UC-NRLF $B 3Db Mfi3 iliiiiii.; '' m w 111 i!!lili!iili!ii|!!;!:|||."'ia I In 11 pliii GIFT or SOLID GEOMETRY A SERIES OF MATHEMATICAL TEXTS EDITED BY EARLE RAYMOND HEDRICK THE CALCULUS By Ellery Williams Davis and William Charles Brenke. PLANE AND SOLID ANALYTIC GEOMETRY By Alexander Ziwet and Louis Allen Hopkins. PLANE AND SPHERICAL TRIGONOMETRY WITH COMPLETE TABLES By Alfred Monroe Kenyon and Louis Ingold. PLANE AND SPHERICAL TRIGONOMETRY WITH BRIEF TABLES By Alfred Monroe Kenyon and Louis Ingold. THE MACMILLAN TABLES Prepared under the direction of Earle Raymond Hedrick. PLANE GEOMETRY By Walter Burton Ford and Charles Ammerman. PLANE AND SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. SOLID GEOMETRY BY WALTER BURTON FORD JUNIOR PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MICHIGAN AND CHARLES AMMERMAN THE WILLIAM MCKINLEY HIGH SCHOOL, ST. LOUIS EDITED BY EARLE RAYMOND HEDRICfC !c THE MACMILLAN COMPANY 1920 All rights reserved Copyright, 1913, By the MACMILLAN COMPANY. Set up and electrotyped. Published November, 1913. J. 8. Cushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE This book contains the Chapters on Solid Geometry from the Plane and Solid Geometry by the same authors. The general nature of the motives that led to the organization of the work are described in the preface of the complete edition, and it does not seem necessary to repeat all of them here. In order to make it possible to refer to theorems proved in Plane Geometry, a complete syllabus of them, together with other necessary quotations, is printed on pages xxix-xlvi of this book. All references made in the text, and any other questions in Plane Geometry concerning which there may be doubt, can there be looked up by the student. An excellent opportunity for a review of Plane Geometry is afforded by this syllabus. The book is distinguished by its acceptance of the principle of emphasis of important theorems laid down by the Commit- tee of Fifteen of the National Education Association in their Keport.* Thus, theorems of the greatest value and importance are printed in bold-faced type, and those whose importance is considerable are printed in large italics. The Report just mentioned has been of great assistance, and its principles have been accepted in general, not in a slavish sense but in the broad manner recommended by the Committee itself. A perusal of the Eeport will give more fully and accurately than could be done in this brief preface, the con- siderations which led to the adoption of these principles, in particular, the principle of emphasis upon important theorems, both by the Committee and by the authors of this book. * Printed as a separate pamphlet with the Proceedings for 1912. Ke- printed also in School Science, 1911, and in The Mathematics Teacher, December, 1912. V 459948 vi PREFACE The great excellence of the figures, particularly the very unusual and effective * phantom ' halftone engravings, deserves mention. These figures should go far toward relieving the unreality which often attaches to the constructions of Solid Geometry in the minds of students. W. B. FORD, CHAS. AMMERMAN, E. R. HEDRICK, Editor. CONTENTS Chapter VI. Lines and Planes in Space . Part 1. General Principles Part II. Perpendiculars and Parallels Part III. Dihedral Angles . Part IV. Polyhedral Angles Miscellaneous Exercises on Chapter VI Chapter VII. Polyhedrons. Cylinders. Cones Part I. Prisms. Parallelepipeds Part II. Pyramids Part III. Cylinders and Cones .... Part IV. General Theorems on Polyhedrons. Similar- ity. Regular Solids. Volumes Miscellaneous Exercises on Chapter VII Chapter VIII. The Sphere Part I. General Properties .... Part II. Spherical Angles. Triangles. Polygons Part III. Areas and Volumes .... Miscellaneous Exercises on Chapter VIII . Tables Table I. Quantities Determined by a Given Angle Table II. Powers and Roots .... Table III. Important Numbers .... Syllabus of Plane Geometry Introduction .... Chapter I. Rectilinear Figures Chapter II. The Circle PAGES 215-237 215-218 219-227 228-231 232-234 235-237 238-283 238-252 253-262 263-273 274-281 282-283 284-321 284-293 296-308 309-317 317-321 i-xxvii ii-vii viii-xxvi xxvii xxix-xlvi XX ix xxxii xxxvii CONTENTS IX Chapter III. Proportion. Similarity . Chapter IV. Areas of Polygons. Pythagorean Theo rem Chapter V. Regular Polygons and Circles . Appendix : Maxima and Minima Index xliii xliv xlv xlvii-xlix The complete text for the theorems listed in the Syllabus of Theorems of Plane Geometry in this book is published separately under the title PLANE GEOMETRY The contents of that book and of this one are published together under the title PLANE AND SOLID GEOMETRY !« SOLID GEOMETRY CHAPTER Yl LINES AND PLANES IN SPACE PAET I. GENERAL PRINCIPLES 239. Definitions. Solid Geometry, or Geometry of Three Di- mensions, treats of figures whose parts are not confined to a plane. A plane is a surface such that if any two points in it are taken, the straight line passing through them lies wholly in the surface. Fig. 163 Thus, m Fig. 163, if A and B are two points of a plane iHfiV, the entire straight line AB lies in the plane MN. Any point O on AB lies inMN. A plane is said to be determined by certain points and lines if that plane and no other plane contains those points and lines. 240. Corollary 1. It is evident from the definition of a plane that if a line has two of its points in a plane^ it lies wholly in that plane. 216 216 LINES AND PLANES IN SPACE [VI, § 241 241. Assumptions, or Postulates. 1. A plane is unlimited in extent. 2. TJirough any straight line an unliinited number of planes may he passed. See Fig. 164. 3. If a plane is revolved about any straight Ime in it as an axis, it may be made to pass through any point in space. 4. One and only one plane can he made to pass through three points not in the same straight line. Fig. 164 Fig. 165 (a) Fig. 165 (&) Fig. 165 {a) represents a plane PQ through three points A, B, C. Fig. 165 (6) represents a plane piece of glass resting on the points of three tacks. 5. Two planes cannot intersect each other in only a single point. 242. Corollary 1. A plane is determined by tiuo intersecting lines. Fig. 166 (a) Fig. 166 (&) [Hint. Consider the point where the lines intersect, and two other points, one on each line ; then apply 4, § 241.] VI, § 244] GENERAL PRINCIPLES 217 243. Corollary 2. A line and a point ivithout the line de- termine a plane. [Hint. Use 4, § 241.] [^^ ' ! Fig. 167 (a) Fia. 167 (&) 244. Corollary 3- Two parallel lines determine a plane. Fig. 168 (a) Fig. 168 (6) [Hint. By the definition of parallel lines (§ 48), two such lines must lie in a plane. Show that this is the only one.] EXERCISES 1. How many planes pass through a given straight line in space ? How many pass through two given points ? 2. In a carpenter's plane the knife-edge lies along a straight line. As soon as any rough surface has been sufficiently planed off, the whole length of the knife-edge keeps on the surface as the plane is moved along. Connect this fact with § 240. 3. Why are cameras, surveyors' transits, etc., mounted on three legs instead of four ? 4. Prove that a straight line can intersect a plane in but one point unless it lies wholly in the plane. See § 240. !i 0^' Fig. 169 218 LINES AND PLANES IN SPACE [VI, § 245 245. Theorem I. The intersection of two ^ilanes is a straight line. Given the two intersecting planes MN and RS. To prove that their intersection is a straight line. Proof. Let A and B be any two points common to both planes. 5, § 241 Draw the straight line AB. Then every point in AB lies in MN and also in RS. § 240 Therefore, AB is common to the two planes. Moreover, no point not on AB can be common to both planes, for the two planes would then coincide. 4, § 241 Therefore, the intersection of the planes MN and RS is a straight line. EXERCISES 1. What is the locus of all points common to two intersect- ing planes ? 2. If a sheet of paper is folded, why is the crease straight ? 3. In how many points (in general) will three planes inter- sect? What can be said of the intersection of four or more planes in space ? 4. Can two pencils be held in such a position that a plane cannot be passed through them ? State the general fact about a plane that is illustrated by your answer. 5. Can a plane be passed (in general) through four or more given points in space? Can a plane be passed (in general) through three lines all of which pass through a common point in space ? 6. Can there be two straight lines that are not parallel and that do not meet ? Find a pair of such lines in Fig. 169. VI, §248] PERPENDICULARS AND PARALLELS 219 PART 11. PERPENDICULARS AND PARALLELS 246. Line Perpendicular to a Plane. The point where a line intersects a plane is called the foot of the line on that plane. A straight line is perpendicular to a plane when it is perpen- dicular to every straight line in the plane drawn through its Fig. 170 (a) Fig. 170 (6) foot. The plane is then also said to be perpendicular to the line. Thus, in Fig. 170 (a), if PQ is perpendicular to the plane MJ^j it is then perpendicular to all the lines QA, QB, QC, etc. ; and PQ is called the distance from P to MN] see Ex. 1 below. 247. Parallel Planes and Lines. A straight line is parallel to a plane if they never meet, however far produced. Two planes are parallel if they never meet, however far produced. It is to be remembered (§ 48) that two lines are parallel only when they lie in the same plane and do not meet. 248. Corollary 1. A plane that contains one of two parallel lines is parallel to the other line. EXERCISES 1. Show, by § 77, that the perpendicular from a point P to a plane MN (Fig. 170 a) is shorter than any other line that can be drawn from P to MN. 2. Show, by § 71, that if two oblique lines from a point P to a plane MN cut off equal distances from the foot of the per- pendicular from P to MN, they are equal. See Ex. 1, p. 63. 220 LINES AND PLANES IN SPACE [VI, § 249 Fig. 171 249. Theorem II. If a line is perpendicular to each of two lines at their point of intersection, it is perpendicular to their plane. Given FB perpendicular at B to each of two straight lines AB and BC ot the plane MN. To prove FB perpendicular to the plane MJ^. Proof. Draw A C, and through B draw any line, as BH, meet- ing AC Sit II. Prolong FB to E so that BE = FB. Join F and E to A, H, and O. Then AB and BG are perpendicular bisectors of FE whence FA = AE, FG = GE. Therefore A AFG ^ A AEG, whence Z HAF = Z HAE. Also A HAF ^ A HAE, and HF=HE. Hence HB±FE ov FB. But ^5 was any line in MN' drawn through ^. Therefore FBA.MK Const. §100 §45 Why? Why? Why? AVhy? §246 250. Corollary 1. At a point in a plane only one perpen- dicular line can he erected. [Hint, Suppose a second perpen- dicular line BC could be erected (Fig. 172). Pass a plane through AB and BC. This plane will intersect MN'm. a straight line, as DE. Then AB and BC are both perpendicular to DE at the same point B. But, since AB, BC, and DE ail lie in the same plane, this is impossible, by 7, § 31.] Fig. 172 VI, §254] PERPENDICULARS AND PARALLELS 221 251. Corollary 2. From a point icithout a plane, only one line can be drawn perpendicular to the plane. [Hint. If two perpendiculars, as PB and PA, could be drawn from P to the plane MN, then A PBA would contain two right angles so that the sum of the angles of APBA would be more than two right angles. But this is impossible. Why ?] Fig. 173 252. Corollary 3. Through a given point in a straight line, only one plane can he drawn perpendicular to the line. [Hint. Draw two different perpendiculars in space to the given line at the given point, and apply §§ 242, 249. If two such planes exist, their intersections with a plane through the given line violate 7, § 31.] 253. Corollary 4. TJirough a given point without a straight line, only one plane can he drawn perpendicular to the line. [Hint. Prove by reduction to an absurdity. Show that the inter- sections of two sucb perpendicular planes with the plane determined by the given line and given point would violate § 58.] 254. Corollary 5. All perpendicular lines that can he drawn to a straight line at a given point in it lie in a plane perpendicular to the line at the given point. [Hint. Show that otherwise two perpendicular lines could be drawn to the given line in the same plane at the given point, thus violat- ing 7, §31.] EXERCISES 1. Show how to determine a perpendicular to a plane by means of two carpenter's squares. 2. Tell how to test whether or not a flagpole is erect. 3. A spoke of a wheel is perpendicular to the axis on which it turns. Show by § 254 that it describes a plane in its rotation. 222 LINES AND PLANES IN SPACE [VI, § 255 255. Theorem III. Two planes per- pendicular to the same line are parallel. [Hint. Show that if the two planes met, say in a point P, § 253 would be violated.] 256. Theorem IV. If a plane intersects two parallel planes, the lilies of intersection are parallel. Given the plane PQ intersect- ing the parallel planes MN and ES in AB and CD, respectively. To prove AB II CD. [Hint. Prove, by reduction to an absurdity, that AB and CD cannot meet.] M f R< 1 ?u 7' U< ^ ^ .../"■^r-^ r Fig. 175 257. Theorem V. • Two lines parallel to a third line (in space) are parallel to each other. Compare § 50. [Hint. Let BB' and CC be two lines parallel to a third line AA' (Fig. 176). The plane determined by OC and the point B on BB' is par- allel to AA' (§ 248) . Therefore (§ 48) the line of intersection of this plane with the plane of the parallels AA' and BB' is parallel to AA'. Hence show, by § 49, that this line of intersection coincides with BB'^ so that BB' and CC lie in a plane. Finally, show that BB' and CC cannot meet ; for, if they did meet, say at a point D, the plane determined by I) and AA' would contain (§ 244) both BB' and CC.^ VI, §258] PERPENDICULARS AND PARALLELS 223 258. Theorem VI. If two angles, not in the same plarie, have their sides respectively imraUel and extending in the same direc- tion, they are equal and their planes are parallel. Fig. 176 Given the angles BAG and B'AC, lying in the planes MN and PQ, respectively, with AB II A'B', and AC II A'C. To prove that Z.A = AAl, and that JfA^II PQ. Proof. Take AB = A'B', and AG = A'G'. Draw AA', BB', GG\ GB, and C'jS'. Since AB is equal and parallel to A'B', it follows that ABB' A is a parallelogram ; hence AA is equal and parallel to BB'. Similarly, A A' is equal and parallel to CO'. Hence BB^ is equal and parallel to GG'. Then BB'C'G is a parallelogram, and GB=G'B'. Therefore A ABG ^ A A'B'G'. Hence Z.A = ZA'. Now PQ II AB. Likewise PQ II AG. Therefore, PQ II ilOT" for, if not, the line of intersection of PQ and MN would meet either AB ot AG (or both) extended ; hence PQ would not be parallel to each of them. Note. The similar theorem for angles that lie in the same plane was proved in § 67. As in § 67, the two angles are supplementary to each other if one pair of corresponding sides extend in opposite directions from the vertices. Why? Why? §257 Why? Why? Why? §248 224 LINES AND PLANES IN SPACE [VI, § 259 Fig. 177 259. Theorem VII. A plane perpendicular to one parallel lines is perpendicular to the other also. Given the two parallel lines AB and CD, and a plane MN perpen- dicular to CD at C. To prove that MN is perpendicu- lar to AB. Proof. The parallel lines AB and CD determine a plane (§ 244) which intersects MN in some line AC. 'Now AC is perpendicular to CD ; whence ACis perpendicular to AB. Draw any line AE in the plane Jf A'' through A. Draw CF in MlSf parallel to AE through C. Then CF is perpendicular to CD. Hence AE is perpendicular to AB. Therefore AB is perpendicular to MN". of two §245 §246 §60 § 246 § 258 §246 260. Corollary 1. Two lines perpendicular to the same plane, are parallel. [Hint. Let AB and CD (Fig. 177) be perpendicular to the plane Mli. Imagine a parallel CD' to AB through C. Then CD' is perpendicular to MN, by § 259. Hence CD' coincides with CD, by § 250.] EXERCISES 1. The legs of a table lie along parallel lines in space. What preceding theorem or corollary is illustrated here ? Mention other similar illustrations. 2. How many lines can be drawn through a given point parallel to a given plane ? If there is more than one such, what is the locus of them all ? • 3. Given a plane and two points without it. When will the line through the two points be parallel to the plane ? VI, §261] PERPENDICULARS AND PARALLELS 225 261. Theorem VIII. If tioo straight lines are inter- sected by three parallel planes, the corresponding seg- ments of these lines are proportional. Given the straight lines AB and CD cut by the parallel and N, planes L, M, To prove that AE/EB = GF/FD. Proof. Draw BQ meet- ing the plane M in G. Draw EF, EG, FG, BD, and AC. Then GF II BD, and EG 11 AC Now AE/EB=CG/GB, and CF/FD Therefore AE/EB = CF/FD. EXERCISES Fig. 178 § 256 CG/GB. §145 Why? 1. Show that if parallel planes intercept equal segments on one line, they will intercept equal segments on any other line. 2. In Fig. 178, AE = 5, EB = 4:, and CF=6. What is the value of FD ? 3. Two ordinary blocks C and D having the respective heights H and h are placed upon each other as shown in the figure. Show that any line AB drawn from the upper sur- face of C to the lower surface of D will be divided in the ratio H'.hhj the point P where AB intersects the common surface of the two blocks. Q 1 /4, ^, He ■'■ .\ —/ -- — f V 226 LINES AND PLANES IN SPACE [VI, § 262 262. Perpendicular Planes. Two planes MN and PQ are said to be perpendicular to each other when any line CD drawn in the one perpendicular to their intersection is perpendicular to the other plane. ^ M AAVC. Proof. If Z AVGis equal to or less than either of the other angles, we know the proposition is true without further proof. If Z.AVG is greater than either of the other angles, lay off any lengths VA and VG on the sides of Z AVG, and draw AG. Then draw VD in the plane AVG, making /.AVD^^ZAVB. Lay off VB = VD, and draw AB and GB. Then AAVB^AA VD. Why ? Therefore AB = AD. Why ? Kow AB -{- BG > AD -{- DG. Why? Whence, subtracting, BG> DG. Ax. 6 Therefore ZBVG>ZDVG. §79 By construction Z. AVB = Z. AVD. Adding, . Z AVB -{- Z BVG > Z AVG, EXERCISES 1. If in the trihedral angle V-ABG, Z AVB = 60°, and Z BVG = 80°, make a statement as to the number of degrees in Z^ Fa 2. Show that any face angle of a trihedral angle is greater than the difference of the other two. 234 LINES AND PLANES IN SPACE [VI, § 273 273. Theorem XIII. The sum of the face angles of any convex polyhedral angle is less than four right angles. V Fig. 186 Given the polyhedral angle V-ABCDE with the edges cut by any plane in the points A, B, C, D, E. To prove that the sum of the face angles of the polyhedral angle is less than four right angles. Proof. Connect any point in the polygon ABODE with the vertices A^ B, O, D, E. The number of triangles with the common vertex O is the same as the number having the vertex V. Now Z VBA + Z VBOZABO + Z OBC, §272 and Z VAB + Z VAE > Z BAO + Z OAE, etc. Therefore the sum of the base angles of the triangles having V for a common vertex is greater than the sum of the base angles of the triangles having for vertex. But the sum of all the angles of all the triangles whose vertex is Fis equal to the sum of all the angles of all the tri- angles whose vertex is 0. Why ? Therefore the sum of the angles about the vertex V is less than the sum of the angles about 0, that is, less than four right angles. VI, §273] MISCELLANEOUS EXERCISES 235 MISCELLANEOUS EXERCISES ON CHAPTER VI 1. Lean one book in a slanting position against another book that lies flat on a table, and hold a stretched string parallel to the cover of the slanting book. Can the string have more than one position ? Can the string be horizontal ? Vertical ? 2. Show that if a half-open book is placed on a table, rest- ing on its bottom edges, the back edge of the book is perpen- dicular to the plane of the table, and the lines of printing are parallel to that plane. 3. Show that the dihedral angle between the pages of an open book is measured by the plane angle between opposite lines of type on the two pages. 4. What is the shape of the end of an ordinary plank after it has been sawed off in a slanting direction, assuming that the opposite faces of the original board are parallel planes ? 5. Prove that the segments of two parallel lines included be- tween parallel planes are equal. [Hint. Pass a plane through the parallel lines and then prove that the given segments form the opposite sides of a parallelogram. 6. Prove that a plane perpendicular to the edge of a dihe- dral angle is perpendicular to both its faces. [Hint. Use § 263.] 7. What is the locus of all the points equidistant from the three faces of a trihedral angle ? 8. Show that the locus of any given point on a line seg- ment of fixed length, whose ends touch two parallel planes, is a third plane parallel to the given planes. 