of a wedge, of whicli the edge is 24 inches, the sides of the base 36 inches and 9 inches, and the altitude 22 inches ? 3. How many solid feet in a wedge, of which the sides of the base are 35 inches and 15 inches, the length of the edge 55 inches, and the altitude 17/^ inches ? Ans. 3 cu. ft. 175f cu. in. Rectangular Prismoid. 670. A RECTANGULAR PRISMOID is a -polyedron bounded by two rectangles, called the bases of the prismoid, and by four trapezoids called the lateral faces of the prismoid. The altitude of a prismoid is the perpendicular distance between its bases. I\ I "\ 1 T "x I'-ja. 1 M 1 _ : "] L \ 288 ELEMENTS OP GEOMETBT. Problem VIII. 671. To find the solidity of a eectangulak peismoid. Add the area of the two bases to four times the area of a parallel section at equal distances from the bases ; mul- tiply the sum by one sixth of the altitude. Let L and B be the length and breadth of the lower base, I and b the length and breadth of the upper base, M and m the length and breadth of the parallel section equidistant from the bases, and h the altitude of the prismoid. If a plane be passed through the opposite edges L and /, the prismoid will be divided into two wedges, having for bases the bases of the prismoid, and for edges L and I. The solidity of these wedges, which compose the pris- moid, is (Prob. VII.), ^BA X (2L + + i«'/tX (2Z + L) = iA(2BL + B/+2^./+^>L). But M being equally distant from L and /, 2 M = L -|- /, and 2m = B + 6 (Prop. VII. Cor., Bk. IV.); conse- quently, 4Mm=(L + Z)X(B + 5) = BL-fBi-fftL + ftZ. Substituting 4Mm for its base, in the preceding equation, we have, as the expression of the solidity of a pi'ismoid, iA(BL + Z>/ + 4Mm). 672. Scholium. This demonstration applies to prismoids of other forms. For, whatever bo the form of the two bases, there may bo inscribed in each such a number of smnll rrctiiiip;los that tlm sum of them in cnoh base shall differ less from tliat base tlian any assignable quantity ; so that tiie sum of the rectangular prismoids that may be BOOK XII. 289 constructed on these rectangles will differ from the given prismoid by less than any assignable quantity. Examples. 1. Required the solidity of a prismoid, the larger base of wliicli is 80 inches by 27 inclies, the smaller base 24: inches by 18 inches, and tlic altitude 48 inches. SO X 27 = 810 ; 24 X 18 = 432 ; -+-- X — t~ X ^ = 2430 ; (810 + 432 -f 2430) X ^* = 29,376 cu. in. = 17 cu. ft., the solidity required. 2. What is the solidity of a stick of timber, whose larger end is 24 inches by 20 inches, the smaller end 16 inches by 12 inches, and the length 18 feet ? 3. Wliat is tlie solidity of a block, whose ends are re- spectively SO by 27 inches and 24 by 18 inches, and whose length is 36 inches ? 4. What is the capacity in gallons of a cistern 47^^ inches deep, wliose inside dimensions are, at the top 8lJ- and 55 inches, and at the bottom 41 and 29|- inches ? Ans. 546.929 gall. Peoblem IZC, 673. To find the sui-face of a regular polyedron. Multiply the area of one of the faces by the number of faces ; or multiply the square of one of the edges of the polyedron by the surface of a similar polyedron whose edges are 1. For, since the faces of a regular polyedron are all equal, it is evident that the area of one face multiplied by tlio aiumber of faces will give the area of the whole surface. Also, since the surfaces of regular polyedrons of the same name are bounded by the same number of similar poly- gons (Prop. I. Blc. VI.), their surfaces are to each other as the squares of the edges of the polyedrons (Prop. I. Cor., Bk. VI.). 25 200 ELEMENTS OF GEOMETRY. 674. Table op Surfaces and Solidities of Poltedeons WHOSE Edge is 1. NAMES. KO. OF PACES. SURFACES. SOLIDITIES. Tetraedron, Hexaedron, Octaedron, Dodecaedron, Icosaedron, 4 6 8 12 20 1.7320508 6.0000000 3.4641016 20.6457288 8.6602540 0.1178511 1.0000000 0.4714045 7.GG31189 2.1816950 The surfaces in the table are obtained by multiplying the area of one of the faces of the polyedron, as given in Art. 632, by the ntimber of faces. Examples. 1. What is the surface of an octaedron whose edge is 16 inches ? IG'^ X 3.4G41016 = 886.81 sq. in., the area required. 2. Required the surface of an icosaedx'on whose edge is 20 inches. 3. Required the surface of a dodecaedron whose edge is 12 feet. Ans. 2972.985 sq. ft. Problem X. 675. To find the solidity of a regular polyedron. MuUiply the surface by one third of the perpendicular distance from the centre to one of the faces ; or muUiply the cube of one of the ed^es by the solidity of a similar polyedron whose edg'e is 1. For any regular polyedron may bo divided into as many equal pyramids as it has faces, the common vertex of the pyramids being the centre of the polyedron ; hence, the* solidity of the polyedron must equal the product of the areas of all its faces by one third the perpendicular dis- tance from the centre to each face of the polyedron. BOOK XII. 291 Also, since similar pyramids arc to each other as the cubes of their liomologous edges (Prop. XXII. Bk. YIII.), two polyedroiis containing the same number of similar pyramids are to each other as the cubes of tlieir edges ; hence, the solidity of a polyedron wliose edge is 1 (Art. 673), may be used to measure otlicr similar polyedrous. Examples. 1. Required the solidity of an octaedron whose edge is 16 inches. le'' X 0.4714045 = 1930.8728 cu.in., solidity required. 2. "Wliat is the solidity of a tctraedroii whose edge is 2 feet ? 3. Required the solidity of an icosaedrou whose edge is 15 inches. Ans. 7363. 2:^06 ca. in. Probleji XI. 676. To find the surface of a cylinder. MuUiply the circumference of its base by Us altitude, and the product will be the convex surface (Prop. I. Bit. X.). To this add the areas of its two bases, and the re- sult ivill be the entire surface. Examples. 1. What is the entire surface of a cylin- der, the altitude of wliich, A B, is 10 feet, and the circumference of the base 20 feet ? 10 X 20 = 200 ; 20^ X 0.07958 X 2 = 63.264; 200 + 63.264= 263.264 sq. ft., the surface required. 