L SOLID GE OMET RY STONE- MiLLIS > BENJ.H, SANBORN f ^ ,f'», f\ ^m^t'^ttfMsi mtumv^ntiJ i iJM*iiijii]i i iJ.ywig8tfjjHiCfN.iii i !i^ ^ IN MEMORIAM FLORIAN CAJORl SOLID GEOMETRY BY JOHN C. STONE, A.M. )) HEAD OF THE DEPARTMENT OF MATHEMATICS, STATE NORMAL SCHOOL MONTCLAIR, NEW .JERSEY, CO-ACTHOR OF THE 80UTHW0RTH-ST0NE ARITHMETICS, STONE-MILLIS ARITHMETICS, SECONDARY ARITHMETIC, ALGEBRAS AND GEOMETRIES AND JAMES F. MILLIS, A.M. HEAD OF THE DEPARTMENT OF MATHEMATICS, FRANCIS W. PARKER SCHOOI CHICAGO, CO-AUTHOR OF THE STONE-MILLIS ARITHMETICS SECONDARY ARITHMETIC, ALGEBRAS ▲ND G£OM£TRI£S ov TToXX' dWa iroXv BENJ. H. SANBORN & CO. CHICAGO NEW YORK BOSTON 1925 STONE-MILLIS MATHEMATICS SERIES LEMENTAKY ARITHMETICS: Primary Primary Intermediate Complete Advanced Secondary Arithmetic Elementary Algebra, First Course Elementary Algebra, Second Course Elementary Algebra, Complete Course Higher Algebra Plane Geometry Solid Geometry Plane and Solid Geometry Benj. H. Sanborn & Co., Publishers OOPTBIGHT, 1916 BY BENJ. H. SANBORN 4 00. PREFACE The Stone-Millis Geometry — Plane, Solid, and Plane AND Solid — published in 1910, was a pioneer in its field, being the first of the group of American textbooks on geometry which in recent years have attempted in various ways to make the teaching of geometry conform to modern thought in education. It marked a wide departure from the traditional Greek geometry after which textbooks for secondary schools had for generations been patterned. This text has met with remarkable success. The educational ideals which it embodied are now recognized as national, and are summarized in the Report of the National Committee of Fifteen on the teaching of geometry. The present geometry, by the same authors, has been prepared in the attempt to produce a text which shall preserve the dis- tinctive features of the older text, but which, if possible, shall be more simple, practical, and teachable. The following are some of the features which distinguish this text: 1. Simplicity. — 1. The subject is graded so that the easier topics come first and so that the student is introduced to only one new difficulty at a time. The grading of geometry is made possible in this text by abandoning the Greek division of geome- try into books and re-grouping the material in chapters. 2. Some of the theorems on fundamental properties of figures are treated informally at the beginnings of many topics. 3. The subject is also abbreviated and simplified by the omis- sion of certain useless traditional theorems and the reduction to a reasonable minimum of the number of theorems, constructions, and corollaries requiring formal treatment. iii Mr?05027 IV PREFACE 4. Tlie use of the theory of limits in the proofs of incommen- surable cases of theorems has been eliminated. II. Practicality. — 1. Geometry is humanized by using as exercises a large number of practical problems. This phase of geometry, which was first introduced into American schools by the Stone-Millis text, is now universally recognized as an integral part of the subject. The Stone-Millis Geometry contains a very large number and a very great variety of simple and genu- ine practical problems. They are selected from many fields of human activity, such as home life, art, architecture, astronomy, engineering, designing, navigation, science, the construction and use of implements and machinery, etc. 2. Directions are given for the construction of many home- made instruments and their use in out-of-door exercises. 3. Geometry is correlated with trigonometry by the introduc- tion of simple work with trigonometric ratios, in the chapter on similar polygons. Application is made to practical problems in the solution of triangles. III. Teachableness. — 1. A concrete approach to formal geometry is provided. This develops a body of experience and imagery as a basis of formal geometry, and the latter is not introduced until need for it is felt. In the approach to demon- strative geometry, familiarity with important geometric figures is secured through their accurate construction with drawing in- struments. This development of clear imagery through accurate drawing of the figures involved in the formal theorems, etc., is continued throughout plane geometry. 2. Use is made of the suggestive method in the treatment of theorems. While complete model proofs of a large number of theorems are given — and whenever a proof is given it is given in complete form, with numbered steps — the proofs of many theorems are left, with suggestions, to the student. The sug- gestions in a large part of the theorems are given in the form of analyses. It is believed that suggestions of this nature are superior to those of the traditional kind, which are mere outlines PREFACE V of tlie proofs, because they give the student training in exactly the kind of thinking which he must do when attacking a proof unaided, and thus they teach method of attack and develop power of originality. 3. Exercises which demand technical knowledge have been eliminated. Many new exercises have been introduced. The exercises are grouped in every case immediately after the theo- rem, construction, or corollary to which they relate. Many mis- cellaneous review exercises are given. 4. Special care has been given to the illustrations in this text. When construction lines are required in drawing a figure, they show in the book. Throughout the Solid Geometry shaded drawings of models are placed by the side of the more compli- cated geometric figures, to aid the student in visualizing the third dimension in the figures while looking at the flat drawings of them. The consistent plan of representing hidden parts of figures by thinner lines than the others is carried out, dotted lines being employed exclusively for auxiliary lines as in plane geometry. Grateful acknowledgment of the authors is due to all those who by timely suggestions have aided in the preparation of this text; especially to Professor H. E. Cobb and Professor A. W. Cavanaugh of Lewis Institute, Chicago ; to Miss Alice M. Lord of the High School, Portland, Maine ; and to Professor Guido H. Stempel of Indiana University. Special acknowledgment is due to Mr. Charles McCauley of Chicago, who has made the excellent illustrations. JOHN C. STONE, JAMES F. MILLIS. Januabt, 1916. CONTENTS CHAPTER PAGE Xil. Lines and Planes in Space ... . . 277 XUI. Prisms and Cylinders . . . . . . . 321 XIV. Pyramids and Cones ........ 355 XV. Regular Polyedrons. Similar Polyedrons. Pris- MATOIDS . . 390 XVI. The Sphere 404 I ▼U SYMBOLS =, is equal to; equals. =, is identically equal to. ~, is similar to. ^, is congruent to. >, is greater than. <, is less than. = , approaches as a limit. II , is parallel to ; parallel. The plural of any symbol representing a noun is obtained by affixing the letter s. Thus, A represents angles. ±, is perpendicular to; perpen- dicular. .*., therefore. •••, and so on. Z., angle. A, triangle. O, parallelogram. O, circle. ABBREVIATIONS Ax.y axiom. Alt., alternate. Comp., complementary. Cor., corollary. Carres., corresponding. Def., definition. Ex.f exercise. Ext., exterior. Hyp., hypothesis. Int., interior. Rect., rectangle. Rt., right. St., straight. Supp., supplementary. SOLID GEOMETRY CHAPTER XII LINES AND PLANES IN SPACE 288. Solid geometry. — It has been seen that plane geometry deals with figures which lie in a flat or plane surface. Solid geometry treats of figures consisting of geometric solids, sur- faces, lines, points, or combinations of them, which are not confined to a plane. Solid geometry, like plane geometry, has been developed by man in the endeavor to meet his practical needs, and dates back to an- cient times. Ahmes, an Egyptian, in his book entitled " Directions for Ob- taining the Knowl- edge of All Dark Things," written about 1700 B.C., gave rules for find- ing the contents of wells and granaries. The principal geometric forms — prismatic, cylindrical, pyramidal, conical, and spherical — appear in nature in endless combinations. They are employed by man in countless ways — in the construction of buildings, monuments, temples, embankments; in the 2:7 Lincoln Memorial, Washington, D. C. 278 SOLID GEOMETRY making of pottery, jewelry, furniture ; in the great fields of engineering, mechanics, architecture, art, astronomy, etc. 289. Geometric drawings and models. — Although the figures of solid geometry are not in one plane, they must be repre- sented by drawings all parts of which do lie in one plane. It is necessary, therefore, for the student of solid geometry to learn to sense thickness, or the third dimension, of a figure while looking at a flat drawing of it. For .assistance in overcoming this difficulty of the third dimension, those parts of a geometric figure which would be invisible if it were a physical object are represented in a drawing by thinner lines than the other lines of the figure. This is illustrated in the drawing of a prism below. When auxiliary lines occur, they are represented by dotted lines as in plane geometry. <_> ^j—b" Peism Also, for assistance in sensing the third dimension in a drawing, sometimes a picture of a model of the figure is given in the text, adjoining the geometric figure, as illus- trated in the case of the prism above. The use of wooden, wire, or cardboard models would be of great aid to the beginning student. Models may be con- structed of cardboard as illustrated in the following exer- cises. LINES AND PLANES IN SPACE 279 EXERCISES 1. On a piece of cardboard, draw a figure similar to the adjoining figure, making the side of each square 3 in. Cut out the pat- tern, and by folding along the dotted lines and pasting, make a model of a cube. How many squares form the surface of this model ? How many edges has it V How many corners has it ? 