9. Prove that if three lines are perpendicular to each other at a common point in space, each line is perpendicular to the plane of the other two, and that the planes of the lines (taken in pairs) are perpendicular to each other. Kote how this is il- lustrated on a cube, or in a corner of a room, or in a corner of an ordiuary box. 236 LINES AND PLANES IN SPACE [VI, § 273 10. The trihedral angle formed when three planes meet each other, so that each is perpendicular to the other two is called a trirectangular trihedral angle. Prove that the edges of a trirectangular trihedral angle are mutually perpendicular by pairs. See §§ 246, 265. 11. Prove that the space about a point is divided into eight congruent trirectangular trihedral angles by three planes mutu- ally perpendicular by pairs at the point. 12. Prove that if a line is parallel to the intersection of two planes, it is parallel to each of the planes. [Hint. Suppose that the hne is not parallel to one of the planes and thus argue to an absurdity.] 13. Prove that if a line is parallel to each of two intersect- ing planes it is parallel to their intersection. 14. Prove that if a line is parallel to a plane, any plane perpendicular to the line is perpendicular to the plane. [Hint. Pass a plane through the given line perpendicular to the given plane and use § 267.] 15. Can a trihedral angle be formed by placing three equi- lateral triangles so that one vertex of each lies at the vertex of the trihedral angle ? [Hint. Use § 273.] 16. Can a convex polyhedral angle be formed as in Ex. 15 by placing at its vertex one vertex of each of four equilateral triangles ? Can this be done with five equilateral triangles ? With six ? With more than six ? 17. Can a convex polyhedral angle be formed by placing at its vertex one vertex of each of three squares ? Four squares ? 18. Can a convex polyhedral angle be formed by placing at its vertex one vertex of each of three regular pentagons ? Four ? 19. Show that just five different convex polyhedral angles can be formed as in Exs. 15-18 by placing at a single point one vertex of each of several similar regular polygons. VI, § 273] MISCELLANEOUS EXERCISES 237 M 20. Show that the sum of the dihedral angles of a trihedral angle lies between two and six right angles. 21. Is there (in general) a point in space that is equidistant from four given points not all of which lie in the same plane ? Give reason for your answer. 22. Given any line I and a plane MN, drop a perpendicular PA from any point P in ? to MN. Prove that I and PA de- termine a plane perpendicular to MN. [This plane is called the projecting plane of I on MN. Its intersection AB with MN is called the pro- jection of I on MN. Define similarly the projection of the segment PQ.] 23. Prove that the projection on a plane MN of the line segment joining two points P and Q (Ex. 22) is the line joining the feet A and jB of the perpendiculars dropped to the plane from P and Q, respectively. 24. If a line I meets a plane MN at a point B, prove that the projection of I on MN is the line joining B to the foot ^ of a perpendicular let fall from any point P in I. [The angle ABP between the line I and its pro- jection is called the angle between the line and the plane j or the inclination of the line to the plane."] 25. Prove that the sides of an isosceles triangle make equal angles with any plane containing its base. *26. Show that the length of the projection of any line segment PQ on any plane is the length of PQ times the cosine of the angle between the line and the plane. CHAPTER VII POLYHEDRONS CYLINDERS CONES PART I. PRISMS 274. Polyhedrons. A polyhedron is a limited portion of space completely bounded by planes. The intersections of the bounding planes are called the edges ; the intersections of the edges, the vertices; and the portions of the bounding planes bounded by the edges, the faces, of the polyhedron. ICOSAHJIDBOX DODEGAHEDEON OCTAHEDRON HEXAHEDRON TeTRAUEDROX A polyhedron of four faces is called a tetrahedron; one of six faces (for example, a cube), a hexahedron; one of eight faces, an octahedron; one of twelve faces, a dodecahedron; one of twenty faces, an icosahedron. A diagonal of a polyhedron is a straight line joining any two vertices not in the same face. 275. Prisms. A prism is a polyhedron, two of whose faces, called its bases, are congruent polygons in parallel planes, and whose other faces, called lateral faces, are parallelograms whose vertices all lie in the bases. A triangular prism is one whose base is a triangle. The sum of the areas of the lateral faces of any prism is called the lateral area of the prism. 238 VII, § 275] PRISMS 239 The intersections of the lateral faces are the lateral edges of the prism. The altitude of a prism is the perpendicular distance between its bases. Right Prisms Fig. 188 Oblique Prisms A right prism is one whose lateral edges are perpendicular to its bases. An oblique prism is one whose lateral edges are oblique to its bases. A regular prism is a right prism whose bases are regular polygons. Any polygon made by a plane which cuts all the lateral edges of a prism, as the polygon A'B'CD'E' in Fig. 189, is called a section of the prism. A right ^. . J 1, 1 Fig. 189 section is one made by a plane perpen- dicular to all the lateral edges of the prism, as ABODE. 240 POLYHEDRONS [VII, § 276 276. Theorem I. The sections of a prism made by parallel planes cutting all the lateral edges are con- gruent polygons. Fig. 190 Given ABODE and AB'CD'M, sections of the prism MN, made by parallel planes. To prove that ABODE ^ A'B' C'D'E'. Proof. The sides of the polygon ABODE are parallel to the sides of the polygon A!B'0'D'E'. § 256 Therefore the polygons are mutually equilateral. § 84 Also the polygons are mutually equiangular. § 258 Therefore polygon ABODE ^ polygon A'B'O'D'E', § 33 EXERCISES 1. How many edges has a tetrahedron ? A hexahedron ? 2. How many diagonals has a hexahedron ? A tetrahedron ? 3. Prove that any two lateral edges of a prism are equal and parallel. 4. Prove that any lateral edge of a right prism is equal to the altitude. 5. Prove that all right sections of a prism are congruent. 6. Prove that a section of a prism parallel to the base is congruent to the base. VII, § 278] PRISMS 241 277. Theorem II. The lateral area A of a prism is equal to the j^enineter j^r ^/ <^ right section multiplied hy the lateral edge e ; that is, A —pr x 6- J' Given the prism ^D' with a right section PQIIST\ let p^ denote the perimeter of the right section, e the lateral edge, and A the lateral area. To prove that A= Pr'X ^• Proof. The lateral area consists of a number of parallelo- grams, each of which has a line equal to A A' for its base. Why ? Each of these parallelograms has one of the sides of the right section PQRST for an altitude. Why ? Therefore the areas of these parallelograms = the perimeter of PQRS T multiplied by AA^. Why ? That is A= Pr y< e. 278. Corollary 1. The lateral area A of a right prism is equal to the perimeter of its base multiplied by a lateral edge ; ov A = Pf, X e, where Pf, denotes the perimeter of the base, and e de- notes a lateral edge. EXERCISES 1. Find the altitude of a regular prism, one side of whose triangular base is 5 in. and whose lateral area is 195 sq. in. 2. Show that the lateral area of a regular hexagonal right prism is 4V3 • a/i, where h is the altitude and a the distance from the center of the base to one of the sides. 242 POLYHEDRONS [VII, § 279 279. Congruent Solids. Any two solids, in particular any two prisms, are said to be congruent when they can be made to coincide completely in all their parts. J J' A, Congruent Prisms 280. Theorem III. Two prisms are congruent if three faces that include a trihedral angle of one are congruent respectively, and similarly placed, to three faces that include a trihedral angle of the other. Given the prisms ^Jand A I' with face ^J^ face AJ\ face AO ^ face A'O', and face AD ^ face A^D\ To prove that prism AI ^ prism J.' J'. Proof. AEAF, FAB, and EAB are equal respectively to A E'A'F', F'A'B', and E'A'B'. Why ? Then trihedral ZA^ trihedral Z A\ § 271 Place the prism AI on the prism A'l' so that the trihedral Z A coincides with its congruent trihedral Z A'. Then the face AJ will coincide with the congruent face A'J'; AG with the congruent face A'G'; and AD with A'D' ; and points O and D will fall on G' and D'. § 33 Since the lateral edges of a prism are parallel and equal, CH coincides with C'H', and DI with D'l'. §§ 257, 49 Therefore the upper bases coincide, and the prisms coincide throughout and are congruent. 281. Corollary 1. Two right prisms having congruent bases and equal altitudes are congruent. VII, § 283] PRISMS 243 282. Truncated Prisms. A truncated prism is a portion of a prism included between the base and a section oblique to the base. Fig. 193 (a) Fig. 193 (6) 283. Corollary 2. Two truncated prisms are congruent if three faces including a trihedral angle of the one are congruent respectively to three faces including a trihedral angle of the other. EXERCISES 1. A wooden beam 10 ft. long has a rectangular right cross section whose dimensions are 12 in. by 16 in. If the beam be sawed lengthwise along one of its diagonal planes, show that the resulting triangular prisms are congruent. 2. What will be the lateral area of one of the triangular prisms of Ex. 1 ? Its total area ? Ans. 40 sq. ft. ; 411 sq. ft. 3. A carpenter is to saw from a given square piece of timber a portion of which one end is to be perpendicular to the lateral edges, while three given lateral edges are to be 3 ft. 6 in., 3 ft. 4 in., and 3 ft. long, respectively. Show that these measure- ments are sufficient to enable him to saw off the desired portion. 4. Show that to make a right prism of any desired shape, it is sufficient to have a pattern of a right section of the desired prism, and the length of one lateral edge. 5. Show that to make a truncated prism of any desired shape, of which one end is a right section, it is sufficient to have a pattern of that end, and the lengths of three given con- secutive lateral edges. 244 POLYHEDRONS [VII, § 284 284. Theorem IV. An oblique prism is equal in volume to a right prism whose base is a right section of the oblique prism and whose altitude is a lateral edge of the oblique prism. Given the oblique prism AD' ; and given a right prism PS' whose base PS is a right section of the prism AD', and whose altitude is equal to a lateral edge A A' of the prism AD'. To prove that prism AD' = prism PS'. Proof. The lateral edges of the prism PS' equal the lateral edges of the prism AD'. Const. Therefore AP = A'P', BQ = B'Q', etc. Why ? Moreover PQ = P'Q', and the face angles at P, Q, P', Q' are right angles. Why ? Hence, by superposition, Face ^Q^ Face .4' Q'. Likewise, Face BR ^ Face B'E', etc. Now, Section PQBST ^ Section P'Q'R'S'T. § 276 Whence, truncated prism AR ^ truncated prism A'R'. § 283 Therefore, truncated prism AR + truncated prism PD' = truncated prism A'R' 4- truncated prism PD'. Ax. 1 It follows that prism AD' = prism PS'. VII, § 286] PARALLELEPIPEDS 245 285. Equivalent Solids. Two solids that have the same volume are said to be equivalent, or equal in volume. Thus we proved, in § 284, that any oblique prism is equiva- lent to a right prism whose base is a right section of the oblique prism and whose altitude is equal to the lateral edge of the oblique prism. 286. Parallelepipeds. A parallelepiped is a prism whose bases are parallelograms. A right parallelepiped is a parallelepiped whose lateral edges are perpendicular to its bases. Rectangular Parallelepiped Cube Oblique Parallelepiped Fig. 195 A rectangular parallelepiped is a right parallelepiped whose bases are rectangles. A cube is a parallelepiped whose six faces are squares. An oblique parallelepiped is one whose lateral edges are oblique to its bases. EXERCISES 1. Show that the lateral faces of a right parallelepiped are rectangles. ' 2. Find the sum of all the face angles of a parallelepiped. 3. Find the diagonal of a cube whose edge is 4 in. ; 20 in. ; a. 246 POLYHEDRONS [VII, § 287 \ 287. Theorem V. The plane passed through two diagonally opposite edges of a parallelepiped divides the parallelepiped into two triangular prisms that are equal in volume. G Given the plane ACOE passing through the opposite edges AE and CO of the parallelepiped AG. To prove that the parallelepiped AG is divided into two equal triangular prisms, ABG-F and ADC-H. Proof. Let IJKL be a right section of the parallelepiped. From the definition of parallelepiped, the opposite faces, AF and DG, and AH and BG, are parallel and congruent. § § 258, 88 Therefore IJ II LK, and IL II JK. § 256 Then IJKL is a parallelogram. Why ? The intersection IK of the right section with the plane ACGE is the diagonal of the O IJKL. Therefore A IJK^ A IKL. Why ? But the prism ABC-F\s equal in volume to the right prism whose base is Z/JS'and altitude AE, and the prism ACD-His equal in volume to the right prism whose base is ILK and al- titude AE. § 284 But since these right prisms are congruent, § 281 it follows that prism ABC-F =ipvism ADC-H. Note. If the faces EFQH and ABCD (Fig. 196) are perpendicular to the edges AE, BF, etc., it is easy to see that the diagonal plane AECG divides the parallelepiped into two congruent triangular prisms. This happens for any rectangular parallelepiped. VII, § 287] PARALLELEPIPEDS 247 EXERCISES 1. Prove tliat the diagonals of a rectangular parallelepiped are equal, and that the square of the diagonal is equal to the sum of the squares of the three edges that meet in any vertex. [Hint. The diagonal is the hypotenuse of a right triangle whose sides are one of the edges and a diagonal of one face ; the diagonal of the face is the hypotenuse of a right triangle whose sides are two of the edges. ] 2. Find the length of the diagonal of a rectangular parallel- epiped whose edges are 8, 10, 12. 3. Find the edge of a cube whose diagonal is 64 in. 4. Prove that the diagonals of a parallelepiped bisect each other. [Hint. Consider each pair of diagonals separately, and apply § 87 to the diagonal plane in which they lie.] 6. Prove that if the right section IJKL (Fig. 196) of a par- allelepiped is a rectangle, the two diagonal planes ACGE and BDHF divide the parallelepiped into four triangular prisms that are equal in volume. 6. A tank in the form of a rectangular parallelepiped that holds 100 gal. is divided into four compartments by two vertical diagonal planes. What is the capacity of each compartment ? 7. A cube each of whose edges is 1 ft. long is called a unit cube; its volume is one cubic foot. If six such cubes are placed side by side in two rows of three each, they form a rec- tangular parallelepiped 2 ft. wide, 1 ft. high, and 3 ft. long. What is the volume of this parallelepiped ? What is the vol- ume of each of the triangular prisms into which it is divided by a diagonal plane ? 8. How many unit cubes are there in a cube each of whose edges is 5 units long ? 9. How many unit cubes are there in a rectangular par- allelepiped 3 units long, 4 units wide, and 2 units high ? What is the volume of this parallelepiped ? 248 POLYHEDRONS [VII, § 288 the ^ ^1 /\ A y y" T y\ y\ A 1 ^y< y\ y\ /'\ Y^ < y\ y\ A\^^ y\ y\ ^ > J- > y y y y y 2 y L B 288. Volume of a Rectangular Parallelepiped. The three edges of a rectangular parallelepiped which meet at a common point are called its dimensions. In Chapter IV (§ 181), we assumed (without proof) the well-known principle that the area of a rectangle is equal to the product of its two dimensions. Simi- larly, we shall now assume that the vol- ume of a rectangular parallelepiped is equal to the product of its three dimensions, that is, to the product of its length, breadth, and height ; i.e. For any rectangular parallelepiped the volume V is V=lxbxh, where I, h, h denote the length, breadth, and height of parallelepiped. The student is reminded that the meaning of the principle is that, if the three dimensions are each measured in terms of the same unit of length, then the volume in terms of the corresponding unit cube is the product of the three dimensions. The following corollaries result at once from this principle : 289. Corollary 1. Two rectangular parallelepipeds having congruent bases are to each other as their altitudes. [Hint. If Z, 6, and h represent the dimensions of the one parallele- piped, then Z, 6, and h' will represent the dimensions of the other. The cor- responding volumes will therefore be to each other in the ratio {lbh)/{lbh')y I I that is, in the ratio h/h'.^ Fia. 199 & D Fig. 198 / ^ / / / h' h / / VII, § 292] PARALLELEPIPEDS 249 290. Corollary 2. Two rectangular parallelepipeds having equal altitudes are to each other as their bases. 291. Corollary 3. The volume of a cube is equal to the cube of its edge. 292. Corollary 4. The volume V of any rectangular parallele- piped is the product of the area of its base B and its altitude h; that is J V=Bxh. EXERCISES 1. Two rectangular parallelepipeds with equal altitudes have bases containing 10 sq. in. and 15 sq. in., respectively. The volume of the first is 56 cu. ft. Find the volume of the second. Ans. 84 cu. ft. 2. Compare the volume of the rectangular parallelepiped whose dimensions are 8 in., 10 in., 11 in. with the one whose dimensions are 1 ft., 1 ft., and 16 in. 3. In a lot 120 ft. long and 66 ft. wide a cellar is to be dug for a building. The cellar is to be 44 ft. long, 36 ft. wide, and 7 ft. deep. The earth removed is to be used to fill the surrounding yard. What depth of fill can be made ? 4. A standard (U.S.) gallon contains 231 cu. in. How many gallons can be put in a tank of the form of a rectangular par- allelepiped that is 2 ft. high, l^ ft. wide, and 3 ft. long ? 5. How many gallons (see Ex. 4) are there in 1 cu. ft. ? 6. Find the size of a cubical tank that will contain 50 gal. 7. Find the edge of a cube whose volume is 1728 cu. in. ; of a cube whose volume is 1500 cu. in. 8. Find the diagonal of a cube whose volume is 521 cu. in. 9. If the volume of one cube is twice that of another, how do their edges compare ? Ans. V 2 : 1. 10. Find the edge of a cube whose total surface is 60 sq. ft. 11. The edge of a cube is a. Find the area of a section made by a plane through two diagonally opposite edges. 250 POLYHEDRONS IVII, § 293 293. Theorem VI. The volume V of any parallelepiped is equal to the product of its base B and its altitude h; that is, V=Bxh. Fig. 200 Given the parallelepiped / with its volume denoted by V, its base by B, and its altitude by h. To prove that V=B x h. Proof. Produce AO and all the other edges of / that are parallel to AC. On the prolongation oi AG take BE = AG, and through D and E pass planes perpendicular to AE, forming the right parallelepiped II whose base is B'. Then /=//. § 284 Prolong FE and all the other edges of II that are parallel toi^^. On the prolongation of FE take MN = FE, and through M and N pass planes perpendicular to FN, forming the rec- tangular parallelepiped III whose base is B". Then II =111 Why? Therefore /=///. Why? Moreover B' = B" Why? and h, the altitude of 7, is equal to the altitude of III. Why ? But the volume of /// is B" x 7i, by § 288; hence the vol- ume of /is V=B'' xh = B X h. VII, § 296] VOLUME FORMULAS 251 294. Theorem VII. The volume V of any triangular prism is equal to the product of its base B and its altitude h; that is,V=Bxh. ^r L;<=r- — ^~~■::::^-/' W i M Fig. 201 Given the triangular prism LMN-N' whose base B is the triangle LMN, and whose altitude is li. To prove that the volume V of LMN-N' = B X h. Proof. Complete the parallelepiped LMPN-P\ [The remainder of the proof is left to the student. Use § 293.] 295. Corollary 1. TJie volume V of any prism is equal to the product of its base B and its altitude h; that is, V—Bxh. Fig. 202 [Hint. Any prism may be divided into triangular prisms by diagonal planes.] 296. Corollary 2. Prisms having equivalent bases and equal altitudes are equal. 252 POLYHEDRONS [VII, § 296 EXERCISES 1. Describe one or more ways in which a parallelepiped may be distorted and yet have its volume remain unchanged. 2. The base of a parallelepiped is a rhombus one of whose diagonals is equal to its side. The altitude of the parallele- piped is a, and is also equal to a side of the base. Find the volume of the parallelepiped. Ans. a? V3/2. 3. The altitude of a parallelepiped is 3 in., and a diagonal of a base divides the base into two equilateral triangles, each side of which is 6 in. Find the volume of the parallelepiped. 4. The volume of a rectangular parallelepiped is 2430 cu. in. and its edges are in the ratio of 3, 5, and 6. Find its edges. 5. The altitude of a prism is 6 in. and its base is a square each side of which is 3 in. Find its volume. 6. Show that two prisms with equal bases are to each other as their altitudes j and that those with equal altitudes are to each other as their bases. 