2. Required the convex surface of a cylinder whose alti- tude is 16 feet, and the circumference of whose base is 21 feet. 3. Wliat is the entire surface of a cylinder whose alti- tude is 10 inches, and whose circumference is 4 feet ? 292 ELEMENTS OF GEOMETRY. 4. How many times must a cylinder 5 feet 3 inches long, and 21 inches in diameter, revolve, to roll an acre ? Ans. 1509.18 times. Problem XII. 677. To find the solidity of a cylinder. Multiply the area of the base by the altitude, and the product will be the solidity (Prop. II. 13k. X.). Examples. 1. What is the solidity of a cylinder, -whose altitude is 10 feet, and the circumference of whose base is 20 feet ? 20^ X 0.07958 X 10 = 318.32 cu. ft., solidity required. 2. Required tlie solidity of a cylindrical log, whose length is 9 feet, and the circumference of whose base is 6 feet. Ans. 25.7831 cu. ft. 3. The Winchester bushel is a hollow cylinder 18J- inches in diameter, and 8 inches deep ; what is its ca- pacity in cubic inches ? Problem XIII. 678. To find the surface of a cone. Multiply the circumference of the base by half the slant heiffht (Prob. III. Bk. X.), and the product will be the convex surface. To this add the area of the base, and the result will be the entire surface. Examples. 1. What is the convex surface of a cono, whose slant height is 28 feet, and the circumference of whose base is 40 feet ? 40 X ^ = 500 sq. ft., the surface required. 2. Required the entire surface of a cone, whose slant height is 14 foot, and the circumfcrcnco of whose base is 92 inches. BOOK XII. 293 3. "What is the surface of a cone, whose slant height is 9 feet, and the diameter of whose base is 36 inclies ? 4. How many yards of canvas are required for the cov- ering of a conical tent, the slant height of which is 30 feet, and the circumference of the base 900 feet ? Ans. 1500 sq. yd. Problem XIV. 679. To find the surface of a frustum of a cone. MuUiply half the sum of the circumferences of its tioo bases by its slant heifrht, and the product ivill be the cotir- vex surface (Prop. IV. Bk. X.). To this add the area of its bases, and the result will be the entire surface. 680. Scholium. The convex surface of a frustum of a cone may also be found by multiplying the slant height by the circumference of a section at equal distances between the two bases (Prop. IV. Cor., Bk. X.). Examples. 1. Required the convex surface of a frustum of a cone, whose slant height is 20 feet, and the circumferences of whose bases are SO feet and 40 feet. 2 — X 20 = 700 sq. ft., the surface required. 2. Required the surface of a frustum of a cone, the di- ameters of the bases being 43 inches and 23 inches, and the slant height 9 feet. 3. What is the convex surface of a frustum of a cone, of which a section equidistant from its two bases is 24 feet in circumference, the slant height of the frustum being 19 feet ? 4. From a cone the circumference of wliose base is 10 feet, and whose slant heiglit is SO feet, a cone has been cut off, whose slant height is 8 feet. What is the convex surface of the frustum ? Ans. 139^ sq. ft. 25* 294 ELEMENTS OP GEOMETRY. Problem XV. 681. To find the solidity of a CONE. MuHipli/ the area of its base by one third of its altitude, and the product will be the solidity (Prop. V. Bk. X.). Examples. 1. What is tlie solidity of a cone ■vrhose altitude is 42 feet, and the diameter of whose base is 10 feet ? 10'' X 0.7854 X ¥ = 1099.56 cu. ft., solidity required. 2. Required the solidity of a cone wliose altitude is Go feet, and the radius of wliose base is 12 feet 6 inches. 3. How many cubic feet in a conical sticlc of timber, whose length is 18 feet, the diameter at the larger end being 42 inches ? Aus. 57.7209 cu. ft. Problem XYI. 682. To find the solidity of the frustum of a cone. Add tog-ether the areas of the tioo bases and a mean proportional betioeen them, and muUiply that sum by one third of the altitude of the frustum ; and the result will be the solidity required (Prop. W. Blc. X.). Examples. 1. Wliat is the solidity of a frustum of a cone, C D E P, whose altitude, A B, is 21 feet, and the area of whose bases, P E, CD, are 80 square feet and 300 square feet ? (80 + 300 + V80 X 300) X V = 3732.96 cu. ft., solidity required. 2. Required the solidity of a frustum of a cone, the dinmeters of the bases being 38 and 27 inches, and the altitude 11 feet. 3. If a cask, which is two equal frustums of cones joined together at the larger bases, have its bung diameter 28 BOOK XII. 295 inches, the head dicameter 20 inches, and length 40 inches, how many gallons of wine will it hold ? Ans. 79.06. Problem XVII. 683. To find the surface of a sphere. Multiply the diameter by the circumference of a great circle of the sphere (Prop. VIII. Bk. X.) ; or multipb] the area of one great circle of the sphere by 4 (Prop. VIII. Cor 1, Bk. X.) ; or multipli/ 3.1416 by the square of the diameter (Prop. VIII. Cor. 4, Bk. X.). D EXAMPLKS. 1. What is the surface of a sphere, whose diameter, ED, is 40 feet, and whose circum- ference, A E B D, is 125.664 ? 125.664 X 40 = 5026.56 sq. £ft., the surface required. 2. Required the surface of a sphere whose diameter is 30 inches. 3. What is the surface of a globe whose diameter is 7 feet and circumference 21.99 feet ? Ans. 153.93. 4. How many square miles of surface has tlie earth, its diameter beinor 7912 miles ? Problem XVIII. 684. To find the surface of a zone or SEGirENT op a SPHERE. Multiply the altitude of the zone or segment by the cir- cumference of a great circle of the sphere (Prop. VIII. Cor. 2, Bk. X.) ; or midtiply the product of the diameter and altitude by 3.1416 (Prop. VIII. Cor. 6, Bk. X.). 296 ELKMENTS OP GEOMETRY. ExAJIPI.KS. 1. What is the surface of a segment of a sphere, the altitude of the segment being 10 feet, and the diameter of tlie sphere 50 feet ? 50 X 10 X 3.1416 = 1570.80 sq. ft., surface required. 2. The altitude of a segment of a spliere is 38 inches, and the circumference of the sphere is 25 feet ; what is the surface of the segment ? 3. Required the surface of a zone or segment, the diam- eter of the sphei-e being 72 feet, and the altitude of the zone 24 feet. Ans. 5428.6848 sq. ft. 4. If the earth be regarded as a perfect sphere whose axis is 7912 miles, and the part of the axis corresponding to each of the frigid zones is 327.192848, to each of tlie temperate zones 2053.468612, and to the torrid zone S150. 