JOHN ALEXANDER JAMESON, Jr. 1903-1934 fcfc4l?*-* ■ * *W tJMsMft a!0 This book belonged to John Alexander Jameson, Jr., A.B., Wil- liams, 1925; B.S., Massachusetts Institute of Technology, 1928; M.S., California, 1933. He was a member of Phi Beta Kappa, Tau Beta Pi, the American Society of Civil Engineers, and the Sigma Phi Fraternity. His untimely death cut short a promising career. He was engaged, as Research Assistant in Mechanical Engineering, upon the design and construction of the U. S. Tidal Model Labora- tory of the University of California. His genial nature and unostentatious effectiveness were founded on integrity, loyalty, and devotion. These qualities, recognized by everyone, make his life a continuing beneficence. Memory of him will not fail among those who knew him. WENTWORTH-SMITH MATHEMATICAL SERIES SOLID GEOMETRY BY GEORGE WEXTWORTir AND DAVID EUGENE SMITH -/■_ < gs; s mit h aa > r; R. V JM r GINN AND COMPANY BOSTON • HEW FOBS • 'III' \«.'» • LONDOH ATLANTA • DALLAS • COLUMBUS • BAM FRANCISCO , \: COPYRIGHT, 1888, 1899, BY G. A. WENTWORTH libra** COPYRIGHT, 1911, 1913, BY GEORGE WENTWORTH AND DAVID EUGENE SMITH ENTERED AT STATIONERS' HALL ALL RIGHTS RESERVED 919.1 ENglNEERlM LIBRAHI gfle atftenaum jgregj GINN AND COMPANY • PRO- PRIETORS • BOSTON • U.S.A. PREFACE Long after the death of Robert Kecorde, England's first great writer of textbooks, the preface of a new edition of one of his works contained the appreciative statement that the book was " entail' d upon the People, ratified and sign'd by the approbation of Time." The language of this sentiment sounds quaint, but the noble tribute is as impressive to-day as when first put in print two hundred and fifty years ago. With equal truth these words may be applied to the Geom- etry written by George A. Wentworth. For a generation it has been the leading textbook On the subject in America. It set a standard for usability that every subsequent writer upon geometry has tried to follow, and the number of pupils who have testified to its excellence has run well into the millions. In undertaking to prepare a work to take the place of the Wentworth Geometry the authors have been guided by certain well-defined principles, based upon an extended investigation of the needs of the schools and upon a study of all that is best in the recent literature of the subject. The effects of these principles they feel should be summarized for the pur- pose of calling the attention of the wide circle of friends of the Wentworth-Smith series to the points of similarity and of difference in the two works. 1. Every effort has been made not only to preserve but to improve upon the simplicity of treatment, the clearness of ex- pression, and the symmetry of page that characterized the successive editions of the W.-nt worth Geometry. It 1ms been the purpose to prepare a book that should do even more than maintain the traditions this work has fostered, 9754 iv SOLID GEOMETRY 2. The proofs have been given substantially in full, to the end that the pupil may always have before him a model for his independent treatment of the exercises. 3. To meet a general demand, the number of propositions has been decreased so as to include only the great basal theo- rems and problems. A little of the less important material has been placed in the Appendix, to be used or not as cir- cumstances demand. 4. The exercises, in some respects the most important part of a course in geometry, have been rendered more dignified in appearance and have been improved in content. The number of simple exercises has been greatly increased, while the diffi- cult puzzle is much less in evidence than in most American textbooks. The exercises are systematically grouped, appear- ing in general in full pages, in large type, and at frequent intervals. They are not all intended for one class, but are so numerous as to allow the teacher to make selections from year to year. 5. The work throughout has been made as concrete as is reasonable. Definitions have been postponed until they are actually needed, only well-recognized terms have been em- ployed, the pupil is led to apply his geometry to practical cases in mensuration, and correlation is made with the algebra already studied. 6. All the references to Plane Geometry that are directly made in the proof of Solid Geometry have been prefixed to this edition so as to be easily accessible. The authors are indebted to many friends of the Wentworth- Smith series for assistance and encouragement in the labor of preparing this edition, and they will welcome any further sug- gestions for improvement from any of their readers. GEORGE WENTWORTH DAVID EUGENE SMITH COXTKXTS REFERENCES TO PLANE GEOMETRY . BOOK VI. LINES AND PLANES IN SPACE Links and Pl \m> . Dihedral Am. les ..... Polyhedral Am. i Exercis] ...... BOOK VII. POLYHEDRONS, CYLj 5, ANlTCONI Polyhedrons . Prisms .... Parallelepipeds Pyramids . . v . Regular Polyhedrons . Cylinders Cones .... Exercises BOOK VIII. THE SPHERE Spheres .... Plane Sections and Tangent , SPHERK \ l. POL! GONS Measurement of Spherical Surfai i - MEASl REMENT OF SPHERK \ I. Sol [DS Exercises APPENDIX POL! II I DRONS ..... SPHERK \ I SEG mini- K'l ,i RE LTIONS «'i ( , BOM BTRY History of < Ieome i ry . Paoi \ii .,-■■ 273 293 308 314 317 317 317 322 337 350 353 376 » / I O 1 L •1. : i 41 4- o INDEX SYMBOLS AND ABBREVIATIONS = equals, equal, equal to, is equal to, or is equivalent to. > is greater than. < is less than. II parallel. _L perpendicular. A angle. A triangle. O parallelogram. □ rectangle. O circle, st. straight. rt. right. V since. .'. therefore. Adj. adjacent. Alt. alternate. Ax. axiom. Const, construction. Cor. corollary. Def. definition. Ex. exercise. Ext. exterior. Eig. figure. Hyp. hypothesis. Idem identity. Int. interior. Post. postulate. Prob. problem. Prop. proposition. Sup. supplementary. These symbols take the plural form when necessary, as in the case of lis, A, A, CD. The symbols +, — , x , ■*- are used as in algebra. There is no generally accepted symbol for " is congruent to," and the words are used in this book. Some teachers use = or =, and some use = , but the sign of equality is more commonly employed, the context telling whether equality, equivalence, or congruence is to be understood. q.e.d. is an abbreviation that has long been used in geometry for the Latin words quod erat demonstrandum, "which was to be proved." q. e. f. stands for quod erat faciendum, "which was to be done." VI REFERENCES TO PLANK GEOMETRY 28. A portion of a plane bounded l»y tlnvi« straight lines is called a triangle. 41. The whole angular space in a plane about a point is • •ailed a perigon. 52. The following are the most important axioms used in geometry : 1. If equals are added to equals, the sums are equal. 2. If equals are subtracted from equals, the remainders are equal. 3. If equals are multiplied by equals, the products are equal. 4. If equals are divided by equals, the quotients are equal. In division the divisor is never zero. 5. Like powers and like positive roots of equals are equal. 6. If unequals are operated on by positive equals in the same way, the results are unequal in the same order. 7. If unequals are added to unequals in the same order, the sums are unequal in the same order ; if unequals are subtracted from equals, the remainders are unequal in the reverse order. 8. Quantities that are equal to the same quantity or to equal quantities are equal to each other. 9. A quantity may be substituted for its equal in an equa- tion or in an inequality. 10. If the first of three quantities is greater than the second, and the second is greater than the third, then the first is great, r than the third. 11. The whole is greater than any of its parts, and is equal to the sum of all its parts. vii viii SOLID GEOMETRY 53. Postulate 5. Any figure may be moved from one place 1 ) another without altering its size or shape. 56. All right angles are equal. 57. From a given point in a given line only one perpendic- ular can be drawn to the line. 60. If two lines intersect, the vertical angles are equal. 66. Definition of congruent figures. 67. Corresponding parts of congruent figures are equal. 88. Two triangles are congruent, if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. 69. Two right triangles are congruent, if the sides of the right angles are equal respectively. 72. Two triangles are congruent, if two angles and the in- cluded side of the one are equal respectively to • • • . 80. Two triangles are congruent, if the three sides of the one are equal respectively to the three sides of the other. 82. Only one perpendicular can be drawn to a given line from a given external point. 84. Of two lines drawn from a point in a perpendicular to a given line, cutting off on the given line unequal segments from the foot of the perpendicular, the more remote is the greater. 89. Two right triangles are congruent, if tile hypotenuse and a side of ths one are equal respectively to the hypotenuse and a side of the other. 93. Lines that lie in the same plane and cannot meet how- ever far produced are called parallel lines. 94. Through a given point only one line can be drawn par- allel to a given line. 95. Two lines in the same plane perpendicular to the same line are parallel. REFERENCES TO PLANE GEOMETRY ix 97. If a line is perpendicular to one of two parallel lines, it is perpendicular to the other also. 112. The sum of any two sides of a triangle is greater than the third side, and the difference between any two sales is less than the third side. 116. If two triangles have two sides of the one equal respec- tively to two sides of the other, but the third side of the ii: triangle greater than the third side of the second, then the angle opposite the third side of the first is greater than the angle opposite the third side of the second. 118. A quadrilateral may be a trapezoid, having two Bides parallel ; a parallelogram, having the opposite sides parallel ; or it may have no sides parallel. 125. The opposite sides of a parallelogram are equal. 126. A diagonal divides a parallelogram into two congruent triangles. 127. Segments of parallel lines cut off by parallel lines are equal. 130. If two sides of a quadrilateral are equal and parallel, then the other two sides are equal and parallel, and the figure is a parallelogram. 131. The diagonals of a parallelogram bisect each other. 132. Two parallelograms are congruent, if two sides and the included angle of the one are equal respectively to t v. and the included angle of the other. 133. Two rectangles having equal bases and equal altitudes are congruent. 136. The line which joins the mid-points of two sides of a triangle is parallel to the third side, and is equal to half the third side. x SOLID GEOMETRY 142. Two polygons are mutually equiangular, if the angles of the one are equal to the angles of the other respectively, taken in the same order ; mutually equilateral, if the sides of the one are equal to the sides of the other respectively, taken in the same order ; congruent, if mutually equiangular and mutually equilateral, since they then can be made to coincide. 145. Each angle of a regular polygon of n sides is equal to 2(n-2) . — * '- right angles. n 146. The sum of the exterior angles of a polygon, made by producing each of its sides in succession, is equal to four right angles. 148. To prove that a certain line or group of lines is the locus of a point that fulfills a given condition, it is necessary and sufficient to prove two things : 1. That any point in the supposed locus satisfies the con- dition. 2. That any point outside the supposed locus does not satisfy the given condition. 150. The locus of a point equidistant from the extremities of a given line is the perpendicular bisector of that line. 151. Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line. 152. The locus of a point equidistant from two given inter- secting lines is a pair of lines bisecting the angles formed by those lines. 159. A closed curve lying in a plane, and such that all of its points are equally distant from a fixed point in the plane, is called a circle. REFERENCES TO PLANE GEOMETRY xi 162. All radii of the same circle or of equal circles are equal; and all circles of equal radii are equal. 167. In the same circle or in equal circles equal arcs subtend equal central angles : and of two unequal arcs the greater sul> tends the greater central angle. 172. In the same circle or in equal circles, if two chords are equal, they subtend equal arcs ; and if two chords are unequal, the greater subtends the greater arc. 174. A line through the center of a circle perpendicular to a chord bisects the chord and the arcs subtended by it. 178. In the same circle or in equal circles equal chords are equidistant from the center, and chords equidistant from the center are equal. 185. A tangent to a circle is perpendicular to the radius drawn to the point of contact. 195. If two circles intersect, the line of centers is the per- pendicular bisector of their common chord. 204. When a variable approaches a constant in such a way that the difference between the two may become and remain less than any assigned positive quantity, however small, the constant is called the limit of the variable. 207. If, while approaching their respective limits, two vari- ables are always equal, their limits are equal. 212. In the same circle or in equal circles two central angles have the same ratio as their intercepted arcs. 213. A central angle is measured by the intercepted arc. 261. In any proportion the product of the extremes is equal to the product of the means. 262. The mean proportional between two quantities is equal to the square root of their product. xii SOLID GEOMETRY 269. In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. 270. Like powers of the terms of a proportion are in pro- portion. 273. If a line is drawn through two sides of a triangle par- allel to the third side, it divides the two sides proportionally. 274. One side of a triangle is to either of its segments cut off by a line parallel to the base as the third side is to its corresponding segment. 275. Three or more parallel lines cut off proportional in- tercepts on any two transversals. 282. Polygons that have their corresponding angles equal, and their corresponding sides proportional, are called similar polygons. 285. Two mutually equiangular triangles are similar. 288. If two triangles have an angle of the one equal to an angle of the other, and the including sides proportional, they are similar. 289. If two triangles have their sides respectively propor- tional, they are similar. 290. Two triangles which have their sides respectively par- allel, or respectively perpendicular, are similar. 292. If two polygons are similar, they can be separated into the same number of triangles, similar each to each, and similarly placed. 298. If a perpendicular is drawn from any point, on a circle to a diameter, the chord from that point to either extremity of the diameter is the mean proportional between the diameter and the segment adjacent to that chord. REFERENCES TO PLANE GEOMETRY xiii 322. The area of a parallelogram is equal to the product of its base by its altitude. 323. Parallelograms having equal bases and equal altitudes are equivalent. 325. The area of a triangle is equal to half the product of its base by its altitude. 326. Triangles having equal bases and equal altitudes are equivalent. 327. Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are to each othei their bases; any two triangles are to each other as the prod- ucts of their bases by their altitudes. 329. The area of a trapezoid is equal to half the product of the sum of its bases by its altitude. 332. The areas of two triangles that have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. 334. The areas of two similar polygons are to each other as the squares on any two corresponding sides. 377. If the number of sides of a regular inscribed polygon is indefinitely increased, the apothem of the polygon approaches the radius of the circle as its limit. 381. The circle is the limit which the perimeters of regular inscribed polygons and of similar circumscribed polygons ap- proach, if the number of sides of the polygons is indefinitely increased. The area of the circle is the limit which the areas of the inscribed and circumscribed polygons approach. 382. Two circumferences have the same ratio as their radii 385. The circumference of a circle equal- 2 irr. 389. The area of a circle = wr*. SOLID GEOMETRY BOOK VI LINES AND PLANKS IN SPACE » • » 421. The Nature of Solid Gecmetry. la plane geometry we deal with figures lying in a §at suiiact , BtjiMj^irjg their proper- ties and relations and measuring div figu2< -. In solid geometry we shall deal with figures not only of two dimensions but of three dimensions, also studying their properties and relations and measuring the figures. 422. Plane. A surface such that a straight line joining any two of its points lies wholly in the surface is called a plane. A plane is understood to be indefinite in extent, but it is conveniently represented by a rectangle seen obliquely, as here shown. 423. Determining a Plane. A plane is said to be determined by certain lines or points if it contains the given lines or points, and no other plane can contain them. When we suppose a plane to be drawn to include given points or lines, we are said to pass the plane through these points or lines. When a straight line is drawn from an external point to a plane, its point of contact with the plane is called its foot. 424. Intersection of Planes. The line that contains all the points common to two planes is called their intersection. 273 274 BOOK VI. SOLID GEOMETRY 425. Postulate of Planes. Corresponding to the postulate that one straight line, and only one, can be drawn through two given points, the following postulate is assumed for planes : One plane, and only one, can be passed through two given intersecting straight lines. For it is apparent from the first figure that a plane may be made to turn about any single straight line AB, thus assuming different positions. But if CD intersects AB at P, as in the second figure, then when the plane through AB turns until it includes C, it must include D, since it includes two points, C and P, of the line (§ 422) . If it turns any more, it will no longer contain C. 426. Corollary 1. A straight line and a point not in the line determine a plane. For example, line AB and point C in the above figure. 427. Corollary 2. Three points not in a straight line deter- mine a plane. For by joining any one of them with the other two we have two inter- secting lines (§ 425). 428. Corollary 3. Two parallel lines determine a plane. M C T D A A B \ N For two parallel lines lie in a plane (§ 93), and a plane containing either parallel and a point P in the other is determined (§ 426). LINES AND PLANES 27") Proposition I. Theorem 429. If two planes cut each other, their niter section is a straight line. p N Given MN and PQ, two planes which cut each other. To prove that the plane* MN and PQ intersect in a straight line. Proof. Let .4 and B be two points common to the two planes. Draw a straight line through the points A and B. Then the straight line AB lies in both planes. § 422 (For it has two points in each plane.) No point not in the line AB can be in both planes ; for one plane, and only one, can contain a straight line and a point without the line. § 426 Therefore the straight line through A and B contains all the points common to the two planes, and is consequently the intersection of the planes, by § 424. q.e.d. Discussion. What is the corresponding statement in plane geometry ? 430. Perpendicular to a Plane. If a straight line drawn to a plane is perpendicular to every straight line that passes through its foot and lies in the plane, it is said to be perpendicular to the plane. When a line is perpendicular to euplane, the plane is also said to be perpendicular to the line. 276 BOOK VI. SOLID GEOMETRY Proposition II. Theorem 431. If a line is perpendicular to each of tivo other lines at their point of intersection, it is perpendicular to the plane of the two lines. Given the line AO perpendicular to the lines OP and OR at 0. To prove that AO is 1. to the plane MN of these lines. Proof. Through draw in MN any other line OQ, and draw PR cutting OP, OQ, Oil, at P, Q, and R. Produce AO to A', making OA' equal to OA, and join A and A ' to each of the points P, Q, and R. Then OP and OR are each _L to A A' at its mid-point. .'. AP = A'P, and AR = A'R. § 150 .'. A APR is congruent to A A' PR. § 80 ,'.ZRPA=Z.A'PR. §67 That is, Z QPA =ZA 'PQ. .'. APQA is congruent to APQA'. §68 .-. AQ = A'Q(§ 67); and OQ is _L to AA' at 0. § 151 .*. AO is _L to any and hence to every line in MN through O. .'. AO is -L to the plane MN, by § 430. Q.e.d. LINKS AND PLANES 277 Proposition III. Theorem 432. All the perpendiculars that can be drawn to a given line at << given point Ha in a plane which is per pendictdar to the given Urn at the given point R 1° N Given the plane MAT perpendicular to the line OY at O. To prove that OP, any line ±to OY at 0, lies in MN. Proof. Let the plane containing OY and OP intersect the plane MN in the line OP'; then OY is ± to OP'. § 430 In the plane POY only one _L can be drawn to OY at 0. § 57 Therefore OP and OP' coincide, and OP lies in MN. Hence every _L to OY at 0, as OQ, OR, lies in MN. Q.e.d. 433. I orollary 1. Through a given point in a given line one plane, and only one, can be passed perpendicular to the Urn , 434. Corollary 2. Through agiven externalpoint oneplaru . and only one, eon be passed perpendicular to a given line. Given the line OY an. 1 the point P. Draw PO ± to OY. and OQ ± to OY. Then OQ and OP determine a plane through H 1 PI to OY. / Q L "-->-~P \ Only one such plane can be drawn: for N only one _L can be drawn to OY from the point P (5 v 435. Oblique Line. A line that meets a plane but is nut per- pendicular to it is said to be oblique to the plane. 278 BOOK VI. SOLID GEOMETRY Proposition IV. Theorem 436. Through a given point in a plane there can he drawn one line perpendicular to the plane, and only one. Given the point P in the plane MN. To prove that there can be drawn one line perpendicular to tlie plane MN at P, and only one. Proof. Through the point P draw in the plane MN any line AB, and pass through P a plane IF 1 to AB, cutting the plane MN in CD. § 433 At P erect in the plane XY the line PQ J_ to CD. The line AB, being _L to the plane XY by construction, is _L to PQ, which passes through its foot in the plane. § 430 That is, PQ is _L to AB ; and as it is _L to CD by construc- tion, it is _L to the plane MN. § 431 Moreover, any other line PR drawn from P is oblique to MN. For PQ and PR intersecting in P determine a plane. To avoid drawing another plane, use XY again to represent the plane of PQ and PR, letting it cut MN in the line CD. Then since PQ is _L to MN, it is _L to Cfi. § 430 Therefore PR is oblique to CD. § 57 Therefore PR is oblique to MN. § 435 Therefore PQ is the only _L to MN at the point P. q.e.d. Discussion. What is the corresponding proposition in plane geometry? LINES AXD PLANES 279 Proposition V. Theorem 437. Through a given external point there can he draion one line perpendicular to a given plane, and only one. Given the plane MN and the external point P. To prove that there can be drawn one line from P perpen- dicular to the plane MX, and only one. Proof. In MN draw any line Eft, and let A'}' be a plan*' through P _L to Elf, cutting MN in AB, and EH in C. Draw PO _L to AB, and in MN draw any line 01) from to Elf. Produce PO, making OP' ~0P, and draw PC, PD } P'C, P'D. Since DC is _L to X Y, A PCD and P'CD are right angles. § 430 Since the side DC is common, and PC =P'C, § 150 .-. rt. A PCD is congruent to rt. A P'CD. .'. PD = P'D. .-. Of) is_Lto PP'at 0. .'.PO is ± to J/.V, being _L to OD and J /;. Moreover, every other line PF from P to .l/.V is oblique to MN. (The proof is left for tin- student.) ..PO is the only _L from f> to MN. q.e. d. 438. Corollarv. 77".' perpendicular is the shortest line from a point to a plan*'. The- length of this J. is called the distance from the point to the plant-. §69 §67 §151 § i.;i 280 BOOK VI. SOLID GEOMETRY Proposition VI. Theorem 439. Oblique lines drawn from a point to a plane, meeting the plane at equal distances from the foot of the perpendicular ', are equal ; and of two oblique lines, meeting the prt ane at unequal distances from the foot of the perpendicular, the more remote is the greater. N Given the plane MN, the perpendicular line PO, the oblique lines PA, PB, PC, the equal distances OB, OC, and the unequal dis- tances OA, OC, with OA greater than OC. To prove that PB = PC, and PA > PC. Proof. In the A OBP and OCP, OP = OP, Iden. OB = OC, Given and Z.BOP = Z.POC. §56 .'.A OBP is congruent to A OCP. § 69 .'.PB = PC. §67 Let A, B, and O lie in the same straight line. Then OA>OC. Given .-. OA>OB. Ax. 9 .-.PA>PB. §84 .'. PA>PC, by Ax. 9. q.e.d. Discussion. Compare the corresponding case in plane geometry. LINKS AND PLANES 281 440. Corollary 1. Equal oblique lines drawn from >> point to a plane meet the plane at equal distances from the foot of the perpendicular ; and of two unequal oblique lines the greatt t meets the plane at the greater distance from the foot of the perpi ndicular. In the figure on page 280, if PB is given equal to PC, then since PO = PO, and the angles at are righl angles, what follows with re- spect to the & OBP and OCP? with respect to OB and OC ? Furthermore, if PA > PC, how does PA compare with PB? Then how does OA compare with OB '.' Why? Then how does OA compare with OC ? 441. Corollary 2. The locus of a point equidistant from all points on a circle is a line through the center, perpen- dicular to the plane of the circle. In the figure on page 280, in order to prove that PO is the required locus what must be proved for any point on PO (§ 148) ? for any point not on PO ? Prove both of these facts. 442. Corollary 3. The locus of a point equidistant fmm the vertices of a triangle is a line through the center of the circumscribed circle, perpendicular to the plane of the triangle* How does this follow from Corollary 2 ? "What locus is the line through the center of the inscribed circle, per- pendicular to the plane of the triangle ? 443. Corollary 4. The locus of a point equidistant from two given points is the plane perpendicular to the line joining them, at its mid-point. For any point C in this plane lies in a JL to AB at 0, its mid-point (§ 430). Hence how do CA and CB compare (§ 150) ? And any point D outside the plane MN cannot lie in a ± to AB at O. What may therefore be sai.l as to the distan from Dto A and B (§ 160)? What is the proposition in plane geotnetry corresponding to Corol- lary 4? In what respect do the two proofs differ ? 282 BOOK VI. SOLID GEOMETKY Proposition VII. Theorem 444. Two lines perpendicular to the same plane are parallel. \ — «r- -^rF -c^ c \D N Given the lines AB and CD, perpendicular to the plane MN. To prove that AB and CD are parallel. Proof. Draw AD and BD, and in MN draw through D EF _L to BD, making DE = DF. Draw BE, AE, BF, AF. Now prove that A BDE and BDF are congruent (§ 69), that AADE and ADF are right angles (§ 80), and that BD, CD, and AD lie in the same plane (§ 432). But AB also lies in this plane, § 422 and AB and CD are both _L to BD. § 430 .