7. A clay cube having a 2-in. edge is molded into the form of a triangular prism of height 3 in. What is the area of its base ? Does it make a difference in the answer whether the prism is made right or oblique ? Explain. 8. Assuming that iron weighs about 450 lb. per cu. ft., find the weight of a rod 3 ft. long, whose cross section is a rectangle 11 by 2 in. 9. With the data of Ex. 8, find the weight of an iron rod 2 ft. 6 in. long, whose cross section is a regular hexagon 1 in. on each side. 10. What must be the length of the side of an equilateral triangle in order that a triangular prism erected upon it and of height 1 ft. shall have a volume of 1 cu. ft. ? Solve the same problem, when it is assumed that the base is a regular hexagon. VII, § 297] PYRAMIDS 253 PART II. PYRAMIDS 297. Pyramids. A pyramid is a polyhedron bounded by a polygon, called its base, and several triangles that have a common vertex. The triangles are called the lateral faces, the common vertex is called the vertex of the pyramid, and the perpendicular dis- tance from the vertex to the base is called the altitude. Fig. 203. Pyramid A pyramid is triangular, quadrangular, etc., according as its base is a triangle, a quadrilateral, etc. A regular pyramid is one whose base is a regular polygon and whose vertex lies in the perpendicular erected at the center of the base. V B C Fia. 204. Regular Pyramid The slant height of a regular pyramid is the altitude of any one of its triangular faces {VH in Fig. 204). 254 POLYHEDRONS [VII, § 297 A truncated pyramid is the portion of a pyramid included between the base and any section made by a plane cutting all the lateral edges. A frustum of a pyramid is a portion included between the base and a section made by a plane parallel to the base. Fig. 205. Fkustum of a Pyramid The altitude of a frustum is the length of the perpendicular between the planes of its bases. The lateral faces of a frustum of a regular pyramid are con- gruent trapezoids. The slant height of the frustum of a regular pyramid is the altitude of one of the trapezoids forming its faces. EXERCISES 1. Of which type are the celebrated Egyptian pyramids ? 2. Prove the equality of the lateral edges of a regular pyramid. Of those of a frustum of a regular pyramid. 3. Prove that the faces of any frustum of a pyramid are trapezoids. 4. Prove the statement made in § 297 that the faces of a frustum of a regular pyramid are congruent trapezoids. 5. Prove that the lateral faces of a regular pyramid are con- gruent isosceles triangles ; hence show that the slant height, measured on any face, is the same as that measured on any other face. 6. Prove that any triangular pyramid is a tetrahedron. State and prove the converse. VII, § 299] PYRAMIDS 255 298. Theorem VIII. The lateral area A of a regular pyra- mid is equal to one half the product of the perimeter of its base p, and its slant height I ; that is, A=p x 1/2. V Given the regular pyramid V-ABGDE with the slant height I, the lateral area A, and the perimeter of the base p. To prove that A=px 1/2. Proof. A = the sum of the areas of the triangles VAB, VBC, etc. Hence A = [AB + BC-^'-]xl/2=pxl/2. Why? 299. Corollary 1. The lateral area of the frustum of a regular pyramid is equal to one half the product of the sum of the perimeters of the bases and its slant height. [See § 191.] EXERCISES 1. The slant height of a regular hexagonal pyramid is 10 ft. Each side of its base is 8 ft. What is its lateral area ? Also, what is its total area ? Ans. 240 sq. ft. ; 406.27 sq. ft. 2. The altitude of a regular quadrangular pyramid is 4 in. One side of its base is 6 in. What is its lateral area ? What is its total area ? Ans. 60 sq. in. ; 96 sq. in. 3. Find the lateral area of the frustum formed by a plane bisecting the altitude of the pyramid mentioned in Ex. 2, Find its total area. 256 POLYHEDRONS [VII, § 300 300. Theorem IX. If a pyramid is cut hy a plane par- allel to the base, (a) The altitude and the lateral edges are divided 2?roportionally ; (h) The section is a polygon shnilaj' to the base, V Given the pyramid V-ABCDE cut by a plane A^D' parallel to the base AD. To prove (a) that VS/VS' = VA/ FA' = VB/ VB' = ..., etc. (b) that the section A'B'CD'E' is similar to the base. Proof of (a) Pass a plane through V parallel to the base and apply § 261. Proof of (h) A VAB ~ A VA'B' ; A VBC ~ A VB'C, etc. Why ? Therefore AB/A'B' = VB/ VB' j VB/ VB' = BC/B'C, etc. ; Why ? and hence AB/AB' = BC/B'C = CD /CD' = • • • Ax. 9 Thus, the polygons ABODE and A'B'C'D'E' have their cor- responding sides proportional. Moreover, the same polygons are mutually equiangular. § 258 Hence ABCDE ^ A'B'C'D'E'. §165 Vn, § 302] PYRAMIDS 257 301. Corollary 1. Parallel sections of a pyramid are to each other as the squares of their distances from the vertex. Proof : In Fig. 207, ABCDE/A'B'C'D' E' = AB^/AJW\ §§ 195, 300 But AB/A'B' =VB/VB', Why? and also VS/ VS' = VB/ VB' ; (a) , § 300 hence AB/A'B' = VS/VS'. Ax. 9 Whence, squaring, AB^/'AB'^ = V^/VS^: Hence ABCDE/A'B'C'D'E' = T^Vl^- 302. Corollary 2. If tivo pyramids that have equivalent bases and equal altitudes are cut by planes parallel to their bases and at equal distances from their vertices, the sections are equivaleyit. [Hint. Represent by B and B' the areas of the two sections, and by B and B' the areas of the bases. Let h be the common altitude of the pyra- mids, and k the distance from the vertex of either pyramid to the section made in it. Then B/B = k^h'^ and B'/B' = k^/h'^ (§ 301) ; hence B/B = B'/B'. But B = B' hy hypothesis ; hence B = B'.} EXERCISES 1. Compare the areas of two sections of a pyramid whose perpendicular distances from the vertex are 3 in. and 4 in. respectively. Does it make any difference in your answer whether the pyramid is of one shape or another ? Ans. 9 : 16. 2. The altitude of a pyramid with a square base is 16 in., the area of a section parallel to the base and 10 in. from the vertex is 5Q\ sq. in. Find the area of the base. 3. The bases of the frustum of a regular pyramid are equi- lateral triangles whose sides are 10 in. and 18 in. respectively ; the altitude of the frustum is 8 in. Find the altitude of the pyramid of which the given figure is a frustum. Ans. 18 in. 4. The altitude of a pyramid is H. At what distance from the vertex must a plane be passed parallel to the base so that the section formed shall be (1) one half as large as the base ? (2) one third ? (3) one ninth ? 258 POLYHEDRONS [VII, § 303 303. Theorem X. Tioo triangular pyramids having eguivalent bases and equal altitudes are equivalent. Fig. 208 Given the pyramids V-ABC and V'-A'B'C having equiva- lent bases ABC, A'B'C, and a common altitude AH. To prove V-ABC = V'-A'B'C. Proof. The proof of the theorem consists in showing that^ the pyramids V-ABC and V'-A'B'C cannot differ in volume by as much as any given amount, however small. This means that the two volumes are actually equal, for if they were unequal, they would differ by as much as some Jixed amount, — in fact, that is what unequal means. We proceed, then, to show that V-ABC and V'-A'B'C^ can- not differ by as much as any given amount, however small. Divide the altitude AH into a number of equal parts, and through the points of division pass planes parallel to the plane of the bases. The corresponding sections made by any one of these planes in the two pyramids are equivalent. § 302 Now inscribe in each pyramid a series of prisms having the triangular sections as upper bases and the distance between the sections as their common altitude. VII, §304] PYRAMIDS 259 Each pair of corresponding prisms in the two pyramids are then equivalent. § 296 Therefore, the sum of the prisms inscribed in V-ABG is equivalent to the sum of the prisms inscribed in V-A'B'C If the number of divisions into which AH is divided is taken sufficiently large, the sum of the prisms in V-ABG may be made to differ from the volume of V-ABG by less than any given amount. Likewise, by taking the number of divisions in AH sufficiently large, the sum of the prisms in V'-A'B'O may be made to differ from the volume of V'-A!B'G' by less than the same given amount, however small it has been taken. Since, by taking the number of divisions in AH sufficiently large, the volumes of V-ABG and V'-A'B'G' differ by less than any given amount from these equal sums, the pyramids must differ from each other by less than the same given amount, and this is what we were to show. Compare § 211. 304. Corollary 1. Ayiy two pyramids having equivalent bases and equal altitudes are equivalent. Fig. 209 [Hint. Divide each pyramid into triangular pyramids.] EXERCISES 1. What is the locus of the vertices of all pyramids whose bases and volumes are the same ? 2. Prove that if the base of a pyramid is a parallelogram, the plane determined by its vertex and either diagonal of its base divides it into two equivalent triangular pyramids. 260 POLYHEDRONS [VII, § 305 305. Theorem XI. The volume V of a triangular pyramid is equal to one third the product of its base B, and its altitude h; that is, V=Bxh/3. Fig. 210 Given the triangular pyramid 0-LMN. To prove that the volume V of 0-LMN equals \ its base B times its altitude }i ; that is, V = 5 X /i/3. Proof. Construct a triangular prism MP having LMN iov its base, and its lateral edges equal and parallel to the edge OM. The prism MP is made up of the triangular pyramid 0-LMN and the quadrangular pyramid 0-LNPQ. Pass a plane through OQ and ON dividing the quadran- gular pyramid into two triangular pyramids, 0-LNQ and 0-NQP. Pyramid 0-LNQ = pyramid 0-NQP. § 303 Pyramid 0-NQP may be read N-QOP. Pyramid N-QOP = pyramid 0-LMN § 303 Therefore, the three triangular pyramids are equal and 0-LMN is one third the prism. But the volume of the prism is equal to the product of its base and its altitude. § 294 Therefore, pyramid 0-LMN = ^ the product of its base and its altitude. VII, § 307] PYRAMIDS 261 306. Corollary 1. The volume V of any pyramid is equal to one third the product of its base B and its altitude h; that is. V=Bh/3. Fig. 211 [Hint. Divide the pyramid into triangular pyramids and apply § 305. ] 307. Theorem XII. If a frustum of a pyramid has bases B a7id B' ayid altitude h, and is cut from a pyramid P whose base is B and whose altitude is H, the volume V of the frustum is given by the formula : y^BH B'(H-h) 3 3 * Given the pyramid P with base B and altitude H, and a frustum of it with bases B and B' and altitude h. To prove that the volume Vof the frus- tum is V= ^^ B'jH-h) '- 3 3 Fig. Proof. The frustum is the difference M 212 between the two pyramids P and P', where P' has tl le base B and the same vertex as P. The volume of P is BH/3. Why? Since the altitude of P' is H — h, its volume is B'(H-h) 3 Why? Hence V-P P = ^^ ^'(^-^>. o o 262 POLYHEDRONS [VII, § 308 308. Corollary 1. The volume V of a frustum of a pyramid of bases B and B' and altitude h is given by the formula : V={B +5' +VBByi /3. Outline of Proof. By § 301, using the notation of § 307. Hence VB'/VB = {H-h)/n=l- h/H, whence H=h Vb/{ VB - VW) . But, by § 307, V= BH/S -B'(H- h)/S = (B- B')H/S + B'h/3. Substituting the value of H just found, we have V = [ (^-^')^ ^B'l h/S =[B + VBB' + B'}h/S. ^ VB- y/B' J EXERCISES 1. Find the altitude of a triangular pyramid whose volume is 50 cu. in. and whose base is 12 sq. in. Ans. 12^ in. 2. If a prism and a pyramid have a common base and alti- tude, what is the ratio of their volumes ? 3. If the base of a pyramid is a square and its altitude is 3 ft., how long must each side of the square be in order that the volume may be 16 cu. ft. ? 4. Show that the volume of the tetrahedron, all of whose edges are equal to a, is V2 ayi2. [Hint. See Th. XXXIII, § 102.] 5. Find the volume of a frustum of the pyramid of Ex. 1 cut off by a plane 6 in. from the base. 6. Find the volume of each of the parts into which the pyr- amid of Ex. 3 is cut by two planes parallel to its base which trisect the altitude. VII, § 309] CYLINDERS 263 PART III. CYLINDERS AND CONES 309. Cylinders. A cylindrical surface is a curved surface generated by a moving straight line, called the generatrix, which moves always parallel to itself and constantly passes through a fixed curve called the directrix. The generatrix in any one position is called an element of the surface. One element and only one can be drawn through a given point on the cylindrical surface. A cylinder is a solid bounded by a cylindrical surface and two parallel planes. The two plane surfaces are called the bases, and the cylindrical surface is called the lateral surface. The altitude of a cylinder is the length of the perpendicular between the bases. A right section of a cylinder is a section made by a plane perpendicular to all its elements. Cylindrical Surface Right Cylinder Fig. 214 Oblique Cylinder A circular cylinder is one whose bases are circles. A right cylinder is one whose elements are all perpendicular to its bases ; otherwise, the cylinder is said to be oblique. A right circular cylinder is a right cylinder whose base is a circle. Such a cylinder can be generated by the revolution of 264 CYLINDERS [VII, § 310 a rectangle about one of its sides as an axis ; for this reason a right circular cylinder is sometimes called a cylinder of revo- lution. 310. Postulate. A prism is inscribed in a cylinder when its lateral edges are elements of the cylinder and its bases are inscribed in the bases of the cylinder. Fia. 215 In order to study the properties of the cylinder the following postulate is needed : A cylinder and a prism inscribed within it may he made to differ by as little as we please, both in lateral area and in volumCj by making the number of sides of the base of the prism sufficiently great, while the length of each of those sides becomes sufficiently small. The length of an edge of the inscribed prism is equal to the length of an element of the cylinder (see Ex. 5, p. 235) ; and, by increasing the number of sides of the inscribed prism, the base of the prism approaches, both in area and in the length of its perimeter, as nearly as we please to the base of the cylinder. This latter fact we assume, as in § 210. We shall now proceed to show that the theorems already proved for prisms can be extended to cylinders by the use of the preceding postulate. VII, § 313] AREAS AND VOLUMES 265 311. Theorem XIII. The lateral area A of any cylinder is equal to the product of an element I and the perim- eter p of a right section ; that is, A = lxp. Fig. 216 Outline of Proof. In order to prove the theorem, inscribe in the cylinder any prism whose base is a polygon of n sides. Then, for this pris7n, by § 277 : (1) A' = lxp', where A' and p' are, respectively, the Jateral area and the pe- rimeter of the right section of the prism ; and where I is the length of an edge of the prism, which is equal to an element of the cylinder. As the number n of sides increases so that the length of each side approaches zero, A' comes to differ from A by as little as we please ; § 310. Ixp' comes to differ from Z xp by as little as we please. § 310. Hence, by (1), A comes to differ from I xp by as little as we please. It follows, as in § 303, that A =1 X p. 312. Corollary 1. TJie lateral area of a right circular cylinder is equal to 2 nrh, where r is the radius of the circular base and h is the altitude of the cylinder. See § 214. 313. Corollary 2. Tlie lateral area of any cylinder whose right section is a circle is equal to 2 rrrrl, where r is the radius of the right section, and I is the length of an element. 266 CYLINDERS [VII, § 314 314. Theorem XIV. The volume V of any cylinder is equal to the product of its base B and its altitude h ; that is, V=Bxh. [The proof is left to the student. Inscribe a prism of n sides in the cyl- inder and use § 295 and § 310. Follow the steps suggested by § 311.] 315. Corollary 1. The volume of a circular cylinder is equal to TrrVi, where r is the radius of the base and h is the altitude of the cylinder. See § 216. EXERCISES [In these exercises, use the approximate value tt = 22/7. J 1. In a steam engine 65 flues, or cylindrical pipes, each 2 in. in outside diameter and 12 ft. long, convey the heat from the fire-box through to the water. How much heating surface is presented to the water ? Ans. 408^ sq. ft. 2. Neglecting the lap, how much tin is required to make a stovepipe 10 ft. long and 8 in. in diameter ? 3. A right circular cylinder has the radius of its base equal to 3 in. How great must its altitude be in order that it shall have a lateral area of 30 sq. in. ? 4. Find the total area, including the ends, of a covered tin can whose diameter is 4 in. and whose height is 6 in. Ans, 1004^ sq. in. 5. Derive a general formula for the total area (including the bases) of a right circular cylinder whose height is h and the radius of whose base is r. 6. What fraction of the metal in a tin can 5 in. wide and 5 in. high is used to make the top and bottom ? What to make the circular sides ? 7. If the diameter of a well is 7 ft. and the water is 10 ft. deep, how many gallons of water are there in it, reckoning H gal. to the cubic foot ? Ans. 2887.5 gals. VII, § 315] AREAS AND VOLUMES 267 8. When a body is placed in a cylindrical tumbler of water 3 in. in diameter, the water level rises 1 in. What is the volume of the body? Note that a method for finding the volume of a body of any shape is here illustrated. 9. Show that the volume Fand the lateral area ^ of a right circular cylinder are connected by the relation V= Ax r/2. 10. One gallon is 231 cubic inches. At what heights on a cylindrical measuring can whose base is 6 in. in diameter will the marks for 1 gallon, 1 quart, 2 quarts, 3 quarts, be made ? 11. Find the total area of the gallon measuring can of Ex. 10. 12. Having given the total surface !r of a right circular cylin- der, in which the height is equal to the diameter of the base, find the volume V- [Hint. —Call the height h.'] 13. Eind the volume of metal per foot of length in a pipe whose outer diameter is 3^ in., and whose inner diameter is 3 in. Hence find the weight per foot of length if the pipe is iron, it being given that iron weighs (about) 450 lb. per cubic foot. ^^^TTJT'-^^ 14. If the outer and inner diameters of a /^^^^^^^ tube are D and d, respectively, show that the '^M_ ^-^^-^^ volum^e in a length I is ttI{D'^ — d2)/4. If the HT'^ "^B thickness of the tube is denoted by t, show '^^^^^^^^i that t—(D — d)/2, and hence that the volume j---^^^^^fl_4 m a length I is i< d ^ '7Tlt{D + d)/2. 15. Show that the volume of the tube of Ex. 14 can also be represented by the formula Trlt{d + 1) ; or by the formula Trlt{D — t), in the notation of Ex. 14. 16. What is the volume of the largest beam of square cross section that can be cut from a circular log 16 in. in diameter and 10 ft. long ? What fraction of the log is wasted ? 268 CONES [VII, § 316 316. Cones. A conical surface is a surface generated by a moving straight line AB, called the generatrix, which passes always through a fixed point V, called the vertex, and constantly Fig. 217 intersects a fixed curve, called the directrix. A conical surface thus consists of two parts, which are. called the upper and lower nappes. The generatrix in any one position is called an element of the surface. A cone is a solid bounded by a conical surface and a plane which cuts all its elements. The plane is then called the base of the cone ; and the conical surface is called the lateral surface of the cone. The altitude of a cone is the perpendicular dis- tance from its vertex to its base. Fig. 218 A circular cone is one whose base is a circle. The axis of a cir- cular cone is the line joining the vertex to the center of the base. VII, § 316] AREAS AND VOLUMES 269 A right circular cone is a circular cone whose axis is perpen- dicular to the base. Such a cone is also called a cone of revolu- tion, since it may be generated by revolving a right triangle about one of its sides as an axis. LV B VM Fig. 219. Right Circular Cone and Section Parallel to Base The slant height of a cone of revolution is the length of one of its elements. A frustum of a cone is the portion of a cone included be- tween the base and a section parallel to the base and cutting all the elements. Fig. 220. Frustum of a Cone The lateral surface of a frustum of a cone is the portion of the lateral surface of the cone included between the bases of the frustum. The slant height of a frustum of a cone of revolution is the portion of any element of the cone included between the bases. 