2. On a piece of cardboard, draw a figure similar to the adjoining figure, making the side of each hexagon 2 in. Cut out the pattern, and by folding along the dotted lines and pasting, make a model of a hex- agonal prism. Models of other kinds of prisms may be made by using other kinds of regular poly- gons instead of the hexagons. 3. Following the method of Exercises 1 and 2, by first making a pattern similar to the pattern here shown, with the side of each triangle 4 in., construct a model of a triangular pyramid. ff=h ri — ' — I i ^r - I t I I III I III j r II N ! I ? 290. Representation of a plane surface — A plane surface is unlimited in extent. Hence it is impossible to show an entire plane to the eye by a drawing. In geometric figures, a plane is represented by some kind of polygon, which incloses a portion ^^ of the plane. Trapezoids and parallelograms are used ex- tensively for representing planes. This figure represents a horizontal plane, seen obliquely. The plane is named MN. 280 SOLID GEOMETRY 291. Fundamental property of a plane surface. — A straight- edge, held so that it touches a flat or plane surface at two points, touches the surface all along the straightedge. Hence it is inferred that : (1) If two points of a straight line lie in a plane^ the entire line lies in the plane. From the fundamental property above, let the student prove the following : (2) A straight line can intersect a plane in only one point. 292. Revolution of a plane. — It ib evident that : By revolving a plane about any straight line in it as an axis, it may be made to contain any point of space that is not on the line, in one and only one position. Thus, ii AB is a straight line in plane MN and C is any point of space that is not on AB, by revolving MN about AB as an axis, it will arrive at a position PQ in which it contains point C. If the plane is revolved far- ther, it will no longer contain point C. 293. Corollary l. — A straight line lies in an unlimited number of planes. For, a straight line AB lies in at least one -plane, by definition of a straight line. (See § 6.*) This plane may be revolved about AB, by § 292. And each successive position of the revolving plane represents a different plane containing AB. r ^C I A ^,^-^ yN M\^ ^ B * See the References to Plane Oeometry at the end of the book. LINES AND PLANES IN SPACE 281 294. Corollary 2. — An unlimited number of straight lines may he drawn perpendicular to a given straight line in space at a given point of the line. For, at any point of the line AB in § 298, a perpendicular to AB may be drawn in each of the planes containing AB. 295. Detennming a plane. — A plane is determined by cer- tain points, lines, or combinations of them, if that plane and no other plane contains all of those points, etc. 296. Theorem. — A straight line and a point not on the line determine a plane. Hypothesis. AB is any straight line and O any point of space not on AB, Conclusion. Line AB and point C determine a plane. Proof. 1. (7 is a point not on straight line AB. Hyp. 2. AB lies in a plane. § 6 3. By revolving this plane about AB as an axis, it may be made to contain point O in one and only one position. § 292 4. Hence line AB and point C determine a plane. § 295 Without the book, draw a figure and write out this proof. 297. Corollary l. — Three points not in the same straight line determine a plane. ^ For, if J., B, and Q are the three / '^^ points, A and B determine a / ^)». -^^ straight line. One and only one plane can contain line AB and point C, by § 296. Hence, one and only one plane can contain J., B, and C. Why ? 282 SOLID GEOMETRY 298. Corollary 2. — Two intersecting straight lines determine a plane. For, if the straight lines AB and QB intersect at 0, let P be any other point on OB, Then one and only one plane can contain / ^ AB and point P, by § 296. This / C^...^^^^^ ^ plane contains the whole line (7i>, by § 291. Why can only this one plane contain the lines AB and QB'l 299. Corollary 3. — Two 'parallel straight lines determine a plane. For, if AB and OB are parallel, / ^ ^p they lie in a plane, by definition of parallel lines. Let P be any point on AB. Then only one plane can contain point P and OB. Why ? Why, then, can only one plane contain AB and OB ? EXERCISES 1. What tool does a carpenter use for reducing the rough surface of a board to a smooth or plane surface ? Explain how the principle in § 291 is involved in the use of this tool. 2. A plasterer, after putting plaster on a wall, smooths it down to a plane surface by rubbing a board or straightedge over it in many directions. Builders of concrete sidewalks do the same thing. Why does this produce a plane surface ? 3. In measuring grain with a half-bushel measure, the grain is first heaped, then it is raked off even with the top of the measure, with a board. Why is the measure then level full ? What kind of surface is thus formed by the grain ? LINES AND PLANES IN SPACE 283 4. Illustrate the principle in § 292 by using the point of a pencil to denote any point of space and a page of an open book to denote the revolving plane. Think of another way in which this principle may be illustrated. 5. Place three tacks on a table so that they are not in a straight line and have their points up. Support a book or a pane of glass on the points of the tacks. How does this illustrate §297? 6. Place a ruler on a table, with its edge up, and place a tack on the table, with its point up. Support a book or a pane of glass on the edge of the ruler and the point of the tack. How does this illustrate § 296 ? 7. Place two rulers in parallel positions on a table, with their edges up. Support a book or pane of glass on the edges of the rulers. How does this illustrate § 299 ? 8. Cameras, telescopes, etc., are mounted on tripods. WUl an object mounted on three legs always rest firmly on the floor? Why? 9. Do four given points lie in one plane ? 10. Why do not all chairs, tables, etc., with four legs rest firmly on the floor ? 11. Find four points in the room through which a plane may be passed. Find four points through which a plane cannot be passed. 12. Hold two pencils in such positions as to show that a plane cannot be passed through any two straight lines taken at random. 13. Show that if a piece of cord is fastened at both ends, and, by grasping it at any third point, the two parts are stretched straight, the stretched cord will lie entirely in one plane. 14. Show that any transversal of two parallel lines must lie in the plane of those lines. 15. Show that a figure made of three straight lines, each intersecting the other two, but not in a common point, must lie in one plane. 300. Intersection of planes. — It is assumed that : If two planes intersect, they have more than one point in common. 284 301. Theorem. line. SOLID GEOMETRY The intersection of two planes is a straight Hypothesis. Planes iHfiV and PQ intersect. Conclusion. The intersection of planes MN and P§ is a straight line. Proof. 1. ilifiV and P§ intersect. Hyp. 2. . •. iHfiV and PQ have at least two common points. Let them be A and B, § 300 3. Then all points of the straight line joining A and B lie in both planes. § 291 4. No point without line AB can lie in both planes, for the two planes would then coincide. § 296 5. Hence, since all common points of the planes, and no other points, must lie in the intersection, line AB is the in- tersection of the planes. That is, the intersection of the planes is a straight line. Write out the proof without using the book. EXERCISES 1. What is the meaning of "Two intersecting planes determine a straight line " ? 2. What is the locus of all points common to two planes? 3. Show that if a sheet of cardboard, or other kind of stiff sheet ma- terial, is folded and creased, the crease must be straight. 