I L METR ICA GEO MET Y. AN ELEMENTARY TREATISE ON MENSURATION BY GEORGE BRUCE HALSTED A.B., A.M., AND EX-FELLOW OF PRINCETON COLLEGE; PH.D. AND EX-FELLOW OF JOHNS HOPKINS UNIVERSITY; INSTRUCTOR IN POST-GRADUATE MATHEMATICS, PRINCETON COLLEGE. BOSTON: PUBLISHED BY GINN, HEATH, & CO. 1881. Entered according to Act of Congress, in the year 1881, by GEORGE BRUCE HALSTED, in the office of the Librarian of Congress, at Washington. GINN, HEATH, & CO.o J. S. CUSHING, PRINTER, I6 HAWLEY STREET, BOSTON. INSCRIBED TO J. J. SYLVESTER, A.2M., Cam.; F.R.S., L. and E.; Cor. 1emm. Institute of France; lefem. Acad. of Sciences in Berlin, GOttingen, Naples, J1filan, St. Petersburg, etc.; LL.D., Univ. of Dublin, U. of E.; D.C.L., Ox:foord; Ion. Fellow of St. John's Col., Cam., IN TOKEN OF THE INESTIMABLE BENEFITS DERIVED FROM TWO YEARS' WORK WITH HIM, BY THE AUTHOR. PREFACE. T HIS book is primarily the outcome of work on the subject while teaching it to large classes. The competent critic may recognize signs of a Berlin residence; but a considerable part, it is believed, is entirely new. Special mention must be made of the book's indebtedness to Dr. J. W. DAVIS, a classmate with me at Columbia School of Mines; also to Prof. G. A. WENTWORTH, who has kindly looked over the proofs. But if the book be found especially accurate, this is due to the painstaking care of my friend H. B. FINE, Fellow of Princeton. Any corrections or suggestions relating to the work will be thankfully received. GEORGE BRUCE HALSTED. PRINCETON, NEW JERSEY, May 12, 1881. CONTENTS. MENSURATION. INTRODUCTION THE METRIC SYSTEM. NOTATION AND ABBREVIATIONS. Page. 1 2. 2.... 3 CHAPTER I. THE MEASUREMENT OF LINES. ~ (A). - STRAIGHT LINES. ILLUSTRATIVE PROBLEMS (a) To measure a line the ends of which only are accessible. (B) To find the distance between two objects, one of which is inaccessible (7) To measure a line when both ends of it are inaccessible. (8) To measure a line wholly inaccessible ~ (B).- STRAIGHT LINES IN TRIANGLES. 5 5 5 6 6 Article. I. RIG-HT-ANGLED TRIANGLES 1. To find hypothenuse 2. To find side. II. OBLIQUE TRIANGLES 3. Obtuse. 4. Acute. 5. Given perpendicular 6. To find medials III. STRAIGIIT LINES IN SIMr: 7. To find corresponding line IV. CHORDS OF A CIRnCTLE 8. To find diameter. 6 6 7 7. 8 8 9 9 R FIGURS...10. 10. 10. 10 ILA Viii CONTENTS. Article. Page. 9. To find height of arc 11 10. Given chord and height.......12 11. Given chord and radius, to find chord of half the arc..12 12. To find circumscribed polygon... 13 (C).- METHOD OF LIMITS. V. DEFINITION OFI A LIMIIT..14 13. Principle of Limits........ 15 14. The length of the curve...... 16 (D). —THE RECTIFICATION OF THE CIRCLE. 15. Length of semicircumference......19 16. Circumferences are as their radii......20 VI. LINES IN ANY CIRCLE.... 21 17. Value of rr.......... 21 18. To find circumference........ 21 19. Value of diameter.......21 CHAPTER II. THE MEASUREMENT OF ANGLES. (E). — THE NATURAL UNIT OF ANGLE. VII. ANGLES ARE ARCS......22 20. Numerical measure of angle. 22 VIII. ANGLE MEASURED BY ARC.....23 (F). CIRCULAR MEASURE OF AN ANGLE. 21. Angle in radians..... 24 22. To find length of arc. 24 23. To find number of degrees in arc.. 25 IX. ARCS wIICIs CORRESPOND TO ANGLES.... 25 24. Angle at center... 25 25. Inscribed angle.. 26 26. Angle formed by tangent and chord 26 27. Angle formed by two chords.. 26 28. Secants and tangents... 26 CONTENTS. ix Article. 29. Given degrees to find circular measure. 30. Given circular measure to find degrees. 31. Given angle and arc to find radius X. ABBREVIATIONS FOR AREAS. CHAPTER III. THE MEASUREMENT OF PLANE AREAS. (G).- PLANE RECTILINEAR FIGURES. Page.. 26. 27. 27. 28 XI. MEASURING AREA OF SURFACE 32. Area of rectangle 33. Area of square 34. Area of parallelogram. 35. Area of triangle (given altitude) 36. Area of triangle (given sides) 37. Radius of inscribed circle 38. Radius of circunscribed circle 39. Radius of escribed circle XII. TRAPEZOID AND TRIANGLE AS TRAPEZOID 40. Area of trapezoid. XIII. COORDINATES OF A POINT. -xr -rv T'r w a r -.a.. c.. A...-........ 29. 29. 32 33. 33 34 35. 36. 37. 38. 38. 39 Al V. lOLYGO-No AS OUi1 oF1 IXrilAPEZOI 41. To find sum of trapezoids 42. Area of any polygon 43. Area of quadrilateral 44. Area of a similar figure XVT. CONGRUENT AND EQUIVALENT XVI. PROPERTIES OF REGULAR POLYGON. 45. Area of regular polygon 46. Table of regular polygons (H).-AREAS OF PLANE CURVILINEAR FIGURES. 47. Area of circle 48. Area of sector 49. Area of segment 50. Circular zone 40 40 43 45 48 48 49 49 50 51 51 52 54 51. Crescent. 54 CONTENTS. x Article. 52. Area of annulus 53. Area of sector of annulus XVII. CNMIes 54. Area of parabola 55. Area of ellipse CHAPTER IV. MEASUREMENT OF SURFACES. XVIII. DEFINITIONS RELATING TO POLYHEDRONS 56. Faces plus summits exceed edges by two (I). —PRISM AND CYLINDER. XIX. PARALLELEPIPED AND NOR.AL 57. Mantel of prism XX. CYLINDRIC AND TRUNCATED 58. Mantel of Cylinder. (J).-PYRAMID AND CONE. Page.. 54. 55. 56.57. 59. 61. 61. 62. 63. 64. 64 XXI. CONICAL AND FIRUSTUM 59. Mantel of pyramid 60. Mantel of cone 61. Mantel of frustum of pyramid 62. Mantel of frustum of cone 63. Frustum of cone of revolution. 66. 66. 67 68. 69. 70. (K).-THE SPHERE. XXII. SPIERE A SURFACE, GLOBE A SOLID 64. Area of a sphere. XXIII. SPHERICAL SEGMENT AND ZONE 65. Area of a zone 66. Theorem of Pappus 67. Surface of solid ring. 71. 71. 73. 73. 74. 74 CONTENTS. xi ~ (L).- SPHERICS AND SOLID ANGLES. Article. XXIV. STEREGON AND STERADIAN 68. Area of a lune XXV. SOLID ANGLE MALDE BY TWO,. PLANES XXVI. SPHERICAL PYRAMID 69. Solid angles are as spherical polygons XXVII. SPHERICAL EXCESS 70. Area of spherical triangle 71. Area of spherical polygon TABLE OF ABBREVIATIONS Page.. 77. 78 THREE, OR MORE. 79. 80. 80. 80. 80. 81. 83 CHAPTER V. THE M/EASUREIENT OF VOLUMES. (M).-PRISM AND CYLINDER. XXVIII. SYMMETRICAL AND QUADER XXIX. VOLUME AND UNIT OF VOLUIE XXX. LENGTHS, AREAS, AND VOLUMES ARE RATIOS Volume of quader XXXI. MAss, DENSITY, WEIGHT To find density Volume of parallelepiped Volume of prism. Volume of cylinder Volume of cylindric shell 72. 73. 74. 75. 76. 77. 84 84 84 85 86 87 88 90 91 92 ~ (N).- PYRAMID AND CONE. XXXII. ALTITUDE OF PYRAMID 78. Parallel sections of pyramid. 79. Equivalent tetrahedra. 80. 'Volume of pyramid. 81. Volume of cone. 93. 93. 9 -95. 97 xii CONTENTS. ~ (0).- PRISMA9 rOID. Article. XXXIII.-XLI. DEFINITIONS RESPECTING PRISMATOID, 82. Volume of prismatoid. 83. Volume of frustum.of pyramid 84. Volume of frustum of cone. 85. Volume of ruled surface 86. Volume of wedge 87. Volume of tetrahedron (P). SPHERE. 88. Volume of sphere 89. Volume of spherical segment XLII. GENERATION OF SPHERICAL SECTOR 90. Volume of spherical sector. XLIII. DEFINITION OF SPHERICAL UNGULA 91. Volume of spherical ungula 92. Volume of spherical pyramid ~ (Q). -THEOREM OF PAPPUS. 93. Theorem of Pappus.. 94. Volume of ring ( (R).-SIMILAR SOLIDS. XLIV. SIMILAR POLYIEDRONS. 95. Volume of similar solids Page 98-101 101 104 106 107 108 109.110.112 113 113.114 114.115. 115.116.118. 118 ~ (S).- IRREGULAR SOLIDS. 96. By covering with liquid.. 119 97. Volume by weighing.. 120 98. Volume of irregular polyhedron... 120 CHAPTER VI. THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 99. Test for applicability. 121 CONTENTS. Xiii (T). —PRISMOIDAL SOLIDS OF REVOLUTION. Article. Page. XLV. EXAMINATION OF THE DIFFERENT CASES 124 (U). —PRISMOIDAL SOLIDS NOT OF REVOLUTION. (V). —ELIMINATION OF ONE BASE. 100. Prismatoid determined from one base.... 130 CHAPTER VII. APPROXIMATION TO ALL SURFACES AND SOLIDS. (W).-WEDDLE'S METHOD. 101. Seven equidistant sections.. 131 CHAPTER VIII. MASS-CENTER. 102-104.......... 136 105-131. SY ETRY........ 136 132. THE MASS-CENTER OF AN OCTAHEDRON.... 143 EXERCISES AND PROBLEMS IN MENSURATION. [These are arranged and classed in accordance with the above 132 Articles]........ 145 LOGARITHMS. 225 A TREATISE ON MENSURATION. AN ELEMENTARY TREATISE ON MENSTURATIONo INTRODUCTION. MENSURATION is that branch of mathematics which has for its object the measurement of geometrical magnitudes. It has been called, that branch of applied geometry which gives rules for finding the length of lines, the area of surfaces, and the volume of solids, from certain data of lines and angles. A Magnitude is anything which can be conceived of as added to itself, or of which we can form multiples. The measurement of a magnitude consists in finding how many times it contains another magnitude of the same kind, taken as a unit of measure. Measurement, then, is the process of ascertaining the ratio which one magnitude bears to some other chosen as the standard; and the measure of a magnitude is this ratio expressed in numbers. Hence, we must refer to some concrete standard, some actual object, to give our measures their absolute meaning. The concrete standard is arbitrary in point of theory, and its selection a question of practical convenience. A discrete aggregate, such as a pile of cannon-balls, or a number, has a natural unit, - "one of them." But in the continuous quantity, space, with which we chiefly have to deal, the fundamental unit, a length, is 2 MENSURATION. defined by fixing upon a physical object, such as a bar of platinum, and agreeing to refer to its length as our standard. That is, we assume some arbitrary length in terms of which all space measurements are to be expressed. The one actually adopted is the Meter, which is the length of a special bar deposited in the French archives. This we choose because of the advantages of the metric system, which l_ applies only a decimal arithmetic, and has a unir"4_ form and significant terminology to indicate the Emultiples and submultiples of a unit. THE METRIC SYSTEM o # convenes to designate multiples by the Greek nu-FE merals, and submultiples by the corresponding Latin words; as follows: '-, \ I MULTIPLE. NAME. DERIVATION. IMEANING. _LK~ ~~~~ i| ~Greek. - | S 10,000 myria- uvpdaS. ten thousand; i S b 1,000 kilo- Xi2?oi. a thousand. 100 hecto- eccar-v. a hundred. XIr 10 deka- dKca. ten. Divisions. Latin. 0 al r deci- decem. ten. co -1- lo1 centi- centum. a hundred. Kz _t__oo_ milli- mille. a thousand. So a millimeter (mm) is one-thousandth of a meter. Thus we are given a number of subsidiary units. For any particular class of measurements, the most convenient of these may be chosen. The kilometer (km) is used as the unit of distance; and along roads and railways are placed kilometric poles or stones. INTRODUCTION. 3 The centimeter (cm) is the most common submultiple of the meter. A chief advantage of this decimal system of measures is, that in it reduction involves merely a shifting of the decimal point. NOTATION AND ABBREVIATIONS TO BE USED IN THIS BOOK. Large letters indicate points; thus, A, B, and C denote the three angular points of a triangle, or C may denote the center of a circle, while A and B are on the circumference; then AB will denote the chord joining A to B; and, generally, AD means a straight line terminated at A and D. In the formulae, small letters are used to denote the numerical measures of lines; so that ab, as in common algebra, denotes the product of two numbers. The following choice of letters is made for writing a formula: [TABLE FOR REFERENCE.] a, b, and c are the sides of any triangle, respectively, opposite the angular points A, B, C. If the triangle is right-angled, a = altitude, b = base, c = hypothenuse. In regard to a circle, c= circumference, d= diameter. e = spherical excess. f= flat angle. g = number of degrees in an angle or arc. g~ means expressed in degrees only, g/ minutes, g1I seconds. h height. i = medial line. j = projection. k = chord. I =length. rn= meter. n? = any number. 6 = perigon. p = perimeter. q = square. r = radius. s = 2(a - b + c). t = tangent. u = circular measure. v = a discrete variable. w = width. =y = coordinates of a point. Z; CHAPTER I. THE MEASUREMENT OF LINES; (A).- STRAIGHT LINES. An accessible straight line is practically measured by the direct application of a standard length suitably divided. If the straight line contain the standard unit n times, then n is its numerical measure. But, properly speaking, any description of a length by counting of standard lengths is imperfect and merely approximate. Few physical measurements of any kind are exact to more than six figures, and that degree of accuracy is very seldom obtainable, even by the most delicate instruments. Thus, in comparing a particular meter with the standard bar, a difference of a thousandth of a millimeter can be detected. In four measurements of a base line at Cape Comorin, it is said the greatest error was 0.077 inch in 1.68 mile, or one part in 1,382,400; and this is called an almost incredible degree of accuracy. When we only desire rough results, we may readily shift the place of the line to be measured, so as to avoid natural obstacles. Still, under the most favorable circumstances, all actual measurements of continuous quantity are only approximately true. But such imperfections, with the devised methods of correction, have reference to the physical measurement of things; to the data, then, which in book-questions we suppose accurately given. THE MEASUREMENT OF LINES. 5 ILLUSTRATIVE PROBLEMS. (a) To measure a line the ends of which only are accessible. Suppose AB the line. Choose a point C from which A and B are both visible. Measure AC, and prolong it >' until CD AC. Measure BC, and prolong it until CE, =BC. Then ED =AB. E- I D Ww. 49 & 106; (Eu. I. 15 &4; Cv. I. 23 & 76). NOTE. Ww. refers to Wentworth's Geometry, third edition, 1879. Eu. refers to Todhunter's Euclid, new edition, 1879. Cv. refers to Chauvenet's Geometry. These parallel references are inserted in the text for the convenience of students having either one of these geometries at hand. References to preceding parts of this Mensuration will give simply the number of the article. (p) To find the distance between two objects, one of which is inaccessible. Let A and B be the two objects, separated by some obstacle, as a river. From A measure any straight line AC. Fix any point D in the \.. ( ( direction A B. Pro- duce AC to F, making CF= AC; and produce D1C to E, making ' W CE =-CD. Then find E- ' the point G at which the directions of BC and FE intersect; that is, find the point from which C and B appear in 6 MENSURATION. one straight line, and E and F appear in another straight line. Then the triangles A CD and CEF are congruent, and therefore ABC and CFG; whence FG =AB. Ww. 106 & 107; (Eu. I. 4 & 26; Cv. I. 76 & 78). Hence, we find the length of AB by measuring FG. (y) Lo measure a line when both ends of it are inacces 1CI I II I,. ~~ I1= ~\ - sible. At a point C, in the accessible part of AB, erect a perpendicular CD, and take DE —CD. At E make FG perpendicular to DE. Find in FG the point F which falls in the line BD, and the point G in the line AD. _FG -AB. Ww. 107; (Eu. I. 26; Cv. I. 78). (8) To measure a line wholly inaccessible. If AB is the line, choose a convenient point C from which A and B are both visible, and measure A C and BC by (/); then AB may be measured by (a). (B). —STRAIGHT LINES IN TRIANGLES. I. RIGHT-ANGLED TRIANGLES. 1. Given the base and perpendicular, to find the hypothenuse. Rule: Square the sides, add together, and extract the square root. Formula: a2 + b2 c2. Proof: Ww. 331; (Eu. I. 47; Cv. IV. 25). THE MEASUREMENT OF LINES. 7 EXAM. 1. The altitude of a right-angled triangle is 3, the base 4. Find the hypothenuse. a2 = 32 = 9. b'2= 42 = 16. a2 + b2 = 32 + 42= 25 =c2... c- 5. Answer. 2. Given the hypothenuse and one side, to find the other side. Rule: Maultipljy thei sum by their difference, and extract the square root. Formula,: c2 -- 2 = (c + a) (c - a) 'b2. EXAM. 2. The hypothenuse of a right-angled triangle is 13, the altitude 12. Find the base. c + a=25 c-a= 1.. c- a= 25'= 2... b -5. Ans. [Foir exercises on 1 arid 2, see table of right-angled triangles.] II. OBLIQUE TnIANGLES. When two lines form an angle, the projection of the first on the second is the line between the vertex and the foot of a A perpendicular let fall from the // extremity of the first on to the second. Thus the projection of _ _ _ AC on BOC is CGD. B c D Given two sides and the lprojection of one on the other, to find the third side 8 MENSURATION. 3. If the angle contained by the given sides be obtuse. Rule: To the sum of the squares of the given sides add twice the product of the projection and the side on which (when prolonged) it falls; then extract the square root. Formula: a2 +b2- 2 bj = c2. Proof: Ww. 336; (Eu. II. 12; Cv. III. 53). B A b C j D a2 = 25 l)= 36 21'j= 48 C. =C 109. EXAM. 3. Given the sides a = 5, b = 6, containing an obtuse angle, and given j =4, the projection of a on 6; find the third. side..'. = 10-44+. Ans. 4. If the angle contained by the given sides is acute. B i II C b j A D Rule: From the sum of the squares of the given sides subtract twice the product of the projection and the side on which it falls; the square root of the remainder gives the third side. Formula: a2 + b2 - 2 bj = c2. Proof: Ww. 335; (Eu. II. 13; Cv. III. 52). THE MEASUREMENT OF LINES. 9 EXAM. 4. Given the sides a =5, b 6, containing an acute angle, and given j 4, the projection of a on b; find the third side. =2 3 61 a a2 + b2= 25 + 3G = 61 2bj 48 lh.. c'2 13 j b.. c- 3-60+. Ans. 5. If two sides and the perpendicular let fall on one from the end of the other, are given, the projection can be found by 2, and then the third side by 3 or 4. If three sides are given, a projection can be found by 3 or 4, and then the perpendicular by 2. 6, Given three sides of a triangle to find its three medials; i.e., the distances from the vertices to the midpoints of the opposite sides. Rule: Fr.om the sum of the squares of any two sides subtract twice the square of half the base; the square root of half the remainder is the corresponding medial. Formula: a2+ c2 - 2 =- 2 i2 Proof: Ww. 338; (Eu. Appen. 1; Cv. III. 62). Corollary: Dividing the difference of the squares of two sides by twice the third side, gives the projection on it of its medial. 2 2 Formula: j a2-c 2b EXAM. 5. Given two sides, a= 7, c= 9, and the base, b = 4. Find the medial. Here a2 + 2 = 49 + 81 = 130 b2 42, '..2 6. = 8.'.2i - =122.. = 61 /.. i= 78. Ans. b 10 IMENSURATION. III. STRAIGHT LINES IN SIMILAR FIGURES. 7. Given two straight lines in one figure, and a line corresponding to one of them in a similar figure, to find the line corresponding to the other. Rule: The like sides of similatr figures are proporlional. a2b1 /tH '\ Formula: b2 --. 1J0 '~\ ~ Ww. 278; (Eu. VI., Def. I.; Cv. III. 24). EXAM. 6. The height of an upright stick is 2. meters,.... 1,' —<H' -~~~ ~-'~~- ~.~ and it casts a shadow 3 ___..... meters' long; the shadow of a flag-staff is 45 meters. Find the height of the staff. 3: 2::45: b2. 2 -- 30 meters. Ans. ~F 3 IV. CHORDS OF A CIRCLE. Suppose AB any chord in a circle. Through the center C a diameter perpendicular to AB.~ r>~ j~ ~ meets it at its middle point ), Av^ B\7 and bisects the arc at -11 DIT is the height of the arc, and 1H ~ A-H the chord of half the arc. 8. Given the height of an arc and the chord of half the arc, to find the diameter of the circle. THE MEASUREMENT OF LINES. 11 Rule: -Divide the square of the chord of half the arc by the height of the arc. 7.2 112 Formula: d - Proof: HAF is a right angle. Ww. 204; (Eu. III. 31; Cv. II. 59).. I. F: IA:: IHA: 1ID. Ww. 289; (Eu. VI. 8, Cor.; Cv. III. 44). EXAM. 7. The height of an arc is 2 centimeters, the chord of half the arc is 6 centimeters. Find the diameter. d_= 6 26 18. Ans. 2 9. Given the chord of an arc and the radius of the circle, to find the height of the arc. Rule: From the radius subtract the square root of the cifference of the squares of the radius and half the chord. Formula: h = r - V/,2 - k2 Proof. HD - D H -DC- ID- h, HC =r, and.DC2=-A C2-AD2=- 2_ (AB)2. EXAM. 8. The chord of an arc is 240 millimeters, the radius 125 millimeters. Find the height of the arc. DC2 = r2 (k)2 = (r + -k) (r- -k) = (125 + 120) (125- 120)..c. vDC-/-' _k/= A/245 X 5 = 125 = 35.. h - r -DC — 125 - 35 90 millimeters 9 centimeters. Ans. 12 MENSURATION. 10. Given the chord and height of an arc, to find the chord of half the arc. Rule: Take the square root of the sum of the squares of the height and half the chord. Formula: Vh -- V/2 + kh2 Proof: AH2 H= D2 + AD2. Ww. 331; (Eu. I. 47; Cv. IV. 25). EXAM. 9. Given the chord = 48, the height - 10. Find the chord of half the arc. 2-100. (1k)2= (24)2 576..-. k2=676... kh=26. Ans. If, instead of the height, the radius is given, substitute in 10 for h its value in terms of r and k from 9, and we have = V(r - Vr - -k2)2 + 22 = V2r2 - 2.rVr2- i2 -From this follows: 11. Given the chord of an arc and the radius of the circle, to find the chord of half the arc. Formula: hk V= 2 r2-.r 4r2 - k2. EXAM. 10. Calculate the length of the side of a regular dodecagon inscribed in a circle whose radius is 1 meter; that is, find /7 when r and k are each 1 meter long, for k is here the side of a regular inscribed hexagon, which always equals the radius. Ww. 391; (Eu. IV. 15, Cor.; Cv. V. 14). THE MEASUREMENT OF LINES. 13 Thus r and k being unity, our formula becomes 7 =- V2-V4- 1 = 0-51763809. Ans. EXAM. 11. With unit radius, find the length of one side of a regular inscribed polygon of 24 sides. 2 = 2 - /4 - (.51763809)2 = 0.26105238. And so on with regular polygons of 48, 96, 192, etc., sides. 12. Given the radius of a circle and the side of a regular inscribed polygon, to find the F side of the similar circumscribed polygon. Formula: t= 2 -/4 r2 -_ k2 Proof: Suppose AB the given side k. Draw the tangent at the middle point H of the arc AB, and produce it both ways to the points E and G, where it meets the radii CA and CB produced; required, t. In the similar triangles CEI, CAD, is G IEG is the side CO: CD:::Ef: AD:: t: k. kr.. t =2 CD2 CD2 = CA2- AD= r2- (.2 ()2 But CD 1 223 k2~i- IVJ esZ) = V,2 _4 k2-. V4 2 -_. EXAM. 12. When r= 1, find one side of a regular circumscribed dodecagon. 14 lMENSURATION. 2jk With radius taken as unity, t 2 -V4- l2 From Exam. 10, k =0-51763809, and substituting this value, 112= 0-535898. Ans. In the same way, by substituting kc24- 0-26105238 from Exam. 11, we find /24 — 0263305, and from k48= 01308062G we get 48 - 0-131087, and so on for 96, 192, 384, etc., sides. 2 (C). METHOD OF LIMITS. V. A variable is a quantity which may have successively an indefinite number of different values. DEFINITION OF A LIIIT. When a quantity can be made to vary in such a manner that it approaches as near as we please and continually nearer to a definite constant quantity, but cannot be conceived to reach the constant by any continuation of the process, then the constant is called the limit of the variable quantity. Thus the limit of a variable is the constant quantity which it indefinitely approaches, but never reaches, though the difference between.the variable and its limit may become and remain less than any assignable magnitude. EXAM. 13. The. limit of the sum of the series, 1 + + i- + -8 + -L- + -3 +3 + etc., is 2. EXAM. 14. The variable may be likened to a convenient ferry-boat, which will bear us just as close as we choose to the dock, - the constant limit, -but which cannot actually reach or touch it; for, if they touch, both explode into the unknown infinite. The bridge, the method for passing, in the order of our knowledge, from variables to their limits, is the THE MIEASUREMENT OF LINES. 15 13. PRINCIPLE OF LIMITS. If, while tending toward their respective limits, two varziable quantities are always in the same ratio to each other, their limits will be to one another in the same ratio as the variab es.* A b c B' B C' c' C C" Let the lines AB and AC represent the limits of any two variable magnitudes which are always in the same ratio to one another, and let Ab, Ac represent two corresponding values of the variables themselves; then Ab: Ac:: AB A C'. If not, then Ab: Ac:: AB: some line greater or less than AAC. Suppose, in the first place, that Ab: Ac:: AB: AC'; AC' being less than AC. By hypothesis, the variable Ac continually approaches A C, and may be made to differ from it by less than any given quantity. Let Ab and Ac, then, continue to increase, always remaining in the same ratio to one another, till Ac differs from A C by less than the quantity C'C; or, in other words, till the point c passes the point C' and reaches some point, as c, between C' and C, and b reaches the corresponding point b'. Then, since the ratio of the two variables is always the same, we have Ab: Ac:: Ab': Ac! By hypothesis, Ab: Ac:: AB AC'; hence, Ab': Ac':: AB: AC,' or, AC' x Ab = Ac' x AB; which is impossible, since each factor of the first member is less than the corresponding factor of the second member. - This principle, and the following demonstration of it, are contained essentially in Eu. XII. 2, though quoted here from Bledsoe. 16 MENSURATION. Hence the supposition that Ab: Ac: AB: A C' or to any quantity less than AC, is absurd. Suppose, then, in the second place, that Ab: Ac:: AB: A C" or to some term greater than AC. Now, there is some line, as AB' less than AB, which is to AC as AB is to AC' If, then, we conceive this ratio to be substituted for that of AB to AC," we have Ab: Ac:: AB': AC; which, by a process of reasoning similar to the above, may be shown to be absurd. Hence, if the fourth term of the proportion can be neither greater nor less than AC, it must be equal to AC; or we must have Ab: Ac::AB: AC. Q.E.D. Cor.: If two variables are always equal, their limits are equal. 14. When their sides tend indefinitely toward zero, the perimeter of the polygon inscribed increascs, circumscribed decreases, toward the same limit, the length of the curve. Proof: Inscribe in a circle any convenient polygon, say, the regular hexagon.. Ww. 391; (Eu. IV. 15; Cv. V. 14). Join the extremities of each side, as AB, to the point of the curve equally distant from them, as Hr; that is, the point of intersection of the arc and a perpendicular at the middle point of the chord. Thus we get sides of a regular dodecagon. Repeat the process with the sides of the dodecagon, and we have a regular polygon of twenty-four sides. So continuing, the number of sides, -always doubling, will increase indefinitely, while the length of a side will tend toward zero. THE MEASUREMENT OF LINES. 17 The length of the inscribed perimeter augments with the number of sides, since we continually replace a side by two which form with it a triangle, and so are together greater. Ww. 96; (Eu. I. 20; Cv. 5). G D Thus AB of the hexagon is replaced by AIT and IIB in the dodecagon. But this increasing perimeter can never become as long as the circumference, since it is always made up of chords each of which is shorter than the corresponding arc, by the axiom, "A straight line is the shortest distance between two points." Therefore, this perimeter increases toward a limit which cannot be longer than the circumference. In doubling the number of sides of a circumscribed polygon, by drawing tangents at the middle points of the arcs, we continually substitute a straight for a broken line; as, TU for TN —NU. So this perimeter decreases. 18 MENSURATION. But two tangents from an external point cannot be together shorter than the included arc. E.g., fiT + TB > arc fIB. Therefore this perimeter decreases toward a limit which cannot be shorter than the circumference. But the limit toward which the circumscribed perimeter decreases is identical with that toward which the corresponding inscribed perimeter increases. For, in a regular circumscribed polygon of any number of sides, n, the perimeter is n times one of the sides. Pn, n. THE MEASUREMENT OF LINES. 19 But, from 12, 2rk, 4r2 2 nrk,, V/4 r2- kn2 But pj, the perimeter of the corresponding inscribed polygon, is nk,. ]. p 2r Tn t/4r -kJn the ratio of the perimeters. Cutting the circumference into n equal parts makes each part as small as we please by taking n sufficiently great. But chords are shorter than their arcs; therefore k,, tends toward the limit zero as n increases. Thus the limit of 2r is 2r 1. The vari4-;2 42 ables Pn and 2 - are always equal. Therefore, by pn -v 4r2-/ k 13, Cor., their limits are equal, and limit of P- - 1... lim. pn = lim. p'. p" But we have shown that lim. tp cannot be longer than c, and lim. pC cannot be shorter than c. Therefore, the common limit is c, the length of the curve. ~(D). THE RECTIFICATION OF THE CIRCLE. 15. In a circle whose radius is unity, to find the length of the semicircumference. From 14, an approximate value of the semicircumference in this circle is given by the semiperimeter of every polygon inscribed or circumscribed, the latter being in excess, and the former in defect of the true value. In examples 10, 11, and 12, we have already calculated 20 MENSURATION. the length of a side in the regular inscribed and circumscribed polygons of 12, 24, and 48 sides. Continuing the same process, and in each case multiplying the length of one side by half the number of sides, we get the following table of semiperimeters: n. I- n- n1t.r 6 3-000 3.4641016 12 3.1058285 3-2153903 24 3,1326286 3-1596599 48 3.1393502 3.14608G2 96 3.1410319 3-1427140 192 3 1414524 3.1418730 384 3.1415576 3.1416627 768 3.1415838 3.1416101 1536 3.1415904 3.1415970 3072 3.1415921 3.1415937 6144 3.1415925 3.1415929 12288 3.1415926 3.1415927 etc. etc. etc. Since the semicircumference, c, is' always longer than nlcn, and shorter than Int,, therefore its value, correct to seven places of decimals, is 31415926. 16. The circumferences of any two circles are to each other in the same ratio as their radii. Proof: The perimeters of any two regular polygons of the same number of sides have the same ratio as the radii of their circumscribed circles. Ww. 374; (Eu. XII. 1, V. 12, Cv. V. 10). The inscribed regular polygons remaining similar to each other when the number of sides is doubled, their THE MEASUREMENT OF LINES. 21 perimeters continue to have the same ratio. Hence, by 13, the limits, the circumferences, have the same ratio as their radii. Cor. 1. Circumferences are to each other as their diameters. Cor. 2. Since c: c':: r: r:: 2r: 2r, C Ct _C 2r 2r' r That is, the ratio of any circumference to its diameter is a constant quantity. This constant, identical with the ratio of any semicircumference to its radius, is denoted by the Greek letter 7r. But, in circle with radius 1, semicircumference we have found 3-1415926+. Therefore, the constant ratio = 3-1415926+. VI. LINES IN ANY CIRCLE. 17. X =e = _C = 3.1415926+. d r Multiplying both sides of this equation by d, gives 18. c= d 7= 2r7r... The diameter of a circle being given, to find the circumference. Rule: Multiply the diameter by 7r. NOTE. In practice, for 7r the approximation 31 or 22_ is generally found sufficiently close. A much more accurate value is 3 5,; easily remembered by observing that the denominator and numerator written consecutively, thus, 113 1 3 55, present the first three odd numbers each written twice. The value most used is r = 3.1416. 19. Dividing 18 by 7r gives c I d= 2r=- c - = cx 0.3183098+. 7r 7' CHAPTER II. THE MEASUREMENT OF ANGLES. s (E). THE NATURAL UNIT OF ANGLE. To say "all right angles are equal," assumes that the amount of turning necessary to take a straight line or direction all around into its first position is the same for all points. Thus the natural unit of reference for angular magnitude is one whole revolution, called a perigon, and equal to four right angles. VII. A revolving radius describes equally an angle, a surface, and a curve. Moreover, the perigon, circle, and circumference are each built up of congruent parts; and any pair of angles or sectors have the same ratio as the corresponding arcs. Ww. 201; (Eu. VI. 33; Cv. II. 51). r any angle its intercepted arc Therefore, = perigon circumference That is, if we adopt the whole circumference as the unit of arc; 20. The numerical measure of an angle at the center of a circle is the same as the numerical measure of its intercepted arc. THE MEASUREMENT OF ANGLES. 23 And this remains true, if, to avoid fractions, we adopt, as practical units of angle and arc, some convenient part of these natural units. The Egyptian astronomers divided the whole circle into 360 equal parts, called degrees; each of these degrees was divided into 60 parts, called minutes; these again into 60 parts, called seconds. These numbers have very convenient factors, being divisible by 1, 2, 3, 4, 5, 6, etc. EXAM. 15. A perigon = 3590 601 of angle. A circumference = 3590 591 6011 of arc. VIII. Hence we say, An angle at the center is measured by its intercepted arc; meaning, An angle at the center is such part of a perigon as its intercepted arc is of the whole circumference. (F). CIRCULAR MEASURE OF AN ANGLE. Half a perigon is a flat angle; hence, halving the denominators in VII., and using ~ to mean angle, gives any 2 _ its intercepted arc flat 4 semicircumference But, from 18, in every circle, c- = r7r. Therefore, dividing denominators by 7r, gives any 4 _length of its arc. flat r 71 If, now, we adopt as unit angle that part of a perigon denoted by flat;. that is, the ~ subtended at the center 7r 24 MENSURATION. of every circle by an arc equal in length to its radius, and hence named a radian, then, by 20, 21. The number which expresses any angle in radians also expresses its intercepted arc in terms of the radius. So, in terms of whatever arbitrary unit of length the arc and radius may be expressed, if u denote the number of radians in an angle, then, for every A, U= 1 r Thus the same angle will be denoted by the same number, whatever be the unit of length employed. u, or the fraction arc divided by radius, is called the circular measure of an A. EXAM. 16. Find the circular measure off. Here, the arc being a semicircumference, its length 1= r-r. r7r.'. A6 = -. Ans. r This is obviously correct, since dividing a flat ~ by 7r first gave us our radian. 22. Given the number of degrees in an angle, to find the length of the arc intercepted by it from a given circumference. Rule: Multiply the length of the circumference by the number of degrees in the angle, and divide the product by 360. Formula: 1= 3eqO 3600 Proof: From VII. we have 3600: g0:: c:l. NOTE. If the 4 be given in minutes, the formula becomes I- e' If in seconds, I= -12960 1296000" THE MEASUREMENT OF ANGLES. 25 ExAM. 17. How long is the arc of one degree in a circumference of 25,000 miles? -250 = 69.4+. Ans. 36 23. Given the length of an arc of a given circumference, to find the number of degrees it subtends at the center. Rule: Multiply the length of the arc by 360, and divide the product by the length of the circumference. 1360~ Formula: g =. NOTE. To find the number of minutes or seconds: 121600' 11296000" Formulae: y'-; and g"=-. C C EXAM. 18. Find the number of degrees subtended in any circle by an arc equal to the radius. 