270 CONES [VII, § 317 317. Postulate. A pyramid is inscribed in a cone when its lateral edges are elements of the cone and its base is inscribed in the base of the cone. I The following postulate, corresponding to that of § 310, is needed in the study of the cone : A cone and pyramid inscribed within it may be made to differ by as little as lue please, both in lateral area and in volume, by making the number of sides of the pyramid sufficiently great, while the length of each side of the base becomes sufficiently small. Fig. 221. Cone with Inscribed Pyramid The base of the inscribed pyramid approaches, both in area and in perimeter, the base of the cone (§ 310); and the alti- tude of any face of the pyramid approaches an element of the cone, as the pyramid approaches the cone. If the cone is a right circular cone, the pyramid can be made a regular pyramid ; then the slant height of the pyramid ap- proaches the slant height of the cone. ' 318. Restriction. The word cone as used hereafter in this book will be understood to refer to a circular cone only. The preceding postulate applies, however, to any kind of cone; and it may be used to obtain results for cones of any form in the manner illustrated below for circular cones only. We proceed to extend to circular cones certain theorems already proved for pyramids. VII, § 321] AREAS AND VOLUMES 271 319.- Theorem XV. The lateral area A of a right circular cone is equal to one half the product of its slant height I and the circumfere7ice p of its base; that is, A = lxp/2. Outline of Proof. Inscribe a regular pyramid of n faces in the cone (see Fig. 221) ; then, by § 298, the lateral area A' of the pyramid is equal to one half the product of its slant height V and the perimeter p' of its base ; that is, A'=irxp'', and this formula is correct no matter how large n may be. By taking n sufficiently large, A' comes to differ by as little as we please from A ; while Z' and p' come to differ by as little as we please from I and p, respectively. § 317. Whence, as in § 311, A = ^lxp. 320. Corollary 1. T7ie lateral area of a right circular cone is irr ' I, where r is the radius of the base and I is the slayit height. See §§ 319 and 214. 321. Corollary 2. The lateral area of a frustum of a right circular cone is equal to one half the product of its slant height and the sum of the circumferences of its bases. Fig. 222 [The proof is left to the student. Inscribe a frustum of a regular pyramid in the given frustum of a cone, and use § 317 and § 299.] 272 CONES [VII, § 322 322. Theorem XVI. Any section of a circular cone paralleX to its base is a circle tvhose area is to that of the base as the squan of its distance from the vertex is to the square of the altitude of the cone. [Hint. To prove that the section is a circle, pass any two planes through the axis of the cylinder, and show that their intersections with the section are equal. Then inscribe a regular pyramid and proceed as in § 319, using §§ 301 and 317.] 323. Theorem XVII. 71ie volume V of a cojie is equal to one third the product of its base B and its altitude h ; that is, V == Bh/S. [Hint. Use §§ 317, 306, and proceed as in § 319.] 324. Corollary 1. Jf from any cone whose base is B and whose altitude is H, a frustum is cut, whose upper base is B' and whose altitude is h, the volume V of the frustum is 'V /h| \ 1 "'"'^p _[jn \ ^ 1 1 r\ Fig. 223 325. Corollary 2. The volume of a frustum of any cone is equal to the sum of three cones ivhose common altitude is the alti- tude of the frustum and whose bases are the two bases and a mean proportional between them. [Hint. Use §§ 322, 324, noting also § 308.] VII, §325] AREAS AND VOLUMES 273 EXERCISES 1. The altitude of a right circular cone is 12 in. and the radius of the base 9 in. Find the lateral area and the total area of the cone. Ans. 424^^ sq. in. ; 678f sq. in. 2. How many square yards of canvas are there in a conical tent 12 ft. in diameter and 8 ft. high ? 3. The total area of a right circular cone whose altitude is 10 in. is 280 sq. in. Find the total area of the cone cut off by a plane parallel to the base and 6 in. from the vertex. 4. The altitude of a right circular cone is H. How far from the vertex must a plane be passed parallel to the base so that the lateral area of the cone cut off shall be one half that of the original cone ? Ans. H/V2. [Hint. First prove that the area of the cross section made by the plane will be one half the area of the base. Then apply § 322.] 5. The slant height and the diameter of the base of a right circular cone are each equal to /. Find the total area, includ- ing the base. 6. The circumference of the base of a circular cone is 11 ft. and its height is 8 ft. What is its volume ? Ans. ^, or 25| cu. ft. 7. If the height of a circular cone is 10 ft., what must be the radius of its base in order that the volume may be 30 cu. ft. ? 8. A frustum of a cone is 1 ft. high and the radii of its bases are respectively 9 ft. and 4 ft. Find its volume. 9. If r and B are the radii of the bases of the frustum of a cone and I is its slant height, find the formula for its volume. 274 POLYHEDRONS [VII, § 326 PART IV. GENERAL THEOREMS ON POLYHEDRONS SIMILARITY REGULAR SOLIDS VOLUMES 326. Theorem XVIII. Two triangular pyramids 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 edges including the equal trihedral angles. h Given the triangular pyramids 0-FGH 2indi O'-F'G'II', with the trihedral Z = trihedral Z 0', and with volumes denoted by V and V, respectively. To prove that V/V =10F • OG • OH^/IO'F' • O'G' - O'lp-]: Proof. Place pyramid O'-FG'H' so that trihedral Z 0' will coincide with trihedral Z. 0. Ytotr H and H' draw HK Sixid H'K' perpendicular to the plane OFG. Then V/r = [A OFG - HKy[A OF'G' - H'K'] = [A OFGy[A OF'G'^ - [HK/H'K^^. Why ? A OFG OF'OG But § 193 A 0F'& OF' . OG' Again, let the plane determined by HK and H'K' intersect plane OFG in line OK'K. Then rt. A OKH ^ rt. A OK'H'. W^hy ? Therefore HK/H'K' = OH/OW Why ? V ^ OF' OG OH ^ O F'OG -OH V OF' 'OG' ' OH' O'F' ' O'G' • O'H'' Therefore VII, § 330] GENERAL THEOREMS 275 327. Corollary 1. Two triangular prisms that have a tri- hedral angle of the one equal to a trihedral angle of the other are to each other as the products of the edges including the trihedral angles. [Hint. Break the prism up into triangular pyramids, and use § 326 and Theorem H, § 144.] 328. Corollary 2. Two parallelepipeds 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 Qdges including the trihedral angle. 329. Similar Tetrahedrons. Two tetrahedrons (that iS; triangular pyramids) are said to be similar if their faces are similar each to each and similarly placed. 330. Theorem XIX. The volumes of two similar tetrahe- drons are to each other as the cubes of any two corresponding edges. ^H Fig. 225 Given the similar tetrahedrons 0-i^G^^and O'-F'G'H' with the volumes denoted by V and V', and with OF and OF' two corresponding edges. To prove that V/ V = 0F'/WF'\ § § 158, 271 Proof. Trihedral Z = trihedral Z 0^ § 329 Therefore V/r= OF 'OG 'OH ^ OF OO OH . 3^5 ^ O'F' . O'G' . O'H' O'F' ' O'G' * O'H'' But OF/O'F' = OG/O'G' = OH/0'H\ Why ? Therefore V/r = {OF/ O'F') {OF/ O'F') {OF/ O'F') = 0F'/WF'\ 276 POLYHEDRONS [VII, § 331 331. Similar Polyhedrons. In general, similar polyhedrons are polyhedrons which have the same number of faces similar each to each and similarly placed, and their corresponding polyhedral angles equal. In the case of similar tetrahedrons, the trihedral angles of the one are necessarily equal to those of the other, if we know only that the faces are similar each to each, since the simi- larity of the faces makes the three face angles at each vertex equal in the two tetrahedrons, by § 158. By § 330 and Theorem H, § 144, w^e can show that any two similar polyhedrons are to each other as the cubes of any two corresponding edges. 332. The Regular Solids. A regular polyhedron is one whose faces are all congruent regular polygons and whose polyhedral angles are all likewise congruent. Five types of such poly- hedrons are represented below. Fig, 226. —The Five Regular Solibs We shall now show that the above types are the only pos- sible types of regular polyhedrons. VII, § 333] GENERAL THEOREMS 277 333. Theorem XX. There exist only Jive different types of regular polyhedrons. Proof. The proof is based upon two facts : (1) that any- polyhedral angle has at least three faces and (2) that the sum of the face angles of any convex polyhedral angle must be less than 360°. (See § 273.) Suppose first that each face is to be a triangle. Then, from the definition of a regular polyhedron, the triangle must be equilateral. Each of its angles will therefore be 60°. Con- sequently, by statement (2) above, polyhedral angles may be formed by combining three, four, or five such angles, but no more than five can be thus used, since six such angles amount to 360°, while seven or more of them exceed 360°. Therefore, not more than three regular polyhedrons are possible having triangles as faces. The three that are possible are the regular tetrahedron, regular octahedron, and regular icosahedron. (See Fig. 226.) Suppose secondly that each face is to be a square. Each face angle must then be 90° and, the sum of four such angles being 360°, it follows that but one regular polyhedron is pos- sible having squares as sides. The cube is the one that is possible. Thirdly, suppose that each face is to be a regular pentagon. Since each of the angles of such a figure is 108°, it follows that no more than one regular polyhedron is possible whose faces are pentagons. The one that is possible is the dodeca- hedron. We can proceed no farther, for the sum of three angles of a regular hexagon is 360°, while the sum of three angles of any regular polygon of more than six sides is greater than 360°. Hence the theorem is proved. Note. These regular solids occur in nature in the forms of a variety of crystals ; but not all crystals are regular solids. 278 POLYHEDRONS [VII, § 334 334. Models. Models of the five possible regular polyhe- drons can be easily constructed as follows : Draw diagrams on cardboard as indicated in the figures below. Cut these out and then cut half way through the dotted lines so as to make it easy to fold along these lines. Fig. 227 Fold on the dotted lines so as to bring the edges together, sub- sequently pasting strips of paper over the edges to hold the solid in position. Models of the tetrahedron, octahedron, and icosahedron may also be made very quickly by hinging to- gether short umbrella wires by means of strong copper wires strung through the end holes, joining together at each corner the proper number of rods. The student may show that each of these models will be quite rigid when completed. VII, § 335] GENERAL THEOREMS 279 335. Theorem XXI. Cavalieri^s Theorem. If two solids are included between the same pair of parallel planes, and if every section of one of the solids by any plane parallel to one of these parallel planes is equal in area to the section of the other solid by the sanie plane, the volumes of the two solids are equal. Fig. 228 Outline of Proof. The two solids may be divided into a large number of thin slices by sections parallel to the two including planes. These slices may be thought of as approximately cylindrical, and the sum of all the slices in either case is the volume of the solid. The bases of two corresponding slices of the two solids between the same two planes are equal in area, by hypothesis. It is therefore apparent that the volumes of the two correspond- ing slices differ as little as we please if their thickness is suffi- ciently small. It can be shown in a precise manner that the total volume V of one of the solids, which is the sum of all such slices, differs from the total volume V of the other solid by as little as we please. Hence, as in §§ 303, 311, it follows that V = V'. Note. Observe also that §§ what precedes. 303 are essentially special cases of 280 POLYHEDRONS [VII, § 336 336. Theorem XXII. The Prismoid Formula. If any solid S of a7iy of the kinds considered in this Chapter is bounded by two parallel plane sections B and T, and if M denotes the area of another section parallel to and midway between B and T, the volume V of S is given by the formula : V=(B-\- T + 4.M)'h/(y, where B, T, and M denote the areas of the sections, and h denotes the distance beticeen B and T. The proof of the preceding formula consists in showing that it reduces in every case to the very formula for volume that has already been proved in the articles above. Outline of Proof for Prisms and Cylinders. In these cases, all parallel sections are equal (§§ 276, 310). Hence B= T= M; and the formula to be proved becomes V=B'h, which we have alread}* proved to be correct (§§ 296, 314). Outline of Proof for Pyramids and Cones. In these cases we know that the area of any section parallel to the base B is pro- portional to the square of its distance from the vertex (§§ 301, 322). Hence, since J^f is at a distance h/2 from the vertex, M (h/2y 1 T. A nr — = ^ ' ' =-, or B = 4:M. B W 4' The top section T is zero, since the top bounding plane meets a pyramid or a cone in just one point on the vertex. Hence the formula to be proved becomes, in this case, V=[_B+ r+ 4 Jf] . V6 = (J5 4- 4- B) - h/6 = Bh/3, which we know to be correct. §§ 306, 323 Outline of Proof for Frustums. Given a frustum of a pyramid or of a cone, let H be the distance from the vertex to the larger of the two bounding sections. Let B represent this larger section. Then H — h is the distance from O to the other bounding section T, and H — h/'2. is the distance from to the middle section M. We know that the volume V of the frustum is V = \_BH - T{H - h)y^. §§307,324 Or, since T/B = {H- hY/m, §§ 301, 322 we know that VII, § 337] GENERAL THEOREMS 281 V = B[H -{H- hY/m^/Z = -^[3 H^- SHh-\- h'^:\h/3. Since T/B ={H- hy^H^, and M/B ={H- h/2f/H^, the formula to he proved may be written, H" 3 This is equivalent to the formula that we know to be correct ; hence the theorem is proved. 337. Uses of the Prismoid Formula. The prismoid formula is a con- venient means of remembering the volumes of a variety of solids. We shall see in Chapter VIII that it holds for spheres and frustums of spheres as well as for the solids of this chapter. It also holds for any solid bounded by two parallel planes, made up by joining together pyramids, prisms, etc. , bounded by the same two planes ; such a solid is called a prismoid. Fig. 229 The formula is used very extensively by engineers to estimate the volumes of various objects, such as the volume of a hill, or the volume of a metal casting. Since the same formula holds for such a large variety of solids, it is reasonably safe to use it without even stopping to see which of these solids really resembles the object whose volume is desired. It is shown in more advanced books that the same formula holds when- ever the area of the section by any plane parallel to the bounding planes is proportional to the square of the distance from some fixed point to that plane. Many solids not mentioned otherwise in elementary geometry satisfy this requirement. 282 POLYHEDRONS [VII, § 337 MISCELLANEOUS EXERCISES. CHAPTER VII 1. Show that every section of a cylinder made by a plane passing through an element is a parallelogram. What is the section when the cylinder is a right cylinder ? 2. Show that every section of a cone made by a plane through the vertex is a triangle. What is the section when the cone is a right cone ? 3. If a, h, and c are the dimensions of a parallelepiped, show that the length of its diagonal is Va^ + &^ + c^ 4. How long an umbrella will go into a trunk measuring 32 in. by 17 in. by 21 in., inside measure, (a) if the umbrella is laid on the bottom ? (6) if it is placed diagonally between opposite corners of the top and bottom ? 5. Find the volume of a pyramid whose base is a rhombus 6 in. on a side and whose height is 6 in., if one angle of the rhombus is 60°. 6. The Great Pyramid in Egypt is about 480 ft. high and its base is a square measuring about 764 ft. on a side. Find approximately its volume in cubic yards. 7. Water is poured into a cylindrical reservoir 25 ft. in diameter at the rate of 300 gallons a minute. Find the rate (number of inches per minute) at which the water rises in the reservoir (1 gal. = 231 cu. in.). 8. A copper teapot is 9|- in. in diameter at the bottom, 8 in. at the top, and 11 in. deep. Allowing 42 sq. in. for locks and waste, how much metal is required for its construction, excluding the cover ? 9. A conical spire has a slant height of 60 ft. and the perimeter of the base is 50 ft. Find the lateral surface. 10. How many cubic inches of lead are there in a piece of lead pipe 2 yd. long, the outer diameter being 2 in. and the thickness of the lead being \ of an inch ? VII, §337] MISCELLANEOUS EXERCISES 283 11. The chimney of a factory has approximately the shape of a frustum of a regular pyramid. Its height is 75 ft. and its upper and lower bases are squares whose sides are 5 ft. and 8 ft. respectively. The flue*is throughout a square whose side is 3 ft. How many cubic feet of material does the chimney contain ? Assuming that a brick is 8 in. long, 3J in. wide, and 2J in. thick (as is ordinarily the case), estimate the number of bricks in such a chimney. 12. Compare the lateral areas, the total areas, and the volumes of (1) a right circular cylinder and a right circular cone having equal bases and altitudes, (2) a regular pyramid and a regular prism having congruent bases and equal altitudes. 13. A standard rain-gauge is made by inclosing a tube B in the interior of a can ACDE and connecting the mouth of the tube to the mouth of the can by a fun- nel FOHI. The amount of water, measured in inches (depth), that has fallen in the vicinity of the gauge is determined by read- ing the height of the water in the tube B. Find a formula for the amount of rain that has fallen in terms of the height h of the water in the tube B, the radius r of the tube, and the radius R of the can. Ans. hr^/R^. 14. If one of the edges of a tetrahedron is 1 in. long, how long will be the corresponding edge of a similar tetrahedron of 8 times the volume? Answer the same question for the case in which the new tetrahedron is to have half the volume of the original. Ans. 2 in. ; l/-v^2 = .79 in. 15. It is usual to state the diameter d of a tube in inches, and the area A of its surface in square feet. Show that the formula used by engineers : ^ = 0.2618 dZ gives very nearly the correct value in square feet, if d is meas- ured in inches, and the length I is measured in feet. CHAPTER VIII THE SPHERE PART I. GENERAL PROPERTIES 338. Spheres. A sphere is a portion of space bounded by a surface such that all straight lines to it from a fixed point within are equal. The fixed point within the sphere is called its center; a line segment joining the center to any point on the surface is Fig. 230 a radius; a line segment drawn through the center and ter- minated at both ends by the surface is a diameter. It follows from these definitions that : (a) The radii of a sphere, or equal spheres, are equal. (b) The diameters of a sphere, or of equal spheres, are equal. (c) Spheres having equal radii, or equal diameters, are equal. (d) A sphere may he generated by the revolution of a semicircle about its diameter. 284 VIII, § 339J GENERAL PROPERTIES 285 EXERCISES 1. What is the locus of the points that are 2 in. from the surface of a sphere whose radius is 4 in. ? 2. Show that the distance from the center of a sphere to a point outside the sphere is greater than the radius. (Use Ax. 10.) State the converse. Is it true ? 3. If two spheres have the same center, they are called concentric. Show that one of two concentric spheres lies wholly within the other. 4. Show that if the center of each one of two given spheres lies on the surface of the other, their radii are equal. 5. Show by § 77, that a plane perpendicular to a diameter of a sphere at its extremity has only one point in common with the sphere. 339. Tangent Planes and Lines. A plane that has only one point in common with a sphere is called a tangent plane to the sphere. A line that has only one point in common with a sphere is called a tangent line to the sphere. In either case, the single common point is called the point of tangency. 286 THE SPHERE [VIII, § 340 Fig. 231 340. Theorem I. A plane perpendicular to a diameter of a sphere at one of its extremities is tangent to the sphere. Outline of Proof. Let MN be a plane perpendicular to a diameter PP' at P. Connect any point A of MN to the center of the sphere 0. Show, by § 77, that 0A> OP; whence, by § 338, A cannot be on the sphere, so that P is the only point of the plane on the sphere. 341. Corollary 1. (Converse of Theorem I.) If a plane is tangent to a sphere, it is perpen- dicular to the radius draivn to the point of contact. [Hint. Show, by Ax. 10, that the radius is shorter than any other line drawn from the center to the plane. Then use § 77.] 342. Corollary 2. A straight line perpendicular to a diameter of a sphere at one of its extremities is tayigent to the sphere; and conversely. [Hint. Use §§ 254, 340 for direct, and § 116 for converse.] 343. Corollary 3. All of the straight lines tangent to a sphere at a given point lie in the plane tangent to the sphere at that point, [Hint. Use § 264.] EXERCISES 1. What is the locus of a point in space at a given distance from a given point ? 