4. When one saws a board in two, why is the edge of the cut made by the saw a straight line? LINES AND PLANES IN SPACE 285 5. In slicing a loaf of bread, if a surface of the loaf is flat, why is the edge of the slice straight? Think of other applications in the home of the principle in § 301. 6. Carpenters, in shingling a roof, often mark off straight chalk lines on the roof to assist in placing the shingles in straight rows. A cord, whitened with chalk, is held firmly at the two ends, and stretched straight. Then grasping the cord at some third point, they pull it away a little from the roof and let it go. It springs back and strikes the roof, leaving a white mark, which is a straight line. Prove that this is an application of § 301. 7. When three planes, not containing the same straight line, intersect in pairs, if the three lines of intersection arn not parallel, they meet at one point. Suggestions. — Let the planes in- tersect in AB, CD, and EF. Prove that AB and CD intersect at some point 0. Prove that is on EF by showing that it is in plane EC and also in plane A E. 8. Illustrate the principle in Exercise 7 by the walls and fseilings of -jj^^^n 1 1 11 II II II 11 II II 1 ll \ 1 1 11 11 11 ' 'yvw 1 IT II II II .1 the schoolroom. Give other illustrations. 302. Line and plane perpendicular. — A straight line which intersects a plane is perpendicular to the plane if it is perpen- dicular to every straight line in the plane drawn through the point of intersection. The plane also is said to be perpendicular to the line. If a line is perpendicular to a plane, the point of inter- section of the line and plane is called the foot of the per- pendicular. If a straight line intersects a plane, but is not perpendicu- lar to it, the line is said to be oblique to the plane. 286 SOLID GEOMETRY 303. Theorem. — If a straight line is perpendicular to each of two intersecting straight lines at their point of intersection^ it is perpendicular to the plane determined hy them. Hypothesis. Lines QJ) and EF intersect at 0, and deter- mine plane MN\ AB ± CD and AB ± EF at 0. Conclusion. Line AB is perpendicular to plane MN. Proof. 1. In plane MN, CD and EF intersect at 0; also AB ± CD and AB ± EF at 0. Hyp. 2. Draw OK, any other straight line through in MN. Draw any straight line intersecting CD at (7, EF at H, and OK At K On AB, take B on the opposite side of from A so that OB = AO. Draw AG, AH, AK, Ba, BE, ^K 3. Then Aa=Ba and AH=^ BE. § 105 4. GE is a common side of A AGE and A BGE. 5. ..AAGE^ABGE. §76 6. .'. ZAGE== ZBGE. Def. congruence 7. In A J. (t^ and A BGK, GK is a common side. 8. .'.AAGK^ABGK § 63 9. .'.AK=BK. Def. congruence 10. .'.AB±OK. § 107 11. Hence, since OK is any other line than CD and EF drawn through in MN, AB is perpendicular to the plane MN. § 302 Write the proof in full without referring to the book. LINES AND PLANES IN SPACE 287 304. Theorem. — Through a given point in a given straight line^ one plane and only one plane can he drawn perpendicular to the line. B M/ " ; Mrr===^=^ B Hypothesis. AB is a given straight line, and is a given point on AB, Conclusion. Through 0, one plane and only one plane can be drawn perpendicular to AB, Proof. 1. AB is a given straight line, and (9 is a given point on AB, Hyp. 2. Two straight lines, OQ and OB (figure at left), can be drawn perpendicular to AB at 0, § 294 3. 0(7 and OB determine a plane MN. § 298 4. Plane MN 1. line AB, § 303 5. Suppose that MN"is not the only plane through per- pendicular to AB, and that PQ (figure at right) is another plane through perpendicular to AB. 6. Then a plane can be drawn containing AB and inter- secting planes iKfZVand PQ in two straight lines OF Siud OE, respectively. § 301 7. OF ± AB and OE i. AB. § 302 8. But this is impossible. § 53 9. .♦. the supposition is false, and hence MN is the only plane through 0, perpendicular to AB. Draw figures, and write the proof in full without consulting the book'. 288 SOLID GEOMETRY 305. Theorem. — Through a given point without a given straight line, one plane and only one plane can he drawn per- pendicular to the line. B I^^\ N A — ^ yr-F - n\\ 1 ' "^ \ pl--2 Suggestions. If AB (figure at left) is tl:ie line and the point, and AB determine a plane, in which OQ can be drawn ± AB, Then another line CD can be drawn ± AB, 0(7 and CD determine a plane MN L AB, Show (figure at right) by using § 54 that no other plane PQ can be drawn through J_ line AB. 306. Theorem. — Through a given point in a given plane, one line and only one line can he drawn perpendicular to the plane. R p E M/c A\0 JO !\ A^ Suggestions. If (figure at left) is a point in plane MN, straight line AB can be drawn through in MN \ then plane FQ can be drawn through 01. AB, intersecting MN in CD ; then line OE can be drawn in P^ ± CD. Then 0^± plane ilOT. Show (figure at right) by using § 53 that no other line Oi^ can be drawn through ± plane MN, LINES AND PLANES IN SPACE 289 307. Theorem. — Through a given point without a given plane, one line and only one line can he drawn perpendicular to the plane. P K / G . 'H-' \ 7^\ H N Q Ok ML N Hypothesis. MNis a plane and a point not in MN. Conclusion. Through (9 one line and only one line can be drawn perpendicular to MN. Proof. 1. MJSf is a plane and a point not in MN, Hyp. 2. Let AB be any straight line in MN (figure at left). 3. A plane FQ can be drawn through ± AB, § 305 4. FQ intersects MJVin a straight line CB, and line AB at a point U. § 301 5. Draw OG in FQ± OB. Let GF be any straight line drawn in iHfiV, meeting AB at F. Produce OG through G to IT, making GH^OG. Draw OE, OF, HE, HF. 6. AB ± OE siud AB ± HE. §302 7. .-. Z OFF and Z HEF are rt. A. Dei. ± 8. OE^HE. §105 9. In A OEF and A ^^^, EF is a common side. 10. .-.A OEF ^ A HEF. § 64 11. .♦. OF = HF. Def. congruence 12. .-. GFl. OG. § 107 13. .-. OG i^ perpendicular to plane MN. § 303 The proof that 0(7 is the only line that can be drawn through perpendicular to MN is left to the student. Use the figure at the right. Employ § 54. 290 SOLID GEOMETRY EXERCISES 1. Explain how to test by means of a carpenter's squarvi whether or not a post on a level surface is vertical. 2. Establish a line perpendicular to the top of the table by using two rectangular pieces of cardboard or two books. 3. How can a carpenter determine if a floor is level by means of a plumb line and a steel square ? 4. Explain how a carpenter could set a timber perpendicular to the floor by using only a ten-foot pole notched at the six-foot and eight-foot points. 5. Show how by § 303 a carpenter is enabled to saw a piece of lumber squarely in two. 6. Prove that any point in the plane perpen- dicular to a given line-segment at its middle point is equidistant from the ends of the line-segment. 7. Prove that any point of space equidistant from the ends of a given line-segment is in the plane per- pendicular to the line-segment at its middle point. 8. Prove that if a plane is perpendicular to a line-segment at its middle point, any point not in the plane is unequally distant from the ends of the line-segment. 9. What is the locus of points in space equidistant from two given points ? 10. Find the locus of points in a given plane that are equidistant from any two given points not in that plane. 11. A ray of light that starts from a given point B and is reflected by a mirror to a given point A travels along the shortest possible path. Find the point of the mirror from which the ray is reflected. Suggestion. — Draw a perpendicular from one of the given points to the plane of the mirror, and extend it to a point an equal dis- tance on the other side of the plane of the mirror. Join this point to the other given point. Prove that the point at which this second line meets the surface of the mirror is the required point. LINES AND PLANES IN SPACE 291 308. Theorem. — All lines perpendicular to a given line at a given point lie in the plane perpendicular to the given line at the given point. A n \N Hypothesis. Plane MN is perpendicular to line AB at ; OOis any line perpendicular to AB at 0. Conclusion. OClies in MN. Suggestions. Suppose that 00 does not lie in MN. Then the plane determined by AB and 00 must intersect MN in another line OB. Show that this violates § 53. Write the proof in full. EXERCISES 1. If a spoke of a wheel is perpendicular to the axle upon which it turns, it describes a plane in its rotation. 2. The locus of all lines perpendicular to a given line at a given point is the plane perjifendicular to the given line at the given point. 3. If a plane is perpendicular to a given line, any line perpendicular to the given line through any point of the given plane must lie in the given plane. 309. Theorem. — 77ie perpendicular to a plane from an ex- ternal point is the shortest line-segment that can be drawn to the plane from the point. The proof is left to the student. Write the proof in full. The perpendicular line-segment drawn from a given point to a given plane is called the distance from the point to the plane. 