13600 rb360~ 180~ Here g becomes -- = -- c 2r7r 7r = 57-2957795~+. Ans. Hence a radian = p = 570 17'44.8"+ = 206264.8"+. IX. The arcs used throughout as corresponding to the angles are those intercepted from circles whose center is the angular vertex. These arcs are said to measure the angles at the center which include them, because these arcs contain their radius as often as the including angle contains the radian. Using nteasured in this sense, we may state the following Theorems: 24. An angle at the center is measured by the arc intercepted between its sides. Ww. 202; (Cv. II. 52). 26 MENSURATION. 25. An inscribed angle is measured by.half its intercepted arc. Ww. 203; (Eu. III. 20; Cv. II. 57). 26. An angle formed by a tangent and a chord is measured by half the intercepted arc. Ww. 209; (Eu. III. 32; Cv. II. 62). 27. An angle formed by two chords, intersecting within a circle, is measured by half the sum of the arcs vertically intercepted. Ww. 208; (Eu. Appen. 2; Cv. II. 64). 28. If two secants, two tangents, or a tangent and a secant intersect without the circle, the angle formed is measured by half the difference of the intercepted arcs. Ww. 210; (Eu. Appen. 3; Cv. II. 65). 29. Given the measure of an angle in degrees, to find its circular measure. Rule: Multiply the number of degrees by wr, and divide by 180. Trq~ org, erg" Formula: u- -qO ' 1800 10800' 648000"' Proof: A flat 4 is 180~ and its circular measure is 7r. Hence, g~ _ u 1800 -' since each fraction expresses the ratio of any given 4 to a flat 4. Therefore, _ 9go 180~' and also _ 1800 V7r TI-IE MEASUREMENT OF ANGLES. 27 This recalls to mind again that the circular measure of any 4 is independent of the lengitA of the radius of the cirele. EXAM. 19. Find the circular measure of 4 of one degree. 7r Here u= - -.0174532925+. Ans. 180 EXAM. 20. Find the circular measure of 4 of one minute. Dividing the last answer by 60 gives *000290888208+. Ans. Of course, this number equally expresses the length of an arc of one minute in parts of the radius, and in the same way we obtain Arc 1'" = r x 0.00000484813681+. 30. Given the circular measure of an angle, to find its measure in degrees. Rule: Multiply the circular measure by 180, and divide by r. u 180~ Formula: g u = -. 7r EXAM. 21. Find the number of degrees in 4 whose circular measure is 10. 1 Here g~= 10 x 10 x 57.2957795+ -r 572-957795+. Ans. 31. Given the angle in degrees and the length of the arc which subtends it, to find the radius. Rule: Divide 180 times the length by wr times the number of degrees. 11800 Formula: r Irff 28 MENSURATION. Proof: 10~. r" 180 0 r rr EXAM. 22. An arc of 6 meters subtends ~ of 10~ find radius. 180 6 Here r= -X 0.6 X 57.2957795+ 7r 10 =34-3775 meters. Ans. X. One ~ is called the complement of another, when their sum equals a rt. I; the supplement, when their sum f=; the explement, when their sum- 6. REFERENCE TABLE OF ABBREVIATIONS FOR AREAS. When used alone, as abbreviations, capital letters denote the area of the figures; to denote volume, a V is prefixed. A = annulus. B = base. C = cylinder. o = circle. D = volume of prismatoid. E= ellipse. F = frustum. G = segment. H== sphere. I = volume of an irregular polyhedron. o= parabola. K== cone. L = lune. M-= midsection. N= polygon of n sides. 0 = solid ring. P = prism. 0= parallelogram. Q = quadrilateral. q = square. 1 = rectangle. = sector. T = trapezoid. A = triangle. U= volume of quader. V= volume. W= volume of wedge. X= volume of tetrahedron. Y= pyramid. Z = zone. v=n Tv == T T+ T, + 3 + *....+ Tn. v=l i = sum of angles. A = spherical A. N= spherical N. Y = volume of spherical Y. s2 = steregon. means equivalent; i.c., equal in size. 11 = parallel. 1 = perpendicular. = similar... = therefore. 4= angle. CHAPTER III. THE MEASUREMENT OF PLANE AREAS. (G). PLANE RECTILINEAR FIGURES. XI. The area of a surface is its numerical measure. Measuring the area of a surface, whether plane or curved, is determining its ratio to a chosen surface called the unit of area. The chosen unit of area is a square whose side is a unit of length. EXAM. 23. If the unit of length be a meter, the unit of area will be called a square meter (qm). If the unit of length be a centimeter, the unit Square of area will be a square centimeter (qcm). Centimeter 32. To find the area of a rectangle. Rule: lMultiply the base by the altitude. Formula: R = ab. Proof: SPECIAL CASE. When the base and altitude, or length and breadth of tke rectangle are commensurable. In this case there is always a line which will divide both base and altitude exactly. If this line be assumed as linear unit, a and b are integral numbers. 30 MENSURATION. In the rectangle ABCD divide AD into a, and AB into b equal parts. Through the points of division draw lines parallel to the sides of the rectangle. These lines divide the...........................-..........-..... rectangle into a num ber of................. squares, each of which is a unit of area. In the bottom row...................................... there are b such squares; and, since there are a rows, we have A B* b squares repeated a times, which gives, in all, ab squares. NOTE. The composition of ratios includes numerical multiplication as a particular case. Ordinary multiplication is a growth from addition. The multiplier indicates the number of additions or repetitions. The multiplicand indicates the thing added or repeated. Therefore, it is not a mutual operation, and the product is always in terms of the unit of the multiplicand. The multiplicand may be any aggregate; the multiplier is an aggregate of repetitions. To repeat a thing does not change it in kind, so the result is an aggregate of the same sort exactly as the multiplicand. When the rule says, Multiply the base by the altitude, it means, Multiply the numerical measure of the base by the number measuring the altitude in terms of the same linear unit. The product is a number which we have shown to be the area of the rectangle; that is, its numerical measure in terms of the superficial unit. This is the meaning to be assigned whenever we speak of the product of one line by another. GENERAL PROOF. Rectangles, being equiangular parallelograms, have to one another the ratio which is compounded of the ratios of their sides. Ww. 315; (Eu. VI. 23; Cv. IV. 5). Let R and R' represent the surfaces or areas of two rectangles. Let a and a' represent their altitudes; b and b' their bases. THE MEASUREMENT OF PLANE AREAS. 31 Thus, R a b ab /l a bl ab1'./= Rtab_. a'b' For the measurement of surfaces, this equation is fundamental. To apply it in practice, we have only to select as a standard some particular unit of area. a R a' B ' b b The equation itself points out as best the unit we have already indicated. If we suppose a' and b' to be, each of them, a unit of length, _R' becomes this superficial unit, and the equation becomes R = ab. This shows that the number of units of area in any rectangle is that number which is the product of the numbers of units of length in two adjacent sides. This proof includes every case which can occur, whether the sides of the rectangle be commensurable or incommensurable with the unit of length; that is, whether a and b are integral, fractional, or irrational. EXAM. 24. Find the area of a ribbon 1 meter long and 1 centimeter wide. 1 meter is 100 centimeters... 100 square centimeters. Ans. 32 MENSURATION. 33. To find the area of a square. Rule: Take the second power of the number denoting the length of its side. Formula: q b2. Proof: A square is a rectangle having its length and breadth equal. NOTE. This is the reason why the product of a number into itself is called the square of that number. Cor. Given the area of a square, to find the length of a side. Rule: Extract the square root of the nunmber denoting the area. EXAM. 25. 1 square dekameter, usually called an Ar (a), contains 100 square meters. Every E l Illlll_ unit of surface is equivalent to 100 of the next lower denomination, because every unit of length is 10 of the next lower order. Thus a square hektometer is a hektar (h). EXAM. 26. How many square centimeters in 10 square millimeters? 100 square millimeters is 1 square centimeter... 10 square millimeters is -15 of 1 square centimeter. Ans. Or, 10 square millimeters make a rectangle 1 centimeter long and 1 millimeter wide. EXAM. 27. How many square centimeters in 10 millimeters square? Y AI, ' 11 ~..-L1 /,0. THE MEASUREMENT OF PLANE AREAS. 33 REMARK. Distinguish carefully between square meters and meters square. We say 10 square kilometers (qkm), meaning a surface which would contain 10 others, each a square kilometer; while the expression 5 kilometers square means a square whose sides are each 5 kilometers long, so that the figure contains 25 square kilometers. EXAM. 28. The area of a square is 1000 square meters. Find its side. Find its side. 000 - 31-623 meters. Ans. 34. To find the area of any parallelogram. Rule: JIMultiply the base by the altitude. Formula: =- ab. Proof: Any parallelogram is equivalent to a rectangle of the same base and altitude. Ww. 321; (Eu. I. 35; Cv. IV. 10). Cor. The area of a parallelogram, divided by the base, gives the altitude; and the area, divided by the altitude, gives the base. EXAM. 29. Find the area of a parallelogram whose base is 1 kilometer, and altitude 1 centimeter. b =1000 meters. a - - meter... ab = 10 square meters. Ans. 35, Given one side and the perpendicular upon it from the opposite vertex, to find the area of a triangle. Rule: Take half the product of the base into the altitude. Formula: A = -ab. 34 MENSURATION. Proof: A triangle is equivalent to half a parallelogram having the same base and altitude. Ww. 324; (Eu. I. 41; Cv. IV. 13). Cor. 1. If twice the number expressing the area of a triangle be divided by the number expressing the base, the quotient is the altitude; and vice versa. Cor. 2. Two A's or C7's, having an equal A, are as the products of the sides containing it. EXAM. 30. One side of a triangle is 35-74 meters, and the perpendicular on it is 6'3 meters. Find the area. 2-b = 17-87 meters... ab =113581 square meters. Ans..J 36. Given the three sides of a triangle, to find the area. Rule: From half the sum of the three sides subtract each side separately; multiply the half sum and the three remainders together: the square root of the product will be the Proof: By 4, 2= b2 + - 2 bj, whence b2 - - a 2b THE MEASUREMENT OF PLANE AREAS. 35 By 2, h,2 =2 _ 2_ _ (b2 +2 - a2)2 4 b2 whence, 4 b'22 = 4 b2c2 - (b2 + c2 - a2)2.-. 2 bh = X/4 b2- (2 + c2 - a2)2... 2 b= V(2 be + b2 + c- a2) (2c - b2 _ C2 + a2)... 2bh = /(a + b + c) (b + c- a) (a + b- c) (a-b + c). But, by 35, 2 bA equals four times the area of the triangle. Cor. To find the area of an equilateral triangle, multiply the square of a side by 0.433+. EXAM. 31. Find the area of an isosceles triangle whose base is 60 meters and each of the equal sides 50 meters. Here, from last formula in Proof, 2bh= b /(2 a + b)(2a - b) = (60 V/0 x 40 = 6O0 l x00 x 4... 2bh60 x 40 x 2.. Area 1200 square meters. Ans. B A D C 37. To find the radius of the circle inscribed in a triangle. Rule: ]Divicle the area of the triawnge by hacf the sum of its sides. Formula i r -. s 36 MENSURATION. Proof. By 35, area of BOC = a_, 2 br area of COA = b, l area of AOB= cr. 2.. by addition, A - (a + + c). Cor. The area of any circumscribed polygon is half the product of its perimeter by the radius of the inscribed circle. EXAM. 32. Find radius of circle inscribed in the triangle whose sides are 7, 15, 20. Here s = 21,.. A= 21 x 14 x 6 = /3. 7. 7. 2. 2. 3..A. =3x7x2=42... -2. Ans. 38. To find the radius of the circle circumscribing a triangle. Rule: JDivide the product of the three sides byfour times the area of the triangle. Formula: -abe 4A Proof: In any triangle the rectangle of two sides is equivalent to the rectangle of the diameter of the circumscribed circle by the perpendicular to the base from the vertex. Ww. 300; (Eu. VI. C.; Cv. III. 65)..ac.=2 9h. l_ /ac _ abc 21 2kbh THE MEASUREMENT OF PLANE AREAS. 37 Cor. The side of an equilateral A, b - V/3 = 2 r /3. EXAM. 33. Find radius of circle circumscribing triangle 7, 15, 20. Here abe= 2100, A 42..'.* ~ 2100 _121. Ans. 168 39. To find the radius of an escribed circle. Rule: Divide the area of the triangle by the difference between half the sum of its sides and the tangent side. A a Proof: Let rl denote the radius of the escribed circle which touches the side a. The quadrilateral O1BAC may be divided into the two triangles, O1AB and OA C;.. by 35, c b *y* ' 35its area= rl + r,. 2 2 But the same quadrilateral is composed of the triangles O1BC and ABC; a.'. its area = -r + A. 21 38 MENSURATION. Thus, c b a -r + - r _ a A. 2 2 2 c + b-a 2 rA. A s-a In the same way, s-b' s-C A, Cor. 1. Since, by 37, r = -, therefore, rrillr, - - A4 ---A A2, by 36. ' s(s -a) (s - b) (s - c) Thus, A = Vrr rr. Cor. 2. 1 1 1 1 r, r,2 3 r EXAM. 34. Find rl, r2, 93, when a =7, b =15, c- 20. 42 42 42 9-14 =3, r-2 =7, 7 1 -42. Ans. 14 6 1 XII. A trapezoid is a quadrilateral with two sides parallel. Cor. A triangle is a trapezoid one of whose parallel sides has become a point. 40. To find the area of a trapezoid. Rule: Multiply the sum of the parallel sides by half their distance apart. 2o+ Y Formula: T= x' 2 THE MEASUREMENT OF PLANE AREAS. 39 Proof: Let E be the midpoint of the side AB. Through B and E draw BII and GF parallel to CD. Then A AEG = A BEE Vw. 107; (Eu. I. 26; Cv. I. 78).. Trapezoid ABCD )= g GFCD. A.. That is, by 34, T= GCD X x; where x is s I the distance of BC from AD. But; B DET= BC, and HG = AG.. D. _ AD +_BC 2 ' and calling AD, yl, and BC, y2, we have v 2 GDl = EK= Y + —c2. 2 K 1c T.= yl + y2 xx. / 2 K Cor. The area of a trapezoid equals J the distance apart of the parallel sides multiplied by the line joining the midpoints of the non-parallel sides. EXAM. 35. Find the area of a trapezoid whose 11 sides or bases are 12-34 meters apart, and 56'78 meters and 90 -meters long. 56-78 + 90 = 146-78 12-34- 2= 6-17 102746 14678 88068 905-6326 square meters. Ans. XIII. COnRDINATES OF A POINT. The ordinate of a point is the perpendicular from it to a fixed base line or axis. The corresponding abscissa is the distance from the foot of this ordinate to a fixed point on the axis called the origin. 40 4MENSUrATION. The coordinates of any point are its abscissa x and its ordinate y. XIV. If to any convenient axis ordinates be dropped from the angular points of any polygon, the polygon is exhibited as an algebraic sum of trapezoids, each having one side perpendicular to the two parallel sides, and hence called. right trapezoids. If triangles occur, as 1 2, 6E5, they are considered trapezoids, yj and y5 being zero. 2 1 5 —. 41. To find the sum of any series of right trapezoids. Rule: eMultiply the distance of EACH intermediate ordinate from the first by the difference between its two acjacent ordinates, always subtracting the one following from the one preceding in order along the broken line. Also multiply distance of last ordinate from first by the sum of last two ordinates. Halve the sum of these products. Formula: fT,-24 [(2-X1) i(y-y3) +(X3-X1) (y2-y4) +.. 4r(X-Xl) (Yn-l-Yn+l)+(Xn+l~Xl) (Y-+Yn+l)]Proof: With 0 as origin, the area of the first trapezoid, by 40, is (x2 - x) y1+ / and of the second is (x3 - ) y2 + y3 2 o 2 THE MEASUREMENT OF PLANE AREAS. 41 Adding the two, we have T + 7 = i-2 [(R2 - r) (Y1 + Y2) + (X3 - X2) (Y2 + Y3)1 -Performing the indicated multiplications, X2y2 is cancelled by -x2y, and 21 + 2 = ~ [x2Y1 - rly - 1y2 + y2 + X 3s 3 - 2y3]. *~ T1 + T2 = I [(- 1) (Y --.v3) + (3 - x1) (Y2 + Y3)], since here xy3 is balanced by- x1y3. Thus we have proved our rule for a pair of trapezoids. Taking three, we get, by 40, T3= (y[-y3) + 3 -24 - *' * T+ T,+ T3= 21[(2-) (Y-Y3) + (3- X1) (Y2+ y3)1]+ 421 [(- 3) (y3+ 4)] As before, replacing the balancing terms x3y3-x y3 by ~1y4 - x1y4, this becomes T,+ T, + T3= 2 [(R2- X1)(Y1-Y3) + (3-x1) (y2-y4) + (X4-X1) (Y3 + Y)1. This proves the rule for three trapezoids; and a generalization of this process proves that if the rule is true of a series of n trapezoids, it is true of n + 1. For, by 40, the area of the (n + l)th trapezoid 1 3?22 I r, I 1 1 I 3 /, lY '!y ~ 7Y 0 x, T +1 = ('X+2 - rn +i) 2y a+ i +y; and, adding this to the first n trapevzoids, as given by formula, therefore v=n+l T = [(X2 - X) (Y1- 3) +.... += (xn+1 - i (- )] + [+2 - X) (n + 1)] + - [(+- ) ( + Yn+2)1. *42 MENSURATION. Replacing x,,+l y,+ - Xn+l yn+ by the balancing terms XlYn+ 2 —Xln+2, this becomes v=n+l Z V= - [(X2 - X1) (y1 - Y3) +..... v=l + (ni- (Xn+1 - 1) ( - yn+2) + (Xn+2 - X2) (ynil + y,,+2)]. The same method proves that if the formula applies to n trapezoids, it must apply to n - 1. Therefore, the rule is true for any and every series whatsoever of right trapezoids. EXAM. 36. Find the right portion of a railroad crosssection whose surface line 3 breaks twice, at the points ~1 — I-;4 4 (X2, Y2), (X3, 3), to the right [ of center line; (origin being c. / Ir on grade in midpoint of roadr bed). - -- -- -- This asks us to find the 0 x, b 5 r ' ~0 2 Ib 5 r' sum of right trapezoids corresponding to the five points (0, c), (x2, Y2), (3, 3), (b + rl r9), (b, 0)... by formula, v=4 - [X2 (C - y/3) + x8 (2- r) + (b + 9') y, + bfr]. Ans. In the same way, the portion to the left of center line, whether without breaks, or with any number of breaks, is given by our formula, which thus enables us at once to calculate all railroad cross-sections, whether regular or irregular. EXAM. 37. To find the area of a triangle in terms of the coordinates of its angular points. THE MEASUREMENT OF PLANE AREAS. 43 Here are three trapezoids, and consequently, four points; but A is both 1 and 4, so X4 - X1 = 0 and the formula becomes v=3 A = T, = [(x2 - x,) (y, - Y3) + (x, - x) (y,- y1)]..'. A= —i [xiy2-X2y1 + X2zy3- y2 + X3y1-X 3]. Ans. Notice the symmetry of this answer. B, C X~~~~~~~~~I~ ~ I For practical computation this is written 2 A = 1 (3 - Y) + X2 (Y1 - 3) + r 3(Y3 - Y, or 2 = Y1 (x - r3) + 2 (X3 - 1) + y3 (T1 - X2)To insure accuracy, reckon the area by each. 42. To find the area of any polygon. Rule: T/ake half the sum of the products of the abscissa of each vertex by the difference between the ordinates of the two adjacent vertices; always making the subtraction in the same direction around the polygon. Formula for a polygon of n sides: N= 2 x(y, - y2) ~ 2,(y, -2/3) + x(y,2- y4).+ * + X-(Yj- - Y1) Proof: This is only that special case of 41, where the broken line, being the perimeter of a polygon, ends where it began. 44 4MENSURATION. Join the vertex 1 of the polygon 1, 2, 3, 4f,.... n,, 1, with the origin 0. Then the area enclosed by the perimeter is the same, whether we consider it as starting and stopping at 1 or at O. But, under the latter supposition, though we 4 2 n-i xoyo xi have n + 3 points, the coordinates (x0, yo) of the first and last are zero, and the second (xl, yi) is identical with the point next to last; so that formula 41 becomes [1(o - Y2) + X (y,1- ) + x3 (y2- ) +.* + (y3n(n-1 Y) + X(y,- 0)]. NOTE. No mention need be made of minus trapezoids, since the rule automatically gives to those formed by the broken line while going forward, the opposite sign to those formed while going backward. Our expression for the area of any rectilinear figure is the difference between a set of positive and an equal number of negative terms. If this expression is negative when the angular points are taken in the order followed by the hands of a watch, then it is necessarily positive when they are taken in the contrary sense, for this changes the order in every pair of ordinates in the formula. Observe that each term is of the form xy, and that there is a pair of these terms, with the minus sign between them, for each vertex of the figure. Thus, for the vertex (xmy,), THE MEASUREMENT OF PLANE AREAS. 45 we have the pair X,((ym-_ - Y+); or, pairing those- terms which have the same pair of suffixes, for every vertex nz, we have (xm+1Y —xyYm+l). Hence, for twice the area write down the pair xy —xy for each vertex, and add symmetrically the suffixes, 1,2 2,1; 2,3 3,2; 3,4 4,3;....,1 1, n. Thus, for every quadrilateral, 2 Q = xy/2 - x2yl + x2y3 - x3Y2 + X3y4 - x43 + x4Y1 - xy4. But, if any point of perimeter be to the left of origin, or if, to shorten the ordinates, the axis be drawn across the figure, then one or more of the coordinates will 1 be essentially negative. Thus, if in o a quadrilateral, we take for axis a diagonal, then X' =, y,=1, y3,=o, and 2 Q = - x3/2 + x3yJ4 -Here, y4 being essentially negative, the two terms have the same sign, and give the ordinary rule: 43. To find the area of any quadrilateral. Rule: Multiply half the diagonal by the sum of the perpendiculars upon it from the opposite angles. EXAM. 38. The two diagonals of a quadrilateral measure 1-492 and 37-53 meters respectively, and are I_ to one another. Find the area. By 43 1.492 x 37-53 55-99476 Area- - - 2 2 - 27.99738 square meters. Ans. 46 MENSURATION. EXAM. 39. Find the area of the polygon 1234567891, the coordinates of whose angular points are (0, 90), (30, 140), (110, 130), (80, 90), (84, 80), (130, 40), (90, 20), (40, 0), (35, 70). By 42, 21V= 30 x-40 +110x50 + 80 x 50 + 84X 50 +130 60 + 90 x 40 + 40 X - 50 + 35 x - 90 = 18750... area -9375. Ans. REMARK. The result of any calculation by coordinates may be verified by a simple change of origin. If the origin is moved to the right through a unit of distance, then the numerical values of all positive abscissae will be diminished by one, and all negative abscissae increased by one. Thus, to verify our last answer, move the origin thirty units to the right, and the question becomes EXAM. 40. To calculate the area of polygon whose coordinates are: x Y 1 -30 90 2 0 140 3 80 130 4 50 190 5 54 80 6 100 40 7 60 20 8 10 0 9 5 70 By 42, twice the area equals the sum of 30x-70= 2100 80 x 50= 4000 50x 50= 2500 54 x 50= 2700 100 x 60= 6000 60X 40 2400 19700 10 X- 50 and 5 X - 90 =- 950 18750.. area = 9375, as before. EXAM. 41. From the data of Exam. 40 construct the figure. Choose a convenient axis and origin, noticing that the THE MEASUREMENT OF PLANE AREAS. 47 polygon will lie wholly above the axis, since there are no minus ordinates. Then, to find first vertex, measure off on the axis 30 units to the left of origin, and at the point thus determined, erect a perpendicular 90 units in length. Its 2 extremity will be the angular point numbered 1. The extremity of a perpendicular at origin 140 units long gives vertex 2, and an ordinate 130 long from a point on axis 80 units to the right of origin gives 3. When all the angular points have been thus determined, join them by straight lines in their order of succession. 48 MIENSURATION. 44. Given the area and one side of a figure, and the corresponding side of a similar figure, to find its area. Rule: 3llfltiply the given area by the squared ratio of the sides. Formula: A - A2a a2 Proof: The areas of similar figures are to one another as the squares of their like sides. Ww. 343; (Eu. VI. 20; Cv. IV. 23). D K X M^ I B If A t JB Cor. The entire surfaces of two similar solids are proportional to the squares of any two homologous lines. EXAM. 42. The side of a triangle containing 480 square meters is 8 meters long. Find area of a similar triangle whose homologous side is 40. A = 480 x 1600 = 480 x 25 = 12000 square meters. Ans. 64 XV. Magnitudes which can be made to coincide are congruent. Magnitudes which agree in size, but not in shape, are equivalent. THE MEASUREMIENT OF PLANE AREAS. 49 XVI. A regular polygon is both equilateral and equiangular. The bisectors of any two angles of a regular polygon intersect in a point equidistant from all the angular points of the polygon, and hence also equidistant from all the sides, / and at once the center of an inscribed and a circumscribed / / circle. Joining this center to every — / angle of the polygon cuts it up into congruent isosceles triangles. Hence the area of the regular - ~ polygon is the area of any one of these triangles multiplied by the number of sides of the polygon. 45. To find the area of a regular polygon. Rule: 3ultiply toyelher one sicde7, tk/c pepceniciularc fromi/ the E K D centeqr, and ihalf he tc?6emtber of sides. Or, in other \words:, __ Take ha/c' the pi'odlct of perimzeter b7y? cpotheem. Formula: N clnV- A G B 2 2 EXAM. 43. The side of a regular hexagon is 98 centimeters, and its apothem 84-87 centimeters; find its area. Area =3 x 98 x 84-87 - 24951-78 square centimeters. Ans. 50 MENSURATION. 46. By the aid of a table of polygons, to find the area of any regular polygon. Rule: Multiply the squctre of one of the sides of the polygon by the area of a similar polygon whose side is unity. Formula: N= 2 N1. Proof: This follows from 44, all regular polygons of the same number of sides being similar. TABLE OF REGULAR POLYGONS. Number z Area when Number Area in Terms of of ides. Nae. Side = 1. of Sides. Square on Side. 3 Triangle 0.4330127 15 17-642363 4 Square 1.0000000 16 20-109358 5 Pentagon 1.7204774 20 3-1-568757 6 Hexagon 2.5980762 24 45-574525 7 Heptagon 3.6339124 25 49-473844 8 Octagon 4.8284271 30 71-357734 9 Nonagon 6.1818242 32 81-225360 10 Decagon 7.6942088 40 127-062024 11 Undecagon 9.3656399 48 138-084630 12 Dodecagon 11-1961524 EXAM. 44. The side of a regular meters; find its area. hexagon is 98 centi Area- 98 x 98 X 2.5980762 - 24951-78 square centimeters. Ans. EXAM. 45. If the side of a regular decagon is 0.6 meters, its area is 0.6 x 0.6 x 7.6942088 = 2.76991524 square meters. Ans. THE MEASUREMENT OF PLANE AREAS. 51 ~(H). AREAS OF PLANE CURVILINEAR FIGURES. 47. To find the area of a circle. Rule: Mnultiply its squared radius by 7r. Formula: 0= - rv. Proof. If a regular polygon be circumscribed about the circle, its area, by 45, is N= rpn; and, by 14, as n increases, p,, decreases toward c as limit, and N toward 0. But the variables Tand fn are always in the constant ratio -2'; therefore, by 13, their limits are in the same ratio, and we have = 2re. By 18, c= 2rr. Therefore, 0 =A rr. EXAM. 46. Find the area of a circle whose diameter is 7-5 meters. Here 2 = 14-0625..-. --- 44-178+ square meters. Ans. 48. To find the area of a sector. < Rule: Jultiply the length of the \ aric by half the racdius. Formula: S= 2 -r u2. Proof: Ww. 382; (Cv. V. 44); MENSURATION. or, as follows: By Eu. VI. 33, S: 0(::: c:: u: 2r. 01 Ott c 2 wr. S — rcl _ r27'1 c 27.. S= lr = U. EXAM. 47. Find the area of a sector whose arc is 99-58 meters long, and radius 86-34 meters. 99-58 x 43-17 = 4298-8686 square meters. Ans. EXAM. 48. Find the area of a sector whose radius is 28 centimeters, and which contains an angle of 50~ 36! Here, by 29, z 0883~+ r2 —282= 784 3532 706,4 6181 2) 692.272.'. S= 346-136 square centimeters. Ans. 49. To find the area of a segment less than a semicircle. Rule: e From the sector hav/ing the same arc as the seg/ / — |\\H merit, subtract the triancgl /,-'\\ formeed by the cho rd and the A two radii from its extremitics.,,-'^: / Formula: \X~~x,,~~,, C,'T / + k)+ 4 (l - k) 4A Proof: The segment AHB is the difference between the sector ATBC and the triangle ABC. THE MEASUREMENT OF PLANE AREAS. 53 By 48, AHB C =- r. By 35, ABB =ABxCD =k(r- h). G = -8-A= -r - 2 (r - h)..2. 2G = lr-kr + kh. But lD x DF = AD2. Ww. 307; (Eu. VI. 13; Cv. III. 47)... h(2r- h) =k. 4 22 + 7b2 2h Substituting this value of r in the expression for 2G, we obtain 22 + 72 + 1 k 2( - ) + h1 - h2k + 2h2k 2G (I- k) m + kh 2h " 2h.1 k2( _- k) i h2(1 k) 4h Cor. The area of a segment of a circle is equal to half the product of its radius and the excess of its arc over half the chord of double that arc. For sector AHB C = Ir, and AABC= — rx BL... segment AHB = r(l - BL). Approximate Rule for Segment: Take two-thirds the product of its chord and height. Approximate Formula: G - = k. EXAM. 49. If the chord of a segment is r x 0-959851+, and its height is r X 0122417+, then an approximation to its area is -.r2 X 0.959851+ X 0.122417+ = -r2 X 0-117502+- = r2 X 0.0783+. 54 - MENSURATION. But if, also, we can measure the arc, and here find it equal to radius, then r2(0 122417+)2(r+rxO 959851+)+-L 2(0.959851+)2(r-rxO 959851+) 4rX 0 122417+.*. G=-r2XO 122417++1r2xO 117502++- r2x7.52603+-T-r2x722387-.G. G=-r2XO.239919++-1Lr'1 2 030216+.-. G=r2 (0 239919++0.07554+)..G=-r2X0'315499+. -. X 0.07886+. Ans. Proceeding directly by Rule 49, instead of Formula 49, we here get S= r2, and A = (r X 0.959851+) (r - r X 0-122417+).. = r X 0-4799255+ X r X 0-877583-.'. A = r2X 0.421174.. G =S- A =r2X 007883. Since, in this example, arc r,.. G is, in any 0, the segment whose ~ is p. 50. A circular zone is that part of a circle included between two parallel chords, and may be found by taking the segment on the shorter chord out of that on the longer. 51. A crescent is the figure included between the corresponding arcs of two intersecting circles, and is the difference between two segments having a common chord, and on the same side of it. 52. To find the area of an annulus; that is, the figure included between two concentric circumferences. Rule: ~lfultiply the sum of the two radii by their difcerence, and the product by 'r. Formula A = (1 -+ -2) (r' -- r) 7r. THE MEASUREMENT OF PLANE AREAS. 55 Proof: By 47, the area of the outer circle is r2=r, and of the inner circle r2%r. Therefore, their difference, the annulus, is (ri - r1) 7r. Cor. The area of the annular figure will be the same whether the circles are concentric or not, provided one circle is entirely within the other. If the two circles intersect, they form two tunes, one on each side of the common chord, and the difference of the two lunes will always be equal to the annulus formed by the same circles. EXAM. 50. The radii of two concentric circles are 39 meters and 11-3 meters. Find the area of the ring between their circumferences. Here A = 503 x 27.7 x r -- 4377-2+ square meters. Ans. 53. To find the area of a sector of an annulus. Rule: JMultiply the suvm of the bounding arcs by half the distance between them. Formula: S. A. = - (i1 — + 12). 56 MENSURATION. Proof. The sectorial area ABEED is the difference between the sector ABC and the sector CDE... by 48, / ^ ---.S\ i.A. = -2 (ra+i ), —21 rl, ~ V /\ I Now, since l, and 12 are arcs subtending the same angle at C, l ^'2,./. by IX, - 1.1 1. 11.-~~. Id2hl,-lr r. Substituting, we have S. A. -i (h1 ~ /212) = 7(6l 1- 12) Co)r. By comparison with 40, we see an annular sector is equivalent to a trapezoid whose parallel sides equal the arcs, and are at the same distance from one another. EXAM. 51. The upper arc of a circular arch is 35-25 meters; the lower, 24-75 meters; the distance between the two is 3-5 meters. How many square meters are there in the face of the arch? Here S. A. = 1'75 x 60 = 105 square meters. Ans. XVII. CONICS. If a straight line and a point be given in position in a plane, and if a point move in the plane in such a manner that its distance from the given point always bears the same ratio to its distance from the given line, the curve traced out by the moving point is called a conic. The fixed point is called the focus, and the fixed line the directrix of the conic. THE MEASUREMIENT OF PLANE AREAS. 57 When the ratio is one of equality, the curve is called a pI-arabola. 54. To find the area of a parabolic segment; that is, the area between any chord of a parabola and the part of the curve intercepted. Rule: cTake two-thircs the product of the chorcd by the height of the segment. Formula: J — h3 k. Pr)oof. A parabolic segment is two-thirds of the triangle made by the chord and the tangents at its extremities. If AB, AC, be two tangents to a parabola, to prove that the area between the curve and the chord BC is two-thirds of the triangle ABC. Parallel to BC draw a tangent DPE. Join A to the point of contact P, and produce AP to cut the chord BC at N. By a property of the parabola, deducible from its definition, AP = PN..B. C-B 2 DE. Ww. 276 & 279; (Eu. VI. 2 & 4; Cv. III. 15 & 25)... by 35, "A BPC= 2ADE. This leaves for consideration the two small triangles PIDB, PEC, each made by a chord and two tangents. Wit h each proceed exactly as with the original triangle: e.g., draw the tangent FQCG parallel to PB; join DQ, and produce it to iMr; then DQ= QU.M.P. P =2 FG... APQB=2-FDG. 58 MENSURATION. This leaves four little tangential triangles, like PFQ. In each of these draw a tangent parallel to the chord, etc., and let this process be continued indefinitely. Then the sum of the triangles taken away within the parabola is double the sum of the triangles cut off without B C it. But the sum of the interior triangles approaches, as its limit, the parabolic segment. For the triangle BPC, since it is half of ABC, is greater than half the parabolic area BQPC, and so successively with the smaller interior triangles. Therefore, the difference between the parabolic segment and the sum of these triangles can be made less than any assignable quantity. Ww. 198; (Eu. XII., Lemma; Cv. V. 28). Therefore, the constant segment is, by definition V., the limit of this variable sum, THE MEASUREMENT OF PLANE AREAS. 59 Again, each outer triangle cut off is greater than half the area between the curve and the two tangents; c. y., AD-E, being half the quadrilateral ABPC, is more than half the area ABQPC. Therefore, the limit of the sum of the outer triangles is the area between the curve and the two tangents AB, AC. But these two variable sums are always to each other in the constant ratio of 2 to 1. Therefore, by 13, their limits are to each other in the same ratio, and the parabolic segment is two-thirds its tangential triangle. But the altitude of this triangle is twice the height of the segment... = hk, and J= a hk. 55. To find the area of an ellipse. Rule: Mulztipzly the pproduct of the semi-axes by -r. Formula: E-= ab7r. Proof: Let ADA']' be a circle of which AC, CD are radii at right angles to one another. In CD let any point B be taken; then, if this point move so as to cut off from all ordinates of the circle the same part that BC is of DC, the curve traced is called an ellipse. In one quadrant of the circle take a series of equidistant ordinates, as Q1 P-1 M1, Q P2 ]J,, Q3 Ps3 23, etc. Draw Pi Rl, QlR', etc., parallel to the axis AA' Then, by 32, area R^: area R',1M1:: Pkli: Q,1M,:: BC: AC; and each corresponding pair being in this constant ratio,.'