2. Prove that if two lines are tangent to a sphere at the same point; their plane is tangent to the sphere. [Hint. Connect this with one of the corollaries on this page.] 3. All lines tangent to a sphere from the same point are equal. [Hint. Connect the center of the sphere with the given point and with two or more points of tangency.] VIII, § 344] GENERAL PROPERTIES 287 344. Theorem II. Every section of a sphere made by a plane is a circle. 7\: A Fig. 232 Given the sphere whose center is 0, cut by a plane in the section AMBN. To prove that section AMBN is a circle. Proof. Draw OQ J- section AMBN\ join Q to C and D, any two points in the perimeter of the section ; draw OC and OD. In the rt. A OQC and OQD, OQ = OQ, and OC = OD. Why ? Therefore , A OQO ^ A OQD. Why ? Therefore QC = QD. Why? Since (7 and D are a^ii/ two points on the perimeter of the section, all points on the perimeter of the section are equally distant from Q. Therefore section AMBC is a circle. § 103 EXERCISES 1. If, in Fig. 232, the radius of the sphere, OC, is 10 in., and the distance OQ from the center to the plane AB is 6 in., find the radius CQ of the circle. 2. If, in Fig. 232, the distances CQ and OC are given, show how to find the distance OQ. 288 THE SPHERE [Vni, § 345 345. Great and Small Circles. A circle on the sphere whose plane passes through the center is called a great circle of the sphere, as CD, Fig. 233. Fig. 233 A circle on the sphere whose plane does not pass through the center is called a small circle of the sphere, as AB, Fig. 233. The axis of a circle of a sphere is the diameter of the sphere which is perpendicular to the plane of the circle. The poles of a circle of a sphere are the extremities of the axis of the circle. 346. Corollary 1. Through any three points on the surface of a sphere one and only one circle of the spheres may he drawn. [Hint. Use 4, § 241.] 347. Corollary 2. Through any two points on the surface of a sphere a great circle may he drawn. It follows from 4, § 241, that there is one and only one such great circle through the two given points, unless they lie at the opposite ends of a diameter. 348. Distance on a Sphere. By the distance between two points on the surface of a sphere is meant the length of the shorter arc of the great circle joining them. It can be shown that this is the shortest path on the surface of the sphere be- tween the two points. VIII, § 348] GENERAL PROPERTIES 289 EXERCISES 1. If we consider the earth as a sphere, what kind of circles are the parallels of latitude? the equator ? the meridians ? 2. Prove that the axis of a small circle of a sphere passes through the center of the circle ; and con- versely, a diameter of the sphere through the center of a small circle is the axis of that small circle. 3. Prove that in the same sphere, or in equal spheres, all great circles are equal. 4. The radius of a sphere is 10 in. Find the area of a section made by a plane 5 in. from the center. 5. The area of a section of a sphere 7 in. from the center is 288 TT sq. in. Find the area of a section 4 in. from the center. 6. Prove that in the same sphere, or in equal spheres, if two sections are equal, they are equally distant from the center, and conversely. 1 •■' 9. Q > \ & (3 t 7. Prove that any two great circles of a sphere bisect each other. 290 THE SPHERE [VIII, § 349 349. Theorem III. All points in the circumference of a circle of a sphere are equally distant from either one of its poles. - A,l :•: . 1 :D / \ -4 ' I o 1 -- j j Fig. 234 (a) Fig. 234 (Jb) Given any two points A and B in the circumference of the circle ABC, and P and P', the poles of ABC. To prove that PA = PB, and P^4 = jFb. Proof. Draw the great circles PAP' and PBP'. Let D be the intersection of the axis PP' with the plane ABC. Draw the straight lines AD, BD, PA and PB. Now PD = PD, and DA = DB, Why ? and Z PDA = Z PDB = 90°. Why ? Hence chord PA = chord PB, Why ? Therefore PA = PB. Why ? In the same way it may be proved that P'A = P'B. Note. The manner in which circles may be drawn on a sphere is illustrated by Fig. 234 (6). If one end of a string is held at any point on the sphere, while a pencil attached to the other end is moved around the sphere, keeping the string taut, the end of the pencil describes a circle on the sphere, by Theorem III. The various figures drawn in this chapter can be reproduced on the surface of an actual sphere, by this method of drawing the circles. VIII, § 351] GENERAL PROPERTIES 291 350. Polar Distance. The polar distance of a circle of a sphere is the distance on the sphere (§ 348) from its nearest pole to any point of the circumference, as PA or PB in Fig. 234. A quadrant is one fourth part of the circumference of a great circle ; i.e. an arc of 90° on a great circle. 351. Corollary 1. quadrant. The polar distance of a great circle is a Fig. 235 [Hint. Let ABC be a great circle. Then its center is also the center of the great circle PBP'. Hence the arc PB measures the right angle POB.'\ EXERCISES 1. What is the locus of all the points on the surface of the earth at a quadrant's distance from the north pole ? from the south pole ? from the equator ? See the figure for Ex. 1, p. 289. 2. The distance of the plane of a certain small circle from the center of a sphere is one half the radius of the sphere. If the diameter of the sphere is 12 in., find the polar distance of the small circle in degrees and in inches. Ans. 60° ; 2 tt in. 3. Show that a great circle on the earth whose poles lie on the equator passes through the north pole. 292 THE SPHERE [VIII, § 352 352. Problem I. To determine the radius of a given material sphere. "' ^-^ ! e ^•"^IfeL .0 / \ 1 Fig. 236 (a) Fig. 236 (&) Given any material sphere, OPQ. To find its radius. Construction. Take any point P on the surface of the sphere as a pole, and describe the circle ABC. Take any three points on this circle, as A, B, C. By means of the compasses construct on paper or on the blackboard the triangle A'B'C congruent to the triangle ABC. Circumscribe a circle around AA'B'C, and let D' be the center of this circle. Draw D''A'' equal to the radius lyA'. Through D" draw an indefinite line P'Q' perpendicular to ly'A". Erom A" lay off with compasses A'^P' equal to line AP. At A" erect a perpendicular to A"P' and extend it to meet P'Q' at Q\ Then P'^ is the diameter and P'Q'/2 is the radius of the given sphere. [The proof is left to the student. ] VIII, § 353] GENERAL PROPERTIES 293 353. Problem II. To construct a sphere through four given points not all in the same playie. Fig. 237 Given the four points A, B, C, D not all in the same plane. To construct a sphere that passes through A, B, C, and D. Construction. At E, the middle point of ABy erect a plane QEP perpendicular to AB. Likewise, let PFR be a plane perpendicular to .BO at its middle point F\ and let QOR be a plane perpendicular to -BZ> at its middle point G. Let be the point common to all three planes QEPj PFR, and QOR. With as center, and OA as radius, draw a sphere. This is the required sphere passing through A, B, C, and D. Proof. The plane QEP is the locus of all points equidistant from A and B. Why ? Likewise, PFR is the locus of points equidistant from B and C; and QGR is the locus of points equidistant from B and D. The planes QEP and PFR meet in a line OP. Why ? The line OP meets the plane QOR in a single point 0. Why? Therefore is equidistant from A, B, C, D. Why ? Moreover, is the only point equidistant from A, B, O, D. Why? 294 THE SPHERE [Vin, § 354 354. Inscribed and Circumscribed Spheres. A sphere is said to be circumscribed about any polyhedron when the ver- tices of the polyhedron all lie on its surface. Fig. 238. Circumscribed Sphere A sphere is said to be inscribed in any polyhedron when it is tangent to each of the faces of the polyhedron. / aD / \ A ''^i "^ Fia. 239. Inscribed Sphere 355. Corollary 1. One and only one sphere may he circum- scribed about any given tetrahedron (triangular pyramid). [Hint. Pass a sphere through the four vertices as in § 353. Notice that the four vertices cannot all lie in one plane.] VIII, §356] GENERAL PROPERTIES 295 356. Problem III. To inscribe a sphere in a given tetrahe^ dron (triangular pyramid). Given the tetrahedron ABCD, Fig. 239. To construct the sphere inscribed in it. Construction and Proof. Bisect the dihedral angles whose edges are BC, CD, and DB, by the planes BOC, CED, and DEB, respectively. The plane BOC is the locus of the points equidistant from the faces BCD and BAC; the plane CED is the locus of the points equidistant from the faces BCD and CAD ; and the plane DEB is the locus of the points equidistant from the faces BCD and DAB. Why ? The intersection of these planes is equidistant from the four faces of the tetrahedron. Hence the sphere whose center is and whose radius is the perpendicular distance OE from O to the face ABC, is tangent to each of the faces; it is therefore inscribed in the tetrahedron. No other sphere exists that is inscribed in the tetrahedron, for no other point than is equidistant from the four faces. Why? EXERCISES 1. By means of an instrument called a spherometer, the distances AD and DP, Fig. 236, can be measured directly. Show, by § 162, how to find the radius from these values. 2. Show that the process of § 352 can be used to find the radius of a sphere, if only a piece of the sphere is available, as in the case of a glass lens. 3. Show that four points in space determine a sphere, pro- vided they do not lie in one plane. 4. Show that a sphere is determined if any circle that lies on it and one pole of that circle are given. 5. Show that any two circles of a sphere completely de- termine the sphere. 296 THE SPHERE [VIII, § 357 PART II. SPHERICAL ANGLES — TRIANGLES — POLYGONS 357. Spherical Angles. The line tangent to a great circle of a sphere at any point is a tangent to the sphere at that point ; for it touches the sphere in only one point (§ 339). The angle formed by the intersection of two great circles is called a spherical angle. It is defined to be equal to the angle formed by the tangents to the two great circles, at their point of intersection, as the angle CPD, Fig. 240. 358. Theorem IV. Tlie angle hetioeen two great circles is measured hy the arc of a great circle described from its vertex as a pole and included hetiveen its sides. Fig. 240 Given the great circles PAP and PBP' intersecting at P, and AB the arc of a great circle described with P as a pole. To prove that ^B is the measure of Z APE. Proof. Draw the radii OAj OB, and the tangents PC, PD. Then OA II PC and OB II PD. Why ? Hence Z. AOB =^ Z. CPD. Why ? But Z. AOB is measured by the arc AB', hence Z CPD is VIII, § 360] TRIANGLES AND POLYGONS 297 measured by the arc AB. It follows that the spherical angle APB, which is equal to Z CPD by definition (§ 357), is meas- ured by the arc AB. 359. Corollary 1. The spherical angle between two great circles is equal to the plane angle of the dihedral angle formed by the planes of the tivo great circles. 360. Spherical Triangles and Polygons. A spherical polygon is a portion of a spherical surface bounded by three or more arcs of great circles; as ABODE, Fig. 241. The bounding arcs of great cir- cles are called the sides of the spherical polygon ; their intersec- tions, the vertices ; and the angles formed by the sides at the ver- tices, the angles of the spherical polygon. A diagonal of a spherical polygon is an arc of a great circle joining any two non-adjacent ver- tices. A spherical triangle is a spherical polygon of three sides, as ABC, Fig. 242. The words isosceles, equilateral, acute, right, and obtuse are applied to spherical triangles in precisely the same way as to plane triangles. Thus, in Fig. 242, the spherical triangle ABC is isosceles if the two sides, as AB and BC, are equal; the triangle is equilateral if AB — BC = AC\ the tri- angle is a right triangle if any one angle is a right angle ; etc. Fig. 241 Fig. 242 298 THE SPHERE [VHI, § 361 361. Relation to Central Polyhedral Angles. The planes of the arcs of the great circles forming the sides of a spher- ical polygon meet at the center of the sphere and form a poly- hedral angle, as 0-ABCD. Fig. 243 This polyhedral angle and the spherical polygon are so closely related that the student can easily prove the following statements : (a) The sides of a spherical polygon have the same measures as the corresponding face angles of the polyhedral ajigle. (b) The angles of the spherical polygon have the same measures as the corresponding dihedral angles of the polyhedral angle. Thus, sides AB, BO, etc., of the spherical polygon ABCD have the same measures as face A AOB, BOO, etc., of polyhe- dral Z O-ABOD; and spherical A ABO, BOD, etc., have the same measures as the dihedral A whose edges are OB, 00, etc. (c) Any angle of a spherical polygon (or, the corresponding dihedral angle of the polyhedral angle) is measured by the arc of a great circle described with the vertex of the angle as pole and terminated by the sides. See §§ 858, 359. In general, any fact proved for the sides and the angles of a spherical polygon is true also for the ' corresponding face angles and dihedral angles of the corresponding central polyhedral angle. VIII, § 363] TRIANGLES AND POLYGONS 299 362. Theorem V. The sum of any two sides of a spherical triangle is greater than the third side. [Compare § 272.] /^: x^.. \ H ; Fig. 244 Given the spherical A ABC. To prove that AB -\- BC > 62. Proof. Z AOB + Z BOO > Z COA, Z AOB is measured by AB, Z BOG is measured by BO, Z COA is measured by CA. Therefore AB -{- BC > CA. 363. Theorem VI. The sum of the sides of any convex spherical polygon is less than 360°. [Com- pare § 273.] Given the spherical polygon ABCD. To prove that AB-{-BC+cb-\-DA< 360°. [Hint. Make use of § 273.] §272 Why? Fig. 245 300 THE SPHERE [Vni, § 364 EXERCISES 1. Show that any side of a spherical polygon is less than 180°. 2. In the spherical A ABC, AB = 35°, and BC tween what limits must CA lie ? 75°. Be- 3. Three of the sides of a spherical quadrilateral are respec- tively 88° IT, 70° 36', and 50° 33'. Between what limits must the fourth side lie ? 364. Polar Triangles. If from the vertices of a spherical triangle as poles arcs of great circles are drawn, these arcs form a second triangle which is called the polar triangle of the first. Fig. 24() Thus, if A, B, G, the vertices of the spherical A ABC, are the poles of the arcs B'C, A'C, A'B', forming the spherical A A'B'C, then A'B'C is the polar triangle of ABC. If the entire circles be drawn, they will intersect so as to form eight spherical triangles, but the polar of the given triangle ABC is that one of the eight triangles whose vertices lie on the same side of the arcs of the given triangle as the corresponding vertices of the given triangle, and no side of which is greater than 180°. VIII, §365] TRIANGLES AND POLYGONS 301 365. Theorem VII. If one spherical triangle is the polar of another y then the second is the polar of the first. Fig. 247 Given A A'B'C, the polar of A ABC. To prove that A ABC is the polar of A A'B'C. Proof. A is the pole of B'C, and (7 is the pole of A'B' ; Given. hence B' is at a quadrant's distance from A and (7, so that B' is the pole of AG. § 249 Similarly, A' is the pole of BC, and C" is the pole of AB. Therefore ABC is the polar triangle of A'B'C\ § 364. EXERCISES 1. Show that if one side of a spherical triangle on the earth's surface is on the equator, one vertex of the polar triangle is either at the north pole or at the south pole. 2. Show that if one vertex of a triangle on the earth is at the north pole, one side of the polar triangle is on the equator. 3. Show that if one vertex of a triangle on the earth is at the north pole, and if one side of the triangle is on the equator, the polar triangle also has one vertex at the north pole and one side along the equator. 302 THE SPHERE [Vni, § 366 366. Theorem VIII. In tivo polar triangles, each angle of the one is measured by the supplement of the side opposite to it in the other. ^/€^~<^- /^c r--x^ ' V / / a\ dV' h X -B), 1 b r . 5>\/'' ^ V'^^-"""''''^ -' -7^ C^-''' ' .^'^' 1 Fig. 248 Given the polar triangles ABC and AB'O, with the sides denoted by a, 6, c, and a', h' , c', respectively. To prove that (a) Z ^ + a' = 180°, Z5 + 6' = 180°, Z(7+c' = 180°; lh)/.A + a =180°, ZJB' + 6 = 180°, ZC" + c = 180°. Proof. Let AB and AC (prolonged, if necessary) intersect B'C at D and E, respectively. Then Therefore That is C^D Why? Why? (c) § 361 90°, andjB;J5' = 90°. O'i) + ^^' = 180°. OE ■\-ED-\-ED + DB' = 180°, or, ^Z> + a' = 180°. But ED is the measure of Z A. Therefore Z^ + a' = 180°. In a similar manner Z J5 + &' = 180°, and Z (7+ c' = 180°. The proof of (6) is left for the student. EXERCISE 1. If the angles of a spherical triangle are 70°, 90°, and 80°, respectively, find the sides of the polar triangle (in degrees). VIII, § 368] TRIANGLES AND POLYGONS 303 367. Theorem IX. The sum of the angles of a spherical tri- angle is greater tha,n 180° and less than 540°. Fig. 249 Given the spherical A ABC with the sides a, b, and c. - To prove that ZA-\-ZB-\-ZO> 180° and < 540°. Proof. Let AA'B'G\ with its sides denoted by a', h', and c', be the polar of A ABC. Then ZA + a' = 180°, Z 5 + 6' = 180°, AC+c' = 180°. Therefore Z. A -\- A B -\- /. C + a' + ?>' + c' = 540°. Why? But ' a' 4- &' + c' < 360°. § 363 Therefore ZA^-ZB-\-ZC> 180°. Why ? Again a' + 5' + c' > 0°. Therefore Z ^ + Z5 + Z C< 540°. Why? 368. Corollary 1. In a spherical triangle there can he one, two, or eveyi three right angles; there can he one, two, or three ohtuse angles. EXERCISES 1. Show that a triangle on the earth's surface whose sides are the equator and two meridians, has two of its angles right angles, and two of its sides quadrants. 2. If, as in Ex. 1, two of the angles of a spherical triangle are right angles, between what limits must the third angle lie ? 304 THE SPHERE [Vni, § 369 369. Birectangular and Trirectangular Triangles. A spherical triangle having two right angles is called a birectangu- FiG. 250 lar spherical triangle. A spherical triangle having all of its angles right angles is called a trirectangular spherical triangle. If Z P in the figure is either acute or obtuse, while A A and B are right, AABP is birectangular; if ZP is also a right angle, A ABP is trirectangular, as in Fig. 230, p. 284, EXERCISES 1. The sides of a epherical triangle are 80°, and 126°, and 175°. How large are the angles of its polar triangle ? 2. Show that in a birectangular triangle the sides opposite the right angles are quadrants. 3. Show that three mutually perpendicular planes through the center of a sphere divide its surface into eight congruent trirectangular triangles. 4. Show that the area of a trirectangular triangle on a sphere is one eighth of the area of the sphere. 5. Show that each of the sides of a trirectangular triangle is a quadrant. Hence show that the polar of a trirectangular tri- angle coincides with it. VIII, §371] TRIANGLES AND POLYGONS 305 370. Symmetric Triangles. Two spherical triangles are symmetric when their parts are equal each to each, but are in opposite order. Thus, in the A ABC and A'B'C (Fig. 251), Fig. 251 if angles A = A', B = B', C= C, and sides AB = A'B', BG = B'C, CA = G'A'j but the order of arrangement is opposite in the two figures, the triangles are symmetric. In general, two symmetric triangles cannot be superposed and hence cannot be said to be congruent. Thus, if A ABC is moved so that side AB coincides with its equal, A'B'j in the symmetric A A'B'C, then the vertices G and (7 lie on opposite sides of A'B'. In plane triangles, A ABC could be revolved about ^B till it coincided with A A' B'C; but this is in general impossible with spherical triangles. 371. Corollary. Two isosceles syinmetric spherical triangles are congrue7it. EXERCISES 1. Prove that the base angles of an isosceles spherical tri- angle are equal. [Hint. Draw an arc bisecting the vertical angle, thus forming two symmetric triangles.] 2. Show that if two sides of a spherical triangle are quad- rants, the triangle is birectangular. 