67708 miles ; what is the surface of each zone ? Ans. Each frigid zone 8132797.39568 ; each temperate zone 51041592.99898; torrid zone 78314115.07768 miles. Problem XIX. 685. To find the solidity of a sphere. Multiply the surface of the sphere by one third of its radius (Prop. IX. Bk. X.) ; or multiply the cube of the di- ameter of the sphere by 0.5236 (Prop. IX. Cor. 5,Bk. X.). Examples. 1. What is the solidity of a sphere whose diameter is 40 inches ? 40' X 0.5236 = 33510.4 cu. in., the solidity required. 2. Required the solidity of a globe whose circumference is 60 inches. 3. Wliat is the solidity of the moon in cubic miles, sup- posing it a perfect sphere with a diameter of 2160 miles ? 4. Required the solidity of the oartli, supposing it to bo a povfoct sphere, whoso diameter is 7912 miles. Ans. 259332805349.80493 cu. miles. 1)00 K XII. 207 Problem XX. 686. To find the surface of a spherical polygon. From the sum of all the angles subtract the product of two right angles by the number of sides less tioo ; divide the remaimler by 90°, and multiply the (quotient by one eighth of the surface of the sphere ; and the result ivill be the surface of the spherical polygon (Prop. XX. Bk. IX.). Examples. 1. Required the surface of a spherical polygou having five sides, described on a splicre wliose diameter is 100 feet, tlie sum of the angles being 720 degrees. 2 X 90° X (5 — 2) = 640° ; (720° — 540°) -=- 90° = 2 ; 100'^ X 3.1416 == 31416 ; 2 X ^^^^ = 7854 sq.ft., the surface required. 2. Wliat is tlie surface of a triangle on a sphere ■nliosc diameter is 20 feet, the angles being 150°, 90°, and 54° ? Problem XXI. 687. To find the solidity of a spherical pyramid or SECTOR. Multiply the area of the polygon or zone which forms the base of the pyramid or sector by one third of the radius (Prop. IX. Cor. 1, Bk. X.) ; or multiply the altitude of the base by the square of the radius, and that product by 2.0944 (Prop. IX. Cor. 7, Bk. X.). Examples. 1. Required the solidity of a E spherical sector, A C B E, the al- titude, E D, of tlie zone forming its base being 5 feet, and the radius, C B, of the sphere being 12 feet. 5 X 24 X 3.1416 = 376.992 ; 376.992 X J# = 1507.968 cu. ft., the solidity required. 298 ELEMENTS OP GEOMETEY. 2. What is the solidity of a spherical pyramid, the area of its base being 3G4 square feet, and the diameter of the sphere 60 feet ? 3. Required the solidity of a spherical sector, whose base is a zone 16 inches in altitude, in a sphere 3 feet in diameter. 4. "What is the solidity of a spherical sector, whose base is a zone 6 feet in altitude, in a sphere 18 feet iu diam- eter ? Ans. 1017.88 cu. ft. Problem XXII. 688. To find the solidity of a spherical segment. When the segment is less than a hemisphere, from the solidity of the spherical sector lohose base is the zone of the segment, take the solidity of the cone v>hose vertex is the centre of the sphere, and whose base is the circular base of the segment; but tohen the segment is greater than a hemisphere, take the sum of these solidities (Prop. IX. Sch., Bk. X.). 689. Scholium. If the segment has two plane bases, its solidity may be foimd by taking the difference of the two segments which lie on the same side of its two bases (Prop. IX. Sch., Bk. X.). Examples. 1. What is the solidity of a segment, ABE, whose altitude, E D, is 5 feet, cut from a sphere whose radius, C E, is 20 feet ? Tlie altitude of the cone A B C is equal to C E — E D, or 20 — 5, which is equal to 15 feet ; and the radius of its base is equal to V C A'^^CD", or a/ •!()■' — W\ whicli is ccjual to 13.2:^ ; consoquently the diameter A B is equal to 26.46 feet ; 5 X 20" X 2.0944 = 4188.8 BOOK XII. 299 cubic feet, the solidity of the sector A C D E (Prob. XXI.); 26.4i;^ X 0.7^''>4 X -'/= 2946.99 cubic feet, tlie solidity of the cone A B - C (Prob. XV.) ; 4188.8 — 2946.99 = 1241.81 cubic feet, the solidity of the segment ABE required. 2. Required the solidity of a segment, whose altitude is 67 inches, the diameter of the sphere being 153 inches. 3. What is tlie solidity of a spherical segment, whose altitude is 13 feet, and the diameter of the sphere S3 feet 6 inches ? 4. Required the solidity of the segments of the earth which are boimded severally by its five zones, the earth's diameter being 7912 miles, and the part of the diameter corresponding to each of the frigid zones being 327.19, to each temperate zone 2053.47, and to the torrid zone 3150.68. Ans. Each frigid zone 1293793463.32, each temperate zone 65013912318.45, and the torrid zone 146717393786.26 cubic miles. The Spheroid. 690. A SPHEROID is a solid which may be described by the revolution of an ellipse about one of its axes, which remains immovable. An oblate spheroid is one described by the revolution of the ellipse about its minor or conjugate axis. A prolate spheroid is one described by the revolution of the ellipse about its major or transverse axis. Problem XXIII. 691. To find the solidity of a spheroid. Multiply the square of the axis of revolution by the fixed axis, and that product by 0.5236. A full demonstration of this, wliioh is based upon the principle that a spheroid is two thirds of its circximscribing 300 ELEMENTS OF GEOMETKT. cylinder, -would require a knowledge of Conic Sections, or of the Differential and Integral Calculi, with neither of which is the learner here supposed to be acquainted. The relation, however, of tlie spheroid to its circumscrib- ing cylinder, is that which the sphere sustains to its cir- cumscribing cylinder (Prop. X. Bk. X.). Now the area of the base of the cylinder is found by multiplying the square of the axis of revolution by 0.785-i, and the solidity of the cylinder by multiplying that pro- duct by the fixed axis (Prop. II. Bk. X.). But the solid- ity of the spheroid is only two thirds of that of the cylin- der ; hence, to obtain the solidity of the former, instead of multiplying by 0.7854, we must use a factor only two thirds as large, which will be 0.52-36. EXAMPLKS. 