. the sum of the rectangles rPf is to the sum of 'JrI as BC AC. But the sum of BRil differs from one-quarter 60 IMENSURATION. of the ellipse by less than the area BM1, which can be made less than any assignable quantity by taking Cl1I, the common distance between the ordinates, sufficiently small. Ds Q D' Hence, A'BC is the limit of the sum of the rectangles lIR; and, in the same way, the quadrant of the circle is the limit of the sum of kR~M. Therefore, by 13, E BC b 0 AC a Ob _ c67 cb E = " = abr. a a Cor. The area of any segment of an ellipse, cut off by a line parallel to the minor axis, will be to the corresponding segment of the circle upon the major axis in the ratio of b to a. EXAM. 52. Find the area of an ellipse whose major axis is 61.6 meters, and minor axis 44-4 meters. E= 30.8 x 22.2 x 3.14159 -2148-09+ square meters. A1s. CHAPTER IV. THE MEASUREMENT OF THE AREAS OF BROKEN AND CURVED SURFACES. XVIII. A polyhedron is a solid bounded by polygons. A polyhedron bounded by four polygons is called a tetrahedron; by six, a hexahedron; by eight, an octahedron; by twelve, a dodecahedron; by twenty, an icosahedron. The faces of a polyhedron are the bounding polygons. If the faces are all congruent and regular, the polyhedron is regular. The edges of a polyhedron are the lines in which its faces meet. The summits of a polyhedron are the points in which its edges meet. A section of a polyhedron is a polygon formed by the intersection of a plane with three or more faces. A convex figure is such that a straight line cannot meet its boundary in more than two points. 56. The number offaces and summits in any polyhedron taken together exceeds by two the number of its edges. Formula: 0 + 2 = (E + 2. Proof: Let e be any edge joining the summits aft and the faces AB, and let E vanish by the approach of P to a. If A and B are neither of them triangles, they both re 62 MENSURATTION. main, though reduced in rank and no longer collateral, and the figure has lost one edge E and one summit,. If B is a triangle and A no triangle, B vanishes with E into an edge through a, but A remains. The figure has lost two edges of B, one face B, and one summit 3. If B and A are both triangles, B and A both vanish with E, five edges forming those triangles are reduced to two through a; and the figure has lost three edges, two faces, and the summit f. In any one of these cases, whether one edge and one summit vanish, or two edges disappear with a face and a summit, or three edges with a summit and two faces, the truth or falsehood of the equation remains unaltered. By causing all the edges which do not meet any face to vanish, we reduce the figure to a pyramid upon that face. Now, the relation is true of the pyramid; therefore it is true of the undiminished polyhedron. (I). PRISM AND CYLINDER. XIX. A prism is a polyhedron two of whose faces are congruent, parallel polygons, and the other faces are parallelograms. AREAS OF BROKEN AND CURVED SURFACES. 63 The bases of a prism are the congruent, parallel polygons. A paralllepiped' is a prism whose bases are parallelograms. A normal is a straight line perpendicular to two or more non-parallel lines. The cltitude of a prism is the normal distance between the planes of its bases. A right prism is one whose lateral edges are normal to its bases. 57, To find the lateral surface or mantel of a prism. Rule: htil4ply a lateral edge by the peirimeter of a eight section. Formula: P = p. Proof: The lateral edges of a prism are all equal. The sides of a right section, being perpendicular to the lateral edges, are the altitudes of the parallelograms which form the lateral area of the prism. Cor. The lateral area of a 9ight prism is equal to its altitude multiplied by the perimeter of the base. EXAM. 53. The base of an oblique prism is a regular pentagon, each side being 3 meters, the perimeter of a right section is 12 meters, and the length of the prism 14 meters. Find the area of the whole surface. 64 MENSURATION. By 46, the area of the pentagonal base is 9 x 1.7204774 = 15-4842966. Doubling this for the base and top together, and adding the lateral area of the prism, which, by 56, is 12 x 14=168, the total surface = 168 + 30-9685932 = 198-9686- square meters. Ans. XX. A cylinclric surface is generated by a straight line so moving that every two of its positions are parallel. The generatrix in any position is called an element of the surface. A cylindcer is a solid bounded by a cylindric surface and two parallel planes. The axis of a cylinder is the straight line joining the centers of its bases. A truncated cylinder is the portion between the base and a non-parallel section. 58. To find the curved surface or mantel of a right circular cylinder. Rule: 2fultiply its length by the circumference of its base. Formula: C = cl= 2rrrl. ----- ~First Proof: Imagine the curved * )} surface slit along an element and then spread out flat. It thus becomes a rectangle having for one side the circumference and for the adjacent side the length of the cylinder. Seconcz Proof' Inscribe in the right cylinder a right prism having a regular polygon as its base. Bisect the AREAS OF BROKEN AND CURVED SURFACES. 65 arcs subtended by the sides of this polygon, and thus inscribe a regular polygon of double the number of sides, and construct on it, as base, an inscribed prism. Proceeding in this way continually to double the number of its sides, the base of the inscribed prism, by 14, approaches the base of the cylinder as its limit, and the prism itself approaches the cylinder as its limit. But, by 56, p= p, and always the variable P bears to the variable p the constant ratio 1. Therefore, by 13, their limits are in the same ratio, and c. Cor. 1. The curved surface of a truncated circular cylinder is the product of the circumference of the cylinder by the intercepted axis. For, by sym- f ---~jmetry, substituting an oblique for a right section through the same point of the axis alters neither w ii the curved surface nor the volume, since the solid between the two sections will be the same above and below the right section. Cor. 2. The curved surface of any cylinder on any curve equals the length of the cylinder multiplied by the perimeter of a right section. EXAM. 54. Find the mantel of a right cylinder whose diameter is 18 meters and length 30 meters. C-= 30 X 18 X 3 14159 = 1696-4586 square meters. Ans. 66 MENSURATION. (J). PYRAMID AND CONE. XXI. A regular pyramid is contained by congruent isosceles triangles whose bases form a regular polygon. A conical surface is generated by a straight line moving so as always to pass through a fixed point called the vertex. A cone is a. solid bounded by a conical surface and a plane. The frustum of a pyramid or cone is the portion included between its base and a cutting plane parallel to the base. o 59. To find the area of the lateral surface or mantel of a regular pyramid.:Rule: Multiply the perimeter of the base by half the slant height. Formula: Y - h2p...... —T --- —A P.roof: The altitude of each of the _^B equal isosceles triangles is the slant height of the pyramid, and the sum of their bases is the perimeter of its base. EXAM. 55. Find the lateral area of a regular heptagonal pyramid whose slant height is 13-56224 meters, and basal edges each 1- meters. One quarter of 13-56224 is 3-39056. Adding these and dividing their sum by 2 gives 8-4764 for the area of one triangular face. The lateral area is 7 times this, or 59-3348 square meters. Ans. AREAS OF BROKEN AND CURVED SURFACES. 67 60. To find the area of the curved surface or mantel of a right circular cone. 0o Rule: Multiply the circumference of its base by half the,lcant he'iht. i Formula: K - - ch =- rrh. f First Proof: The distance A - from the vertex of a cone of revolution to each point on the circumference of its base is the slant height of the cone. Therefore, if the surface of the cone be slit along a slant height and spread out flat, it becomes the sector of a circle, with the slant height as radius and the circumference of cone's base as arc.. by 48, its area is - ch. Second Proof: About the base of the cone circumscribe a regular polygon, and join its vertices and points of contact to the vertex of the cone. Thus is circumscribed about the cone a regular pyramid whose slant height equals the slant height of the cone. By drawing tangents, cir- \ - cumscribe a regular polygon of double the number of sides, and construct on it, as before, a circumscribed regular pyramid. Thus proceeding continually to double the number of sides, the base of the circumscribed pyramid, by 14, approaches the base of the cone as its limit, and the pyramid itself approaches the cone as its limit. 68 MENSURATION. But, by 58, Y-p, and always the variable Y has to the variable p the constant ratio - h... by 13, their limits are in the same ratio, and K ch. Cor. 1. In the Proof of 47 we find O = icr... the slant height of a right circular cone has the same ratio to the radius of the base that the curved surface has to the base, or K:: h:. Cor. 2. Calling X the sector angle of the cone, we have A: 360:: r: h. EXAM. 56. Given the two sides of a right-angled triangle. Find the area of the surface described when the triangle revolves about its hypothenuse. Calling a and b the given altitude and base, and x the length of the perpendicular from the right angle to the hypothenuse, by 59, the area described by a is 7rxa, and described by b is 7rxb. Thus the whole surface of revolution is 7r(Ca + b)x. But a: x:: a2/ b2: b. Eu. I. 47 & VI. 8. ab 7 (a + ) abA x'=,6 and r(a + b)x L+b) Ans. Va2/a + (c' + Y)i 61. To find the lateral surface or mantel of the frustum of a regular pyramid. AREAS OF BROKEN AND CURVED SURFACES. 69 Rule: ltaifiply the slant height of the frustum by half the sum of the perimeters of its bases. Formula: F -= h (pi + 192). Proof: The base and top being similar regular polygons, the inclined faces are congruent trapezoids, the height of each being the slant height of the frustum. If n be the number of faces, by 40, area of each face = h 1 +.'. area of lateral surface = F = - h (p1 + P) A B EXAM. 57. Find the lateral area of a regular pentagonal frustum whose slant height is 11-0382 meters, each side of its base being 2- meters, and of its top 1~ meters. The sum of a pair of parallel sides is 1-. 11-0382 x 13 = 143-4966. 143-4966 - 6= 23-9161, the area of one trapezoidal face. The lateral area is five times this, or 119 5805 square meters. Ans. 62. To find the curved surface or mantel of the frustum of a right circular cone. 70 I MENSURATION. Rule: Multiply the slant height of the frustum by half the sum of the circumferences of its bases. Formula: F = - h (c1 + 2)=-h (r + r2). Proof: Completing the cone and slitting it along a slant height, the curved surface of the frustum develops into the difference of two similar sectors having a common angle, the arcs of the sectors being the circumferences of the bases of the frustum. By 53, the area of this annular sector = F = h (c1 + c2). EXAM. 58. Find the mantel area of the frustum of a right cone whose basal diameter is 18 meters; top diameter, 9 meters; and slant height, 171-0592 meters. 3'1416 X 9 28-2744 = circumference of top. Twice top circumference = 56-5488 = circumference of base. Half their sum is 42-4116, and this multiplied by the slant height 171-0592, gives, for the curved surface, 7254-89 square meters. Ans. 63. To find the curved surface of a frustum of a cone of revolution. Rule: Multiply the projection of the frustum's slant height on the axis by twice rt times a perpendicular erected at the midpoint of this slant height and terminated by the axiss. Formula: F - 2,-aj. Proof. By 62, the curved surface of the frustum whose slant height is PR and axis 2MJC is F= r x PR (P + RN). AREAS OF BROKEN AND CURVED SURFACES. 71 But, by 40, Cor., Pl + RN= 2QO... F=2rx PRxQO. But the triangle RPL is equiangular to CQO, since the three sides of one are perpen- p M dicular to the sides of the other. ---.. Px QO=PL xQC. Ww. 279 & 259; (Eu. VI. 4 & 16; - ------ Cv. III. 25 & 5)..F. F=2ir(LPx CQ)=27r(fMNx CQ) -N /! __ -_-_, = 2rj a. Coor. This remains true, if either PM or RN vanish, or if they become equal; that is, true for a cone or cylinder of revolution. 2 (K). THE SPHERE. XXII. A sphere is a closed surface all points of which are equally distant from a fixed point within called its center. A globe is the solid bounded by a sphere. 64. To find the area of a sphere. Rule: Multiply four times its squared radius by ir. Formula: H- 4r2 r=. Proof: In a circle inscribe a regular polygon of an even number of sides. Then a diameter through one vertex passes through the opposite vertex, halving the polygon symmetrically. Let PR be one of its sides. Draw PM, RN perpendicular to the diameter BD. From the center C the perpendicular CQ bisects PR. Ww. 183; (Eu. III. 3; Cv. II. 15). Drop the perpendiculars PL, QO. 72 MENSURATION. Now, if the whole figure revolve round B s as axis, the semicircle will generate a sphere, while each side of the. inscribed polygon, as PR, will gene-p - rate the curved surface of the frustur of a cone. By 63, this Pi Y 1 'F = 2 MlF21riNT X CQ; LR \~ ~ and the sum of all the frustums, that is, the surface of the solid generated by the revolving semi-polygon, equals 27rCQ into the sum of the projections... = 2 FriCQ X BD = 2 -Ca X 2 = 4ar7r. v=l As we double n the number of sides of the inscribed polygon, by 14, D its semiperimeter approaches the semicircumference as limit, and its solid of revolution approaches the sphere as limit, while CQ or a, its apothem, approaches 9r the radius of the sphere as limit. But the variable sum bears to the variable a the constant ratio 4r7r. Therefore, by 13, their limits have the same ratio, and ff = 4 r2rr. Cor. 1. A sphere equals four times a circle with same radius. Cor. 2. A sphere equals the curved surface of its circumscribing cylinder. EXAM. 59. Considering the earth as a sphere whose radius is 6-3709 X 108 centimeters, find its area. H = 4(6-3709)2 x 1016 x,r. = = 4 x 40-58836681 x 3'14159265 x 1016. f-= 5,100,484,593,831,997,860 square centimeters. Arts. Or, about 510 million square kilometers. AREAS OF BROKEN AND CURVED SURFACES. 73 XXIII. A sphereical segment is the portion of a globe cut off by a plane, or included between two parallel planes. A zone is the curved surface of a spherical segment. The Proof of 64 gives also the following rule for the area of a zone: 65. To find the area of a zone. Rule: luiltiply the ltitucde of the segment by twice 7r times the rcadius of the sphere. Formula: Z= 2 7rra. Cor. 1. Any zone is to the sphere as the altitude of its segment is to the diameter of the sphere. B Cor. 2. Let the arc BP generate a calot or zone of a single base. By 65, its area Z= 27r XBC B = 7TBD X Bf = =r X BP2. Ww. 289; (Eu. VI. 8, Cor.; Cv. III. 44). Hence, a calot or zone of one base is equivalent to a circle whose radius is the chord of the generating arc.,-, I I I, k I i " \ 'Xxx C D EXAM. 60. Find the area of a zone of one base, the diameter of this base being 60 meters, and the height of the segment 18 meters. Using Cor. 2, the square of the chord of the generating arc is arc 1S (30)2 + (18)2 = 1224, which, multiplied by wr, gives, for the area of the calot, 3845-31 square meters. Ans, 74 MIENSURATION. 66. THEOREM OF PAPPUS. If a plane curve lies wholly on one side of a line in its own plane, and revolving about that line as axis generates thereby a surface of revolution, the area of the surface is equal to the product of the length of the revolving line into the path described by its center of mass. Scholium. The demonstration given under the next rule, though fixing the attention on a single representative case, applies equally to all cases where the generatrix is a closed figure, has an axis of symmetry parallel to the axis of revolution, and so turns as to be always in a plane with the axis of revolution, while its points describe circles perpendicular to both axes. EXAM. 61. Use the Theorem of Pappus to find the distance of the center of mass of a semicircumference from the center of the circle, by reference to our formula for the surface of a sphere. By 64, H = 4 r2T. By 66, II= rT X 2 xr. Equating, we get 2r= x7r. 2r 2 ~^. Ans. 71 67. To find the area of the surface of a solid ring. Rule: Multiply the generating circumference by the path of its center. Formula: 0 = 4 7r2 r r2. Proof: Conceive any plane to revolve about any straight line in it. Any circle within the plane, but without the axis, will generate a solid ring. AREAS OF BROKEN AND CURVED SURFACES. 75 Draw the diameter BC1D parallel to the axis AO. Divide the semicircumference BPD into n equal arcs, and call their equal chords each k. From the points of division drop perpendiculars to the axis, thus dividing the /) D other semicircumference BQD into n corresponding parts. Let BP, BQ, be a pair of arcs. If we draw their chords, we have a pair of right-angled trapezoids, ABPN and ABQV, which, during the revolution, describe frustums whose curved surfaces, by 62, are F, =rk(BA+PN), and F= 7rk(BA + QN). If JILGF is the medial line, then, by Proof to 40, F, = 2 k FM, and F= 27rk xLM.'. F + F2 = 2 rk(FMl + LM) = 2 rk(FG + GM + GM-GL). But the diameter BCD is an axis of symmetry, and GM'= CO = r 76 MENSURATION. the radius of the path of C..-. +FT,=-27rk2r,. This is the expression for each pair; and, as we have n pairs, therefore, the whole surface generated by a symmetrical polygon of 2n sides equals 2 nk 2 r'2 = 1) 2 rr, since 2nk is p the whole perimeter. But, as we increase 2n the number of sides of the inscribed polygon, by 14, p approaches c as its limit, and the sum of frustral surfaces approaches the surface of the ring as limit. But the variable sum bears to the variable perimeter the constant ratio 27rr2. Therefore, by 13, their limits have the same ratio, and O = c 2' 2 T2rr1 2i2rr2, where rl is the radius of the generating circle, and r2 the radius of the path of its center. EXAM. 62. Find the surface of a solid ring, of which the thickness is 3 meters, and the inner diameter 8 meters. Here r1 is 21 meters, and r2 is 5~ meters... 4 r21'lr2- 31416 X 3 x 3'1416 x 11 -9'4248 x 34-5576 = 325-698 square meters. Ans. EXAM. 63. Find the area of the surface of a square ring described by a square meter revolving round an axis parallel to one of its sides, and 3 meters distant. Here the length of the generating perimeter is 4 meters. The path of its center is 77r, since r2 is 3' meters..0. O = 28 r = 87-9648 square meters. Ans. AREAS OF BROKEN AND CURVED SURFACES. 77 EXAMr. 64. A circle of 1-35 meters radius, with an inscribed hexagon, revolves about an axis 6-25 meters from its center and parallel to a side of the hexagon. Find the difference in area of the generated surfaces. Here Here r1,=135 and r2=-625. Therefore, area of circular ring is 4r2r r2 == 72 X 2-7 X 12-5 = 9 8696 X 33-75 = 333-099. For the hexagonal ring Ww. 391; (Eu. IV. 15, Cor.; Cv. V. 14) the length of the generating perimeter is 6 x 135 8.1. The path of its center is T X 12.5 = 39-27-. Therefore, its area is 39-27 x 8.1 =318-087. Thus the difference in area is 15-012 square meters. Ans..(L). SPHERICS AND SOLID ANGLES. XXIV. A great circle is a section of a globe made by a plane passing through the center. A lune is that portion of a sphere comprised between two great semicircles. The angle of two curves passing through the same point is the angle formed by the two tangents to the curves at that point. A spherical angle is the angle included between two arcs of great circles. 78 MENSURATION. - A plane angle is the amount of divergence between two straight lines which meet in a point. A solid angle is the amount of spread between two or more planes which meet at a point. Two polyhedral angles, having all their parts congruent, but arranged in reverse order, are symmetrical. A steregon, the natural unit of solid angle, is the whole amount of solid angle round about a point in space. As a perigon corresponds to a circle and its circumference, so a steregon corresponds to a globe and its sphere. The steregon is divided into 360 equal parts, called spherical degrees of angle, and these divide the whole sphere into 360 equal parts, each called a degree of spherical surface. A steradian is.the angle subtended at the center by that part of every sphere equal to the square of its radius. 68, To find the area of a lune. Rule: JMiultiply its angle in radians by twoice its squared radius. Formula: L= 2r2iu. Proof: Let PAQBP, PBQCP be two lunes having equal angles at P; then one of these lunes may be placed on the other so as to coincide exactly with it: thus lunes h/aving equal angles are congruent. Then, by the process of Eu. VI. 33; (WVw. 766; Cv. VIII. 95), it follows that a lune is to the splhere as its angle is to a perigon; t4-=w; L=W2riU. 4 r27 27 ' AREAS OF BROKEN AND CURVED SURFACES. 79 Cor. 1. A lune contains as many degrees of spherical surface as its angle contains degrees. p Cor. 2. A lune measures twice as many steradians as its angle contains radians. A B c EXAM. 65. Find the area comprised between two meridians one degree apart on the earth's surface. Q Assuming as the earth's surface 196,625,000 square miles, dividing by 360, gives for the lune, 546,180-5+ square miles. Ans. XXV. Suppose the angular point of a solid angle is made the center of a sphere; then the planes which form the solid angle will cut the sphere in arcs of great circles. Thus a figure will be formed on the sphere, which is called a spherical triacngle if it is bounded by three arcs of great circles, each less than a semicircumference. If the solid angle be formed by the meeting of vmore than three planes, the corresponding figure on the sphere is bounded by more than three arcs of great circles, and is called a spherical polygon. The solid angle made by only two planes corresponds to the lune intercepted on any sphere whose center is in the common section of the two planes. The dihedral angle of two planes is the amount of rotation which one plane must make about their intersection in order to coincide with the other. The angles of a spherical polygon equal the dihedral angles of its solid angle. The sides of the polygon measure the face angles of this polyhedral angle. so80 0MENSURATION. From any property of polyhedral angles we may infer an analogous property of spherical polygons. Reciprocally, from any property of spherical polygons we may infer a corresponding property of polyhedral angles. XXVI. A spherical pyramid is a portion of a globe bounded by a spherical polygon and the planes of the sides of the polygon. The center of the sphere is the vertex of the pyramid; the spherical polygon is its base. 69. Just as plane angles at the center of a circle are proportional to their intercepted arcs, and also sectors; so solid angles at the center of a sphere are proportional to their intercepted spherical polygons, and also spherical pyramids. XXVII. The spherical excess of a spherical triangle is the excess of the sum of its angles over a flat angle. The spherical excess of a spherical polygon is the excess of the sum of its angles above as many flat angles as it has sides less two. 70. To find the area of a spherical triangle. Rule:;Multiply its spherical excess in radians by its squared radius. Formula: A = er2. Proof: Let ABC be a spherical triangle. Produce the arcs which form its sides until they meet again, two and two. The A ABC now forms a part of three lunes, namely, ABDCA, BCEAB, and CAFBC. Since the A's CDE and FAB subtend vertical solid AREAS OF BROKEN AND CURVED SURFACES. 81 angles at 0, they are equivalent, by 69. Therefore, the lune CAFBC equals the sum of the two triangles ABC and CDE. Thus the lunes whose angles are A, B, and C are together equal to a hemisphere plus twice A ABC. 4/ F - E Subtracting the hemisphere, / which equals a lune whose angle is a flat angle, we have 2A ABC= \, lune whose 4 is (A + B + C-f). B.. A = lune whose. is -e..b. by 68, A= er2. Cor. 1. A A contains half as many degrees of spherical surface as its e contains degrees. (Cor. 2. A A measures as many steradians as its e contains radians. Cor. 3. Every ~ of a A is > i e. EXAM. 66. Find the area of a tri-rectangular A. Here e=rt 4 =- 7..*. A= -2 rr2; or, a tri-rectangular triangle is one-eighth of its sphere. By Cor. 1, a tri-rectangular A contains forty-five degrees of spherical surface. 71. To find the area of a spherical polygon. Rule: Multzply its spherical excess in radians by its squared radius. Formula: N -= [I -(n-2) 7r] r2. 82 MENSURATION. Proof. From any angular point divide the polygon into (n —2) A's... by 70, N= =[ -(n-2) 7]?'2. This expression is true even when the polygon has reentrant angles, provided it can be divided into A's with each 4 less than f. Cor. 1. On the same or equal spheres, n-gons of equal angle-sum are equivalent; or, nT,1=1 if CT=" Cor. 2. To construct a dihedral solid 4 equal to any polyhedral; that is, to transform into a lune any spherical polygon; add its angles, subtract (n - 2), and halve the remainder. EXAM. 67. Find the ratio of the vertical solid angles of two right cones of altitude a, and a2, but having the same slant height h. These solid angles are as the corresponding calots on the sphere of radius h. Therefore, from 65, the required ratio is 2 rrh(h - ac) _ h - a 2 rh (th - cc2) I - a2 the ratio of the calot-altitudes. For the equilateral and right-angled cones this becomes 2- V 2-V2 AREAS OF BROKEN AND CURVED SURFACES. 83 THIRD REFERENCE TABLE OF ABBREVIATIONS. = angles. y J = density. E = edge.; = V. of paraboloid. 7 = V. of ellipsoid. 0 = V. of prolate spheroid. I = - 1. A = = of cone. [I = mass. = approximation. c d p = radian. = V. of oblate spheroid. T= distance. = V. of spherical ungula. = function. = V. of hyperboloid. = V. of mid F. of spindle. = weight. = varies as. = congruent. CHAPTER V. THE MEASUREMENT OF VOLUMES. (M). PRISM AND CYLINDER. XXVIII. Two polyhedrons are sym/tmetrical whose faces are respectively congruent, and whose polyhedral angles are respectively symmetrical; e.g., a polyhedron is symmetrical to its image in a mirror. A quacder is a parallelepiped whose six faces are rectangles. ii: A cube is a quader whose ii six faces are squares. XXIX. The volume of a solid is its ratio to an assumed unit. The unit for measurement of volume is a cube whose edge is the unit of length. Thus, if the linear unit be a meter, the unit of volume, contained by three square meters at right angles to each other, is called a cubic meter (cbm). XXX. Using length of a line to mean its numerical value, lengths, areas, and volumes, are all three quantities of the same kind, namely ratios. All ratios, whether expressible as numbers or not, combine according to the same simple laws as ordinary numbers and fractions. Ww. Bk. III.; (Eu. Bk. V.; Cv. Bk. II.). THE MEASUREMENT OF VOLUMES. 85 Therefore, we may multiply lengths, areas, and volumes together promiscuously, or divide one by the other in any order. If we ever speak of multiplying a line, a surface, or a solid, we mean always the length of the line, the area of the surface, or the volume of the solid. 72. To find the volume of a quader. Rule: iMultiply together the length, breadth, and height of the quader. Or, in other words, Xulltiply together the lengths of three adjacent edges. Formula: U- abl. Proof: By 32, the number of square units in the base of a quader is the product of two adjacent edges, bl. If on each of these square units WAe place a unit cube, for every unit of altitude we have a layer of bl cubic units; so that, if the altitude is a, the quader contains abl cubic units. Cor. 1. The volume of any cube is the third power of the length of an edge; and this is why the third power of a number is called its cube. Cor. 2. Every unit of volume is equivalent to a thousand of the next lower order. Cor. 3. The arithmetical or algebraic extraction of cube root makes familiar the use of the equation (a + a)3 = C3 - 3 ca2b + 3 ab2 + b3 86 8MENSURATION. Its geometric meaning and proof follow from inspection of the figure of a cube on the edge a +, cut by three planes into eight quaders. ^ 7 The cube a3 of the longer rod a, taken out, had faces a2 in common with three quaders of altitude b; had edges a in common with three quaders of base b2, and one corner the corner point of the smaller cube b3. XXXI. MASS, DENSITY, WEIGHT. The unit of capacity is a cubic decimeter, called the liter (l). The quantity of matter in a body is termed its mass. The unit of mass is called a gram (g). Pure water at temperature of maximum density is 1-000013 gram per cubic centimeter (ccm). So, in physics, the centimeter is chosen as the unit of length, because of the advantage of making the unit of mass practically identical with the mass of unit-volunme of water; in other words, of making the value of the density of water practically equal to unity; density being defined as mass er uniti-volume. THE MEASUREMENT OF VOLUMES. 87 The second is the fundamental unit of time adopted with the centimeter and the gram. Though the weight of a body, that is, the force of its attraction toward the earth, varies according to locality, yet weight being proportional to mass, the number expressing the mass of a body expresses also its weight in terms of the weight of the mass-unit at the same place. Thus, in terms X'l lNOEW ' & Cubic Gram Liter = Cubic Decimeter. Centimeter. Weight. Liter (common form). of the gr am and centimeter, or of the kilogram (kg) and liter, the mass, weight, and volume of water are expressed by the same num ber. So the density of any substance is the numzber of times the weight of the substance contains the weight of an equal bulk of water. Therefore, the density of a substance is the weight of a cubic centimeter of that substance in grams, or the weight of a liter in kilograms. Hence, 73. To find the density of a body. Rule: Divide the weight in grams by the bulk in cubic centimeters. Formula: 8 =. — Vccm V1 EXAM. 68. If 65 cubic centimeters of gold weigh 1251-77 grams; find its density. 1.7. -.oQ A_09Ci1.F77 * AA-? -OKO AO 1( 1L/,-L' I I -. -- - U mo. -Ol7. 88 MENSURATION. EXAM. 69. How many cubic centimeters (ccm) in one hektoliter (') )? Since 1 liter = 1000 cubic centimeters,.'. 1 hektoliter= 100,000 cubic centimeters. Ans. EXAM. 70. If the density of iron is 7-788, find the mass of a rectangular iron beam 7 meters long, 25 centimeters broad, and 55 millimeters high. The volume of'the beam in cubic centimeters is 700 X 25 x 5.5 = 96,250 cubic centimeters. Therefore, its mass is 96,250 X 7-788= 749,595 grams. AAns. 74. To find the volume of any parallelepiped. Rule: lfultiply its cltituce by the area of its base. Formula: V. P = abl. Proof: Any parallelepiped is equivalent to a quader of equal base and altitude. For, supposing AB an oblique parallelepiped on an oblique base, prolong the four edges parallel to AB, and cut them normally by two parallel planes whose distance apart, CD, is equal to AB. This gives us the parallelepiped ODE, which is still oblique, but on a rectangular base. Prolong the four edges parallel to DE, and cut them normally by two planes whose distance apart FG is equal to DE. This gives us the quader JG. Now, the solids AC and BDE are congruent, having all their angles and edges respectively equal. Subtracting each in turn from the whole solid ADE leaves CIDE equivalent to AB, THE MEASUREMENT OF VOLUMES. 89 Again, the solids CDJF and EG are congruent. Taking from each the common part EF leaves CDE equivalent D G to FG. Therefore, the parallelepiped AB is equivalent to the quader FG of equal base and altitude. EXAM. 71. The square of the altitude of a parallelepiped is to the area of its base as 121 to 63, and it contains 1,901,592 cubic centimeters. Find its altitude. Here aB=1,901,592 and 63aS= 121 B... 63 a3 121 x 1,901,592 = 230,092,632. a3 = 230,092,632 - 63 = 3,652,264... a = 154 cubic centimeters. Ans. 75. To find the volume of any prism. Rule: AJultip ly the altitude of the /prism by the area of its base. Formula: V. P ---c B. 90 MENSURATION. Proof. For a three-sided prism this rule follows from 74, since any three-sided prism is half a parallelepiped of the same altitude, the base of the prism being half the base of this parallelepiped. To show this, let ABCa3Py be any three-sided prism. Extend/ing the planes of its bases, fi?-; s ^and through the edges Aa, \^ / < ^ i-' (C7y, drawing planes parallel to -! -~f \ the sides, By, AP, we have the parallelepiped ABCDa/3y8, whose base ABCOD is double B1/ |/ // the base ABCY of the prism. S,..l Ww. 123; (Eu. I. 34; Cv. I. 105). A~^ Also, this parallelepiped itself is twice the prism. For, its two halves, the prisms,are congruent if its sides are all rectangles. If not, the prisms are symmetrical and equivalent. For, draw planes perpendicular to Aa at the points A and a. Then the prism ABC'ay is equivalent to the right prism AEVa-qfp, because the pyramid AEBC7if is congruent to the pyramid arlPy/. In the same way, ADCa8Zy equals ALliaXA. But AE3fa-fqL and ALlfiaX/p. are congruent. Therefore, ABCa/3y and ADCa8y are equivalent, and the parallelepiped ABCDaP/3y is double the prism ABCaP3y. Thus, the rule is proved true for triangular prisms, and consequently for all prisms; since, by passing planes through any one lateral edge, and all the other lateral edges, excepting the two adjacent, we can divide any prism into a number of triangular prisms of the same altitude, whose triangular bases together make the given polygonal base. THE MEASUREMENT OF VOLUMES. 91 Cor. 1. The volume of any prism equals the product of a lateral edge by the cross-section normal to it. Cor. 2. Every parallelepiped is halved by each diagonal plane. Cor. 3. Every plane pass- \ / ing through two opposite corners, halves the paral- /:: i lelepiped. Cor. 4. The volume of \i' - -"' a truncated parallelepiped '. /, equals half the sum of two opposite lateral edges multiplied by the cross-section normal to them. EXAM. 72. The altitude of a prism is 5 meters, and its base a regular triangle. If, with density 4, it weighs 1836 kilogrammes, find a side of its base. Its volume is 1836 + 4 = 449 cubic decimeters. The area of its base = 459 + 50 = 9-18 square decimeters. By 36, Cor., the square of a side of this regular triangle is 9.18 - 0-433 = 21-2 square decimeters. Therefore, a side equals 4604 decimeters. An X 4.604+ decimeters. Ans. 76. To find the volume of any cylinder. Rule: Multiply the altitude of the cylincler by the area of its base. Formula when Base is a Circle: V. C = a2Sr. 92 MENSURATION. Proof. In 58, Second Proof, we saw the cylinder to be the limit of an inscribed prism when the number of sides of the prism is increased indefinitely, and the breadth of each side indefinitely diminished, the base of the cylinder being consequently the limit of the base of the prism. But, by 75, always V. P is to B in the constant ratio a; hence, by 13, their limits will, be to one another in the same ratio; and V. C = aB. Scholium. This applies to all solids whose cross-section does not vary, whatever be the shape of the cross-section. Cor. 1. Between any two parallel planes, the volume of. —'... ---., any cylinder equals the product of its axis by the cross-section normal to it. Cor. 2. By 58, Cor. 1, the volume of any truncated circular cylinder equals the product of its axis by the circle normal to it. EXAM. 73. A gram of mercury, density 13-6, fills a cylinder 12 centimeters long; find the diameter of the cylinder. Volume of cylinder = =073529+ cubic centimeters. Area of base of cylinder= r2rr = -073529 - 12 = ~006127412 square centimeters. Therefore r2=.006127412 - 3.1416 = 001950. Thus r == 044 centimeters = -44 millimeters, and d = 2r = 88 millimeters. Ans. 77. To find the volume of a cylindric shell. Rule: _31uiltply the sum of the inner and outer radii by their dcfference, and this product by T times the altitude of the shell, THE MEASUREMENT OF VOLUMES. 