*. AB is II to CD, by § 95. Q.e.d. 445. Corollary 1. If one of two parallel lines is perpen- dicular to a plane, the other is also perpendicular to the plane. m For if through any point of CD a line is drawn ± to / MN, how is it related to AB (§ 444) ? Now apply § 94. 446. Corollary 2. If two lines are parallel o B b N A C M , Q±Z1 N to a third line, they are parallel to each other. For a plane MN ± to CD is _L to AB and EF (§ 445) 447. Line and Plane Parallel. If a line and plane cannot meet, however far produced, they are said to be parallel. LINES AND PLANKS 283 EXERCISE 74 1. Why does folding a sheet of paper give a straight edge ? 2. If equal oblique lines are drawn from a given external point to a plane, they make equal angles with lines drawn from the points where the oblique lines meet the plane to tin- toot of the perpendicular from the given point. 3. If from the foot of a perpendicular to a plane a line is drawn at right angles to any line in the plane, the line drawn from its intersection with the line in the plane to any point of the perpendicular is perpendicular to the line of the plane. 4. If two perpendiculars are drawn from a point to a plane and to a line in that plane respectively, the line joining the feet of the perpendiculars is perpendicular to the given line. 5. From two vertices of a triangle perpendiculars are let fall on the opposite sides. From the intersection of these perpen- diculars a perpendicular is drawn to the plane of the triangle. Prove that a line drawn to any vertex of the triangle, from any point on this perpendicular, is perpendicular to the line drawn through that vertex parallel to the opposite side. 6. Find the point in a plane to which lines may be drawn from two given external points on the same side of the plane so that their sum shall be the least possible. From cue point A suppose a _L AO drawn to the plane and produced to .4'. making OA' — OA. Connect A' and the other point B by a line cutting the plane at P. Then BPA is the shortest line. 7. If three equal oblique lines are drawn from an externa] point to a plane, the perpendicular from the point to the plane meets the plane at the center of the circle circumscribed about the triangle having for its vertices the feet of the oblique lines. 8. State and prove the propositions of plane geometry cor- responding to §§ 444, 44o. and 446. Why do not the proofs of those propositions apply to these sections ? 284 BOOK VI. SOLID GEOMETRY Proposition VIII. Theorem 448. If two lines are parallel, every plane containing one of the lines, and only one, is parallel to the other line. Given the parallel lines AB and CD, and the plane MN contain- ing CD but not AB. To prove that the plane MN is parallel to AB. Proof. AB and CD are in the same plane, AD. § 93 This plane AD intersects the plane MN in CD. Given Now AB lies in the plane AD, however far produced. § 422 Therefore, if AB meets the plane MN at all, the point of meeting must be in the line CD. § 422 But since AB is II to CD, Given .'. AB cannot meet CD. § 93 .*. AB cannot meet the plane MN. .'. MN is II to AB, by § 447. q.e.d. 449. Corollary 1. Through either of two lines not in the same plane one plane, and only one, can he passed parallel to the other. A B For if AB and CD are the lines, and we pass a plane through CD and a line CE which is drawn ^f- parallel to AB, what can be said of the plane MN determined by CD and CE, with respect to the line \q. AB? Why can there be only one such plane ? 2£ LINKS AND PLANES 285 450. Corollary 2. Through a given point one plane, and only one, can be passed parallel to any tir<> given lines in space. Suppose P the given point and AB and CD the given lines. If, now, we draw through 1' the line A A'B' parallel to AB, and the line CD' parallel to CD, (f/ p these lines will determine the plane MX (§ 425), Then what may be said of the plane MX with re- / i P/ B speet to the lines AB and CD? Why can only one / C /l' l) ' plane be so passed through P ? N Discussion. Proposition VIII might of course be made more general by allowing both of the parallels to lie in the plane MX. That is, If two lines are parallel, a plane containing one of the lines cannot intersect the other, although the other line might lie in it. In the figure of Corollary 2 the Z D'PB' is sometimes spoken of as the angle between the nonintersecting lines AB and CD, although this is not commonly done in elementary geometry. 451. Parallel Planes. Two planes which cannot meet, how- ever far produced, are said to be parallel. EXERCISE 75 1. What is the locus of a point in a plane equidistant from two parallel lines? What is the corresponding locus in space, given two parallel planes instead of two parallel lines ? Draw the figure, without proof. 2. Find the locus in a plane of a point at a given distance from a given external point. What is the corresponding case of plane geometry ? 3. If a given line is parallel to a given plane, the intersection of the plane with aiiy plane passed through the given line i> parallel to that line. 4. If a given line is parallel to a given plane, a line parallel to the given line drawn through any point of the plane lies in the plane. 280 BOOK VI. SOLID GEOMETRY Proposition IX. Theorem 452. Two planes perpendicular to the same line are parallel. Given the planes MN and PQ perpendicular to the line AB. To prove that the planes MN and PQ are parallel. Proof. If MN and PQ are not parallel, they must meet. If they could meet, we should have two planes from a point of their intersection _L to the same straight line. But this is impossible. § 434 .*. MN and PQ are parallel, by § 451. q.e.d. EXERCISE 76 1. What is the locus of a point equidistant from two given points A, B, and also equidistant from two other given points C, D? 2. What is the locus of a point at the distance d from a given plane P, and at the distance aV from a given plane P' ? 3. What is the locus of a point at the distance d from a given plane P, and equidistant from two given points A, B? 4. Find a point at the distance d from a given plane P, at the distance d' from a given plane P\ and equidistant from two given points A, B. Can there be more than one such point ? Draw the figure, without proof. LINKS AM) PLANES *2ST Proposition X. Theorem 453. The intersections of 'two parallel planes by a third plane are parallel lines. Q s Given the parallel planes MN and PQ, cut by the plane RS in AB and CD respectively. To prove that the intersections AB and CD an- parallel. Proof. AB and CD are in the same plane RS. Given If .17? and CD meet, the planes .V.Vand PQ must meet, since AJJ is always in MN and CD is always in PQ. § 122 But MN and PQ cannot meet. § 451 .-.AB is II to CD, by § 93. Q.e.d. 454. Corollary 1. Parallel lines included between par- allel planes are equal. In the above figure, suppose AC II to BD. Then the plane of A C and 111) will intersect MN and PQ in lines that are how related to each other? Then what kind of a figure is ACDB? 455. Corollary 2. Two parallel planes are everywhere equidistant from each other. Drop perpendiculars from anypoints in MN to PQ. Prove that these perpendiculars are parallel and hence (§ 464) thai they are equal, 288 BOOK VI. SOLID GEOMETRY Proposition XI. Theorem 456. A line perpendicular to one of two parallel planes is perpendicular to the other also. 15 Given the line AB perpendicular to the plane MN, and the plane PQ parallel to the plane MN. To prove that AB is perpendicular to the plane PQ. Proof. Pass through AB two planes AE, AF, intersecting MN in AC, AD, and intersecting PQ in BE, BF, respectively. Then AC is II to BE, and AD is II to BF. § 453 But AB is ± to AC and AD. § 430 .'. AB is _L to BE and BF. § 97 .*. AB is _L to the plane PQ, by § 431. q.e.d. 457. Corollary 1. Through a given point one plane, and only one, can be passed jmrallel to a given plane. How is a plane through A, _L to AB, related to PQ ? Now use § 433. 458. Corollary 2. The locus of a point equidistant from two parallel planes is a plane perpendicular to a line which is perpendicular to the planes and which bisects the segment cut off by them. 459. Corollary 3. The locus of a point equidistant from two parallel lines is a plane perpendicular to a line which is perpendicular to the given lines and which bisects the segment cut off by them. LINKS AND PLANES 289 Proposition NIL Theorem 460. If two intersecting lines are <<"■// jtantllrl t<> ut A C and BE lie in the same plane, Const. and .1 C cannot meet BE without meeting the plane PQ, which is impossible. Similarly §447 .'. BE is li to AC. §93 BF is II to AD. AB is _L to AC and to AD. § ( .»7 AB is JL to the plane MN. §431 MN is 11 to PQ. by § 152. Q.E. D. Discussion. It is evident that this proposition does not depend upon the position of A. For example, C and /> might remain where they are and A might recede a long distance, AC and J.D becoming more nearly- parallel. So long as the lines intersect, and only so long, are we certain that the planes are parallel. 290 BOOK VI. SOLID GEOMETRY Proposition XIII. Theorem 461. If two angles not in the same plane have their sides respectively parallel and lying on the same side of the straight line joining their vertices, the angles are equal, and their planes are parallel. Given the angles A and A\ in the planes MN and PQ respec- tively, and their corresponding sides parallel and lying on the same side of AA' . To prove that ZA = ZA', and that MN is II to PQ. Proof. Take AD and A'D' equal, also AC and A'C equal. Draw DD', CC, CD, CD'. Since AD is equal and II to A'D', .'. AA' is equal and II to DD'. § 130 In like manner A A' is equal and II to CC. .'. DD' and CC' are equal, Ax. 8 and DD' and CC' are parallel. § 446 .'.CD=C'D'. §130 .'.A ADC is congruent to AA'D'C. § 80 .-. AA=Z.A'. §67 But MN is II to each of the lines A'C and A'D'. § 448 .'. MN is II to PQ, by § 460. Q.e.d. Discussion. Why does not the proof of the corresponding proposition in plane geometry apply here ? LINES AND PLANES 291 Proposition XIV. The<>i;i:m 462. If two lines are cut by three parallel planes > their correspond ing segments are proportional. Given the lines AB and CD, cut by the parallel planes MN, PQ, RS, in the points A, E, B, and C, F, D, respectively. To prove that AE : EB = CF : FD. Proof. . Draw AD cutting the plane PQ in G. Pass a plane through AB and AD, intersecting PQ in the line EG, and intersecting RS in the line BD. Also pass a plane through AD and CD, intersecting PQ in the line GF, and intersecting MN in the line A C. Thru EG is II to BD, and <7Fis II to AC. §453 .'. .4/;: EB = AG: GD, and CF: FD = A G : GD. § 27.°. .-. .1 /:: F/?= CF: 77>. by Ax. 8. q.e.d. Discussion. This is a generalization of §275. It may be stated still more generally, If tvjo lines are cut by any number of parallel planes, (I. * "i-responding segments are proportional. In particular, tin- cum- might be considered in which -1 B ami lr , approaches as a limit, Z.ABD ll AABD and the variable . " ' approaches ^ \_i> ( <_d aS a limit " „ Z.lTTE . ' A A'-B'C'-E , .,_,__ But Z ^^ D is always equal to -^ — — > as Z ,1 S£ varies in value and approaches A A'B'D 1 as a limit. Case 1 AA'-B'C'-D' A A'B'D' nv ^ ■> by § 20 i. q.e.d. Z A-BC-D AABD 473. Corollary. 77"' p&ww "//.\ Will. Theorem 477. If a Vim is perpendicular to a plane, even/ plane passed through this line is perpendicular to tJu plane. Given the line CD perpendicular to the plane MN at the point Z), and PQ any plane passed through CD intersecting MN in AB. To prove that the plane PQ is perpendicular to the plane MN. Proof. Draw l)E in the plane JAVJLto AB. CD is J_ to MX, .'. CD is _L to AB. . .'.Z.EDC measures Z X-AB-P. But Z EDC is "a right angle. .-. PQ is J. to MN, by §467. Since Given §430 § 173 §430 Q.E.D. EXERCISE 78 1. A plane perpendicular to the edge of a dihedral angle is perpendicular to each of its faces. 2. If one line is perpendicular to another, is any plane passed through the first line perpendicular, to the second '.' Prove it. 3. If three lines are perpendicular to one another at a com- mon point, what is the relation. to out* another of the tin ■ planes determined by the three pairs of lines? Prove it. 300 BOOK VI. SOLID GEOMETRY Proposition XIX. Theorem 478. If tivo intersecting planes are each perpendicular to a third plane, their intersection is also perpendicular to that plane. Given two planes BC and BD, intersecting in AB y and each per- pendicular to the plane PQ. To prove that AB is perpendicular to the plane PQ. Proof. Let the plane BC intersect the plane PQ in BF, and let the plane BD intersect the plane PQ in BE. From any point A on AB draw AX _L to BE, and from A draw A Y _L to BF. Then AX and A Y are both J_ to the plane PQ. § 474 But it is impossible to draw two Js to the plane PQ from a point outside the plane PQ, § 437 or from a point in the plane PQ. § 436 .*. AX and AY must coincide. But AX and A Y can coincide only if they lie in both planes. And all points common to both planes lie in AB. § 429 .*. AX and A Y coincide with AB. .'. AB is _L to the plane PQ. q.e.d. Discussion. How does it appear from this proof that AB cannot be parallel to PQ ? The proposition is illustrated in the intersection of two walls of a room with the floor or the ceiling. - - DIJIKDKAL ANCLKS 301 Proposition XX. Theorem 479. The locus of a point equidistant from tJu faces <■/" a il'ilii'ili'itl <P bisects Z /'"/:. § 173 .*. OP is the locus of a point equidistant from "/'and OE. § L62 .-. 4ilf, wln.-h contains all point- /•. is the locus of a point equidistanl from the planes AD and AC. q.e.d. 302 BOOK VI. SOLID GEOMETRY Proposition XXI. Theorem 480. Through a given line not perpendicular to a given plane, one plane and only one can he passed perpen- dicular to the plane. Given the line AB not perpendicular to the plane MN. To prove that one plane can be passed through AB perpen- dicular to the plane MN, and only one. Proof. From any point X of AB draw ATI to the plane MN, and through AB and XY pass a plane A P. § 425 The plane AP is _L to the plane MN, since it passes through AT, a line _L to MN. § 477 Moreover, if two planes could be passed through AB _L to the plane MN, their intersection AB would be JL to MN. § 478 But this is impossible, since AB is not _L to MN Given Hence one plane can be passed through AB ± to the plane MN, and only one. q.e.d. 481. Projection of a Point. The foot of the line from a given point perpendicular to a plane is called the projection of the point on the plane. 482. Projection of a Line. The locus of the projections of the points of a line on a plane is called the projection of the line on the plane. DIHEDRAL ANGLES :\\):\ Proposition XXII. Theorem 483. Hie projection of a straight line riot perpendicu- lar to a plane, upon that plane, is a straight Urn . § 125 § 177 Given the straight line AB not perpendicular to the plane MN, and A'B' the projection of AB upon MN.- To prove that A'B' is a straight line. Proof. From any point X of AB draw A'}' _L to MX, and pass a plane AP through XY and AB. The plane .IP is _L to the plane MX, and contains all the Js drawn from AB to MX. § 176 Hence A'B' must be the intersection of these two planes. Therefore A'B' is a straight line, by § 429. q.e.d. 484. Corollary. The projection of a straight line perpen- dicular to a plane, upon that plane, is a point. 485. Inclination of a Line. The angle which a line makes with its projection on a plane is considered as the angle which it makes with the plane, and is called the inclination of tin- line to the plane. Therefore a line ordinarily makes an acute angle with a plane, since it makes an acute angle with its projection on the plane. The case- perpendicular and parallel lines have already been considered. 304 BOOK VI. SOLID GEOMETRY Proposition XXIII. Theorem 486. The acute angle which a line makes with its projection upon a plane is the least angle which it makes with any line of the plane. N Given the line AB meeting the plane MN at A , AB' being the pro- jection of AB upon the plane MN, and AD being any other line drawn through A in the plane MN. To prove that Z B'AB is less than Z DAB. Proof. Make AD equal to AB', and draw BB' and BD. Then in A BAB' and BAD, AB = AB, Iden. AB' = AD, Const. and BB' 6 in., a circle is described in the plant-: at any point Con this circle a tangent CD is drawn 24 in. in length. Find the distance from .1 to I>. 4. Equal lines drawn from a given external point to a given plane are equally inclined to the plane. 5. If three equal lines are drawn to a plane from an exter- nal point, the perpendicular from the point to the plant- deter- mines the center of the circle circumscribed about the triangle determined by the planes of the three lines. 6. Three lines not in the same plane meet in a point. How shall a line be drawn so as to make equal angles with all thi of these lines '.' 7. From a point /'two perpendiculars PX and PFare drawn to two planes MN and A C which intersect in All. From }' a perpendicular YZ is drawn to MX. Prove that the line XZ is perpendicular to AB. 8. If the length of the shadow of a tree standing on level ground exceeds the height ofthe tree, the angle made by the sun above the horizon must be less than what known angL 9. Find the locus of a point at a given distance from a given plane and equidistant from two given points not in the plane. 306 BOOK VI. SOLID GEOMETRY Proposition XXIV. Theorem 487. Between two lines not in the same plane there can he one common perpendicular, and only one. Given AB and CD, two lines not in the same plane. To prove that there can be one common perpendicular, and only one, between AB and CD. Proof. Through any point A of AB draw AG II to DC. Let MN be the plane determined by AB and AG. § 425 Then the plane MN is II to DC. § 448 Through DC pass the plane PQ _L to the plane MN. § 480 Then DC cannot meet D'C, since it is II to the plane MN and lies in the plane PQ. § 422 .'. DC is II to D'C. § 93 .-. if AB is II to D'C it must be II to DC. § 446 But AB is not II to DC, for they are not in the same plane. Given .*. AB must intersect D'C at some point as C. Draw C'C 1 to the plane MN. Then C'C is _L to AB and to D'C. § 430 Since C'C is _L to D'C, and lies in plane PQ, § 475 .-.C'C is _L to DC. §97 Therefore one common perpendicular can be drawn. It remains to be proved that no other can be drawn. DIHEDRAL ANGLES 307 If it were possible that another common perpendicular could be drawn, we might suppose /•;.! to be _L to both 42? and CD. Then EA would be _L to A G } § ( .»7 and therefore EA would be _L to the plane MX. § 431 BtslwEE'± to D'C. Then EE' is J_ to the plane MN. §474 But this is impossible, if EA is also J_ to the plane MN. § 437 Hence the supposition that there is a second common per- pendicular, EA, leads to an absurdity. Therefore there can be one common perpendicular, and only one, between AB and CD. q. e.d. 488. Corollary. The common perpendicular between two lines not in the same plane is the shortest line joining them. How does CC compare in length with EE' ? Why ? How does EE' compare in length with EA ? EXERCISE 80 1. Parallel lines have parallel projections on a plane. 2. If two planes are perpendicular to each other, any linp perpendicular to one of them is how related to the other '.' 3. If three lines passing through a given point P are cut by a fourth line that does not pass through P, the four lines all lie in the same plane. 4. Seven lines, no three of which lie in the same plane, pass through the same point. How many planes are deter- mined by these lines ? 5. A cubical tank 10 in. deep contains water to a depth of 7 in. A foot rule is placed obliquely <>n the bottom so as just to reach the top edge of the tank. Make a sketch of the tank, and compute the length of the rule covered by water. 308 BOOK VI. SOLID GEOMETRY 489. Polyhedral Angle. The opening of three or more planes which meet at a common point is called a polyhedral angle. The common point V is called the vertex of the angle ; V the intersections VA, VB, etc., of the planes are called yf\ the edges; the portions of the planes lying between the // \ edges are called the faces; and the angles formed by "4f~y£. \ adjacent edges are called the face angles. f- — — -4(7 Every two adjacent edges form a face angle, and every two adjacent faces form a dihedral angle. The face angles and dihedral angles are the parts of the polyhedral angle. 490. Size of a Polyhedral Angle. The size of a polyhedral angle depends upon the relative position of its faces, and not upon their extent. 491. Convex and Concave Polyhedral Angles. A polyhedral angle is said to be convex or concave according as a section made by a plane that cuts all its edges at other points than the vertex is a convex or concave polygon. Only convex polyhedral angles are considered in this work. 492. Classes of Polyhedral Angles. A polyhedral angle is called a trihedral angle if it has three faces, a tetrahedral angle if it has four faces, and so on. Other names, like pentahedral, hexahedral, heptahedral, etc., for angles with 5, 6, 7, etc., faces, are rarely vised. A polyhedral angle is designated by a letter at the vertex, or by let- ters representing the vertex and all the faces taken in order. Thus, in the above figure the trihedral angle is designated by V or by V-ABC. A tetrahedral angle would be designated by V or by V-ABCD. 493. Equal Polyhedral Angles. If the corresponding parts of two poly- hedral angles are equal and are ar- ranged in the same order, the poly- ' l£ ^Jl- ' Y---~--\n' hedral angles are said to be equal. Thus the angles V-ABC and V'-A'B'C are equal. Equal polyhedral angles may evidently be made to coincide by superposition. POLYHEDRAL ANGLES 309 Proposition XXV. THEOREM 494. The sum of any two face angles of a trihedral angle is greater than the third fain angle* v 'r Given the trihedral angle V-XYZ, with the face angle XVZ greater than either of the face angles XVY or YVZ. To prove that Z XVY+ Z YVZ is greater them Z XVZ. Proof. In the Z X VZ draw VW, making Z XV W = Z XVY. Through any point D of VW draw ADC in the plane XVZ. On VY take VB equal to VD. Pass a plane through the line AC and the point /;. Then since A V = A V, VD = VB, and Z A VD = Z A VB, .'. A A VD is congruent to A A VB. § 68 .\AD = AB. §67 In the A. 4 £C, AB + BOAC. §112 Since \B=AD, .\BC>DC. Ax. 6 In the A £T'C and DFC, PC= TC, and F£ = TO, but BC > DC. .-. ZBVCis greater than Z.DVC. §11< ! .'. Z^ VB + Z.BVC is greater than Z.I FD + ZD7C. Ax. 6 But Z.AVD + /.DVC = Z.AVC. Ax. 11 .'. Z A VB + Z.BVC is greater than Z.I I \ '.» That is, Z XVY + Z FKZ is greater than Z \T/ q.e.d. 310 BOOK VI. SOLID GEOMETRY Proposition XXVI. Theorem 495. The sum of the face angles of any convex poly- hedral angle is less than four right angles. Given a convex polyhedral angle V, all of its edges being cut by a plane making the section ABCDE. To prove that AAVB + ABVC, etc., is less than four rt. A. Proof. From any point P within the polygon draw PA, PB, PC, PD, PE. The number of the A having the common vertex P is the same as the number, having the common vertex V. Therefore the sum of the A of all the A having the common vertex V is equal to the sum of the A of all the A having the common vertex P. But in the trihedral A formed at A, B, C, etc., Z EA V + Z.BAV is greater than Z BAE, Z.VBA + Z CBV is greater than Z CBA , etc. § 494 Hence the sum of the A at the bases of the A whose com- mon vertex is V is greater than the sum of the A at the bases of the A whose common vertex is P. Ax. 7 Therefore the sum of the A at the vertex V is less than the sum of the A at the vertex P. Ax. 7 But the sum of the A at P is equal to 4 rt. A. § 41 Therefore the sum of the A at V is less than 4 rt. A. q.e.d. POLYHEDRAL ANGLES 311 496. Symmetric Polyhedral Angles. It" the faces of a poly- hedral angle V-ABCD arc produced through the vertex l\ another polyhedral angle V-A'B'C'D* is formed, symmetric with respect to /- V-A Bt I >. The face angles .I VB, BVC, etc., are equal respectively to the face angles A'VB', /M'< . etc. (§ 00). Also the dihedral angles VA, VB, etc., are equal respectively to the dihedral angles VA', VB', etc. (§470). (The second figure shows a pair of these vertical dihedral angles.) Looked at from the point V. the edges of Z. V-A1UD are arranged from left to right (counterclockwise) in the order VA, VB, \'<\ PD,bat the edges of AV-A'B'C'T)' are arranged from right to left (clockwi in the order VA', VB', VC, VD'; that is. in an order the reverse of the order of the edges in Z V-ABCD. Therefore, Two symmetric polyhedral angles have all their parts equal each to each hut arranged in reverse order. 497. Symmetric Polyhedral Angles not Superposable. In gen- eral, two symmetric polyhedral angles are not superposable. Thus, if the trihedral angle V-A'B'C is made to turn 180° about AT, the bisector of the angle CVA', then VA 1 will coincide with VC, VC" with VA, and the face A'VC" with AVC ; but the di- hedral angle VA, and hence the dihedral angle VA', not being equal to VC, the plane A'VB' will not coincide with BVC] and. for a similar reason, the plane C'VB' will not coincide with -1 VB. Hence the ed \'B' takes some position VB" not coincident with VB\ that is, the trihedral angles are not superposable. An analogous case is seen in a pair of gloves. All the pai one are equal to the corresponding parts of the other, but the right-hand glove will not fit the left hand. 312 BOOK VI SOLID GEOMETRY Proposition XXVII. Theorem 498. Tioo trihedral angles are equal or symmetric when the three face angles of the one are equal respec- tively to the three face angles of the other. B' E' A' B E A A' E'B' Given the trihedral angles V and V\ the angles BVA } CVA, CVB being equal respectively to the angles B'V'A', C'V'A', CVB'. To prove that the angles V and V are equal or symmetric. Proof. On the edges of these angles take the six equal seg- ments VA, VB, VC, V'A', V'B', V'C. Draw AB, BC, CA, A'B', B'C, C'A'. The isosceles ABA V, CA V, CBV are congruent respectively to the isosceles AB'A'V, C'A'V, C'B'V. § 68 .'. AB, BC, CA are equal respectively to A'B', B'C, C'A'. § 67 .'.ABAC is congruent to AB'A'C. § 80 From any point D in VA draw DE in the face A VB and DF in the face AVC, each _L to VA. These lines meet AB and AC respectively. (For the AVAB and VAC are acute, each being one of the equal A of an isosceles A.) Draw EF. On A'V take A'D' equal to AD. POLYHEDRAL ANGLES 313 Draw D'E' in the face . 1 ' V'B' and D'F 1 in the face .1 ' \''C, each _L to I'M', and draw E'F\ Then since AJ> = A'l>', Const. and Z />.!/•; = Z I>'.\ 'E', §67 .'. rt. A .4/)£ is congruent to rt, A I 'D'E'. § 71' .-. .1 B = A'E', and 7>/^ = ///•;'. § 67 In like manner A E=A'E', and DF= D'j". Furthermore, since it has been proved that ABAC is congruent to AB'A'C, .'. £CAB = Z.C'A'B'. § 67 .'. AA EE is congruent to A A'F'E'. § 68 .\EF=E'E'. §67 .'.A EDE is congruent to AE'D'F'. § 80 .-. ZEDE = ZE'D'E'. § 67 .". dihedral Z F.4 = dihedral Z I'M '. § 47:; (For A FDE and FITE', the measures of these dihedral A, are equal.) In like manner it may be proved that the dihedral angles VB and VC are equal respectively to the dihedral angles V'B 1 and V'C. .