1. What is the solidity of the ob- late spheroid AOBD, whose fixed axis, C D, is 30 inches, and the axis of revolution, A B, 40 inches. 40= X 30 X 0.5236 = 25132.8 cubic inches, the solidity required. 2. Eequired the solidity of a prolate spheroid, whose fixed axis is 50 feet, and the axis of revolution 36 feet. 3. What is the solidity of a prolate, and also of an oblate spheroid, the axes of each being 25 and 15 inches ? Ans. Prolate, 2945.25 cu. in. ; olilate, 4908.75 cu. in. 4. What is the solidity of a prolate, and also of an ob- late spheroid, the axes of each being 3 feet 6 inches and 2 feet 10 inches ? 5. Required the solidity of the eai-th, its figure being that of an oblate spheroid whoso axes are 7025.3 and 7898.9 miles. Ans. 259774684886.834 cubic miles. BOOK XIII. MISCELLANEOUS GEOMETRICAL EXERCISES. 1. If the opposite angles formed by four lines meeting at a point are equal, these lines form but two straight lines. 2. If the equal sides of an isosceles triangle are pro- duced, the two extei'ior angles formed "with the base will be equal. 3. The sum of any two sides of a triangle is greater than the third side. 4. If from any point within a triangle two straight lines are drawn to the extremities of eitlicr side, they will in- clude a greater angle than that contained by the other two sides. 5. If two quadrilaterals have the foyr sides of the one equal to the four sides of the other, each to each, and tlie angle included by any two sides of the one equal to the angle contained by the corresponding sides of the other, the quadrilaterals are themselves equal. 6. The sum of the diagonals of a trapezium is less than the sum of any four lines which can be drawn to the four angles from any point within the figure, except from the intersection of the diagonals. 7. Lines joining the corresponding extremities of two equal and parallel straight lines, are themselves equal and parallel, and the figure formed is a parallelogram. 8. If, in the sides of a square, at equal distances from the four angles, points be taken, one in each side, the straight lines joining these points will form a square. 26 302 ELEMENTS OP GEOMETRY. 9. If one angle of a parallelogram is a right angle, all its angles are riglit angles. 10. Any straight line drawn through the middle point of a diagonal of a parallelogram to meet the sides, is bi- sected in that point, and likewise bisects the parallelogram. 11. If four magnitudes are proportionals, the first and second may be multiplied or divided by the same magni- tude, and also the third and fourth by the same magni- tude, and the resulting magnitudes will be proportionals. 12. If four magnitudes are proportionals, the first and third may be multiplied or divided by the same magni- tude, and also the second and fourth by the same magni- tude, and the resulting magnitudes will be proportionals. 13. If there be two sets of proportional magnitudes, the quotients of the corresponding terms will be proportionals. 14. If any two points be taken in the circiimference of a circle, the straight line joining them will lie wholly within the circle. 15. The diameter is the longest straight line that can be inscribed in a circle. 16. If two straight lines intercept equal arcs of a circle, and do not cut each other within the circle, the lines will be parallel. 17. If a straight line be drawn to touch a circle, atfd be parallel to a chord, the point of contact will be the middle point of the arc cut off by that chord. 18. If two circles cut each other, and from either point of intersection diameters be drawn, the exti-omities of these diameters and the other point of intersection will be in the same straight line. 19. If one of the equal sides of an isosceles triangle be the diameter of a circle, the circuniferciico of the circle will bisect tlio base of the triangle. 20. If the opposite angles of a q\iadrilateral be together equal to two right angles, a circle may bo circumscribed about the quadrilateral. BOOK xiir. 303 21. Parallelograms "wliicli liavo two sides and the in- cluded angle equal in each, are themselves equal. 22. Equivalent ti-iangles \ipoii the same base, and upon tlie same side of it, are between the same parallels. 23. If the middle points of the sides of a trapezoid, which are not parallel, be joined by a straight line, that line will be parallel to each of the two parallel sides, and be equal to half their sum. 24. If, in opposite sides of a parallelogram, at equal distances from opposite angles, points be taken, one in each side, the straight line joining these points will bisect the parallelogram. 25. The perimeter of an isosceles" triangle is greater than the perimeter of a rectangle, which is of the same altitude with, and equivalent to, the given triangle. 26. If tlie sides of tlie square described upon the hypoth- enuse of a right-angled triangle be produced to meet the sides (produced if necessary) of the squares described upon the other two sides of the triangle, the triangles thus formed will be similar to the given triangle, and two of them will be equal to it. 27. A square circumscribed about a given circle is double a square inscribed in the same circle. 28. If the sum of the squares of the four sides of a quadrilateral be equivalent to the sum of the squares of the two diagonals, the figure is a parallelogram. 29. Straight lines drawn from the vertices of a triangle, so as to bisect the opposite sides, bisect also the triangle. 30. The straight lines which bisect the three angles of a triangle meet in the same point. 31. The area of a triangle is equal to its perimeter mul- tiplied by half the radius of the inscribed circle. 32. If the points of bisection of the sides of a given tri- angle be joined, the triangle so formed will be one fourth of the given triangle. S3. To describe a square upon a given straight line. 304 ELEMENTS OP GEOMETRY. 34. To find in a given straight line a point equally dis- tant from two given points. 35. To construct a triangle, the base, one of tlie angles at the base, and the sum of the other two sides being given. 36. To trisect a right angle. 