292 SOLID GEOMETRY 310. Theorem. — (1) Oblique line-segments drawn from a point to a plane^ meeting the plane at equal distances from the foot of the perpendicular from the pointy are equal ; (2) of two oblique line-segments meeting the plane at unequal distances from the foot of the perpendicular^ the more remote is the greater. ^R Hypothesis. Line AB is perpendicular to plane MN"^ meet- ing MN at B ; oblique line-segments from A meet MN at (7, D, B ; BI)=^BCdnd BB > BO. Conclusion. AB = AO, and AB > A O. Suggestions. It may be proved that AI) = AO if it is proved that A ABB ^AABO. For i^roYing AB>AO, on BB mark o^BF=BO, and draw AF. Point F lies between B and B. Why ? It can be proved that AB > AO il it is shown that AO = AF and AB > AF. But AB > AF ii Z. AFB > Z ASF, etc. Write the proof in full. 311. Theorem. — (1) Bqual oblique line-segments drawn from a point to a plane meet the plane at equal distances from the foot of the perpendicular from the point; (2) of two unequal line-segments^ the greater meets the plane at the greater distance from the foot of the perpendicular. Suggestions. This theorem, which is the converse of the theorem in § 810, may be proved easily by the indirect method, making use of § 310. The proof is left to the student. Write the proof in full, LINES AND PLANES IN SPACE 293 EXERCISES 1. If a derrick stands vertically on level ground, guy ropes reaching from the top of the derrick to stakes in the ground at equal distances from the foot of the derrick are equal. 2. If a line-segment AB is perpendicular to a plane at B, it subtends equal angles at all points of a circle lying in the plane and having B as center. 3. If the line-segment AB in perpendicular to a plane, and intersects the plane at a point C on AB produced, AB subtends equal angles at all points of a circle lying in the plane and having C as center. 4. If the line-segment AB is perpendicular to a plane, and intersects the plane at a point C between A and B, AB subtends equal angles at all points of a circle lying in the plane and having C as center. 5. If a line-segment AB is ]-)erpendicular to a plane at B, and a circle is drawn in the plane, with center at J5, then the angle subtended by AB B,t a, point of the plane within the circle is greater, and at a point without the circle less, than the angle subtended at a point on the circle. Suggestion. — Prove ZAEB>ZACB and ZADB, because it would be in both planes. 5. But AB cannot meet CD. Def. II 6. .-. AB cannot meet MN. 7. .-. AB is parallel to MN. § 312 EXERCISES 1. State the converse of the theorem iu § 313. Is it a true theorem? Illustrate. 2. Through a given point without a given straight line, an unlimited number of planes can be passed which are parallel to the given line. 3. If two straight lines are not in the same plane, a plane can be drawn containing one of them and parallel to the other. 4. Through a given point without a given plane, any number of lines can be drawn parallel to the plane. 5. If two intersecting planes are drawn through two parallel lines, theii line of intersection is parallel to each of the lines. 296 SOLID GEOMETRY 314. Theorem. — If a straight line is parallel to a plane^ the intersection of this plane and any plane containing the given line is parallel to the given line. Mz D .N Hypothesis. Line AB is parallel to plane MN \ plane AD contains AB^ and intersects plane MN in line QD. Conclusion. OD II AB. Suggestion. If AB and QD intersect, AB must intersect plane MN^ which is impossible. Write the proof in full. 315. Theorem. — If a plan^ ^dtersects two parallel planes^ the lines of intersection are parallel. M P^<^7\. y*^ ^^\ ^o .y ^ ^/^/' c N Hypothesis. Plane P§ is parallel to plane RS ; plane MN intersects plane PQ in AB and plane ^aS' in CD. Conclusion. AB II CD. Suggestion. AB and CD lie in the same plane and cannot intersect. Why ? Write the proof in full. LINES AND PLANES IN SPACE 297 EXERCISES 1. Segments of parallel lines which are cut off by two parallel planes are equal. 2. If a line is parallel to a plane, a plane can be drawn containing the given line and parallel to the given plane. 3. If a straight line and plane are parallel, any straight line drawn through a point of the plane and parallel to the given line lies wholly in the given plane. Suggestions. — The two parallel lines determine a plane, which in- tersects the given plane in a line through the given point. 4. If a line is parallel to each of two intersecting planes, it is parallel to their line of intersection. 5. If a straight line is parallel to a plane, any straight line which is parallel to the given line and not in the given plane is also parallel to the given plane. Suggestion. — The two parallel lines determine a plane. Consider the two cases when this plane is parallel to the given plane and when it intersects it. In the latter case, what is known about the line of in- tersection ? 6. If each of two intersecting straight lines is parallel to a given plane, the plane determined by these lines is parallel to the given plane. Suggestion. — Use indirect proof. Apply § 314, then § 46. 7. If two lines of one plane are paral- lel respectively to two intersecting lines of another plane, the planes are parallel. Suggestions. — If ^5 II ^F and CD II GH, what relation has plane MN to EF and to GH'i Now apply Exercise 6. 8. Through a given point a plane can be passed parallel to any two given straight lines in space. Suggestion. — Through the given point, draw lines parallel to the given lines. 9. If a straight line is parallel to each of two given planes, the lines of intersection which any plane passing through the given line makes with the given planes are parallel. 298 SOLID GEOMETRY 316. Theorem. — Two straight lines which are parallel to a third straight line not in their plane are parallel to each other. GF Hypothesis. CD II AB and EF II AB. Conclusion. OB II EF. Proof. 1. OB II AB and EF II AB, • Hyp. 2. AB and OB determine a plane AD^ and AB and EF determine a plane AF. § 299 3. OD and point E determine a plane OGr. § 296 4. Planes OG- and ^^ intersect in a line J57(^. § 301 5. (7i) II plane ^^. §313 6. .-. OB II ^a. § 314 7. AB II plane Oa, § 313 8. :•. AB II J5;(7. § 314 9. .-. Ea and jE:^ coincide. § 46 10. .-. (7i> II EF. Step 6 Write out the proof without the book. EXERCISES 1. If a straight line and plane are parallel, the lines of intersection of th^ given plane and all planes which contain the given line are parallel. 2. In a rectangular room the intersection of the ceiling and a wall is parallel to the intersection of the floor and the opposite wall. 3. In a quadrilateral whose sides are not all in the same plane (called a gauche quadrilateral), the line-segments which join the middle points of the adjacent sides form a parallelogram. LINES AND PLANES IN SPACE 299 317. Theorem. — If two angles in different planes have their sides parallel each to each and extending in the same direction from the vertices^ the angles are equal and their planes are parallel. Hypothesis. Z ABO is in plane MN, and Z DEF is in plane PQ ; BA II ED and BQ II EF, and the parallel lines ex- tend in the same directions from the vertices. Conclusion. Z. ABO = Z. DEF and MN II P Q, Proof. 1. BA WED and BOW EF, Hyp. 2. Mark off on the sides of the angles BA = ED and BO=EF. Draw BE, AD. OF, AO, DF. 3. Then EDAB and EFOB are parallelograms. § 90 4. .-. AD II BE and OF H BE. Dei O 5. .-. AD 11 OF. § 316 6. Also AD = BE and (TF = BE, § 82 7. .'.AD^OF. Ax. I 8. .*. ADFO = CJ, § 90 9. .'.AO=DF. §82 10. .'.AABO^ADEF. §76 11. .-. Z ^5(7 = Z i)^jP. Def. congruence 12. Now BA II plane P§ and BOW plane P^. § 313 13. If MZVand P§ intersected, their line of intersection would meet either BA or B or both. § 46 14. But this is impossible. § 312 15. .-. MN and PQ do not intersect ; and therefore MNWPQ. §312 300 SOLID GEOMETRY 318. Theorem. — A straight line perpendicular to one of two parallel planes is perpendicular to the other also. m/ -------£p \ y ---^ E N^- Hypothesis. Plane MNW plane PQ ; line AB ± plane PQ. Conclusion. Line AB X plane MN. Proof. 1. Plane MN II plane PQ ; and line AB ± plane PQ. Hyp. 2. Let two planes containing AB intersect ilifiVin ^(7 and AT)., respectively, and PQ'\w BE and BP., respectively. 3. Then AOWBE and AB 11 BF. § 315 4. But AB X BE and AB ± BF. § 302 5. r.ABA.AO?^ndiAB^^AD. §34 6. .-. line ^^ J- plane MN. § 302 Draw a figure and write the proof without the book. 319. Theorem. — Two planes which are perpendicular to the same straight line are parallel. ./ A \^ ./ B M( Hypothesis. Plane il!fiV± line J.5 at A\ plane P§±line AB at ^. Conclusion. MNW PQ. Suggestion. Suppose MN and PQ not parallel, and from any point of their line of intersection, lines drawn to A and B. Write the proof in full. LINES AND PLANES IN SPACE 301 320. Theorem. — If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to the plane, C Mi B D .N Hypothesis. Lines AB and CI) meet plane MJV at B and Z>, respectively ; ^5 ± plane MN; AB II CD. Conclusion. OB ± plane MK Proof. 