*. the trihedral angles V and V' are equal, § 493 or else they are symmetric, by § 496. o.p.D. This demonstration applies to either of the two figures denoted by Y'-A'B'C which are symmetric -with respect to each other. If the first of these figures is taken, V and V are equal. If the second is taken, V and V are symmetric. 499. Corollary. If two trihedral angles }>-,' the tin-,, /> three face angles of the other, thin the dihedral angles of the one are equal respectively to th>- dihedral angles of the other. For whether the trihedral angles are equal or symmetric, - • ir»- as to be constantly 5 in. from each of these points. Find the length of the locus of P. Answer to three decimal plac< 5. Two parallel planes MN and PQ are cut by a third plane RS so as to make one of the dihedral angles 27° 15' 30". Find the other dihedral angles. 6. Two lines are cat by three parallel planes. The segments cut from one line are 3 in. and 5^ in., and these cut from the other line are 7^ in. and .'•. Find the value of .>■. 7. Two given planes I angles to each other. A point A' is 8 in. from each plan . How far is .V from the e.i of the right dihedral igle ' 8. What is the lengtl I the projection on a plane of a line whose length is 10 ~v ~. the inclination of tne line to the plane being I.V ? 9. From the external point P a perpendicular /'/''. 9 in. long. is drawn to a plane M V. From P the line PQ is drawn to the plane making the angle P'PQ equal to 30°. Find the length of the projection of PQ on the plane M V. 316 BOOK VI. SOLID GEOMETRY EXERCISE 83 Review Questions 1. How many and what conditions determine a straight line ? How many and what conditions determine a plane ? 2. What simple numerical test, following the measurement of certain lengths, determines whether or not one line is perpen- dicular to another ? a line is perpendicular to a plane ? 3. How many planes can be passed through a given line perpendicular to a given plane ? Is this true for all positions of the given line ? 4. Through a given point how many lines can be drawn parallel to a given line ? parallel to a given plane ? Through a given point how many planes can be passed parallel to a given line ? parallel to a given plane ? 5. What is the locus, in a line, of a point equidistant from two given points ? in a plane ? in a space of three dimensions ? 6. What is the locus, in a plane, of a point equidistant from two intersecting lines ? State a corresponding proposition for solid geometry. 7. What may be said of two lines in one plane perpendicular to the same line ? State two corresponding propositions for solid geometry. Does one of these propositions state that two planes perpendicular to the same plane are parallel ? 8. What may be said of a line perpendicular to one of two parallel lines ? State two corresponding propositions for solid geometry. Is a plane perpendicular to one of two parallel planes perpendicular to the other ? 9. If a line is perpendicular to a plane, what may be said of every plane passed through this line ? Does a true prop- osition result from changing the word "perpendicular" to " parallel" in this statement? BOOK VII POLYHEDRON *. CYLINDERS, WD CONES 500. Polyhedron. A solid bounded by planes Is called & poly- hedron. For example, the figures on pages :i 1 7 and 318 are polyhedrons. The bounding planes are called the faces of the polyhedron, the in- tersections of the faces are called the edges <»f tin- polyhedron, and the intersections of the edges are called the vertices of the polyhedron. A line joining any two vertices not in the same fair is called a diiHjonal of the polyhedron. The plural of polyhedron \& polyhedrons or polyhedra. 501. Section of a Polyhedron. It a plane passes through a polyhedron, the intersection of the plane with such laces as it cuts is called a section of the polyhedron. 502. Convex Polyhedron. If every section oi a polyhedron is a convex polygon, tin- polyhedron is said t<> be convex. Only convex polyhedrons are considered in this work. 503. Prism. A polyhedron of which two faces are congruent polygons in parallel planes, tin- other lac, being parallelograms, is called a prism. The parallel polygons are called the bases of the prism, the parallelograms arc called the lateral faces, and the intersections of the lateral fac< are called the lateral edges. The sum of the areas of the lateral faces is called the lateral area of the prism. The lateral edges of a prism are equal (£ 125). 504. Altitude of a Prism. The perpendicular distance tween the planes of the l»;i>es of a prism is called its altitude. ::i7 318 BOOK VII. SOLID GEOMETRY 505. Right Prism. A prism whose lateral edges are per- pendicular to its bases is called a right prism. The lateral edges of a right prism are equal to the altitude (§455). 506. Oblique Prism. A prism whose lat- eral edges are oblique to its bases is called an oblique prism. 507. Prisms classified as to Bases. Prisms are said to be triangular, quadrangular, and so on, according as their bases are triangles, quadrilaterals, and so on. Right Prism 508. Right Section. A section of a prism made by a plane cutting all the lateral edges / and perpendicular to them is called a right 'section. Oblique Triangular Prism In the case of oblique prisms it is sometimes necessary to produce some of the edges in order that the cutting plane may intersect them. Truncated Prism Eight Section of a Prism 509. Truncated Prism. The part of a prism included between the base and a section made by a plane oblique to the base is called a truncated prism. PRISMS 319 Proposition I. Theorem 510. Tin sections of a prism modi by parallel planes cutting all the lateral edges are congruent polygon Given the prism PR and the parallel sections AD, A'D' cutting all the lateral edges. To prove that AI) is congruent to A'D'. Proof. -AB is II to A'B', BC is II to B'( '. ( h is to C'D\ and so on for all the corresponding sidi § 153 .-. .1/; = A'B', BC = B'(", CD= CD', and so on for all the corresponding sid ? 1-7 and Z.CBA = ZC'B'A' i Z.DCB = Z D'C'B\ and so on for all the corresponding angL 161 .-. AD is congruent to A'D\ by § 1 l_ q.e. d. Discussion. I- the proof the same whether or not the two parallel planes are parallel to the basi If the sections are all parallel to the bases, are they also congruent to the bases ? Would the proposition be true if the prism were concave instead of convex ? Suppose the bases were squares, what would be known as to the form of the sectioi 511. Corollary. Every section of a prism made by a plane parallel to the bas* is congruent to th* ■•■ .' and all right sections of a prism an wngruent 320 BOOK VII. SOLID GEOMETRY Proposition II. Theorem 512. Hie .lateral area of a prism is equal to the product of a lateral edcje bij the perimeter of a rigid section. Given VWXYZ a right section of the prism AD', I the lateral area, e a lateral edge, and p the perimeter of the right section. To prove that I = ep. Proof. A A' = BB' = CC' = DD' = EE' = e. § 503 Furthermore, VW is _L to BB', WX to CC', XY to DD', YZ to EE', and ZV to A A'. § 508 .'. the area of O AB' = BB' X VW = e X VW y § 322 the area of OBC'=CC'x WX = e X WX, the area of O CD' = DD' x XY = e X XY, and so on. But I is equal to the sum of these parallelograms. § 503 .-. I == e (VW + WX + XY+YZ + ZV). But VW + WX + X Y+YZ + ZV =p. .-. 1= ep, by Ax. 9. 513. Corollary. The lateral area of a right prism is equal to the product of the altitude by the perimeter of the base. Tor how would p then compare with AB + BC + CD + DE + EA ? The truth of the corollary is easily seen hy imagining the right prism laid on one of its lateral faces, and the surface as it were unrolled. Ax. 1 Ax. 11 Q.E.D. PRISMS 323 EXERCISE 84 Find the lateral areas of the right prisms whose altitudes and i" rirm U rs of basi s an as follows : 1. a=18in.,jp =29 in. 4. a = 1 ft. 7 in., p = 2 ft. 9 in. 2. a = 22 in., p = .">7 in. 5. a = •"» ft. 8 in../' = 5 ft. 7 in. 3. a = 4.25 in../> = 6.75 in. 6. a =12 ft. '1 in../, == L'7 ft. 9 in. Find the lateral areas of the prisms whose lateral edges and perimeters of right sections are as follows: 7. e =17 in., j? = 27 in. 10. e=l ft. 3 in., p = 2 ft. 3 in. 8. i =23 in.,^? = 35 in. 11. e=2 ft. 7 in., p = 3 ft. ( .» in. 9. e = 2| in.,// = 4; in. 12. e= 6ft. 1\ in., p = 8ft. 9 J in. JV^/7 the lateral edges of the prisms whose lat ral areas and perimeters of right sections are as follows: 13. £=187 sq. in.,]) =11 in. 14. £ = 357 sq. in., p = 21 in. 15. 2=169 sq. in..y/=l ft. 1 in. 16. The lateral surface of an iron bar 5 ft. long is to be gilded. The right sectioo is a square whose area is 2.89 sq. in. How many square inches of gilding are required '.' 17. A right prism of glass is 2\ in. long. Its righl ion is an equilateral triangle whose altitude is 0.866 in. i .1 \ 3 in. Find the lateral surface. 18. Find the total area of a right prism whose base -pun.' with area 5.29 sq. in., and whose length is twice its thicknefi 19. What is the total area of a right prism whose altitude is 32 in., and whose base is a righl triangle with hypotenuse 106 in. and with one side 84.8 in.? 20. Every section of a prism mad.- by a plane parallel t.» the lateral edges is a parallelogram. 322 BOOK VII. SOLID GEOMETRY 514. Parallelepiped. A prism whose bases are parallelograms is called a parallelepiped. The word is also, with less authority, spelled parallelopiped.' 515. Right Parallelepiped. A parallelepiped whose edges are perpendicular to the bases is called a right parallelepiped. 516. Rectangular Parallelepiped. A right parallelepiped whose bases are rectangles is called a rectangular parallelepiped. By §§ 430 and 453 the f our lateral faces are also rectangles. Rectangular Parallelepiped Cube Oblique Parallelepiped 517. Cube. A parallelepiped whose six faces are all squares is called a cube. We might also say that a hexahedron whose six faces are all squares is a cube, because such a figure would necessarily be a parallelepiped. 518. Unit of Volume. In measuring volumes, a cube whose edges are all equal to the unit of length is taken as the unit of volume. Thus, if we are measuring the contents of a box of which the dimen- sions are given in feet, we take 1 cubic foot as the unit of volume. If the dimensions are given in inches, we take 1 cubic inch as the unit. 519. Volume. The number of units of volume contained by a solid is called its volume. 520. Equivalent Solids. If tAvo solids have equal volumes, they are said to be equivalent. 521. Congruent Solids. If two geometric solids are equal in all their parts, and their parts are similarly arranged, the solids are said to be congruent. PARALLELEPIPEDS 323 Proposition III. Theorem 522. Two prisms an congrw nt if the thn e faa s which include << trihedral angle of the one an respectively con- gruent to three faces which includi a trihedral angh oj the other, and are similarly placed. Given the prisms AI and AT, with the faces AD, AG, ^/re- spectively congruent to A'D', A'G', A'J', and similarly placed. To prove thai AI is congruent to .17'. Proof. The face ABAE, BAF, EAF are equal to the face . B'A'E', B'A'F', E'A'F' respectively. § 1 12 Therefore the trihedral angles .1 and .1' are equal. 198 Apply tin- trihedral angle .1 to its equal A'. Then the face AD coincides with A'D\ AG with A'G 1 , and A.i with .!'./'; and C falls al C", and D at D'. The lateral edges of the prisms are parallel. § I L6 Therefore CH falls along C'lV, and Dl along D'l'. § 94 Since the points /', G, and J coincide with /■''. G\ and J\ each to each, the planes of the upper bases coincide. § 4l'7 Hence // coincides with //', and / with /'. Hence the prisms coincide and are congruent, hy § 521. q. e. d. 523. Corollary 1. Two truncated prisms are congruent under the conditions giv< n in Proposition III. 524. Corollary 2. Two right prisms having congruent /"/.sfs and rijual altitudes an congruent 824 BOOK VII. SOLID GEOMETRY Proposition IV. Theorem 525. An oblique prism is equivalent to a right prism whose base is equal to a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. Given a right section FI of the oblique prism AD\ and FI' a right prism whose lateral edges are equal to the lateral edges of AD'. To prove that AD' is equivalent to FI 1 . Proof. If from the equal lateral edges of AD' and FI' we take the lateral edges of FD', which are common to both, the remainders AF and A'F', BG and B'G', etc., are equal. Ax. 2 The bases FI and F'l' are congruent. § 510 Place A I on A 'I' so that FI shall coincide with F'l'. Then FA, GB, etc., coincide with F'A', G'B', etc. § 436 Hence the faces GA and G'A ', IIB and H'B', coincide. But the faces FI and F'l' coincide. .'. the truncated prisms A I and A' I' are congruent. § 523 .'. AI + FD' = A'l' + FD'. Ax. 1 But . AI+FD' = AD', and A 'I' + FD' == FI 1 . Ax. 11 Therefore AD' is equivalent to FI', by Ax. 9. q.e.d. PARALLELEPIPEDS 325 Proposition V. Theorem 526. The opposite fares of a parallelepiped are con- gruent and paralUI. B Given a parallelepiped ABCD-A'B'C'D' . To prove that the opposite faces AB' and DC are con- gruent and parallel. Proof. AB is II to DC, § 118 and AB=DC. §125 Likewise J . I ' is II and equal to DD\ ,\ZBAA' = ZCDD'. § 461 .-. AB' is II to DC'. § 161 .-. AB' is congruent to DC', by § 132. q.e.d. EXERCISE 85 1. If in the above figure the three plane angles at A are 80°, 70°, 75°, what are all the other angles in the faces ? 2. Given a parallelepiped with the three plane angles at one of the vertices 85°, 75°, 60°, to find all the other angles in the faces. 3. Given a rectangular parallelepiped lettered as in the fig- ure above, and with AB= 4, BC = 3, and CC' = 3j, to find the length of the diagonal AC 1 . 4. The four diagonals of a rectangular parallelepiped are equal 5. Compute the lengths of the diagonals of a rectangular parallelepiped whose edges from any vertex are a, b, a. 326 BOOK VII. SOLID GEOMETRY Proposition VI. Theorem 527. Hie plane passed through two diagonally oppo- site edges of a parallelepiped divides the parallelepiped into tivo equivalent triangular prisms. Given the plane ACCA 1 passed through the opposite edges A A' and CC of the parallelepiped AC. To prove that the parallelepiped AC is divided into tivo equivalent triangular prisms ABC-B' and AC D-D 1 . Proof. Let WXYZ be a right section of the parallelepiped. The opposite faces AB' and DC' are parallel and equal. § 526 Similarly, the faces AD 1 and BC' are parallel and equal. .'. WX is II to ZY, and WZ to XY. § 453 Therefore WXYZ is a parallelogram. § 118 The plane ACC'A' cuts this parallelogram WXYZ in the diagonal WY. § 429 .'.A WXY is congruent to A YZW. § 126 Mow shall it be proved that prism ABC-B' is equivalent to a right prism with base WXY and altitude .4.4' ? How shall it be proved that prism CDA-D' is equivalent to a right prism with base YZW and altitude A A' ? How are these two right prisms known to be equivalent ? How does this prove the proposition ? Discussion. What is the corresponding proposition of plane geometry ? PARALLELEPIPEDS 327 EXERCISE 86 1. The lateral luces of a right prism are rectangles. 2. The diagonals of a parallelepiped bisect one another. 3. The three edges of the trihedral angle at one of the ver- tices <>f a rectangular parallelepiped are 5 in.. 6 in., and 7 in. respectively. Required the total area of the six faces of the parallelepiped. 4. The three lace angles at one vertex of a parallelepiped an- each 60°, and the three edges of the trihedral angle with that vertex are 3 in., 2 in., 1 in. respectively. Required the total area of the six faces. Answer to two decimal places. 5. In a rectangular parallelepiped the square on any diag- onal is equivalent to the sum of the squares on any three edges that meet at one of the vertices. 6. In a box 3 in. deep and 6 in. wide a wire 1 ft. long can be stretched to reach from one corner to the diagonally oppo- site corner. Required the length of the box. Answer to two decimal places. 7. The diagonal of the base of a rectangular parallelepiped is 31f in. and the height of the parallelepiped is 23.7 in. Required the length of the diagonal of the parallelepiped. 8. The total area of the six faces of a cube is 18 sq. in. Find the diagonal of the cube. 9. The diagonal of the face of a cube equals Vl4. Find the diagonal of the cube. 10. The diagonal of a rube equals 2.75 V§. Find the diagonal of a face of the cube. 11. .V water lank is 3 ft. long, 2 ft. 6 in. wide, and 1 ft. 9 in. deep. How many square feet of zinc will be required to line the four sides and the base, allowing 1? sq. ft. for overlapping and for turning t lie top edge '.' 328 BOOK VII. SOLID GEOMETRY Proposition VII. Theorem 528. Tivo rectangular parallelepipeds having con- gruent bases are to each other as their altitudes. Given two rectangular parallelepipeds P andP', with congruent bases and with altitudes AB and A'B'. To prove that P : P'= AB : A'B'. Case 1. When AB and A'B' are commensurable. Proof. Suppose a common measure of AB and A'B' to be contained m times in AB, and n times in A'B'. Then AB : A'B' = m : n: Apply this measure to AB and A'B', and through the several points of division pass planes perpendicular to these lines. These planes divide the parallelepiped P into m parallele- pipeds and the parallelepiped P' into n parallelepipeds, con- gruent each to each. § 524 .'. P :P' — m'.n. .-. P:P' = AB: A'B', by Ax. 8. Q.E.D. The proof for the incommensurable case is similar to that in other propositions of this nature. It may be omitted at the discretion of the teacher without destroying the sequence, if the incommensurable cases are not being considered by the class. PARALLELEPIPEDS Cask '2. When AB and A'B' are incommensurable, r :\'1V Proof. Divide AB into any number of equal parts, and apply one of these parts to J'/;' as a unit of measure as many tin, as A'B' will contain it. Sine.' AB and A'B' are incommensurable, a certain number of these parts will extend from A' to a point D, leaving a remainder hi'.' less than <>n.- of the parts. Through I> pass a plane X to. 17;', and lei Q denote the parallelepiped whose base is the same as that of J'', ami who altitude is .1 'D. Then Q:P=:A'D: AB. < lase t If the number of parts into which M> is divided is indefi nitely increased, the ratioQ:P approaches P':P as a limit, and the ratio A 'D: AB approaches A'B': M> as a limit § 204 The remainder of the proof of the incommensurable case is substantially as in the proof given on page297, and it is therefore left for the student. 529. Dimensions. The lengths of the three edges of a rec- tangular parallelepiped which meet at a common vertex are called its dimensions. 530. Corollary. Two rectangular parallelepipeds which have two dim % in common an to each other as their third dimensions. 330 BOOK VII. SOLID GEOMETRY Proposition VIII. Theorem 531. Two rectangular parallelepipeds having equal altitudes are to each other as their hases. Given two rectangular parallelepipeds, P and P f , and a, &, c, and a\ b\ c, their three dimensions respectively. P ab To prove that P'" a'V Proof. Let Q be a third rectangular parallelepiped whose dimensions are a', b, and c. Now Q has the two dimensions b and c in common with P, and the two dimensions a' and c in common with P\ Therefore and P . > Q.E.D. 534. Corollary 1. Th volurrn of a rectangular parallel piped is equal to th product of its thra dimensions. For in the abovi b'= r' = 1. thm /" = 1 x 1 • l- 1 (§ 518). But the volume of P(§ 519) ia P : /' . and P P abc : 1 (§ There- fore the volume of P is a&c. 535. Corollary 2. '/'//.■ /•„/// m« of a rectangular parallele- piped is equal to the product and altitude. r r the volume of P is oi . and aft equals the base and c the altitude. 332 BOOK VII. SOLID GEOMETRY Proposition X. Theorem 536. The volume of any parallelepiped is equal to the product of its base by its altitude. Given an oblique parallelepiped P of volume v, with no two of its faces perpendicular, with base b and with altitude a. To prove that v = ba. Proof. Produce the edge EF and the edges II to EF, and cut them perpendicularly by two parallel planes whose distance apart GI is equal to EF. We then have the oblique parallele- piped Q whose base c is a rectangle. Produce the edge IK and the edges II to IK, and cut them perpendicularly by two planes whose distance apart MN is equal to IK. We then have the rectangular parallelepiped R. Now P = Q, and Q = R. § 525 .\P = R. Ax. 8 The three parallelepipeds have a common altitude a. § 455 Also b = c, § 323 and c = d. § 133 .'.b = d. Ax. 8 But the volume of it = da. § 535 Putting P for R, and b for d, we have v = ba, by Ax. 9. Q. e. d. 537. Corollary. The volume of any parallelepiped is equal to that of a rectangular parallelepiped of equivalent base and equal altitude. PAEALLELEPIPEDS 333 EXERCISE 87 1. Find the rati<» of two rectangular parallelepipeds, If their dimensions are •">. I. 5, and 9, 8, 1<> respectively. 2. Find the ratio of two rectangular parallelepipeds, if their altitudes are each <*» in., and their bases 5 in. by 1 in., and 1<> in. by 8 in. respectively. 3. Find the volume of a rectangular parallelepiped 2 It. (') in. long, 1 ft. s in. wide, and 1 ft. 6 in. high. 4. Find the volume of a rectangular parallelepiped win base is 27 sq. in. and whose altitude is 13| in. 5. The volume of a rectangular parallelepiped is 1152 cu. in. and the area of the Last- is half a square foot. Find the altitude. 6. The volume of a rectangular parallelepiped with a square base is 'J7'-kS cu. in. and the altitude is 5 in. Find the dimen- sions. 7. A rectangular tank full of water i^ 7 ft. ."> in. long h\ 4 ft. G in. wide. How many cubic feel of water must he drawn <>\\' in order that the surface may he lowered a fool '.' 8. Find to two decimal places tin- length of each side of a cubic reservoir that will contain exactly a gallon i-'.'A cu. in. i. 9. .V box has as its internal dimensions IS in.. 9| in., and 4] in. The box and cover are made of steel | in. thick, [f steel weighs 490 11>. per cubic font, what is the weight of the box? 10. A steel rod 4 ft. 8 in. long is - in. wide and ]\ in. thick. How much does it weigh, at 490 11». per cubic l"<»t '.' 11. If 3 eu. in. of gold beaten into ,u r <>ld leaf will o- 75,000 aq. in. of surface, find the thickness of the leaf. 12. The sum of the squares on tin- four diagonal a par- allelepiped is equivalent t<> the sum of tl on the twelve edges. 334 BOOK VII. SOLID GEOMETRY Proposition XL Theorem 538. The volume of a triangular prism is equal to the product of its base by its altitude. D' Given the triangular prism ABC-B', with volume v> base 6, and altitude a. To prove that v = ba. Proof. Upon the edges AB, BC, BB' construct the parallele- piped ABCD-B'. Then ABC-B' = J ABCD-B'. § 527 The volume of ABCD-B' = ABCD x a. § 536 But ABCD = 2b. §126 .'. v = ^(2 ba) = ba, by Ax. 9. Q.E.D. EXERCISE 88 Find the volumes of the triangular prisms whose bases and altitudes are as follows : 1. 17 sq. in., 8 in. 2. 15.75 sq. ft., 3 ft. 3. 3i sq. ft., 1 ft. 8 in. 4. 51 sq. ft., 2 ft. 9 in. 5. 15.84 sq. ft., 3 ft. 10 in. 6. 16f sq. in., 2| in. 7. 221 S q. in., 41 in. 8. 33i sq. in., !\ in. 9. 42| sq. in., 3| in. 10. 27f sq. in., 3| in. 11. 12 sq. ft. 75 sq. in., 2 ft. 7 in. PRISMS Pkoposition XII. Theorem 335 539. Tin volutin of any prism is equal to tin product of its /"/» by its alt if ml, . Given the prism AC with volume v, base b } and altitude a. To prove thai v = ba. Proof. It is possible to divide any prism in general into what kind of simpler prisms '.' 1 [ow is this done '.' What is the volume of each of these simpler prisn 538 What is the sum of the volumes of these simpler prisms '.' Whal is the sum of their bases '.' How does the common altitude of these simpler prisms compare with a, the altitude of the given prism '.' What conclusion can be drawn from these statements? Write the proof in full. 540. Corollary 1. Prisms having equivalent bases are to each other as their altitudes; prisms having equal altitudes an to each other as their ba% Write thr proof in full. 541. Corollary 2. Prisms having equivalent vnd ,'i>i,il altitudes an equivalent, Writr the proof in full. 336 BOOK VII. SOLID GEOMETRY EXERCISE 89 1. If the length of a rectangular parallelepiped is 18 in., the width 9 in., and the height 8 in., find the total area of the surface. 2. Find the volume of a triangular prism, if its height is 15 in. and the sides of the base are 6 in., 5 in., and 5 in. 3. Find the volume of a prism whose height is 15 ft., if each side of the triangular base is 10 in. 4. The base of a right prism is a rhombus of which one side is 20 in., and the shorter diagonal 24 in. The height of the prism is 30 in. Find the entire surface and the volume. 5. How many square feet of lead will be required to line an open cistern which is 4 ft. 6 in. long, 2 ft. 8 in. wide, and con- tains 42 cu. ft.? 6. An open cistern 6 ft. long and 4^ ft. wide holds 108 cu. ft. of water. How many square feet of lead will it take to line the sides and bottom ? 7. One edge of a cube is e. Find in terms of e the surface, the volume, and the length of a diagonal of the cube. 8. The diagonal of one of the faces of a cube is d. Find in terms of d the volume of the cube. 9. The three dimensions of a rectangular parallelepiped are a, b, c. Find in terms of a, b, and c the volume and the area of the surface. 10. Find the volume of a prism with bases regular hexagons, if the height is 10 ft. and each side of the hexagons is 10 in. 11. An open cistern is made of iron \ in. thick. The inner dimensions are : length, 4 ft. 6 in. ; breadth, 3 ft. ; depth, 2 ft. 6 in. What will the cistern weigh when empty ? when full of water ? (A cubic foot of water weighs 62i lb. Iron is 7.2 times as heavy as water ; that is, the specific gravity of iron is 7.2.) PYRAMIDS 337 542. Pyramid. A polyhedron of which one face, called the base, is a polygon of any Dumber of Bides and the other faces are triangles having a common vertex is called a pyramid. The triangular fares having a common vertex are called the lateral /wis. their intersections are called the int> /•"/ edges, ami their common vertex is called tli' ex of the pyramid. The base of a pyramid may he any kind of a polygon, but usually a convex polygon is taken. 543. Lateral Area. The sum <>1 the mens of the lateral taces of a pyramid is called the lateral area of the pyramid. 544. Altitude. The perpendicular distance from the vertex t»t the plane of the base is called the altitude of the pyramid. 545. Pyramids classified as to Bases. Pyramids are said to l>c triangular, quadrangular, and so on, according as their bases are triangles, quadrilaterals, ami s<» on. A triangular pyramid has four triangular faces ami is called a tetra- hedron. Any one of its faces may he taken as the base. 546. Regular Pyramid. If the base of a pyramid is a regular polygon whose center coincides with the foot of the perpendicular let fall from the vertex to the base, the pyramid is called a regular pyramid. A regular pyramid is also called a right pyramid. 547. Slant Height of a Regular Pyramid. The altitude of any one of the lateral Paces of a regular pyramid, drawn from the vertex <>t' the pyramid, is called the slant height. The slant height Lb the same whatever face La taken I ■>■ < >nly a regular pyramid can have a slant height. 