37. To divide a triangle into two parts by a line drawn parallel to a side, so that these parts sliall be to each other as two given straiglit lines. 38. To divide a triangle into two parts by a line drawn perpendicular to the base, so that these parts shall be to each otlier as two given lines. 39. To divide a triangle into two parts by a line drawn from a given point in one of the sides, so that the pai-ts shall be to each other as two given lines. 40. To divide a triangle into a square number of equal triangles, similar to each other and to the original triangle. 41. To trisect a given straight line. 42. To inscribe a square in a given right-angled isosceles triangle. 43. To inscribe a square in a given quadrant. 44. To describe a circle that shall pass through a given point, have a given radius, and touch a given straight line. 45. To describe a circle, the centre of which sliall be in the perpendicular of a given right-angled triangle, and the circumference of which shall pass througli the right angle and touch tlie hypothenuse. 46. To describe tliree circles of equal diameters which shall touch each other, and to describe anotlier circle which sliall touch the three cii'cles. 47. If, on tlie diameter of a semicii-cle, two equal circles be described, and in the curvilinear space included by the tlirce circumferences a circle bo inscribed, its diameter will be to that of the equal circles in the ratio of two to three. 48. If two points bo taken in the diameter of a circle, BOOK XIII. 305 equidistant from the centre, the sum of the squares of two lines drawn from these points to any point in the circum- ference will always be the same. 49. Given the vertical angle, and the radii of the in- scribed and circumscribed circles, to construct the triangle. 60. If a diagonal cuts off three, five, or any odd number of sides from a regular polygon, the diagonal is parallel to one of the sides. 51. The area of a regular hexagon inscribed in a circle is double that of an equilateral triangle inscribed in the same circle. 52. The side of a square circumscribed about a circle is equal to the diagonal of a square inscribed in the same circle. 63. To describe a circle equal to half a given circle. 64. A regular duodecagon is equivalent to three fourths of the square constructed on the diameter of its circum- ncribed circle ; or is equal to the square constructed on the side of the equilateral triangle inscribed in the same circle. 55. If semicircles be described on the sides of a right- angled triangle as diameters, the one described on the hypothenuse will be equal to the sum of the other two. 66. If on the sides of a triangle inscribed in a semi- circle, semicircles be described, the two crescents thus formed will together equal the area of the triangle. 67. If the diameter of a semicircle be divided into any number of parts, and on them semicircles be described, their circumferences will together be equal to the circum- ference of the given semicircle, 68. To divide a circle into any number of parts, which shall all be equal in area and equal in perimeter, and not have the parts in the form of sectors. 59. To draw a straight line perpendicular to a plane, from a given point above the plane. 60. Two straight lines not in the same plane being 26* 306 ELEMENTS OP GEOMETRY. given in position, to draw a straight line which shall be perpendicular to them both. 61. The solidity of a triangular prism is equal to the product of the area of either of its rectangular sides as a base multiplied by half its altitude on that base. 62. All prisms of equal bases and altitudes are equal in solidity, whatever be tiie figure of their bases. 63. The convex surface of a regular pyramid exceeds the area of its base in the ratio that the slant height of the pyramid exceeds the radius of the circle inscribed in its base. 64. If from any point in the circumference of the base of a cylinder, a straight line be drawn perpendicular to the plane of the base, it will be wholly in the surface of the cylinder. 65. A cylinder and a parallelopipedon of equal bases and altitudes are equivalent to each other. 66. If two solids have the same height, and if their sec- tions made at equal altitudes, by planes parallel to the bases, have always the same ratio which the bases have to one another, the solids have to one another the same ratio which their bases have. 67. The side of the largest cube that can be inscribed in a sphere, is equal to the square root of one third of the square of the diameter of the sphere. 68. To cut off just a square yard from a plank 14 feet 3 inches long, and of a uniform width, at what distance from the edge must a line be struck ? Ans. 7-f^ in. 69. How much carpeting a yard wide will be required to cover the floor of an octagonal hall, whose sides ai-e 10 feet each 1 70. The perambulator, or s\irvoying-wheel, is so con- structed as to turn just twice in the length of a rod ; what is its diameter ? Ans. 2.626 ft. 71. Wliat is the excess of a floor 50 feet long by 30 broad, iihoxo two otlicrs, each of half its dimensions ? BOOK XIII. 307 72. The four sides of a trapezium arc 13, 13.4, 24, and 18 feet, and the first two contain a riglit angle. Required the area. Ans. 253.38 sq. ft. 73. If an equilateral triangle, whose area is equal to 10,000 square feet, be surrounded with a walk of uniform width, and eqiial to the area of the inscribed circle, what is the width of the walk ? Ans. 11.701 ft. 74. A right-angled triangle has its base 16 rods, and its perpendicular 12 rods, and a triangle is cut off from it by a line parallel to its base, of whicli the area is 24 rods. Re- quired the sides of that triangle. Ans. 8, 6, and 10 rods. 75. There is a circular pond whose area is 6028f square feet, in the middle of which stood a pole 100 feet high ; now, the pole having been broken off, it was observed that the top portion resting on the stump just reached the brink of the pond. What is the height of the piece left stand- ing ? Ans. 41.9968 ft. 76. The area of a square inscribed in a circle is 400 square feet ; required the diagonal of a square circum- scribed about the same circle. 