1. AB ± plane MN and AB II OB. Hyp. 2. In ifiV^ draw any line DF from i), and draw BU II BF. Then Z ABE = Z OBF. § 317 But J.5 ± BE. § 302 .-. Z^^^ = rt. Z. Def. ± .'. Z CDF =rLZ, Ax. I Hence, since BF is awy line in MN' through 2>, Ci) J- plane MN, § 302 321. Theorem. — Two lines perpendicular to the same plane are parallel to each other. mL B' lE D .N Hypothesis. Line ^5 _L plane MN-, line (7i)± plane MN. Conclusion. AB 11 CB. Suggestion. Suppose AB not parallel to CB, but EB II CB. Apply § 320. Write out the complete proof. 322. Distance between parallel planes. — A line-segment con- necting points in two parallel planes and perpendicular to each plane (See § 318) is called the distance between the par- allel planes. 302 SOLID GEOMETRY 323. Theorem. — Two parallel planes are everywhere equally) distant . M^ C M P^ D vO Hypothesis. Plane MNW plane PQ, Conclusion. MN and PQ are everywhere equally distant. Suggestion. Let AB and CD be any two line-segments be- tween il^ZVand PQ^ each being perpendicular to both planes. Prove AB = OB, Write the proof in full. EXERCISES 1. If three equal line-segments which are not in one plane are paral- lel, the triangles formed by joining their corresponding end points are congruent and lie in parallel planes. 2. Two points on the same side of a plane and equally distant from it determine a line that is parallel to the plane. 3. Through a given point only one plane can be passed parallel to a given plane. 4. Two planes each parallel to a third plane are parallel to each other. 5. A straight line and a plane, both perpendicular to the same straight line, are parallel. 6. What is the locus of points equidistant from two given parallel planes ? Prove it. 7. What is the locus of points equidistant from anj two given points and also equidistant from two given parallel planes ? Prove it. 8. What is the locus of points at a given distance from a given plane ? Prove it. 9. Find a point equidistant from two given points, equidistant from two given parallel planes, and at a given distance from a third given plane. 10. If two angles in different planes have their sides parallel each to each, one pair of parallel sides extending in the same direction and the other pair in opposite directions from the vertices, the angles are_ supple- mentary. LINES AND PLANES IN SPACE 303 324. Theorem. — If two straight lines intersect three parallel planes^ their corresponding segments cut off hy the planes are proportional. X Hypothesis. Planes MN, PQ, and MS are parallel, and cut off segments AB and BO from line AC, and corresponding segments BU and UF from line BF, n . ' AB BE Conclusion. — — = — — • BO EF Suggestions. Draw AF, intersecting PQ at G-^ and draw BGi, OF^ AB, and GrB. The conclusion may be established if it is first proved that — — and — — - are each equal to -;r— -• BO EF CrF This may be proved if it is first proved that BQ- II OF, etc. Write the proof in full. EXERCISES 1. A line meets three parallel planes in the points A, B, and C, re- spectively. A second line meets the planes in the corresponding points D, E, and F, respectively. AB = 6 in., BC = 8 in., and DF= 18 in. Find the lengths of DE and EF. 2. The distances between four shelves are 6 in., 8 in., and 10 in., re- spectively. A diagonal brace 36 in. long reaches from the top shelf to the bottom shelf. Find the segments of the brace between the shelves, making no allowance for the thickness of the shelves. 3. If three parallel planes intercept equal segments on one straight line, they intercept equal segments on any other straight line which they intersect. 4. If two straight lines intersect any number of parallel planes, their corresponding segments are proportional. D 304 SOLID GEOMETRY 325. Diedral angles. — When two planes intersect, the line of intersection divides each plane into two parts. One such part of one plane and one such part of the other plane to- gether form a figure called a diedral angle. The parts of planes forming a diedral angle are called the faces of the diedral angle, and the line of intersection of the planes is called the edge of the diedral angle. Thus, in the figure, the planes ylC and BD are the faces, and the line AB is, the edge of A the diedral angle. A diedral angle is named by naming a point in one face, then the edge, then a point in the other face. When confusion would not result, a diedral angle may be named by merely naming its edge. Thus, the diedral angle above is named "angle C-AB-D^' or merely " angle AB^ 326. The plane angle of a diedral angle. — The angle formed by straight lines drawn in the two faces of a diedral angle, perpendicular to the edge at the same point of the edge, is called the plane angle of the diedral angle. Thus, if PE ±AB and PE lies in face BC, and if PF±AB and PF lies in face AD, of diedral angle C-AB-D, then Z EPF is the plane angle of diedral angle C-AB-D. It is evident from § 317 that the plane angle of a diedral angle has the same size at whatever point of the edge its vertex is taken. Let the student draw a figure representing any two posi- tions of the plane angle of a diedral angle, and reason out this truth. LINES AND PLANES IN SPACE 305 327. Theorem. — Two diedral angles are equal if their plane angles are equal. Hypothesis. /. MON and Z PQB are plane angles of diedral angles C-AB-B and G-EF-H, respectively; and Z MO]Sr= Z PQR. Conclusion. Angle C-AB-D = angle G-EF-H. Proof. 1. Z MONwmX Z PQR are plane angles of diedral angles C-AB^D and G-EF-ff, respectively ; and Z MOJSf = Z PQR. Hyp. 2. .-. /LMON can be superposed on Z P§i2 so that they coincide. Let it be so placed. § 13 3. AB _L OM, AB X Oi\r; also EF JL QP, EF± QR. § 326 4. .-.^^Xplaneof Zil[fOiV;and^PJ-planeof ZP^jR. § 303 5. But plane of Z il[fOiVand plane of Z P^i2 coincide. § 298 6. .-. AB and ^P coincide. § 306 7. .-. the faces of angle C-AB-B and the faces of angle G-EF-H coincide. § 298 8. .-. angle C-AB-B = angle G-EF-H^ because they coincide throughout. 328. Theorem. ^-If two diedral angles are equals their plane angles are equal. The proof of this theorem is left to the student. Use superposition, as in § 327. Write the proof in full. 306 SOLID GEOMETRY 329. Special diedral angles. — It follows from §327 and § 328 that the plane angle of a diedral angle may be taken as the measure of the diedral angle. A diedral angle is called a right, acute, or obtuse diedral angle according as its plane angle is right, acute, or obtuse. Two diedral angles are called adjacent, vertical, complementary, or supplementary, according as their plane angles are adjacent, vertical, etc. If two planes are intersected by a third plane, the terms cor- responding diedral angles^ etc., are applied to the pairs of diedral angles in the same sense that the corresponding terms are applied to plane angles. EXERCISES 1. Name a pair of vertical diedral angles in the figure of § 329. Name a pair of adjacent diedral angles. Name a pair of supplementary diedral angles. 2. Draw a figure representing two complementary diedral angles. 3. Prove that any two vertical diedral angles are equal. 4. Prove that any two right diedral angles are equal. 5. Prove that diedral angles which are complenjents of the same diedral angle or equal diedral angles are equal. 6. Prove that diedral angles which are supplements of the same diedral angle or equal diedral angles are equal. Prove that if two parallel planes are cut by a third plane : 7. The corresponding diedral angles formed are equal. 8. The alternate interior diedral angles formed are equal. 9. The alternate exterior diedral angles formed are equal. 10. The consecutive interior diedral angles formed are supplementary. 330. Perpendicular planes. — Two intersecting planes which form a right diedral angle are called perpendicular planes. LINES AND PLANES IN SPACE 307 331. Theorem. — If a straight line is perpendicular to a plane^ any "plane which contains the line is perpendicular to the plane. I ID m/ ^ "^ / c Hypothesis. Line AB ± plane MN at A ; plane PQ con- tains AB. Conclusion. Plane PQ 1. plane MN. Proof. 1. Line ^^ ± plane MN ?it A; plane PQ con- tains AB. Hyp. 2. .-. PQ and ife?iV intersect in a straight line OB. § 301 3. Draw AU ± OB in plane MN 4. AB±CB. § 302 5. Then Z UAB is the plane angle of diedral angle N-CB-Q. . § 326 6. But AB± AH. § 302 7. .-. Z BAB = rt. Z. Def. ± 8. .