306 THE SPHERE [VIII, § 372 372. Theorem X. Tivo triangles on the same sphere, or 07i equal spheres, are either congruent or symmetric, if two sides and the included angle of the one are equal, respectively, to two sides and the included angle of the other. / Fig. 252 (a) Fig. 252 (b) Given the spherical A ABC and A'B'C on the same sphere or equal spheres, having AB = A'B', AC=A'C', Z A = Z. A'. To prove that ^ ABC and A'B'C are either congruent or else symmetric. Proof. If the equal parts of the two triangles are in the same order, A ABC can be placed on A A'B'C as in the corre- sponding case of plane triangles. See Fig. 252 (a). If the equal parts of the two triangles are not in the same order,construct A^'^'O" symmetric to A^'^'C". (Fig.252 (6).) In A ^5(7 and A!B'C', AC = A'C", AB = A'B', Sind Z A = Z B'A'C". Since these parts are arranged in the same order, A ABC and A'B' C" are congruent. Therefore spherical A ABC is symmetric to spherical A A'B'C. Why? 373. Theorem XI. Two triangles on the same sphere, or on equal spheres, are either congruent or symmetric, if two angles and the included side of the one are equal, respectively, to two angles and the included side of the other. [Proceed as in § 372.] VIII, §375] TRIANGLES AND POLYGONS 307 374. Theorem XII. Two triangles on the same sphere, or on equal spheres, are either congruent or symmetric, if the three sides of the one are equal, respectively, to the three sides of the other. [The proof is left to the student. ] 375. Theorem XIII. Two triangles on the same sphere, or on equal spheres, are either congruent or symmetric, if the three ' angles of the one are equal, respectively, to the three angles of the other. Fig. 253 Outline of Proof. If ABO and A'B'C are the two given spherical triangles so that ZA = ZA',ZB = ZB', ZC = AC', their polar triangles LMN and VM^N^ have the three sides of one equal to the three sides of the other, respectively. §366 Then, by § 374, A LMN and L'M'N' are either congruent or symmetric. In either case, the three angles of A LMN are equal to the three angles of A VM^N, respectively ; and there- fore the three sides of A ABC are equal to the three sides of A^'^'C, respectively. § 366 It follows, by § 374, that A ABC and AB'G' are either con- gruent or symmetric. Note. Theorems analogous to those of §§ 41, 43, 44, etc., may be proved in a manner similar to §§ 372-375. 308 THE SPHERE [VHI, § 375 EXERCISES 1. Show that two trihedral angles are congruent if they intercept congruent triangles on the surfaces of two equal spheres whose centers are at their vertices, respectively. 2. If two trihedral angles intercept symmetric spherical tri- angles on the surface of a sphere whose center is at their ver- tices, respectively, show that the face angles and the dihedral angles of one trihedral angle are equal to those of the other, but taken in reversed order. [Such trihedral angles are called symmetric] 3. Prove the following theorem, which states for trihedral angles (§ 361) a theorem analogous to that of § 372 : Two trihedral angles are either congruent or symmetric if two face angles and the included dihedral angle of the one are respec- tively equal to two face angles and the included dihedral angle of the other. [Hint : Consider the spherical triangles cut out by the two trihedral angles on the surfaces of two equal spheres whose centers lie at the ver- tices of the two trihedral angles, and apply § 372.] 4. Prove the following theorem, analogous to § 373 : Two trihedral angles are either congruent or symmetric if two dihedral angles and the included face angle of the one are respec- tively equal to two dihedral angles and the included face angle of the other. 5. State and prove theorems for trihedral angles similar to Theorems XII-XIII, § 374-375. 6. Prove Theorem XI, § 373, by first considering, as in § 375, the polars of the given triangles, and applying § 372. 7. Show that any trirectangular triangle on the earth's sur- face is congruent to the trirectangular triangle formed by the equator and two meridians whose longitude differs by 90°. VIII, § 376] AREAS AND VOLUMES 309 PAET III. AEEAS AND VOLUMES 376. Theorem XIV. The area of the surface generated by a straight line revolving about an axis in its plane is equal to the product of the projection of the line on the axis and the length of the circle whose radius is a perpendicular erected at the middle point of the line and terminated by the axis. Given EF, the projection upon XF of AB revolving about XY, and OF A. AB at its mid-point, and meeting XY at 0. To prove that the area generated by AB = EF x 2 ttOP. Proof. Draw PD J. XY, and AG II XY Since the surface generated by AB is the lateral surface of the frustum of a cone, the area generated by AB is f(: Now Therefore Then And AB x2 ttPD = EF x 2 ttOP. That is, the area generated by AB is EF x 2 ttOP. If AB meets XY, the surface generated is a conical surface whose area again =z EF x 2 ttOP. § 320 If AB is parallel to XY, the surface generated is a cylin- drical surface whose area again = EF x 2 it OP. § 313 2 irAE + 2 'jtBF^ - ^B X 2 tt . M^ + ^^A = ABx2Tr'PD. §321 A ABC ^ APOD. § 157 AB:OP=AC:PD. Why? ABx PD = AC X 0P= EFx OP. Why? 310 THE SPHERE [VIII, § 377 377. Theorem XV. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. C Given a sphere generated by the revolution of the semicircle ABCDE about the diameter AOE, S 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 ABCDE, of any number of sides, and draw BF, CO, DO, per- pendicular to AE. From draw 0P1.AB. Then OP bisects AB, Why? and is equal to each of the Js drawn from to the equal Why? § 376 chords BC, CD, DE. Now the area generated hy AB = AF x 2 tt • OP, the area generated hj BC = FO x 2 tt • OP, the area generated by CD = OG X 2 tt - OP, the area generated hy DE = GE x 2 tt • OP. Therefore, if JS' denotes the surface generated by the semi- polygon, S' = (AF+FO-^OG+GE)2TrOP=AEx2 7rOP. Let the number of sides of the semipolygon be now indefi- nitely increased. Then OP has for its limit r, the semipolygon for its limit the semicircle, and S' for its limit S. Hence, as in § 303, S = AEx2 7rr. 378. Corollary 1. to 4 TTJ^. The area of the surface of a sphere is equal VIII, § 382] AREAS AND VOLUMES 311 379. Corollary 2. The area of the surface of a sphere is equal to the sum of the areas of four great circles. For S = 2rx2 7rr = 4:7rr^ §378 and Trr^ is the area of a great circle. 380. Corollary 3. The areas of the surfaces of two spheres are to each other as the squares of their radii; or, as the squares of their diameters. 381. Zones. A zone is a portion of the surface of a sphere bounded by the circumferences of two circles whose planes are parallel. Fig. 256. Zones on the Earth's Surface The circumferences forming the boundary of a zone are its bases. If the semicircle NES is revolved about NS as an axis, arc AB will generate a zone, while points A and B will generate the bases of the zone. The altitude of a zone is the perpendicular distance between the planes of the bases. 382. Corollary 4. The area of a zone of a sphere is equal to the product of the altitude h of the zone and the circumference of a great circle; or 2Trrh, where r is the radius of the sphere. 312 THE SPHERE [VIII, § 383 383. Lunes. A lune is a portion of a spherical surface bounded by two semicircumferences of great circles ; as \ ^' i Fig. 257 ABGDA (Fig. 257). The angle of a lune is the angle formed by its bounding arcs. Thus BAD is the angle of the lune ABCDA. 384. Theorem XVI. Tlie area of a lune is to the area of the surface of the sphere as the angle of the lune is to four right angles. Fig. 258 Given the lune PAP'B, let L denote the area of the lune, S the area of the surface of the sphere, and a the angle of the lune. VIII, § 386] AREAS AND VOLUMES 313 To prove that L/S = Z a/4 rt. A. Proof. With P as a pole describe the great circle ABCD. Then the arc AB measures Z a of the lune. Why ? Therefore arc ^B/circle ABCD = Z a/4 rt. A. If AB and ABCD are commensurable, let their common measure be contained m times in ^S and n times in ABCD. Then arc ^5/circle ABCD = m/n. Therefore a/4 rt. A = m/n. § 358 Pass arcs of great circles through each point of division of ABCD and the poles P and P'. These arcs will divide the entire surface into w equal lunes, of which PAPB will contain m. Therefore L/S = m/n, or, L/S = a/4 rt. A. If AB and ABCD are incommensurable, the theorem can be proved as in § 130. The details are left to the student. 385. Corollary 1. Hie area of a lune whose angle is 1° is 4 7rrV360 = 7rrV90. 386. Corollary 2. The area of a lune whose angle is k° is 4 7rr^k/S60 = 7rr^k/90. EXERCISES 1. If the surface of a sphere is 10 sq. ft., what is the area of a lune whose angle is 40°? What is the radius of the sphere ? Ans. li sq. ft. ; 0.89+ ft. 2. Show that two lunes on the same sphere or equal spheres have the same ratio as their angles. 3. What is the angle of a lune which has the same area as a trirectangular triangle ? 4. Show that the area of a lune is one ninetieth of the area of a great circle multiplied by the number of degrees in the angle of the lune. 314 THE SPHERE [Vni, § 387 387. Theorem XVII. Two symmetric triangles are equal m area. Given the two symmetric spherical triangles ABC and A^B'O. To prove that AABC^AA'B^a. Proof. Let P be the pole of the small circle passing through the points A, B, C, and draw the great circle arcs PA, PB, and PC. Then PA = PB = PC. Fig. 259 Why? Now place the two triangles diametrically opposite to each other and draw the diameter POP. Also draw the great circle arcs PA', P'B', and PC. Then the triangles PBC and PB'C are symmetrical and isosceles and therefore congruent. § 371. Similarly APCA^APGA', and APAB^APA'B'. That is, the three parts of ABC are respectively congruent to the three parts of ABC Therefore A ABC = A A'B'C. *i Fig. 260 388. Corollary 1. If tivo semi- circumferences of great circles BCB' and AC A intersect on the surface of a hemisphere, the sum of the areas of the two opposite spheri- cal triangles ACB and A'CB' is equal to the area of a lune whose angle is equal to ACB. [Hint. Show tliat the triangle ABC is symmetric to the triangle A'B'C. Hence show that A ACB + A A'CB' = lune A' CB'C] VIII, § 391] AREAS AND VOLUMES 315 389. Spherical Degree. The area of a lune whose angle is 1° is 4 7rr2/360, or 7rrV90 (§ 385). Half this area, that is, 47rrY720 or Trr^/lSO, is often taken as a unit of area on the sphere, and it is called a spherical degree. 390. Measure of Solid Angles. A trihedral angle whose vertex is at the center of a sphere cuts out a spherical triangle on the surface of the sphere. The area of the spherical tri- angle, in spherical degrees, is called the measure of the trihedral angle. Likewise any polyhedral angle is measured by the area, in spherical degrees, that it cuts out upon the surface of a sphere whose center is at its vertex. EXERCISES 1. Show that the area of a lune whose angle is 1° is 2 spher- ical degrees. 2. Show that the area of the entire sphere is 720 spherical degrees 3. Show that the area of a birectan- gular triangle whose third angle is 1° is 1 spherical degree. 4. Show that the area of a trirectan- gular triangle is 90 spherical degrees, or one eighth of the entire surface. 391. Spherical Excess. The excess of the sum of the angles of a spherical triangle over 180° is called the spherical excess of the triangle. If, for example, the angles of a spherical triangle are 80°, 100°, and 125°, the spherical excess of the triangle is 125°. Likewise, the spherical excess of any spherical polygon is the excess of the sum of its angles above the sum of the angles of a plane polygon of the same number of sides. 316 THE SPHERE [Vin, § 392 392. Theorem XVIII. The area of a spherical triangle is equal to the area of a Uine luhose angle is half the spherical excess of the triangle. Fig. 261 Given the spherical A ABC. To prove that A ABC is equal to a lune whose angle is i(Z^ + ^-B + Z (7-180°). Proof. Complete the great circles by producing the sides of the A ABC, as in Eig. 261. Since A AB'C and A'BC are symmetric, they are equal in area § 387 Therefore lune ABA'C= A ABC + A AB'C. § 388 But, denoting the area of the whole sphere by S, A CB'A -j- A ACB + A ABC-^A AB'C = iJS. Why ? Therefore (lune BCB'A - A ABC) + (lune CAC'B - A ABC) + lune ABA'C= ^ S. Why ? Therefore, transposing, we obtain 2 A ABC = lune .^1B^'C+ lune BCB'A + lune CAC'B - I S. But -1- S is the area of a lune whose angle is 180°. Therefore A ABC is equal to a lune whose angle is i(AA+ZB-{-ZC- 180°). § 384 VIII, § 396] AREAS AND VOLUMES 317 393. Corollary 1. The area of a spherical triangle, measured in spherical degrees, is numericallij equal to its spherical excess. Note. This result enables us to compute the area of any spherical tri- angle in ordinary units of area, when we know its angles and the radius of the sphere. Thus, if r denotes the radius of the sphere, E the spheri- cal excess, and A the required area, we have, by § 385 ^-^""l^Q- 180* 394. Corollary 2. The area of a trirectangular triangle is 90 spherical degrees. 395. Corollary 3. Tlie area of any spherical triaiigle is to the area of the entire sphere as its spherical excess is to 720°. 396. Solid Angles. In general, if any closed polygon or curve is drawn on the surface of a sphere, the figure formed by all radii of the sphere that join the center to the points of this figure on the spherical surface is called a solid angle. The area on the surface of the sphere cut out by such a solid angle, in spherical degrees, is the measure of the solid angle. EXERCISES 1. What is the measure of a hemisphere in spherical degrees ? 2. The radius of a sphere is 2 ft. Find the area of a triangle on its surface whose angles are 75°, 35°, 105°, respectively. Solve first by § 392 ; then by § 394. Ans. 7 7r/9 sq. ft. 3. The radius of the earth is approximately 4000 miles. Find the entire area. Show that the area in square miles of one spherical degree is approximately 278,000 square miles. 4. Find how large a triangle on the earth's surface would have the total sum of its three angles equal to 181°. 5. Show that a region containing about 270,000 sq. mi. on the earth contains no triangle whose spherical excess is 1°. 6. What is the area of the state in which you live ? What is its measure in spherical degrees ? 318 THE SPHERE [VIII, § 397 397. Theorem XIX. The volume V of a sphere is equal to the product of its surface by one third of its radius ; or, • .'i? A 1 A. B Fig. 2G2 Given a sphere whose center is O; let S denote its surface, r its radius, and V its volume. To prove that V= S x r/3 = 4 ttt^/S. Proof. Circumscribe about the sphere any polyhedron as D-ABC, and denote its surface by S' and its volume by V\ Form pyramids, as 0-ABC, etc., having the faces of the polyhedron as bases and the center of the sphere as a common vertex. These pyramids will have a common altitude equal to r, and the volume of each pyramid is equal to its base X r/3. Why? Therefore V ==S'x r/3. Why ? If the number of pyramids is indefinitely increased by pass- ing planes tangent to the sphere at points where the edges of the pyramids cut the surface of the sphere, as in Fig. 262, the difference between S and S' becomes as small as we please ; the difference between V and V becomes as small as we please. VIII, § 400] AREAS AND VOLUMES 319 But however great the number of pyramids, r=/S'Xr/3. Therefore, as in § 303, V=Sx r/3. Since /S = 4 Tr?'^, it follows that F = 4 Trr^ x r/3 = 4 7r?V3. 398. Corollary 1. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. 399. Corollary 2. The volmne of the pyramidal piece cut out of a sphere by any polyhedral angle whose vertex is at the center is equal to one third the area of the spherical polygon cut out of the surface times the radius. 400. Corollary 3. The prismoid formula (§ 336) holds for a sphere. [Hint. Two parallel planes that include the entire sphere are tan- gent planes at the ends of a diameter ; these cut the sphere in only one point each. A plane parallel to these two and halfway between thera cuts the sphere in a great circle. Hence, in the notation of § 336, B — 0^ T = 0, ilf = 4 Trr^, h = 2r ; hence the prismoid formula would give V=h r ^+ r + 4Jf -1 ^ 2 r r O + + 4 Trr ^-i ^Airr^^ which, by § 397, is correct.] [Note, As a matter of fact, the prismoid formula holds for the por- tion of a sphere intercepted between any two parallel planes.] EXERCISES 1. Assuming that the earth is a sphere whose radius is 4000 mi., find its volume. 2. Show that a cube circumscribed about a sphere has a volume 8 r^. Hence show that the sphere occupies a little more than half this volume 320 THE SPHERE [VIH, § 400 3. Find the volume of the material in a hollow sphere, if the radius of the outer surface is 6 in. and that of the inner surface is 5 in. 4. Show that the volume of a hollow sphere whose outer and inner radii are B and r, respectively, is 4 tt {R^ — r^)/3. 5. Find the volume of the material in a hollow sphere whose outer radius is 10 in., if the material is ^ in. thick. 6. Show that the volume of a sphere in terms of its diame- ter, d, is 7rdy6. 7. If the radius of one sphere is twice that of another, how do their volumes compare ? 8. If the volume of one sphere is twice that of another, how do their radii compare ? 9. Find approximately the radius of a sphere whose volume is 100 cu. in. 10. How many shot -^^ in. in diameter can be made from 10 cu. in. of lead ? 11. If oranges 3 in. in diameter sell for 30 cents per dozen, and those 4 in. in diameter sell for 50 cents per dozen, which are the cheaper by volume ? 12. If the skins are of equal thickness, which of the oranges of Ex. 11 has the greater percentage of skin to the cubic inch of volume ? 13. Assuming that raindrops are practically spherical, if the diameter of one drop is half that of another, how do their volumes compare ? their areas ? 14. Which of the two drops of Ex. 13 has the greater ratio of area to volume ? How much greater ? Which will fall the more rapidly through the air ? [Hint. The greater the ratio of area to volume, for the same material, the slower the body will fall through the air. ] VIII, §400] AREAS AND VOLUMES 321 15. Explain, by the principle of Ex. 14, why very small dust particles remain floating in the air for a long time. 16. The same amount of material, in the form of a cube, is melted and cast into a sphere. Is the surface area less in the form of the cube or in that of the sphere ? [Hint. Assume the cube to be 1 unit on each edge ; find the radius of the resulting sphere,] 17. If a surveyor wishes to be certain that the sum of the angles in any triangle in a region on the earth's surface shall be equal to 180° to within one minute, how large may the re- gion be ? Ans. About 4600 sq. mi. 18. Demonstrate the existence of spherical triangles with three obtuse angles from the existence of triangles whose sides are very short. IT TABLES TABLE I Ratios of the Sides of Right Triangles and Chords and Arcs of a Unit Circle TABLE II Squares and Square Roots of Numbers Cubes and Cube Roots of Numbers TABLE III Values of Important Numbers including Units of Measurement TABLE I RATIOS OF THE SIDES OF RIGHT TRIANGLES AND LENGTHS OF CHORDS AND ARCS OF A UNIT CIRCLE EXPLANATION OF TABLE I 1. Ratios of the Sides of Right Triangles. If an angle given in the Angle Column is one acute angle of a right triangle: The Sine Column gives the ratio of the side opposite the angle to the hypotenuse ; The Tangent Column gives the ratio of the side opposite the angle to the side adjacent to the angle. To find the Cosine of any angle, take the sine of the comple- ment of that angle. 