338 BOOK VII. SOLID GEOMETRY 548. Properties of Regular Pyramids. Among the properties of regular pyramids the following are too evident to require further proof than that referred to below : (1) The lateral edges of a regular 'pyramid are equal (§ 439). (2) The lateral faces of a regular pyramid are congruent isosceles triangles (§ 80). (3) The slant height of a regular pyramid is the same for all the lateral faces (§ 439). 549. Frustum of a Pyramid. The portion of a pyramid in- cluded between the base and a section parallel to the base is called a frustum of a pyramid. The base of the pyra- mid and the parallel section are called the bases of the frustum. A more general term, including frustum as a special case, is truncated pyramid, the portion of a pyramid included between the base and any section made by a plane that cuts all the lateral edges. This term is little used. 550. Altitude of a Frustum. The perpendicular distance between the bases is called the altitude of the frustum. E.g. C'C is the altitude of the frustum in the above figure. 551. Lateral Faces of a Frustum. The portions of the lateral faces of a pyramid that lie between the bases of a frustum are called the lateral faces of the frustum. In the case of a frustum of a regular pyramid the lateral faces are congruent isosceles trapezoids. The sum of the areas of the lateral faces is called the lateral area of the frustum. 552. Slant Height of a Frustum. The altitude of one of the trapezoid faces of a frustum of a regular pyramid is called the slant height of the frustum. Thus JOT in the above figure is the slant height. PYRAMIDS ■ Pboposihoh Mil. Theorem 553. Tin lateral area of a regular pyramid is equal to half th product of its slant height by tJu perimeter of its bo Given the regular pyramid V-ABCDE, with / the lateral area, s the slant height, and p the perimeter of the base. gruenl To prove that / = .\ sp. Proof. The .1.1/:. VBC, VCD, VDE, and YEA are ron- 548 The area of each A= \& x its ba The sum of the I • of the triangles =p .-. the sum of the areas of these A = }, sp But the sum of the areas of these A = /. .*. I = \ >/'. by Ax. 554. Corollabt. The lateral area . rust, mi of a regular pyramid equal to half the sum of the perim- ,-.< ,,,' the bases multiplied by t/u jf slant height of th frustum. How ia the an oid found (| ru.nT ? What ia the sum of their !"■ 1 their a] : What is the sum of their areas I the formu] 325 \ . 11 \-.. 1 543 Q.E.D. B 340 BOOK VII. SOLID GEOMETRY Proposition XIV. Theorem 555. If a pyramid is cut by a plane parallel to the base : 1. The edges and altitude are divided proportionally \ 2. The section is a polygon similar to the base. Given the pyramid V-ABCDE cut by a plane parallel to its base, intersecting the lateral edges in A\ B\ C\ D\ E\ and the alti- tude VO in O'. - ' • „, VA' VB' VO' 1. To prove that -j^— = -=-- = • • . = — -• Proof. Since the plane A'D' is II to the plane AD, Given .'.A'B' is WtoAB,B'C'is II to BC, • • ., and A'O' is II to ^4 0. §453 VA 1 VB' VO' , . _, .-. = = ...= ? by §274. q.e.d. VA VB VO J 2. To prove the section A 1 B'C'D'E' similar to the base AB CDE. Proof. Since AVA'B' is similar to AVAB, AVB'C similar to AVBC, and so on (why ?), how can the corresponding sides of the polygons be proved proportional ? Since A'B' is II to AB, B'C to BC, etc. (why ?), how can the corresponding angles be proved equal f Then why is A'B'C'D'E' similar to ABODE? PYRAMIDS 341 556. Cobollabt 1. Any tectum of a pyramid parallel to the base is to the base as the square of the distance from the vertex is to the %quan of the altitude of the j/t/ntmiif. VO r.i Fur VO Therefore YO'~ V A .\n ~AIi a i; A ir But. from similar polygons, A'B'C'jyE' ABCDE Hence, by substituting, A'B'C'irE' ABCDE A'B IS V6 2 a S 288 §27" §334 Ax. 8 557. Cobollaby 2. Tf two pyramids have equal atiitu and equivalent bases, sections made by planes parallel to the ,. and at equal distances from the vertices, >><: equival Wha1 is the ratio of A /; ' D W to ABCDE? A II- .w can this be shown to equal VO f ■ VO ' What is the ratio of X'Y '// to ZTZ? How can this bi sho wn to equ al W P* WP ? An- the ratio* VO**: ro*and WP*\ WF* equal? - ooeit Isghren that ABCDE ZrZ,whatcan be said ol A'&C 1>'E' and X'Y'Z'l 342 BOOK VII. SOLID GEOMETRY Proposition XV. Theorem 558. Two triangular pyramids having equivalent bases and equal altitudes are equivalent. B B' Given two triangular pyramids, V-ABC and 7'-^ f B r C', having equivalent bases and equal altitudes. To prove that V-ABC and V'-A'B'C are equivalent. Proof. Suppose the pyramids are not equivalent, and V'-A'B'C' > V-ABC. Place the bases in the same plane, and suppose the altitude divided into n equal parts, calling each of these parts h. Through the points of division pass planes parallel to the base, cutting the pyramids in DEF, GHI, • • • , D'E'F', G'H'I', On A'B'C, D'E'F', G'H'I', and other parallel sections, if any, construct prisms with lateral edges parallel to A'V, and with altitude h. In the figure these are represented by X', Y', and Z'. On DEF, GHI, and other parallel sections, if any, as upper bases, construct the prisms Y, Z, with lateral edges parallel to VA, and with altitude h. Then since DEF = D'E'F', § 557 and h = h, Idem .*. prism Y= prism Y'. § 541 Similarly prism Z = prism Z'. PYRAMIDS 343 But A' + )" + /'> I"- 17;'' '. and V+Z< V-ABC. Ax. 11 .-. v'-A'B'C — V-ABC < X' + V + Z' — ( Y+Z), or V'-A'B'C V- iBC< ' \ . That is, tlic difference between the pyramids mnsl be l< than the difference between the 9< ta of prisms. Now by increasing n indefinitely, and consequently de- creasing // indefinitely, A' can be made less than any assigned quantity. Hence whatever difference we suppose to exist between the pyramids, A Van be made smaller than thai supposed difference. But this is absurd, since we have shown thai A' is greater than the difference) if any exist Hence it Leads to a manifest absurdity to suppose that I '-.17;'''; . V-ABC. In the same \\a\ it leads to an absurdity to suppose that V-ABO V'-A'B'C. .-. V-ABC = V'-A'B'C. . q.e.d. EXERCISE 90 1. Tli" slant height of a regular pyramid is 6 in., and the base is an equilateral triangle of altitude 2Vo in. Find the lateral area of the pyramid. 2. The slant height of a regular triangular pyramid equals the altitude of the base. The area of the base is ^ 3 s. Prove the pyramids equivalent. 344 BOOK VII. SOLID GEOMETRY Proposition XVI. Theorem 559. The volume of a triangular pyramid is equal to one third the product of its base by its altitude. Given the triangular pyramid E-ABC, with volume v, base &, and altitude a. To prove that v ~\ oa ' Proof. On the base AB C construct a prism ABC-DEF. Through DE and EC pass a plane ODE. Then the prism is composed of three triangular pyramids E-ABC, E-CFD, and E-A CD. Now the pyramids E-CFD and E-A CD have the same altitude and equal bases CFD and A CD. § 126 . ' . E-CFD = E-A CD. § 558 But pyramid E-CFD is the same as pyramid C-DEF, which has the same altitude as pyramid E-ABC, and has base DEF equal to base ABC. § 511 .'. E-CFD = E-ABC. § 558 . ' . E-ABC = E-CFD = E-A CD. Ax. 8 .'. pyramid E-ABC = ^ prism ABC-DEF. But the volume of ABC-DEF =ba. § 539 .-. v = i ba, by Ax. 4. Q-E.d. 560. Corollary. The volume of a triangular pyramid is equal to one third the volume of a triangular prism of the same base and altitude. PYRAMIDS Proposition XVII. THEOREM 345 561. The volunu of any pyramid is equal to om third tl,, product of its base by its altitudi . Given the pyramid V-ABCDE, with volume v } base b, and alti- tude a. To prove that v = J ba. Proof. Through the edge VD and the diagonals of the ba l>.\ , />/;, pass planes. These planes divide the pyramid V-ABCDE into three tri- angular pyramids. What can he said as to the altitudes of the original pyramid and of the triangular pyramids *.' What ran be Baid as to the hase of the original pyramid in relation to the bases of the triangular pyramids '.' What is the volume of each triangular pyramid? What is the sum of the volumes of the triangular pyramids '.' ( omplete the proof. 562. Corollary. The volumes of two pyramids an tch other as the products of their bases and altitudes; pyramids having equivalent bases are to >tfi> r ,i.< their altitudi pyramids having >'/i"f/ altitudes are '■> each other as their bases; pyramids having equivalent ba fual altitudes art equivalent 346 BOOK VII. SOLID GEOMETRY EXERCISE 91 Find the lateral areas of regular pyramids, given the slant heights and the perimeters of the bases, as follows : 1. s — 34 in., p = 57 in. 3. s = 2 ft. 7 m.,p = 4 ft. 6 in. 2. 5 = 8f in., p = 17| in. 4. s=127ft. 5 in., _p = 63 ft. 2 in. Find the lateral areas of frustums of regular pyramids, given the slant heights of the frustums and the perimeters of the bases, as follows : 5. s = 4 in., p = 8 in., p' = 6 in. 6. s = 5£ in., ^j> = 9| in., £>' = 7| in. 7. s = 2 ft. 3 in., p = 4 ft. 8 in., 79' = 3 ft. 9 in. Find the volumes of pyramids, given the altitudes and the areas of the bases, as follows : 8. a = 7 in., h = 9 sq. in. 11. a = 3| in., h = 5| sq. in. 9. a = 6 in., & = 23 sq. in. 12. a = 4| in., 6 = 19 sq. in. 10. a = 17 in., 6=51 sq. in. 13. a = 27.5 ft., b = 325 sq. ft. .Fu'wrf ^/^ lateral areas of regular pyramids, given the slant heights, the number of sides of the bases, and the length of each side, as follows : 14. s = 2.3 in., n = 4, 1= 2.1 in. 15. s = 3.7 in., n = 6, 1= 2.9 in. 16. s = 5.33 in., n = 8, Z= 3 in. i^m^ //* and the total surface t. 33. show how to find the volume of any polyhedron by dividing the polyhedron into pyramids. 348 BOOK VII. SOLID GEOMETRY Proposition XVIII. Theorem 563. The frustum of a triangular pyramid is equiva- lent to the sum of three pyramids whose common altitude is the altitude of the frustum and whose bases are the loiver base, the upper base, and the mean proportional between the two bases of the frustum. Given the frustum of a triangular pyramid, ABC-DEF, having ABC, or b, for its lower base ; DEF, or b\ for its upper base ; and the altitude a. To prove that ABC-DEF = \ab + \aV + \a y/bb'. Proof. Through A, E, and C, and also through C, D, and E, pass planes dividing the frustum into three pyramids. Then E-ABC = £ ab, and C-DEF=±ab'. §559 It therefore remains only to prove that E-ACD = i a^/bb'. We see by the figure that we may speak of E-ABC as C-ABE, and of E-A CD as C-AED. But C-ABE : C-AED = A ABE : A AED. § 562 Since A. 4 BE and AED have for a common altitude the altitude of the trapezoid ABED, .'. A ABE : A AED = AB : DE. § 327 .'. C-ABE : C-AED = AB : DE, Ax. 8 or E-ABC : E-A CD = AB : DE. Ax. 9 PYRAMIDS 349 In like manner E-ACD and E-CFD have a common vertex E and have their bases in the same plane, .1 ( 'FD, so that E-ACD : E-CFD = A ACD: A CFD. § 562 Since A A CD and ( 'FD have for a common altitude the alti- tude of the trapezoid .1 ( 'FD, .-.A ACD: A CFD — AC:DF. § 327 .*. E-ACD: E-CFD — AC : DF. Ax. 8 1 Jut A D£F is similar to A A EC. § 555 .*. I/; : DE = AC:DF. §'282 .'. E-ABC : 7-;--4 CD = AC:DF. A x. 8 .'. /•;,!/;' ' : /•;-.! < 7; = E-4< 7> : E-< 'FD. Ax. 8 But E-CFD is the same as C-DEF, which has been shown to equal ^ «&'. .'. J o£ : E-A CD = ]■:-. VCD: \ ab'. Ax. 9 .-. E-ACD= V.\ oh x £ itb' § 262 .*. 7wl£C + C-D-EF -f E-A CD= : i : afl + J o£' + i a VST'. Ax. 1 That is, ABC-DEF— \ ab + % ab 1 + J a VST', by Ax. 9. Q.e.d. 564. Corollary 1. !Z%e volume of a frustum of a tri- angular pyramid may I" expressed as -J a {b -f- b' -f- V A// ). For we may factor by a a. 565. Corollary 2. 27ie volume of a frustum of any pyramid is equal to the sum of tin 1 volumes of three pyramid* "•hose common attitude is the altitude of the frustum, and tr/mxe bases are the lower base, the upper base, and the mean proportional between the l>0°, and so on. Hence only live regular convex polyhedrons are possible. The regular polyhedrons are the regular tetrahedron, the regular hexahedron, or cube, the regular octahedron, the regular dodecahedron, and the regular icosahedron. q.e.f. It adds greatly to ;i clear understanding of the five regular poly- hedrons if they are constructed from paper as suggested in § ">'">7. Since these ><>li = 24 sq. ft. 72 sq. in., b' = 2 sq. ft. 6. A pyramid 2 in. nigh, with a base whose area is 8 sq. in., is cut by a plane parallel to the base 1 in. from the vertex. Find the volume of the frustum. 7. A pyramid 3 in. high, with a base whose area is 81 sq. in., is cut by a plane parallel to the base 2 in. from the base. Find the volume of the frustum. 8. The lower base of a frustum of a pyramid is a square 1 in. on a side. The side of the upper base is half that of the lower base, and the altitude of the frustum is the same as the side of the upper base. Find the volume of the frustum. 9. The lower base of a frustum of a pyramid is a square 3 in. on a side. The area of the upper base is half that of the lower base, and the altitude of the frustum is 2 in. Find to two decimal places the volume of the frustum. 10. A pyramid has six edges, each 1 in. long. Find to two decimal places the volume of the pyramid. 11. A regular tetrahedron has a volume 2 V2 cu. in. Find to two decimal places the length of an edge. 12. The base of a regular pyramid is a square I ft. on a side. The slant height is s ft. Find the area of the entire surface. 13. Consider the formula v = £ a (b + V + V^), of § 564, when V ±= 0. Discuss the meaning of the result. Also discuss the case in which b = b'. CYLINDERS 353 569. Cylindric Surface. A surface generated by a straight line which is constantly parallel to a fixed Btraight line, and touches a fixed curve not in the plane of the straight line, is called a cylindric surface^ or a cylindrical surface. The moving line is called the generatrix and the fixed curve the directrix. In the figure ABC is the directrix. 570. Element. The generatrix in any position is called an ele- ment of the cylindric surface. 571. Cylinder. A solid bounded by a cylindric surface and two parallel plane surfaces is called a cylinder. It follows, therefore, that all the elements of a cylinder are equal. The terms bases, lateral surface, and altitude are used as with prisms. 572. Right and Oblique Cylinders. A cylinder whose element s are perpendicular to its bases is called a right cylinder; other- wise a cylinder is called an oblique cylinder. 573. Section of a Cylinder. A figure formed by the intersec- tion of a plane and a cylinder is called ^section of the cylinder. 354 BOOK VII. SOLID GEOMETKY Proposition XX. Theorem 574. Every section of a cylinder made by a plane passing through an element is a parallelogram. Given a cylinder AC, and a section ABCD made by a plane pass- ing through the element AB. To prove that ABCD is a parallelogram. Proof. Through D draw a line in the plane ABCD II to AB. This line is an element of the cylindric surface. § 570 Since this line is in both the plane and the cylindric surface, it must be their intersection and must coincide with DC. Hence DC coincides with a straight line parallel to AB. Therefore DC is a straight line II to AB. Also AD is a straight line II to BC. § 453 .*. ABCD is a parallelogram, by § 118. q.e.d. 575. Corollary. Every section of a right cylinder made by a plane passing through an element is a rectangle. 576. Circular Cylinder. A cylinder whose bases are circles is called a circular cylinder. A right circular cylinder, being generated by the revolution of a rec- tangle about one side as an axis, is also called a cylinder of revolution. CYLINDERS Proposition- XXI. Tiikokkm 577. Tin bases of a cylinder are congruent, G C 355 Given the cylinder AC, with bases ABE and DCG. To j>r<"'r that ABE is congruent f>> DCG. Proof. Let A, B, E be any three points in the perimeter of the lower base, and AD, EC, Ed be elements of the surface. Draw AB, AE, El',, DC, 1><;, GC. Then AD, BC, EC are equal, § 571 and parallel. § 569 .-. AB = Dc, AE= DG, EB= GC. § 130 .*. A A HE is congruent to A DCG. § 80 Place the lower base on the upper base so that HlqAABE shall fall on the A DCG. Then A, B, E will fall on D, C, G. Therefore all points in either perimeter will coincide with points in the other, and the bases are congruent, by § 66. q. e. d. 578. Cmkollary 1. A//// two parallel sr<-ti<>iis of a cylinder, cutting all the elements, are congruent 579. Corollary 2. Any section of a cylinder parallel to the l>ase is congruent t<> the base. 580. Corollary 3. The straight line joining the center* of tin hoses of a circular cylindt r jesses through thr centers of all sections of the cylindi r parallel f>> the bat 356 BOOK VII. SOLID GEOMETRY 581. Tangent Plane. A plane which contains an element of a cylinder, but does not cut the surface, is called a tangent plane to the cylinder. 582. Construction of Tangent Planes. From a consideration of the nature of a tangent plane and of the construction of a cylindric surface it is evident that : A plane passing through a tangent to the base of a circular cylinder and the element drawn through the point of contact is tangent to the cylinder. If a plane is tangent to a circular cylinder, its intersection with the plane of the base is tangent to the base. 583. Inscribed Prism. A prism whose lateral edges are ele- ments of a cylinder and whose bases are inscribed in the bases of the cylinder is called an inscribed prism. In this case the cylinder is said to be circumscribed about the prism. Inscribed Prism Circumscribed Prism 584. Circumscribed Prism. A prism whose lateral faces are tangent to the lateral surface of a cylinder and whose bases are circumscribed about the bases of the cylinder is called a circumscribed prism. In this case the cylinder is said to be inscribed in the prism. CYLINDERS 357 585. Right Section. \ Bection of a cylinder made by a plane that cuts all the elements and is perpendicular to them is called a right section of the cylinder. 586. Cylinder as a Limit. From the work already done in connection with limits, and from the nature of the inscribed and circumscribed prisms, the following properties of the cylinder may new be assumed without further proof than that given below : If a prism whose base is a regular polygon is inscribed in or circumscribed about a circular cylinder^ and if the number of sides of the prism is indefinitely increased^ 1. The volume of the cylinder is the limit of the volume of the prism. 2. The lateral area of the cylinder is the limit of the lateral area of the prism. 3. The perimeter of a right section of the cylinder is the limit of the perimeter of a right section of the prism. For as we increase the number of sides of the base of the inscribed or circumscribed prism whose base is a regular polygon, the perimeter of the base approaches the circle as its limit ($ 381). This brings the lateral surface of each prism nearer and nearer the lateral surface of the cylinder. It also brings the volume of each prism o-er and nearer the volume of the cylinder. In the same way it bril the right Bection of each prism nearer and nearer the right section of the cylinder. 358 BOOK VII. SOLID GEOMETRY Proposition XXII. Theorem 587. The lateral area of a circular cylinder is equal to the product of an element by the perimeter of a right section of the cylinder. Given a circular cylinder C, / being the lateral area, p the perim- eter of a right section, and e an element. To prove that I = ep. Proof. Suppose a prism with base a regular polygon to be inscribed in C, V being its lateral area and p' being the perim- eter of its right section. Then V = ep'. § 512 If the number of lateral faces of the prism is indefinitely increased, , .. Z' approaches £ as a limit, anc [ y approaches j? as a limit, § 586 and consequently ep' approaches ep as a limit. .-. l = ep, by §207. Q-e.d. 588. Corollary. The lateral area of a cylinder of revolu- tion is equal to the product of the altitude by the circum- ference of the base. In the case of a right circular cylinder of altitude a, lateral area I, total area t, and radius of base r, we have I = 2 7rra, and t = 2 irra + 2 7rr 2 = 2 7rr (a + r). CYLINDERS 359 Proposition Will. Tiikokkm 589. The volume of a circular cylinder is equal to the product Of its l>'/sr bif its (lltitudi . Given a circular cylinder C, b being the base, v the volume, and a the altitude. To prove that v = ba. Proof. Suppose a prism with base a regular polygon to be inscribed in C, />' being its base and v* being its volume. Then c' = h'u. §539 If the number of lateral faces of the prism is indefinitely increased, v' approaches v as a limit, § 586 // approaches b as a limit, § 381 and consequently b'a approaches ba as a limit. But u'= b'a, whatever the number of sides. § 539 .-. c = ba, by § 207. q.e.d. 590. Corollary. The volume of a cylinder of revolution with radius r and altitude a is irr^a. What is the area of the base ? By what is this to be multiplied ? 591. Similar Cylinders. Cylinders generated by the revolu- tion of similar rectangles about corresponding sides are called similar cylinders of revolution. §§ 601 and 592 may be omitted without destroying the sequence. 360 BOOK VII. SOLID GEOMETRY Proposition XXIV. Theorem 592. The lateral areas, or the total areas, of similar cylinders of revolution are to each other as the squares of their altitudes or as the squares of their radii ; and their volumes are to each other as the cubes of their altitudes or as the cubes of their radii. Given two similar cylinders of revolution, / and V denoting their lateral areas, f and V their total areas, v and v f their volumes, a and a 1 their altitudes, and r and r' their radii. To prove that l:l'=t:t' = a 2 : a' 2 = r 2 : r'% and that v : v' = a 3 : a rd = r 3 : r'\ Proof. Since the generating rectangles are similar, § 591 a r a-\-r '''~a 1 = ? = 'a~ r +r r § 269 Also we have by this proportion and § 588, I V 2 irra 2 irr'a' ra r'a' a r 12 a i-y But t = 2 irra -f 2 irr 2 (§ 588), and v = irr^a t 2 irija + r) r(a + r) r '''t l ~~~~ 2 irr' (a' + r*) ' ' r'(a' + r') ' ' ? 590 a' ,'2 a 12' and irv^a v inra v a r a v' = ^rTV = 7 2 X a' ~ 7 s == a 1 *' Q.E.D. CYLINDERS 361 EXERCISE 93 1. The diameter of a well is 6 it. and the water is 7 ft. deep. How many gallons of water are there in the well, reck- oning 7^- gal. to the cubic fool '.' 2. When a body is placed under water in a right circular cylinder 60 centimeters in diameter, tin- level of the water ri 40 centimeters. Find the volume of the body. 3. I I<:>\v many cubic yards of earth must be removed in constructing a tunnel 100 yd. long, the section being a semi- circle with a radius of 18 ft. ? 4. How many square feet of sheet iron are required to make a pipe 18 in. in diameter and 40 ft. long ? 5. Find the radius of a eylindric pail 14 in. high that will hold exactly 2 CU. ft. 6. The height of a eylindric vessel that will hold 20 liters is equal to the diameter. Find the altitude and the radius. 7. If the total surface of a right circular cylinder is t and the radius of the base is r, find the altitude '/. 8. If the lateral surface of a right circular cylinder is / and the volume is r, find the radius /• and the altitude a. 9. If the circumference of the base of a right circular cyl- inder is c and the altitude is a, find the volume v. 10. If the circumference of the base of a right circular cylinder is c and the total surface is t, find the volume v. 11. If the volume of a right circular cylinder is y and the altitud ■. find the total surface t 12. If v is the volume of a right circular cylinder in which altitude equals the diameter, find the altitude •) 588) find the value of r. (Omit unless quadratics have been studied.) 362 BOOK VII. SOLID GEOMETRY 593. Conic Surface. A surface generated by a straight line which constantly touches a fixed plane curve and passes through a fixed point not in the plane of the curve is called a conic surface or a conical surface. The moving line is called the generatrix, the fixed curve the directrix, and the fixed point the vertex. Hold a pencil by the point and let the other end swing around a circle, and the pencil will generate a conic surface. We may also swing a blackboard pointer about any point near the middle, so that either end shall touch any fixed plane curve, and thus generate a conic surface. Such a surface is rep- resented in the annexed figure. 594. Element. The generatrix in any position is called an element of the conic surface. If the generatrix is of indefinite length, the surface consists of two portions, one above and the other below the vertex, which are called the upper nappe and lower nappe respectively. The two nappes are shown in the above figure. 595. Cone. A solid bounded by a conic surface and a plane cutting all the elements is called a cone. The conic surface is called the lateral surface of the cone, and the plane surface is called the base of the cone. The vertex of the conic sur- face is called the vertex of the cone, and the elements of the conic surface are called the ele- ments of the cone. The perpendicular distance from the vertex to the plane of the base is called the altitude of the cone. (ONKS :]i;:>, 596. Circular Cone. circular cone. A cone whose base is a circle is called a The straight line joining the vertex of a circular cone and the center of the base is called the axis of the cone. 597. Right and Oblique Cones. A circular cone whose axis is perpendicular to the base is called a right cone ; otherwise a circular cone is called an oblique cone. 598. Cone of Revolution. Since a right circular cone may be generated by the revolution of a right triangle about one of the sides of the right angle, it is called a cone of revolution. In this case the hypotenuse corresponds to an clement of the surface and is called the slant Jieight. 599. Conic Section. A section formed by the intersection of a plane and the conic surface of a cone of revolution is called a coiii' 1 section. Fig. 1 Fig. 2 Fig. 3 Fig. 4 Fig. 5 In Fig. I the conic section is two intersecting straight lines, and this Liscussed in § 000. This is true for all kinds of com-. In Fi_ r . 2 the conic section is a circle, and this is discussed in § 601. In Fig. 3 the conic section is called an ellipse, the form a circle seems to take when Looked at obliquely. The orbit of a planet is an elli] - In Fig. 4 the conic section is & parabola, the path of a projectile (in a vacuum). Here the cutting plane is parallel to an element. In Fig. 5 the conic section is an hyperbola. The general study of conic sections is not a part of elementary geome- try, but the names of the sections may profitably be known. 