77. The four sides of a field, whose diagonals are equal, are known to be 25, 35, 31, and 19 rods, in a successive order ; what is the area of the field ? Ans. 4A. IR. 38^ p. 78. The wheels of a chaise, each 4 feet high, in turning within a ring, moved so that the outer wheel made two turns while the inner made one, and their distance from one another was 5 feet ; what were the circumferences of the tracks described by them ? Ans. Outer, 62.8318 ft.; inner, 31.4159 ft. 79. The girt of a vessel round the outside of the hoop is 22 inches, and the hoop is 1 inch thick ; required the true girt of the vessel. 80. If one of the Egyptian pyramids is 490 feet high, having each slant side an equilateral triangle and the base a square, what is the area of the base ? Ans. 11 A. 3 rd. 223^ ft. 308 ELEMENTS OP GEOMETRY. 81. An ellipse is surrounded by a wall 14 inches thick ; its axes are 840 links and 612 links ; required the quan- tity of ground enclosed, and the quantity occupied by the wall. Ans. 4 A. 6 rd. enclosed, and 1760.49 sq. ft., area oc- cupied by the wall. 82. There is a meadow of 1 acre in the form of a square ; what must be the length of the rope by which a horse, tied equidistant from each angle, can be permitted to graze over the entire meadow ? 83. A gentleman has a rectangular garden, whose length is 100 feet and breadth 80 feet ; what must be the uni- form width of a walk half-way round the same, to take up just half the garden ? Ans. 25.9688 ft. 84. Two trees, 100 feet asunder, are placed, the one at the distance of 100 feet, and the other 60 feet from a wall ; what is the distance that a person must pass over in rxm- ning from one tree to touch the wall, and then to the other tree, the lines of distance making equal angles with the wall ? Ans. 173.205 ft. 85. There is a rectangular park 400 feet long and 300 feet broad, all round which, and close by the wall, is a border 10 feet broad ; close by the border there is a walk, and also two others, crossing each other and the park at right angles, in the middle of the garden. The walks are all of one breadth, and their area takes up one tenth of the whole park ; required the breadth of the walks. Ans. 6.2375 fit. 86. A farmer borrowed a cubical pile of wood, which measured 6 feet every way, and repaid it by two cubical piles, of which the sides were 3 feet each ; what part of the quantity borrowed has he returned ? 87. A board is 10 feet long, 8 inches in breadth at the greater end, and 6 inches at the less ; how much must bo cut off from the less end to make a square foot ? Ans. 23.2493 in. 8.'i. A piece of timber is 10 foot long, each side of the BOOK XIII. 309 greater base 9 inches, and each side of the less 6 inches ; how much must be cut off from the less end to contain a solid foot ? Ans. 3.39214 ft. 89. What must be the inside dimensions of a cubical box to hold 200 balls, each 2^ inches in diameter ? 90. Near my house I intend making a hexagonal or six- sided seat around a tree, for which I have procured a pine planic 16J- feet long and 11 inches broad ; what must be the inner and outer lengths of each side of the seat, that there may be the least loss in cutting up tlie plank ? Ans. 26.64915 in. inner, and 39.36085 in. outer length. 91. Required the capacity of a tub in the form of a frustum of a cone, of which the greatest diameter is 48 inches, the inside length of the staves 30 inches, and the diagonal between tl>e farthest extremities of the diameters 50 inches. Ans. 165.34 gals. 92. The front of a hoiise is of such a height, that, if the foot of a ladder of a certain length be placed at tlie dis- tance of 12 feet from it, tlie top of the ladder will just reach to the top of the house ; but if the foot of the ladder be placed 20 feet from the front, its top will fall 4 feet be- low the top of the house. Required the height of the house, and the length of the ladder. Ans. 34 feet, the height of the building ; 36.0555 feet, the length of the ladder. 93. A sugar-loaf in form of a cone is 20 inches higli ; it is required to divide it equally among three persons, by sec- tions parallel to the base ; what is the height of each part ? Ans. Upper 13.8672, next 3.6044, lowest 2.6284 in. 94. Within a rectangular court, whose length is four chains, and breadth three chains, there is a piece of water in the form of a trapezium, whose opposite angles are in a direct line with those of the court, and the respective dis- tances of the angles of the one from those of the other are 20, 25, 40, and 45 yards, in a successive order ; required the area of the water, Ans. 960 sq. yd. 310 ELEMENTS OP GEOMETRY. 95. Wliat will the diameter of a sphere be, when its solidity and the area of its surface are expressed by the same numbers ? Ans. 6. 96. There is a circiilar fortification, which occupies a quarter of an acre of ground, surrounded by a ditch coin- ciding with the circumference, 24 feet wide at bottom, 26 at top, and 12 deep ; how much water will fill the ditcli, if it slope equally on both sides ? Ans. 135483.25 cu. ft. 97. A father, dying, left a square field containing CO acres to be divided among his five sons, in such a manner that the oldest son may have 8 acres, the second 7, the third 6, the fourth 5, and the fifth 4 acres. Now, the division fences are to be so made that the oldest sou's share shall be a narrow piece of equal breadth all around the field, leaving the remaining four shares in the form of a square ; and in like manner for each of the other shares, leaving always the remainders in form of squares, one within another, till tlie share of the youngest be the inner- most square of all, equal to 4 acres. Required a side of each of the enclosures. Ans. 17.3205, 14.8324, 12.2474, 9.4868, and 6.3246 chains. 98. Required the dimensions of a cone, its solidity be- ing 282 inches, and its slant height being to its base diam- eter as 5 to 4. Ans. 9.796 in. the base diameter ; 12.246 in. ilie slant height ; and 11.223 in. the altitude. 