2. Chords and Arcs of a Unit Circle. If an angle given in the Angle Column is an angle at the center of a circle of unit radius : The Chord Column gives the length of the chord that subtends that angle ; The Arc Column gives the length of the arc that subtends that angle. To find the lengths of chords or arcs of any circle of radius r, multiply the values given in the table by that radius. The table is limited to angles less than 90° ; but to find the chord that subtends an obtuse angle, first take half the angle, find the sine of this half angle, and multiply by 2. This follows from the fact that the chord of any angle is twice the sine of half that angle. I] Quantities Determined by a Given Angle iii Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord Arc 0°00' .0000 .0000 .0000 .0000 9° 00' .1564 .1584 .1569 .1571 10 .0029 .0029 .0029 .0029 10 .1593 .1614 .1598 .1600 20 .0058 .0058 .0058 .0058 20 .1622 .1644 .1627 .1629 30 .0087 .0087 .0087 .0087 30 .1650 .1673 .1656 .1658 40 .0116 .0116 .0116 .0116 40 .1679 .1703 .1685 .1687 50 .0145 .0145 .0145 .0145 50 .1708 .1733 .1714 .1716 1°00' .0175 .0175 .0175 .0175 10° 00' .1736 .1763 .1743 .1745 10 .0204 .0204 .0204 .0204 10 .1765 .1793 .1772 .1774 20 .0233 .0233 .0233 .0233 20 .1794 .1823 .1801 .1804 30 .0262 .0262 .0262 .0262 30 .1822 .1853 .1830 .1833 40 .0291 .0291 .0291 .0291 40 .1851 .1883 .1859 .1862 50 .0320 .0320 .0320 .0320 50 .1880 .1914 .1888 .1891 2° 00' .0349 .0349 .0349 .0349 11°00' .1908 .1944 .1917 .1920 10 .0378 .0378 .0378 .0378 10 .1937 .1974 .1946 .1949 20 .0407 .0407 .0407 .0407 20 .1965 .2004 .1975 .1978 30 .0436 .0437 .0436 .0436 30 .19^)4 .2035 .2004 .2007 40 .0465 .0466 .0465 .0465 40 .2022 .2065 .2033 .2036 50 .0494 .0495 .0494 .0495 50 .2051 .2095 .2062 .2065 3° 00' .0523 .0524 .0524 .0524 12° 00' .2079 .2126 .2091 .2094 10 .0552 .0553 .0553 .0553 10 .2108 .2156 .2119 .2123 20 .0581 .0582 .0582 .0582 20 .2136 .2186 .2148 .2153 30 .0610 .0612 .0611 .0611 30 .2164 .2217 .2177 .2182 40 .0640 .0641 .0640 .0640 40 .2193 .2247 .2206 .2211 50 .0669 .0670 .0669 .0669 50 .2221 .2278 .2235 .2240 4° 00' .0698 .0699 .0698 .0698 13° 00' .2250 .2309 .2264 .2269 10 .0727 .0729 .0727 .0727 10 .2278 .2339 .2293 .2298 20 .0756 .0758 .0756 .0756 20 .2306 .2370 .2322 .2327 30 .0785 .0787 .0785 .0785 30 .2334 .2401 .2351 .2356 40 .0814 .0816 .0814 .0814 40 .2363 .2432 .2380 .2385 50 .0843 .0846 .0843 .0844 50 .2391 .2462 .2409 .2414 6° 00' .0872 .0875 .0872 .0873 14° 00' .2419 .2493 .2437 .2443 10 .0901 .0904 .0901 .0902 10 .2447 .2524 .2466 .2473 20 .0929 .0934 .0931 .0931 20 .2476 .2555 .2495 .2502 30 .0958 .0963 .0960 .0960 30 .2504 .2586 .2524 .2531 40 .0987 .0992 .0989 .0989 40 .2532 .2617 .2553 .2560 50 .1016 .1022 .1018 .1018 50 .2560 .2648 .2582 .2589 6° 00' .1045 .1051 .1047 .1047 15°00' .2588 .2679 .2611 .2618 10 .1074 .1080 .1076 .1076 10 .2616 .2711 .2639 .2647 20 .1103 .1110 .1105 .1105 20 .2644 .2742 .2668 .2676 30 .1132 .1139 .1134 .1134 30 .2672 .2773 .2697 .2705 40 .1161 .1169 .1163 .1164 40 .2700 .2805 .2726 .2734 50 .1190 .1198 .1192 .1193 50 .2728 .2836 .2755 .2763 7° 00' .1219 .1228 .1221 .1222 16° 00' .2756 .2867 .2783 .2793 10 .1248 .1257 .1250 .1251 10 .2784 .2899 .2812 .2822 20 .1276 .1287 .1279 .1280 20 .2812 .2931 .2841 .2851 30 .1305 .1317 .1308 .1309 30 .2840 .2962 .2870 .2880 40 .1334 .1346 .1337 .1338 40 .2868 .2994 .2899 .2909 50 .1363 .1376 .1366 .1367 50 .2896 .3026 .2927 .2938 8° 00' .1392 .1405 .1395 .1396 17° 00' .2924 .3057 .2956 .2967 10 .1421 .1435 .1424 .1425 10 .2952 .3089 .2985 .2996 20 .1449 .1465 .1453 .1454 20 .2979 .3121 .3014 .3025 30 .1478 .1495 .1482 .1484 30 .3007 .3153 .3042 .3054 40 .1507 .1524 .1511 .1513 40 .3035 .3185 .3071 .3083 50 .1536 .1554 .1540 .1542 50 .3062 .3217 .3100 .3113 9° 00' .1564 .1584 .1569 .1571 18° 00' .3090 .3249 .3129 .3142 IV Quantities Determined by a Given Angle [I Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord Arc 18°00' .3090 .3249 .3129 .3142 27° 00' .4540 .5095 .4669 .4712 10 .3118 .3281 .3157 .3171 10 .4566 .5132 .4697 .4741 20 .3145 .3314 .3186 .3200 20 .4592 .5169 .4725 .4771 30 .3173 .3346 .3215 .3229 30 .4617 .5206 .4754 .4800 40 .3201 .3378 .3244 .3258 40 .4643 .5243 .4782 .4829 50 .3228 .3411 .3272 .3287 50 .4669 .5280 .4810 .4858 19° 00' .3256 .3443 .3301 .3316 28° 00' .4695 .5317 .4838 .4887 10 .3283 .3476 .3330 .3345 10 .4720 .5354 .4867 .4916 20 .3311 .3508 .3358 .3374 20 .4746 .5392 .4895 .4945 30 .3338 .3541 .3387 .3403 30 .4772 .5430 .4923 .4974 40 .3365 .3574 .3416 .3432 40 .4797 .5467 .4951 .5003 50 .3393 .3607 •3444 .3462 50 .4823 .5505 .4979 .5032 20° 00' .3420 .3640 .3473 .3491 29° 00' .4848 .5543 .5008 .5061 10 .3448 .3673 .3502 .3520 10 .4874 .5581 .5036 .5091 20 .3475 .3706 .3530 .3549 20 .4899 .5619 .5064 .5120 30 .3502 .3739 .3559 .3578 30 .4924 .5658 .5092 .5149 40 .3529 .3772 .3587 .3607 40 .4950 .5696 .5120 .5178 50 .3557 .3805 .3616 .3636 50 .4975 .5735 .5148 .5207 21° 00' .3584 .3839 .3645 .3665 30° 00' .5000 .5774 .5176 .5236 10 .3611 .3872 .3673 .3694 10 .5025 .5812 .5204 .5265 20 .3638 .3906 .3702 .3723 20 .5050 .5851 .5233 .5294 30 .3665 .3939 .3730 .3752 30 .5075 .5890 .5261 .5323 40 .3692 .3973 .3759 .3782 40 .5100 .5930 .5289 .5352 50 .3719 .4006 .3788 .3811 50 .5125 .5969 .5317 .5381 22° 00' .3746 .4040 .3816 .3840 31° 00' .5150 .6009 .5345 .5411 10 .3773 .4074 .3845 .3869 10 .5175 .6048 .5373 .5440 20 .3800 .4108 .3873 .3898 20 .5200 .6088 .5401 .5469 30 .3827 .4142 .3902 .3927 30 .5225 .6128 .5429 .5498 40 .3854 .4176 .3930 .3956 40 .5250 .6168 .5457 .5527 . 50 .3881 .4210 .3959 .3985 50 .5275 .6208 .5485 .5556 23° 00' .3907 .4245 .3987 .4014 32° 00' .5299 .6249 .5513 .5585 10 .3934 .4279 .4016 .4043 10 .5324 .6289 .5M1 .5614 20 .3961 .4314 .4044 .4072 20 .5348 .6330 .5569 .5643 30 .3987 .4348 .4073 .4102 30 .5373 .6371 .5597 .5672 40 .4014 .4383 .4x01 .4131 40 .5398 .6412 .5625 .5701 50 .4041 .4417 .4130 .4160 50 .5422 .6453 .5652 .5730 24° 00' .4067 .4452 .4158 .4189 33° 00' .5446 .6494 .5680 .5760 10 .4094 .4487 .4187 .4218 10 .5471 .6536 .5708 .5789 20 .4120 .4522 .4215 .4247 20 .5495 .6577 .5736 .5818 30 .4147 .4557 .4244 .4276 30 .5519 .6619 .5764 .5847 40 .4173 .4592 .4272 .4:^5 40 .5544 .6661 .5792 .6876 50 .4200 .4628 .4300 .4334 50 .5568 .6703 .5820 .5905 25° 00' .4226 .4663 .4329 .4363 34° 00' .5592 .6745 .5847 .5934 10 .4253 .4699 .4357 .4392 10 .5616 .6787 .5875 .5963 20 .4279 .4734 .4386 .4422 20 .5640 .6830 .5903 .5992 30 .4305 .4770 .4414 .4451 30 .5664 .6873 .5931 .6021 40 .4331 .4806 .4442 .4480 40 .5688 .6916 .5959 .6050 50 .4358 .4841 .4471 .4509 50 .5712 .6959 .5986 .6080 26° 00' .4384 .4877 .44^)9 .4538 35° 00' .5736 .7002 .6014 .6109 10 .4410 .4913 .4527 .4567 10 .57(50 .7046 .6042 .61.38 20 .4436 .4950 .4556 .4596 20 .5783 .7089 .6070 .6167 30 .4462 .498() .4584 .4625 30 .5807 .7133 .6097 .6196 40 .4488 .5022 .4612 .4654 40 .5831 .7177 .6125 .6225 50 .4514 .5059 .4641 .4683 50 .5854 .7221 .6153 .6254 27° 00' .4540 .5095 .4669 .4712 36° 00' .5878 .7265 .6180 .6283 1] Quantities Determined by a Giyen Angle V Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord 1 Arc 36° 00' .5878 .7265 .6180 .6283 45° 00' .7071 1.0000 .7654 .7854 10 .5901 .7310 .6208 .6312 10 .7092 1.0058 .7681 .7883 20 .5925 .7355 .6236 .6341 * 20 .7112 1.0117 .7707 .7912 30 .5948 .7400 .6263 .6370 30 .7133 1.0176 .7734 .7941 40 .5972 .7445 .6291 .6400 40 .7153 1.0235 .7761 .7970 50 .5995 .7490 .6318 .6429 50 .7173 1.0295 .7788 .7999 37° 00' .6018 .7536 .6346 .6458 46° 00' .7193 1.0355 .7815 .8029 10 .6041 .7581 .6374 .6487 10 .7214 1.0416 .7841 .8058 20 .6065 .7627 .6401 .6516 20 .7234 1.0477 .7868 .8087 30 .6088 .7673 .6429 .6545 30 .7254 1.0538 .7895 .8116 40 .6111 .7720 .6456 .6574 40 .7274 1.0599 .7922 .8145 50 .6134 .7766 .6484 .6603 50 .7294 1.0661 .7948 .8174 38° 00' .6157 .7813 .6511 .6632 47° 00' .7314 1.0724 .7975 .8203 10 .6180 .7860 .6539 .6661 10 .7333 1.0786 .8002 .8232 20 .6202 .7907 .6566 .6690 20 .7353 1.0850 .8028 .8261 30 .6225 .7954 .6594 .6720 30 .7373 1.0913 .8055 .8290 40 .6248 .8002 .6621 .6749 40 .7392 1.0977 .8082 .8319 50 .6271 .8050 .6649 .6778 50 .7412 1.1041 .8108 .8348 39° 00' .6293 .8098 .6676 .6807 48° 00' .7431 1.1106 .8135 .8378 10 .6316 .8146 .6704 .6836 10 .7451 1.1171 .8161 .8407 20 .6338 .8195 .6731 .6865 20 .7470 1.1237 .8188 .8436 30 .6361 .8243 .6758 .6894 30 .7490 1.1303 .8214 .8465 40 .6383 .8292 .6786 .6923 40 .7509 1.1369 .8241 .8494 50 .6406 .8342 .6813 .6952 50 .7528 1.1436 .8267 .8523 40° 00' .6428 .8391 .6840 .6981 49° 00' .7547 1.1504 .8294 .8552 10 .6450 .8441 .6868 .7010 10 .7566 1.1571 .8320 .8581 20 .6472 .8491 .6895 .7039 20 .7585 1.1640 .8347 .8610 30 .6494 .8541 .6922 .7069 30 .7604 1.1708 .8373 .8639 40 .6517 .8591 .69.50 .7098 40 .7623 1.1778 .8400 .8668 50 .6539 .8642 .6977 .7127 50 .7642 1.1847 .8426 .8698 41° 00' .6561 .8693 .7004 .7156 50° 00' .7660 1.1918 .8452 .8727 10 .6583 .8744 .7031 .7185 10 .7679 1.1988 .8479 .8756 20 .6604 .87% .7059 .7214 20 .7698 1.20.59 .8505 .8785 30 .6626 .8847 .7086 .7243 30 .7716 1.2131 .8531 .8814 40 .6648 .8899 .7113 .7272 40 .7735 1.2203 .8558 .8843 50 .6670 .8952 .7140 .7301 50 .7753 1.2276 .8584 .8872 42° 00' .6691 .9004 .7167 .7330 51° 00' .7771 1.2349 .8610 .8901 10 .6713 .9057 .7195 .7359 10 .7790 1.2423 .8636 .8930 20 .6734 .9110 .7222 .7389 20 .7808 1.2497 .8663 .8959 30 .6756 .9163 .7249 .7418 30 .7826 1.2572 .8689 .8988 40 .6777 .9217 .7276 .7447 40 .7844 1.2647 .8715 .9018 50 .6799 .9271 .7303 .7476 50 .7862 1.2723 .8741 .9047 43° 00' .6820 .9325 .7330 .7505 52° 00' .7880 1.2799 .8767 .9076 10 .6841 .9380 .7357 .7534 10 .7898 1.2876 .8794 .9105 20 .6862 .9435 .7384 .7563 20 .7916 1.2954 .8820 .9134 30 .6884 .9490 .7411 .7592 30 .7934 1.3032 .8846 .9163 40 .6905 .9545 .7438 .7621 40 .7951 1.3111 .8872 .9192 50 .6926 .9(301 .7465 .7650 ' 50 .7969 1.3190 .8898 .9221 44° 00' .6947 .9657 .7492 .7679 53° 00' .7986 1.3270 .8924 .9250 10 .6967 .9713 .7519 .7709 10 .8004 1.3351 .8950 .9279 20 .6988 .9770 .7546 .7738 20 .8021 1.3432 .8976 .9308 30 .7009 .9827 .7573 .7767 30 .8039 1.3514 .9002 .9338 40 .7030 .9884 .7600 .7796 40 .8056 1.3597 .9028 .9367 50 .7050 .9942 .7627 .7825 50 .8073 1.3680 .9054 .93^)6 45° 00' .7071 1.0000 .7654 .7854 54° 00' .8090 1.3764 .9080 .9425 vi Quantities Determined by i a Given Angle [1 Angle Sine Tan- gent Chord - Angle Sine Tan- gent Chord Arc 64° 00' .801X) 1.3764 .9080 .9425 63° 00' .8910 1.9626 1.0450 1.0996 10 .8107 1.3848 .9106 M64: 10 .8923 1.9768 1.0475 1.1025 20 .8124 1.3934 .9132 .9483 20 .8936 1.9912 1.0500 1.1054 30 .8141 1.4019 .9157 .9512 30 .8949 2.0057 1.0524 1.1083 40 .8158 1.4106 .9183 .9541 40 .8962 2.0204 1.0549 1.1112 50 .8175 1.4193 .9209 .9570 50 .8975 2.0353 1.0574 1.1141 65^00' .8192 1.4281 .9235 .9599 64° 00' .8988 2.0503 1.0598 1.1170 10 .8208 1.4370 .9261 .9628 10 .9001 2.0(>55 1.0623 1.1199 20 .8225 1.4460 .9287 .9657 20 .9013 2.0809 1.0648 1.1228 30 .8241 1.4550 .9312 .9687 30 .9026 2.0965 1.0672 1.1257 40 .8258 1.4641 .9338 .9716 40 .9038 2.1123 1.0697 1.1286 50 .8274 1.4733 .9364 .9745 50 .9051 2.1283 1.0721 1.1316 56° 00' .8290 1.4826 .9389 .9774 65° 00' .9063 2.1445 1.0746 1.1345 10 .8307 1.4919 .9415 .9803 10 .9075 2.1609 1.0771 1.1374 20 .8323 1.5013 .9441 .9832 20 .9088 2.1775 1.0795 1.1403 30 .8339 1.5108 .9466 .9861 30 .9100 2.1t)43 1.0820 1.1432 40 .8355 1.5204 .9492 .9890 40 .9112 2.2113 1.0844 1.1461 50 .8371 1.5301 .9518 .9919 50 .9124 2.2286 1.0868 1.1490 67° 00' .8387 1.5399 .9543 .9948 66° 00' .9135 2.2460 1.C893 1.1519 10 .8403 1.5497 .9569 .9977 10 .9147 2.2637 1.0917 1.1548 20 .8418 1.5597 .9594 1.0007 20 .9159 2.2817 1.0<)42 1.1577 30 .8434 1.5697 .9620 1.0036 30 .9171 2.2998 1.09()6 1.1606 40 .8450 1.5798 .9645 1.0065 40 .9182 2.3183 1.0990 1.1636 50 .8465 1.5900 .9671 1.0094 50 .9194 2.3369 1.1014 1.1665 68° 00' .8480 1.6003 .9696 1.0123 67° 00' .9205 2.3559 1.1039 1.1694 10 .8496 1.6107 .9722 1.0152 10 .9216 2.3750 1.10(33 1.1723 20 .8511 1.6212 .9747 1.0181 20 .9228 2.3945 1.1087 1.1752 30 .8526 1.6319 .9772 1.0210 30 .9239 2.4142 1.1111 1.1781 40 .8542 1.6426 .9798 1.0239 40 .9250 2.4342 1.1136 1.1810 50 .8557 1.6534 .9823 1.0268 50 .9261 2.4545 1.1160 1.1839 59°00' .8572 1.6643 .9848 1.0297 68° 00' .9272 2.4751 1.1184 1.1868 10 .8587 1.6753 .9874 1.0327 10 .9283 2.4960 1.1208 1.1897 20 .8601 1.6864 .9899 1.0356 20 .9293 2.5172 1.1232 1.1926 30 .8616 1.6977 .9924 1.0385 30 .9304 2.5386 1.1256 1.1956 40 .8631 1.7090 .9950 1.0414 40 .9315 2.5605 1.1280 1.1985 50 .8646 1.7205 .9975 1.0443 50 .9325 2.5826 1.1304 1.2014 60° 00' .8660 1.7321 1.0000 1.0472 69° 00' .9336 2.6051 1.1328 1.2043 10 .8675 1.7437 1.0025 1.0501 10 .9346 2.6279 1.1352 1.2072 20 .8689 1.7556 1.0050 1.0530 20 .9356 2.6511 1.1376 1.2101 30 .8704 1.7675 1.0075 1.0559 30 .9367 2.6746 1.1400 1.2130 40 .8718 1.7796 1.0101 1.0588 40 .9377 2.6985 1.1424 1.2159 50 .8732 1.7917 1.0126 1.0617 50 .9387 2.7228 1.1448 1.2188 61° 00' .8746 1.8040 1.0151 1.0647 70° 00' .9397 2.7475 1.1472 1.2217 10 .8760 1.8165 1.0176 1.0676 10 .9407 2.7725 1.1495 1.2246 20 .8774 1.8291 1.0201 1.0705 20 .9417 2.7980 1.1519 1.2275 30 .8788 1.8418 1.0226 1.0734 30 .9426 2.8239 1.1543 1.2305 40 .8802 1.8546 1.0251 1.0763 40 .9436 2.8502 1.1567 1.2334 50 .8816 1.8676 1.0276 1.0^2 .50 .9446 2.8770 1.1590 1.2363 62° 00' .8829 1.8807 1.0301 1.0821 71°00' .9455 2.^)042 1.1614 1.2392 10 .8843 1.8940 1.0326 1.0850 10 .9465 2.9319 1.1638 1.2421 20 .8857 1.9074 1.0351 1.0879 20 .9474 2.9600 1.1661 1.2450 ;io .8870 1.9210 1.0375 1.0908 30 .9483 2.9887 1.1685 1.2479 40 .8884 1.9347 1.0400 1.0937 40 .9492 3.0178 1.1709 1.2508 50 .8897 1.9486 1.0425 1.0966 50 .9502 3.0475 1.1732 1.2537 63° 00' .8910 1.9626 1.0450 1.0996 72° 00' .9511 3.0777 1.1756 1.2566 I] Quantities Determined by a Given Angle vii Angle Sine Tan- gent Chord Arc Angle Sine Tan- gent Chord Arc 72° 00' .9511 3.0777 1.1756 1.2566 81° 00' .9877 6.3138 1.2989 1.4137 10 .9520 3.1084 1.1779 1.2595 10 .9881 6.4348 1.3011 1.4166 20 .9528 3.1397 1.1803 1.2625 20 .9886 6.5606 1.3033 1.4195 30 .9537 3.1716 1.1826 1.2654 30 .9890 6.6912 1.3055 1.4224 40 .9546 3.2041 1.1850 1.2683 40 .9894 6.8269 1.3077 1.4254 50 .9555 3.2371 1.1873 1.2712 50 .9899 6.9682 1.3099 1.4283 73° 00' .9563 3.2709 1.1896 1.2741 82° 00' .9903 7.1154 1.3121 1.4312 10 .9572 3.3052 1.1920 1.2770 10 .9907 7.2687 1,3143 1.4341 20 .9580 3.3402 1.1943 1.2799 20- .9911 7.4287 1.3165 1.4370 30 .9588 3.3759 1.1966 1.2828 30 .9914 7.5958 1.3187 1.4399 40 .9596 3.4124 1.1990 1.2857 40 .9918 7.7704 1.3209 1.4428 50 .9605 3.4495 1.2013 1.2886 50 .9922 7.9530 1.3231 1.4457 74° 00' .9613 3.4874 1.2036 1.2915 83° 00' .9925 8.1443 1.3252 1.4486 10 .9621 3.5261 1.2060 1.2945 10 .9929 8.3450 1.3274 1.4515 20 .9628 3.5656 1.2083 1.2974 20 .9932 8.5555 1.3296 1.4544 30 .9636 3.6059 1.2106 1.3003 30 .9936 8.7769 1.3318 1.4573 40 .9644 3.6470 1.2129 1.3032 40 .9939 9.0098 1.3339 1.4603 50 .9652 3.6891 1.2152 1.3061 50 .9942 9.2553 1.3361 1.4632 75° 00' .9659 3.7321 1.2175 1.3090 84° 00' .9945 9.5144 1.3383 1.4661 10 .9667 3.7760 1.2198 1.3119 10 .9948 9.7882 1.3404 1.4690 20 .9674 3.8208 1.2221 1.3148 20 .9951 10.078 1.3426 1.4719 30 MSI 3.8667 1.2244 1.3177 30 .9954 10.385 1.3447 1.4748 40 .9689 3.9136 1.2267 1.320(5 40 .9957 10.712 1.3469 1.4777 50 .9696 3.9617 1.2290 1.3235 50 .9959 11.059 1.3490 1.4806 76° 00' .9703 4.0108 1.2313 1.3265 85° 00' .9962 11.430 1.3512 1.4835 10 .9710 4.0611 1.2336 1.3294 10 .9964 11.826 1.3533 1.4864 20 .9717 4.1126 1.2359 1.3323 20 .9967 12.251 1.3555 1.4893 30 .9724 4.1653 1.2382 . 1.3352 30 .9969 12.706 1.3576 1.4923 40 .9730 4.2193 1.2405 1.3381 40 .9971 13.197 l.;%97 1.4952 50 .9737 4.2747 1.2428 1.3410 50 .9974 13.727 1.3619 1.4981 77° 00' .9744 4.3315 1.2450 1.3439 86° 00' .9976 14.301 1.3640 1.5010 10 .9750 4.3897 1.2473 1.3468 10 .9978 14.924 1.3661 1.5039 20 .9757 4.4494 1.2496 1.3497 20 .9980 15.605 1.3682 1.5068 30 .9763 4.5107 1.2518 1.3526 30 .9981 16.350 1.3704 1..5097 40 .9769 4.5736 1,2541 1.3555 40 .9983 17.169 1.3725 1.5126 50 .9775 4.6382 1.2564 1.3584 50 .9985 18.075 1.3746 1.5155 78° 00' .9781 4.7046 1.258(i 1.3614 87° 00' .9986 19.081 1.3767 1.5184 10 .9787 4.7729 1.2609 1.3643 10 .9988 20.206 1.3788 1.5213 20 .9793 4.8430 1.2632 1.3672 20 .9989 21.470 1..3809 1.5243 30 .9799 4.9152 1.2654 1.3701 30 .9990 22.904 1.3830 1.5272 40 .9805 4.9894 1.2677 1.3730 40 .9992 24.542 1.3851 1.5301 50 .9811 5.0658 1.2699 1.3759 50 .9993 26.432 1.3872 1.5330 79° 00' .9816 5.1446 1.2722 1.3788 88° 00' .9994 28.636 1.3893 1.5359 10 .9822 5.2257 1.2744 1.3817 10 .9995 31.242 1.3914 1.5388 20 .9827 5.3093 1.2766 1.3846 20 .9996 34.368 1.3935 1.5417 30 •9833 5.3955 1.2789 1.3875 30 .9997 38.188 1.3956 1.5446 40 .9838 5.4845 1.2811 1.3904 40 .9997 42.964 1.3977 1.5475 50 .9843 5.5764 1.2833 1.3934 50 .9998 49.104 1.3997 1.5504 80° 00' .9848 5.6713 1.2856 1.3963 89° 00' .9998 57.290 1.4018 1.5533 10 .9853 5.7694 1.2878 1.3992 10 .9999 68.750 1.4039 1.5563 20 .9858 5.8708 1.2900 1.4021 20 .9999 85.940 1.4060 1.5592 30 .9863 5.9758 1.2922 1.4050 30 1.0000 114.59 1.4080 1.5621 40 .9868 6.0844 1.2945 1.4079 40 1.0000 171.89 1.4101 1.5650 50 .9872 6.1970 1.2967 1.4108 50 1.0000 343.77 1.4122 1.5679 81° 00' .9877 6.3138 1.2989 1.4137 90° 00' 1.0000 1.4142 1.5708 TABLE II — POWERS AND ROOTS EXPLANATION OF TABLE II 1. Squares and Cubes. The squares of numbers between 1.00 and 10.00 at intervals of .01 are given in column headed n^ To find the square of any other number, divide (or mul- tiply) the given number by 10 to reduce it to a number between 1 and 10; find the square of this last number; multiply (or divide) the square thus found by 10 twice as many times as you did the given number. The cube is given in the column headed n^. To find the cube of any number not between 1 and 10, first reduce that number to a number between 1 and 10 by dividing (or multiplying) by a power of 10. Multiply (or divide) the result found by three times as high a power of 10 as was used to reduce the given number. 2. Square Roots. The square roots of numbers between 1 and 10 are found in the column headed Vn. The square roots of numbers between 10 and 100 may be found in the column headed VlO n. The square roots of numbers between 100 and 1000 may be found in the column headed Vn by multiplying the given root by 10, since VlOOw = 10 Vn. Other square roots may be found in a similar manner. 3. Cube Roots. The column headed : Vn gives cube roots of numbers between 1 and 10 ; VlOn gives cube roots of numbers between 10 and 100 ; VlOOn gives cube roots of numbers between 100 and 1000. To find the cube root of a number between 1000 and 10000, take 10 times the value found in the column headed Vn, since VlOOOn = 10 Vn. Other cube roots may be found similarly. viii 11] Table II — Powers and Roots ix n n^ Vn VIO^ n^ 1.8769 1.^)044 1.9321 1.14455 1.14891 1.15326 1.15758 1.16190 1.16619 1.17047 1.17473 1.17898 3.61939 3.63318 3.64692 3.66060 3.67423 3.68782 3.70135 3.71484 3.72827 2.24809 2.29997 2.35264 2.40610 2.46038 2.51546 2.57135 2.62807 2.68562 1.09418 1.09()fK) 1.09972 1.10247 1.10521 1.10793 1.11064 1.11334 1.11602 2.35735 2.36333 2.36928 2.37521 2.38110 2.38697 2.39280 2.39861 2.40439 5.07875 6.09164 5.10447 5.11723 5.12993 5.14256 5.15514 5.16765 5.18010 1.40 1.9600 1.18322 3.74166 2.74400 1.11869 2.41014 5.19249 1.41 1.42 1.43 1.44 1.45 1.46 1.47 1.48 1.49 1.9881 2.0164 2.0449 2.0736 2.1025 2.1316 2.1609 2.1904 2.2201 1.18743 1.19164 1.19583 1.20000 1.20416 1.20830 1.21244 1.21655 1.22066 3.75500 3.76829 3.78153 3.79473 3.80789 3.82099 3.83406 3.84708 3.86005 2.80322 2.86329 2.92421 2.98598 3.04862 3.11214 3.17652 3.24179 3.30795 1.12135 1.12399 1.12662 1.12924 1.13185 1.13445 1.13703 1.13960 1.14216 2.41587 2.42156 2.42724 2.43288 2.43850 2.44409 2.44966 2.45520 2.46072 5.20483 5.21710 5.22932 5.24148 5.25359 5.26564 5.27763 5.28957 5.30146 X P( )wers and Roots [11 n n^ Vn VlOw 1 VlOn n^ ^n S/lOOw 1.50 2.2500 1.22474 3.87298 3.37500 1.14471 2.46621 5.31329 1.51 1.52 1.53 1.54 1.55 1.56 1.57 1.58 1.59 2.2801 2.3104 2.3409 2.3716 2.4025 2.4336 2.4649 2.4964 2.5281 1.22882 1.23288 1.23693 1.24097 1.24499 1.24900 1.25300 1.25698 1.20095 3.88587 3.89872 3.91152 3.92428 3.93700 3.94968 3.96232 3.97492 3.98748 3.44295 3.51181 3.58158 3.65226 3.72388 3.79642 3.86989 3.94431 4.01968 1.14725 1.14978 1.15230 1.15480 1.15729 1.15978 1.16225 1.16471 1.16717 2.47168 2.47712 2.48255 2.48794 2.49332 2.49867 2.50399 2.509i50 2.51458 5.32507 5.33680 5.34848 5.36011 5.37169 5.38321 5.39469 5.40612 5.41750 1.60 2.5600 1.26491 4.00000 4.09600 1.169(51 2.51984 5.42884 1.61 1.62 1.63 1.64 1.65 1.66 1.67 1.68 1.69 2.5921 2.6244 2.6569 2.6896 2.7225 2.7556 2.7889 2.8224 2.8561 1.26886 1.27279 1.27671 1.28062 1.28452 1.28841 1.29228 1.29615 1.30000 4.01248 4.02492 4.03733 4.04969 4.06202 4.07431 4.08(556 4.09878 4.11096 4.17328 4.25153 4.33075 4.41094 4.4i)2l2 4.57430 4.65746 4.74163 4.82681 1.17204 1.17446 1.17687 1.17927 1.181(57 1.18405 1.18642 1.18878 1.19114 2.52508 2.5;?030 2.53549 2.54007 2.54582 2.55095 2.55607 2.56116 2.56623 5.44012 5.45136 5.46256 5.47370 5.48481 5.49586 5.50688 5.51785 5.52877 1.70 2.8900 1.30384 4.12311 4.91300 1.19348 2.57128 5.53966 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 2.9241 2.9584 2.9929 3.0276 3.0625 3.0976 3.1329 3.1684 3.2041 1.30767 1.31149 1.31529 1.31909 1.32288 1.32665 1.33041 1.33417 1.33791 4.13521 4.14729 4.15933 4.17133 4.18330 4.19524 4.20714 4.21900 4.23084 5.00021 5.08845 5.17772 5.26802 5.35938 5.45178 5.54523 5.63975 5.73534 1.19582 1.19815 1.20046 1.20277 1.20507 1.20736 1.20964 1.21192 1.21418 2.57631 2.58133 2.58632 2.59129 2.59625 2.60118 2.60610 2.61100 2.61588 5.55050 5.56130 5.57205 5.58277 5.59344 5.60408 5.61467 5.62523 5.63574 1.80 3.2400 1.34164 4.24264 5.83200 1.21644 2.62074 5.64622 1.81 1.82 1.83 1.84 1.85 1.86 1.87 1.88 1.89 3.2761 3.3124 3.3489 3.3856 3.4225 3.4596 3.4969 3.5344 3.5721 1.34536 1.34907 1.35277 1.35647 1.36015 1.36382 1.36748 1.37113 1.37477 4.25441 4.26615 4.27785 4.28952 4.30116 4.31277 4.32435 4.3351K) 4.34741 5.92974 6.02857 6.12849 6.22950 6.33162 6.43486 6.53920 6.64467 6.75127 1.21869 1.22093 1.22316 1.22539 1.227(50 1.22981 1.23201 1.23420 1.23639 2.(52559 2.63041 2.63522 2.64001 2.