93 Formula: V. C0 - V. C2 = Car (r1 + r'2) (' -- r2). Proof: Since a cylindric shell is the difference between two circular cylinders of the same altitude, its volume equals 'aar2 7T - ar22 7r = a (r2 - r2). EXAM. 74. The thickness of the lead in a pipe weighing 94-09 kilograms is 6 millimeters, the diameter of the opening is 4-8 centimeters; taking 7r — 2,2 and density 11, find the length of the pipe. Here '2 = 24 centimeters and r = 3 centimeters. 94,090-11 ir (r1 + 2') (r, - r) = ll1 2-25.4 X. -242 3.24 - 178408; 658,630 784.081..'. = 658,630 - 784-08 - 840 centimeters 84 meters. Ans. (N). PYRAMID AND CONE. XXXII. The altitude of a pyramid is the normal distance from its vertex to the plane of its base. 78. Parallel plane sections of a pyramid are similar figures, and are to each other as the squares of their distances from the vertex. Proof: The figures are similar, since their angles are respectively equal, Ww. 462; (Eu. XI. 10; Cv. VI. 32). and their sides proportional. Ww. 279; (Eu. VI. 4; Cv. III. 25). _ I 94 MENSURATIOiN. By 44, they are to each other as the squares of homologous sides, and hence as the squares of the normals from the vertex. Ww. 469; (Eu. XI. 17; Cv. VI. 37)..... ' Scholium. This is the reason why the strength of gravity, light, heat, magnetism, electricity, and sound, de-....... creases as the square of the distance from the source. Image part of the beams from a luminous point as a pyramid of light. If a cutting plane is moved away parallel to itself, the number of units of area illuminated increases as the square of the distance. But the number of rays remains unchanged. Therefore, the number of beams striking a unit of area must decrease as the square of the distance. 79. Tetrahedra (triangular pyramids) having equivalent bases and equal altitudes are equivalent. Proof. Divide the equal altitudes a into n equal parts, and through each point of division pass a plane parallel to the base. By 78, all the sections in the first tetrahedron are triangles equivalent to the corresponding sections in the second. Beginning with the base of the first tetrahedron, construct on each section as lower base a prism ~a high with lateral edges parallel to one of the edges of the tetrahedron. In the second, similarly construct prisms on each section as upper base. Since the first prism-sum is greater than the first tetrahedron, and the second prism-sum less than the second THE MEASUREMENT OF VOLUMES. 95 tetrahedron, therefore the difference of the tetrahedra is less than the difference of the prism-sums. But, by 75, each prism in the second tetrahedron is equivalent to the prism next above it on the first tetrahedron. So the difference of the prism-sums is simply the lowest prism of the first series, whose volume, by 75, is an As n increases this decreases, and can be made less than any assignable quantity by taking n sufficiently great. Hence the tetrahedra can have no assignable difference; and, being constants, they cannot have a variable difference. Therefore the tetrahedra are equivalent. Scholium. This demonstration indicates a method of proving that any two solids having equivalent bases and equal altitudes are equivalent, if every two plane sections at the same distance from the base are equivalent. 80. To find the volume of any pyramid. Rule: Miultiply one-third of its altitude by the area of its base. Formula: V. Y = a-B. 96 IMENSURATION. Proof. Any triangular prism, as ABC-FDE, can be divided into three tetrahedra, two (B-DEF and D-ABC) having the same altitude as the prism, and its top and bottom respectively as bases, while the third c A B (BCDF) is seen to have an altitude and base equal to each of the others in turn by resting the prism first on its side CE and next on its side AF. Hence, by 79, these three tetrahedra are equivalent, and therefore, by 75, the volume of each is ~ aB. The rule thus proved for triangular pyramids is true for all pyramids, since, by passing planes through any one lateral edge, and all the other lateral edges excepting the two adjacent to this one, we can exhibit any pyramid as a sum of tetrahedra having the same altitude whose bases together make the given polygonal base. EXAM. 75. If the altitude of the highest Egyptian pyramid is 138 meters, and a side of its square base 228 meters, find its volume. Here V. Y 138 (228)2 — 46 x 51,984 = 2,391,264 cubic meters. Ans. 81. To find the volume of any cone. THE MIIEASUREMENT OF VOLUMES. 97 Rule: n utltiply one-third its altitude lby the area of its base. Formula when Base is a Circle: V. K- Iar27. Proof. In 60, Second Proof, we saw that the base of a cone was the limit of the base of the circumscribed or inscribed pyramid, and therefore the cone. itself the limit of the pyramid. But, by 80, always the variable pyramid is to its variable base in the constant ratio a. Therefore, by 13, their limits are to one 1. ". another in the same ratio and V. K = a aB. Scholium. This applies to all solids determined by an elastic line stretching from a fixed point to a point describing any closed plane figure. Cor. The volume of the solid generated by the revolution of any triangle about one of its sides as axis is onethird the product of the triangle's area into the circumference described by its vertex. V= f TrA. EXAM. 76. Find the volume of a conical solid whose altitude is 15 meters and base a parabolic segment 3 meters high from a chord 11 meters long. By 54, here V. -K — 15 X -33 x 11 — 5x2x 11 -110 cubic meters. Ans. 98 MENSURATION. EXAM. 77. Requirec the volume of an elliptic cone, the major axis of its base being 15-2 meters; the minor axis, 10 meters; and the altitude, 22 meters. By 55, here V. K - 22767r5 =- 387r73 r2783 =875-45+ cubic meters. Ans. /__ EEXAM. 78. The section of a right circular cone by a plane ------------ through its vertex, perpendicular to the base is an equilateral triangle, each side of which is 12 meters; find the volume of the cone. Here a= - /122 _62 = VO8.. i. a27r= -V108r36 - 391-78 cubic meters. Ans.. (0). PRISMATOID. XXXIII. If, in each of two parallel planes is constructed a polygon, in the one an m-gon, e.g., ABC/D; in the other, an n-gon, e.g., A'B;C'; then, through each side of one and each vertex of the other polygon a plane may be passed. Thus, starting from the side AB, we get the n planes ABA' ABB' ABC'; again, with the side BC, the n planes BCA', B(OB' BCOO etc. Using thus all m sides of the polygon ABCD, we get mwn planes. Also, combining each of the n sides A'BI B'C'D' with each of the mn points A, B, C, D, gives nm planes A 'B'A, A'rB. THE MEASUREMENT OF VOLUMES. 99 A'B'C, A'B'D; B'C'A, B'C'B, etc.; so that, altogether, 2 mn connecting planes are determined by the two polygons. Among these are m + n outer planes, which toc' A' c gether enclose the rest. These outer planes form the sides, and the given polygons the bases of a solid called a prisrmatoid. Our figure is a case of this body when m =4 and n =3. The midcross-section IJ is given to show the seven sides. XXXIV. A prismatoid is a polyhedron whose bases are any two polygons in parallel planes, and whose lateral faces are determined by so joining the vertices of these bases that each line in order forms a triangle with the preceding line and one side of either base. REMARK. This definition is more general than XXXIII., and allows dihedral angles to be concave or convex, though neither base contain a reentrant angle. Thus, BB' might have been joined instead of A'C. 100 MENSURATION. From the prismatoids thus pertaining to the same two bases, XXXIII. chooses the greatest. c XXXV. The altitude of a prismatoid is the normal distance between the planes of its bases. Passing through the middle point of the altitude a plane parallel to the bases gives the midcross-section. Its vertices halve the lateral edges of the prismatoid. Hence, its perimeter is half the sum of the basal perimeters. But, if one base reduces to a straight line, this line must be considered a digon, i. e., counted twice. XXXVI. In stereometry the prism, pyramid, and prismatoid correspond respectively to the parallelogram, triangle, and trapezoid in planimetry. XXXVII. Though, in general, the lateral faces of a prismatoid are triangles, yet if two basal edges which form, THE MEASUREMENT OF VOLUMES. 101 with the same lateral edge, two sides of two adjoining faces are parallel, then these two triangular faces fall in the same plane, and together form a trapezoid. XXXVIII. A prismoid is a pris- / \ matoid whose bases have the same number of sides, and every corresponding pair parallel. I ------- it\ XXXIX. A frustum of a pyramid is a prismoid whose two bases are similar. Cor. Every three-sided prismoid is the frustum of a pyramid. XL. If both bases of a prismatoid become lines, it is a tetrahedron. XLI. A wedge is a prismatoid whose lower base is a rectangle, and upper base a line parallel to a basal edge. 82. To find the volume of any prismatoid. Rule: Add the areas of the two bases and four times the midcross-section; multiply this sum by one-sixth the altitude. Prismoidal Formula: D = -- (B1 + 4 1 + B2). Proof: In the midcross-section of the prismatoid take a point N, which join to the corners of the prismatoid. These lines determine for each edge of the prismatoid a 102 MIENSURATION. plane triangle, and these triangles divide the prismatoid into the following parts: G 1. A pyramid whose vertex is N and whose base is B2, the top of the prismatoid. Since the altitude of this pyramid is half that of the prismatoid, therefore, by 80, its volume is l aB2. 2. A pyramid whose vertex is N and whose base is B1, the bottom of the prismatoid. Since the altitude of this pyramid also is 1 a, therefore its volume is I aB1. TIlE MEASUREMENT OF VOLUMES. 103 3. Tetrahedra, like ANIFG, each of which can have its volume expressed in terms of its own part of the midcrosssection. For, let NXi and EVK be the lines in which the two sides ANB, ANGG of the tetrahedron cut the midcross-section; and consider the part AXNIJK of the tetrahedron ANFG~. This part ANiLH is a pyramid whose base is the triangle NHiK, and whose altitude is - a, half the altitude of the prismatoid. Hence, by 80, the volume of ANIHK is 1 a(iNHK). But, drawing KF, by 79, ANHK = - AN FIC, and ANFK = 1- ANTG. Therefore, ANFG = a(NIIK). In like manner, the volume of every such tetrahedron is 4a times the area of its own piece of the midcross-section, and their sum is aX. Now, combining 1, 2, and 3, which together make up the whole volume of the prismatoid, we find D = laB, + IaB, + 4 aM = ia(B + 4 1 + B,). EXAM. 79. Given the plan of an embankment cut perpendicularly by the plane AEIID, its top the pentagon EFGfHI, its bottom the D- c trapezoid ABCUD, with " — -- H I the following measurements: For the lower ------------ base, AB=90 m.eters, CD -110 meters, AD L.................... = 65 meters; for the Ts upper base EF - 70 A B meters, EI = 30 meters, MF = iE'lf= — JIG = 15 meters; the breadth of the scarp AE=- 20 meters, DI= 15 meters; the altitude of the embankment a c 15 meters. Find its volume. 104 MENSURATION. Here, for the midcross-section we get TS = 80 meters, L = 87-5 meters, TpP = 90 meters, TL - 7-5 meters, LI= 325 meters, IN = 75 meters. Thus the areas are B = 6500 square meters, B2 = 2325 square meters, lf= 4337-5 square meters, and for the whole volume we get f1)= -65,437-5 cubic meters. Ans. NOTE. In a prismoid the midcross-section has always the same angles and the same number of sides as each base, every side being half the sum of the two corresponding basal edges. The rectangular prismoid has its top and bottom rectangles; hence, by 32, its volume D = -a(B + 4B + -R2) I -(wlbt + 4w - 2x + b 22) 2 2 I - (2v wb + w b2 + w2b, -I 2 wb,). an>'. CorG. If a prism has trapezoids for bases, its volume equals hlalf the sum of its two parallel side-faces multiplied by their normal distance apart. 83. To find the volume of a frustum of a pyramid. Rule: To the ctreas of the two ends of the frustum add the square root of their product; multjily this sum by one-third the altitude. Formula: V. F = i3 a(B1 + V-B1B2 + B2). THE MEASUREMENT OF VOLUMES. 105 Proof: If w1 and w2 are two corresponding sides of the bases -B and B2, then a side of the midsection is W1 + W2 2 Since in a frustum B1, B2, and Xf are similar, by 44, we have T1 ~e W2::v 2 and 2: w::+ 2 2 whence w1 +W WV1 ~2 V*1 + v'2i Hence 2 VI = v/ + v ', and 4 =B1 + 2 VB1+ B2. Substituting this in 82 gives V. F = a (2 B, + 2 B,B2 + 2 B). Cor. By 44, B — B2= 2 WI Substituting this for B2 gives V. F = aB(1+ W1 +_+ w,2 2 /1 1 106 MENSURATION. EXAM. 80. The area of the top of a frustum is 160 square meters; of the bottom, 250 square meters; and its altitude is 24 meters. Find its volume. Here V. F = 8 (250 + 200 + 160) = 4880 cubic meters. Ans. If, instead of the top, we are given wI: W2:: 5:4, then, by our Corollary, F 2000(1 + + ) v. F = 2000(1 +1 6+) = 4880 cubic meters. Ans. 84. To find the volume of a frustum of any cone. Rule: To the areas of the two ends add the square root of their product; mnultiply this sum by one-third the / altitude. Formula for Circular Cone: V. F = a -Tr (7'2 + rr2 + r22). Proof: As in 81, so the frustum of a cone is the limit of the frustum of a pyramid. EXAM. 81. The radius of one end is 5 meters; of the other, 3 meters; the altitude, 8 meters. Find the volume. Here V. F = i 8 (25 + 15 + 9) 3-1416 = 410-5024 cubic meters. Ans. 85. To find the volume of any solid bounded-terminally by two parallel planes, and laterally by a surface generated by the motion of a straight line always intersecting the planes, and returning finally to its initial position. THE MEASUREMENT OF VOLUMES. 107 Rule: Add the areas of the two ends to four times the midsection; multiply this sum by one-sixth the altitude. Prismoidal Formula: XD = ac (B1 + 4 X + B2). Proof: Join neighboring points in the top perimeter of such a solid to form a polygon, likewise in the perimeter of the bottom. Take the two polygons so formed as bases of a prismatoid. Then when the number of basal edges is indefinitely increased, each edge decreasing indefinitely in length, as thus its bases approach to coincidence with the bases of the solid, the sides of the prismatoid approach the ruled surface, and its volume and midsection approach the volume and midsection of the solid as limit. But always the variable volume is to the variable sum (B1 + 4 f -+ -B2) in the constant ratio h a. Therefore, by 108 MENSURATION. 13, their limits will be to one another in the same ratio; and D=1a(B +4J1f+B2) for the prismoidal solid. EXAM. 82. The radius of the minimum circle in a hyperboloid is 1 meter. Find the volume contained between this circle of the gorge and a circle 3 meters below it whose radius is 2 meters. Solution: The hyperboloid of revolution of one nappe is a ruled surface generated by the rotation of a straight line about an axis not in the same plane with it. All points of the generatrix describe parallel circles whose centers are in the axis. The shortest radius, a perpendicular both to axis and generatrix, describes the circle of the gorge, which is a plane of symmetry. Hence, taking this circle as midsection, and for altitude twice the distance to the base below it, the Prismoidal Formula gives twice the volume sought in Exam. 82. 2 V7= D = - 6[22r + 4 (1)2 + 22r] == 12. Therefore, V=: 67r cubic meters. Ans. 86, To find the volume of any wedge. Rule: To twice the length of the base add the opposite edge; multiply the sum by the width of the base, and this product by one-sixth the altitude of the wedge. Formula: TW = aw (2 bl + b b). Proof. Since the upper base of a wedge is a line, so, by the Prismoidal Formula, W= *a(B, + 4 M). THE MEASUREMENT OF VOLUMES. 109 Therefore, by 32, W = (wb + 4 1 b, 2 x __ ')Ca(b+l b -1 b). Cor. 1. If the length of edge equals the length of base;.," bi = b2, then TV= 1 awb, the simplest form of wedge. Cor. 2. The volume of any truncated triangular prism is equal to the product of its right section by one-third the sum of its lateral edges. /^\^I7 y v~~ EXAM. 83. Find the volume of a wedge, of which the length of the base is 70 meters; the width, 30'meters; the length of the edge, 110 meters; and the altitude, 24-8 meters. Here TI= (140 + 110) 30 24-8 -(140 + 110) 10 x 12-4 = 2500 x 12.4 = 31,000 cubic meters. Ans. 87. To find the volume of any tetrahedron. Rule: fMultiply double the carea of a parallelogram whose vertices bisect any four edges by one-third the perpendicular to both the other edges. Formula: X=23- - X -- b eProof. When, the bases being lines, B1=B2=0, then D=X=-.a4 if = aM. 110 MENSURATION. Since i bisects the line perpendicular to the two basal edges, it bisects the four lateral edges; and is a parallelogram. EXAM. 84. The line perpendicular to both basal edges of a tetrahedron is 2r long; the length of the top edge is 2r, and of the bottom edge, 27rr. The midsection is a rectangle. Find the volume of the tetrahedron. Here a = 2r, and 2M= r2r. Therefore, X — r4r3. Ans. (P). SPHERE. 88. To find the volume of a sphere. Rule: Miultiply the cube of its radius by 4'1888-. Formula: V. H = 7rr3. Proof: Any sphere is equal in volume to a tetrahedron whose midsection is equivalent to a great circle of the sphere, and whose altitude equals a diameter. D G K H Let the diameter DC be normal to the great circle AB at C. Let Q be the point in which the midsection LN bisects the altitude JK at right angles. In both solids THE MEASUREMENT OF VOLUMES. 111 take any height CI -- QR, and through the points I and R the sections PS parallel to AB, and MO parallel to LN. Then, in the sphere, by 47, OAB: OPS:: AC2: PI2, or, by Ww. 289; (Eu. VI. 8, Cor.; Cv. III. 44), OAB: PS:: AC2: TI ID.... (1). In the tetrahedron, by Ww. 279 & 469; (Eu. VI. 4, & XI. 17; Cv. IV. 25, & VI. 37), LU:MW:: EL: EM:: JQ -:JR; LV: MZ:: GL:GM::KQ:KR; and since, by Ww. 315; (Eu. VI. 23; Cv. IV. 5), LN: EMO:: L U X L V: MW x MZ, therefore, LN: fH'O:: JQ x QK: JR x RK.. (2). But now, by hypothesis and construction, in proportions (1) and (2), the first, third, and fourth terms are respectively equal, therefore 0 PS = E7MO; and since these are corresponding sections at any height, therefore, by Scholium to 79, the sphere and tetrahedron are equal in volume. Thus, by 87, V. H = X= all = = 3 2 rr27r = 4 7->3. V. H =X=aM== 22r r = r3. EXAM. 85. If, in making a model of the tetrahedron EFG-TI, we wish the midsection LN to be a square, and the four lateral edges equal, find in terms of radius their length and that of the two basal edges. By hypothesis, square LN = r2i;. L U r /r. 112 1ME NSURAIATION. But Gf= 2L U= 2 rv/, and EF = 2LV= 2LU = 2r-. For any one of the equal lateral edges, E2 = GKK-2 + _-2 = UZ72 + Ey'2 + y -~ 2 = - 2 + ~P'~. -2 72 - -2 2 2 2 EG GGK A KE~=GK f KJ +JE =LU D iT ~LI..E. EG = Vr2 + 4r + r ==4r2 (1 =2r + +. So it is not a regular tetrahedron. 89. To find the volume of any spherical segment. Rule: To three times the sum of the squared radii of the two ends add the squared altitude; multiply this sum by the altitude, and the product by -5236-. Formula: V. G = - ar [3 (912 + r22) + a2]. Proof: In 88, we proved any spherical segment equal in volume to a prismoid of equivalent bases and alti21~ A tude. Therefore, by 82, V. G = a (r2 + 4r2 + r27). To eliminate r2 call x the distance from center of sphere to bottom of segment, and r the radius of sphere; then, by Eu. II. 10, (a+x)+ =\2 +2 2 Doubling and subtracting both members from 4 r2, gives 2r2 - 2(a +- )2 + 2r2 - 224r- 4( + ) - aa2 or, by 2, 2r22 + 2r12 + a2 = 4r32. Substituting, V. G = aT (3 12 + 3 r;2 a2). THE MEASUREMENT OF VOLUMES. 113 Cor. In a segment of one base, since r2 = 0, we have V. G = -a (3 r1 + a2). But now, by Ww. 289; (Eu. VI. 8, Cor.; Cv. III. 44), r2 = a(2r-a). Substituting, V. G = -a r[3a(2r-a) + a2] =- ar(Gar- 3 a2 + a2) = a2 (r - I-a). EXAM. 86. If the axis of a cylinder passes through the center of a sphere, the sphere-ring so formed is equal in volume to a sphere of the same altitude. For, since the bases of a middle segment are equidistant from the center, V. G = 1 ar (6 r2 + 2) = a7rr2 + i 7Wa3. But, by 76, the volume of the cylinder cut out of the segment is arl2r, and the remaining ring ir ra3 is, by 88, the volume of a sphere of diameter a. XLII. When a semicircle revolves about its diameter, the solid generated by any sector of the semicircle is called a spherical sector. 90. To find the volume of any spherical sector. Rule: lMfultiply its zone by one-third the radius. Formula: V. S = irar2. Proof: If one radius of the generating sector coincides with the axis of revolution, the spherical sector is the sum of a spherical segment of one base and a cone on same base, with vertex at center of sphere. 114 MENSURATION. By 89, Cor., V. G a27(r- a). By 81, V.K = (r - a)r2; or, substituting for r12, its value used in 89, Cor., V. K = - (r -a)a(2r - )7r = r(2r2- 3ra2 + a3). Adding, we have V. S = V. G + V. K = a7rr2. Any other spherical sector is the difference of two such sectors. V.S = V. SI- V. S, = ra, - r2a = r2 (a- a). But al - - = a, the altitude of S's zone, whose area, by 65, is 27ra. Thus, for every spherical sector the volume is zone by r. Cor. If r, and r2 are the radii of the bases of the zone, its altitude, a = Vr2 - r- /r2 - r. 2 EXAM. 87. Find the diameter of a sphere of which a sector contains 7'854 cubic meters when its zone is 0'6 meters high. V. S = 6grr2-04ir2 = 7854. Dividing by 7r- 3'1416, 0-4r=25..'. 4r2 =25... 2r= 5 meters. Ans. XLIII. A spherical ungula is a part of a globe bounded by a lune and two great semicircles. 91. To find the volume of a spherical ungula. THE MEASUREMENT OF VOLUMES. 115 Rule: JPlultiply the area of its lune by one-thilrd the radius. Formula: v_. 3u. Proof: By 69, V:V.H:: L:H ff..: 4- r3:: 2 r: 4r2 r... V=- r3U. Cor. On equal spheres, ungulae are as their angles. 92. To find the volume of a spherical pyramid. Rule: fudltiply the area of its base by one-third the radius. Formula: Y = r3e. Proof: By 69, Y: V. H:: N: H. Y: 4 r33: er2: 4 r2... Y=r3e. (Q). THEOREM OF PAPPUS. 93. If a plane figure, lying wholly on the same side of a line in its own plane, revolves about that line, the volume of the solid thus generated is equal to the product of the revolving area by the length of the path described by its center of mass. Scholium. As for 66, so we give under 94, by a single representative case, the general demonstration for all figures having an axis of symmetry parallel to the axis of revolution. 116 MENSURATION. EXAM. 88. Find the distance of the center of mass of a semicircle from the center of the circle. By 88, V = 3 By 93, V.H=r2T2x7. Equating, we get 4rr3 r -r2'z..r. 7Xr=7ra. r X - -. Ans. 3r 94. To find the volume of a ring. Rule: Multiply the generating area by the path of its center. Formula for Ellipse: V. 0= 27r2abr. Proof: Conceive any ellipse to revolve about an exterior axis parallel to one of its axes. Divide the axis of symmetry AE into n equal parts, as AV=VT=z, and from these points of division drop perpendiculars on the axis of revolution PO. Join the points where these perpendiculars cut the ellipse by chords FD, DA, AG, GH, etc. The volume generated by one of the trapezoids thus formed, as D)G-HF, is the difference between the frustums generated by the right-angled trapezoids QDFW and QGHfW. Therefore, by 84, V. by DGHF z7rr(FW2+FW, D Q+DQ2) -z(TW% 1 TW+ W, GQ+GQ2) = zr [(CO + FT)2 + (CO + FT) (CO + D V) + (CO + D V)2 -(CO-FT)2-(CO-FT) (CO-D V) - (CO-D V))2] = zr(6CO, FT + 6 CO, D V) = 2, CO, z(FT+ D V) 2 FI: + DG 2 THE MEASUREMENT OF VOLUMES. 117 Thus, by 40, the volume generated by the polygon EGADF, etc., equals its area multiplied by the path of the center. But, as we increase n, and thus decrease z indefinitely, as shown in 55, the area of the polygon ap preaches the area of the ellipse as its limit. But always the variable volume is to the variable area in the constant ratio 27rr; therefore, by 13, their limits will be to one another in the same ratio; and V. 0 = 2rab7r. EXAM. 89. Find the volume of the ring swept out by an ellipse whose axes are 8 and 16 meters, revolving round an axis in its own plane, and 10 meters from its center. Here V. O -4 x 8 x 10 x 2r2 =640 r2 = 6316-5 cubic meters. Ans. 118 MENSURATION. ~(R). SIMILAR SOLIDS. XLIV. Similar polyhedrons are those bounded by the same number of faces respectively similar and similarly placed, and which have their solid angles congruent. 95. Given the volume and one line in a solid, and the homologous line in a similar solid, to find its volume. Rule: lultiply the given volume by the cubed ratio of homologous lines. Formula: i V 23. a2 Proof: The volumes of two similar solids are as the cubes of any two corresponding dimensions. Ww. 590; (Eu. XI. 33; Cv. VII. 73). Thus, for the sphere, by 88, 1,3' i 4 7-3 3 I2 4- 'r23 72 NOTE. If a tetrahedron is cut by a plane parallel to one of its faces, the tetrahedron cut off is similar to the first. If a cone be cut by a plane parallel to its base, the whole cone and the cone cut off are similar. THE MEASUREMENT OF VOLUMES. 119 EXAM. 90. The edge of a cube is 1 meter; find the edge of a cube of double the volume. The cube of the required number is to the cube of 1 as 2 is to 1; or, x3:1:: 2:1.... =2..-. x = 2 125992+. Ans. Thus a cube, with its edge 1-26 meters, is more than double a cube with edge 1 meter. EXAM. 91. The three edges of a quader are as 3, 4, 7, and the volume is 777,924; find the edges. By 72, the volume of the quader, whose edges are 3, 4, 7, is 84; then 84: 777,924:: 33: 250,047, /250,047 = 63; and 3:4::63: 84, 3: 7:: 63:127. Therefore, the edges are 63, 84, 127. Ans. (S). IRREGULAR SOLIDS. 96. Any small solid may be estimated by placing it in a vessel of convenient shape, such as a quader or a cylinder, and pouring in a liquid until the solid is quite covered; then noting the level, i removing the solid, and again noting the level at which the liquid stands. The volume of the solid is equal to the volume of the vessel between the two levels. 120 MENSURATION. 97. If the solid is homogeneous, weigh it. Also weigh a cubic centimeter of the same substance. Divide the weight of the solid by the weight of the cubic centimeter. The quotient will be the number of cubic centimeters in the solid. From 73, we have the Formula: Vcc= -= g ExxA. 92. A ball 5 centimeters in diameter weighs 431-97 grams. An irregular solid of the same substance weighs 13*2 grams; find its volume. The volume of the ball is 53 x 0.5236 =65-45.. 43197 - 65-45 = 66 grams, the weight of a cubic centimeter... 13-2 - 6-6 = 2 cubic centimeters. Ans. 98. To find the volume of any irregular polyhedron. Rule: Cut the polyhecron into prisnmatoids by passing parallel planes through all its summits. Formula for n consecutive prismatoids: I- [x (B, - B3) + XI - B2,4) + etc. + x (B_1- BP+l) + X"+1 (B7 + B,,+l)] +3 [X2 x + (X3 - X2)Ji + (X4 -,3) -+ etc. + (xn+ - x,,)J]. NOTE. X2 is the distance of B2 from B1, and x3 is the distance of B3 from B1, etc. Proof: This formula is obtained directly by the method of 41. CHAPTER VI. THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 99. To find whether the volume of any solid is determined by the Prismoidal Formula. Rule: The Prismoidal Formula applies exactly to ALL SOLIDS contained between two parallel planes, OF WHICH the area of any section parallel to these planes can be expressed by a rational integral algebraic function, of a degree not higher than the third, of its distance -romz either of these bounding planes or bases. Test: A = q +mx + nx2 +fA. NOTE. Ax is the area of any section of the solid at the distance x from one of its ends. The coefficients q, m, n,f, are constant for the same solid, but may be either positive or negative; or any one, two, or three of them may be zero. Proof: Measuring x on a line normal to which the sections are made, let g) (x) be the area of the section at the distance x from the origin. The problem then is, What function f will fulfil the conditions of the Prismoidal Formula? For any linear unit, the segment between 0 (0) and q (4) is the sum of the segments between < (0) and 4 (2) and between 4 (2) and c (4). Therefore, if 4 is such a function 122 MENSURATION. as to fulfil the requirements of the Prismoidal Formula, we have identically l[(0))+40(2). [ 0(4)]= [ () 40 (1)+ (2)]+ 2 [0(2)+4 (3)+ (4)].... (0)- 4(1) + 6 (2)-4 (3) + 4 (4)= 0. But for (x) = q + n + 1n2; +-f3 + gx4, c(x) - 40(1) + 60(2) - 40(3) + 0(4) becomes + q -4q- 4m- 4n — 4f - 4g +Gq+12m+24n-+ 48f+ 96g -4- 12m-36n- 108f- 324g + q+ 4m +16n 64f+256g O O 0 0 + 24 So the conditions are satisfied only by functions which have no fourth and higher powers. Hence 0 (x) must be an algebraic expression of positive integral powers not exceeding the third degree. Thus, in general, the cubic equation Ax = qmx + r nx2 + fx3 expresses the law of variation in magnitude of the plane generatrix of prismoidal spaces; i.e., solids to which the Prismoidal Formula universally applies. Cor. 1. Since for prismoidal solids (x) = n + nix n2x2 + n3x3, therefore, 2(0) + 4(2- a) + (a) = n,t +4no+2anl+ a2n+ l +a3)3 = n + 3 an + 2 a2 +- a343a =no + 3 an1 + 2 a2n2 + a3n3 THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 123 Thus, D = a(B + 4 M+ B,) = a [0(0) + 40(4 a) + i(a)] a= an0+ + I a a + - a-a3 + 4 a4n3. Cor. 2. Of any solid whose Ax = 0(x) = nx +?1 + n 2 33 + n4x4.+ mn Xm,. the volume is ano a +I3 + an + 3 +a4n3 + L a5, +....+ 1 am+1 nm. m+1 For the volume of the prism whose base is the cross-section b (x), and whose altitude is the nth part of the altitude of the whole solid, is a 6(x). n The limit of the sum of all the prisms of like height a(0()+91 a+0 2a). +0 n —l a n \n \n n when n becomes indefinitely great, is the volume of the whole solid. But IBut tlr 2r..+ ( -)r 1 when - f-l —+-i ri-i, when n co.?r+l + 1 (T). PRISMOIDAL SOLIDS OF REVOLUTION. The general expression A, = q + mx + nXz + f3, has as many possible varieties as there are combinations of four things taken one, two, three, and four together; that is, 24- 1, or 15 varieties. Corresponding to each of these there will be at least one solid of revolution generated by the curve whose equation is, in the general case, 2ry = q + mx + x,22 + f3. For, if y be the revolving ordinate of any point in the curve, then try2 is the area of the section at distance x from one end of the solid. 124 MENSURATION. XLV. EXAMINATION OF THE DIFFERENT CASES. (1) Let r/y2. y;.' y is constant, and the solid is a circular cylinder. (2) Let ry2- mx; '. y2 c x, and the solid is a paraboloid of revolution; for, in a para1 M ^ ~ bola, the square of the ordi- -I - i llfnate varies as the abscissa. = ( ll'l (3) Let 7ry2=-n;.'. y oc x, and -_I_ the solid is a right circular cone. (4) Let - ry23; *' yoc x, and the solid is a semicubic paraboloid of revolution. (5) Let wry2- +m;.'. y2oc (A+x) where h is constant, and the solid is a frustum of a paraboloid of revolution, h being the height of the segment cut off. (6) Let 7ry2 = q + n2; supposing q and n positive, this is the equation to a hyperbola, the conjugate axis being the axis of x, and the center the origin. Hence, the solid is a hyperboloid of one nappe. (7) Let Try2= q +qf3. In this case, the solid is generated by the revolution of a curve, somewhat similar in form to the semicubic parabola, round a line parallel.: to the axis of x, and at a constant distance from it. (8) Let ry2 = mx - x2. In this case, the solid may be a sphere, a prolate spheroid, an oblate spheroid, a hyperboloid of revolution, or its conjugate hyperboloid. THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 125 (9) Let ry2= q + mx + nx. In this case, the solid will be afrustum of a circular cone, or of the sphere, / spheroids, or hyperbo- loids of revolution, made / ' by planes normal to the axis. In the frustum of -- the cone q, m, and n are all positive. The other solids in (8) and (9) are \ distinguished by the values and signs of the constants m and n. (10) Let 7ry2 - q + mx + nx2 +fx3. In this case, the solid is a frustum of a semicubic paraboloid of revolution. For, if x be the distance of the section from the smaller end of the frustum, and h the height of the segment cut off,.'. y2 (A + X)3... Ty2 is of the formi q + mx + nx2 +fx3. (11) Let rry2 - tx +f3. (12) Let 7ry2 nx2 +fx3. (13) Let 7ry =q + fmx +fx3. (14) Let 7ry2 -= q + nx +fx3. (15) Let Try2 = 1x + nx2 +f3. EXAM. 93. Since, for an oblate spheroid, B,=0, B2=0, 4M =4 ra2, and h=2b, therefore its volume ( = 34 Ta b. Similarly, the volume of a prolate spheroid 0= raab2. Thus each is, like the sphere, two-thirds of the circumscribed cylinder. 126 MENSURATION. EXAM. 94. The volume of the solid generated by the revolution round the conjugate axis of an arc of a hyperbola, cut off by a chord = and 11 to the conjugate axis, is twice the spheroid generated by the revolution of the ellipse which has the same axes. Making the conjugate the axis of x, y2 = (b + 2). B1 is the value of Try2 when x b;.'. B = 2 ra2. B2 is the value of 7y2 when = - b;.'. B 2 ra2. LM is the value of 7ry2 when = 0;.'. 4 M= 4ra2. Since the conjugate axis is the height of the solid,.'. h =2b. Hence its volume X= -7ra2b. Ans. ExAM. 95. To find the volume of a paraboloid of revolution. Let A be its height, that is, the length of the axis, r the radius of its base, and p the parameter of the generating parabola, y2 px. Then B,=O, B= irr2= rph, h.. — = 2Trph2. Ans. Cor. Since r2 = ph,.., = Trphh = ~ 1rTh, or a paraboloid is half its circumscribing cylinder. EXAM. 96. To find the volume of any frustum of a paraboloid contained between two planes normal to the axis. F. ( = rph22 - - rph12 = I — rp (h22- h2) = ip (h2 + h) (h2 - h) THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 127 But r = ph1, and '2 = ph, h + h2 = (r12 +r2); and h - h = a, the altitude of the frustum.. F. p = -l(r1 + r2)a = -- Cra (.' 12 + r22). Ans. (U). PRISMOIDAL SOLIDS NOT OF REVOLUTION. Let us now consider the same fifteen possible varieties when A, is not of the form Try2. (1) Let A= q. In this case all the transverse sections are constant. This is the property of all prisms and cylinders; also, of all solids uniformly twisted, e.g., the square-threaded screw. (2) Let Ax mx. This is a property of the elliptic paraboloid, or the solid generated by the motion of a variable ellipse whose axes are the double ordinates of two parabolas which have a common axis and a 128 MENSURATION. common vertex, the plane of the ellipse being always normal to this axis; for, in this solid, the area of the section at distance x from the vertex will be 7ryyI where y and y' are the ordinates of the two parabolas, and since both y2 and y'2 vary as x,.'