364 BOOK VII. SOLID GEOMETRY Proposition XXV. Theorem 600. Every section of a cone made oy a plane pass- ing through its vertex is a triangle. Given a cone, with AVB a section made by a plane passing through the vertex V. To prove that A VB is a triangle. Proof. AB is a straight line. § 429 Draw the straight lines VA and VB. The lines VA and VB are both elements of the surface of the given cone. § 594 These lines lie in the cutting plane, since their extremities are in the plane. • § 422 Hence VA and VB are the intersections of the conic surface with the cutting plane. But VA and VB are straight lines. Const. Therefore the intersections of the conic surface and the plane are straight lines. Therefore the section AVB is a triangle, by § 28. q.e.d. CONES 365 Proposition XXVI. Theorem 601. In a circular cone a section made by a plane parallel to the base is a circle. Given the circular cone V-ABCD, with the section A'B'C'D' parallel to the base. To prove that A'B'C'D 1 is a circle. Proof. Let be the center of the base, and let O 1 be the point in which the axis VO pierces the plane of the conic section. Through TO and any elements r.l. 17;. pass planes cutting the base in the radii OA, OB, and cutting the section A'B'C'D' in the straight lines O'A', O'B'. Then O'A' and O'B' are II respectively to OA and OB. r>:> Therefore the A AOV and OBV are similar respectively to the AA'u'V and O'B'V. § 285 OA VO OB § 282 § 162 ' 'O'A 1 VO' <>'B' But OA — OB. .-. oM'=07J'(§263),;nid A'B'C'D' is a rircle ? by §159. Q.e.d. 602. < Sobollaby. The axis of a circular cone pass( s through t/ (t center oj 7/ seciioint of contact is tangent to the cone. |Jf If a pAane is tangent to a circular |jj cone its intersection with the plane of the base is tangent to the base. 605. Inscribed Pyramid. A pyramid whose lateral edges are elements of a cone and whose base is inscribed in the base of the cone is called an inscribed pyra mid '. In this case the cone is said to be circumscribed about the pyramid. Inscribed Pyramid Circumscribed Pyramid 606. Circumscribed Pyramid. A pyramid whose lateral faces are tangent to the lateral surface of a cone and whose base is circumscribed about the base of the cone is called a cir- cumscribed pyramid. In this case the cone is said to be inscribed in the pyramid. COXES 367 607. Frustum of a Cone. The portion of a cone included be- tween the base and a section parallel to the base is called a frustum of a cone. The base of the cone and the parallel section are together called the bases of the frustum. The terms altitude and lateral area of a frus- tum of a cone, and slant height of a frustum of a right circular cone, are used in substan- tially the same manner as with the frustum of a pyramid (§^ •">•">(), 551, 552). 608. Cones and Frustums as Limits. The following proper- ties, similar to those of § 586, are assumed without proof : If a pyramid whose base is a regular polygon is inscribed in or circumscribed about a circular cone, and if the number of sides of the base of the pyramid is indefinitely increased, the volume <>/ the cone is the limit of the volume of the pyramid^ and Hie lateral area of the cone is the limit of the lateral area of tin' pyramid. The volume of a frustum of a ranr is the limit of the volumes of the frustums of the inscribed and circumscribed 'pyramids^ if the number <>f lateral faces is indefinitely increased^ on, I the lateral area of the frustum of a <■>>/"■ is the limit of the Intern/ areas of the frustums of the inscribed and circumscribed pyramids, the bases being regular polygons* 368 BOOK VII. SOLID GEOMETRY Proposition XXVII. Theorem 609. The lateral area of a cone of revolution is equal to half the product of the slant height by the circumfer- ence of the base. Given a cone of lateral area /, circumference of base c, and slant height s. To prove that 1= ~^sc. Proof. Suppose a regular pyramid to be circumscribed about the cone, the perimeter of its base beings and its lateral area V. Then V = \ sp. § 553 If the number of the lateral faces of the circumscribed pyra- mid is indefinitely increased, V approaches / as a limit, § 608 p approaches casa limit, § 381 and consequently \ sp approaches ^ «c as a limit. But V = % sj), whatever the number of sides. § 553 .-. I =i 2 sc, by § 207. q.e.d. 610. Corollary. If I denotes the lateral area, t the total area, s the slant height, and r the radius of the base of a cone of revolution, then I = |(2 irr X s) = irrs ; t = irrs + tt^ 2 = 7rr (s + r). cnXES 369 EXERCISE 94 Find the lateral areas of cones of revolution, given the slant heights and the circumferences of th bases respectively as follows : * 1. 2|in.,5fin. 4. 3.7 in., 5.8 in. 7. 2 ft. 6 in., 4 ft. 8 in. 2. I; in., sj in. 5. 5.3 in., 9.7 in. 8. 3 ft. 7 in., 8 ft. 6 in. 3. ('»,";. in., lo.i, J,,. 6. 6.5 in., 11.6 in. 9. 5 ft. 8 in., 12 ft. 4 in. Find the lateral areas of cones of revolution, given the slant Jieights and the radii of the bases respectively as follows: 10. 3fin.,2£in. 13. 6.4 in., 4.8 in. 16. 2 ft. 3 in., 8 in. 11. 2£in.,lfin. 14. 7.2 in., 5.3 in. 17. '4 ft. 6 in., 2 ft. 12. \\ in...".] in. 15. 8.9 in^ 5.6 in. 18. 6 ft. 9 in., 3 ft. 2 in. Find the total areas of cones of revolution, given the slant heights and the radii of the bases respectively as follows: 19. 3 in., 2 in. 21. 7 in., 4 in. 23. 6 It.. 1 it. 20. 5 in.. 3 in. 22. 9 in., 5 in. 24. 1 2 ft., 5 ft. 25. Deduce a formula for finding the lateral area of a cone of revolution in terms of the radius of the base and the altitude. 26. Deduce a formula for finding the slant height in terms of the lateral area and the circumference of the base. 27. Deduce a formula for finding the slant height in terms of the lateral area and the radius of the base. 28. Deduce a formula for finding the radius of the base in terms of the Intend area and the slant height. 29. Deduce a formula for finding the slant height in terms of the total area and the radius of the bas 30. Deduce a formula for finding the circumference of the base in terms of the lateral area and the slant height. 370 BOOK VII. SOLID GEOMETRY Proposition XXVIII. Theorem 611. The volume of a circular cone is equal to one third the product of its base by its altitude. Given a circular cone of volume v, base b, and altitude a. To prove that v = ^ba. Proof. Suppose a pyramid with base a regular polygon to be inscribed in the cone, V being its base and v' its volume. Then v' = J b'a. § 561 If the number of lateral faces of the pyramid is indefinitely increased, v' approaches v as a limit, § 608 b' approaches b as a limit, § 381 and consequently b'a approaches ba as a limit. .-. v = ±ba, by § 207. Q.e.d. 612. Corollary. In a circular cone of radius r and alti- tude a, v = 1 7rr 2 a. Eor the area of the base is 7rr 2 (§ 389). 613. Similar Cones. Cones generated by the revolution of similar right triangles about corresponding sides are called similar cones of revolution. In case § 614 is omitted this definition may also be omitted. CONES 371 EXERCISE 95 Find the volumes of circular mnrx, t/iren the } in.. 9f sq. in. 5. 7.8 in., 6.9 sq. in. 3. 5| in., 10V sq. in. 6. 9.3 in., 16.8 sq. in. Find the e<>/ii/nes of circular cones, f/ireii the, altitudes and the radii of the bases respectively as follows: 7. 4 in., 3 in. 10. 9.8 in., 4.3 in. 8. 6 in., 4 in. 11. 10.5 in., 0.2 in. 9. 8 in., 5 in. 12. 14.*) in., 9.6 in. 13. How many cnbic feet in a conical tenl 10 It. in diameter and 7 ft. high? 14. How many cubic feet in a conical pile of earth 15 ft. in diameter and 8 ft. high ? 15. Deduce a formula for finding the altitude of a circular cone in terms of the volume and the area of the base. 16. Deduce a formula for finding the area of the base of a circular cone in terms of the volume and the altitude. 17. Deduce a formula for finding the altitude of a circular cone in terms of the volume and the radius of the base. 18. Deduce a formula for finding the radius of the base of a circular cone in terms of the volume and the altitude. 19. Deduce a formula for finding the volume of a cone of revolution in terms of the slant height and the radius of the base. 20. Deduce formulas for finding the slant heighl and the altitude of a cone of revolution in terms of the volume and the radius of the base. 372 BOOK VII. SOLID GEOMETRY Proposition XXIX. Theorem 614. The lateral areas, or the total areas, of two sim- ilar cones of revolution are to each other as the squares of their altitudes, as the squares of their radii, or as the squares of their slant heights ; and their volumes are to each other as the cubes of their altitudes, as the cubes of their radii, or as the cubes of their slant heights, Given two similar cones of revolution, with lateral areas / and /', total areas t and V, volumes v and v' y altitudes a and a', radii r and r\ and slant heights s and s' respectively. To prove that I: V = t: t r = a 2 : a' 2 = r 2 : r 1 ' 2 = s 2 : s' 2 , and that v : v' = a z : a' z = r s : r' s = s 8 : s'\ Proof. - - = - = 4^ * § § 282 > 269 a r s s + r I irrs r s r 2 s 2 a 2 Z 7 = = ir7V == ? X ^'"V 2 ^^^^ 2 " §610 Jt ~ ^'(3' + r') ~ ? * s' + r' " r' 2 ~7 2 ~~ a' 2 ' * v I irr 2 (i r 2 a r 3 a s s s . _ _ . _. v' ±7rr' 2 a' r' 2 a' r' 3 a ,z s' 3 §§ 613 and 614, like §§ 591 and 592, are occasionally demanded in college entrance examinations. They are not needed for any exercises and they may be omitted without destroying the sequence. « ONES 373 Proposition XXX. Theorem 615. Tin lateral area of a frustum of a cone of revo lution is equal to half the sum of the circumferences <>/' its bases multiplied by the slant height. Given a frustum of a cone of revolution, with lateral area /, circumferences of bases c and c\ and slant height s. T>> 'prove that I = \ (c + c')s. Proof. Suppose a frustum of it regular pyramid circum- scribed about the frustum of the cone, as a pyramid is cir- cumscribed about a cone. Lei the lateral area of the circumscribed frustum be /'_. and let p and p' be the perimeters of the basi corresponding to c and c' respectively. The slant height is s } the same as that of the frustum of the cone. Then /' =i, <>+/). v. §554 If the number of lateral faces of the circumscribed frustum is indefinitely increased, what limits do V and p -\-p' approach '.' Therefore what limit does l( j> + />'<■■< approach ? What conclusion may be drawn, as in § 587 '.' ( !omplete the proof. 616. Corollary. The lateral area of a frustum of a cone of revolution is equal to the circumference of a section equi- distant from it* bases multiplied by its slant height. How can it 1"' proved thai J (c + <■') equals the circumference of this section ? How are the radii related '.' 374 BOOK VII. SOLID GEOMETKY Proposition XXXI. Theorem 617. A frustum of a circular cone is equivalent to the sum of three cones ivhose common altitude is the altitude of the frustum and ivhose bases are the loiver base, the upper base, and the mean proportional between the bases of the frustum. Given a frustum of a circular cone, with volume v, bases &and b\ and altitude a. To prove that v = 1 a (b + V -f y/W). Proof. Suppose a frustum of a pyramid with base a regular polygon to be inscribed in the frustum of the cone, as a pyramid is inscribed in a cone. Let v' be the volume, and let x and x' be the bases corre- sponding to b and b' respectively. The altitude is a, the same as that of the frustum of the cone. Then v' = £ a(x + x' + Vxx'). § 565 If the number of lateral faces of the inscribed frustum is in- definitely increased, what limits do v', x, x', and xx' approach ? Therefore what limit does \ a(x -\- x' -\- -Vxx') approach ? What conclusion may be drawn ? Complete the proof. 618. Corollary. In a frustum of a cone of revolution, r and r' being the radii of the bases, v = ^ 7ra(r 2 + r' 2 + rr'). For b = 7rr 2 , b' = 7rr' 2 . .-. Vw — VVr 2 x 7rr' 2 = irrr'. CONES 375 EXERCISE 96 Find the lateral areas of frustums of cones, given the cir- cumferences of the f>asr* •nnl th,' slant heights respectively as follows : 1. r = 4 in., c' = 3 in., s = 0.5 in. 2. c = (') in., <■' = 5 in., s = 1.4 in. 3. r = T\m., c' = 5| in., s = 2| in. 4. c = 23 in., c' = 18 in., s = 1 (J in. .Fmd to /"■" decimal places the volumes of frustums of <• <■■ ,-. #///■. // £fo altitudes ami the areas of the bases respectively as follow* : 5. a = 3 in., b = 4^ sq. in., b' = 2 sq. in. 6. n — 4 in., /> = S 1 , sq. in., b' = ."> sq. in. 7. t/ = 5.], in., 5 = 16 sq. in., &' = 9 sq. in. 8. a = 6 in., b = 17 sq. in., // = 1 1 sq. in. iwW to few decimal places the volumes of frustums of cones of revolution, given the altitudes and the radii of the baSi 8 /•< spectively as follows : 9. a = 4 in., r = 3 in., r' = 2 in. 10. a = 5 in., r = 3^ in., /•' = 2J in. 11. a = G in., r = 3.7 in., r' = 3.1 in. 12. a = 7.V in., r = 4£ in., r' = 3£ in. 13. Deduce a formula for finding the altitude of a frustum of a circular cone in terms of the volume and the areas of the bases. 14. Deduce a formula for finding the altitude of a frustum of a cone of revolution in terms of tin- volume and the radii of the bases. 376 BOOK VII. SOLID GEOMETRY EXERCISE 97 Industrial Problems 1. There is a rule for calculating the strongest beam that can be cut from a cylindric log, as follows : Erect perpendiculars MD and NB on opposite sides of a diameter A C, at the trisection points M a and N, meeting the circle in D and B. Then A BCD is a section of the beam. Calculate the dimensions, the log being 16 in. in diameter. 2. A cylindric funnel for a steamboat is 4 ft. 3 in. in diam- eter. It is built up of four plates in girth, and the lap of each joint is 1| in. Find one dimension of each plate. 3. A tubular boiler has 124 tubes each 3| in. in diameter and 18 ft. long. Required the total tube surface. Answer to the nearest square foot. 4. A room in a factory is heated by steam pipes. There are 235 ft. of 2-inch pipe and 26 ft. 3 in. of 3-inch pipe, besides 2 ft. 8 in. of 4|-inch feed pipe. Required the total heating surface. Answer to the nearest square foot. 5. A triangular plate of wrought iron | in. thick is 2 ft. 7 in. on each side. If the weight of a plate 1 ft. square and I in. thick is 5 lb., find to the nearest pound the weight of the given triangular plate. 6. The water surface of an upright cylindric boiler is 2 ft. 8 in. below the top of the boiler, and is 12.57 sq. ft. in area. What is the volume of the steam space ? 7. A cylinder 16 in. in diameter is required to hold 50 gal. of water. What must be its height, to the nearest tenth of an inch, allowing 231 cu. in. to the gallon ? 8. How many square feet of tin are required to make a funnel, if the diameters of the top and bottom are 30 in. and 15 in. respectively, and the height is 25 in.? EXK1M ISES 377 holes 9. Find to two decimal places the weight of a steel plate 4 ft. by 3 it. 2 in. by If in., allowing 190 Lb. per cubic foot. 10. A steel plat.- for a steamship is 5 ft. long, 3 ft. 6 in. wide, and \ in. thick. A porthole 10 in. in diameter is cut through the plate. Required the weight of the finished p] allowing 0.29 lb. per cubic inch. Ajaswer. to two decimal pla 11. A cast-iron base for a column i- in the form of a frus- tum of a pyramid, the lower base being a square 2 ft. on a side, and the upper base having a fourth of the area of the lower base. The altitude of the frustum is «.» in. Required tin- weight to the nearest pound, allowing 460 lb. per cubic foot. 12. A cylinder head for a steam engine has the shape shown in the figure, where the dimensions in niches are: " = 12. b = 3, c = 2, cZ = 6,e=3,/=£,^= |,andA= There are six f-inch holes for bob Compute the weight of the plate, allowing 41 lb. for the weight of a steel plate 1 ft. square and 1 in. thick. Answer to the nearest tenth of a pound. 13. A steel beam L0 in. by 5 in., in the form here shown, is is ft. long. The thickness of the beam is § in. and th.« average thickness of the flanges is § in. Find the weight of the beam to the nearest pound, allow- ing 0.29 lb. per cubic inch. 14. A hollow steel shaft 12 ft. long is L8 in. in exterior diameter and 8 in. in interior diameter, hind tin- weight to the nearest pound, all. .win- 0.29 lb. per cubic inch. 15. Find the expense, a1 7<> cents a square foot, of polishing the curved surface of a marble column in the shape of the & turn of a right circular cone whose slant height is 12 ft. and the radii of whose bases are 3 ft. 6 in. and 2 ft. 4 in. respectively. -t.( 378 BOOK VII. SOLID GEOMETRY EXERCISE 98 Miscellaneous Problems 1. The slant height of the frustum of a regular pyramid is 25 ft., and the sides of its square bases are 54 ft. and 24 ft. respectively. Find the volume. 2. If the bases of the frustum of a pyramid are regular hexagons whose sides are 1 ft. and 2 ft. respectively, and the volume of the frustum is 12 cu. ft., find the altitude. 3. From a right circular cone whose slant height is 30 ft., and the circumference of whose base is 10 ft., there is cut off by a plane parallel to the base a cone whose slant height is 6 ft. Find the lateral area and the volume of the frustum. 4. Find the difference between the volume of the frustum of a pyramid whose altitude is 9 ft. and whose bases are squares, 8 ft. and 6 ft. respectively on a side, and the volume of a prism of the same altitude whose base is a section of the frustum parallel to its bases and equidistant from them. 5. A Dutch stone windmill in the shape of the frustum of a right cone is 12 meters high. The outer diameters at the bottom and the top are 16 meters and 12 meters, the inner diameters 12 meters and 10 meters. How many cubic meters of stone were required to build it ? 6. The chimney of a factory has the shape of a frustum of a regular pyramid. Its height is 180 ft., and its upper and lower bases are squares whose sides are 10 ft. and 16 ft. respectively. The flue throughout is a square whose side is 7 ft. How many cubic feet of material does the chimney contain ? 7. Two right triangles with bases 15 in. and 21 in., and with hypotenuses 25 in. and 35 in. respectively, revolve about their third sides. Find the ratio of the total areas of the solids generated and find their volumes. EXERCISFS 379 EXERCISE 99 Equivalent Solids 1. A cube each edge of which is 12 in. is transformed into a right prism whose base is a rectangle K'> in. long and 12 in. wide. Find the height of the prism and the difference between its total area and the total area of the cube. 2. The dimensions of a rectangular parallelepiped are a, b } c. Find the height of an equivalent right circular cylinder, having a for the radius of its base; the height of an equivalent right circular cone having a for the radius of its base. 3. A regular pyramid 12 ft. high is transformed into a regu- lar prism with amequivalent base. Find the height of the prism. 4. The diameter of a cylinder is 14 ft. and its height s ft. Find the height of an equivalent righl prism, the base of which is a square with a side 4 ft. long. 5. If one edge of a cube is e, whal is the height A of an equivalent right circular cylinder whose radius is ,■'.' 6. The heights of two equivalent right circular cylinders are in the ratio 4:9. If the diameter of the first is 6 ft., what is the diameter of the second *.' 7. A right circular cylinder 6 ft. in diameter is equivalent to a right circular cone 7 ft. in diameter. If the height of the cone is 8 ft., what is the height of the cylinder '! 8. The frustum of a regular pyramid ft. high lias for bases squares 5 ft. and 8 ft. on a side. Find the height of an equiva- lent regular pyramid whose base is a square 1- ft. on a side. 9. The frustum of a cone of revolution is 5 ft. high and the diameters of its bases are 2 ft. and 3 ft. respectively. Find the height of an equivalent right circular cylinder whose base is equal in area to the section of the frustum made by a plane parallel to the bases and equidistant from them. 380 BOOK VII. SOLID GEOMETRY EXERCISE 100 Review Questions 1. Define polyhedron. Is a cylinder a polyhedron ? 2. Define prism, and classify prisms according to their bases. 3. How is the lateral area of a prism computed ? Is the method the same for right as for oblique prisms ? 4. Define parallelepiped ; rectangular parallelepiped ; cube. Is a rectangular parallelepiped always a cube ? Is a cube always a rectangular parallelepiped ? 5. Distinguish between equivalent and congruent solids. Are two cubes with the same altitudes always equivalent ? always congruent? Is this true for parallelepipeds? 6. What are the conditions of congruence of two prisms ? of two right prisms ? of two cubes ? 7. The opposite angles of a parallelogram are equal. What is a corresponding proposition concerning parallelepipeds ? 8. How do you find the volume of a parallelepiped ? What is the corresponding proposition in plane geometry ? 9. How do you find the volume of a prism ? of a cylinder ? of a pyramid ? of a cone ? 10. Define pyramid. How many bases has a pyramid ? Is there any kind of a pyramid in which more than one face may be taken as the base ? 11. How do you find the lateral area of a pyramid ? of a right cone ? of a frustum of a pyramid ? of a frustum of a right cone ? 12. How many regular convex polyhedrons are possible ? What are their names ? 13. Given the radius of the base and the altitude of a cone of revolution, how do you find the volume ? the lateral area ? the total area ? BOOK VIII THE SPHERE 619. Sphere. A solid bounded by a surface all points of which are equidistant from a point within is called a sphere. The point within, from which all points on the surface are equally distant, is called tin' center. The surface is called the spherical surface, and sometimes the sphere. Half of a sphere is called a hemisphere. The terms radius and diameU r are used as in the case of a circle 620. Generation of a Spherical Surface. By the definition of sphere it appears that a spherical surface may be generated by the revolution of a semicircle about its diameter as an axis. Thus, if the semicircle ACB revolves about AB, a spherical surface is generated. It is therefore assumed that a sj>htr< may bv described with una 'ilc n point as 'i center and any given line as a radius. m 621. Equality of Radii and Diameters. It follows that: All radii of the same sphere >>>■<■ equal, and "II diameters of t/,r same sphere are equal. Equal spheres have equal radii, and spheres having equal radii "/■<■ > qual. 382 BOOK VIII. SOLID GEOMETRY Proposition I. Theorem 622. Every intersection of a spherical surface by a plane is a circle. Given a sphere with center 0, and ABD any section of its surface made by a plane. To prove that the section ABD is a circle. Proof. Draw the radii OA, OB, to any two points A, B, in the section, and draw OC _L to the plane of the section. Then in A OCA and OCB, A OCA and OCB are rt. A, § 430 OC is common, and OA = OB. § 621 .'.A OCA is congruent to A OCB. § 89 .*. CA = CB. § 67 .*. any points A and B, and hence all points, in the section are equidistant from C, and ABD is a O, by §159. q. e.d. 623. Corollary 1. The line joining the center of a sphere and the center of a circle of the sphere is perpendicular to the plane of the circle. 624. Corollary 2. Circles of a sphere made by planes equidistant from the center are equal; and of two circles made by planes not equidistant from the center the one made by the plane nearer the center is the greater. PLANE SECTIONS 383 625. Great Circle. The intersection of a spherical surface by a plane passing through the miter is called a §reat circle of the sphere. 626. Small Circle. The intersection of a spherical surface by a plane which does nol pass through the center is called a small circle of the sphere. 627. Poles of a Circle. If a diameter of a sphere is perpen- dicular to the plane of a circle of the sphere the extremities are railed the jioles of the circle. 628. Corollary 1. Parallel circles have flw same poles. 629. Corollary 2. ^47/ great circles of a sphere are equal. 630. (iirollary 3. Every great circle bisects the spherical surface. 631. Corollary 4. Two great circles bisect each other. The intersection of the planes passes through what point ? 632. Corollary 5. If the planes of two great circles are perpendicular, each circle passes through the poles of the other. Draw the figure and state the reason. 633. Corollary 6. Through two given points on the sur- face of a sphere a a '//-<■ of a great circle wag always be drawn. Do these two points, together with the center of the sphere, generally determine a plane? Consider the special case in which the two points are ends of a diameter. 634. Corollary 7. Through three given points on the sur- face of a sphere one circle and "///// one can be tlnurn. H<>w many points determine a plane ? 635. Spherical Distance. The length of the smaller arc of the great circle joining two points on the surface of a sphere is called the spherical distance between the points, or, where no confusion is likely to arise, simply the distant 384 BOOK VIII. SOLID GEOMETRY Proposition II. Theorem 636. The spherical distances of all points on a circle of a sphere from either pole of the circle are equal. . ;;||j|||j||||i! Given P, P\ the poles of the Circle ABC, and A, 2?, C, any points on the circle. To prove that the great-circle arcs PA, PB, PC are equal. Proof. The straight lines PA, PB, PC are equal. § 439 Therefore the arcs PA, PB, PC are equal, by § 172. q.e. d. In like manner, the great-circle arcs P'A, P'B, P'C may be proved equal. 637. Polar Distance. The spherical distance from the nearer pole of a circle to any point on the circle is called the polar distance of the circle. The spherical distance of a great circle from either of its poles may be taken as the polar distance of the circle. 638. Quadrant. One fourth of a great circle is called a quadrant. 639. Corollary 1. The polar distance of a great circle is a quadrant. 640. Corollary 2. The straight lines joining points on a circle to either pole of the circle are equal. PLANE SECTIONS AND TANGENT PLANKS 385 Proposition III. Theorem 641. .1 point on a sphere, which is at the distance of a quadrant from each of two other points, not the ex- tremities of a diameter, is a poh of th great circL passing through thesi points. Given a point P on a sphere, PA and PB quadrants, and ABC the j^reat circle passing through A and B. To {>/•">■<■ that P is the poL of OABC. Proof. What kind of angles an.* tin- .J A<>P and BOP? How is PO related to the plane of OABC ? Does this prove that P is the pole of OABC ? 642. Describing Circles on a Sphere. This proposition proves that we may describe a great circle on a sphere of a given radius so that it shall pass through two given points. Open the compasses the length of chord PA = Vr- + r 2 = r\2. 643. Tangent Lines and Planes. A line or plane that has one point and only one point in common with a sphere, however far produced, is said to be tangent to the sphere, and the sphere to be tangent to the line or plane. 644. Tangent Spheres. Two spheres whose surfaces have one point and only one point in common are said to be tangent. 386 BOOK VIII. SOLID GEOMETRY Proposition IV. Theorem 645. A 'plane perpendicular to a radius at its extrem- ity is tangent to the sphere. Given the plane MN perpendicular to the radius OA at A. To prove that MN is tangent to the sphere. Proof. Let P be any point except A in MN. Then which is longer, OP or OA, and why ? Therefore, is P inside, on, or outside the sphere, and why ? What does this tell us concerning all points, except A, onMN? How, then, do we know that MN is tangent to the sphere ? 646. Corollary. A plane tangent to a sphere is perpen- dicular to the radius drawn to the point of contact. What are the proposition and corollary of plane geometry corre- sponding to §§ 645 and 646 ? Do they suggest the proof of this corollary ? 647. Inscribed Sphere. If a sphere is tangent to all the faces of a polyhedron, it is said to be inscribed in the polyhedron, and the polyhedron to be circumscribed about the sphere. 