99. A gentleman has a piece of ground in form of a square, the difference between whose side and diagonal is 10 rods. He would convert two thirds of the area into a garden of an octagonal form, but would have a fish-pond at the centre of the garden, in the form of an equilateral triangle, whose area must equal five square rods. Re- quired the length of each side of the garden, and of each side of the pond. Ans. 8.9707 rods, each sido of the giU-den, and 3.398 rods, each side of the pond. BOOK XIV. APPLICATIONS OF ALGEBEA TO GEOMETRY. 692. When it is proposed to solve a geometrical prob- lem by aid of Algebra, draw a figure which shall represent the several parts or conditions of the problem, both known and required. Represent the known parts by the first letters of the alphabet, and the required jjarts by the last letters. Then, observing the geometrical relations that the parts of the figure have to each other, make as many indepen- dent equations as there are unknown quantities intro- duced, and the solution of these equations will determine the unknown quantities or required parts. To form these equations, however, no definite rules can be given ; but the best aids may be derived from experi- ence, and a thorough knowledge of geometrical principles. It should be the aim of the learner to effect the simplest solution possible of each problem. Problem I. 693. In a right-angled triangle, having given the hy- pothenuse, and the sum of the other two sides, to deter- mine these sides. C Let A B be the triangle, right-an- gled at B. Put A C = a, tlie sum A B + B C = 5, AB = .7;, and B C = ?/. A B 312 ELEMENTS OP GEOMETET. Then, x-\-y = s. and (Prop. XI. Bk. IV.), X- + ;/ = a\ From the first equation, x = a — y. Substitute in second equation this vahie of a;, s^—2sy + 2y^ = a'. Or, 2 2/' — 2sy = a" — s^ Or, y^ — sy==ia!' — ^ s^. By completing the square, y' — sy + ls'^ia-' — ^s^ Extracting sq. root, y — ^ 5 = ± v'j- a- — \ s^, Or, y = xs ± Via^—is\ If A C = 5, and the sum AB + BC = 7,2/ = 4or3, and X ^ 3 or 4. Problem II. 694. Having given the base and perpendicular of a tri- angle, to find tJie side of an inscribed sqtmre. Let ABC be the triangle, and H E F G the inscribed square. Put A B = 6, C D = a, and GForGH = DI = x; then will CI = CD — DI = a — X. Since the triangles ABC, A G F C are similar, or Hence, or. AB:CD::GF:CI, b \ a X : X : a — x, ab — b X = (f .('. ah DE B BOOK XIV. 313 that is, the side of the inscribed square is equal to the pro- duct of the base by the altitude, divided by their sum. Problem III. 695. Having" given the lengths of two straight lines draion from the acute angles of a ^ight-angled trian- gle to the middle of the opposite sides, to determine those sides. Let A B C be the given triangle, ^ and AD, BE the given lines. Put A D = o, B E = 6, CD or ^ C B = .-c, and C E or ^ C A = y ; then, since C D^ + C A^ = A D^ and /" /^e C E'^ + C B^ = B W, we have x^ -\- 4 y^ ^ a", and y' -\- 4 x^ t^ b^. b D ^C By subtracting the second equation from four times the first, 15y^=ia^-~ b\ g^. y .-a^-h^. 15 by subtracting the first equation from four times the second, 15 a;2 = 4 &2 — a% g 4 J2 _ 02 ^15 which values of x and y are half the base and perpendic- ulars of tlie triangle. Problek IV. 696. In an equilateral triangle, having given the lengths of the three perpendiculars draxcn from a point within to the three sides, to determine these sides. 27 314 ELEMENTS OF GEOMETEY. Let A B C be the equilateral triaii- gie, and D E, D F, D G the given per- pendiculars from the point D. Draw D A, D B, D C to the vertices of the three angles, and let fall the perpen- dicular, CH, on th|[ base, AB. Put D E = a, D P = 6, D G = c, and A H or B H, half the side of the equilaterial triangle, = X. Then A or B G = 2 a;, and C H = VAC' — AH^ = V 4 a;* — 3? = V 8 x'' = a; V 3. Now, since the area of a triangle is equal to the product of half its base by its altitude (Prop. VI. Bk. IV.), The triangle ACB = J ABxCH=a;Xa-V3=a:' VS. ABD=iABxDG = a;Xc =cx. BCD = iBCxDE=a;Xa =ax. ACD = iACxDP = a;X6 =hx. But the three triangles ABD,BCD,ACD are together equal to the triangle A C B. Hence, a;' Vs = ax '\-hx-\- cx== x (a-\- b -\- c), or, X Vs = « + 6 + c ; a -\-h -\-c or. X = yr Hence each side, or 2 .•?; ^ 2 (g + 6 -f- c) 7^ ■ 697. Cor. Since the perpendicular, CH, is eqxial to X Vs, it is equal to a + 6 + c ; tliat is, the whole per- pendicular of an equilateral triangle is equal to the sum of all the perpendiculars let fall from any point in the tri- angle to each of its sides. Problem V. 698. To determine the radii of three equal circles de- scribed vnthin and tangent to a given circle, and also tan- gent to I'dch other. BOOK XIV. 815 Let AF be the radius of the given circle, and B E the radius of one of tlie equal circles de- scribed within it. Put AF = a, and B E = a; ; then each side of the equilateral triangle, BCD, formed by joining the centres of the required circles, will be rep- resented by 2 X, and its altitude, C E, by V4a;' — 2;% or x VS. The triangles B C E, A B E are similar, since the angles B C E and ABE are equal, each being half as great as one of the angles of the equilateral triangle, and the angle B E C is common. Hence, or and But hence, or or Hence, .1 CE: BE:: BO: AB, X V 3 : X : : 2x: AB, AB = 2x AB + BF = AF; 2x y~3 + x = a, 2 X + a; V 3 = a V 3, (2 + V3) x = a Vs. = ax 0.4641. 2-1-^3 2.1547 Problem VI. 699. In a right-angled triangle, having given the base, and the sum of the perpendicular and hypothenuse, to find these two sides. Problem VII. 700. In a rectangle, having given the diagonal and perimeter, to find tlic sides. 