. ryy' varies as x. (3) Let A, = znx2. This is a property of all pyramids and cones, whatever may be their bsi3es. (4) Let Ax =fx. The solid will be an elliptic semicubic paraboloid. Substitute semicubic for common parabolas in (2). (5) Let A, =g q- + x. This is a prop-:-._I. erty of a fr'ustum of an elliptic paraboloid. (6) Let A,= q +- nx2. This is a property of a groin, a simple case of which is the square groin seen in the vaults of large buildings. This solid may be generated by a variable square, which moves parallel to itself, with the midpoints of two opposite sides always in a semicircumference, the plane of which is perpendicular to that of the square. If y is a side of the square when at a distance x from the centre;.'. y2=4(r2 -x2). (7) Let A=mx + nx2. This is a property of the ellipsoid, and of the elliptic hyperboloid. (8) Let A = q - mx + nx2. This is a property of a prismoid, and of any frustum of a pyramid or cone, whatever may be the base; also, of any frustum of an ellipsoid, or elliptic hyperboloid made by planes perpendicular to the axis of the hyperboloid, or to any one of the three axes of the ellipsoid. THE APPLICABILITY OF THE PRISMOIDAL FORMULA. 129 EXAM. 97. To find the volume of an ellipsoid. Let a, b, c lie the three semi-axes, a the greatest;.. h=2a, B = 0, B =O, 4M =47rbc... 7=r7rabc. Ans. (V). ELIMINATION OF ONE BASE. For all solids whose section is a function of degree not higher than the second, or A2 = q + mx + nx2, q, m, n, and consequently Ax, for all values of x, are determined if the value of AS for three values of x is known. Measuring x from B1 we have A = B =. Supposing we know the section at 1 the height of the z solid above B1, we have for determining m and n the two equations, B2 = B1 + ma + na2, a a2 Az = B, + m- +? z2 z 12 Hence, z2A _- (z2 - 1)B - B2 qr7, = -— ^ ---— X- - I - (z-)a zB2 + z(z- 1)B1?- z1As (z- 1)a2 For the volume of the solid we have V = Bla + ~ ma2 + -I naa, or = a [(2 - 3) -(z -1) (z - 3)B, + 2A]. 6 (z -1) 2 130 MENSURATION. For z = 3, this gives V=, a(B + 3As). 4 Again, for z= 1-, V= a (B/ + 3A~). These give the following theorem: 100. To find the volume of a prismatoid, or of any solid whose section gives a quadratic: Rule: Jiultiply one fourth its altitude by the sum of one base and three times a section distant from that base twothirds the altitude. Cor. If B2 reduces to an edge or a point, V==4aAs. CHAPTER VII. APPROXIMATION TO ALL SURFACES AND SOLIDS. ~ (W). WEDDLE'S METHOD. 101. To find the content between the first and seventh of equidistant sections: Weddle's Rule: To five times the sum of the even sections add the middle section, and all the odd sections; multiply this sum by three-tenths of the common distance between the sections. Tormula: 4 = -Al r5 (y2+ y +3y6) + Y4 +Y 1+ 3 + 5 + 7] Proof: If we take the origin midway between the ends (of the solid or plane figure), and suppose every section y perpendicular to the axis expressible as an algebraic function of positive integer powers of x not exceeding the seventh degree, then for every y- = (z) = A + Bx + CX2 + Dx3 + E tx - + x + Gx, + EHv, we must have a y'- ~ (-_ ) = A - + + -1Ex -F F L + Gx6 - IH7. *. y + y'= 2 (A + CV2 + ExC4 + Gill). 132 MENSURATION. Therefore the whole content of the figure is = r r74 76 6 mq x ^(A+32C2- +34E4 3666), n=l_2 m4 in which equals h(A - 3 C7h2 + -8Eh4 -9 Gh6). Now Weddle's Rule gives for the same figure 5(Y2 G)= 10(A+ 4Ch2+ 16Eh4+ 64Gh6) 6y4 = 6A 1 + Y7 - 2(A+- 9CAh2+ 81Eh4 + 729Gh6) Y3 -+ Y = 2 (A + Ch2 + Eh4+ Ch6).. = -3h [20A+ 60C7h2 + 324Eh4 + 2100Gh6].'. f = 6h(A + 3Ch2 + E —.h4 + t7-5h6). Therefore the value given by our formula is in excess by the quantity - Gh7. Therefore the error in the last term is w_ = 6 = 0-0082+, 6X 29Gh7 J2 or the error is less than T —o- of the last term. Hence the error in the whole quantity will be very much smaller than this, since the most important terms in the expression for y are the earlier terms. Change of origin does not change the degree of an equation; hence, we have demonstrated that by means of Weddle's Rule we can find exactly the contents of any surface or solid in which the sections may be expressed by a rational integral algebraic function, of a degree not higher than the fifth, of its distance from either end; and very approximately when the expression is of the sixth or seventh degree. Cor. Suppose there are (6n + 1) equidistant sections Y1,?2, Y3, Y4,.... Y6n+l, APPROXIMATION TO ALL SURFACES AND SOLIDS. 133 y3 and Y6,~+l being the extreme or bounding sections; and let h be their common distance; then, 3 0h [/Yodd + 5 zYeven + zyevery third], observing not to take either of the two extreme sections twice; that is, begin Sy every third with Y4, and end it with y3.-2. EXAM. 98. Between yl and y19 l+ A/ [yl + Y3 + 3/ + Y7 + Y9 ++ /11 + fl3 + Y15 + Y17 + Y19+ + 5 (Y2 + Y4 + Y6 + Y/8 + Y1o + Y12 + Y14 + Y16 +- 18)+ + Y4 + 3/7 + /10 + Y1'3 + /16]. EXAM. 99. Find the volume of the middle frustum of a parabolic spindle. A spindle is a solid generated by the revolution of an arc of a curve round its chord, if the curve is symmetric about the perpendicular bisector of the chord. Hence a parabolic spindle is generated by the revolution of an arc of a parabola round a chord perpendicular to the axis. Let the altitude of the middle frustum be divided into six parts each equal to A, and let A1, A2, A3, A4, A6, A, be the areas of the sections. Taking origin at center and r the longest radius, or radius of mid-section, by the equation to a parabola, we have for any point on the curve x2 == a (r - y), x2.'. ~r --—; Ca will be the area of the section at the distance x from C; and this being a rational integral function of x of the 184 M[ENSURATION. fourtA degree, Weddle's Rule will determine the volume exactly. Now, if rl is the shortest radius, or radius of the two bases, we have A A ( 2 4 72 1 ) A, =.2 2r + arA=4 T2. Therefore, by 101, =-3h 5r (3r2-16r- r 3) + r) 22 rl 27rr2+ r2 -.2rht + ) a a2 \ a ]6~2 r- Tr But (3 7)2 = a(r-);... ra 9 '. = - hT [18 2 + 2 r,2 - --. (- r,) + 2(r- r1)2]. "- t= 10[32r72+16rr+12r,2]. 3.. = -2 h((Sr 2 + 4 rri + 3 12). Aqz. Making r1- 0, gives for the volume of the entire spindle, B156 1w,2 But 6 hTrr2 is the volume of the circumscribing cylinder; hence volume of spindle is A — circumscribing cylinder. The middle frustum of a parabolic spindle is a very close approximation to the general form of a cask, and hence is used in cask-gauging. Ex AMv. 100. The interior length of a cask is 30 decimeters; the bung diameter, 24 decimeters; and the head diameters, 18 decimeters. Find the capacity of the cask. Here r == 12) r 9, 79 h=5;.. k= -r (2304 + 864 + 486) =7r X 3654. Ans. APPROXIMATION TO ALL SURFACES AND SOLIDS, 135 EXAM. 101. The two radii which form a diameter of a circle are bisected, and ordinates are raised at the points of bisection. Find approximately the area of that portion of the circle between them. Here 6h = r= y. y, = y7 - / 2 = /_ (_r2 - r (3 h)2 / r Y2 = Y: 2 _( h) Vr2 _ ) - 28 =r yy x1/rV -19 V62- - )2 = 3 r2 35. Hence, by 101, <X = ox6 ' -\/8 +r) + r + r ~ 4:L TVs/8 + 6 + v+3 = 6 [ 18 + 20 2 + 3 3 + -35]..'..= " 2 X 0'956608. Ans. But the exact area is the difference between a semicircle and the segment whose height is half the radius. Taking tr =31415927 and V/3= 1-7320508 gives for this area r2 X 0-956612, so that our approximation is true to five places of decimals. In all approximate applications, it is desirable to avoid great differences between consecutive ordinates; applied to a quadrant of a circle, Weddle's Rule leads to a result correct to only two places of decimals. CHAPTER VIII. MASS-CENTER. 102. The point whose distances from three planes at right angles to one another are respectively equal to the mean distances of any group of points from these planes, is at a distance from any plane whatever equal to the mean distance of the group from the same plane. 103. The mass-center of a system of equal material points is the point whose distance is equal to their average distance from any plane whatever. 104. A solid is homogeneous when any two parts of equal volume are exactly of the same mass. The determination of the mass-center of a homogeneous body is, therefore, a purely geometrical question. Again, in a very thin sheet of uniform thickness, the masses of any two portions are proportional to the areas. In a very thin wire of uniform thickness, the masses of different portions will be proportional to their lengths. Hence we may find the mass-center of a surface or of a line. SYMMETRY. 105, Two points are symmetric when at equal distances on opposite sides of a fixed point, line, or plane. 106, If a body have a plane of symmetry, the mass-center /C lies in that plane. Every particle on one side cor MASS-CENTER. 137 responds to an equal particle on the other. Hence the /'C of every pair is in the plane, and therefore also the /C of the whole. 107. If a body have two planes of symmetry, the /C lies in their line of intersection; and if it have three planes of symmetry intersecting in two lines, the /C is at the point where the lines cut one another. 108. If a body have an axis of symmetry, the /C is in that line. 109. If a body have a center of symmetry, it is the CO. 110. The /tC of a straight line is its midpoint. 111. The PC of the circumference or area of a circle is the center. 112. The /C of the perimeter or area of a parallelogram is the intersection of the diagonals. 113. The /C of the volume or surface of a sphere is the center. 114. The /C of a right circular cylinder is the midpoint of its axis. 115. The C0 of a parallelepiped is the intersection of two diagonals. 116, The PC of a regular figure coincides with the IO of its perimeter, and the /C of its angular points. 117. The yC of a trapezoid lies on the line joining the midpoints of the parallel sides. 138 MENSURATION. 118. Since in any triangle each medial bisects every line drawn parallel to its own base, therefore the (C of any triangle is the intersection of its medials. By similar triangles, this point lies on each medial two-thirds its length from its vertex, and so coincides with the /C of the three vertices. 119. The YC of the perimeter of any triangle is the center of the circle.inscribed in the triangle whose vertices are the midpoints of the sides. Proof: The mass of each side is proportional to its length, and its PC is its midpoint. So the [C of ]W and Ms is at a point D, such that DgMk c =i AB 1iM DM~3 b A AG MC M Hence, the 0C of the whole perimeter is in the line D11; and since D111 divides the base AL.sl into parts proportional to the sides, it bisects 4 Xlf. Similarly the [UC is in 'the line bisecting 4 Ml., 120. The [C of the surface of any tetrahedron is the center of the sphere inscribed in the tetrahedron whose summits are the I/C's of the faces. 121, If the vertex of a triangular pyramid be joined with the [C of the base, the [C of the pyramid is in this line at three-fourths of its length from the vertex. (Proof by similar triangles.) 122, The /LC of a tetrahedron is also the /C of its four summits. 123. The YC of any pyramid or cone is in the line joining the 'C of the base with the apex at three-fourths of its length from the apex. MASS-CENTER. 139 124. A sect is a limited line or rod. 125. The opposite to a point P on a sect AB is a point P', such that P and P' are at equal distances from the center of AB, but on opposite sides of it. 126. The "C of any quadrilateral is the /LC of the triangle whose apices are the intersection of its two diagonals and the opposites of that intersection on those two diagonals respectively. Proof. Construct the C(7's E and y of the triangles AB C and A. CD made by the diagonal A C of the quadrilateral ABCD; then the point on the line EF, which divides it inversely as the areas of these triangles, will be the /C of the quadrilateral. But if the diagonal BD is cut by AC in the point G, then ABC: ACD = BG: GD. So the sought point is the /C of BG x E and DG x F; that is, of BG xA, BG xB, BGx C and GD XA, GD x D, GD x C. But we may substitute BD X A for BG x A and GD x A; also BD x C for BG x C and GD x C For BG x B and GD X D we may substitute ]BD x Ai, where K is the opposite of G on the sect BD. Therefore the sought point is the /C of A, C, and K, that is, of G, J, and K, where J is the opposite of G on AC. Cor. Calling the /C of the quadrilateral L, we have ACL = ACF — AEC = - (A CD-ABC). 127. The P/C of the four angular points of a quadrilateral is the intersection of the lines joining the midpoints of pairs of opposite sides. Let this point O be called the midcenter, and G the intersection of the two diagonals be called the cross-center; then the /'C of the quadrilateral is in the line joining these two centers produced past the mid .140 MENSURATION. center, and at a distance from it equal to one-third of the distance between the two centers. That is, LOG will be a straight line, and OL= OG. 128. In any body between parallel planes, we can reckon the distance of its /C above its base, if every cross-section is a given function (x) of its distance x from the base. The prism whose base is the section A(x), and whose height is the nth part of the altitude a of the body, has for volume -c (x). Its 0C is x+ - from the base of the body, n 2n and has the coefficient a- (x). Now suppose the MC of the n body distant r from its base, and form the product of r with the sum D of the values which a b (x) takes when x equals 1 2 n —1 I 0, -a, -a,..., a; also form the sum S of the values n n n which (x4 —i>- ) (x) takes for the same. worths of x. The 2 2nn product rD equals the sum S when the arbitrary number n is taken indefinitely great. rD= S, for n - oo. But for n - oo, the sum D expresses the volume of the body. r The sum 8 consists of the sum C of terms from -x (x), 92 and of the sum E of terms from 2a -a(x). But the sum E has the value - D, and vanishes for = oo. 2n Therefore, for the determination of r, we have the equation D = C. MASS-CENTER. 141 If (x) is of degree not higher than the second, then x q (x), which we will call (x), is not higher than the third. Therefore, by 99, D== a [ 0(O) + 40 Q a) + (a)]. C= a f/(0) + 4/(l- a) +/f ()]. But /(0) 0, /- a)= a (- a), f(a)= a(a). rm, r. 1= T, ~ ~2o(i a) +o(a) Therefore, a 2T() + 2a (a) a (0) + 40((a) +- (a) a' [ (a) - 9 (0)] Therefore, r = a + [1(a)- ( 12D or a + 12D 129. For the applicability of the Mass-Center Formula: = a+2(B-,) 12D the test is the = is A +mx+nx2. For an examination of the possible varieties, see 99, ~ (T), (1), (2), (3), (5), (6), (8), (9); and ~ (U), (1), (2), (3), (5), (6), (7), (8). Of course it applies also to the corresponding plane figures. EXAM. 102. For trapezoid a2(b2- bl) a(2b2+ b) 12 (b + b,) 2 3(b, + b2 EXAM. 103. For pyramid or cone fr=a + a 1a 12(I aB2) from B1, or i4- a from B2. 142 MENSURATION. EXAM. 104. The t'C of a pyrimidal frustum is in the line joining the /C's of the parallel faces, and _ I a + a (B2-B1) 2 4 (B + BB2 + B2) So, if any two homologous basal edges are as I to X, the distance of the frustral 'C from one base will be as 12 + 22 +2 A+ 3 \ 12 +21 A3 ), and from the other 47-_ 2 _- 2 EXAM. 105. From Exam. 95, for paraboloid I a 27'7 3__ o r = a + (a == 2 a. 6 7w r2a For frustum of this we obtain an expression similar to that for trapezoid. It is c t2 r,- (? 2 ) C (222r +r 2 1 (r 2 ) 3 (22 + 2) EXAM. 106. For PC of half-globe, from center, 4r 1 rr 130. The average haul of a piece of excavation is the distance between the /C of the material as found and its /C as deposited. 131. The PC of a series of consecutive equally-long quadratic shapes may be found by assuming the PC' of each shape to be in its mid-section, then compounding, and to the distance of the point thus found from A2 adding a ( Bl) 12-D MASS-CENTER. 143 NOTE. It is a singular advantage of the PC formula that its second or correction term remains as simple for any number of shapes in the series as for one. In consequence, the error of the assumption that the /C of each shape is in its mid-section, is less as the series is longer. No error whatever results from this assumption when the end areas B1 and B2 are equal. For instance, in finding /-C of a spherical sector whose component cone and segment have equal altitude, we may assume that the PCof each is midway between its bases. 132. THE MASS-CENTER OF AN OCTAHEDRON. Let AF, BiG, CIIbe three sects (finite lines) not meeting. By an octahedron understand the solid whose eight faces are AB_ ACG,, AGR, AHB, A FBC, FCG, FGH, FHB. The solid is girdled by the perimeters of three C skew quadrilaterals, BCGG_, CAH F, //,. G ABFG. Now the mid-points of the B. / sides of any skew quadrilaterals are\ in one plane. Draw, then, three / planes bisecting the sides of these F quadrilaterals, and let them meet in a point K, which call the cross-center. Let, also, 1(mizd-center) be the mean point of the six vertices A, B, C F, G, f-; it is the /C of the triangle formed by the mid-points of AF, BG, CH. To find, the /C of the solid, join KJf, and produce it to S so that Proof: The sol.id is the sum of the four tetrahedra AFBC, AFCG, AFGH, AFffB. Now the PC of a tetrahedron is the PC or mean point of its vertices; consequently the line joining the /C of AFB C to the mid-point of GIH is divided by the point il in the 144 MENSURATION. ratio 1: 2. The same is true of the other three tetrahedra and the mid-points of JB, BC, CG. Hence, the masscenters of the four tetrahedra are in one plane passing through the point S found by the above construction, and therefore the tC of the whole solid is in this plane. So, also, it is in the other two planes determined by dividing the solid into tetrahedra having the common edge BG and the common edge C(f respectively. Therefore it coincides with the point S. Cor. By making the pairs of faces ABE, AHIG; A CG, CFG; CBF, BHF to be respectively copular, we pass to a truncated triangular pyramid. If we join its cross-center K with its mid-center M, and produce XKM to S, making M/ = -2 KM, then S will be the YC of the trunc. NOTE. This corollary was Sylvester's extension of the geometric method of centering the plane quadrilateral, and suggested to Clifford the above. EXERCISES AND PROBLEMIS IN MENSURATION. PROBLEMS AND EXERCISES ON CHAPTER I. 1. a2+b2 c. EXERCISE 1. Find the diagonal of a square whose side is unity. V/2 141421+. Answer. 2. Find the diagonal of a cube whose edge is unity. V3 =1-732050+. Ans. 3. To draw a perpendicular to a line at the point C. Measure CA =3 meters, and fix at A and C the ends of a line 9 meters long; which stretch by the point B taken 4 meters from C and 5 meters from A. BC is I to AC because 32 + 42 = 52. 4. The whole numbers which express the lengths of the sides of a right-angled triangle, when reduced to the lowest numbers possible by dividing them by their common divisors, cannot be all even numbers. Nor can they be all odd. For, if a and b are odd, a2 and b2 are also odd, each being an odd number taken an odd number of times. a'2 + b is even, and.'. c is even. 2. C2_ 2- (c + a- ) (c- c) - h. 5. To obtain three whole numbers which shall represent the sides of a right-angled triangle. 146 MENSURATION. n2 - 1 Rule of Pythagoras. Take n any odd number, then the second number, and n = the third number. 2 2 Plato's Rule. Take any even number mn, then also — 1, 72 4 and4 1. Euclid's Rule. Take x and y, either both odd or both even, such that xy= b2, a perfect square; then ac= -- Y 2+v 2 and c = -X Rule of Maseres. Of any two numbers take twice their product, the difference of their squares, and the sum of their squares. Proof: Let the whole numbers which express the lengths of the two sides of a rt. A be a and rm. If c be a whole number, it must = a + n, where n is a whole number. By 1,.', a + m=c2= a2 an+ n2..-. m2 = 2 an + n2.. 2an = m2 - n2. m2 - n2 2n m2 - n2+ 22 n m + n2.'. a + n- c= - 2n 2n ms - n2- 2 mn Therefore, the three sides, a, m, c, are -—, -, and m2n ~n2 2n 2n m ~2 n Magnifying the rt. A 2n times, its sides are expressed by the whole numbers -n2 2n, and pressed by the whole numbers mn2 -- n, 2mn, and m2 +.n2 EXERCISES AND PROBLEMS. 147 TABLE I. -DISSIMILAR RIGHT-ANGLED Sides. Area. Sides. TRIANGLES. 3 4 5 5 12 13 8 15 17 7 24 25 9' 40 41 12 35 37 20 21 29 11 60 61 16 63 65 13 84 85 28 45 53 15 112 113 33 56 65 20 99 101 17 144 145 48 55 73 36 77 85 39 80 89 19 180 181 24 143 145 21 220 221 65 72 97 44 117 -125 60 91 109 28 195 197 23 264 265 51 140 149 25 312 313 32 255 257 52 165 173 88 105 137 27 364 365 6 30 60 84 180 210 210 330 504 546 630 840 924 990 1,224 1,320 1,386 1,560 1,710 1,716 2,310 2,340 2,574 2,730 2,730 3,036 3,570 3,900 4,080 4,290 4,620 4,914 57 176 85 132 36 323 29 420 60 221 119 120 31 480 84 187 104 153 95 168 40 399 69 260 33 544 68 285 133 156 44 483 35 612 105 208 75 308 96 247 140 171 120 209 37 684 76 357 48 575 115 252 39 760 52 675 87 416 160 231 136 273 161 240 Area. 185 5,016 157 5,610 325 5,814 421 6,090 229 6,630 169 7,140 481 7,440 205 7,854 185 7,956 193 7,980 401 7,980 269 8,970 54.5 8,976 293 9,690 205 10,374 485 10,626 613 10,710 233 10,920 317 11,550 265 11,856 221 11,970 241 12,540 685 12,654 365 13,566 577 13,800 277 14,490 761 14,820 677 17,550 425 18,096 281 18,480 305 18,564 289 19,320 148 MENSURATION. TABLE I.-Continued. Sides. Are'a. Sides. Area. 56 783 785 93 47G 485 207 224 305 120 391 409 135 352 377 92 525 533 175 288 337 204 253 325 152 345 377 180 299 349 60 899 91.0 145 408 433 225 272 353 100 621 629 105 608 617 189 340 389 64 1,023 1,025 252 275 373 168 425 457 155 468 493 228 325 397 111 680 689 108 725 733 68 -1,155 1,157 203 396 445 165 532 557 297 304 425 72 1,295 1,297 184 513 545 116: 837 845 280 351 449 217 456 505 261 380 461 21,924 22,134 23,184 23,460 23,760 24,150 25,200 25,806 26,220 26,910 26,970 29,580 30,600 31,050 31,920 32,130 32,736 34,650 35,700 36,270 37,050 37,740 39,150 39,270 40,194 43,890 45,144 46,620 47,196 48,546 49,140 49,476 49,590 319 360 481 124 957 965 231 520 569 200 609 G41 279 440 521 185 672 697 336 377 505 80 1,599 1,601 308 435 533 195 748 773 396 403 565 259 660 709 368 465 593 336 527 625 315 572 653 273 736 785 400 561 689 364 627 725 455 528 697 407 624 745 301 900 949 468 595 757 432 665 793 369 800 881 429 700 821 315 988 1,037 555 572 797 540 629 829 451 780 901 504 703 865 329 1,080 1,129 420 851 949 464 777 905 533 756 925 57,420 59,334 60,060 60,900 61,380 62,160 63,336 63,960 66,990 72,930 79,794 85,470 85,560 88,536 90,090 100,464 112,200 114,114 120,120 126,984 135,450 139,230 143,640 147,600 150,150 155,610 158,730 169,830 175,890 177,156 177,660 178,710 180,264 201,474 76 1,443 1,445 548,34 EXERCISES AND PROBLEMS. 149 TABLE I. -Concluded. Sides. Area. Sides. Area. 616 663 905 473 864 985 580 741 941 496 897 1,025 615 728 953 330 644 725 557 840 1,009 696 697 985 660 779 1,021 645 812 1,037 476 1,107 1,205 620 861 1,061 585 928 1,097 731 780 1,069 504 1,247 1,345 704 903 1,145 660 989 1,189 612 1,075 2,237 765 868 1,157 705 992 1,217 832 855 1,193 532 1,395 1,493 799 960 1,249 204,204 204,336 214,890 222,456 223,860 107,226 234,780 242,556 257,070 261,870 263,466 266,910 271,440 285,090 314,244 317,856 326,370 329,950 332,010 349,680 748 1,035 1,277 893 924 1,285 560 1,551 1,649 884 987 1,325 792 1,175 1,417 684 1,363 1,525 740 1,269 1,469 931 1,021 1,381 833 1,056 1,345 969 1,120 1,481 720 1,519 1,681 836 1,325 1,565 780 1,421 1,621 936 1,127 1,465 1,036 1,173 1,565 988 1,275 1,613 880 1,479 1,721 1,113 1,184 1,625 1,140 1,219 1,669 1,040 1,431 1,769 1,248 1,265 1,777 1,148 1,485 1,877 1,312 1,425 1,937 387,090 412,566 434,280 436,254 465,300 466,830 469,530 474,810 481,474 542,640 546,840 554,014 554,190 556,738 607,614 629,850 650,760 658,896 694,830 744,120 799,480 852,390 863,550 355,680 371,070 383,520 Sides. Area. 4,059 4,060 5,741 23,660 23,661 33,461 207,000 207,151 292,849 159,140,519 159,140,520 225,058,681 150 MENSURATION. 3. a2 +b2 2bj = 2. 6. Two sides, a- 6'708, b = 5, contain an obtuse angle. If j= 3, find c. 10. Ans. 7. If a =13, b 11, c =20, find j. 5. Ans. 4. (62+ b2- 2 j- =c2. 8. The three sides of a triangle are 2.5 meters, 4-8 meters, 3-2 meters; find the projections of the other two sides on b= 48 meters. 1-985 meters and 2-815 meters. Ans. 9. Two sides of a triangle, 3 meters and 8 meters long, enclose an angle of 60~ Find the third side. 7 meters. Ans. HINT. Joining the midpoint of 3 with the vertex of the rt. 4. made in projecting 3 on 8 gives two isosceles triangles, and.. j-1. 10. Two sides of a triangle are 13 meters and 15 meters. Opposite the first is an angle of 60~ Find the third. 7 or 8 meters. Ans. HINT. Drop I to 15-meter side. The segment adjacent the third side is half the third side. 11. Two sides of a triangle are 9'6 meters and 12 8 meters; the perpendicular from their vertex on the third side is 7-68 meters; find that side. 4. Ans. 6. a2+c2-ib2=2i2. 12. Given ii, the medial from A to a; i2, from B to b; and is, from C to c. Find a, b, and c. From 6, a + -b2= 2i2 a2 +2 - ==2i 2 a2 + b2 - C2 2i32, b2 + c2 _ a'2-2 i2. EXERCISES AND PROBLEMS. 151 Taking the last equation from twice the sum of the two former gives 4a2 + a2 = 2(2i22 + 2 i32 -i).. a- -2V2i,2+ 2 i-i2. Ans. Symmetrically, find b and c. 13. If i1 = 18, i -= 24, i= 30; find a, b, and c. a - 34-176, b = 28-844, = 20. Ans. 14. Prove i+2 +- 2 + i232 = (a 2+ b2 + C2). 15. In any right-angled triangle prove 1a= 5 42- -i. 15 16. The locus of a point, the sum of the squares of whose distances from two fixed points is constant, is a circumference whose center is the midpoint of the straight line joining the two fixed points. 17. The locus of points, the difference of the squares of whose distances from two fixed points is constant, is a straight line perpendicular to that which joins the two fixed points. 18. The sum of any two sides of a triangle is greater than twice the concurrent medial. 19. In every quadrilateral the sum of the squares of the four sides exceeds the sum of the squares of the two diagonals by four times the square of the straight line joining the middle points of the diagonals. 20. The sum of the squares on the four sides of a parallelogram is equivalent to the sum of the squares on the diagonals. 21. The sum of the squares of the diagonals of a trapezoid is equal to the sum of the squares of the non-parallel sides augmented by twice the product of the parallel sides. 152 MENSURATION. 22. On the three sides of a triangle squares are described outward. Prove that the three lines joining the ends of their outer sides are twice the medials of the triangle and perpendicular to them. 23. Prove that the medials of a triangle cut each other into parts which are as 1 to 2. 24. The intersection-point of medials is the center of gravity of the triangle. 25. Prove an 4 in a A, acute, rt., or obtuse, according as the medial through the vertex of that 4 is >, =, or < half the opposite side. 7. a2babl 26. The three sides of a triangle are 17-4 meters, 23-4 meters, 31-8 meters. The smallest side of a similar triangle is 5-8 meters. Find the other sides. 7'8 meters and 10.6 meters. Ans. 27. Find the height of an object whose shadow is 37-8 meters, when a rod of 2-75 meters casts 1-4 meters shadow. 74-25 meters. Ans. 28. The perpendicular from any point on a circumference to the diameter is a mean proportional between the two segments of the diameter. 29. Every chord of a circle is a mean proportional between the diameter drawn from one of its extremities, and its projection on that diameter. 30. From a hypothenuse of 72-9 meters, a perpendicular from the right angle cuts a part equal to 6-4 meters; find the sides. a --- 21-6 meters, b = 69-62 meters. Ans. EXERCISES AND PROBLEMS. 153 31. The hypothenuse is 32-5 meters, and the perpendicular on it 15-6 meters; find the segments. 11-7 and 20-8. Ans. 32. In a rt. A, c = 36-5 meters; a + b 51-1 meters; find a and b. a - 21*9 meters, b =29-2 meters. Ans. 33. The sides of a triangle are 4-55, 6-3, and 4-445; the perimeter of a similar triangle is 4-37. Find its sides. 1-3, 1.8, and 1-27. Ans. 34. A lamp-post 3 meters high is 5 meters from a man 2 meters tall; find the length of his shadow. 35. The sides of a pentagon are 12, 20, 11, 15, and 22; the perimeter of a similar pentagon is 16 meters. Find its sides. 2-4, 4, 2'2, 3, and 4-4. Ans. 36. Through the point of intersection of the diagonals of a trapezoid a line is drawn parallel to the parallel sides. Prove that the parallel sides have the same ratio as the parts into which this line cuts the non-parallel sides. 8. d= h 9. Ah -~-r - r ~~ 10. k V22+- V i. 37. Find the side of a regular decagon. If a radius is divided in extreme and mean ratio, the greater segment is equal to a side of the inscribed regular decagon. Ww. 394; (Eu. IV. 10; Cv. V. 17)..'. r(r- ko)- k,,) lo2, r2 - rk =:2 k2 + rk = r2, -1 + rk + r2= - r'2. k'. klo=r. Ans. 2 154 MENSURATION. 11. 1/:L -^-z\2 2 /rV4 7 2-k. 38. Prove that the sides of the regular pentagon, hexagon, and decagon will form a right-angled triangle; or kI102 + k62 k52. 39. Show that k2 =- r2 -- 2. k48 = r \ 2-2 + V2 V, k96 = r2 - 2 + \2 + \V2 + 3, etc. 40. Show that k4 = rV2. k = r2-, k,1 = rAl22 + V2, 32=r - 2 + 2 t2, etc. Vr k2 12. t = 2 v\4_ --- 41. If a. is the apothegm of a regular inscribed n-gon, prove a2n = HINT. Use 7 and 2. 42. If also p', be perimeter of inscribed polygon, np of ncircumscribed, prove M p' 2= _lnr43. pn: 7 Pn a.n44. Indicate how to reckon wr by 40, 41, and 42, or by the following P2n == - l - and p2n = - 2n. 13. 45. When a quantity increases continually without becoming greater than a fixed quantity, it has a limit equal or inferior to this constant. EXERCISES AND PROBLEMS. 155 46. When a quantity decreases continually without becoming less than a fixed quantity, it has a limit equal or superior to this constant. 47. When two variables are always equal, if one of them has a limit, so has the other also. 14. Lim p lim p' = c. 48. Show that p' 2 - P',np2. 2p,7 1 2pRnp n 49. Prove p'2, p i-n4 F, and also p2=,,, 4 Pn rove n P nppn 50. If 9, r. be the radii of the circumscribed and inscribed circles of a regular polygon of n sides, and 92n, r2n the corresponding radii for a regular 2n-gon of the same perimeter as the n-gon; then 9r2,= 2, and 9n + r — 2 r2. 51. If i, 2, 2;3, etc., 2n, be the angles of a polygon of 2n sides, inscribed in a circle, then il + f3 + etc. + 2,n-1 -= + ~4 + etc. + 4-. 52. The greater the number of sides of a regular polygon, the greater is the magnitude of each of its angles. The limit is f, which can never be reached since the sum of the exterior angles is always 6. 15.?r- 3-141592653589793238462643383279+. 16. cl: C2:: '1 r2. 17. r 2.-C= 31416-. d r 18. c = dr 2 rr. 53. Find c when r 14, taking 7r = 2. 88. Ans. 19. d=2r= — c x 0.3183098+. 7r 54. Find the diameter of a wheel, which, in a street 19,635 meters long, makes 3,125 revolutions. 2 meters. Ans. 156 MENSURATION. 55. The hypothenuse is 10, and one side is 8; semicircles are described on the three sides. Find the radius of the semicircle whose circumference equals the circumferences of the three semicircles so described. 12. Ans. 56. Find the radius of a sphere in which a section 30 centimeters from the center has a circumference of 251-2 centimeters. r 32 (+ 251 2 An \2 X/ s. EXERCISES AND PROBLEMS ON CHAPTER II. 20. Arc measures 4 at center. 57. Find the third angle of a triangle whose first angle is 12~56" and second angle 114~48". 53~58'16". Ans. 58. The angles of a triangle are as 1: 2: 3. Find each. 30~, 60~, 90~. Ans. 59. Of the three angles of a triangle 4 a is 12~ 20' smaller than 4 /3, and 4 y is 5043' smaller than /3. Find each. a =53~41', 6 = 66~1', =60~ 18'. Ans. 60. The sum of two angles of a triangle is 174~48'24"; the difference of the same two is 48~24'50". Find all three. 110~ 36'37", 63~ 11'47", 5~11 36". Ans. 61. Three angles of a quadrilateral are 125~48'32", 127~58'45", 85~37'27"; find the fourth. 20~35'16". Ans. 62. The angles of a quadrilateral are as 2: 3: 4: 7; find each. 45~, 67~30', 90~, 157~30'. Ans. 63. In what regular polygon is every angle 168~45'? A 32-gon. Ans. 64. The vertical angle of an isosceles triangle is 148~47"; find the others. 15~59'36,". Ans. EXERCISES AND PROBLEMS. 157 65. The sum of two angles adjacent to one side of an isosceles triangle is 35~23'48"; find the three angles. 44~ 36'12", 44~ 36'12", and 90~ 47' 36". Ans. 21. u=-. r 66. Find the circular measure of the angle subtended by a circular watch-spring 3 millimeters long and radius 1~ millimeters. 67. If a perigon be divided into n equal parts, how many of them would a radian contain? n Ans. 27r -22. 1=- 3600 68. Find the arc pertaining to a central angle of 78~ when r = 1-5 meters. = 2-042 meters. Ans. 69. Find the arc intercepted by a central angle of 36~ 25' when r = 8-5 meters. I - 5-39 meters. Ans. 70. Find an arc of 112~ which is 4 meters longer than its radius. 1= 8 189 meters. Ans. *23. o_ 13600 71. When dc= 11-5, find arc 4-6 meters long. = 45~57' 14". Ans. 72. Calling 7r= -2, find r when 64~ measure 70-4 meters. r 63 meters. Ans. * HINT. 27rr: 70'4:: 360: 64. 24, 25, 26, 27, and 28. 73. Find the complement, supplement, and explernent of 30~. 158 MENSURATION. 74. Find the angle between the bisectors of any two adjacent angles. 75. If a medial equals half its base; its angle is right. 76. If one angle in a right-angled triangle is 30~, one side is half the hypothenuse. 77. In every isosceles right triangle half the hypothenuse equals the altitude upon it. 78. One angle in a right triangle is 30~; into what parts does the altitude divide the right angle? 79. How large is the angle between perpendiculars on two sides of an equilateral triangle? 80. Find the inscribed angle standing on an arc of 116 271 38". 58013'49". Ans. 81. Find the inscribed angle cutting out one-tenth circumference. 18~. Ans. 82. Find the angle of intersection of two secants which include arcs of 100~ 48' and 54~ 12'. 23~ 18'. Ans. 83. An angle made by two tangents is measured by the difference between 180~ and the smaller intercepted arc. 84. From the same point in a circumference two chords are drawn which cut off respectively arcs of 120~ and 80~; find the included angle. 85. The four angle-bisectors of any quadrilateral from a quadrilateral whose opposite angles are supplemental. 29. u= -0 1800' 86. Find the circular measure of 42,. 73303. Ans. 87. Find the circular measure of 45~, -785398. Ans. EXERCISES AND PROBLEMS. 159 88. Find the circular measure of 30~. -523598. Ans. 89. Find the circular measure of 60~. *. Ans. 3 7-2 90. Find the circular measure of 71~. 10. Ans. 180 91. Express seven-sixteenths of a right angle in circular measure. 92. Express in circular measure an angle of 240~. 47r. Ans. 3 93. Find in circular measure the angle made by the hands of a watch at 5:15 o'clock. 94. Find u of 4 made by watch-hands at a quarter to 8. 95. Find u of watch 4 at 3:30 o'clock. 96. Find u of watch 4 at 6:05 o'clock. 97. The length of an arc of 45~ in one circle is equal to that of 60~ in another. Find the circular measure of an angle which would be subtended at the center of the first by an arc equal to the radius of the second. 