648. Circumscribed Sphere. If all the vertices of a polyhedron lie on a spherical surface, the sphere is said to be circumscribed about the polyhedron, and the polyhedron to be inscribed in the sphere. PLANE SECTIONS AND TANGENT PLANES 387 Proposition V*. Theorem 649. A sphere may be inscribed in any given tetra- hedron. A Given the tetrahedron ABCD. To prove that a sphere may be inscribed in ABCD. Proof. Bisect the dihedral A at the edges AB, BC, and CA by the planes OAB, OBC, and OCA respectively. Every point in the plane OAB is equidistant from the fai ABC andABD. § 479 Pot a like reason every point in the plane OB< ' is equidistant from the faces ABC and BBC ; and every point in the plane OCA is equidistant from the faces ABC and ADC. Therefore the point 0, the common intersection of these three planes, is equidistant from the four faces of the tetra- hedron and is the center of the sphere inscribed in the tetra- hedron, by § 647. Q e.d. Discussion. What is the corresponding proposition in plane geom* - try ? Is the line of proof similar ? It is shown in plane geometry that the three Lines which bisect the three angles of a triangle meet in a point. What is the corresponding proposition with reference to planes in a tetrahedron ".* Is it substan- tially proved in this proposition ? It is proved in plane geometry that a circle may be inscribed in what kind of a polygon ? What corresponding proposition may be interred in solid geometry ? 388 BOOK VIII. SOLID GEOMETRY Proposition VI. Theorem 650. A sphere may be circumscribed about any given tetrahedron. Given the tetrahedron ABCD. To prove that a sphere may be circumscribed about ABCD. Proof. Let P, Q respectively be the centers of the circles circumscribed about the faces ABC, ABD. Let PR be _L to the face ABC, and QS _L to the face ABD. Then PR is the locus of a point equidistant from A, B, C, and QS is the locus of a point equidistant from A, B, D. § 442 Therefore PR and QS lie in the same plane, the plane J_ to AB at its mid-point. § 443 If QS were II to PR, it would be _L to the face ABC. § 445 But this is impossible, for QS is J_ to the face ABD which intersects the face ABC. Given Since Pit and QS cannot be II, and since they lie in the same plane, they must therefore meet at some point 0. .' . is equidistant from A, B, C, and D, and is the center of the required sphere, by § G48. q. e. d. 651. Corollary. Through four points not in the same plane one spherical surface and only one can be passed. The center of any sphere whose surface passes through the four points must be in the planes mentioned in the proof, and since there is only one point of intersection, there can be only one sphere. PLANE SECTIONS AND TANGENT PLANES 389 Proposition VII. Theorem 652. The intersection of two spherical surfaces is a circle whose plane is perpendicular to the line which joins the centers of th spheres and whose center is in that line. Given two intersecting spherical surfaces, with centers O and 0' . To prove that the spherical surfaces intersect in a circle whose plane is /» rpendicular to <><)', and whose center is in 00'. Proof. Let the two great circles formed by any plane through and 0' intersect in .1 and B. Then 00' is a _L bisector of AB. § 195 If this plane revolves about 00', the circles generate the spherical surfaces, and .4 describes their line of intersection. But during the revolution AC remains constant in length and perpendicular to 00 % . Therefore A generates a circle with center C, whose plane is perpendicular to 00', by § 432. Q-e.d. 653. Spherical Angle. The opening between two great-circle arcs that intersect is called a spherical angle. A spherical angle is considered equal to the plane angle formed by the tangente to the arcs at their point of intersection. Draw a figure Illustrating this definition. In elementary geometry we do not consider angles formed by arcs of small circles. 390 BOOK VIII. SOLID GEOMETRY EXERCISE 101 1. The four perpendiculars erected at the centers of the circles circumscribed about the faces of a tetrahedron meet in the same point. 2. The six planes perpendicular to the edges of a tetra- hedron at their mid-points intersect in the same point. 3. The six planes which bisect the six dihedral angles of a tetrahedron intersect in the same point. 4. Circles on the same sphere having equal polar distances are equal. 5. Equal circles on the same sphere have equal polar dis- tances. 6. Eind the locus of a point in a plane at a given distance from a given point. Also of a point in a three-dimensional space. 7. A line tangent to a great circle of a sphere lies in the plane tangent to the sphere at the point of contact. 8. An}- line in a tangent plane drawn through the point of contact is tangent to the sphere at that point. 9. One plane and only one plane can be passed through a given point on a given sphere tangent to the sphere. 10. Eind a point in a plane equidistant from two intersecting lines in the plane, and at a given distance from a given point not in the plane. Discuss the solution. 11. How many points determine a straight line? a circle? a spherical surface ? Prove that two spherical surfaces coin- cide if they have this number of points in common. 12. If two planes which intersect in the line AB touch a sphere at the points C and D respectively, the line CD is perpendicular to AB in the sense mentioned in the discussion under § 450, — that a plane can be passed through CD per- pendicular to AB. PLANE SECTIONS AND TANGENT FLANKS 391 Proposition VIII. Theobem 654. A spherical angle is measured Inj the arc of the great circle described from Its vertex as a pole and included between its sides, produced if necessary. Given PA and PB, arcs of great circles intersecting at P ; PA 1 and PB\ the tangents to these arcs at P; AB, the arc of the great circle described from P as a pole and included between PA and PB. To p rove that the spherical A A I'll is measured by arc AB. §185 §213 §05 Proof. In the plane POD, PB 1 is _L to PO, and OB is _L to PO. .'. PB' is I! to OB. Similarly PA 1 is II to OA. .'.ZA'PJJ' = ZAOB. § h">l But /. A OB is measured by arc AB. §213 .\Z A'PB' is measured by arc -I /•'. ,\Z -1/v; is measured by arc .1 B, by § 653. q.e.d. 655. Corollary 1. A spherical angle has the same meas- ure as the dihedral angle formed by the planes of the two circles. 656. Corollary 2. All arcs of great circles drawn through the pole of a given great circle arc j>> rpendicular to the given circle. 392 BOOK VIII. SOLID GEOMETRY 657. Spherical Polygon. A portion of a spherical surface bounded by three or more arcs of great circles is called a spherical polygon. The bounding arcs are called the sides of the polygon, the angles between the sides are called the angles of the polygon, and the points of intersection of the sides are called the vertices of the polygon. 658. Relation of Polygons to Polyhedral Angles. The planes of the sides of a spherical polygon form a polyhedral angle whose vertex is the center of the sphere, whose face angles are measured by the sides of the polygon, and whose dihedral angles have the same numerical measure as the angles of the polygon. Thus the planes of the sides of the polygon ABCD form the polyhedral angle O-ABCD. The face angles BOA, COB, and so on, are measured by the sides AB, BC, and so on, of the polygon. The dihedral angle whose edge is OA has the same measure as the spherical angle BAD, and so on. Hence from any property of polyhedral angles ive may infer an analogous property of spherical polygons ; and conversely. 659. Convex Spherical Polygon. If a polyhedral angle at the center of a sphere is convex (§ 491), the corresponding spherical polygon is said to be convex. Every spherical polygon is assumed to be convex unless the contrary is stated. 660. Diagonal. An arc of a great circle joining two non- consecutive vertices of a spherical polygon is called a diagonal. 661. Spherical Triangle. A spherical polygon of three sides is called a spherical triangle. A spherical triangle may be right, obtuse, or acute. It may also be equilateral, isosceles, or scalene. 662. Congruent Spherical Polygons. If two spherical polygons can be applied, one to the other, so as to coincide, they are said to be congruent SPHERICAL POLYGONS 393 Proposition IX. Theorem 663. Each side of a spherical triangle is less than the sum of the other two sides. Given a spherical triangle ABC, CA being the longest side. To prove that CA ints .1 './;'.' '. the two spherical tri- angles ABC and A'B'C'siTe called sym- metric spherical triangles. In the same way we may Eorm two sym- metric polygons of any number of sidi . Having thus formed the symmetric polygons, we may place them in any position we cho upon the surface of the sphere. 672. Relation of Symmetric Triangles. Two symmetric tri- angles are mutually equilateral and mutually equiangular: yel in general they are not congruent, for the}' cannot be made to coincide by superposition. If in the above figure the triangle ABC is made to slide on the surface of the sphere until the rtex A falls on A', it is evident that the two triangles cannot be made to coincide for the reason that the corresponding parts of the triangles occur in reverse ordi To try to make two symmetric spherical polygons coincide is very much like trying to put the right-hand glove on the left hand. The rela- tion of two symmetric spherical triangles may be illustrated by cutting them out of the peel of an orange or an apple. 673. Symmetric Isosceles Triangles. If, however, we have two symmetric triangles ABC and A'B'C' such that AB = ACj and A'B' = A'C' } that is, if the two sym- metric triangles are isosceles i then because AB, AC } A'B', A'C are all equal and the angles A and A' are equal, being originally formed by vertical dihedral angles i § 671), the B ir two triangles can be made to coincide Therefore, If two symmetric spherical triangles are isosceles^ they "re superpasable and tft congruent. 400 BOOK VIII. SOLID GEOMETRY Proposition XIV. Theorem 674. Tioo symmetric spherical triangles are equivalent. Given two symmetric spherical triangles ABC, A'B'C y having their corresponding vertices opposite each to each with respect to the center of the sphere. To prove that the triangles ABC, A'B'C are equivalent. Proof. Let P be the pole of a small circle passing through the points A, B, C, and let POP' be a diameter. Draw the great-circle arcsP.4, PB, PC, P'A', P'B', P'C'. Then PA=PB = PC. §636 Now P'A' = PA, P'B' = PB, P'C = PC. § 672 .'. P'A' = P'B' = P'C. Ax. 8 .'. the two symmetric A PC A and P'C A' are isosceles. .'. A PC A is congruent to A P'C 'A'. § 673 Similarly . A PAB is congruent to A P'A'B', and A PBC is congruent to A P'B'C. Now AABC = A PC A + A PAB -f A PBC, and A A 'B'C = A P'C A' + A P'A'B' + A P'B'C. Ax. 11 .'.A ABC is equivalent to A A'B'C, by Ax. 9. q.e.d. Discussion. If the pole P should fall without the A ABC, then V would fall without A A'B'C, and each triangle would be equivalent to the sum of two symmetric isosceles triangles diminished by the third ; so that the result would be the same as before. SPHERICAL POLYGONS 401 Proposition XV. Theorem 675. Two triangles on the sarru sphere or on equal spheres an either congruent or symmetric if two sides and the included angle of the one <.V> .'. the A are either congruent or symmetric, by § , ''7<">. q.e.d. Discussion. In the figures the parts are arranged in the same order, so that the triangles are congruent. They might be arranged as in the figures of § 676. I>iscuss the proposition when the triangles are equilateral and each side is a quadrant. Discuss the proposition when two sides of each triangle air quadrants. What is the corresponding proposition in plane geometry, and why does not the form of proof there given hold here ? 404 BOOK VIII. SOLID GEOMETRY Proposition XVIII. Theorem 678. Two mutually equiangular triangles on the same spliere or on equal spheres are mutually equilateral, and are either congruent or symmetric. Given two mutually equiangular spherical triangles T and T' on equal spheres. To prove that T and T are mutually equilateral, and are either congruent or symmetric. Proof. Let the AP be the polar triangle of A T, and the AP r be the polar triangle of A T'. Since the A T and T' are mutually equiangular, Given . * . the polar A P and P' are mutually equilateral. § 667 .". the polar AP and P' are mutually equiangular. § 677 But .the A T and T' are the polar A of A Pand P'. § 666 .". the A T and T' are mutually equilateral. § 667 Therefore the A T and V are either congruent or symmetric, by §677. Q.e.d. Discussion. The statement that mutually equiangular spherical tri- angles are mutually equilateral, and are either congruent or symmetric, is true only when they are on the same sphere or on equal spheres. When the spheres are unequal, the spherical triangles are unequal. In this case, however, their sides have the same arc measure, and therefore have the same ratio as the circumferences or as the radii of the spheres (§ 382). SPHERICAL POLYGONS 405 Proposition XIX. Theorem 679. In an isosceles spherical triangle the angles op- posite the equal sides are equal. Given the spherical triangle ABC, with AB equal to AC. To prove that ZB = ZC. Proof. Draw the arc AD of a great circle, from the vertex A to the mid-point of the base BC. Then AABD and ACD are mutually equilateral. .'. A ABD and A CD are mutually equiangular. § 677 .'.AB = /-C. Q.e.d. EXERCISE 104 1. The radius of a sphere is 4 in. From any point on the sur- face as a pole a circle is described upon the sphere with an open- ing of the compasses equal to 3 in. Find the area of this circle. 2. The edge of a regular tetrahedron is is»- the vertical angle, is perpendicular to the base, and divides the triangle into two svmmetric triangles. 406 BOOK VIII. SOLID GEOMETRY Proposition XX. Theorem 680. If two angles of a spherical triangle are equal, the sides opposite these angles are equal and the tri- angle is isosceles. Given the spherical triangle ABC, with angle B equal to angle C. To prove that AC ' = AB. Proof. Let A A'B'C be the polar triangle of A ABC. Since ZB = ZC, .-. A'C = A'B'. . §667 .■.ZB' = ZC. §679 .'.AC = AB j by § 667. Q.e.d. EXERCISE 105 1. To bisect a given spherical angle. 2. To construct a spherical triangle, given two sides and the included angle. 3. To construct a spherical triangle, given two angles and the included side. 4. To construct a spherical triangle, given the three sides. 5. To construct a spherical triangle, given the three angles. 6. To pass a plane tangent to a given sphere at a given point on the surface of the sphere. 7. To pass a plane tangent to a given sphere through a given straight line without the sphere. SPHERICAL POLYGONS 407 Proposition XXI. Theorem 681. If two angles of a sph rical triangle are urn qual, the sides i>j>i><>.<<'it< ihesi angles are unequal, and the sidi opposite the greater angh is the greaii r; and if two sides (//■>■ unequal, the angles opposite these sides are unequal, and the angl opposite th greater side is the greater. Given the triangle ABC, with angle C greater than angle B. To prove that AB > AC. Proof. Draw the arc CD of a great circle, making A D CB equal to A B. Then DB = DC. § G80 Now AD + DC> AC. §663 .'. AD+ DB>AC, or AB>AC, by Ax. 9. q.e.d. Given the triangle ABC, with AB greater than AC. To prove that Z C is greater than Z /;. Proof. The Z C must be equal to, less than, or greater than the Z /;. If AC = AB, th.n AB = AC; § 680 and if Z C is less than Z /?, then ABAC + BC. §663 Take away AF from the left member of the inequality, and its equal A C from the right member. Then BF>BC. Ax. 6 Therefore BF>BG, the equal of BC. Ax. 9 Hence F lies outside the circle whose pole is B, and the arcs DCF and GCE have only the point C in common. SPHERICAL POLYGONS 409 Now let A DEB be any line from A to B on the surface of the sphere, which does not pass through C. This line will cut the arcs DCF and GCE in separate points D and E\ and if we revolve the line AD about A as a iixed point until D coincides with C, we shall have a line from A to C equal to the line A D. In like manner, we can draw a line from B to C equal to the line BE. Therefore a line can be drawn from A to B through C that is equal to the sum of the lines AD and BE, and hence is less than the line A DEB by the line DE. Therefore no line which does not pass through C can be the shortest line from A to B. Therefore the shortest line from A to B passes through C. But C is any point in the arc A B. Therefore the shortest line from A to B passes through every point of the arc AB, and consequently coincides with the arc AB. Therefore the shortest line from A to B is the great-circle arc A B. q.e.d. EXERCISE 106 1. The three medians of a spherical triangle are concurrent. 2. To construct with a given radius a spherical surface that passes through three given points. 3. To construct with a given radius a spherical surface that passes through two given points and is tangent to a given plane. 4. To construct with a given radius a spherical surface that passes through two given points and is tangent to a given sphere. 5. The smallest circle on a given sphere whose plane passes through a given point within the sphere is the circle whose plane is perpendicular to the radius through the given point. 410 BOOK VIII. SOLID GEOMETRY 683. Zone. A portion of a spherical surface included be- tween two parallel planes is called a zone. Thus on the earth we have the torrid zone included between the planes of the tropics of Cancer and Capricorn. The circles made by the planes are called the bases of the zone, and the distance between the planes is called the altitude of the zone. If one of the planes is tangent to the sphere and the other plane cuts the sphere, the zone is called a zone of one base. If both planes are tangent to the sphere, the zone is a complete spherical surface. 684. Generation of a Zone. If a great circle revolves about its diameter as an axis, any arc of the circle generates a zone. Thus, in the figure of § 683, if the great circle PACQ revolves about its diameter PQ as an axis, the arc AC generates the zone A J), of which the altitude is the distance between the parallel planes. Similarly, the arc AP generates the zone ABP, and the arc CQ generates the zone CDQ, these both being zones of one base. 685. Lune. A portion of a spherical surface bounded by the halves of two great circles is called a lune. Thus PAQB is a lune. A lune is evidently generated by the partial or complete revolution of half of a great circle about its diameter as an axis. 686. Angle of a Lune. The angle between the semicircles bounding a lune is called the angle of the lune. Thus Z APB is the angle of the lune PA QB. A lune is usually taken as having an angle less than a straight angle. This is not necessary, for we may consider a hemispherical surface as a lune with an angle of 180°. We may also conceive of lunes with angles greater than a straight angle, and we may even think of an entire spherical surface as a lune whose angle is 360°. MEASUREMENT OF SPHERICAL SURFACES 411 Proposition XXIII. Theorem 687. The area of the surface generated by a straight Urn n wiving about an axis in its plane is equal to tin product of the projection of the line on the axis by tin circle ivhose radius is a perpendicular erected at tin midrpomt of the line ami terminated by the axis. A 31 X I C O "TZ)"*" 1 \ ■-I b. i 1 1 \ -I L_l. C OR i __L_ D Given an axis XY about which a line AB in the same plane with XY revolves, M being the mid-point of AB, CD being the projec- tion of AB on XY, MO being perpendicular to XY, MR being per- pendicular to AB, and a being the area generated by AB. To prove that a = CD x 2 ttMR. Proof. 1. If AB is II to IF, CD = AB, MR coincides with MO, and a is the lateral area of a right cylinder. § 588 2. If AB is not II to A'}', and does not cut AT, a is the lateral area of the frustum of a cone of revolution. .-. a = AB x 2 it MO. Draw A E II to XY. The A AEB and MOR are similar. .-. MO: AE = MR : AB. .'. AB X MO = AE X MR, or AB X MO = CD x .1//,'. Substituting, a = CD x 2 77.1//,'. 3. If A lies in the axis XY, then AE and CD coincide, and a = CD x 2ttMB, by § 609. q.e.d. §616 §290 § 282 §261 Ax. 9 412 BOOK VIII. SOLID GEOMETRY Proposition XXIV. Theorem 688. The area of the surface of a sphere is equal to the product of the diameter by the circumference of a great circle. E D' B' a Given a sphere generated by the semicircle ABCDE revolving about the diameter AE as an axis, 5 being the area of the surface, r being the radius, and d being the diameter. To prove that s=2 irrd. Proof. Inscribe in the semicircle half of a regular polygon having an even number of sides, as ABCDE. From the center O draw _k to the chords AB, BC, CD, DE. These J§ bisect the chords (§174) and are equal. § 178 Let I denote the length of each of these _k From B, C, and D drop perpendiculars to AE. Then area of surface generated by AB = AB' x 2 nrl, § 687 area of surface generated by BC = B'O x 2 ttI, etc. .*. area of surface generated by ABCDE = AE x 2 irl Ax. 1 = 2 irld. Ax. 9 Denote the area of the surface generated by ABCDE by s', and let the number of sides of ABCDE be indefinitely increased. Then s' approaches s as a limit, / approaches r as a limit, and consequently 2 irld approaches 2 irrd as^ limit. But s' = 2 irld, always. • '.s = 2Trrd, by §207. §377 §687 Q.E.D. MEASUREMENT OF SPHERICAL SURFACES 413 689. Corollary 1. The area of the surface of " sphere is equivalent to the area of four great circles, or to 4 irr 1 . In s = 2-rrrd, what is the value of d in terms of r ? Then what is the value of s in terms of r? Fur example, if the radius is 10 in., the area of the surface of the sphere is 4 7T- 100 sq. in., or 1266.64 sq. in. 690. ( Orollary 2. The areas of the surfaces of two spheres are to each other as the squares on their radii, or as the squares on their diameters. If the radii are r and r', the diameters d and d', and the surfa s and $', then what is the ratio of .s to s', according to § 680 ? Show that this also equals r 2 : r' 2 , and d- : d" 1 . 691. Corollary 3. The area of a zone is equal to the product of the altitude by the circumference of a great circle. If we apply the reasoning of § G88 to the zone generated by the revo- lution of the arc BCD, we obtain the area of zone BCD = B'T/ x 2 7rr, where B'T/ is the altitude of the zone and 2 7n- the circumference of a great circle. For example, if the radius is 10 in., and the altitude is 5 in., the area of the zone is 5 • 27r • 10 sq. in., or 314.16 sq. in. 692. Corollary 4. The area of a zone of one base is equiv- alent to the area of a circle whose radius is the chord of th>- generating arc. The arc AB generates a zone of one base. .-. the area of the zone AB = AB' x 2 -nr = ttAB' x AE. But AB' x AE = AB 2 . § 298 .-. the area of the zone AB = irAB". 693. Spherical Excess of a Triangle. The excess of the sum of the angles of a spherical triangle over 180° is called the spin rim! excess of the triangle. For example, if the angles of a spherical triangle are 80°, 90°, and 100°, the spherical excess of the triangle is 90°. 414 BOOK VIII. SOLID GEOMETRY Proposition XXV. Theorem 694. The area of a lime is to the area of the surface of the sphere as the angle of the lune is to four right angles. Given a lune PAQB, the great circle ABCD whose pole is P, a the value in degrees of the angle of the lune, / the area of the lune, and s the area of the surface of the sphere. To prove that I : s = a : 4 rt. A. Proof. The arc AB measures the Z a of the lune. § 654 Hence arc AB : circle ABCD = a : 4 rt. A. § 212 If AB and ABCD are commensurable, let their common meas- ure be contained m times in AB, and n times in ABCD. Then arc AB : circle ABCD = m : n. . ' . a : 4 rt. A = m : n. Pass an arc of a great circle through the poles P and Q and each point of division of ABCD. These arcs will divide the entire surface into n equal lunes, of which the lune PA QB will contain m. .'. 1: s = m : n. . • . I : s = a : 4 rt. A. Ax. 8 If AB and ABCD are incommensurable, the theorem can be proved by the method of limits as in § 472. q.e.d. MEASUREMENT OF SPHERICAL SURFACES 415 EXERCISE 107 T~*hiiy it = 3.1416 for all examples in this exercise, find the areas of spheres whose radii are as follows: 1. 2 in. 3. 3J in. 5. 2 ft. 1 in. 7. 48.8 in. 2. 7 in. 4. 5§ in. 6. 3 ft. G in. 8. 4000 mi. Find tlic radii of spheres whose areas are a* follows: 9. 12.5664 sq. in. 11. 1 sq. ft. 13. s. 10. 50.2656 sq. in. 12. 100 tt sq. in. 14. 4 7T 3 . On a sphere whose radius is 20 in., find the areas of zones whoSi piienlent to a lime whose angle is the angle between the arcs. 697. Computation of Area. To illustrate the computation in- volved in §695, findthearea of a triangle whose angles are 110°, 100°, and 95°, on the surface of a sphere whose radius is G in. Spherical excess = 110° + 100° + 95° - 180 3 = 125°. .-. angle of lime = <;•_>' . 02 1 .-. area of lime = — - of the spherical surface. 360 .-. area of lime = — - x 4 x 3.1416 x 36 sq. in. 360 .-. area of triangle = 78.54 sq. in. 698. Spherical Excess of a Polygon. The excess of the sum of the angles of a spherical polygon of n sides over (n — 2) x 180° is called the spherical excess of the polygon. EXERCISE 108 Compute the areas of triangles on spheres of the given diam- eters, the angles being as follows: 1. 100°, 120°, 140°, il =16 in. 4. 115°, 124°, 85°, d = 30 in. 2. 105°, 130°, 125°, d =10 in. 5. 135°, 110°, 92°, d = 40 in. 3. 127°, 132°, 90°, d = 20 in. 6. 148°, 93°, 68°, d = 25.8 in. 7. 115° 27' 30", 102° 32' 48", 08° 27 ; 39", d = 8000 mi. Compute the areas of triangles on spheres of the given radii, the angles being as follows: 8. 120°, 100°, 90°, r = 9 in. 11. 115°, 102°, 30°, r = 36 in. 9. 130°, 90°, 80°, r=10 in. 12. 1 10°, 120°, 85°, r = 90in. (10. 105°, 75°, (35°, r = 18 in. 13. 130°, 117°, 93°, r = 1.8 in. 418 BOOK VIII. SOLID GEOMETRY' Compute the areas of triangles on spheres of the given cir- cumferences, the angles being as folloivs : 14. 93°, 94°, 120°, c = 31.416 in. 15. 82°, 105°, 98°, c = 62.832 in. 16. 148°, 27°, 125°, c = 15.708 in. 17. 162°, 39°, 120°, c = 78.54 in. 18. 149°, 41°, 116°, c = 39.27 in. 19. 126° 30' 42", 105° 26' 15", 63° 15' 3", c = 314.16 in. 20. What is the area of a triangle on the earth's surface the vertices of which are the north pole and two points on the equator, one at 37° W. and the other at 16° E., the earth being considered a sphere with a radius of 4000 mi. ? 21. If the radii of two spheres are 6 in. and 4 in. respec- tively, and the distance between the centers is 5 in., what is the area of the circle of intersection of the spheres ? 22. Find the radius of the circle determined on a sphere of 5 in. diameter by a plane 1 in. from the center. 23. If the radii of two concentric spheres are r and r', and if a plane is passed tangent to the interior sphere, what is the area of the section made in the other sphere ? 24. Two points A and B are 8 in. apart. Find the locus in space of a point 5 in. from A and 7 in. from B. 25. Two points A and B are 10 in. apart. Find the locus in space of a point 7 in. from A and 3 in. from B. 26. The radii of two parallel sections of the same sphere are a and b respectively, and the distance between the sections is d. Find the radius of the sphere. 