(54479 2.64954 2.65428 2.65901 2.66371 5.65665 5.66705 5.67741 5.68VV3 5.69802 5.70827 5.71848 5.72865 5.73879 1.90 3.6100 1.37840 4.35890 6.85900 1.23856 2.66840 5.74890 1.91 1.92 1.93 1.94 1.95 1.96 1.97 1.98 1.99 3.6481 3.6864 3.7249 3.7636 3.8025 3.8416 3.8809 3.9204 3.9601 1.38203 1.38564 1.38924 1.39284 1.39642 1.40000 1.40357 1.40712 1.41067 4.37035 4.38178 4.39318 4.40454 4.41588 4.42719 4.43847 4.44972 4.46094 6.96787 7.07789 7.1890(J 7.30138 7.41488 7.52954 7.64537 7.76239 7.880(50 1.24073 1.24289 1.24.505 1.24719 1.24933 1.25146 1.25a59 1.25571 1.25782 2.67307 2.67773 2.68237 2.68700 2.69161 2.69620 2.70078 2.70534 2.70989 5.75897 5.76900 5.77900 5.78896 5.79889 5.80879 5.81865 5.82848 5.83827 H] Powers and Roots xi n n^ v^ VlOn n^ ^ <^10n i44 4.50555 4.51664 4.52769 4.53872 4.54973 4.56070 4.57165 8.120(^0 8.24241 8.36543 8.48966 8.61512 8.74182 8.86974 8.99891 9.12933 1.26202 1.26411 1.26619 1.2(5827 1.27033 1.27240 1.27445 1.27650 1.27854 2.71893 2.72344 2.72792 2.73239 2.73(585 2.74129 2.74572 2.75014 2.75454 5.85777 5.86746 5.87713 5.88677 5.89637 5.90594 5.91548 5.92499 5.93447 2.10 4.4100 1.44914 4.58258 9 26100 1.28058 2.75892 5.94392 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 4.4521 4.4944 4.5369 4.5796 4.6225 4.6656 4.7089 4.7524 4.7961 1.45258 1.45602 1.45945 1.46287 1.46629 1.46969 1.47309 1.47648 1.47986 4.59347 4.60435 4.61519 4.62601 4.63681 4.64758 4.65833 4.66905 4.67974 9.39393 9.52813 9.663(50 9.80034 9.93838 10.0777 10.2183 10.3602 10.5035 1.28261 1.28463 1.28665 1.28866 1.2fX)66 1.29266 1.29465 1.29664 1.29862 2.76330 2.76766 2.77200 2.77633 2.78065 2.78495 2.78924 2.79352 2.79779 5.95334 5.96273 5.97209 5.98142 5.99073 6.00000 6.00925 6.01846 6.02765 2.20 4.8400 1.48324 4.69042 10.6480 1.30059 2.80204 6.03(581 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 4.8841 4.9284 4.9729 5.0176 5.0625 5.1076 5.1529 5.1984 5.2441 1.486(51 1.48997 1.49332 1.49666 1.50000 1.50333 1.50665 1.50997 1.51327 4.70106 4.71169 4.72229 4.73286 4.74342 4.75395 4.76445 4.77493 4.78539 10.7939 10.9410 11.0896 11.2394 11.3906 11.5432 11.6971 11.8524 12.00<)0 1.302.56 1.30452 1.30648 1.30843 1.31037 1.31231 1.31424 1.31617 1.31809 2.80628 2.81050 2.81472 2.81892 2.82311 2.82728 2.83145 2.83560 2.83974 6.04594 6.05505 6.06413 6.07318 6.08220 6.09120 6.10017 6.10911 6.11803 2.30 5.2{)00 1.51658 4.79583 12.1670 1.32001 2.84387 6.12693 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 5.3361 5.3824 5.4289 5.4756 5.5225 5.5696 5.6169 5.6644 5.7121 1.51987 1.52315 1.52643 1.52971 1.53297 1.53623 1.53948 1.54272 1.54596 4.80(525 4.81664 4.82701 4.83735 4.84768 4.85798 4.86826 4.87852 4.88876 12.3264 12.4872 12.6493 12.8129 12.9779 13.1443 13.3121 13.4813 13.6519 1.32192 1.32382 1.32572 1.32761 1.32950 1.33139 1.33326 1.33514 1.33700 2.84798 2.85209 2.85618 2.86026 2.86433 2.86838 2.87243 2.87646 2.88049 6.13579 6.14463 6.15345 6.16224 6.17101 6.17975 6.18846 6.19715 6.20582 2.40 5.7600 1.54919 4.89898 13.8240 1.33887 2.88450 6.21447 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 5.8081 5.8564 5.9049 5.9536 6.0025 6.0516 6.1009 6.1504 6.2001 1.55242 1.55563 1.55885 1.56205 1.56525 1.56844 1.57162 1.57480 1.57797 4.90918 4.91935 4.92950 4.93964 4.^)4975 4.95984 4.96991 4.97996 4.98999 13.9975 14.1725 14.3489 14.5268 14.7061 14.8869 15.0692 15.2530 15.4382 1.M072 1.34257 1.34442 1.34626 1.34810 1.34993 1.35176 1.35358 1.35540 2.88850 2.89249 2.89647 2.90044 2.90439 2.90834 2.91227 2.91620 2.92011 6.22308 6.23168 6.24025 6.24880 6.25732 6.26583 6.27431 6.28276 6.29119 xii Powers and Roots Cn n n^ v^ ^/10n n^ ^n ^10 n s/lOOn 2.50 6.2500 1.58114 5.00000 15.6250 1.35721 2.92402 6.29961 2.51 2.52 2.53 2.54 2.55 2.56 2.57 2.58 2.59 6.3001 6.3504 6.4009 6.4516 6.5025 6.5536 6.6049 6.6564 6.7081 1.58430 1.58745 1.59060 1.59374 1.59687 1.60000 1.60312 1.60624 1.60935 5.00999 5.01996 5.02991 5.03984 5.04975 5.05964 5.06952 5.07937 5.08920 15.8133 16.0030 16.1943 16.3871 16.5814 16.7772 16.9746 17.1735 17.3740 1.35902 1.36082 1.30202 1.36441 1.36020 1.30798 1.30970 1.37153 1.37330 2.92791 2.93179 2.93507 2.93953 2.94338 2.94723 2.95106 2.95488 2.95869 6.30799 6.31636 6.32470 6.33303 6.34133 6.34960 6.35786 6.36610 6.37431 2.60 6.7600 1.61245 5.09902 17.5700 1.. 37507 2.96250 6.38250 2.61 2.62 2.63 2.64 2.65 2.66 2.67 2.68 2.69 6.8121 6.8644 6.9169 6.9696 7.0225 7.0756 7.1289 7.1824 7.2361 1.61555 1.61864 1.62173 1.62481 1.62788 1.63095 1.63401 1.03707 1.04012 5.10882 5.11859 5.12835 5.13809 5.14782 5.15752 5.16720 5.17687 5.18052 17.7796 17.9847 18.1914 18.3997 18.0096 18.8211 19.0342 19.2488 19.4051 1.37083 1.37859 1.38034 1.38208 1.38383 1.38557 1.38730 1.38903 1.39070 2.90029 2.97007 2.97385 2.97761 2.98137 2.98511 2.98885 2.99257 2.99629 6.39068 6.39883 6.40696 6.41507 0.42316 6.43123 6.43928 6.44731 6.45531 2.70 7.2900 1.64317 5.19f)15 19.0830 1.39248 3.00000 6.46330 2.71 2.72 2.73 2.74 2.75 2.76 2.77 2.78 2.79 7.3441 7.3984 7.4529 7.5076 7.5625 7.6176 7.6729 7.7284 7.7841 1.64621 1.64924 1.65227 1.65529 1.65831 1.66132 1.66433 1.60733 1.07033 5.20577 5.21536 5.22494 5.23450 5.24404 5.25357 5.26308 5.27257 5.28205 19.9025 20.1236 20.3464 20.5708 20.7969 21.0246 21.2539 21.4850 21.7176 1.39419 1.39591 1.39701 1.39932 1.40102 1.40272 1.40441 1.40610 1.40778 3.00370 3.00739 3.01107 3.01474 3.01841 3.02206 3.02570 3.02934 3.03297 6.47127 6.47922 6.48715 6.49507 6.5029882 2.07123 6.48845 6.49(515 6.50384 6.51153 6.51920 6.52687 6.53452 6.54217 6.54981 74.6185 75.1514 75.6870 76.2250 76.7656 77.3088 77.8545 78.4028 78.9536 1.61471 1.61599 1.61726 1.61853 1.61981 1.62108 1.62234 1.62361 1.62487 3.47878 3.48154 3.48428 3.48703 3.48977 3.49250 3.49523 3.49796 3.50068 7.49481 7.50074 7.50666 7.51257 7.51847 7.52437 7.53025 7.53612 7.54199 4.30 18.4900 2.07364 6.55744 79.5070 1.62613 3.50340 7.54784 4.31 4.32 4.33 4.34 4.35 4.36 4.37 4.38 4.39 18.5 18.f 18.7 18.8 18.fl 19.C 19.C 19.1 19.2 761 «24 489 356 225 K)96 969 844 721 2.07605 2.07846 2.08087 2.08327 2.08567 2.08806 2.09045 2.09284 2.09523 6.56506 6.57267 6.58027 6.58787 6.59545 6.60303 6.61060 6.61816 6.62571 80.0630 80.6216 81.1827 81.7465 82.3129 82.8819 83.4535 84.0277 84.6045 1.62739 1.62865 1.62991 1.63116 1.63241 1.63366 1.63491 1.63619 1.63740 3.50611 3.50882 3.51153 3.51423 3.51692 3.519G2 3.52231 3.52499 3.52767 7.55369 7.55953 7.56535 7.57117 7.57698 7.68279 7.58858 7.59436 7.60014 4.40 19.3600 2.09762 6.63325 85.1840 1.63864 3.53035 7.60590 4.41 4.42 4.43 4.44 4.45 4.46 4.47 4.48 4.49 19.4481 19.5364 19.6249 19.7136 19.8025 19.8916 19.9809 20.0704 20.1601 2.10000 2.10238 2.10476 2.10713 2.10950 2.11187 2.11424 2.11660 2.11896 6.64078 6.64831 6.65582 6,66333 6.67083 6.67832 6.68581 6.69328 6.70075 85.7661 86.3509 86.9383 87.5284 88.1211 88.7165 89.3146 89.9154 90.5188 1.63988 1.64112 1.64236 1.64.359 l.(i4483 1.64606 1.64729 1.64851 1.64974 3.53302 3.53569 3.53835 3.54101 3.54367 3.54632 3.54897 3.55162 3.55426 7.61166 7.61741 7.62315 7.62888 7.63461 7.64032 7.64603 7.65172 7.65741 xvi Powers and Roots [11 n W-2 V^ VIO** n^ ■^n ^IQn S/lOOw 4.50 20.2500 2.12132 6.70820 91.1250 1.65096 3.55689 7.66309 4.51 4.52 4.53 4.54 4.55 4.56 4.57 4.58 4.59 20.3401 20.4304 20.5209 20.6116 20.7025 20.7936 20.8849 20.9764 21.0681 2.12368 2.12603 2.12838 2.13073 2.13307 2.13542 2.13776 2.14009 2.14243 6.71565 6.72309 6.73053 6.73795 6.74537 6.75278 6.76018 6.76757 6.77495 91.7339 92.3454 92.9597 93.5767 94.1964 94.8188 95.4440 96.0719 96.7026 1.65219 1.65341 1.65462 1.65584 1.65706 1.65827 1.65948 1.66069 1.66190 3.55953 3.56215 3.56478 3.56740 3.57002 3.57263 3.57524 3.57785 3.58045 7.66877 7.67443 7.68009 7.68573 7.69137 7.69700 7.70262 7.70824 7.71384 4.60 21.1600 2.14476 6.78233 97.3360 1.66310 3.58305 7.71944 4.61 4.62 4.63 4.64 4.65 4.66 4.67 4.68 4.69 21.2521 21.3444 21.4369 21.5296 21.6225 21.7156 21.8089 21.9024 21.9961 2.14709 2.14942 2.15174 2.15407 2.15639 2.15870 2.16102 2.16333 2.16564 6.78970 6.79706 6.80441 6.81175 6.81909 6.82642 6.83374 6.84105 6.84836 97.9722 98.6111 99.2528 99.8973 100.545 101.195 101.848 102.503 103.162 1.66431 1.66551 1.66671 1.66791 1.66911 1.67030 1.67150 1.67269 1.67388 3.58564 3.58823 3.59082 3.59340 3.59598 3.59856 3.60113 3.60370 3.60626 7.72503 7.73061 7.73619 7.74175 7.74731 7.75286 7.75840 7.76394 7.76946 4.70 22.0900 2.16795 6.85565 103.823 1.67507 3.60883 7.77498 4.71 4.72 4.73 4.74 4.75 4.76 4.77 4.78 4.79 22.1841 22.2784 22.3729 22.4676 22.5625 22.6576 22.7529 22.8484 22.9441 2.17025 2.17256 2.17486 2.17715 2.17945 2.18174 2.18403 2.18632 2.18861 6.86294 6.87023 6.87750 6.88477 6.89202 6.89928 6.90652 6.91375 6.92098 104.487 105.154 105.824 106.496 107.172 107.850 108.531 109.215 109.<)02 1.67626 1.67744 1.67863 1.67981 1.68099 1.68217 1.68334 1.68452 1.68569 3.61138 3.61394 3.61649 3.61903 3.62158 3.62412 3.6?665 3.62919 3.63172 7.78049 7.78599 7.79149 7.79697 7.80245 7.80793 7.81339 7.81885 7.82429 4.80 23.0400 2.19089 6.92820 110.592 1.68687 3.63424 7.82974 4.81 4.82 4.83 4.84 4.85 4.86 4.87 4.88 4.89 23.1361 23.2324 23.3289 23.4256 23.5225 23.6196 23.7169 23.8144 23.9121 2.19317 2.19545 2.19773 2.20000 2.20227 2.20454 2.20681 2.20907 2.21133 6.93542 6.94262 6.94982 6.95701 6.96419 6.97137 6.97854 6.98570 6.99285 111.285 111.980 112.679 113..380 114.084 114.791 115.501 116.214 116.930 1.68804 1.68920 1.69037 1.69154 1.69270 1.69386 1.69503 1.69619 1.69734 3.63676 3.63928 3.64180 3.64431 3.64682 3.64932 3.65182 3.65432 3.65681 7.83517 7.84059 7.84601 7.85142 7.85683 7.86222 7.86761 7.87299 7.87837 4.90 24.0100 2.21359 7.00000 117.649 1.69850 3.65931 7.88374 4.91 4.92 4.93 4.94 4.95 4.96 4.97 4.98 4.99 24.1081 24.2064 24.3049 24.4036 24.5025 24.6016 24.7009 24.8004 24.9001 2.21585 2.21811 2.22036 2.22261 2.22486 2.22711 2.22935 2.23159 2.23383 7.00714 7.01427 7.02140 7.02851 7.03562 7.04273 7.04982 7.05691 7.063<)9 118.371 119.095 119.823 120.554 121.287 122.024 122.763 123.506 124.251 1.69965 1.70081 1.70196 1.70311 1.70426 1.70540 1.706.55 1.70769 1.70884 3.66179 3.6fi428 3.66676 3.66924 3.67171 3.67418 3.67665 3.67911 3.68157 7.88909 7.89445 7.89979 7.90513 7.91046 7.91578 7.92110 7.92641 7.93171 n] Powers and Roots xvii n n2 Vn VlO^i n^ ^ ^100 n 20 186.1(59 187.149 188.133 189.119 irK).109 191.103 192.100 193.101 194.105 1.78736 1.78840 1.78944 1.79048 1.79152 1.79256 1.79360 1.79463 1.79567 3.85075 3.85300 3.85524 3.85748 3.85972 3.86196 3.86419 3.86642 3.868(55 8.2<)619 8.30103 8.30587 8.31069 8.31552 8.32034 8.32515 8.32995 8.33476 5.80 33.6400 2.40832 7.61577 195.112 1.7i¥570 3.87088 8.33955 5.81 5.82 5.83 5.84 5.85 5.86 5.87 5.88 5.89 33.7561 33.8724 33.9889 34.1056 34.2225 34.3396 34.4569 34.5744 34.6921 2.41039 2.41247 2.41454 2.41661 2.41868 2.42074 2.42281 2.42487 2.42693 7.62234 7.62889 7.63544 7.64199 7.64853 7.65506 7.66159 7.66812 7.67463 196.123 197.137 198.155 199.177 200.202 201.230 202.262 203.297 204.3.36 1.79773 1.79876 1.79979 1.80082 1.80185 1.80288 1.80390 1.80492 1.80595 3.87310 3.87532 3.87754 3.87975 3.88197 3.88418 3.88639 3.88859 3.89080 8.34434 8.34913 8.35390 8.35868 8.36345 8.36821 8.37297 8.37772 8.38247 5.90 34.8100 2.42899 7.(>8115 205.379 1.80697 3.89;X)0 8.38721 5.91 5.92 5.93 5.94 5.95 5.96 5.97 5.98 5.99 34.9281 35.0464 35.1649 35.283fi 35.4025 35.5216 a5.6409 .35.7604 35.8801 2.43105 2.43311 2.43516 2.43721 2.43926 2.44131 2.44336 2.44540 2.44745 7.68765 7.69415 7.70065 7.70714 7.71362 7.72010 7.72658 7.7a305 7.73951 206.425 207.475 208.528 209.585 210.(545 211.709 212.776 213.847 214.V)22 1.80799 1.80i)01 1.81003 1.81104 1.81206 1.81307 1.81409 1.81510 1.81611 3.89519 3.89739 3.89958 3.90177 3.90396 3.90615 3.90833 3.91051 3.912(59 8.39194 8.39667 8.40140 8.40612 8.41083 8.41554 8.42025 8.42494 8.42964 11] Powers and Roots xix n n^ V^ V10»* n^ 4.00624 4.00832 4.01039 4.01246 4.01453 4.01660 4.01866 8.62222 8.62671 8.63118 8.63566 8.64012 8.64459 8.64904 8.65350 8.65795 XX Powers and Roots [n n n^ Vn VIO^ n3 ^n i/lOn \ 446 4.29627 4.29807 4.2<^)987 4.30168 4.30348 4.30528 4.;W07 9.24823 9.25213 9.25602 9.25991 9.26380 9.26768 9.27156 9.27544 9.27931 n] Powers and Boots XXlll n W,2 Vn VIO^ n^ ^ ^10^ 2 2.02144 4.35153 4.35329 4.35505 9.37510 9.37889 9.38268 8.27 8.28 8.29 68.3929 68.5584 68.7241 2.87576 2.87750 2.87924 9.09395 9.09945 9.10494 565.609 567. (J64 569.723 2.02225 2.02307 2.02388 4.35681 4.35856 4.36032 9.38646 9.39024 9.39402 8.30 68.8900 2.88097 9.11043 571.787 2.02469 4.36207 9.39780 8.31 8.32 8.33 69.0561 69.2224 69.3889 2.88271 2.88144 2.88617 9.11592 9.12140 9.12688 573.856 575.930 578.010 2.02551 2.02632 2.02713 4.36382 4.36557 4.36732 9.40157 9.40534 9.40911 8.34 8.35 8.36 69.5556 69.7225 69.8896 2.88791 2.88964 2.89137 9.13236 9.13783 9.14330 580.094 582.183 584.277 2.02794 2.02875 2.02956 4.36907 4.37081 4.37256 9.41287 9.41663 9.42039 8.37 8.38 8.39 70.0569 70.2244 70.3921 2.89310 2.89482 2.89655 9.14877 9.15423 9.15969 586.376 588.480 590.590 2.03037 2.03118 2.03199 4.37430 4.37604 4.37778 9.42414 9.42789 9.43164 8.40 70.5600 2.89828 9.16515 592.704 2.03279 4.37952 9.43539 8.41 8.42 8.43 70.7281 70.8964 71.0649 2.90000 2.90172 2.90345 9.17061 9.17606 9.18150 591.823 59(3.948 599.077 2.03360 2.03440 2.03521 4.38126 4.38299 4.38473 9.43913 9.44287 9.44661 8.44 8.45 8.46 71.2336 71.4025 71. .5716 2.90517 2.90689 2.90861 9.18695 9.19239 9.19783 601.212 603.351 605.496 2.03601 2.03682 2.03762 4.-38646 4.38819 4.38992 9.45034 9.45407 9.45780 8.47 8.48 8.49 71.7409 71.9104 72.0801 2.91033 2.91204 2.91376 9.20326 9.20869 9.21412 607.645 609.800 611.960 2.03842 2.03923 2.04003 4.39165 4.39338 4.39510 9.46152 9.46525 9.46897 xxiv Powers and Roots [11 n W2 y/n VlOw n3 2 9.25203 9.25743 9.26283 9.26823 616.295 618.470 620.650 622.836 625.026 627.222 629.423 631.629 633.840 2.04163 2.04243 2.04323 2.04402 2.04482 2.04562 2.04641 2.04721 2.04801 4.39855 4.40028 4.40200 4.40372 4.40543 4.40715 4.40887 4.41058 4.41229 9.47640 9.48011 9.48381 9.48752 9.49122 9.49492 9.49861 9.50231 9.50600 8.60 73.9600 2.93258 9.27362 636.05(5 2.04880 4.41400 9..50t)69 8.61 8.62 8.63 8.64 8.65 8.66 8.67 8.68 8.69 74.1321 74.3044 74.4769 74.6496 74.8225 74.9956 75.1689 75.3424 75.5161 2.93428 2.93598 2.93709 2.93939 2.94109 2.94279 2.94449 2.94618 2.94788 9.27901 9.28440 9.28978 9.29516 9.30054 9.30591 9.31128 9.31665 9.32202 638.277 640.504 642.736 644.973 647.215 649.462 651.714 653.972 656.235 2.04959 2.0r.039 2.05118 2.05197 2.05276 2.05355 2.05434 2.05513 2.05592 4.41571 4.41742 4.41913 4.42084 4.42254 4.42425 4.42595 4.42765 4.42935 9.51337 9.51705 9.52073 9.52441 9.52808 9.53175 9.53542 9.53908 9.54274 8.70 75.6900 2.f)4958 9.32738 658.503 2.05671 4.43105 9.54(540 8.71 8.72 8.73 8.74 8.75 8.76 8.77 8.78 8.79 75.8641 76.0384 76.2129 76.3876 76.5625 76.7376 76.9129 77.0884 77.2641 2.95127 2.95296 2.95466 2.95635 2.95804 2.95973 2.96142 2.96311 2.96479 9.33274 9.33809 9.34345 9.34880 9.35414 9.35949 9.36483 9.37017 9.37550 660.776 663.055 665.339 667.628 669.922 672.221 674.526 (576.836 679.151 2.05750 2.05828 2.05907 2.05986 2.06064 2.06143 2.06221 2.06299 2.06378 4.43274 4.43444 4.43613 4.43783 4.43952 4.44121 4.44290 4.44459 4.44627 9.55006 9.55371 9.55736 9.56101 9.56466 9.56830 9.57194 9.57557 9.57921 8.80 77.4400 2.96648 9.38083 681.472 2.06456 4.44796 9.58284 8.81 8.82 8.83 8.84 8.85 8.86 8.87 8.88 8.89 77.6161 77.7924 77.9689 78.1456 78.3225 78.4996 78.6769 78.8544 79.0321 2.96816 2.96985 2.97153 2.97321 2.97489 2.97658 2.97825 2.97993 2.98161 9.38616 9.39149 9.39681 9.40213 9.40744 9.41276 9.41807 9.42338 9.42868 683.798 686.129 688.465 690.807 693.154 695.506 697.864 700.227 702.595 2.06534 2.06(512 2.06690 2.06768 2.06846 2.06924 2.07002 2.07080 2.071.57 4.44964 4.45133 4.45301 4.45469 4.45(537 4.45805 4.45972 4.46140 4.46307 9.58(547 9.59009 9.59372 9.59734 9.60095 9.60457 9.60818 9.61179 9.61540 8.90 79.2100 2.98329 9.43398 704.969 2.072:55 4.46475 9.61900 8.91 8.92 8.93 8.94 8.95 8.96 8.97 8.98 8.99 79.3881 79.5664 79.7449 79.92.36 80.1025 80.2816 80.4609 80.6404 80.8201 2.98496 2.98664 2.98831 2.98998 2.99166 2.99333 2.99500 2.99(566 2.998.33 9.43928 9.44458 9.44987 9.45516 9.4(K)44 9.46573 9.47101 9.47629 9.48156 707.348 709.732 712.122 714.517 716.917 719.323 721.734 724.151 726.573 2.07313 2.073f)0 2.07468 2.07545 2.07622 2.07700 2.07777 2.07854 2.07931 4.46642 4.46809 4.46976 4.47142 4.47309 4.47476 4.47642 4.47808 4.47974 9.62260 9.(52620 9.62980 9.63339 9.(5;?698 9.64057 9.64415 9.64774 9.65132 11] Powers and Roots XXV n n^ v^ VlOri n^ ^ ^10 n ^100^ 9.00 81.0000 3.00000 9.48683 729.000 2.08008 4.48140 9.65489 9.01 9.02 9.03 9.04 9.05 9.06 9.07 9.08 9.09 81.1801 81.3604 81.5409 81.7216 81.5)025 82.0836 82.2649 82.4464 82.6281 3.00167 3.00,333 3.00500 3.00666 3.00832 3.00998 3.01164 3.01330 3.01496 9.49210 9.49737 9.50263 9.50789 9.51315 9.51840 9.52365 9.52890 9.53415 731.4.33 733.871 736.314 738.763 741.218 743.677 746.143 748.613- 751.089 2.08085 2.08162 2.08239 2.08316 2.08393 2.08470 2.08546 2.08623 2.08699 4.48306 4.48472 4.48638 4.48803 4.48969 4.49134 4.49299 4.49-164 4.49(529 9.65847 9.66204 9.66561 9.66918 9.67274 9.67630 9.67986 9.68.342 9.68697 9.10 82.8100 3.01662 9.53939 753.571 2.08776 4.49794 9.69052 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 82.9921 83.1744 83.3569 83.5396 83.7225 83.9056 84.0889 84.2724 84.4561 3.01828 3.01993 3.02159 3.02324 3.02490 3.02655 3.02820 3.02985 3.03150 9.54463 9.54987 9.55510 9.5G033 9.565.56 9.57079 9.57601 9.58123 9.58645 756.058 758.551 761.048 763.552 766.061 768.575 771.095 773.621 776.152 2.08852 2.08929 2.09005 2.09081 2.09158 2.09234 2.09310 2.09386 2.09462 4.49959 4.50123 4.50288 4.50452 4.50616 4.50781 4.50945 4.51108 4.51272 9.69407 9.69762 9.70116 9.70470 9.70824 9.71177 9.71531 9.71884 9.722.36 9.20 84.6400 3.03315 9.59166 778.688 2.09538 4.51436 9.72589 9.21 9.22 9.23 9.24 9.25 9.26 9.27 9.28 9.29 84.8241 85.0084 85.1929 85.3776 85.5625 85.7476 85.9329 86.1184 86.3041 3.03480 3.03645 3.03809 3.03974 3.04138 3.04302 3.04467 3.04631 3.04795 9.5<)687 9.60208 9.60729 9.61249 9.61769 9.62289 9.62808 9.63328 9.(53846 781.230 783.777 786.330 788.889 791.4.53 794.023 796.598 799.179 801.765 2.09614 2.09690 2.09765 2.09841 2.09917 2.09992 2.10068 2.10144 2.10219 4.51599 4.51763 4.51926 4.52089 4.-52252 4.52415 4.52578 4.52740 4.52i)03 9.72941 9.73293 9.73645 9.73996 9.74348 9.74699 9.75049 9.75400 9.75750 9.30 86.4fX)0 3.04959 9.64365 804.357 2.10294 4.53065 9.76100 9.31 9.32 9.33 9.34 9.35 9.36 9.37 9.38 9.39 86.6761 8(5.8624 87.0489 87.2356 87.4225 87.6096 87.7969 87.9844 88.1721 3.05123 3.05287 3.05450 3.05614 3.05778 3.05941 3.06105 3.06268 3.06431 9.64883 9.65401 9.65919 9.66437 9.66954 9.67471 9.67988 9.68504 9.69020 806.954 809.558 812.166 814.781 817.400 820.026 822.657 825.294 827.9.36 2.10370 2.10445 2.10520 2.10595 2.10671 2.10746 2.10821 2.10896 2.10971 4.53228 4.5335)0 4.53552 4.53714 4.53876 4.54038 4.54199 4.54361 4.54522 9.76450 9.76799 9.77148 9.77497 9.77846 9.78195 9.78543 9.78891 9.79239 9.40 88..3600 3.06594 9.695,36 830.584 2.11045 4.54684 9.79586 9.41 9.42 9.43 9.44 9.45 9.46 9.47 9.48 9.49 88.5481 88.7364 88.9249 89.1136 89.3025 89.4916 89.6809 89.8704 90.0601 3.0(>757 3.00920 3.07083 3.07246 3.07409 3.07571 3.07734 3.07896 3.08058 9.70052 9.705(57 9.71082 9.71597 9.72111 9.72625 9.73139 9.73653 9.74166 833.238 835.897 838.562 841.232 843.909 846.591 849.278 851.971 854.670 2.11120 2.11195 2.11270 2.11344 2.11419 2.11494 2.11568 2.11642 2.11717 4.54845 4.55006 4.55167 4.55328 4.55488 4.55649 4.55809 4.55970 4.56130 9.79933 9.80280 9.80627 9.80974 9.81320 9.81666 9.82012 9.82357 9.82703 xxvi Powers and Roots [11 n n^ y/n VWn n^ 4638 9.94975 9.95311 9.95648 9.95984 9.96320 9.81 9.82 9.83 9.84 9.85 9.86 9.87 9.88 9.89 96.2361 96.4324 96.6289 96.8256 97.0225 97.2196 97.4169 97.6144 97.8121 3.13209 3.13369 3.13528 3.13688 3.13847 3.14006 3.14166 3.14325 3.14484 9.90454 9.<)0959 9.91464 9.91968 9.92472 9.92975 9.93479 9.93982 9.94485 944.076 946.96(5 949.862 952.764 955.672 958.585 961.505 964.430 967.362 2.14070 2.14143 2.14216 2.14288 2.14361 2.14433 2.14506 2.14578 2.14651 4.61200 4.61357 4.61514 4.61670 4.61826 4.61983 4.62139 4.622{)5 4.62451 9.90 98.0100 3.14(343 9.94987 970.2t)9 2.14723 4.62607 9.96655 9.91 9.92 9.93 9.94 9.95 9.96 9.97 9.98 9.99 98.2081 98.4064 98.6049 98.8036 99.0025 99.2016 9<).4009 SW.6004 i)9.8001 3.14802 3.14 greater than struction < less than Cor. Corollary ^ is congruent to Def. Definition JL perpendicular, or is perpendic- Hyp. Hypothesis, or by hypoth- ular to esis II parallel, or is parallel to Iden. being identical ~ similar, or is similar to Prop. Proposition A angle rt. right A angles St. straight A triangle Th. Theorem ^ triangles Prob. Problem O parallelogram Fig. Figure or diagram HJ parallelograms .... and so on O circle hence or therefore © circles • ^ arc The signs 4- , — , x , -4- , are used with the same meanings as in algebra. The following agreements are also made : a X b = a ' b = ab, a-^-b = a/b = a : 6 xxvm SYLLABUS OF PLANE GEOMETRY* INTEODUCTION PART I. DRAWING SIMPLE FIGURES PART II. FUNDAMENTAL IDEAS PART III. STATEMENTS FOR REFERENCE 27. Axioms. 1. If equals are added to equals^ the sums are equal. Thus, if a = 6 and c = d, then a -\- c = b + d. 2. If equals are subtracted from equals, the remainders are equal. Thus, if a = b and c = d, then a — c = b — d. 3. If equals are multiplied by equals, the products are equal. Thus, if a = b and c = d, then ac = bd. 4. If equals are divided by equals, the quotients are equal. Thus, if a = b and c = d, then - = -• In applying this axiom it is supposed that c a G and d are not equal to zero. 5. If equals are added to unequals, the results are unequal and in the same order. Thus, if a = b and c > d, then a -\- c^b + d. 6. If equals are subtracted from unequals, the results are unequal and in the same order. Thus, if a > & and c = d, then a— c>b — d. 7. If unequals are added to unequals in the same sense, the results are unequal in the same order. Thus, if a>b and c^d, then a -i- c>b + d. 8. If unequals are subtracted from equals, the results are unequal in the opposite order. Thus, if a = b and c > d, then a — c