4. Ans. 3O.0 u 1800 30. g~ —. 7r 98. The angle whose circular measure equals one-half is 28~ 38' 52" 24"'. 99. Find the number of degrees in an angle whose u —2. g 23(57-2957795+) = 38-197186+. Ans. 100. Find 4 whose u =. 42~09718346. Ans. 101. Find the number of degrees in an angle whose u= -. 71-61972439. Ans. 4' a 160 MENSURATION. 102. Find the number of degrees, minutes, and seconds in an angle whose u= 1. 30~0'43"45. Ans. 103. The circular measure of the sum of two angles is 7trr, and their difference is 17~; find the angles. 27~15' and 10 15'. Ans. 104. Express in degrees the angle whose u = 7 r. 120~. Ans. 105. How many degrees, minutes, and seconds are there in the angle whose circular measure is? 47~44'47" nearly. Ans. 106. Express in degrees and circular measure the vertical angle of an isosceles triangle which is half of each of the angles at the base. 107. How many times is the angle between two consecutive sides of a regular hexagon contained (1) in a right angle? (2) in a radian? ) (2) 3 (1), (2) - Ans. 4 27w 108. Two wheels with fixed centers roll upon each other, and the circular measure of the angle through which one turns gives the number of degrees through which the other turns in the same time; find the ratio of the radii of the wheels. 180. A. Ans. 7r 1180o 31. 1180 7rg 109. The length of an arc of 60~ is 363; find the radius. r 35. Ans. 110. Find circumference where 4 30~ is subtended by arc 4 meters. 111. If 14 be the length of an arc of 45~ to radius rl, and 12 the length of arc 60~ to r2, prove 3rl - =- 4r2 1. EXERCISES AND PROBLEMS. 161 EXERCISES AND PROBLEMS ON CHAPTER III. 32. R- ab. 112. The area of a rectangle is 2,883 square meters; the diagonal measures 77-5 meters. Find the sides. b = 46-5 meters, a — 62 meters. Ans. HINT. a + b2= (77.5)2.... I. ab = 2,883.. II. Substitute in I. the value of a from II. This gives a biquadratic soluble as a quadratic. Or, to avoid the solution of a quadratic, multiply II. by 2, then add and subtract from I. This gives a2 + 2ab b2= 11,77225, a - 2 ab + b2 = 240-25... a+b = 108.5, a-b= 15-5. 113. Find the perimeter and area of a rectangle whose altitude a - 1,843-02 meters and base b = 845-6 meters. p = 5,377-24 meters, R= 1,558,457-712 square meters. Ans. 114. Find the number of boards 4 meters long, 0-5 meters broad, necessary to floor a rectangular room 16 meters long and 8 meters broad. 64. Ans. 115. Of two equivalent rectangles, one is 4-87 meters long and 2-84 meters broad, the other is 4-26 meters long. How broad is it? 3-246 meters. Ans. 116. The perimeter of a rectangle is 24-54 meters; the base is double the altitude. Find the area. 1R = 33-4562 square meters. Ans. 117. The difference of two sides of a rectangle is 1*4 meters; their sum, 8-2 meters. Find its area. R = 16-32 square meters. Ans. 162 MENSURATION. 118. The base of a rectangle of 46-44 square meters is 3-2 meters longer than the altitude; find these dimensions. b = 8-6 meters, a = 54 meters. Ans. HINT. b = a + 3-2 meters... a (a + 32) =46-44, etc. 119. The perimeter of a rectangle is 13 meters longer than the base; the area is 20-88 square meters. Find the sides. 5-8 and 3-6 meters, or 7-2 and 2-9 meters. Ans. 120. The perimeter of a rectangle is 3-78 meters; its diagonal, 1-35. Find the area. R = 0-8748 square meters. Ans. 121. A rectangular field is 60 meters long by 40 meters wide. It is surrounded by a road of uniform width, the whole area of which is equal to the area of the field. Find the width of the road. 10 meters. Ans. 122. A rectangular court is 20 meters longer than broad, and its area is 4,524 square meters; find its length and breadth. 78 and 58 meters. Ans. 33. qb2. 123. Find the area of a square whose side is 15-4 meters. 237-16 square meters. Ans. 124. The area of a square is - square meter; find its side. 0-79057 meters. Ans. 125. The side of a square is a; find the side when the area is n times as great. al — = n. Ans. 126. The sum of two squares is 900 square meters, the difference 252 square meters; find the sides. 24 and 18 meters. Ans. HINT. aa1 + a2 = 909, 2 a 2 = 252. 2 = 576. al EXERCISES AND PROBLEMS. 163 127. The sides of two squares differ by 12 meters; their areas by 240 square meters. Find the side and area of each. Sides, 4 and 16 meters; Areas, 16 and 256 square meters. Ans. HINT. (a + 12)2 - a2 = 240. 128. The perimeter of a square is 48 meters longer than the diagonal; find the area. 344-544375 square meters. Ans. HINT. 4 a =a2 -+48. 129. The sum of the diagonal and side of a square is 100 meters; find the area. 1715-6164 square meters. Ans. HINT. 2 a2 = (100 - a)2. 34. EZ= ab. 130. The area of a parallelogram is 120 square meters, two sides are 12 and 14 meters; find both diagonals. 24 and 10-2 meters. Ans. 131. The altitudes al and a2 of a parallelogram are 5 and 8 meters; one diagonal is 10 meters. Find the area. 40-285 or 100-65 square meters. Ans. HINT. b = 5 b, b, = 102 + b2 2b2j. Also, j2=100-64=36..-. j=6. 132. One side of a parallelogram is 8 meters longer than the corresponding altitude, and = 384 square meters; find this side. 24 meters. Ans. 35. A= 2ab. 133. The area of a rhombus is half the product of its diagonals. 164 MENSURATION. 134. From any point in an equilateral triangle the three perpendiculars on the sides together equal the altitude. 135. Find the area of a right triangle whose two sides are 248-2 and 160-5 meters. 19,918-05 square meters. Ans. 136. If 18-4 meters is the altitude of a A = 125-36 square meters, find b. 13-626 meters. Ans. 137. The two diagonals of a rhombus are 8-52 and 6-38 meters; find the area. 27-1788 square meters. Ans. 138. The altitude of a triangle is 8 meters longer than its base, and area is 44-02 square meters; find b. 6-2 meters. Ans. HINT. ( b(+ 8)= 44.02. 2 139. The altitude of a right triangle cuts the hypothenuse into two parts, 7-2 and 16-2 meters long; find the area. 126-36 square meters. Ans. 36. a= s (s - a) (s- b) (s - c). 140. 2 log logs +log (s-a) + log (s - b) + log (s - ). 141. If b is the base of an equilateral triangle, find the area. Here a=ib2 b2 b.. A=ab = 142. The altitude of an equilateral triangle is 8-5 meters; find the area. A - /-3 = 41-71 square meters. Ans. 143. The area of an equilateral triangle is 5-00548 square meters; find the side. 3-4. Ans. EXERCISES AND PROBLEMS. 165 144. The side of an equilateral triangle is 4 meters longer than the altitude; find both. a-25-856, b 29-856. Ans. HINT. (b 4)2b2 4 b'2- 32x - -64. 145. The three sides of a triangle are 10-2, 13-6, and 17 meters; find the area. 69-36 square meters. Ans. Here = 20-4... A = V20.4(20.4 - 102) (204 - 13-6) (204 - 17). 146. By measurement, a- 37-18 meters, b =48-72 meters, c = 56-46 meters; find A. Here s =71-18;.'. logs = 1'85236 s- a =34-00;.. log(s-a) = 153148 s-b = 22-46;. log(s - b)= 135141 s-c=1472;.. log(s- c)= 1-16791.. log'A2 = 590316.. logA = 295158.. A - 894-50 square meters. Ans. 147. If three arcs, whose radii are 3, 2, 1, at their centers subtend angles of 60~, 90~, 120~, and intersect each other at their extremities, prove that the sides of the triangle formed by their chords are 3, 2V2, V3; and its area HINT. The perpendicular from vertical 4 120' of isosceles A equals half a side, since joining its foot with midpoint of side makes an equilateral A. 148. The area of a triangle is 1012; the length of the side a is to that of b as 4 to 3, and c is to b as 3 to 2. Required the length of the sides. a =52-470, b = 39-353, c= 59-029. Ans. 166 MENSURATION. Here a= 4b, c= 3b;.2. 2s= -b + b + =-b-+6-9- b 263b... s a= 3- b... -a = 23l-,16 b = 72 b, and s - b = 1b, and s-c 23 —1 b b..Are.. Tf. 4Area= - V. T 15 = 2 1012.. V8855 b2 = 1012 x 144 = 145,728... 94-101 b2 145,728.. b = /457 28 = 39-353. 94.101 149. The area of a triangle is 144, and one of two equal sides is 24; find the third side, or base. Here s=24+, and s-a=s-c =~b. s-b=24 —~b. s=24 + and s2 2,.. 144 = -V(242 - b2)4b'.. 576 = v(2304 b2)b2. 150. Show that, in terms of its three medials, A =- IV2 ii2 + i22 i32 + 2i32?2-2 i - -. Proof: 4(i,2 + i2 + i) = 3 (a2 + 2 + c2), 16(i,2i22 + i2,2 i22i)= 9(ab2 + a2 +c b2c), 16 (i14 + 24 + i34)= 9 (a4 + b4 - c4). But, by multiplying out 36, we have A =1 /2(a2b2 + a292 + b22) _ (a4 + b4 + c4). 151. Prove that the triangle whose sides equal the medials of a given triangle is three-fourths of the latter. EXERCISES AND PROBLEMS. 167 TABLE II.-SCALENE TRIANGLES. Sides. Area. Sides. Area. 4 13 15 3 25 26 9 10 17 7 15 20 6 25 29 11 13 20 5 29 30 13 14 15 8 29 35 10 17 21 12 17 25 19 20 37 16 25 39 13 20 21 15 28 41 11 25 30 11 90 97 13 40 51 15 26 37 10 35 39 13 30 37 12 55 65 7 65 68 17 25 28 9 73 80 15 41 52 13 37 40 9 65 70 33 34 65 15 37 44 25 51 74 20 37 51 24 36 36 42 60 66 72 84 84 84 90 114 120 126 126 132 132 156 156 168 180 198 210 210 216 234 240 252 264 264 300 306 25 33 52' 11 100 109 17 39 44 24 35 53 25 29 36 13 68 75 20 51 65 25 39 56 21 85 104 26 35 51 21 61 65 19 60 73 35 44 75 25 39 40 8 123 125 29 35 48 51 52 101 29 60 85 28 65 89 25 51 52 25 52 63 36 91 125 26 51 55 25 92 113 29 52 69 17 105 116 32 53 75 34 65 93 25 63 74 39 41 50 21 89 100 35 52 73 330 330 330 336 360 390 408 420 420 420 420 456 462 468 480 504 510 522 546 624 630 630 660 690 690 714 720 744 756 780 840 840 168 MENSURATION. TABLE II.-Continued. [ Sides. Area. Sides. Area..~Rir~i I 25 84 101 840 43 259 300 1,806 14 157 165 924 26 145 153 1,836 35 53 66 924 51 75 78 1,836 33 56 65 924 80 91 165 1,848 22 85 91 924 55 84 125 1,848 40 51 77 924 45 85 104 1,872 31 156 185 930 45 91 116 1,890 23 140 159 966 53 75 88 1,980 34 61 75 1,020 65 66 109 1,980 57 60 111 1,026 48 85 91 2,016 36 61 65 1,080 65 72 119 2,016 31 97 120 1,116 17 260 267 2,040 39 62 85 1,116 92 117 205 2,070 25 101 114 1,140 61 69 100 2,070 38 65 87 1,140 65 68 105 2,142 51 98 145 1,176 60 73 91 2,184 35 78 97 1,260 61 74 87 2,220 16 195 205 1,288 55 136 183 2,244 41 66 85 1,320 19 289 300 2,280 40 111 145 1,332 68 75 77 2,310 23 123 130 1,380 58 85 117 2,340 46 75 109 1,380 45 133 164 2,394 51 74 115 1,380 29 182 195 2,436 44 75 97 1,584 87 119 200 2,436 35 100 117 1,638 35 174 197 2,436 39 85 92 1,656 41 169 200 2,460 50 69 73 1,656 85 123 202 2,460 41 84 85 1,680 65 89 132 2,574 56 61 75 1,680 31 193 210 2,604 57 65 68 1,710 39 145 164 2,610 39 110 137 1,716 65 87 88 2,640 29 - 150 169 1,740 61 91 100 2,730 29 125 136 1,740 21 340 353 2,856 52 73 75 1,800 49 200 241 2,940 I EXERCISES AND PROBLEMS. 169 TABLE II.-Continued. Sides. Area. Sides. Area. 27 275 292 2,970 105 124 205 5,208 35 197 216 3,024 75 176 229 5,280 76 85 105 3,196 51 233 260 5,304 37 195 212 3,330 65 173 204 5,304 87 112 185 3,360 45 296 325 5,328 45 164 187 3,366 91 125 174 5,460 78 95 97 3,420 104 111 175 5,460 57 122 125 3,420 99 113 140 5,544 65 109 116 3,480 47 250 267 5,640 73 102 145 3,480 55 244 273 6,006 65 126 173 3,484 105 116 143 6,006 65 119 156 3,570 100 217 303 6,510 40 231 257 3,696 91 145 180 6,552 69 113 140 3,864 153 185 328 6,660 65 119 138 3,864 43 520 555 6,708 60 145 161 3,864 119 150 241 7,140 89 99 100 3,960 50 369 401 7,380 57 148 175 3,990 89 170 189 7,560 75 109 136 4,080 65 297 340 7,722 85 99 140 4,158 37 525 548 7,770 91 100 159 4,200 85 234 293 7,956 90 97 119 4,284 123 133 200 7,980 40 291 325 4,290 65 272 303 8,160 87 100 143 4,290 111 200 281 8,880 68 87 145 4,350 140 143 157 9,240 39 280 305 4,368 68 273 275 9,240 89 111 170 4,440 111 175 176 9,240 55 207 244 4,554 89 208 231 9,240 66 175 221 4,620 116 231 325 9,240 143 168 305 4,620 111 175 232 9,324 61 155 156 4,650 74 277 315 9,324 37 411 440 4,884 117 164 175 9,450 41 337 360 4,904 116 181 225 10,440 123 208 325 4,920 91 253 300 10,626 170 MENSURATION. TABLE II.-Concluded. Sides. Area. Sides. Area. 148 153 175 10,710 190 231 377 17,556 113 195 238 10,920 175 221 318 18,564 149 156 175 10,920 175 221 276 19,320 66 389 425 11,220 125 312 323 19,380 123 187 200 11,220 143 296 375 19,536 85 293 336 11,424 186 221 275 20,460 170 171 305 11,628 212 225 247 22,230 75 403 452 12,090 260 287 519 22,386 93 325 388 12,090 129 377 440 22,704 130 185 231 12,012 205 286 411 27,060 113 225 238 12,600 221 346 525 27,300 157 165 184 12,144 123 595 676 29,274 87 340 385 13,398 253 260 315 31,878 164 225 349 14,760 277 304 315 38,304 125 253 312 15,180 255 407 596 41,514 225 287 496 15,624 217 404 495 42,966 195 203 356 15,834 175 527 600 44,268 144 221 275 15,840 273 425 628 46,410 126 269 325 16,380 37... 152. Any right triangle equals the rectangle of the segments of the hypothenuse made by a perpendicular from center of inscribed circle. 153. In any triangle ABC, let 1fbe the mid-point of the base AC, I the point of contact of the inscribed circle, ff and K the points where the perpendicular from the vertex B, and the bisector of the angle B meet AC; prove the relation 11-. II-I- <lIf. Kt EXERCISES AND PROBLEMS. 171 154. Each tangent from A equals - a; from B equals s - b; from C equals s- c. 155. In a right triangle, CD is the perpendicular from C on hypothenuse; prove that the circles inscribed in triangles CA]D, CBD) have the same ratio as these triangles. 156. If A/, A2, h3 be the perpendiculars from the angles of a triangle upon the opposite sides, and r the radius of the inscribed circle, prove 1 1 1 1 hA h6?3 I a I b HINT. - and -= - etc. 2 and h2 2A' (a +6 b 157. Prove h hh3.+ )3c S 158. If A'l, A'2, A'3 be the perpendiculars from any point within a triangle, upon the sides, prove + 3 = 1. hl h2 h3 159. If rT, r2, 73 be the distances from the angles of a triangle to the points of contact of the inscribed circle, prove ( l 72 T73s ~1 + 2+ 160. If r4, T5, Tr be the distances from the angles of a triangle to the center of the inscribed circle, prove 74 7 (a + b + c). abc 161. Prove abc arT4 + 61- + ci-. 162. Prove 742 + r2 -+ T2 - ab + ac + be 6 abc. 163. The radius of a circle is 8 meters. Find the side of an inscribed equilateral triangle. b = 13'8564 meters. Ans. 172 MENSURATION. 164. Find the radius circumscribing the equilateral triangle whose base equals 8-66 meters. 5 meters. Ans. 165. In every triangle, the sum of the perpendiculars from the center of the circumscribed circle on the three sides is equal to the sum of the radii of the inscribed and circumscribed circles. 166. If from the vertices of an equilateral triangle perpendiculars be drawn to any diameter of the circle circumscribing it, the perpendicular which falls on one side of this diameter will be equal to the sum of the two which fall on the other side. 167. If the altitude of an isosceles triangle is equal to its base, b = St. 168. If A', B', C' be the feet of the perpendiculars from the angles of a triangle upon the sides, prove that the radius circumscribing ABC is twice the radius circumscribing A'B' C'. 169. If a, /, y be the perpendiculars from the center of the circumscribing circle upon a, b, c, the sides of a triangle, prove abc 4(a+b +) A A A 39. 9'1 —; 2- r3 — = s - a s-b s - c 170. If the sides of a triangle be in arithmetical progression, the perpendicular on the mean side from the opposite angle, and the radius of the circle which touches the mean side and the two other sides produced, are each three times the radius of the inscribed circle. 171. Each of the common outer tangents to two circles equals the part of the common inner tangent intercepted between them. EXERCISES AND PROBLEMS. 173 172. Each tangent from A to the circle escribed to a equals s; from B to circle escribed to a equals s - c; from C to circle escribed to a equals s - b. Similar theorems hold for the escribed circles which touch b and c. 173. The area of a triangle of which the centers of the abo escribed circles are the angular points is — b 174. If a; b, c denote the sides of a triangle; h, 2, h3 the three altitudes; q1, q2, s the sides of the three inscribed squares, prove the relations 1 1 1 1 1 1 1 1 1 -=-+-; -= -+; - t --- Zl Al a? 2 h2 b ) 3 h3 c 1 1 1 1 175. Prove - -+ -+ r 1 Ai hA2 3 1,1 1 1 - + ---+ -; etc. 9-2 1 h 1 h1 2 1 1 1 1 176. Prove -- - - Al 9'-1 9U 2 23 2 1 21 1 A2 r r2 93 r' 2 1 1 1 1 h3 r 9^3 rlt r2 TTr1 T,'2 9'3 177. Prove A= -- = rr -- ri - r 92 - r3 178. If T, r7, r8, T9 are the distances from the center of the circumscribed circle to the centers of the inscribed and escribed circles, prove the relations R2 = T2 + 2 rT = 772 - 2 T.-T82 - 2r2 _ 2- r 2 7 2 77 8 - + 797 + + 9 -2 73T' - 2 12 HINT. r2 = - 2r'; r7 = S2 + r2 r9; 82 = R2 + 2r2,; 2 = 92 + 2r3. 174 MENSURATION. abe 179. Prove abc = 2 (a + b + c)' abe abe abc 4(s - a)' 4(s-b)' 4(s c) 180. Prove r1+r2+-3= r+4a. 181. In any triangle prove S2 - i2 = rr1; etc. 40. T=xY +Y2 2 182. The base of a triangle is 20 meters, and its altitude 18 meters. It is required to draw a line parallel to the base so as to cut off a trapezoid containing 80 square meters. What is the length of the line of section, and its distance from the base of the triangle? Calling b2 the line of section, and x its distance from b,, T=8= - (20 +b)x. Now, b2: 20::18 - x:18. 2 = 0 (18 - x) = (18 -- -) = 20 - — x.... 80 (20 + 20 - 1Qx)x = (20 - x)x.. 720=180x - 5x. x2- 36x - 144. x.. - 18 = v/324 - 144..x. a= 18-v/180..x x = 18 - 13-416 = 4-584, and b2 = 20 - -1-Q(4 584) = 20 - 5'093 = 14-907. 183. In a perpendicular section of a ditch, the breadth at the top is 26 feet, the slopes of the sides are each 45~, and the area 140 square feet. Required the breadth at bottom and the depth of the ditch. EXERCISES AND PROBLEMS. 175 Here T==140 = [26 +26-2xx=[26-xx. 140 =26x -x2. x. x = 13 = /1(9 - 140 = 13 = -/29 = 13 = 5-385 7-615..b. b=26- 15230= 1077. 184. The altitude of a trapezoid is 23 meters; the two parallel sides are 76 and 36 meters; it is required to draw a line parallel.to the parallel sides, so as to cut off from the smaller end of the trapezoid a part containing 560 square meters. What is the length of the line of section, and its distance from the shorter of the two parallel sides? Let x equal altitude of required part. T1= 2 [76 + 36] 23 = 1288, T2 = 728 = [76 + 1] [23 + x]. 1456 1456 =*5 _=23 - x...x = 23- 145 76 + 1 76 +t 1 Also, 3 = 560 = [36 + ] Lx. 1120 36 + 1120 3 1456 23 -36 + 1 76 + I.. 137,536 + 2576.1 = 62928 + 2576. 1 + 23. l2, 62,928 74,608 = 23. 12..-. =3243-8... =56-95. = 1120 = 12-048. 36 + 56-95 185. The two parallel sides of a trapezoid are 83-2 and 110-4 meters; the altitude, 50'4 meters. Find the area. T= 5227'2 square meters. Ans. 186. The perimeter of a trapezoid is 122 meters. The non-parallel sides are 36 and 32 meters; the altitude, 30'4 meters. Find the area. T =820'8 square meters. Ans. 176 MENSURATION. 187. T= 151-9 square meters, c = 12-4 meters, b = 18-6 meters. Find the other parallel side. b2 = 5'9 meters. Ans. 188. The altitude and two parallel sides of a trapezoid are 2: 3: 5, and T= 1270-08 square meters. Find the parallel sides. l - 63 meters; 2 =37'8 meters. Ans. 189. The triangle formed by joining the mid-point of one of the non-parallel sides of a trapezoid to the extremities of the opposite side is equivalent to half the trapezoid. 190. The area of a trapezoid is equal to half the product of one of its non-parallel sides, and the perpendicular from the mid-point of the other upon the first. 191. The line which joins the mid-points of the diagonals of a trapezoid is parallel to the bases, and equals half their difference. 192. Cutting each base of a trapezoid into the same number of equal parts, and joining the corresponding points, divides the trapezoid into that number of equivalent parts. 193. If the mean line of a trapezoid be divided into n equal parts, and through these points lines, not intersecting within the trapezoid, be extended to its bases, they cut the trapezoid into n equal trapezoids. 194. In every trapezoid, the difference of the squares of the diagonals has to the difference of the squares of the nonparallel sides the same ratio that the sum of the parallel sides has to their difference. 195. Let bl be the longer, b2 the shorter, of the two parallel sides in any trapezoid, zl and z2 the other two sides, and take A= V(b1 — b2+Z1+Z2) (62-bl+Zl+Z2) (-b2-Zl+Z2) (b1-b2+Z1-2). EXERCTSES AND PROBLEMS. 177 Prove T -= -b b A. 4 (& - -2) From the intersection point of b2 and Z2 draw a line parallel to zl; the base of the triangle so formed is (6 — b2), and its other sides are zl and z2.. by 36, A='A. 2. ~ -, (b4-82) 2(bl -b2) 196. The two parallel sides of a trapezoid are 184 and 68 meters; the two others, 84 and 72 meters. Find the area. 6536 square meters. Ans. 197. The diagonal of a symmetric trapezoid is /z" 4-2 bb12. 198. The altitude of a trapezoid is 80 meters; the two diagonals 110 and 100 meters. Find the area. 5419'6 square meters. Anzs. HINT. T -a-(b, + b,) - a (/C~ -- a2 + Vc/~2-c ). 199. In a trapezoid a = 140, 6b= 160, b2 — 120 meters; if the area is halved by a line parallel to the bases, find its length and distance below the shorter base. I= 141-42 meters, d =- 74-97 meters. Ans. HINT I. - b2: b - b2:: a, IL Tib ' ~8 b- 9o II. dI =+ bd_ = + — b =9800 2 2 4 or, I.7 -- 2 d= 840, II. 120 d + Id = 19,600. Substitute the value of d from I. in II. 200. In a trapezoid b = 312, b2 39, -- 350, 2 - 287 meters; if cut by parallels to b into three similar trapezoids, 178 MENSURATION. find where the two parallels cut the sides, and find the areas of the three trapezoids. If 2- n, then 12=nb2. 52.. 1 =n12 = 22b2..b. b1 = 1n21 = 'L n3b2 = 39n3= 312..n. = 2.. z3:z:z7::2 l: b: 1: 2: 4. *. Z = = 50 meters. 37 z = = 41 meters, etc. 7 v=l 41. - -- [(x2 - X1l) (y1 - /3) + (3 - ^1) (/2 /4) +* ---=1 + (n - x1) (/-1 - yn+l) + (x+1 ~- X1) (y1 + /Y+1)] 201. Find the distance between the points 1 and 2. Between two points (x1yl), (x2y2) the distance = V( 2 - x1)2 + (Y2 - y1)2. 202. Find the sum of the two right trapezoids determined by the ordinates of the three points (12-3, 45-6), (78-9, 13), (24, 57). 203. If the cross section of an excavation is a trapezoid, b breadth of top, h depth, with side slopes rn and n in 1, which means that one side falls m meters vertically for one meter of horizontal distance; then show T -= bh - - t h2. 2 rnn 2. i-= - [X1 (yn - /2) + 2 (y1 - 3/3) + -'3 (Y2 - 3/4)..... + n (Yn- -- Y/1)] - 204. Prove that a polygon may be constructed when all but three adjacent parts (1 side and 2 4s, or 2 sides and 1 4) are given. What theorem for congruence of polygons follows from this? EXERCISES AND PROBLEMS. 179 205. Find the area of a heptagon from the coordinates of its vertices, measured as follows: y 1 0 1.72 2 10-48 16-84 3 16-26 14-36 4 32-54 4.84 5 50-02 10-32 6 50-02 0 7 0 0 N — 476-21 square meters. Ans. 206. Find the measurements. area of an enneagon from the following x y 1 i 0 16-96 2 26-36 20-04 3 58-02 2216 4 104-00 11-24 5 97-48 248 6 i 92-22 - 11-86 7 61-00 - 2-36 8 35-46 - 4-10 9 9-84 -14-22 Also draw the figure. JV — 2429'16 square meters. Ans. 207. Find the area of a pentagon, the coordinates of whose vertices are as follows: (133, 917), (261, 325), (486, 916), (547, 325), (828, 916). 208. MODEL SOLUTION. ORDINATES. ABSCISSAE. DOUBLE AREA., 0 IDIFFERENCE. DIFFERENCE. - m V I m Sym-l-2/m~^+i-.-1 +1 Xn+1 (y —l m+lA~ Meters. Mete. eters. etMeters. + Sq. Meters. - Sq. Meters. + Sq. Meters. - Sq. Meters. 1 00 - 837064 0 - 1,080-911 2 837-064 -1,386-464 - 183722 + 766-893 641,938-6 254,7933 88,5409 3 1,386-464 - 115-454 + 766893 + 1,598-809 2,216,690-0 73,683-2 4 + 952-518 -- 52-073 + 1,415-037 + 1,435-171 1,367,025-6 5 + 1,438-537 + 182-240 + 2,202-064 + 1,452-368 2,089,277-1 401,304-1 6 + 770-278 + 1,268218 + 2,867-405 + 493-805 380,367-1 3,636,495-3 7 + 170-319 + 339-356 + 2,695-869 - 817-984 139,285-8 914,859-6 8 + 430-922 + 549-886 +2,049-421 - 884-471 381,569-2 1,126,949-6 9 - 379-567 + 955-946 +1,810'398 - 841-719 319,488-7 1,730,642-1 10 - 525-024 - 379-567 - 1,207702 - 913-259 479,482-7 458,401-8 11 0 - 525-024 a- 897139 -1,207-702 471,005 I471,001 0 co N 0 FQ Therefore, the square meters. 7,494,270 520,855 b52(j,55 8,0(j5,044 1,091,631 i, ty', o1 i area of the Hendecagon is 3,486,707 6,973,415 = double area= 6,973,413 EXERCISES AND PROBLEMS. 181 209. Find the area of a hexagon from its coordinates (719, 313), (512, 852), (719, 454), (513, 116), (720, 242), (513, 993). 43. 2 Q =- Z12 — 2 y1+ -2 23- X3 Y2+ X3y4- X4y3+ XZ4y1 —1 4. 210. The area of a quadrilateral inscribed in a circle is at+b+c+d = \(S - Ca) (s-b)(s -- c) (s-d) where s = 2 c _2=p. 211. Ihr h th -pnt f through the id-point E of the diagonal of a quadrilateral ABCD, FEG~ be drawn parallel to the other diagonal AC, prove that the straight line A G divides the quadrilateral into two equivalent parts. 212. Show that two quadrilaterals whose diagonals contain the same angle are as the products of their diagonals. 213. A circle of r is inscribed in a kite, and another of r' in the triangle formed by the axis of the kite and the two unequal sides; show that, if 21 be the length of the other kite-diagonal, 1 1 1 r r I 214. To find the area of any quadri- Mr lateral from one side, and the distances c from that side of the other two vertices, and the intersection-point of the / X diagonals. / Given the side AB= b, and the ordinates from C B), and E; namely, y3, A4 - y4, and 5. Parallel to BD draw CJM to intersection with AD prolonged, and drop y6, the ordinate of I. Then ABCD = AAfB = 1 by6. But y,6: Y4= Af: AD = AC: AE= y: y5; y = 3___4;.. Q IY4 ' y.~ ' " ' 2 y/ 182 MENSURATION. 44. Al 2 a2 215. The area of a triangle equals 3259'6 square meters; one side equals 112-4 meters. Find the area of a similar triangle whose corresponding side equals 28-1 meters. 203'725 square meters. Ans. 216. The sides of a triangle are 389-2, 486-5, and 291-9 meters. The area of a similar triangle is 2098-14 square meters. Find its sides. 74-8, 56-1, 93-5 meters. Ans. 217. The areas of two similar triangles are 24-36 and 182-7 square meters. One side of the first is 8-5 meters shorter than the homologue of the second. Find these sides. 4-88 and 13-38 meters. Ans. HINT. 24-36: 1827:: (x - 85)2: x2. 218. Two triangles are 21-66 and 43-74 square meters, and have an equal angle whose including sides in the first are 7'6 and 5-7 meters. The corresponding sides in second differ by 2-7 meters. Find them. 10-8 and 8-1 meters. Ans. 219. The areas of two similar polygons are 46'37 and 185-48 square meters. A side of the first is 15 meters smaller than the corresponding side of the other. Find these sides. 15 and 30 meters. Ans. HINT. 4637: 185-48: x2: (x + 15)2.. aln = ap 45. NA= - 2 2 220. The sum of perpendiculars dropped from any point within a regular polygon upon all the sides is constant. 221. In area, an inscribed 2n-gon is a mean proportional between the inscribed and circumscribed n-gons, EXERCISES AND PROBLEMS. 183 222. With what regular polygons can a vestibule be paved? 223. If a regular n-gon is revolved about its center through the X 5, it coincides with itself. n. 6 224. In a regular Argon, each =/ — 46. %.- n Xi. 225. A hexagon is inscribed in a circle, and the alternate angles are joined, forming another hexagon. Find its area. 2. Ans. 2 226. What is the area of a regular dodecagon whose side is 54 feet? (54)2=2916, and 2916 X 11-1961524 = 3647-980+. Ans. 184 MENSURATION. 47. O - 2r. 227. There are three circles whose radii are 20, 28, and 29 meters respectively. Required the radius of a fourth circle whose area is equal to the sum of the areas of the other three. 01=4 4Or,,= 784r,,3 = 841;. 4. -- 2025 r = r2w;.-. r =- /2025 = 45. Ans. 228. If a circle equals 34-36 square meters, find its radius. 3-3 meters. Ans. 229. Two 0's together equal 740-4232 square meters, and differ by 683-8744 square meters. Find radii. — = 15-056 meters and r' = 3 meters. Ans. I. rr2 ~ Trrl = 740-4232. II. irr2- I 'r2 == 683-8744. 230. If 0 be the area of the inscribed circle of a triangle, 01, 02, 03 the areas of the three escribed circles, prove 1 1_ 1 1 V0S1V09\(O2 \/D3 0/(3 231. If from any point in a semicircumference a perpendicular be dropped to the diameter, and semicircles described on these segments, the area between the three semicircumferences equals the circle on the perpendicular as diameter. 232. The perimeters of a circle, a square, and an equilateral triangle are each of them 12 meters. Find the area of each of these figures to the nearest hundredth of a square meter, 11-46, 9, 6.93 square meters. Ans. EXERCISES AND PROBLEMS. 185 233. Find the side of a square inscribed in a semicircle. 2 V V. Ans. 234. An equilateral triangle and a regular hexagon have the same perimeter; show that the areas of their inscribed circles are as 4 to 9. 235. How far must the diameter of a circle be prolonged, in order that the tangent to the circle from the end of the prolongation may be m long? 2 (Vd2- 4 2 -). Ans. 48. S = l- = ur2. 236. Find the area of a sector of 68~ 36' when r = 7-2. 31-03398 square meters. Ans. 237. When circle equals 432 square meters, find sector of 84~ 12'. 100-8 square meters. Ans. HINT. 432: S:: 360: 841. 238. Find the number of degrees in the arc of a sector equivalent to the square of its radius. 239. In different circles, sectors are equivalent whose angles have a ratio inverse to that of the squared radii. 240. Find radius when sector of 7~ 12' is 2 square centimeters. 241. Find sector whose radius equals 25, and the circular measure of whose angle equals I. 234-375. Ans. 242. The length of the arc of a sector of a given circle is 16 meters; the angle of the sector at center is - of a right angle. Find sector. 488-9 square meters. Ans. 186 MENSURATION. 49. G -= ( ) -) 4 A 243. AB is a chord of a given circle; if on the radius CA, which passes through one of its extremities, taken as diameter, a circle be described, the segments cut off from the two circles by the chord AB are in the ratio of 4 to 1. 244. Show that, if a is the angle or arc of a segment, for a- 600, G-2 (27-3V3); 1.2 a 120, = (4 - 3V3); a= 90~, G = -( - 2); 7r a= 36~, - (47 - 5VIO ---2 V5); a= 72~, - = - (87r - 5 i/ - 2 ). 245. In a segment of 60~, to how many places of decimals is our approximation correct? 246. Prove that there can be no segment with k -120, =- 156, h =12. 247. In a circle, given two parallel chords 7ci and k2, and their distance apart T; find the diameter. d = k k — 2 + 2 (1 + k,2 2T2). Ans. HINT. If x = r' - k2 and y = v?2_ - k 2 then I. z -y =r. II. x2 - y2= (.2 k2). EXERCISES AND PROBLEMS. 187 50. 248. What is the area of a circular zone, one side of which is 96 and the other 60, and the distance between them 26 (r = 50), when the area of the larger sector is 3217-484, and of the smaller 1608-736? 2136-75. Ans. 51. 249. HIPPOCRATES'S THEOREM. The two crescents made by de- scribing semicircles outward on the two sides of a right triangle, and a semicircle toward them on the hypothenuse, are equivalent to the right triangle. 250. The crescent made by describing a semicircle on the chord of a quadrant equals the right triangle. 52. A= (9r + 9(2) (rIl- 92)7r. 251. A circle of 60 meters diameter is divided into seven equal parts by concentric circles; find the parts of the liarneter. r2, = 900 x 3'14159 = 2827-431..'. outer annulus =2827431 = 403-9186 = (900 - r22)rr. 403.9186... r2 = 900 - 43 = 900 - 128-575 = 771-425. 3.14159.'. % = 27'77+... =55-55.. 1st part = 60 - 55-55 = 4-450+. In the same way, 2d part = 4-840, 3d part = 5-353+, 4th part = 6-076+, etc, 188 MIENSURATION. 252. Find the annulus between the concentric circumferences, c1 = 21-98 meters and c2 =18-84 meters, taking 7r = 3-14. A = 10-205 square meters. Ans. 53. S. A-= — (7l,+ 2). 253. To trisect a sector of an annulus by concentric circles. 54. J — hk. 254. What is the area of a parabola whose base is 18 meters and height 5 meters? 60 square meters. Ans. 255. What is the area of a parabola whose base is 525 meters and height 350 meters? 122,500 square meters. Ans. 55. E = ab-. 256. The area of an ellipse is to the area of the circumscribed circle as the minor axis is to the major axis. 257. The area of an ellipse is to the area of the inscribed circle as the major axis is to the minor axis. 258. The area of an ellipse is a mean proportional between the inscribed and circumscribed circles. 259. What is the area of an ellipse whose major axis is 70 meters, and minor axis 60 meters? 3298-67 square meters. Ans. 260. What is the area of an ellipse whose axes are 340 and 310? 82,780-896. Ans. EXERCISES AND PROBLEMS. 189 EXERCISES AND PROBLEMS ON CHAPTER IV. POLYHEDRONS. 56. ~+ =S- +2. 261. The number of plane angles in the surface of any polyhedron is twice the number of its edges. HINT. Each face has as many plane angles as sides. Each edge pertains, as side, to two faces. 262. The number of plane angles on the surface of a polyhedron is always an even number. 263. If a polyhedron has for faces only polygons with an odd number of sides, e.g., trigons, pentagons, heptagons, etc., it must have an even number of faces. 264. If the faces of a polyhedron are partly of an even, partly of an odd number of sides, there must be an even number of odd-sided faces. 265. In every polyhedron 2- (. HINT. The number of plane angles on a polyhedron can never be less than thrice the number of faces. 266. In every polyhedron 3 - (S. 267. In any polyhedron ( -- 6 3 3. 268. In any polyhedron E- + 6 3 3 ~. 269. In every polyhedron ( < 3 A, and E < 3 ~. 270. In a polyhedron not all the summits are more than five-sided; nor have all the faces more than five sides. 271. There is no seven-edged polyhedron. 190 MENSURATION. 272. For every polyhedron s -6= (C -- j), that is, the sum of the plane angles is as many perigons as the difference between the number of edges and faces. 273. For every polyhedron -- 6 ( - 2), just as for every polygon 3 =.(n - 2). 274. How many regular convex polyhedrons are possible? 275. In no polyhedron can triangles and three-faced summits both be absent; together are present at least eight. 276. A polyhedron without triangular and quadrangular faces has at least twelve pentagons; a polyhedron without three-faced and four-faced summits has at least twelve five-faced. 57. P=lp. 277. In a right prism of 9 meters altitude, the base is a right triangle whose legs are 3 and 4 meters. Find the mantel. EXERCISES AND PROBLEMS. 191 278. The base of a right prism 12 meters tall is a triangle whose sides are 12, 14, and 15 meters. Find its surface. 279. To find the mantel of a truncated prism. Rule: Multiply each side of the perimeter of the right section by the sum of the two edges in which it terminates. The sum of these products will be twice the area. 280. The mantel of a truncated prism equals the axis multiplied by perimeter of a right section. 281. A right prism 4 meters tall has for base a regular hexagon whose side is 1-2 meters. Find its surface. 282. In a right triangular prism the lateral edges equal the radius of the circle inscribed in the base. Show that the mantel equals the sum of the bases. 58. C-cl= 2 7rgl. 283. In a right circular cylinder, (i) Given a and C; find r.. Ans. (2) Given B and C; find a. 2 (3) Givan C and a = 2r; find surface. (4) Given surface and a - r; find C. (5) Given a and B - C; find r. 284. The mantel of a right cylinder is equal to a circle whose radius is a mean proportional between the altitude of the cylinder and the diameter of its base. 285. The bases of a circular cylinder together are to the mantel as radius to altitude. 