27. The diameter of a certain sphere is V2. The chords of the arcs that form the sides of a triangle on the surface of the sphere are respectively 1,1, and £ V2. Find the area of the spherical triangle. MEASUREMENT OF SPHERICAL SURFACES 419 Proposition XXVII. Theorem 699. A spin rical polygon is > quivalent to a lune ivhose angle is half tin spJierical excess of the polygon. Given a spherical polygon P of n sides, the sum of the angles being s. To prove that P is equivalent to a lune whose angle is ^( s _^T^2xl80°). Proof. Draw all the diagonals from any vertex. Since there is a distinct triangle for each side except those meeting at the vertex chosen, there are (n — 2) triangles. Since each triangle is equivalent to a lune whose angle is half the excess of the sum of its angles over 180°, § 695 therefore the (?i — 2) triangles are equivalent to a lune whose angle is half the excess of the sum of all the angles of the polygon over (n — 2) X 180°. .*. P = a lune whose angle is J(s — n — 2 x ISO ). Q. B. d. 700. Computation of Area. Find the area of a spherical poly- gon whose angles are 100°, 110°, 120°, and 170°, r bring 6 in. Spherical excess = 100° + 110° + 120° + 170° - 2 x 180° = 140°. -, i angle of lune =: 70 area of lune = jfc of 4 irr' 2 = ^ of 4 x 3.1410 x 30 sq. in. = 87.9048 sq. in. 420 BOOK VIII. SOLID GEOMETRY EXERCISE 109 Find the areas of spherical polygons on spheres of the given areas, the angles being as follows : 1. 30°, 90°, 120°, 130°, a =2 sq. ft. 2. 45°, 60°, 100°, 165°, a = 288 sq. in. 3. 70°, 168°, 92°, 120°, a = 500 sq. in. 4. 68° 30 r , 149° 50', 96° 54', 136° 52', a = 750 sq. in. 5. 122°27'40", 130°32'50", 98°31'30", 96° 48', a = 600sq.in. 6. 132°, 96°, 154°, 120°, 150°, a = 3 sq. ft. 120 sq. in. 7. 130°, 156°, 172°, 95°, 120°, 100°, a = 157.2 sq. in. Find the areas of spherical polygons on spheres of the given radii, the angles being as follows : 8. 130°, 150°, 80°, 90°, r = 10 in. 9. 148°, 157°, 90°, 100°, 120°, r = 20 in. 10. 172°, 169°, 86°, 141°, 100°, 90°, r = 24 in. 11. 135° 30', 148° 42', 96° 37', 102° 11', r = 10 in. Find the areas of spherical polygons on spheres of the given diameters, the angles being as follows: 12. 148°, 92°, 60°, 120°, d = 10 in. 13. 172°, 168°, 93°, 37°, 100°, ^ = 22 in. 14. 102°, 162°, 139°, 141°, 138°, 126°, d = 20 in. 15. 82°50'42", 120° 29' 18", 98°37'15", 141°22'45", d = 20 in. Find the areas of spherical polygons on spheres of the given circumferences, the angles being as follows : 16. 39°, 148°, 172°, 168°, c = 3.1416 in. 17. 128°, 92°, 168°, 109°, c = 31.416 in. 18. 146°, 129°, 102°, 137°, 100°, c = 6.2832 in. 19. 128°, 145°, 139°, 82°, 161°, 137°, c = 18.8496 in. MEASUREMENT OF SPHERICAL SOLIDS 421 701. Spherical Pyramid. A portion of a sphere bounded by a spherical polygon and the planes of its sides is called a spherical pyramid. The center of the sphere is called the vert< x of the spherical pyramid, and the spherical polygon is called the base. Thus O-ABC is a spherical pyramid. 702. Spherical Sector. A portion of a sphere generated by the revolution of a circular sector about any diameter of the circle of which the sector is a part is called a spherical sector. Thus if the sector A OB revolves about the diameter MX as an axis, it generates the spherical sector AB-O-A'B'. The zone generated by the arc of the generating sector is called the base of the spherical sector. 703. Spherical Segment. A portion of a sphere contained between two parallel planes is called a spherical segment. The sections of the sphere made by the parallel planes are called the bases of the spherical segment, and the distance between these bases is .ailed the altitude of the spherical segment. If one of the parallel planes is tangent to the sphere, the segment is called a spherical segment of one base. A spherical segment of one base may be generated by the revolution of a circular segment about the diameter perpendicular to its bi 704. Spherical Wedge. A portion of a sphere bounded by a lune and the planes of two great circles is called a spherical wedge. 422 BOOK VIII. SOLID GEOMETRY Proposition XXVIII. Theorem 705. The volume of a sphere is equal to the product of the area of its surface by one third of its radius. Given a sphere of radius r, area of surface s, volume z;, and center 0. v = sx\r. To prove that Proof. We may imagine a cube of edge 2 r circumscribed about the sphere. Connect O with each of the vertices of this cube. These connecting lines are the edges of six pyramids whose bases are the faces of the cube and whose altitudes all equal r. The volume of each pyramid is a face of the cube multiplied by ^r, and the volume of the six pyramids, or of the whole cube, is the area of the surface of the cube multiplied by ^ r. Xow imagine planes drawn tangent to the sphere, at the points where the edges of the pyramids cut its surface. We then have a circumscribed solid whose volume is nearer that of the sphere than is the volume of the circumscribed cube, but is still greater than the sphere. Ax. 11 Proceeding as before, connect with the vertices of the new polyhedron. These connecting lines are the edges of pyramids whose bases are together equal to the bases of the polyhedron and whose common altitude is r. § 646 MEASUREMENT OF SPHERICAL SOLIDS 423 Then the sum of the volumes of these pyramids is again the area of the surface of the polyhedron multiplied by £ r. De- noting this volume by v' and the area of the surface by s', we have v' = s' x J r. • If we continue to draw tangent planes to the sphere, we con- tinue to diminish the circumscribed solid. By continuing this process indefinitely we can make the difference between the volume of the sphere and the volume of the circumscribed solid less than any assigned positive quantity, however small, the difference between the surface of the sphere and the surface of the circumscribed solid becoming and remaining less than any assigned value, however small. .\ v is the limit of v', and s is the limit of s'. § 204 And since it has been shown that v' = s r X% i', always, ,\v = sx£ r, by § 207. q.e.d. 706. < < .rollary 1. The volume of a sphere of radius r and diann ter d is equal to |- 7T/* 3 or 1 ird^. For in v = s x | r what is the value of s in terms of r ? What is the value of d in terms of r '? Then what is the value of v in terms of d ? 707. Corollary 2. The volumes of two spheres are to each other as the cubes of their radii. What is the ratio of f 7TT 3 to f irr' z ? By the same reasoning, the volumes are to each other as the cubes of the diameters. 708. Corollary 3. The volume of a spherical sector is equal to one third the product of the area of the zone which forms its base multiplied 1>ij the rad'mx of the sphere. Suppose the base divided into spherical triangles. The planes deter- mined by their vertices are the bases of triangular pyramids with ver- tices at O. What is the limit of the sum of the volumes of these pyramids as the bases decrease in size ? 424 BOOK VIII. SOLID GEOMETRY EXERCISE 110 Problems of Computation Find the volumes of spheres whose radii are : 1. 3 in. 4. 2i in. 7. 20.7 ft. 2. 5 in. 5. 4| in. 8. 2 ft. 3 in. 3. 7 in. 6. 9$ in. 9. 4000 mi. Find the volumes of spheres whose diameters are :. 10. 24 in. 13. 2.8 in. 16. 2 ft. 1 in. 11. 36 in. 14. 3.4 in. 17. 3 ft. 4 in. 12. 48 in. 15. 4.5 in. 18. 8 ft. 6 in. Find the volumes of spheres whose circumferences are : 19. 6.2832 in. 20. 12.5664 in. 21. 18.8496 in. Find the volumes of spheres whose surface areas are : 22. 12.5664 sq. in. 23. 50.2656 sq. in. 24. 113.0976 sq. in. Find the radii of spheres whose volumes are : . 25. 4.1888 cu. in. 26. 33.5104 cu. in. 27. 113.0976 cu. in. 28. The circumference of a hemispherical dome is 66 ft. How many square feet of lead are required to cover it ? 29. If the ball on the top of St. Paul's Cathedral in London is 6 ft. in diameter, how much would it cost to gild it at 9 cents per square inch ? 30. The dihedral angles made by the faces of a spherical pyramid are 80°, 100°, 120°, and 150°, and the length of a lateral edge is 42 ft. Pind the area of the base. 31. The dihedral angles made by the faces of a spherical pyramid are 60°, 80°, and 100°, and the area of the base is 4 7r sq. ft. Pind the radius. EXERCISES 425 32. What is the area of the surface of the earth ? Assume that the earth is a sphere with a radius of 4000 mi., and make the same assumption in subsequent examples relating to the earth. 33. The altitude of the torrid zone is 3200 mi. Find its area. 34. What is the area of the north temperate zone if its altitude is 1800 mi. ? 35. Find the number of square miles of the earth's surface that can be seen from an aeroplane 1500 ft. above the surface. 36. How far in one direction can a man see from the deck of an ocean steamer if his eye is 40 ft. above the water? 37. To what height must a man be raised above the earth in order to see one sixth of its surface ? 38. How much of the earth's surface w^ould a man see if he were raised to the height of the radius above it ? 39. If the atmosphere extends 50 mi. above the surface of the earth, find the volume of the atmosphere. 40. If an iron ball 4 in. in diameter weighs 9 lb., find the weight of a spherical iron shell 2 in. thick, the external diame- ter being 20 in. 41. What is the angle of a spherical wedge if its volume is li cu. ft. and the volume of the entire sphere is 8?- cu. ft. ? 42. The inside of a washbasin is in the shape of the segment of a sphere. The distance across the top is 16 in. and its greatest depth is 8 in. How many pints of water will it hold, allowing 7 gal." to the cubic foot ? 43. Prove that the volume of a spherical pyramid is equal to the product of the base by one third of the radius, and find the volume if the base is one eighth of the surface of a sphere of radius 10 in. 44. Find the volume of a spherical sector whose base is a zone of area .-.. the radius of the sphere being >■, following a process of reasoning similar to that in § 705. 426 BOOK VIII. SOLID GEOMETKY EXERCISE 111 Formulas 1. Find the area z of the zone of a sphere of radius r, illu- minated by a lamp placed at the height h above the surface. 2. Find the volume v of a sphere in terms of c, the cir- cumference. 3. Find the radius r of a sphere in terms of v, the volume. 4. Find the diameter d of a sphere in terms of s, the area of the surface. 5. Find the circumference c of a sphere in terms of s, the area of the surface. 6. What is the altitude a of a zone, if its area is z and the volume of the sphere is v ? 7. Show that in a spherical pyramid v = ^ br. Find r in terms of v and b ; also b in terms of v and r. 8. Find a formula for the volume of the metal in a spher- ical iron shell, the inside radius being r and the thickness of the metal being t. 9. Find a formula for the weight of a spherical shell, the inside radius being r, the thickness of the metal being t, and the weight of a cubic unit of metal being w. 10. If the area of a zone z equals 2 irra (§ 691), find a for- mula for a in terms of z and r. 11. If the area of a zone is expressed by the formula z = 2 irra, what is the diameter of the sphere upon which a zone z has an altitude a ? 12. Find the area z of a zone of altitude a on a sphere whose area of surface is 5. 13. Find a formula for the area a of that part of the surface of a sphere of radius r seen from a point at a distance d above the surface. EXERCISES 427 EXERCISE 112 Problems of Loci Find the locus of a point : 1. At a given distance from a given point. 2. At a given distance from a given straight line. 3. At a given distance from a given plane. 4. At a given distance from a given cylindric surface. 5. At a given distance from a given spherical surface. 6. Equidistant from two given points. 7. Equidistant from two given planes. 8. At a given distance from a given point and at another given distance from a given straight line. 9. At a given distance from a given point and at another iven distance from a given plane. 10. At a given distance from a given point and equidistant from two other given points. 11. At a given distance from a given point and equidistant from two given planes. Find one or more points : 12. At a distance d 1 from a given point, at a distance d 2 from a given straight line, and at a distance and A < ' F. hut they have the same altitude, since their vertices E and />' are in the line EB which is parallel to the plane of their bases. .'. pyramid E-CFD = pyramid B-ACF. § 558 But the pyramid B-A CF may be regarded as having the base ABC and the vertex F; that is, as pyramid F-ABC. Therefore the truncated triangular prism ABC-DEF is equivalent to the sum of the three pyramids E-ABC, D-ABC, and F-ABC. q.e.d. 712. Corollary 1. The volume of a truncated right tri- angular prism is equal to the product of its base by one third the sum of its lateral edges. For the lateral edges DA, EB, FC (Fig. 1), being perpendicular to the base ABC, are the altitudes of the three pyramids whose sum is equivalent to the truncated prism. It is interesting to consider the spe- cial case in which ADEF is parallel to A ABC. .713. Corollary 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. For the right section DEF (Fig. 2) divides the truncated prism into two truncated right prisms. 434 APPENDIX TO SOLID GEOMETRY Proposition II. Theorem 714. The volumes of two tetrahedrons that have a trihedral angle of the one equal to a trihedral angle of the other are to each other as the products of the three edges of these trihedral angles. Given the two tetrahedrons S-ABC and S'-A'B'C, having the trihedral angles S and S' equal, v and v' denoting the volumes. rr + -l , v SAxSBxSC To prove that — = — — - F v' S'A' x S'B' x S'C Proof. Place the tetrahedron S-ABC upon S'-A'B'C so that the trihedral Z S shall coincide with the equal trihedral Z S ! . Draw CD and CD' J_ to the plane S'A'B', and let their plane intersect S'A'B' in S'DD'. The faces S'AB and S'A'B' may be taken as the bases, and CD, C'D' as the altitudes, of the triangular pyramids C-S'AB and C-S'A'B' respectively. Then But and v 1 S ' AB x CD v' ~ ' S'A'B' x CD' S'AB S'A'B' CD S'AB CD X S'A'B' " CD' S'A X S'B S'A' X S'B'' S'C CD' S'C ? i n i §562 §332 §282 S'A X S'B X S'C S'A' X S'B' X S'C S'A' X S'B' X S'C SAxSBxSC , . ft — ; j by Ax. 9. Q.E.D. POLYHEDRONS Proposition III. Theorem i:;;> 715. In any polylinlron the number of edges increased by two is equal to the number of vertices increased by the number of faces. Given the polyhedron AG, e denoting the number of edges, v the number of vertices, and / the number of faces. To prove that e + 2 = v +/. Proof. Beginning with one face BCGF, we have e — v. Annex a second face A B CD by applying one of its edges to a corresponding edge of the first face, and there is formed a surface of two faces having one edge BC and two vertices B and C common to the two faces. Therefore for two faces e = v -f-1. Annex a third face ABFE, adjoining each of the first two faces. This face will have two edges AB, BF and three ver- tices A, B, F in common with the surface already formed. Therefore for three faces e = v + 2. In like manner, for four faces, e = c + 3, and so on. Therefore for (/- 1) faces e = v + (/- 2). But/— 1 is the number of faces of the polyhedron when only one face is lacking, and the addition of this face will not increase the number of edges or vertices. Hence for / faces e= v+f—2, ore + 2= v+f. q.e.d. This theorem is due to the gnat Swiss mathematician, Euler. 436 APPENDIX TO SOLID GEOMETKY Proposition IV. Theorem 716. The sum of the face angles of any polyhedron is equal to four right angles taken as many times, less two, as the polyhedron has vertices. Given the polyhedron P, e denoting the number of edges, v the number of vertices, / the number of faces, and 5 the sum of the face angles. To prove that s = (y — 2) 4 rt. A. Proof. Since e denotes the number of edges, 2 e will denote the number of sides of the faces, considered as independent polygons, for each edge is common to two polygons. If an exterior angle is formed at each vertex of every poly- gon, the sum of the interior and exterior angles at each vertex is 2 rt. A ; and since there are 2 e vertices, the sum of the interior and exterior angles of all the faces is 2 e x 2 rt. A, or e x 4 rt. A. But the sum of the ext. A of each face is 4 rt. A. § 146 Therefore the sum of all the ext. A of / faces is / x 4 rt. A. Therefore s = e X 4 rt. A — f X 4 rt. A = (e _/) 4 rt. A. But e + 2 = v +/; ■ § 715 that is, e— f=v — 2. Ax. 2 Therefore s = (v — 2) 4 rt.. A. Q. e. d. POLYHEDRONS 437 EXERCISE 115 Find the volumes of truncated triangular prisms, given the fntsi's />, mill tin distances of the three vertices p, q, r from the planes of the bases, as follows: 1. h = 8 sq. in., p = ."> in., q = 4 in., r — 5 in. 2. 1) = ( .) sq. in., y> = (5 in., */ = 3 in., r = 4^ in. 3. A = 15 sq. in.. /- == 7 in., y = 9 in., r = 8.1 in. 4. 6 = 32 sq. in., y> = 9 in., <[ = VI in., v = 9.3 in. 5. // = 48 sq. in., /> = 16 in., «/ = 15 in., r = 18 in. 6. A triangular rod of iron is cut square off (i.e. in right section) at one end, and slanting at the othef end. The right section is an equilateral triangle li in. on a side. The edges of the rod are 3 ft. 2 in., 3 ft. 3 in., and 3 ft. 3 in. Find the weight of the rod, allowing 0.28 lb. per cubic inch. 7. Two triangular pyramids with a trihedral angle of the one equal to a trihedral angle of the other have the edges of these angles 3 in., 4 in., 3^ in., and 5 in., b\ in., 6 in. respec- tively. Find the ratio of the volumes. 8. Make a table giving the number of edges, vertices, and faces of each of the five regular polyhedrons, showing that in every case the number conforms to Euler's theorem (§ 715). 9. Make a table similar to that of Ex. 8, giving the sum of the face angles in each of the five regular polyhedrons, showing that in every case s = (v — 2) 4 rt. Zs (§ 716). 10. There can be no seven-edged polyhedron. 11. Can there be a nine-edged polyhedron ? 12. What is the sum of the face angles of a six-edged poly- hedron ? 13. What is the sum of the face angles of a polyhedron with five vertices? with four vert ices ? Consider the possi- bility of a polyhedron with three vertices. 438 APPENDIX TO SOLID GEOMETEY Proposition V. Theorem 717. Two similar polyhedrons can be separated into the same number of tetrahedrons similar each to each and similarly placed. Given two similar polyhedrons P and P 1 . To prove that P and P' can be separated into the same number of tetrahedrons similar each to each and similarly placed. Proof. Let G and G ' be corresponding vertices. Divide all the faces of P and P', except those which include the angles G and G', into corresponding triangles by drawing corresponding diagonals. Pass a plane through G and each diagonal of the faces of P ; also pass a plane through G' and each corresponding diagonal of P'. Any two corresponding tetrahedrons G-ABC and G'-A'B'C' have the faces ABC, GAB, GBC similar respectively to the faces A'B'C', G'A'B', G'B'C. § 292 AG _ AB_ _ AC_ __ BC_ GC_ 6 A'G'~ A'B'~ A'C'~ B'C' G'C 1 ' * .-. the face GAC is similar to the face G'A'C § 289 POLYHEDRONS 439 They also have the corresponding trihedral . s equal. § 498 .-. the tetrahedron G-ABC is similar to G'-A'B'C'. §710 If G-ABC and G'-A'B'C are removed, the polyhedrons re- maining continue similar ; for the oew faces GAC and G'A'C have just been proved similar, and the modified faces .1 GF&nd A'G'F', GCHatod G'C'H', are similar (§ 292); also the iii.»di!i.Ml polyhedral AG and G', .1 and A' } C and C" remain equal each to each, since the corresponding parts taken from these angles are equal. The process of removing similar tetrahedrons can be carried on until the polyhedrons are separated into the same number of tetrahedrons similar each to each and similarly placed, q. e.d. 718. Corollary 1. The corresponding edges of similar poly- hedrons are proportional. For the corresponding faces are similar. Therefore their correspond- ing sides are proportional (§ 282). 719. Corollary 2. Any two corresponding lines in two similar polyhedrons hare the same rutin as any two corre- sponding edges. For these lines may be shown to be sides of similar polygons, and hence § 282 applies. 720. Corollary 3. Two corresponding faces ,,f similar polyhedrons are proportional to the squares on ami two corre- sponding edges. For they are similar polyhedrons, and hence they are to each other as the squares on any two corresponding sides (§ 334). 721. Corollarv 4. The entire surf ires of two similar poly- hedrons are proportional to the squares on any two correspond ing edges. For the corresponding faces are proportional to the squares i l- I- vl = GB : G'B' , and so on. •'. >\+>- 2 +i\+---:, bases b and &', mid-section m, and altitude a. To prove that v = J a (b -f- V + 4 rri). Proof. If any lateral face is a trapezoid, divide it-into two triangles by a diagonal. Take any point P in the mid-section and join P to the vertices of the polyhedron and of the mid-section. Separate the prismatoid into pyramids which have their vertices at P, and for their respective bases the lower base b, the upper base b\ and the lateral faces of the prismatoid. The pyramid P-XAB, which we may call a lateral pyra- mid, is composed of the three pyramids P-XQR, P-QBR, and P-QAB. Now P-XQR may be regarded as having vertex X and base PQR, and P-QBR as having vertex B and base PQR. Hence the volume of P-XQR is equal to i a ■ PQR, and the volume of P-QBR is equal to £ a • PQR. § 559 POLYHEDRONS 443 The pyramids P-QAB and P-QBR have the same vertex P. The base QAB is twice the base QBR (§ 327), since the A QAB has its base AB twice the base QR of the A QBR (§ 13G), and these triangles have the same altitude (§ 724). Hence the pyramid P-QA B is equivalent to twice the pyramid p-qbr. i :><;:; Hence the volume of P-QAB is equal to § a • PQR. Therefore the volume of P-XAB, which is composed of P-XQR, P-QBR, and P-QAB, is equal to f. a PQR. In like manner, the volume of each lateral pyramid is equal to | a x the area of that part of the mid-section which is included within it ; and therefore the total volume of all these lateral pyramids is equal to f am. The volume of the pyramid with base b is ^ ab, and the volume of the pyramid with base V is \ ab'. § 559 Therefore v = \ a (Jj -f- V A- 4 in). q.e. d. EXERCISE 116 Deduce from the formula for the volume of a prinmatoi > I . r = l a (b -f- b' + 4 ni), the following formulas : 1. Cube, v = a 3 . 3. Pyramid, v = ^ ha. 2. Prism, v = ba. 4. Parallelepiped, r = ba. 5. Frustum of a pyramid, v = £ a-(fi + &' -f- v AA'). 6. A prismatoid lias an upper base 3 sq. in., a lower base 7 sq. in., an altitude 3 in., and a mid-section 4 sq. in. What is the volume ? 7. A wedge has for its base a rectangle / in. long and w in. wide. The cutting edge is e in. long, and is parallel to the base. The distance from e to the base is d in. Deduce a f< irmula for the volume of the wedge. Apply this formula to the case in which I = 6. w = 1, e = 5, d = 3. 444 APPENDIX TO SOLID GEOMETEY Proposition IX. Theorem * 726. The volume of a spherical segment is equal to the product of one half the sum of its bases by its altitude, increased bij the volume of a sphere having that altitude for its diameter. Given a spherical segment of volume v, generated by the revo- lution of ABQP about MN as an axis, r being the radius of the sphere, AP being represented by r XJ BQ by r 2 , and PQ by a. To prove that v = 1 a (yrr* -f- 7rr 2 2 ) + 1 ira z . Proof. We shall first find the volume of the spherical seg- ment with one base, generated by AMP. Area of zone AM— 2 irr ■ PM. § 691 . ' . volume of sector generated by OA M=±rx2irr- PM. §708 But the cone generated by OA P = \ irr} (r—PM ). § 611 .'. volume AMP = * r X.2 irr • PM- — i Trr*(r—PM). Ax. 2 But r* = PM x NP = PM (2 r - PM). § 297 .'. volume AMP = i r x 2 irr • PM = tt-PM ( i 7T • PM (2 r — PAQ (r — PM ) Ax. 9 2 In the same way, volume BMQ = it • Qilf (r — i QAf) .-. v = volume ^4ilfP — volume BMQ A = 7T • Pilf • r — 1 7T • 7W — 7T • (23/ • r + £ 7T • (2 AT = irr(PM 2 — QM 2 ) - | 7r(iW 8 - OM" 3 ). SPHERICAL SEGMENTS 445 But PM — QM = a. Given r.v = 7rra(PM+QM)-$ira(PM 2 +PMQM+QM*). Ax. 9 But a* = PM* — 2PM> QM+ QM 2 . Ax. 5 ... a * + 3PM> QM=PM +PM- QM+QM . Ax. 1 .-. r = irra ( PM + QM) - } tt" (a a + 3 PM • QM). Ax. 9 Furthermore (2 r - PM) PM = >- 2 , ami (2r-QM)QM=r|. §297 '. 2 r • PAf -f- 2 r • QM - PM - QM = rf + rl Ax. 1 ,\r>PM+rQM= T H 77 Axs. 1, 4 .. v== ,ra( — — + g --PM.QM = tt« ^-±^ + |* + PM- QM - f - PM- QM = |ffi (7T7'! 2 -f 7ivf) + I 7Ta* Q- E. D. EXERCISE 117 Find the volumes of sj)herical segments having bases b and b', and altitudes a, as follows : 1. b = ±,b' = 5,a = l. 4. b = 6, b'= 8, a = If 2. ft = 4, b' = 6, a = 1 -\. 5. b = 8, b'= 12, a = 2. 3. b = 5, 6' = 7, a = 2J. 6. 5 = 12, &'= 15, a = 3£. 7. b = 27 s<|. in., // = 32 sq. in., a = 2.33 in. Find the volumes of spherical segments ha ring radii of ba±> % r and r 2 , and altitudes a, as follows: 8. r x = 3, r a = 4, a = 2. 11. r t = 5, r 2 = 3, a = l£. 9. r, = 4, /•., = 7, a = 3. 12. r 1= G, r 2 = 5, a = 1 j . 10. r, = 8, r a = 5, a = 4£. 13. r 1= 9, r = 10, a = 2f . 14. /-j = 9 in., r 2 = 7 in., « = 4.75 in. 446 APPENDIX TO SOLID GEOMETRY EXERCISE 118 Examination Questions 1. A pyramid 6 ft. high is cut by a plane parallel to the base, the area of the section being | that of the base. How far from the vertex is the cutting plane ? 2. Find the area of a spherical triangle whose angles are 100°, 120°, and 140°, the diameter of the sphere being 16 in. 3. Two angles of a spherical triangle are 80° and 120°. Find the limits of the third angle, and prove that the greatest possible area of the triangle is four times the least possible area, the sphere on which it is drawn being given. 4. An irregular portion, less than half, of a material sphere is given. Show how the radius can be found, compasses and ruler being allowed. 5. Find the volume of a cone of revolution, the area of the total surface of which is 200 it sq. ft., and the altitude of which is 16 ft. 6. The volumes of two similar polyhedrons are 64 cu. ft. and 216 cu. ft. respectively. If the area of the surface of the first polyhedron is 112 sq. ft., find the area of the surface of the second polyhedron. 7. A solid sphere of metal of radius 12 in. is recast into a hollow sphere. If the cavity is spherical, of the same radius as the original sphere, find the thickness of the shell. 8. The stone spire of a church is a regular pyramid 50 ft. high on a hexagonal base each side of which is 10 ft. There is a hollow part which is also a regular pyramid 45 ft. high, on a hexagonal base of which each side is 9 ft. Find the number of cubic feet of stone in the spire. 9. The volumes of a hemisphere, right circular cone, and right circular cylinder are equal. Their bases are also equal, each being a circle of radius 10 in. Find the altitude of each. EXERCISES 447 10. A sphere of radius 5 ft. and a right circular cone also of radius 5ft. stand on a plane If the heighl of the cone is equal to a diameter of the sphere, find the position of the plane that cuts the two solids in equal circular sections. 11. The vert ices of one regular tel rahedron are at the centers of the faces of another regular tetrahedron. Find the ratio of the volumes. 