316 ELEMENTS OP GEOMETRY. Problem VIII. 701. In a right-angled triangle, having given the base, and the diiFerence between the hypothenuse and perpen- dicular, to find both these two sides. Problem IX. 702. Having given the area of a rectangle inscribed in a given triangle, to determine the sides of the rectangle. Problem X. 703. In a triangle, having given the ratio of the two sides, together with both the segments of the base, made by a perpendicular from the vertical angle, to determine the sides of the triangle. Problem XI. 704. In a'triangle, having given the base, the sum of the other two sides, and the length of a line drawn from the vertical angle to the middle of the base, to find the sides of the triangle. Problem XII. 705. In a triangle, having given the two sides about the vertical angle together with the line bisecting that angle, and terminating in the base, to find tlie base. Problem XIII. 706. To determine a right-angled triangle, having given the perimeter and the- radius of its inscribed circle. Problem XIV. 707. To determine a triangle, having given the base, the perpendicular, and the ratio of the two sides. Problem XV. 708. To determine a right-angled triangle, having given the hypothenuse, and the side of the inswibed square. BOOK XIT. 317 Problem XVI. 709. In a right-angled triangle, having given the perim- eter, or sum of all the sides, and the perpendicular let fall from the right angle on the hypothenuse, to determine the triangle, that is, its sides. Problem XVII. 710. To determine a right-angled triangle, having given the hypothenuse, and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle. Problem XVIII. 711. To determine a triangle, having given the base, the perpendicular, and the difference of the two other sides. Problem XIX. 712. To determine a triangle, having given the lengths of three lines drawn from the three angles to the middle of the opposite sides. Problem XX. 713. In a triangle, having given all the three sides, to find the radius of the inscribed circle. Problem XXI. 714. To determine a right-angled triangle, having given the side of the inscribed square^ and the radius of the inscribed circle. Problem XXII. 715. To determine a triangle, having given the base, the perpendicular, and the rectangle of the two otlier sides. 27* '318 ELEMENTS OP GEOMETEY. Problem XXIII. 716. To determine a right-angled triangle, having given the hypothenuse, and the radius of the inscribed circle. Problem XXIV. 717. To determine a right-angled triangle, having given the hypothenuse and the difference between a side and the radius of the inscribed circle. Problem XXV. 718. To determine a triangle, having given the base, the line bisecting the vertical angle, and the diameter of the circumscribing circle. Problem XXVI. 719. There are two stone pillars in a garden, whose perpendicular heights are 20 and 30 feet, and the distance between them 60 feet. A ladder is to be placed at a cer- tain point in the line of distance, of such a length, that it may just reach the top of both the pillars. What is the length of the ladder, and how far from each pillar must it be placed ? Ans. 39.5899 feet, length of the ladder ; 34^ feet, dis- tance of the foot of the ladder from the bottom of the lower pillar ; and 25|^ feet, distance of the foot of the ladder from the bottom of the higher pillar. Problem XXVII. 720. There is a cistern, the sum of the length and breadth of which is 84 inches, the diagonal of the top 60 inches, and the ratio of the breadth to the depth as 25 to 7. What are its dimensions, provided it has the form of a rectangular parallelopipedon ? Ans. Length 48 inches ; width 86 inches ; depth 10.08 inches. BOOK XIV. 319 Problem XXVIII. 721. The three distances from an oak, growing in an open plain, to tlie tliree visible corners of a sqiiare field, lying at some distance, are known to be 78, 59.161, and 78 poles, in successive order. "What are the dimensions of the field, and its area ? Ans. Side of the square 24 rd. ; area 3 A. 2 R. 16 rd. Problem XXIX. 722. There is a house of three equal stories in height. Now a ladder being raised against it, at 20 feet distance from the foot of the building, reaches the top ; whilst another ladder, 12 feet shorter, raised from the same point, reaches only to the top of the second story. What is the height of tlie building ? Ans. 41.696 ft. Problem XXX. 723. The solidity of a cone is 2513.28 cubic inches, and the slant side of a frustum of it, whose solidity is 2474.01, is 19.5 inches. Required the dimensions of the cone. Ans. Altitude 24 inches ; base diameter 20 inches. Problem XXXI. 724. Within a rectangular garden containing just an acre of ground, I have a circular fountain, T\'hose circum- ference is 40, 28, 52, and 60 yards distant from the four angles of the garden. From these dimensions, the length and breadth of the garden, and likewise the diameter of the fountain, are required. Ans. Length 94.996 yds. ; width 50.949 yds. ; diameter of the fountain 20 yds. Problem XXXII. 725. There is a vessel in the form of a frustum of a cone, standing on its lesser base, whose solidity is 8.67 feet, the depth 21 inches, its greater base diameter -ft) that 320 ELEMENTS 0I<' GEOMETRY. of the lesser as 7 to 5, into which a globe had accidentaHy been put, whose solidity was 2^ times the measure of its surface. Required the diameters of the vessel and of the globe, and how many gallons of water would be requisite just to cover the latter within the former. A.ns. 35 and 25 inches, top and bottom diameters of the frustum ; 15 inches, diameter of the globe ; and 34.2 gallons, the water required. Problem XXXIII. 726. Three trees. A, B, C, whose respective heights are 114, 110, and 98 feet, are standing on a horizontal plane, and the distance from A to B is 112, from B to C is 104, and from A to C is 120 feet. What is the distance from the top of each tree to a point in the plane which shall be equally distant from each ? Ans. 126.634 ft. Problem XXXIV. 727. A person possessed a rectangular meadow, the fences of which had been destroyed, and the only mark left was an oak-iree in the east corner ; he however recol- lected the following particulars of the dimeusious. It had once been resolved to divide the meadow mto two parts by a hedge running diagonally ; and he recollected that a segment of the diagonal intercepted by a perpen- dipular from one of the corners was 16 chains, and the same perpendicular, produced 2 chains, met the other side of the meadow. Now the owner has bequeathed it to four grandchildren, whose shares are to be bounded by the diagonal and perpendicular produced. What is the area of tlie meadow, and what are the several shares ? Ans. Area of the whole meadow, 16 acres ; shares, 1 R. 24 rd. ; 1 A. 2 R. 16 rd. ; 6 A. 1 R. 24 rd. ; 7 A. 2 R. 16 rd. THE END '/"'i!^Hk ..JSlI :'j'>'!'•:^^l4:<^!^t■;A:■