286. If the altitude of a right circular cylinder is equal to the diameter of its base, the mantel is four times the base. 192 MENSURATION. 287. Find a cylinder equivalent to a sum of right circular cylinders of the same height. HINT. Find a radius whose square equals squares of the n radii. 288. How much must the altitude of a right circular cylinder be prolonged to make its mantel equal its previous surface? 289. A plane perpendicular to the base of a right cylinder cuts it in a chord whose angle at the center is a; find the ratio of the curved surfaces of the two parts of the cylinder. 59. Y =-hp. 290. The surface of any regular tetrahedron is to that of the cube on its edge as 1 to 2. 291. Each edge of a regular tetrahedron is 2 meters. Find mantel. 292. Each edge of a regular square pyramid is 2 meters. Find surface. 293. From the altitude a and basal edge b of a regular hexagonal pyramid, find its surface. 3 b (- V3 + Va2 + t a2). Ans. 294. In a regular square pyramid, given p, the perimeter of the base, and the area A of the triangle made by a basal diagonal and the two opposite lateral edges; find the surface of the pyramid. _61-p2 + 2 V32A2 + l-p4. Ans. 60. = ch =7rrh. 295. The convex surface of a right cone is twice the area of the base; find the vertical angle. Here cx -h=2c x r.. h=2r=d. EXERCISES AND PROBLEMS. 193 Thus the section containing the axis is an equilateral triangle; so the angle equals 60~. Ans. 296. Find the ratio of the mantels of a cone and cylinder whose axis-sections are equilateral. 297. Find the locus of the point equally distant from three given points. 298. In a right cone (1) Given a and r; find K. r- Va2 + 2. Ans. (2) Given a and h; find K. rx/h2- a. Ans. (3) Given K and h; find r.. Ans. 7rh 299. In an oblique circular cone, given Al, the longest slant height, h2, the shortest, and a, the altitude; find r, the radius of the base. -V/ (h2 + h2) - a2. Ans. 300. How many square meters of canvas are required to make a conical tent which is 20 meters in diameter and 12 meters high? Here ere K= rrrh = 314159 x 10 x V144 + 100. K= 31-4159 x 15-6205 490'7320+ square meters. Ans. 61. F= Ah(, +2). 301. Given a basal edge b, and a top edge b2, of the frustum of a regular tetrahedron; also a, the altitude of the frustum. Find h, its slant height, and F, its mantel. A- \/iVi (b -- b2)2 +4a(2. F - I (bl +2) V(3 ( —b 2)2 4 a2. Ans. 302. Same for a regular four-sided pyramid. A = -(bl 2) + 4a2. Ans. FF= (h, - b2) V(b - b2) + 4a2 194 MENSURATION. 62. F =-21 (l + C2) = h (r + r2). 303. In the frustum of a right circular cone, given rl, r2, and a; find h. 304. In the frustum of a right circular cone, on each base stands a cone with its vertex in the center of the other base; from the basal radii sl and r2 find the radius of the circle in which the two cones cut. 'r2 A rl + r2 305. Given 61, b2, the basal edges, and I, the lateral edge of a frustum of a regular square pyramid; the frustum of a cone is so constructed that its upper base circumscribes the upper base of this pyramid-frustum, while its lower base is inscribed in its lower base. Find the slant height of the cone-frustum. \/2 - b f + (1- I- 2)612. Ans. 306. How far from the vertex is the cross-section which halves the mantel of a right cone? - a.V2. Ans. 307. Reckon the mantel from the two radii when the inclination of a slant height to one base is 45~. _(r27-2) r V2. Ans. 308. If in the frustum of a right cone the diameter of the upper base equals the slant height, reckon the mantel from the altitude a and the perimeter p of an axial section. p (p2 + 12 a2 p Vp2 - 12 a2). Ans. 63. F — 2raj. 309. In the frustum of a cone of revolution, given rl, r2, h; find a. 310. Find the altitude of the frustum of revolution from the mantel k and the bases B, and B2. a- k- -(B1 -B 2)2 Ans. 7r (-\I+ V B2) EXERCISES AND PROBLEMS. 195 311. A right-angled triangle is revolved about an axis parallel to, and at the distance r from its side a; the areas of the circles described by its base are as m to n. Find the whole surface described by the triangle. 4 2 -- ) + 2a+ - + l a2+ 2 As. 64. Rz=4 2Tr. 312. Find the surface of a cube inscribed in a sphere whose surface is f. 313. A sphere is to the entire surface of its circumscribing cylinder as 2 is to 3. 314. Given r- and r'2, the radii of two section-circles of a sphere, and the ratio (m: n) of their distances from its center. Find its radius. r2 _r - n2r2 A n t r r 2 315. Find the sphere whose radius is 12-6156 meters. 2000. Ans. 316. Find the sphere whose radius is 19 1 2 meters. 5000. Ans. 317. A sphere is 50-265 square meters; find its radius. 2 meters. Ans. 318. Find a sphere from a section-circle c whose distance from the center is r. ( + 47-r. Ans. 319. What will it cost to gild a sphere of 22-6 centimeters radius, if 100 square centimeters cost 872 cents? $56-16. Ans. 320. Find the ratios of the mantel of the cone, described by rotating an equilateral triangle about its altitude, to the sphere generated by the circle inscribed in this triangle. 3:2. Ans. 196 MENSURATION. 65. Z=27rra. 321. Cut a sphere into n equal parts by parallel circles. 322. In a calot, (1) Given r and a; find r. (2) Given r and ri; find Z1. (3) Given a and cl; find Z1. (4) Given a and HE; find Z1. a-V/r.. Ans. 323. In a segment 6 centimeters high, the radii of base and top are 9 and 3 centimeters. Find area of the zone. 36 r V10 square centimeters. Ans. 324. In a segment of altitude a, and congruent bases, calling the top and base radii rl, find the zone. ra VA4 rf + a. Ans. 325. How far above the surface of the earth must a person be raised to see one-third of its surface? Here a = - d = -r; and, by similar triangles, x + r: r= ': - - a. r( + r)=r2... ( r) =r... x= 2, —d. Ans. 326. A luminous point is distant r from a sphere of radius r; how large is the lighted surface? 2 2rr2 As. 327. Find a zone from the radii of its bases rl, r2, and the radius of the sphere r. 2 rr [r2- + Vr2 - - ri2 — ]. Ans. 328. How far from the center must a plane be passed to divide a hemisphere into equal zone and calot? EXERCISES AND PROBLEMS. 197 66. THEOREM OF PAPPUS. 329. An acute-angled triangle is revolved about each side as axis; express the ratio of the surfaces of the three doublecones in terms of a, 6, c, the sides of the triangle. a+ b a+c.b+ Ans. c b a 330. The sides of a symmetric trapezoid are bi, b2, and z. Express the surface generated by rotating the trapezoid about one of the non-parallel sides. (6f + 62 + b1 + b2Z) Vz - ( -- 2)2. Ans. 67. 0 = 4 ~21l2. 331. An equilateral triangle 'rotates about an axis without it, parallel to, and at a distance a from one of its sides b. Find the surface thus generated. br (b 6/3 - 6 a). Ans. 332. A rectangle with sides a and b is revolved about an axis through one of its vertices, and parallel to a diagonal. Find the generated surface. 4 abw (a + b) Ans. Vca2 + b2 Q (L). SPHERICS AND SOLID ANGLES. 68. =2r2u. 333. Find the area of a lune whose angle is 36~. 2 r2v. Ans. 334. Find lune of 36~ when r = 1-26156. 2. Ans. 69. 335. A conical sector is one-fourth of a globe; find its solid angle. 90~. Ans. Find the vertex-angle of an axial section. 120~. Ans. HINT. By 65, Cor. 2, generating arc = 60~. 198 MENSURATION. 70. A er. 336. If two angles of a spherical triangle be right, its area varies as the third angle. 337. In a cube each solid angle is one-eighth of a steregon. (For eight cubes may be placed together, touching at a point.) 338. Find the ratio of the solid angle of a regular right triangular prism to the solid angle of a quader. 2: 3. Ans. 339. Find the ratio of the trihedral angles of two regular right prisms of m and n sides. (- 2) n Ans (n- 2) m 340. Find the area of a spherical triangle from the radius r, and the angles a =20~ 91 30", /3 55~ 53'3211, y =114~ 20' 14". 0-1813r2. Ans. 341. Given r, and a =73~ 12' 8", = 85 3'14", y= 32~ 9' 16"; find A. 0-18593r2. Ans. 342. Given r, and a = 114~ 20' 5"'92, - = 30~ 57' 18"-41, y =90~ 9' 41"-67; find A.,0-9678r2. Ans. 343. Spherical triangles on the same base are equivalent if their vertices lie in a circumference passing through the opposite extremities of sphere-diameters from the ends of the base. 344. All trihedral angles having two edges common, and their third edges prolongcyions of elements of a right cone containing the two common edges, are equivalent. Proof: On the edges of a trihedral angle take SA, S-B, SC equal; and pass through the three extremities a circle ABC' of center O. Join SO, and suppose three planes to start from SO and to pass one through each edge of the trihedral angle. These planes form three new trihedrals EXERCISES AND PROBLEMIS. 199 having a common summit S, and one common edge SO. In each of these are a pair of equal dihedral angles, since each stands on an isosceles triangle with vertex at 0. Thus, BAO= ABO. (1) CBSO = BO. (2) ACO = CAO,. (3) Therefore, ABsC+ CA,B-BCA =ABsO +.CB,O +CAsO +BAO -BC O -ACO =AB,0 —BAO ------ =2BAs,.. (4) Now, 2BASO remains constant as long as the B summit S of the trihedral, the two edges BS and AS, and the center 0 of the circle are unchanged; and equation (4) holds as long as the edge SC passes through the circumference. But ABSC =Q- - ABC... (5) BAsC= = - BAsC...... (6) BC5sA BC'sA....... (7) Making the substitutions (5), (6), (7), equation (4) becomes ABs, + BAOC + BCI'A Q - 2 BAsO. The second member is constant; therefore, in the trihedral SABC', the sum of the three interior angles, and consequently the area of its intercepted spherical triangle, is constant. 200 MENSURATION. 345. Equivalent spherical triangles upon the same base, and on the same side of it, are between the same parallel and equal lesser circles of the sphere. 346. The locus of B, the vertex of a spherical triangle of given base and area, is a lesser circle equal to a parallel lesser circle passed through A and C, the extremities of the given base. 347. Find the spherical excess of a triangle from its area and the radius. 18. Ans. e — _A 1800. Ans. 9"27r 348. Find the ratio of the spherical excesses of two equivalent triangles on different spheres. el: e2 =: r'12. Ans. 349. A spherical triangle whose a 91~ 12' 17", P = 120~ 9' 41", y =100~ 42' 2", contains 3,962 square meters. Find the sphere. 21,600 square meters. Ans. 71. -['4-(- 2) r] 2.2 350. Find the ratio of the vertical solid angles of two regular pyramids of nm and n sides, having the inclinations of two contiguous faces respectively, a and f3. 2 7r -- m(7-) Ans. 27r-n (7r-/) 351. What is the area of a spherical pentegon on a sphere of radius 5 meters, supposing the sum of the angles 640? 43-633 square meters. Ans. EXERCISES AND PROBLEMS. 201 EXERCISES AND PROBLEMS ON CHAPTER V. 72. U= abl. 352. The diagonal of a cube is n; find its volume. 353. Find the volume of a cube whose surface is 3-9402 square meters. 354. The edge of a cube is n; approximate to the edge of a cube twice as large. 355. Find the edge of a cube equal to three whose edges are a, b, Z. 356. Find the cube whose volume equals its superficial area. 357. If a cubical block of marble, of which the edge is 1 meter, costs 1 dollar, what costs a cubical block whose edge is equal to the diagonal of the first block. 3 V3 dollars. Ans. 358. In any quader, (1) Given a, b, and mantel; find U. (2) Given a, b, U; find 1. (3) Given U, B, and (ab); find I and b. (4) Given U, (), (b); find a and b. (5) Given (a6), (al), (bl); find a and b. 359. If 97 centimeters is the diagonal of a quader with square base of 43 centimeters side, find its volume. HINT. a2 = (97)2- 2(43)2. 202 MIENSURATION. 360. What weight will keep under water a cork quader 55 centimeters long, 43 centimeters broad, and 97 centimeters thick, density 0-24? 229-405 - 55-0572 kilograms. Ans. 361. The volume of a quader whose basal edges are 12 and 4 meters is equal to the superficial area. Find its altitude. 362. In a quader of 360 square meters superficial area the base is a square of 6 meters edge. Find the volume. 363. A quader of 864 cubic centimeters volume has a square base equal to the area of two adjacent sides. Find its three dimensions. 364. In a quader of 8 meters altitude and 160 square meters surface the base is square. Find the volume. 365. The volume of a quader is 144 cubic centimeters; its diagonal 13 centimeters; the diagonal of its base 5 centimeters. Find its three dimensions. 366. In a quader of 108 square meters surface the base equals the mantel. Find volume. 367. If in the three edges of a quader, which meet in an angle, the distances of three points A, B, and 0 from that angle be a, b, c; then triangle ABC= -- -/ Va2b2+a2c2+b2c2. 368. How many square meters of metal will be required to construct a rectangular tank (open at top) 12 meters long, 10 meters broad, and 8 meters deep. 472. Ans. 369. The three external edges of a box are 3, 2-52, and 1-523 meters. It is constructed of a material 0-1 meters in thickness. Find the cubic space inside. 8-594208 cubic meters. Ans. EXERCISES AND PROBLEMS. 203 7 _ g _kg V ccm VI' 370. A brick 11 centimeters long, 3 centimeters broad, 2 centimeters thick, weighs 45 grams; find its density. 371. A cube of pine wood of 12 centimeters edge weighs 1 kilogram; find the density of pine. 0'57. Ans. 372. If a mass of ice containing 270 cubic meters weighs 229,000 kilograms, find the density of ice. 0-92. Ans. 373. If a cubic centimeter of metal weighs 6-9 grams, what is its density? 74. V. P = abl. 374. If the base of a parallelepiped is a square, find the altitude a and basal edge b from the volume and mantel. 75. V. P = aB. 375. The base of a prism 10 meters tall is an isosceles right triangle of 6 meters hypothenuse; find volume. 376. In a prism whose base is 210 square meters, the three sides are rectangles of 336, 300, 204 square meters; find volume. 377. Find altitude of a right prism of 480 cubic centimeters volume, standing upon an isosceles triangle whose base is 10 centimeters and side 13 centimeters. 378. In a right prism of 54 cubic centimeters volume, the mantel is four times the base, an equilateral triangle; find basal edge. 204 MENSURATION. 379. The vertical ends of a hollow trough are parallel equilateral triangles, with 1 meter in each side, the bases of the triangles being horizontal. If the distance between the triangularends be 6 meters, find the number of cubic meters of water the trough will contain. 2-598 cubic meters. Ans. 76. V. C -= aCr2r. 380. In a right circular cylinder, (1) Given a and c; find V. C. C. Ans. (2) Given a and C; find V. C. (3) Given (V. C) and C; find r. V. C -ar2=,. a=-. C rT2 C= a2rr,.. a 2r7r V. C C 2V. C.'. -Y --.'. 7' ---.Ans. '27r 2r7r C Ans. 381. If C= 91-84 square meters, and V. C 145 cubic meters, find a. a =4-628986 meters. Ans. 382. A right cylinder of 50 cubic centimeters volume has a circumference of 9 centimeters; find mantel and volume. 383. In a right cylinder of 8 cubic centimeters the mantel equals the sum of the bases; find altitude. 384. If, in three cylinders of the same height, one radius is the sum of the other two, then one curved surface is the sum of the others, but contains a greater volume. 385. Find the ratio of two cylinders when the radius of one equals the altitude of the other. 386. Find the ratio of two cylinders whose mantels are equivalent. EXERCISES AND PROBLEMS. 205 387. If 1728 cubic meters of brass were to be drawn into wire of one-thirtieth of a meter in diameter, determine the length of the wire. Here 1728 = ar2r=a( 1)2r= aaw 1728 3600 a 1728 x 3600 1,980,145 meters. Ans. 7r 388. What must be the ratio of the radius of a right cylinder to its altitude, in order that the axis-section may equal the base?.2: wr. Ans. 389. A cylindric glass of 5 meters diameter holds half a liter; find its height. 390. A rectangle whose sides are 3 meters and 6 meters is turned about the 6-meter side as axis; find the volume of the generated cylinder. 391. The diagonal of the axis-section of a right cylinder is 5 centimeters; the diameter of its base is three-fourths its height. Find its volume. 392. In a right cylinder, from A, the area of the axissection, reckon the area of that section which halves the basal radius normal to it. 2 AV/3. Ans. 393. The longest side of a truncated circular cylinder of 1.5 meters radius is 2 meters; the shortest, 1-75 meters. Find volume. 394. If a room be 40 meters long by 20 meters broad, what addition will be made to its cubic contents by throwing out a semicircular bow at one end? 2513.28 cubic meters. Ans. 395. The French and German liquid measures must be cylinders of altitude twice diameter. Find the altitude for measures holding 2 liters, 1 liter, and 1 liter. 216-7, 172-1, and 136-5 millimeters. Ans. 206 MENSURATION. 396. The German dry measures must be cylinders of altitude two-thirds diameter. Find diameter of a measure containing 100 liters. 575-9 millimeters. Ans. 397. In the French grain measure the altitude equals diameter. Find for hectoliter. 503-7 millimeters. Ans. 77. V. C1 - V. CV -= a7 (r + r2)(r - 2). 398. How many cubic meters of iron are there in a roller which is half a meter thick, with an outer circumference of 61 meters, and a width of 37 meters? (r -2-7). 1353 cubic meters. Ans. 399. Find the amount of metal in a pipe 3-1831 meters long, with r'1 — 12 meters and 2 = 8 meters. 800 cubic meters. Ans. 400. The amount of metal in a pipe is 175-9292 cubic meters, its length is 3-5 meters, and its greater radius is 5 meters. Find its thickness. 2 meters. Ans. 78. SECTIONS SIMILAE. 401. A regular square pyramid, whose basal edge is b, is so cut parallel to the base that the altitude is halved; find the area of this cross-section. 402. A section parallel to the base of a cone (base-radius r), its altitude in the ratio of m to n. Find the area of this 2 1.2 section. rr Ans. (m + n)2' 403. On each of the bases of a right cylinder, radius r, stands a cone whose vertex is the center of the other base. Find the circumference in which the cone-mantels cut. rr. Ans. EXERCISES AND PROBLEMS. 207 79. EQUIVALENT TETRAHEDRA. 404. If a plane be drawn through the points of bisection of two opposite edges of a tetrahedron, it will bisect the tetrahedron. 80. V.Y -- aB. 405. A pyramid of 9 decimeters altitude contains 154 cubic meters; find its base. 52.5 square meters. Ans. 406. The pyramid of Memphis has an altitude of 73 Toises; the base is a square whose side is 116 Toises. If a Toise is 1-95 meters, find the volume of this pyramid. About 2,427,780 cubic meters. Ans. 407. A goldsmith uses up a triangular pyramid of gold, density 19-325, and charges $900 a kilogram. What is his bill if the altitude of the pyramid is 4 centimeters, the altitude of its base 4 millimeters, and the base of its base 1-5 centimeters. $6.972. Ans. 408. Find the volume of a pyramid of 30 meters altitude, having for base a right triangle of 25 meters hypothenuse and 7 meters altitude. 81. V. K 1 ar2r-. 409. In a right circular cone, (1) Given r and A; find V. K. (2) Given a and A; find V. K. (3) Given r and K; find V. K. (4) Given h and K; find V. K. (5) Given a and K; find V. K. - ar (/ - _ 2).. e v/K2 -.42 [K +-~ '2 3 hrC6r h2I i -r a,7r2+ 4 2. Ans. Ans. Ans. Ans. Ans. 208 IMENSURATION. 410. A cone and cylinder have equal surfaces, and their axis-sections are equilateral; find the ratio of their volumes. Surface of cylinder = -- + d2 r = d'2 2 2 h2 tr h,2r 3 h2t. Surface of cone =- + - = 4 2 4.h = dV2. 411. In a triangular prism of 9 meters altitude, whose base has 4 square meters area and 8-85437 meters perimeter, a cylinder is inscribed. Find the base and altitude of an equivalent cone whose axial section is equilateral. B 17-1236 square meters, a = 4-043738 meters. Ans. 412. Find the edge of an equilateral cone holding a liter. 16-4 centimeters. Ans. 413. Halve an equilateral cone by a plane parallel to the base. 414. Find the ratio of the volumes of the cones inscribed and circumscribed to a regular tetrahedron whose edge is n. En /n82 \2 3 Ans. \n^l\J Ans. 82. PRISMOIDAL FORMULA: XD-1 a(B + M4 J/ B2). 82. PRISMOIDAL FORMULA: D - = a(/~- a 4+4M+/ B~2). 415. Find the volume of a rectangular prismoid of 12 meters altitude, whose top is 5 meters long and 2 meters broad, and base 7 meters long and 4 meters broad. 220 cubic meters. Ans. 416. In a prismoid 15 meters tall, whose base is 36 square meters, the basal edge is to the top as 3 to 2. Find the volume. 380 cubic meters. Ans, EXERCISES AND PROBLEMS. 209 417. Every regular octahedron is a prismatoid whose bases and lateral faces are all congruent equilateral triangles. Find its volume in terms of an edge b. ib3 -. Ans. 418. The bases of a prismatoid are congruent squares of side b, whose sides are not parallel; the lateral faces are eight isosceles triangles. Find the volume. ab (2 + V/2). Ans. 419. If, from a regular icosahedron, we take off two five-sided pyramids whose vertices are opposite summits, there remains a solid bounded by two congruent regular pentagons and ten equilateral triangles. Find its volume from an edge b. & M3(5 + 2V5.) Ans. 420. Both bases of a prismatoid of altitude a are squares; the lateral faces isosceles triangles; the sides of the upper base are parallel to the diagonals of the lower base, and half as long as these diagonals; b is a side of the lower base. Find the volume. a6b2. Ans. 421. The upper base of a prismatoid of altitude a = 6 is a square of side b2 = 7-07107; the lower base is a square of side b, - 10, with its diagonals parallel to sides of the upper base; the lateral faces are isosceles triangles. Find volume. Ic (a + 6 2 V2 + b2) = 500. Ans. 422. Every prismatoid is equivalent to three pyramids of the same altitude with it, of which one has for base half the sum of the prismatoid's bases, and each of the others its midcross section. D l (B, + B 2fA V 9 210 MENSURATION. 423. Every prismoid is equivalent to a prism plus a pyramid, both of the same altitude with it, whose bases have the same angles as the bases of the prismoid; but the basal edges of the prism are half the sum, and of the pyramid half the difference, of the corresponding sides of both the prismoid's bases. 424. If the bases of a prismoid are trapezoids whose midlines are b, and b2, and whose altitudes are al and a2, I) =( a( + a b 2 al- a b,-b 2 2 ' 2 + 2 -2 83. V. F = a(B +V V B2+ B2). 425. A side of the base of a frustum of a square pyramid is 25 meters, a side of the top is 9 meters, and the height is 240 neters. Required the volume of the frustum. Here V. F = a 240 (625 + 225 + 81) = 80 X 931 74,480 cubic meters. Ans. 426. The sides of the square bases of a frustum are 50 and 40 centimeters; each lateral edge is 30 centimeters. Find the volume. 59-28 liters. Ans. 427. In the frustum of a pyramid whose base is 50 square meters, and altitude 6 meters, the basal edge is to the corresponding top edge as 5 to 3. Find volume. 588 cubic meters. Ans. 428. Near Memphis stands a frustum whose height is 142-85 meters, and bases are squares on edges of 185-5 and 3-714 meters. Find its volume. 429. In the frustum of a regular pyramid, volume is 327 cubic meters, altitude 9 meters, and sum of basal and top edge 12 meters. Find these. 7 meters and 5 meters. Ans. EXERCISES AND PROBLEMS. 211 430. In the frustrum of a regular tetrahedron, given a basal edge, a top edge, and the volume. Find the altitude. 84. V. F = 1 tar (9ri2 + 9f,., +,22). 431. Divide a cone whose altitude is 20 into three equivalent parts by planes parallel to the base. Volume of whole cone = r2-r 20. Volume of midcone = r,2Tr (20 - a)..-. '2 (20 -a) = r220. But r: 20 = r:: 20 - a. 20 -- a *. r = r. 20 (20 -- a)3 40 400 3 (20 - a)3 = 16000 = 5333333+. 3.. 20-a 5=5333-333+= 17-471+.. a=2-528+. In the same way, a1= 3-604+. THEOREM OF CLAVIUS. 432. The frustum of a cone equals the sum of a cylinder and cone of frustral altitude whose radii are respectively the half-sum and half-difference of the frustral radii. V. F = a7w + r2) 2 la+ (Iri 7r2)2 2 3 This is a formula convenient for computation. 433. A frustum of 8 meters altitude, with 1 = 4 and '-2 = 2, is halved by a plane parallel to the base. Find radius of section and its distance from top of frustum. -3 = -36; a3 =- 4 -36 - 8. Ans. 212 MENSURATION. 434. In a frustum of 3 meters altitude and 63 cubic meters volume, r1- = 29r2; find q2. Ans. 435. In the frustum where a = 8 meters, 9r -- 10 meters, '2-= 6 meters, the altitude is cut into four equal parts by planes parallel to the base. Find the radii of these sections. HINT. The altitude of the completed cone is 8 + 12 = 20, and of the others, 18, 16, 14..'. by similar triangles, 20: 18: 16: 14:: 10: 9:8: 7. 7, 8, 9 meters. Ans. 436. The frustum of an equilateral cone contains 2 hectoliters, and is 40 centimeters in altitude. Find the radii. 27-785 and 50-879 centimeters. Ans. 85. PRISMOIDAL FORMULA: D - 1 G(B1 + 4 MX+ B2). I-, i i ' F =7 437. A solid is bounded by the triangles ABC, C2BD, the parallelogram A CDPE, and the skew quadrilateral B A ED whose elements are parallel to the plane BCD. Find its volume.- -. ABC. Ans. The skew quadrilateral is part of a warped surface called the hyperbolic paraboloid. 438. A tetrahedron is bisected by the hyperbolic paraboloid whose directrices are two opposite edges, and whose plane directer is parallel to another pair of opposite edges. 439. A solid is bounded by a parallelogram, two skew quadrilaterals, and two parallel triangles; find its volume. A (A1+ A- ). Ans. EXERCISES AND PROBLEMS. 213 440. Twice the volume of the segment of a ruled surface between parallel planes is equivalent to the sum of the cylinders on its bases, diminished by tha cone whose vertex is in one of the parallel planes, and whose elements are respectively parallel to the lines of the ruled surface. 86. W- aw (2 b + b2). 441. A wedge of 10 centimeters altitude, 4 centimeters edge, has a square base of 36 centimeters perimeter. Find volume. 442. The three parallel sides of a truncated prism are 8, 9, and 11 meters. The section at right angles to them is a right-angled triangle, with hypothenuse 17 meters, and one side 15 meters. Find volume. 443. The volume of a truncated regular prism is equal to the area of a right section multiplied by the axis or mean length of all the lateral edges. V = AI + 2- + 1...In n 444. To find the volume of any truncated prism. Rule: Multiply the length of each edge by the sum of the areas of all the triangles in the right section which have an angular point in that edge. The sum of the products will be three times the volume. Formula: 3 V- SAl. 87. X= -l-aM 445. Given V, the volume of a parallelepiped; in each of two parallel faces draw a diagonal, so that the two diagonals cross. Take the ends of these as summits of a tetrahedron, and find its volume. i V. Ans. 214 MENSURATION. 88. V. H = 43 r3. 446. Find the volume of a sphere whose superficial area is 20 meters. 447. Find the radius of a sphere equal to the sum of two spheres whose radii are 3 and 6 centimeters. 3$/9 centimeters. Ans. 448. Find the radius of a golden globe, density 19-35, weighing a kilogram. 449. A solid metal globe 6 meters in diameter is formed into a tube 10 meters in external diameter and 4 meters in length. Find the thickness of the tube. 1 meter. Ans. 450. If a cone and hemisphere of equal bases and altitudes be placed with their axes parallel, and the vertex of the cone in the plane of the base of the hemisphere, and be cut by a plane normal to their axes, the sum of the sections will be a constant. ARCHIMEDES' THEOREM. 451. Cone, hemisphere, and cylinder, of same base and altitude, are as 1:2:3. 452. The surfaces and the volumes of a sphere, a circumscribed right cylinder, and a circumscribed right cone whose axial section is an equilateral triangle, are as 4: 6: 9. Therefore, the cylinder is a geometric mean between the sphere and cone. HINT. Hi= 4r2r. V. 1H r3r. C +2B =6 2.. V. C =-3. >+ B = 9r2-r. V. K = Er3ir. 453* A quader having a square base of 5 centimeters edge, is filled 23 centimeters high with water. Into it is EXERCISES AND PROBLEMS. 215 put an iron ball going fully under the water, which rises 13-3972 centimeters. Find the diameter of the sphere. d = 52 X 13-3972. 6 89. V. G = - tar [3 (r2 + r22) + 2]. 454. If a heavy globe, whose diameter is 4 meters, be let fall into a conical glass, full of water, whose diameter is 5 meters and altitude 6 meters, it is required to determine how much water will run over. The slant height of the cone h = V6 + 625 = V42-25 = 65. If a is the altitude of the dry calot, 6-5: 2-5 =6-(2-a): 2. 13 = 10 + 25a..a. =12. Butdry segment equals c27r (r - a) = 144 7(2-0 4)= 1-44 1 6 = 2-304 r = 72382233+. But V. H = - d3c-T = 33-5104... volume of segment immersed is 26-272+ cubic meters. Ans. 455. A section parallel to the base of a hemisphere bisects its altitude; find the ratio of the parts of the hemisphere. 5: 11. Ans. 456. A sphere is divided by a plane in the ratio 5: 7. In what ratio is the globe cut? 325: 539. Ans. 457. A calot 8 centimeters high contains 1200 cubic centimeters; find radius of the sphere. 75 - cubic centimeters. Ans. 7r 216 MIENSURATION. 458. Find the volume of a segment of 12 centimeters altitude, the radius of whose single base is 24 centimeters. r = 30 centimeters; V. G= -014976 r cubic meters. Ans. 459. In terms of sphere-radius, find the altitude of a calot n times as large as its base. _ n-1)2 An a= -( ---- j2. Ans. 460. Find the ratio of the volume of a sphere to the volumne of its segment whose calot is n times its base. V. H: to V. G:: n: (n -— 1) (n + 2). Ans. 461. Find volume of a segment whose calot is 15-085 square meters, and base 2 meters, from sphere-center. V. G= 5-737025 cubic meters. Ans. 462. In a sphere of 10 centimeters radius, find the radii 9r1 and r2 of the base and top of a segment whose altitude is 6 centimeters, and base 2 centimeters, from the spherecenter. r 1= 4\/6 centimeters, qr2 = 6 centimeters. Ans. 463. Out- of a globe of 12 centimeters radius is cut a segment whose volume is one-third the globe, and whose bases are congruent and 8 centimeters apart; find the radius of bases. I'l = - 4 4I242 2 3 r1 centimeters. Ans. 90. V. S -2 ria2. 464. In a spherical sector, (1) Given r, r1, r2; find V. S. (2) Given a, l, r2; find V. S. 465. In a sphere of radius r, find the altitude of a segment which is to its sector as n to m. a =r -3 2 -. Ans. EXERCISES AND PROBLEMIS. 217 466. A sector is 1 of its globe, whose diameter is d; find the volume of its segment. V. G = - - n — ) Ans. 467. A sphere of given volume V is cut into two segments whose altitudes are as m to n; find both calots z1 and z2, and the segments. zl = i- /t36r; Z2 = /36r V2; ~n+n ' r n+n ' G_(m + 3 n); G 2(3mn)F Ans. (m + n)3 (37 + n)3 91. v = 3U. 468. Find the volume of a spherical ungula whose radius is 7-6 and ~ 18~12'. 92-958. Ans. 469. -26 6', r - 13-2. Find 9. 698-45. Ans. 470. A lune of 192 square meters has radius 15 meters; find volume of the ungula. 960 cubic meters. Ans. 471. Given L and rq; find v. v = —. Ans. 3 472. Given i and r; find. 2= 70 Ans. r 37r 473. Given L and 4; find v. v = LE. Ans. 92. Y= -I'e. 474. In a spherical pyramid given the angles of its triangular base, a =78~ 15', / —144~30', y7 108015', and given r= 108; find Y. 1106-61. Ans. 475. Given a, /, y, and A; find Y. 2 5. Ans. ~e7r 218 MENSURATION. 476. A = 486, a = 84~ 13', = 96~ 27', y = 112~ 20'; find Y 2543-06. Ans. 477. Given r=8-8, a 106~ 30', /= 120~ 10', y 150~ 15', 8 - 112~ 5'; find the four-faced Y. 511-433. Ans. 93. THEOREM OF PAPPUS. 478. If an equilateral triangle whose sides are halved by a straight line rotates about its base, the two volumes generated are equivalent. 479. A trapezoid rotates first about the longer, then about the shorter, of its parallel sides; the volumes of the solids generated are as m to n. Find the ratio of the parallel sides. 2n - Ans. 2' - n 94. V. 0 = 2r2abr. 480. Find the volume of a solid generated by rotating a parallelogram about an axis exterior to it; given the area of the parallelogram 7, and the distance r of the intersection-point of its diagonals from the axis. 2 7rr7. Ans. 481. The volume of a spiral spring, whose cross-section is a circle, equals the product of this generating circle by the length of the helix along which its center moves. The helix is the curve traced upon the surface of a circular cylinder by a point, the direction of whose motion makes a constant angle with the generating line of the cylinder. 482. A regular hexagon rotates about i, one of its sides; find the volume generated. 137T. Ans. EXERCISES AND PROBLEMS. 219 95. V1= _ 483. Any two similar solids may be so placed that all the lines joining pairs of homologous points intersect in a point. Every two homologous lines or surfaces in the two solids are then parallel. 484. Any two symmetric solids may be so placed that all the lines joining pairs of homologous points intersect in a point. This point bisects each sect. Every two homologous lines are then parallel. 485. Three persons having bought a sugar-loaf, would divide it equally among them by sections parallel to the base. It is required to find the altitude of each person's share, supposing the loaf to be a cone whose height is 20. 13-8672, 3-6044, and 2.5284. Ans. Let altitude of upper cone equal x, and its volume equal 1. Now 1: 3= x3: 203..'. x = /2666-666 = 13-867+. 96. IRREGULAR SOLIDS. 486. When a solid is placed in a square quader of basal edge 6 meters, the liquid, rising 3-97 meters, covers it; find its volume. 97. Vccm — 8 487. How much mercury, density 13'60, will weigh 7-59 grams? 488. If the density of zinc is 7-19, find how much weighs 3-83 kilograms. 220 MENSURATION. 98. 1 - [x2 ((B1- B3) + x3 (-B2 - 4) + etc. + x, (Bn-1 - Bn+l) + Xn+l (Bn + Bn+l)] + 2 [x,2 + (X3 - 2)-2 + (4 -X 3)J3 + etc. + (X:+1- X -n)2fJ]. 489. If the areas of six parallel planes 2 meters apart are 1, 3, 5, 7, 9, 11 square meters, and of the five mid-sections 2, 4, 6, 8, 10 square meters, find the whole volume. 99. A - q+ mx+nx2 +fx3. 490. Find an expression for the volume of a semicubic paraboloid generated by the revolution of a semicubic parabola round its axis. In this curve y2 oc X3, the revolving ordinate being y. 491. A paraboloid and a semicubic paraboloid have a common base and vertex; show that their volumes are as 2:1. 492. A vessel, whose interior surface has the form of a prolate spheroid, is placed with its axis vertical, and filled with a fluid to a depth h; find the depth of the fluid when the axis is horizontal. 493. A square-threaded screw, with double thread, is formed upon a solid cylinder 3 meters in diameter; the thread projects from the cylinder 5%- of meter, and the screw rises 3 meters in four turns. Find the volume, if the screw be 9 meters in length. 494. Find the volume of a square groin, the base of which is 15 meters square, and the guiding curve a semicircle. 