-. plane PQ± plane MN. § 330 332. Theorem. — A straight line drawn in one of two perpen- dicular planes, perpendicular to their line of intersection., is perpendicular to the other plane. Suggestion. If perpendicular planes MN and PQ intersect in line (72>, and line AB 1. CB and lies in P§, draw line AE 1. CB in plane MN. Then the analysis is as follows: AB ± plane MN if AB J_ AE. And AB 1.AE \i Z EAB = rt. Z. And Z EAB = rt. Z if Z EAB is the plane angle of diedral angle N-CB-Q and angle N-CB-Q is a right diedral angle, etc. Write the proof in full. 308 SOLID GEOMETRY 333. Corollary l. — If two planes are perpendicular^ a line perpendicular to one of them at any point in their intersection lies in the other. Use indirect proof. Apply § 306 and § 332. 334. Corollary 2. — If two planes are perpendicular, a line perpendicular to one of them from any point of the other that is not in the intersection lies in the other plane. Use indirect proof. Apply § 307 and § 332. EXERCISES 1. In sawing a board in two with a polished handsaw, the image of the board is seen in the surface of the saw. When the image of the surface of the board and the surface of the board itself appear to form one plane, the saw is cutting the board squarely in two, i.e. at right angles. Why? 2. The edge of a diedral angle is perpendicular to the plane of its plane angle. 3. If a plane is perpendicular to the intersection of two planes, it is perpendicular to each of the planes. 4. The plane of the plane angle of a diedral angle is perpendicular to the faces of the diedral angle. 5. If a plane is perpendicular to a line in another plane, it is perpen- dicular to that plane. 6. Through a given straight line not perpendicular to a given plane, one and only one plane can be passed perpendicular to the given plane. 7. If three lines are perpendicular to each other at a common point, the planes of the lines are perpendicular to each other. 8. If a line and a plane are parallel, any plane perpendicular to the line is also perpendicular to the plane. 9. If three or more planes intersect in a common line, the lines per- pendicular to them from any common external point lie in one plane. LINES AND PLANES IN SPACE 309 10. If two planes are perpendicular to each other, a line perpendicular to one of the planes and not contained in the other is parallel to the other. 11. If a line is parallel to one plane and perpendicular to another, the two planes are perpendicular to each other. 12. Between two straight lines not in the same plane, one, and only- one common perpendicular can be drawn. , SuGOESTiONS. — Let AB and CD be the given lines. Pass plane MN through CD parallel to AB. Through AB, pass a plane AE perpendicular to MN, inter- secting MN in PE and CD at P. In M. plane AE draw PF ±PE. Prove PF a common perpendicular to AB and CD. Suppose that GQ is a second common perpendicular, draw GH±PE, draw QK WAB, and show that an absurdity results. 13. The common perpendicular between two lines not in the same plane is the shortest line-segment that can be drawn between the lines. 335. Theorem. — If each of two intersecting planes is perpen- dicular to a third plane, their line of intersection is perpendicu- lar to that plane, R\ B Hypothesis. Plane P^± plane MN\ plane T^aS'X plane MN ', and planes PQ and RS intersect in line AB, which meets iHfiV at A. Conclusion. Line AB 1. plane MN. Suggestions. Draw line AO A. plane MN. Then AB ± plane MN if AB and AO are proved to coincide. AB and AO co- incide if J. (7 is proved to lie in P^ and in RS. Hence begin by proving the latter. Write the proof in full. 310 SOLID GEOMETRY 336. Theorem. — Every point in the plane which bisects a diedral angle is equally distant from the faces of the diedral angle. Hypothesis. Plane QR bisects diedral angle M-QP-N \ A is any point in QR ; line AB A. plane NP and line AC 1. plane MP. Conclusion. AB = AQ. Proof. 1. Plane QR bisects diedral angle M-QP-N \ A is any point in QR ; line AB ± plane NP and line AQ 1. plane MP. Hyp. 2. AB and J. (7 determine a plane. § 298 3. This plane intersects planes NP., QR., and MP in lines DB, DA, and BO, respectively. § 301 4. The plane which is determined by AB and ACis per- pendicular to NP and MP. § 331 5. .-. the plane which is determined by AB and AC is perpendicular to edge QP. § 335 6. .-. DB ± QP, DA ± QP, BC± QP. § 302 7. .'. Z BBA 'dnd Z.ABO are plane angles of diedral angles N-QP-R and R-BP-M, respectively. § 326 8. .-. ZBDA^ZABO. § 328 9. In A ABB and A A CD, AD > is common, and Z ABD and Z ACD are right angles. § 302 10. .'.AABD^AACD. §68 11. .-. AB = AC. Def. congruence LINES AND PLANES IN SPACE 311 337, Theorem. — Ani/ point within a diedral angle^ and equally distant from its faces y lies in the plane which bisects the diedral angle. Suggestion. If.^ is a point within diedral angle M-QP-N^ equally distant from MP and iVP, let QR be the plane de- termined by A and edge QP. Prove that plane QB bisects diedral angle M-QP-N. Write the proof in full. EXERCISES 1. Any point not in the bisecting plane of a diedral angle is unequally distant from its faces. 2. The locus of points within a diedral angle and equidistant from the faces is the bisecting plane of the diedral angle. 3. Find the locus of points equidistant from two given points and also equidistant from the faces of a diedral angle. 4. Find the locus of points equidistant from two parallel planes and also equidistant from the faces of a diedral angle. 5. Find the locus of points at a given distance from a given plane and also equidistant from the faces of a given diedral angle. 6. Find the locus of points equidistant from the faces of a given die- dral angle and also equidistant from the faces of another given diedral angle. 7. Find a point in a given plane that is equidistant from the faces of a given diedral angle and also at a given distance from a given plane. 8. Find a point in a given plane that is equidistant from two given points and also equidistant from the faces of a given diedral angle. 9. The bisecting plane of any diedral angle, if produced through the edge, also bisects the vertical diedral angle. 10. If two planes intersect, forming two adjacent diedral angles, the planes which bisect these adjacent angles are perpendicular to each other. 338. Projections. — The projection of a point upon a plane is the foot of the perpendicular from the point to the plane. The projection of a given line upon a plane is the line con- taining the projections upon the plane of all points of the given line. 312 SOLID GEOMETRY 339. Theorem. — The projection of a straight line upon a plane is the straight line determined hy the projections of any two of its points upon the plane. Hypothesis. A and B are any two points of straight line AB ; O and D are the projections of A and B^ respectively, upon plane MN. Conclusion. The projection of AB upon MN is straight line CD, Proof. 1. C and D are projections of A and B^ respec- tively, upon MN. Hyp- 2. .'. AOl. plane ifiV and BD ± plane MK § 338 3. .'.AQWBB. § 821 4. .*. ^C'and BD determine a plane AD. § 299 5. Planes AD and MN intersect in straight line CD. § 301 6. Plane AD _L plane MN. § 331 7. .•. OD contains the projections upon MN of all points of AB, § 334 8. .*. the projection of AB upon MNis straight line OD. §338 340. Inclination of a line to a plane. — A straight line which intersects a plane intersects its projection upon the plane. Why? The acute angle formed by a A straight line which intersects a / — ^^ '^ \ plane, but is not perpendicular to ^^-— ^ it, and its projection upon the plane is called the inclination of the line to the plane, as /. OB A, LINES AND PLANES IN SPACE 313 EXERCISES 1. Show that in the special case when a line is perpendicular to a plane, its projection upon the plane is a point. 2. If a trough is placed directly beneath the edge of a roof, all water dripping from the edge of the roof falls into it. Show how this illus- trates the theorem in § 339. 3. If a line-segment is parallel to a plane, its projection upon the plane is a line-segment equal to the given line-segment. 4. A line-segment 12 in. long has an inclination of 30° with a plane. Find the length of its projection upon the plane. Find its length if the inclination is 45°. If it is 60°. 5. If a straight line is not perpendicular to a plane, the given line and its projection upon the plane determine a second plane perpendicular to the first. Suggestion. — From a point on the given line, draw a line perpen- dicular to the given plane. Does the plane determined by the given line and its projection contain this perpendicular ? 6. The projections upon a plane of two parallel lines which are not perpendicular to the plane are parallel. Suggestion. — From a point on each of the given parallel lines draw a line perpendicular to the' given plane. What relation have these per- pendiculars ? Seek to apply § 317. 7. If two parallel lines intersect a plane, they are equally inclined to it. 8. Parallel line-segments are proportional to their projections upon a plane. 9. If a straight line intersects two parallel planes, it is equally in- clined to the two planes. 