12. Find the area of a spherical triangle, if the perimeter of its polar triangle is L".)7° and the radius of the sphere is 10 centimeters. 13. The radii of two spheres are 13 in. and 15 in. respec- tively, and the distance between the centers is 14 in. Find the volume of the solid common to both spheres, — a spherical lens. 14. The radius of the base of a right circular cylinder is r and the altitude of the cylinder is a. Find the radius and the volume of a sphere whose surface is equivalent to the lateral surface of the cylinder. 15. If the polyhedral angle at the vertex of a triangular pyramid is trirectangular, and the areas of the lateral faces are a, 5, and c respectively, and the area of the base is '/. then a 2 + V 1 + c 2 = d 2 . 16. If the earth is a sphere with a diameter of 8000 mi., find the area of the zone bounded by the parallels 30° north latitude and 30° south latitude Show that this zone and the planes of the circles include {£ of the volume of the earth. 17. The altitude of a cone of revolution is 12 centimeters and the radius of it-- base is 5 centimeters. Compute the radius of the sector of paper which, when rolled up, will just cover the convex surface of the cone, and compute the size of the central angle of this sector in degrees, minutes, and seconds. 18. The volume of any regular pyramid is equal to one third of its lateral area multiplied by the perpendicular dis- tance from the center of its base to any lateral face. 448 APPENDIX TO SOLID GEOMETRY 19. If the area of a zone of one base is n times the area of the circle which forms its base, the altitude of the zone is : (n — 1) times the diameter of the sphere. Discuss the special case when n = 1. 20. If the four sides of a spherical quadrilateral are equal, its diagonals are perpendicular to each other. 21. Find the volume of a pyramid whose base contains 30 square centimeters if one lateral edge is 5 centimeters and the angle formed by this edge and the plane of the base is 45°. 22. On the base of a right circular cone a hemisphere is constructed outside the cone. The surface of the hemisphere equals the surface of the cone. If r is the radius of the hemi- sphere, find the slant height of the cone, the inclination of the slant height to the base, and the volume of the entire solid. 23. Find the total, surface and the volume of a regular tetra- hedron whose edge equals 8 centimeters. 24. If a spherical quadrilateral is inscribed in a small circle, the sum of two opposite angles is equal to the sum of the other two angles. 25. By what number must the dimensions of a cylinder of revolution be multiplied to obtain a similar cylinder of revo- lution with surface n times that of the first ? with volume n times that of the first ? 26. A pyramid is cut by a plane parallel to the base midway between the vertex and the plane of the base. Compare the volumes of the entire pyramid and the pyramid cut off. 27. The height of a regular hexagonal pyramid is 36 ft. and one side of the base is 6 ft. What are the dimensions of a similar pyramid whose volume is J^ that of the first ? 28. One of the lateral edges of a pyramid is 4 meters. How far from the vertex will this edge be cut by a plane parallel to the base, which divides the pyramid into two equivalent parts ? RECREATIONS 449 727. Recreations of Geometry. The following simple puzzles and recreations of geometry may serve the double purpose of adding interest to the study of the subject and of leading the student to exercise greater care in his demonstrations. They have long been used for this purpose and are among the best known puzzles of geometry. EXERCISE 119 1. To prove that every triangle is isosceles. Let ABC be a A that is not isosceles. Take CP the bisector of ZACB, and ZP the ± bisector of AB. These lines must meet, as at P, for otherwise they would be II, which would require CP to be _L to AB, and this could happen only if A ABC were isosceles, which is not the case by hypothesis. From P draw PX _L to BC and PY ± to CA, and draw PA and PB. Then since ZP is the J_ bisector of AB, .-. PA = PB. And since CP is the bisector of Z ACB, .-. PA' = PY. .-. the rt. 4 PBX and PA Y are congruent, and BX = AY. But the rt. & PXC and PYC are also congruent, and .-. XC =YC. Adding, we have BX + AC = A Y + YC, or BC = A C. .-. A ABC is isosceles, even though constructed as not isosceles. 2. To prove that part of an angle equals the whole angle. Take a square ABCD, and draw MWP, the _L bisector of CD. Then A/ATP is also the _L bisector of AB. From B draw any line BX equal to AB. Draw BX and bisect it by the A XP. Since DA" intersects CD, Js to these lines can- not be parallel, but must meet as at P. Draw PA, PD, PC, PX. and PB. Since _VP is the A bisector of CD, PD = PC. Similarly PA = PB, and PD = PX. .-. PX = PD = PC. But BX = BC by construction, and PB is common to & PBX and PBC. .-. A PBX is congruent to A PB( '. and Z XBP = Z CBP. .-. the whole Z XBP equals its i art, the Z.CBP. -iX 450 APPENDIX TO SOLID GEOMETRY 3. To prove that part of an angle equals the whole angle. Take a right triangle ABC and con- struct upon the hypotenuse BC an equi- lateral triangle BCD, as shown. On CD lay off CP equal to CA. Through X, the mid-point of AB, draw PX to meet CB produced at Q. Draw QA. Draw the _L bisectors of QA and QP, as YO and ZO. These must meet at some point because they are ± to two intersecting lines. Draw OQ, OA, OP, and OC. Since is on the ± bisector of QA, .-. OQ = OA. Similarly OQ= OP, and .-. OA = OP. But CA = CP, by construction, and CO = CO. .-. A A OC is congruent to A POC, and ZACO=Z PCO. 4. To prove that part of a line equals the whole line. Take a triangle A BC and draw CP ± to AB. c From C draw CX, making Z ACX = Z B. Then A ABC and ACX are similar. .-. AABC:AACX= BC -.CX\ Furthermore A ABC: A ACX = AB : AX. ^2 or But and BC'-.CX =AB:AX, BC 2 :AB= CX 2 :AX. BC= AC + AB' - 2 AB ■ AP, CX 2 = AC 2 + AX 2 - 2 AX ■ AP. AC + AB--2AB- AP AC + AX' -2AX- AP AB or AC ;2 or AB ' /1X) — £ ^XJT •'• ZC 2 AX Zcr J -^B ■ AX AC AX AC 2 AX AX + AX-2AP. - AB. AC - AB • AX AB AX AB = AX. RECREATIONS 451 5. To show geometrically that 1=0. Take a square that is 8 units on a side, and cut it into three parts, A, B, C, as shown in the right-hand figure. Fit these parts together as in the left-hand figure. Now the square is 8 units on aside, and therefore contains 8 x 8, or 04, small squares, while the rec- tangle is 13 units long and 5 units high, and there- fore contains i i i H-+-+— ' — i- H + - + - I I I 5 x 13, or 65, -4—4--+— }-- H — -4— +-^--4—1— 4— I C| — I — I — h— \—A— A — \- i i i I i I i -H — 4- — | — f~J — I — I — I < I I I I small squares. But the two figures are each made up of A + B+ C i Ax.ll ), and therefore areequal(Ax.8). 65 = 64, and by subtracting 64 we have 1 = (Ax. 2). 6. To prove that any point on a line bisects it. Take any point P on AB. On AB construct an isosceles A ABC, having AC = BC; and draw PC. Then in AAPC and PBC, we have Z A = Z B, AC= BC, and PC = PC. Iiavin° r >\ ' ! \ » » t \ 1 t \ \ §74 / / / / t \ \ i \ i \ i \ Const. t t / \ \ A' ! Iden. ./' /; Three independent parts (that is, not merely the three angles) of one triangle are respectively equal to three parts of the other, and the tri- angles are congruent ; therefore AP = BP (§ 07). 7. To prove that it is possible to let fall two perpendiculars to a line from an external point. Take two intersecting (D with centers and C . Let one point of intersection be P. and draw the diameters PA and PD. Draw AB cutting the circumferences at B and C. Then draw PB and PC. Since Z. PC A is inscribed in a semicircle, it is a right amrle. In the same way, since ZIjP>P is inscribed in a semicircle, it also is '' a right angle. .-. PJJand PC are both L to AD. 452 APPENDIX TO SOLID GEOMETRY 8. To prove that if two opposite sides of a quadrilateral are equal the figure is an isosceles trapezoid. Given the quadrilateral ABCD, with BC = DA. To prove that AB is || to DC. Draw MO and NO, the ± bisectors of AB and CD, to meet at 0. If AB and DC are parallel, the proposition is already proved. If AB and DC are not parallel, then MO and NO will meet at 0, either inside or outside the figure. Let be supposed to be inside the figure. Draw OA, OB, 0(7, OD. Then since OM is the JL bisector of AB, .-. OA = OB. Similarly OD = OC. But DA is given equal to BC. .-. A A OD is congruent to ABOC, ZD0A = ZB0C. * rt. A OCN and ODiV are congruent, ZN0D = ZC0N. rt. A AM and Z>3/0 are consruent, Z.10AT = ZiY0£. .-. Z NOD + Z DCvl + Z A OM = Z CON + Z BOC + Z MOB, or Z NOM = Z MON = a st. Z. Therefore the line M ON is a straight line, and hence AB is II to DC. If the point is outside the quadrilateral, as in the second figure, the proof is substantially the same. Tor it can be easily shown that Z DON - Z DOA - Z A OM = Z NOC - Z BOC - Z MOB, which is possible only if ZD0N=ZD0M, or if ON lies along OM. But that the proposition is not true is evident from the third figure, in which BC = DA, but AB is not II to DC. and Also, and Similarly and HISTORY OF GEOMETRY 453 728. History of Geometry. The geometry of very ancient peoples was largely the mensuration of simple arras and volumes such as is taught to children in elementary arithmetic to-day. They learned how to tind the area of a rectangle, and in the oldest mathematical records that we have there is some discussion of triangles and of the volumes of solids. The earliest documents that we have, relating to geometry, come to us from Babylon and Egypt. Those from Babylon were written about 2000 B.C. on small clay tablets, some of them about the size of the hand, these tablets afterwards having been baked in the sun. They show that the Baby- lonians of that period knew something of land measures, and perhaps had advanced far enough to compute the area of a trapezoid. For the mensuration of the circle they later used, as did the early Hebrews, the value it = 3. The first definite knowledge that we have of Egyptian math- ematics comes to us from a manuscript copied on papyrus, a kind of paper used about the Mediterranean in early times. This copy was made by one Aah-mesu (The Moon-born), com- monly called Ahmes, who. probably nourished about 1700 b.c. The original from which he copied, written about 2300 b.c, has been lost, but the papyrus of Ahmes, written nearly four thousand years ago, is still preserved and is now in the British Museum. In this manuscript, which is devoted chiefly to frac- tions and to a crude algebra, is found some work on mensu- ration. Among the curious rules are the incorrect ones that the area of an isosceles triangle equals half the product of the base and one of the equal sides; and that the area of a trapezoid having bases h, b\ and nonparallel sides each equal to ". is I a(b-\-b'). One noteworthy advance appears however. Ahmes gives a rule for finding the area of a circle, substan- tially as follows: Multiply the square on the radius by (\f') 2 , which is equivalent to taking for it the value 3.1605. Long before the time of Ahmes, however, Egypt had a good working 454 APPENDIX TO SOLID GEOMETEY knowledge of practical geometry, as witness the building of the pyramids, the laying out of temples, and the digging of irrigation canals. Erom Egypt and possibly from Babylon geometry passed to the shores of Asia Minor and Greece. The scientific study of the subject begins with Thales, one of the Seven Wise Men of the Grecian civilization. Born at Miletus about 640 B.C., he died there in 548 B.C. He spent his early manhood as a merchant, accumulating the wealth that enabled him to spend his later years in study. He visited Egypt and is said to have learned such elements of geometry as were known there. He founded a school of mathematics and philosophy at Miletus, known as the Ionic School. How elementary the knowledge of geometry then was, may be understood from the fact that tradition attributes only about four propositions to Thales, substantially those given in §§ 60, 72, 74, and 215 of this book. The greatest pupil of Thales, and one of the most remark- able men of antiquity, Avas Pythagoras. Born probably on the island of Samos, just off the coast of Asia Minor, about the year 580 B.C., Pythagoras set forth as a young man to travel. He went to Miletus and studied under Thales, probably spent several years in study in Egypt, very likely went to Babylon, and possibly went even to India, since tradition asserts this and the nature of his work in mathematics confirms it. In later life he went to southern Italy, and there, at Crotona, in the southeastern part of the peninsula, he founded a school and established a secret society to propagate his doctrines. In geometry he is said to have been the first to demonstrate the proposition that the square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides (§ 337). The proposition was known before his time, at any rate for special cases, but he seems to have been the first to prove it. To him or to his school seems also to have been due the construction of the regular pentagon (§§ 397, 398) HISTORY OF GEOMETRY 455 ar.d of the five regular polyhedrons. The construction of the regular pentagon requires the dividing of a line in extreme and mean ratio (§ 311), and this problem is commonly assigned to the Pythagoreans, although it played an important part in Plato's school. Pythagoras is also said to have known thai six equilateral triangles, three regular hexagons, or four squari . «-an be placed about a point so as jusl to till the 360°, bul thai no other regular polygons can be so placed. To his school is also due the proof thai the sum of the angles of a triangle equals two right angles (§ 107), and the construction of at least one star-polygon, the star-pentagon, which became the badge of his fraternity. For two centuries after Pythagoras geometry passed through a period of discovery of propositions. The state of the science may be seen from the fact that (Enopides of Chios, who flourished about 465 B.C., showed how to let fall a perpendicu- lar to a line (§ 227), and how to construct an angle equal to a given angle (§ 232). A few years later, about 440 B.C., Hippoc- rates of Chios wrote the first Greek textbook on mathematics. He knew that the areas of circles are proportional to the squares on their radii, but was ignorant of the fact that equal central angles or equal inscribed angles intercept equal arcs. About 430 B.C. Antiphon and Bryson, two Greek teachers. worked on the mensuration of the circle. The former attempted to find the area by doubling the number of sides of a regular inscribed polygon, and the latter by doing the same for both in- scribed and circumscribed polygons. They thus substantially exhausted the area between the circle and the polygon, and hence this method was known as the Method of Exhaustions. During this period the great philosophic school of Plato 129-348 b.c.) flourished at Athens, and to this school is due the first systematic attempt to create exact definitions, axioms, and postulates, and to distinguish between elementary and higher geometry. At this time elementary geometry became 456 APPENDIX TO SOLID GEOMETRY limited to the use of the compasses and the unmarked straight- edge, which took from this domain the possibility of con- structing a square equivalent to a given circle ("squaring the circle"), of trisecting any given angle, and of constructing a cube with twice the volume of a given cube ("duplicating the cube "), these being the three most famous problems of antiquity. Plato and his school were interested in the so-called Pythagorean numbers, numbers that represent the three sides of a right triangle. Pythagoras had already given a rule to the effect that J (m 2 + l) 2 = m 2 + J (m 2 - l) 2 . The school of Plato found that [(|m) 2 + l] 2 =m 2 i[(|m) 2 -l] 2 . By giving various values to m, different numbers will be found such that the sum of the squares of two of them is equal to the square of the third. The first great textbook on geometry, and the most famous one that has ever appeared, was written by Euclid, who taught mathematics in the great university at Alexandria, Egypt, about 300 b.c. Alexandria was then practically a Greek city, having been named in honor of Alexander the Great, and being ruled by the Greeks. Euclid's work is known as the M Elements," and, as was the case with all ancient works, the leading divisions were called books, as is seen in the Bible and in such Latin writers as Csesar and Vergil. This is why we speak of the various books of geometry to-day. In this work Euclid placed all the leading propositions of plane geometry as then known, and arranged them in a logical order. Most subsequent geometries of any im- portance since his time have been based upon Euclid, improving the sequence, symbols, and wording as occasion demanded. Euclid did not give much solid geometry because not much was known then. It was to Archimedes (287-212 b.c), a famous mathematician of Syracuse, on the island of Sicily, that some of the most important propositions of solid geometry are due, particularly those relating to the sphere and cylinder. HISTORY OF GEOMETRY 457 He also showed how to find the approximate value of w by a method similar to the one we teach to-day (§404), proving that the real value lies between 3} and 3i f >. Tradition says that the sphere and cylinder were engraved upon his tomb. The Greeks contributed little more to elementary geometry, although Apollonius of Perga, who taught at Alexandria be- tween 250 and 200 n.c. wrote extensively on conic sect ions; and Heron of Alexandria, about the beginning of the Christian era, showed that the area of a triangle whose sides are »i. /,. ,-. equals V* (*■ — a) (n — b) (s — c), where s = ±(a+b + c) (see p. 211). The East did little for geometry, although contributing considerably to algebra. The first great Hindu writer was Aryabhatta, who was born in 47G a.d. He gave the very close approximation for 77-, expressed in modern notation as 3.1416. The Arabs, about the time of the Arabian Nights Tales (800 a.d.), did much for mathematics, translating the Greek authors into their own language and also bringing learning from India. Indeed, it is to them that modern Europe owes its first knowledge of Euclid. They contributed nothing of importance to geometry, however. Euclid was translated from the Arabic into Latin in the twelfth century, Greek manuscripts not being then at hand, or b.-ing neglected because of ignorance of the language. The leading translators were Athelhard of Bath (1120), an English monk who had learned Arabic in Spain or in Egypt; Gerhard of Cremona, an Italian monk; and Johannes Campanus, chap- lain to Pope Urban IV. In the Middle Ages in Europe nothing worthy of note was added to the geometry of the Greeks. The first edition of Euclid was printed in Latin in 1482, the first one in English appearing in 1570. Our symbols are modern, + and — first appearing in a German work in 1489; = in Recorded "Whet- >ne of Witte" in 1557; > and < in the works of Harriot (1560-1621); and x in a publication by Oughtred (1574-1660). 458 APPENDIX TO SOLID GEOMETRY 729. Areas of Solid Figures. The following are the more important areas of solid figures : Prism, l=ep (§512). Regular pyramid, l—'Lsp (§ 553). Frustum of regular pyramid, / — \ (jp +^')s (§ 554). Cylinder of revolution, I = ac '= 2 irra (§ 588). Cone of revolution, / = \sc = irrs (§ 609). Frustum of cone of revolution, / = i (c -f- c') s (§ 615). Sphere, s = 4 tt/- 2 (§ 689). Zone, 5 = 2 7rra (§ 691). Z,4 2 Z,4 360 90 Lune, s = — • 4 tt/- 2 = — — • 7T7- 2 (§ 694). 730. Volumes. The following are the more important volumes : Rectangular parallelepiped, liva (§ 534). Prism or cylinder, ba (§§ 539, 589). Pyramid or cone, \ba (§§ 561, 611). Frustum of pyramid or cone, \ a (p -\-b' -\- vW) (§§ 565, 617). Right-circular cylinder, irra (§ 590). Cone of revolution, \ iri^a (§ 612). Frustum of cone of revolution, \ 7r«(V 2 + r' 2 -f- rr 1 ) (§ 618). Prismatoid, J a (6 + &' + 4 m) (§ 725). Sphere, f vrr 3 = J ird 3 (§ 706). Spherical pyramid, \ br. Spherical sector, i zr (§ 708). Spherical segment, \ a (irrl -j- 7rr|) -+- £ 7ra 3 (§ 726). INDEX PAGl PA( Altitude of cone 362 Classes of polyhedral angles . 308 of cylinder ; >-">:; Concave polyhedral angle . 3 Cone 362 altitude of 362 axis Of :;,;; ' base of 362 circular 363 circumscribed 366 element of '■'>'''- frustum of 367 inscribed 366 lateral surface of . . . . 302 oblique : '> lV ■'• of revolution 363 right \ . . 363 slant height of cone of revo- lution '•'>'''■'> vertex of 362 (Ones 363 . 367 of frustum of cone . . . 367 of frustum of pyramid . . 338 of prism 317 of prismatoid 441 of pyramid 337 of spherical segment . .421 of zone 410 Angle, dihedral 293 of lune 410 polyhedral 308 spherical 389 tetrahedral 308 trihedral 308 Axis of circular cone .... 363 Base of cone 302 . 421 . 421 . 353 . 367 . 338 of spherical pyramid of spherical sector . . bs of cylinder .... of frustum of cone . . of frustum of pyramid . of prism 317 of spherical segment . . .421 of zone 41 1 > prisms classified as to . . 318 pyramids classified as to . 337 ( Circles, describing, on sphere ■ - ( !ircumscribed prism .... 356 pyramid : sphere 386 and frustums as limits . similar 370 Congruent solids 322 spherical polygons . . . 392 ( tonic surface 362 directrix <>t* 362 dement of '■'>''>- neratrix of 362 lower nappe of 362 upper nappe of 362 section ;;,;; > ( tonstruction of tangent plan to cones to cylinders 400 INDEX PAGE Convex polyhedral angle . . 308 polyhedron 317 spherical polygon .... 302 Cube 322 Cylinder 353 altitude of 353 as a limit 357 bases of 353 circular 354 circumscribed 356 inscribed 356 lateral surface of ... . 353 oblique 353 of revolution 354 right 353 right section of 357 section of 353 tangent plane to .... 356 Cylinders, similar 350 Cylindric surface 353 directrix of 353 element of 353 generatrix of 353 Describing circles on sphere . 385 Diagonal, spherical 302 Diameter of sphere 381 Dihedral angle 203 acute 203 edge of 203 faces of 203 obtuse 203 plane angle of 204 reflex 203 right 203 size of 203 straight 203 Dihedral angles, adjacent . . 203 complementary 203 conjugate 203 PAGE Dihedral angles, relation to plane angles 204 supplementary 203 vertical 203 Dimensions 320 Distance from a point to a plane 270 polar 384 spherical -. . 383 Dodecahedron, regular . . . 351 Edge of dihedral angle . . . 203. Edges of polyhedral angle . . 308 polyhedron 317 Element of conic surface . . 362 cylindric surface .... 353 Ellipse 363 Equal polyhedral angles . . 308 Equivalent solids 322 Euler's theorem 435 Excess, spherical, of poly- gon 417 of triangle 413 Face angle of polyhedral angle 308 Faces of dihedral angle . . . 203 polyhedral angle .... 308 polyhedron 317 polyhedrons classified as to 350 Foot of line 273 Frustum of cone 367 altitude of 367 bases of 367 lateral area of 367 slant height of 367 Frustum of pyramid .... 338 altitude of 338 lateral area of 338 lateral faces of 338 slant height of 338 INDEX 461 PAGE Generation of spherical surface 381 zone 410 Hemisphere 381 Bexahedron, regular .... 3">l Hyperbola 363 [cosahedron, regular .... 351 Inclination of a line .... 303 Inscribed polyhedron . . . . 386 prism 356 pyramid 366 sphere 386 Lateral ana of frustum . . . 338 prism 317 pyramid 337 edges of prism 317 pyramid 337 faces of frustum .... 338 prism 317 pyramids 337 surface of cone 3G2 cylinder 353 frustum of cone .... 367 Limit, cylinder as a ... . 357 Limits, cones and frustums as 367 Line and plane parallel . . . 282 foot of 273 inclination of 303 oblique 277 projection of 302 Lune 410 angle of 410 Mid-section of prismatoid . . 441 Nappes of conic surface . . 362 ( iblique line ^77 and right cylinders . . . 353 Octahedron, regular .... 351 PAGE Parabola 363 Parallel line and plane . . . 2S2 planes 285 Parallelepiped 322 rectangular 322 right 322 Perpendicular planes .... 293 to a plane -7"> Plane 273 angle of dihedral angle. . 2 ( .»1 determining 273 perpendicular to ... . 27"> Planes, intersection of . . . 273 parallel 285 perpendicular 293 postulate of 274 Point, projection of a ... 302 Polar distance 384 triangle . 394 Poles of a circle 383 Polygon, angles of 392 excess of 417 sides of 392 spherical 392 vertices of 392 Polyhedral angle 308 concave 308 convex 308 edges of 308 face angles of 308 faces of 308 parts of 308 size of 308 vertex of ....... 308 Polyhedral angles 308 classes of 308 equal 308 symmetric 311 Polyhedron 317 convex 317 462 INDEX TAGE Polyhedron, edges of . . . .317 faces of 317 regular 350 section of 317 vertices of 317 Polyhedrons classified as to faces 350 .similar 431 Postulate of planes .... 274 Prism 317 altitude of 317 bases of 317 circumscribed 356 inscribed 356 lateral area of 317 lateral edges of 317 lateral faces of 317 oblique 318 right 318 right section of 318 truncated 318 Prismatoid 441 altitude of 441 mid-section of 441 Prisms classified as to bases . 318 Projection of a line .... 302 of a point 302 Pyramid 337 altitude of 337 base of . 337 circumscribed 366 frustum of 338 inscribed 366 lateral area of 337 lateral edges of 337 lateral faces of 337 quadrangular 337 regular 337 right 337 slant height of regular . .337 PAGE Pyramid, spherical 421 triangular 337 vertex of 337 Pyramids classified as to bases 337 properties of regular . . . 338 Quadrant 384 Radius of sphere 381 Regular polyhedron .... 350 pyramid 337 Relation of dihedral angles to plane angles 294 polygons to polyhedral angles 392 Revolution of cone 363 Right and oblique cones . . . 363 cylinders 353 Right prism ....... 318 Right section of cylinder . . 357 of prism 318 Section of a cone 363 cylinder ........ 353 polyhedron 317 right, of a cylinder . . . 357 prism 318 Sector, spherical 421 Segment, spherical 421 Similar cones 370 cylinders 359 polyhedrons 431 Size of polyhedral angle . . 308 Slant height of cone of revolu- tion 363 frustum of cone of revolu- tion 367 frustum of pyramid . . . 338 regular pyramid .... 337 Solids, congruent 322 INDEX 4»;:; PAGE Solids, equivalent 322 Sphere 381 center of 381 circumscribed 380 describing circles on . . . 385 diameter of 381 great circle of 383 inscribed 380 poles of a circle 383 radius of 381 small circle of 383 Spheres, tangent 385 Spherical angle 389 distance 383 excess of a triangle . . . 413 polygon 417 convex 392 diagonal of 302 polygons, congruent . . . 302 pyramid 421 base of 421 vertex of 421 sector 421 base of 421 segment 421 altitude of 421 bases of 421 of one base 421 surface, generation of . . 381 Spherical triangle 302 diagonal of 302 wedge 421 Surface, conic 3-2 excess of A\:\ Triangles, birectangular . . 397 classified as to right angles 397 relation of symmetric . . 399 symmetric isosceles . . . 399 spherical 399 trirectangular 397 Trihedral angle 308 Truncated prism 318 Unit of volume 322 Vertex of polyhedral angle . 308 of spherical pyramid . . 421 Vertices of a polyhedron . .317 Volume 322 unit of 322 Wedge, spherical 42] Zone 410" altitude of 410 bases of 410 aeration of 41<> of one base 4 10 UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. MY 2 1 1948 ARY FEB k 1950 J N 19 195: f 2 :' APR ^A?54 MAY 29 1954 \j LD 21-100m-9,'47(A5702sl6)476 9754 3 Ubrwff THE UNIVERSITY OF CALIFORNIA LIBRARY