495. The prismoidal formula applies to any shape contained by two parallel bases, and a lateral surface generated EXERCISES AND PROBLEMS. 221 by the motion of a parabola or cubic parabola whose plane is always parallel to a given plane, but whose curvature may pass through any series of changes in amount, direction, and position. 496. No equation of finite degree, representing a bounding sufcace, can define the limits of applicability of the prismoidal formula, because surfaces of higher degrees enclose prismoidal spaces. 100. TV= (B1 + 3A2c) = (B + 3Aa). 4 4 497. Show how existing rules for the estimation of railroad excavation may be improved. 101. f = — io h[5 (y2 + 4 Y6) + Y4 + yl + y 3I + y3 Y7] 498. If a parabolic spindle is equal in volume to one-fifth of the sphere on its axis as diameter, show that its greatest diameter is equal to half its length. 499. A parabolic spindle is placed in a cylinder half-full of water, the greatest diameter of the spindle being equal to that of the interior of the cylinder; find the height of the cylinder so that the water may just rise to the top. 500. A vessel, laden with a cargo, floats at rest in still water, and the line of flotation is marked. Upon the removal of the cargo every part of the vessel rises 3 meters, when the line of flotation is again marked. From the known lines of the vessel the areas of the two planes of flotation and of five intermediate equidistant sections are calculated and found to be as follows, the areas being expressed in square meters: 3918, 3794, 3661, 3517, 3361, 3191, 3004. Find the weight of the cargo removed. 222 MENSURATION. EXERCISES AND PROBLEMS ON CHAPTER VIII. 501. A square on the line b is divided into four equal triangles by its diagonals which intersect in C; if one triangle be removed, find the /C of the figure formed by the three remaining triangles. CLE Ans. 9 HINT. For such problems let L be the /C of the part left, and 0 of the part cut out; then CL x area left = CO X area cut out. 502. If a heavy triangular slab be supported at its angles, the pressure on each prop will be one-third the weight of the slab., 503. A weight o is placed at any point 0 upon a triangular table ABC (supposed without weight). Show that the pressures on the three props (viz., A, B, C) are proportional to the areas of the triangles BOC, AOC, AOB respectively. g... - k: \ 0Draw the straight lines AOF, BOI, COE; and let A', B', C' be the pressures at A E B A, B, Crespectively. Then C'x CE = x OE. C' AOB U ABC Similarly, for A' and B'. 504. The mid-point of one side of a square is joined with the mid-points of the adjacent sides, and the triangles thus formed are cut off; find the /C of the remainder. EXERCISES AND PROBLEMS. 223 505. If two triangles stand on the same base, the line joining their PC's is parallel to the line joining their vertices. 506. Find the distance from the base of the C' of four uniform rods forming a trapezoid, the two parallel sides of which are respectively 12 meters and 30 meters long, and the other sides each 15 meters long. 54- meters. Ans. 507. The altitude of the segment of a globe is a; find height of /C of its zone. i a. Ans. 508. Find /C of a hemisphere. 509. Find 'C of cylinder-mantel. 510. Find /C of cone-mantel. 511. If a body of density 8 weighs o, express the distance of its /C from its midcross-section. a B2(B2- B) As. 12w 512. Find the /Cf of a portion of a parabola cut off by a line perpendicular to the axis at a distance h from the vertex. 3 h. Ans. 513. Find the 0C' of the segment of a globe at a distance b from the center. 3 (1 +V ) Ans. 4 (2r+ b)' 514. Find the distance from vertex of the C0 of half a prolate spheroid. 515. A right circular cone, whose vertical angle is 60~, is constructed on the base of a hemisphere; find the /C of the whole body..516. Show that the compound body of the last exercise will rest in any position on its convex spherical surface. 517. Every body or system of particles has a P/C, and cannot have more than one. 224 MENSURATION. 518. Find the IC of any polygon by dividing it into triangles. 519. If the sides of a triangle be 3, 4, and 5 meters, find the distance of C from each side. 1,, meter. Ans. MISCELLANEOUS. 520. Find both sides of a rectangle from their ratio m: n, and its area R. mR In. a=. —; b = —. Ans. \ n \ m 521. If two triangles have one angle of the one equal to one angle of the other, and the sides about a second angle in each equal, then the third angles will be either equal or supplemental. 522. Two triangles are congruent, if two sides and a medial in the one are respectively equal to two sides and a corresponding medial in the other. 523. Two triangles are congruent, if three medials in one equal those in the other. 524. On a plane lie three tangent spheres of radius r; upon these lies a fourth of radius r'. How high is its center above the plane, and how large at least is r', since the sphere does not fall through? LOGARITHMS. 133. The logarithm a of a number n to a given base b is the index of the power to which the base must be raised to give the number: So, if b= n, then blogn= a, or the 6-logarithm of n is a. 134. blogb = 1. blog 1= 0. 135. blogmn = b blog + blogn. 136. blogn b log - b ogn. 137. blogP p X blogn. 138. blogn -P=1 X b1log. 139. blogn blogx, 1 1391 b log bl' b lgbl is called the modulus or multiplier for transformblog b' ing the log of a number to base b to the log of same number to base '. 140. The base of the common system of logarithms is 10. 1~log(n X 10P) = loggn +2 Y. 141. 10log (i +- 10P) -= 0logn - 2. 142, The mzcntissa is the decimal part of a logarithm. The czaracteristic is the integral part of a logarithm. 226 MENSURATION. The logs of all numbers consisting of the same digits in the same order have the same mantissa. 143. The characteristic of the log of a number is one less than the number of digits in the integral part. 144. When the number has no integral figures, the characteristic of its log is negative, and is one more than the number of cyphers which precede the first significant digit; that is, the number of cyphers (zeros) immediately after the decimal point. LOGARITHMS. 227 N O 1 2 3 4 5 6 7 8 9 PP 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4.8 11 0114 0453 0492 0531 0569 0607 0645 0682 0719 9755 4.8 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3.7 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3.6 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 3.6 15 1761 1790 1818 18-17 1875 1903 1931 1959 1987 2014 3.6 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 3.5 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2.5 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 2.5 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2.4 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2.4 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2.4 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 2.4 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 2.4 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 2.4 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2.3 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2.3 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 2.3 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 2.3 29 4624 4639 4651 4669 4683 4698 4713 4728 4742 4757 1.3 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1.3 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1.3 32 5051 5065 5079 5092 51050 5119 5132 5145 5159 5172 1.3 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 1.3 34 5315 5328 5340 5353 5366 5378 5391 5403 5116 5428 1.3 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 1.2 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 1.2 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 1.2 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 1.2 39 5911 5922 5933 5944 5955 59066 5977 5988 5999 6010 1.2 40 6021 6031 6042 6053 6004 6075 6085 6096 6107 6117 1.2 41 6128 6138 6149 6160 61700 6106191 6201 6212 6222 1.2 42 6232 6243 6253 6263 6274 628(4 6294 6304 6314 6(325 1.2 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 1.2 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 1.2 228 MENSURATION. N 0 1 2 3 4 5 6 7 8 9 PP 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 1.0 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 1.0 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 1.0 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 1.0 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 1.0 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1.0 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1.0 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1.0 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 1.0 54 7324 7332 7340 73,8 7356 7364 7372 7380 7388 7396 1.0 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 0.8 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 0.8 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 0.8 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 0.7 5979 77 7716 7723 7731 7738 7745 7752 7760 7767 7774 0.7 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 0.7 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 0.7 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 0.7 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 0.7 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 0.7 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 0.7 66 8195 8202 8209 8215 822 8228 8235 8241 8248 8254 0.7 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 0.6 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 0.6 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 0.6 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 0.6 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 0.6 72 8573 8579 8585 859 8597 8603 8609 8615 8621 8627 0.6 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 0.6 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 0.6 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 0.6 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 0.6 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 0.G 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 0.6 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 0.5 LOGARITHMS. 229 N 0 1 2 3 4 5 6 7 8 9 PP 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 0.5 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 0.5 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 0.5 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 0.5 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 0.5 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 0.5 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 0.5 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 0.5 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 0.5 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 0.5 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 0.5 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 0.5 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 0.5 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 0.5 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 0.5 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 0.5 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 0.5 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 0.4 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 0.4 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 0.4 N O 1 2 3 4 5 6 7 8 9 100 0000 0004 0009 0013 0017 0022 0026 0030 0035 0039 101 0043 0048 0052 0056 0060 0065 0069 0073 0077 0082 102 0086 0090 0095 0099 0103 0107 0111 0116 0120 0124 103 0128 0133 0137 0141 0145 0149 0154 0158 0162 0166 104 0170 0175 0179 0183 0187 0191 0195 0199 0204 0208 105 0212 0216 0220 0224 0228 0233 0237 0241 0245 0249 106 0253 0257 0261 0265 0269 0273 0278 0282 0286 0290 107 0294 0298 0302 0306 0310 0314 0318 0322 0326 0330 108 0334 0338 0342 0346 0350 0354 0358 0362 0366 0370 109 0374 0378 0382 0386 0390 0394 0398 0402 0406 0410 230 MENSURATION. N 0 1 2 3 4 5 6 7 8 9 110 0414 0418 0422 0426 0430 0434 0438 0441 0445 0449 111 0453 0457 0461 0465 0469 0473 0477 0481 0484 0488 112 0492 0496 0500 0504 0508 0512 0515 0519 0523 0527 113 0531 0535 0538 0542 0546 0550 0554 0558 0561 0565 114 0569 0573 0577 0580 0584 0588 0 059 0596 0599 0603 115 0607 0611 0615 0618 0622 0626 0630 0633 0637 0641 116 0645 0648 0652 0656 0660 0663 0667 0671 0674 0678 117 0682 0686 0689 0693 0697 0700 0704 0708 0711 0715 118 0719 0722 0726 0730 0734 0737 0741 0745 0748 0752 119 0755 0759 0763 0766 0770 0774 0777 0781 0785 0788 120 0792 0795 0799 0803 0806.0810 0813 0817 0821 0824 121 0828 0831 0835 0839 0842 0846 0849 0853 0856 0860 122 0864.0867 0871 0874 0878 0881 0885 0888 0892 0896 123. 0899 0903 0906 0910 0913 0917 0920 0924 0927 0931 124 0934 0938 0941 0945 0948 0952 0955 0959 0962 0966 125 0969 0973 0976 0980 0983 0986 0990 0993 0997 1000 126 1004 1007 1011 1014 1017 1021 1024 1028 1031 1035 127 1038 1041 1045 1048 1052 1055 1059 1062 1065 1069 128 1072 1075 1079 1082 1086 1089 1092 1096 1099 1103 129 1106 1109 1113 1116 1119 1123 1126 1129 1133 1136 130 1139 1143 1146 1149 1153 1156 1159 1163 1166 1169 131 1173 1176 1179 1183 1186 1189 1193 1196 1199 1202 132 1206 1209 1212 1216 1219 1222 1225 1229 1232 1235 133 1239 1242 1245 1248 1252 1255 1258 1261 1265 1268 134 1271 1274 1278 1281 1284 1287 1290 1294 1297 1300 135 1303 1307 1310 1313 1316 1319 1323 1326 1329 1332 136 1335 1339 1342 1345 1348 1351 1355 1358 1361 1364 137 1367 1370 1374 1377 1380 1383 1386 1389 1392 1396 138 1399 1402 1405 1408 1411 1414 1418 1421 1.424 1427 139 1430 1433 1436 1440 1443 1446 1449 1452 1455 1458 140 1461 1464 1467 1471 1474 1477 1480 1483 1486 1489 141 1492 1495 1498 1501 1504 1508 1511 1514 1517 1520 142 1523 1526 1529 1532 1535 1538 1541 1544 1547 1550 143 1553 1556 1559 1562 1565 1569 1572 1575 1578 1581 144 1584 1587 1590 1593 1596 1599 1602 1605 1608 1611 LOGARITHMS. 231 N 0 1 2 3 4 5 6 7 8 9 145 1614 1617 1620 1623 1626 1629 1632 1635 1638 1641 146 1644 1647 1649 1652 1655 1658 1661 1664 1667 1670 147 1673 1676 1679 1682 1685 1688 1691 1694 1697 1700 148 1703 1706 1708 1711 1714 17i7 1720 1723 1726 1729 149 1732 1735 1738 1741 1744 1746 1749 1752 1755 1758 150 1761 1764 1767 1770 1772 1775 1778 1781 1784 1787 151 1790 1793 1796 1798 1801 1804 1807 1810 1813 1816 152 1818 1821 1824 1827 1830 1833 1836 1838 1841 1844 153 1847 1850 1853 1855 1858 1861 1864 1867 1870 1872 154 1875 1878 1881 1884 1886 1889 1892 1895 1898 1901 155 1903 1906 1909 1912 1915 1917 1920 1923 1926 1928 156 1931 1934 1937 1940 1942 1945 1948 1951 1953 1956 157 1959 1962 1965 1967 1970 1973 1976 1978 1981 1984 158 1987 1989 1992 1995 1998 2000 2003 2006 2009 2011 159 2014 2017 2019 2022 2025 2028 2030 2033 2036 2038 160 2041 2044 2047 2049 2052 2055 2057 2060 2063 2066 161 2068 2071 2074 2076 2079 2082 2084 2087 2090 2092 162 2095 2098 2101 2103 2106 2109 2111 2114 2.117 2119 163 2122 2125 2127 2130 2133 2135 2138 2140 2143 2146 164 2148 2151 2154 2156 2159 2162 2164 2167 2170 2172 165 2175 2177 2180 2183 2185 2188 2191 2193 2196 2198 166 2201 2204 2206 2209 2212 2214 2217 2219 2222 2225 167 2227 2230 2232 2235 2238 2240 2243 2245 2248 2251 168 2253 2256 2258 2261 2263 2266 2269 2271 2274 2276 169 2279 2281 2284 2287 2289 2292 2294 2297 2299 2302 170 2304 2307 2310 2312 2315 2317 2320 2322 2325 2327 171 2330 2333 2335 2338 2340 2343 2345 2348 2350 2353 172 2355 2358 2360 2363 2365 2368 2370 2373 2375 2378 173 2380 2383 2385 2388 2390 2393 2395 2398 2400 2403 174 2405 2408 2410 2413 2415 2418 2420 2423 2425 2428 175 2430 2433 2435 2438 2440 2443 2445 2448 2450 2453 176 2455 2458 2460 2463 2465 2467 2470 2472 2475 2477 177 2480 2482 2485 2487 2490 2492 2494 2497 2499 2502 178 2504 2507 2509 2512 2514 2516 2519 2521 2524 2526 179 2529 2531 2533 2536 2538 2541 2543 2545 2548 2550 232 MENSURATION. N 0 1 2 3 4 5 6 7 8 9 180 2553 2555 2558 2560 2562 2565 2567 2570 2572 2574 181 2577 2579 2582 2584 2586 2589 2591 2594 2596 2598 182 2601 2603 2605 2608 2610 2613 2615 2617 2620 2622 183 2625 2627 2629- 2632 2634 2636 2639 2641 2643 2646 184 2648 2651 2653 2655 2658 2660 2662 2665 2667 2669 185 2672 2674 2676 2679 2681 2683 2686 2688 2690 2693 186 2695 2697 2700 2702 2704 2707 2709 2711 2714 2716 187 2718 2721 2723 2725 2728 2730 2732 2735 2737 2739 188 2742 2744 2746 2749 2751 2753 2755 2758 2760 2762 189 2765 2767 2769 2772 2774 2776 2778 2781 2783 2785 190 2788 2790 2792 2794 2797 2799 2801 2804 2806 2808 191 2810 2813 2815 2817 2819 2822 2824 2826 2828 2831 192 2833 2835 2838 2840 2842 2844 2847 2849 2851 2853 193 2856 2858 2860 2862 2865 2867 2869 2871 2874 2876 194 2878 2880 2882 2885 2887 2889 2891 2894 2896 2898 195 2900 2903 2905 2907 2909 2911 2914 2916 2918 2920 196 2923 2925 2927 2929 2931 2934 2936 2938 2940 2942 197 2945 2947 2949 2951 2953 2956 2958 2960 2962 2964 198 2967 2969 2971 2973 2975 2978 2980 2982 2984 2986 199 2989 2991 2993 2995 2997 2999 3002 3004 3006 3008 : 24 GIZNv, HEA TH, &6 CO.'S PUBLICA TIONS. MATHEMATICS. Wentworth's Elements of Plane and Solid GeOMETR( Y. By GEORGE A. WENTWORTH, Phillips Academy, Exeter, I2mo. Half morocco. 400 pages. Mailing price, $1.45; Introduction, $I.oo; Exchange, 60 cts. Wentworth's Elements of Plane Geometry. I21m. 250 pages. Mailing Price, 85 cts.; Introduction, 75 cts.; Exchange, 40 cts. This work is based upon the assumption that Geometry is a branch of practical logic, the object of which is to detect, and state clearly and precisely, the successive steps from premise to conclusion. In each proposition, a concise statement of what is given is printed in one kind of type, of what is required in another, and the demonstration in still another. The reason for each step is indicated in small type, between that step and the one following, thus preventing the necessity of interrupting the process of demonstration by referring to a previous proposition. The number of the section, however, on which the reason depends, is placed at the side of the page; and the pupil should be prepared, when called upon, to give the proof of each reason. A limited use has been made of symbols, wherein symbols stand for words, and not for operations. Great pains have been taken to make the page attractive. The figures are large and elegant, and the propositions have been so arranged that in no case is it necessary to turn the page in reading a demonstration. A large experience in the class-room convinces the author that, if the teacher will rigidly insist upon the logical form adopted in this work, the pupil will avoid the discouraging difficulties which MA THET A TICS. I25 usually beset the beginner in geometry; that he will rapidly develop his reasoning faculty, acquire facility in simple and accurate expression, and lay a foundation of geometrical knowledge which will be the more solid and enduring from the fact that it will not rest upon an effort of the memory simply. Strong evidence of the merit of this book is found in the fact that since the beginning of the schoolyear, 877-78, it has been introduced into Thirty-six Colleges and nearly Four Hundred Preparatory Schools. Teachers should not fail to examine this book before forming new classes. TESTIMONIALS. Joseph Ficklin, Prof. of Math., treated in a way which will help both Univ. of Missouri. I have examined, the teacher and the taught. with considerable care, Wentworth's (Nov. 22, I879.) Geometry, and the result is a decidedly favorable opinion of the book. Profes- Selden J. Coffin, Prof. of Math., sor Wentworth is evidently a practical Lafayette Coll., Pa.. We are pleased teacher. He has shown in the execu- with the simplicity and clearness of the tion of his work that he knows just demonstrations in Wentworth's Geomwhere beginners in Geometry encoun- etry, and have adopted it as our textter difficulties, and, in my judgment, book in that subject. he has been eminently successful in his attempt to make those difficulties dis- E. Otis Kendall, Prof. of Math., appear. Univ. of Pennsylvania: I have no hesitation in saying that it is the best book Samuel Hart, Prof. of Math., for beginners that I have ever seen. Trinity College. There are some The demonstrations are rigorous, the things in Wentworth's Geometry which language clear, and I have not discovmakes it specially well adapted for ered any defect in the reasoning. use in Schools and Academies. The clear method in which each W. C. Esty, Prof. of Math/., Amproposition is presented as a whole; herst Coll.. It is a step in advance of the manner of reproducing the state- the text-books of its kind. It must ments of former theorems to which make the course in Geometry easier for reference is made, so that the student both teacher and pupil. is almost forced to recall them; the careful and precise use of the method John R. French, PrIof: of Math., of limits - these are among the things Syracuse Univ... The distinctness of which will tend to make the work es- its statements, and the clearness and pecially serviceable. And I am satis- compactness of its demonstrations, are fled that it has a sufficiently extensive admirable. We purpose to test it in view of Geometry for most students, recitation-room with our next class. 142 GINN, HEA 'TH, &' CO.'S PUBLICA TIOANS. Wheeler's Elements of Plane and Spherical Trigonometry. By H. N. WVHEELER, A.M., of Harvard University. I2mo. Cloth. 211 pages. Mailing Price, $i.Io; Introduction, 94 cents. Exchange, 50 cents. Peirce's Mathematical Tables, First Series, are bound with this edition of the Trigonometry. These Tables are so arranged, and so fully explained, as to be readily intelligible to the beginner in the use of logarithms. Two pages, containing the functions of angles less than 6- and greater than 84'', given to every miizute, have been added, since the publication of the first edition. PLANE TRIGONOMETRY. The author believes that a student can get a comprehensive and thorough knowledge of Trigonometry most quickly and easily, if, at the outset, such definitions are given to the trigonometric functions as will apply to all angles; with this idea for a basis, he has endeavored to prepare an elementary text-book for general use. By beginning with an explanation of the use of the negative sign as applied to lines and angles, and then giving general definitions to the trigonometric functions, he has been able to demonstrate all the fundamental formulas in a perfectly general yet simple manner. While he has tried to present the subject from an elementary point of view, he has not lost sight of the fact, that, to most students, Trigonometry is merely a stepping-stone to something higher; and for this reason he has also tried to present the results in such a light as will make them effective tools for the student in his future work. SPHERICAL TRIGONOMETRY. The author has endeavored to prepare a book for the use of schools and colleges which, while brief and simple, shall yet be thorough, and suggestive both of the theoretical and practical bearings of the subject. Such applications to Geometry and Astronomy, and such problems involving these applications, have been given, as will interest the student, and show him that Spherical Trigonometry is not a mere mass of meaningless formulas, but an easy means of solving many practical problems of great importance. MA T7HEIMA TICS. I43 -Wheeler's Trigonometry has been already introduced into the following institlltions: - Harvard University, Mass. Cornell University, N.Y. Vassar College, N.Y. Wesleyan University, Conn. Tufts College, Mass. Colby University, Me. Middlebury College, Vt. Beloit College, Wis. Lawrence Univ., Appleton, Wis. Rutgers College, N.J. Columbian Univ., Wash., D.C. St. John's College, Md. University of North Carolina. Haverford College, Pa. Dickinson College, Pa. Irving Female College, Pa. Wabash College, Ind. Sheffield Scientific School, Ct. Phillips Exeter Academy, N.H. Phillips Andover Acad., Mass. Wesleyan Acad.,Wilbraham, Mass. Thayer Academy, Braintree, Mass. Hopkins Grammar School, Ct. Buckley High Sch., N. London, Ct. Collinsville High School, Ct. New Haven High School, Ct. Newport High School, R.I. Norwich Free Academy, Ct. Betts Military Academy, Ct. Centenary Institute, N.J. Dickinson Seminary, Pa. Tilton Academy, N.H. Taunton High School, Mass. Westfield High School, Mass. Homer Academy, N.Y. Park Institute, Rye, N.Y. La Fayette High School, Ind. Galion High School, O. Byerly's Elements of the Differential Calculus. With Numerous Examples and Applications. Designed for Use as a College Text-Book. By W. E. BYERLY, A.M., Harvard University. Mailing Price, $2.30; Introduction, $2.oo. This book embodies the results of the author's experience in teaching the Calculus at Cornell and Harvard Universities, and is intended for a text-book, and not for an exhaustive treatise. Its peculiarities are the rigorous use of the Doctrine of Limits, as a foundation of the subject, and as preliminary to the adoption of the more direct and practically convenient infinitesimal notation and nomenclature; the early introduction of a few simple formulas and methods for integrating; a rather elaborate treatment of the use of infinitesimals in pure geometry; and the attempt to excite and keep up the interest of the student by bringing in throughout the whole book, and not merely at the end, numerous applications to practical problems in geometry and mechanics. I 44 GIVN, HEA TH, s CO.'S PLUBLICA TIONS. Samuel Hart, Priof: of MatA. in that in selection, arrangement, and Trinity Coll.: ' he student can hardly treatment, it is, on the whole, in a very fail, I think, to get from the book an high degree, wise, able, marked by a exact, and at the same time, a satisfac- true scientific spirit, and calculated to tory explanation of the principles on develop the same spirit in the learner. which the Calculus is based; and the.. The book contains perhaps all of introduction of the simpler methods of the integral calculus, as well as of the integration, as they are needed, enables differential, that is necessary to the orapplications of those principles to be dinary student. And with so much introduced in such away as to be both of this great scientific method, every interesting and instructive, thorough student of physics, and every general scholar who feels any interest James Mills Peirce, Prof. of in the relations of abstract thought, and llait/h., Ha-rvard Univ. (From the HaI- is capable of grasping a mathematical vard Register). In mathematics, as in idea, ought to be familiar. One who other branches of study, the need is aspires to technical learning must supnow very much felt of teaching which is plement his mastery of the elements by general without being superficial; lim- the study of the comprehensive theoited to leading topics, and yet, within retical treatses... But he who is tlorits limits, thorough, accurate, and prac- oughly acquainted with the book before tical; adapted to the communication us has made a long stride into a sound of some degree of powver, as well as; |and practical knowledge of the subject knowledge, but free from details. which If the calculus. He has begun to be a are important only to the specialist. real analyst Professor Byerly's Calculus appears to be designed to meet this want... Such H. A. Newton, Priof of AM/ath/. in a plan leaves much room for the exer- Yale Coll., New 1-aven. I have looked cise of individual judgment; and dif- it through with care, and find the subferences of opinion will undoubtedly ject very clearly and logically develexist in regard to one and another point oped. I am stronglv inclined to use it of this book. But all teachers will agree in my class next year. Peiroe's Three and Four Place Tables of Loqaritmic and Trigonometric Functions. By JAMIES MILLS PEIRCE, University Professor of Mathematics in Harvard University. Quarto. Cloth. Mailing Price, 60 cts.; Introduction, 50 cts. Four-place tables require, in the long run, only half as much time as five-place tables, one-third as much time as six-place tables, and one-fourth as much as those of seven places. They are sufficient for the ordinary calculations of Surveying, Civil, Mechanical, and Mining Engineering, and Navigation; for the work of the Physical or Chemical Laboratory, and even for many computations of Astronomy. They a are also especially suited to be used in teaching, as they illustrate principles as well as the larger tables, and with far less iMA THEMA TICS. I45 expenditure of time. The present compilation has been prepared with care, and is handsomely and clearly printed. Peirce's Mathematical Tables Chiefly to Four Figures. First Series. By Professor JAMES MILLS PEIRCE, of Harvard University. I2mo. Cloth. Mailing Price, 60 cts.; Introduction, 50 cts.; Paper, 30 cts. This little volume contains tables of Logarithms of Numbers, of Sums and Differences, of all the six Circular Functions, and of Hyperbolic Functions (a table which may be used to great advantage in certain computations), and also of Inverse Circular Functions, and of Natural Sines and Cosines, Tangents and Cotangents, and Secants and Cosecants. It is provided, moreover, with a very detailed and fully illustrated Explanation of the Tables. The collection is published both in a duodecimo and in an octavo form, in a clear, handsome, and good-sized type, cast for the purpose. It is adapted to the needs of schools and colleges, and is suitable for the use of engineers and computers. The book may be had either separately or bound with the author's "Elements of Logarithms," or with " Wheeler's Trigonometry." The Second Series of these Tables will contain tables of Reciprocals, Squares. Cubes, and Fourth Powers, Probabilities, Traverses, Meridional Parts, the Comparison of the Metric and English Systems, etc. Peirce's Elements of Loqarithms. With an explanation of the Author's Three and Four Place Tables. By Professor JAMIES MILLS PEIRCE, of Harvard University. I2mo. Cloth. 89 pages. Mailing Price, 60 cts.; Introduction, 50 cts. This book is intended as an introduction to the study and use of Logarithms. The elementary theory of the subject is fully and thoroughly set forth, the main facts of the history of logarithms are stated, and many practical hints are given for the benefit of the young computer. The design of the author has been to give to students a more complete and accurate knowledge of the nature and use of logarithms than they can acquire from the cursory study com 145 GIVIV, IHEA TH, & CO.'S PUBLICAT IONTS. monly bestowed on this subject. The book contains an explanation of the Three and Four Place Tables; and it may be used either with that collection or with the Smaller Four-Place Tables. David Murray, Ratgers Col/.. Literarisches Centralblatt: It is the most compact set of tables I This collection can be fully recoInhave seen. I already have found it mended as a model of handiness and so useful that I could not dispense with judicious arrangement, as well as of it. solid value. J. H. C. Coffin, L. S. N.: They Jahrbuch tiber die Fortsupply a want decidedly felt by corn- schritte der Mathemnatik: The puters in astronomical and trigonomet- tables are distinguished by perspicuity rical work. and convenience. Searle's Outlines of Astronomy. By ARTHUR SEARLE of Harvard College Observatory. I6mo. Cloth. 433 pages. Mailing Price, $I.60; Introduction, $1.40; Exchange, 75 cts. It is designed to serve as a short course of astronomical study, suited to the wants of those who have no knowledge of mathematics. The attempt has been made to place facts distinctly before the reader, with as little use of technical terms as practicable. Chapters I. to VI. contain a summary of the opinions now generally accepted with respect to the chief subjects of astronomical inquiry, such as the general structure of that part of the universe within reach of human observation, the constitution of the Sun and of the Stars, the structure and movements of the Earth, Moon, and other bodies of the Solar System, and the nature of nebula, comets, and meteors. Chapter VII. describes the principal phenomena which are accounted for by the facts already stated. Chapters VIII. and IX. contain rudimentary geometrical and optical information intended to aid in the comprehension of the remainder of the book. Chapters X. to XIII. give an account of the methods and instruments used in practical Astronomy, giving distinct information about the work of astronomers, not instruction in that work. Chapter XIV. explains the application of mechanical principles, including the law of gravitation, to the principal movements of the planets. The simple mathematical propositions are stated with ful MA THHEMA TICS. 147 ness enough to enable those who have forgotten most of their algebra and geometry to follow the reasoning. Chapter XV. contains a short history of Astronomy. New York Tribune: It is one the palm, both for exactness of method, of the clearest, most comprehensive, solidity of information, and appropriand most informing expositions of the ateness of expression. principles and facts of Astronomy that are to be found in the language. The New York Nation: It is full student who has made himself master without being heavy, precise without of its contents, will be in possession of being pedantic, and explanatory withan amount of knowledge for the attain- out being diffuse. It nowhere disapment of which he might cheerfully pay points, and least of all by any enthusia high price, and for which he would astic ascribing of pre-eminence to its have been envied by the late most own subject-matter. As regards the graceful American historian, who could body of the work we have not detected never clearly comprehend the differ- an error of statement in any part of it; ence between the ecliptic and the equa- which is a good deal to say, or would tor. Compared with the popular sci- be of any author but an astronomer... entific treatises which swarm to such an The most readable treatise upon Asextent in England, the present bears tronomy that we have ever seen. Questions and Exercises on Stewart's Elementary Physics, wizit Azswzers and occasional Solu/ions. By GEORGE A. HILL, formerly Assistant Professor of Physics in Harvard University. I8mo. Boards. 192pages. Mailing Price, 40 cts.; Introduction,35 cts. These Questions and Exercises have been drawn up with the aim of making Mr. Stewart's excellent work more useful in elementary teaching. Part I. consists of questions upon the text of Mr. Stewart's book which are intended to be direct and exhaustive. These will be found useful for review and examination purposes. Parts II. and III., which form the principal part of the work, have been written with two objects in view: First, to stimulate original thought on the part of the student, and to give the teacher the means of testing thoroughly the student's knowledge of principles; Secondly, to make certain needful additions to the felicitous but cursory sketch of Mechanics, Hydrostatics, and Pneumatics, contained in the first two chapters of Mr. Stewart's book.