10. If two planes are not perpendicular, the projection upon one of any parallelogram in the other is a parallelogram. 11. If two planes are parallel, the projection upon one of any polygon in the other is a congruent polygon. 314 SOLID GEOMETRY 12. If a straight line intersects a plane, but is not perpendicular to it, the inclination of the line to the plane is the least angle made by the given line and any line drawn in the plane through its foot. Suggestion. — Let BC be the pro- jection of BA , and C the projection of ^< A, upon MN; and let BD be any other jection oi nJi , ana o ine projection oi ^^ A, upon MN; and let BD be any other y ^ * \N line in MN than BC that passes through / <^^ — J~ — I \ B. Mark off BD = BC and draw AD- ^ / ^ ^^^' \ Prove Z CBA < Z DBA . 13. If a line intersects a plane, and is not perpendicular to it, the ob- tuse angle which it makes with its projection upon the plane is the greatest angle that it makes with any straight line drawn in the plane, through the intersection. Suggestion. — In the figure of Ex. 12, produce BC and BD through B. 341. Polyedral angles- — The figure formed by three or more rays which are drawn from the same point, not more than two being in one plane, together with the portions of planes determined by the pairs of adjacent rays and included between them, is called a polyedral angle. The rays are called the edges of the polyedral angle. The point from which the rays are drawn is called the vertex. The portions of planes deter- mined by pairs of adjacent rays and included between them are called the faces. And the angles formed in the faces by the pairs of adjacent edges are called the face angles of the polyedral angle. A polyedral angle whose vertex is and whose edges are OA, OB, 00, and OD is named 0-ABOD. A polyedral angle is convex if a plane which intersects all of the faces intersects them in line-segments which form a convex polygon. A polyedral angle of three faces is called a triedral angle. LINES AND PLANES IN SPACE 315 342. Theorem. — Any face angle of a triedral angle is less than the sum of the other two face angles. Hypothesis. Z A 00 is the greatest face angle of triedral angle 0-ABC. Conclusion. Z AOO < Z AOB -\- Z BOO. Proof. 1. Z AOO is the greatest face angle of triedral angle O-ABO. Hyp 2. Draw OB in face AOO, making Z AOB = Z AOB. And let a plane cut off OB = OB and meet 00 at and OA at A. 3. AO is n common side of A AOB and A AOB. 4. .'.aAOB^aAOB. §63 5. .-. AB = AB. Def. congruence 6. But AO Z PBA + Z PBO, ZOCB + Z 001) > Z POB + Z POD, etc. § 342 4. .-. by* adding, Z OBA + Z OBO + Z OOB + etc. > Z PBA + Z PBO+ Z POB 4- etc. Ax. IX That is, the sum of the base angles of the triangles whose common vertex is is greater than the sum of the base angles of the triangles whose common vertex is P. 5. But the sum of the angles of each triangle is a st. Z. §48 6. .-. since the number of triangles whose vertex is equals the number of triangles whose vertex is P, the sums of all angles of the two sets are equal. Ax. II 7. .-. the sum of the angles whose vertex is is less than the sum of the angles whose vertex is P. Ax. VIII 8. But the sum of the angles at P = 4 rt. A § 17 9. .-. ZvlO^ + Z^Oa+etc. <4rt.A Ax. XII LINES AND PLANES IN SPACE 317 EXERCISES 1. The sum of the face angles of a triedral angle is 320°. What is the greatest value that any one of the face angles may have ? 2. If the face angles of a triedral angle are equal, what is the greatest value that each of the face angles may have ? 344. Equal and symmetrical polyedral angles. — Two polye- dral angles are equal if they may be made to coincide. Hence to each face angle or diedral angle in one of two equal polyedral angles, there corresponds a like angle equal to it in the other, and these correspondingr parts are ar- ^ ° ^ Equal Polyedral Angles ranged in the same order. Two polyedral angles are called symmetrical if to each face angle or diedral angle in one there corresponds a like angle equal to it in the other, and these corresponding parts are arranged in reverse order. In general, symmetrical polyedral angles cannot be made to coincide. ^ .,,,,, Symmetrical Polyedral Two symmetrical polyedral angles may * be compared to a pair of gloves. To each part of one glove there corresponds a like part of the other, but these like parts are arranged in reverse order, so that the right hand cannot be put into the left glove. EXERCISES 1. Construct from cardboard, or stiff paper, models of two symmetri- cal triedral angles whose face angles are respectively 60°, 45'", and 30°. Cut as indicated in the pattern, fold along dotted lines, and paste. Can V 5r~7f^\ /^^ i^ they be made to coincide ? \ /'^j.v vV /&yj45*^ 2. If the faces of a polyedral angle ^^ \ J> "C / \ are produced through the vertex, the given angle and the vertical polyedral angle formed are symmetrical. 318 SOLID GEOMETRY 345. Theorem. — If two triedral angles have the face angles of one equal respectively to the face angles of the other ^ their corresponding diedral angles are equal. M Hypothesis. In triedral angles M-ABQ and N-DEF, Z AMB = Z BNE, A BMO= A ENF, Z. AMO = ZBNF. Conclusion. Angle B-AM-G = angle E-BN-F, etc. Proof. 1. Z AMB = Z BNE, etc. Hyp. 2. Mark off equal distances MA, MB, MO, NB, NE, NF, on the edges of angle M~ABO and angle N-BEF. Draw AB, BO, AC, BE, EF, BF. On AM-dnd BN, respectively, take AX= BY. Draw XG ± AM in face AMB, meeting AB Sit G; XH ±AM in face AMC, meeting AC at R; YI± BN in face BNE, meeting BE at 7; YJ A. BNin face BNF, meeting BF at J. Draw (7^ and IJ. 3. Then A AMB ^ A BNE, etc. § 63 4. .-. AB = BE, and Z BAM^ Z EBN, etc. Def. cong. 5. .'.AABC^ABEF. , § 76 6. .'. ZBAO=Z EBF. ' Def. cong. 7. Also A ^(5^X^ A D/r and A ^^X^ A i>e7F. §67 8. .-. Xa = YI, XE= YJ; also Aa = BI,AH= BJ. Def. cong. 9. .'.aAGH^ABIJ. §63 10. .-. aH= IJ. Def. cong. 11. .-. A aXH ^ A lYJ. § 76 12. .-. Z aXH:= Z lYJ Def. cong. (To be completed by the student.) LINES AND PLANES IN SPACE 319 346. Theorem. — If two triedral angles have the face angles of one equal respectively to the face angles of the other^ and arranged in the same order^ the triedral angles are equal. Suggestion. Use superposition. Show that by placing the triedral angles so that one pair of equal face angles coin- cide, the triedral angles may be made to coincide through- out. Apply § 345. Write out the complete proof. EXERCISES 1. If two triedral angles have the face angles of one equal respec- tively to the face angles of the other, but arranged in reverse order, the triedral angles are symmetrical. 2. If two triedral angles have two face angles and the included diedral angle of one equal respectively to two face angles and the included die- dral angle of the other, and the equal parts arranged in the same order, the two triedral angles are equal. Suggestion. — Superpose one triedral angle upon the other. 3. If two triedral angles have two diedral angles and the included face angle of one equal respectively to two diedral angles and the included face of the other, and arranged in the same order, the two triedral angles are equal. 4. If two triedral angles have two face angles and the included die- dral angle of one equal respectively to two face angles and the included diedral angle of the other, but the equal parts arranged in reverse order, the triedral angles are symmetrical. Suggestion. — Construct a third triedral angle symmetrical to one of the given triedral angles. Compare the parts of the three angles. 5. If two triedral angles have two diedral angles and the included face angle of one equal respectively to two diedral angles and the included face angle of the other, but the equal parts arranged in reverse order, the triedral angles are symmetrical. 6. If two face angles of a triedral angle are equal, the diedral angles opposite them are equal. 7. If two face angles of a triedral angle are equal, the triedral angle is equal to the symmetrical triedral angle. 320 SOLID GEOMETRY MISCELLANEOUS EXERCISES 1. In how many positions must a spirit level be observed in order to determine if the surface on which it rests is horizontal ? Why ? 2. A room is 10 ft. high, 16 ft. wide, and 20 ft. long. Find the length of the shortest line that can be drawn on the floor and walls from a lower corner to the diagonally opposite upper corner. 3. Can a triedral angle have for its faces a regular decagon and two equilateral triangles? Two regular octagons and a square? Why? 4. If the inclination of a line-segment to a plane is 45° and the pro- jection upon the plane is 16 in. long, how long is the line-segment? 5. If each of a series of parallel planed intersects all faces of a trie- dral angle, the intersections form a series of similar triangles. 6. A plane perpendicular to each of the faces of a diedral angle inter- sects them in the sides of the plane angle of the diedral angle. 7. If from any point within a diedral angle perpendicular lines are drawn to the faces, the angle between these lines is the supplement of the plane angle of the diedral angle. 8. Find the locus of points equidistant from three given points that are not in one straight line. 9. Prove that a line cannot be perpendicular to each of two inter- secting planes. 10. In the first figure of § 307, prove that OG Jl_GF by means of the Theorem of Pythagoras, using the equations OG^ + GE^ — 0E\ OE^ + EF^ = 0F'\ GF^ = GE^ + EF\ Add these three equations. 11. If the edges of one polyedral angle are perpendicular to the faces of another, the edges of the second are perpendicular to the faces of the first. 12. In any triedral angle the three planes bisecting the three diedral angles intersect in one line, all points of which are equally distant from the three faces. 13. All points within a triedral angle and equally distant from its three faces are in the intersection of the bisecting planes of the diedral angles. 14. The locus of points within a triedral angle and equally distant from its three faces is the line of intersection of the three planes which bisect the diedral angles of the triedral angle. CHAPTER XIII PRISMS AND CYLINDERS 347 Geometric solids. — A geometric solid is a portion of space which is completely inclosed or separated from the rest of space by some kind of surface. 348. Polyedrons. — A polyedron is a solid whose bounding surface consists of portions of intersecting planes. The figure adjoining is a polyedron. It is bounded by portions of six planes. The portions of planes which form the bounding surface of ., ' Polyedron a polyedron are called its faces. The intersections of the facfes are called its edges. The intersections of its edges are called its vertices. How many edges are there in the polyedron above ? How many ver- tices are there ? A line-segment which joins any two vertices of a polye- dron that do not lie in the same face is called a diagonal of the polyedron. If a plane intersects a poly- edron, the polygon formed by the intersections of the plane and the faces is called a sec- "" "^ ^^ Section ^p tion of the polyedron. ' Polyedron If every section of a poly- edron is a convex polygon, it is called a convex polyedron. Only convex polyedrons will be considered in this volume. 321 322 SOLID GEOMETRY A polyedron of four faces is called a tetraedron ; one of six faces, a hexaedron ; one of eight faces, an octaedron ; one of twelve faces, a dodecaedron; one of twenty faces, an icosaedron. The following drawings show models of these polyedrons. Tbtraedbon Hexaedron Octaedron Dodecaedron Icosaedron EXERCISES 1? What is the least number of faces that a polyedron can have? Edges ? Vertices ? What kind of polyedron is it ? 2. How many edges has a hexaedron? An octaedron? A dodeca- edron ? An icosaedron ? 3. How many vertices has a hexaedron ? An octaedron ? A dodec- aedron? An icosaedron ? 4. If E is the number of edges, F the number of faces, and V the num- ber of vertices of a polyedron, showthat in each of the five polyedrons named above E + 2 = F + V. This, principle is known as Euler's Theorem, 5. Show that in a tetraedron, if S equals the sum of the face angles and V equals the number of vertices, 5 = 4(F — 2) right angles. Is this formula true for a hexaedron? For an octaedron? For a dodecaedron ? For an icosaedron ? 349. Prisms. — A prism is a polyedron of which two faces are congruent polygons in parallel planes, and of which the other faces are parallelograms each of which has sides of the polygons as two of its opposite sides. --^mm^^ mm- mi ^ ^ , . -^^.-w^-- - Right Prism Ine two congruent polygons in parallel planes are called the bases, and the other faces are PRISMS AND CYLINDERS 323 called the lateral faces. The intersections of the lateral faces are called the lateral edges. The perpendicular distance between the bases of a prism is called the altitude of the prism. The sum of the areas of the lateral faces of a prism is called the lateral area. The section of a prism which is made by a plane which intersects all of the lateral edges and is perpendicular to them is called a right section of the prism. A prism whose lateral edges are perpendicular to its bases is called a right ic:^ Drism. ^^ii^^^to^ Oblique Prism A prism whose lateral edges are not perpendicular to its bases is called an oblique prism. A prism is called triangular, quadrangular, hexagonal, etc., according as its bases are triangles, quadrilaterals, hexagons, etc. 350. Fundamental properties of a prism. — The following important properties of a prism are easily deduced from the definitions above. The student should draw figures and reason out the correctness of each. (1) The lateral edges of a prism are parallel and equal. (2) The altitude of a right prism is equal to a lateral edge of the prism. (3) The lateral faces of a right prism are rectangles. (4) Sections of a prism made hy parallel planes cutting all lateral edges are congruent polygons. (5) Right sections of a prism are congruent polygons. 324 SOLID GEOMETRY EXERCISES 1. On a piece of cardboard, draw a figure similar to the adjoining figure, making each side of each triangle 3 in. and the rectangles 3 in. by 5 in. Cut out the pattern, and, by fold- ing along the dotted lines and pasting, make a model of a triangular prism. 2. What kind of polyedrons are the cells which contain the honey in the comb of the bee? What is the advantage of building the cells in this form ? ■\/ 3. In the form of what kind of polyedron are lead pencils sometimes made? 4. A rectangular room or box is what kind of prism ? 5. Glass prisms, such as that shown in the drawing, are used in optical instruments for changing the direction of light passing through them. Abeam of white light passing through this prism is dispersed or separated into a rainbow-colored band, gradually chang- ing from red at one end, through orange, yellow, etc., to violet at the other end. Prismatic pieces of glass are used chandeliers to disperse the light. 6. The drawing shows a model of a polyedron of which two faces are con- gruent polygons in parallel planes, and of which the other faces are all parallel- ograms. Why is it not a prism ? 7. A section of a prism made by a plane parallel to a base is con- gruent to the base. 8. Any section of a prism made by a plane parallel to a lateral edge is a parallelogram. 9. The lateral faces of a right prism are perpendicular to the bases. What kind of prism is it ? also as pendants or fringes on PRISMS AND CYLINDERS 325 351. Truncated prisms. — The part of a prism contained between a base and a plane which is not parallel to the base is called a truncated "^ /^ Truncated prism. 1 352. Parallelopipeds. — A prism whose bases are parallel- ograms is called a par- allelepiped. A parallelopiped whose lateral edges are PabaiIII^pkd perpendicular to the bases is a right parallelopiped. A right parallelopiped whose bases are rectangles is a rectangular parallelopiped. A rectangular parallelopiped all of whose faces are squares is a cube. 353. Fundamental properties of a parallelopiped. — The fol- lowing properties of a parallelopiped follow from the defini- tions above. The student should draw figures and reason out the correctness of each. (1) The opposite lateral faces of a parallelopiped are con- gruent and parallel. (2) Any two opposite faces of a parallelopiped may he taken as the bases. (3) Any right section of a parallelopiped is a parallelogram, EXERCISES 1. Find the sum of all of the face angles of any parallelopiped. 2. The diagonals of a rectangular parallelopiped are equal. 3. How, by measuring the diagonals of a piece of house framing, wsaii a carpenter tell when it is truly rectangular? 326 SOLID GEOMETRY 4. The square of a diagonal of a rectangular parallelopiped is equal to the sum of the squares of three concurrent edges. 5. Find the diagonal of a cube whose edge is 2 in. 6. If the edge of a cube is e, find the length of a diagonal of the cube. 7. The diagonal of a face of a cube is 3 V2. Find the diagonal of the cube. 8. Are the diagonals of a cube perpendicular to each other ? 9. A suitcase is 26 in. long, 15 in. high, and 7 in. thick. Can an umbrella which is 32 in. long be packed inside of it ? 10. Show that the edge, diagonal of a face, and diagonal of a cube are in the ratio of 1 : \/2 : \/3. 11. The sum of the squares of the four diagonals of any parallelopiped is equal to the sum of the squares of the twelve edges. 12. The diagonals of any parallelopiped all meet at one point, which is the middle point of each. 13. The intersection of the diagonals of any parallelopiped and the intersections of the diagonals of two opposite faces are in a straight line. 354. Theorem. — TTie lateral area of a prism is equal to the product of a lateral edge and the perimeter of a right section. /^ Hypothesis. MNOP "- is a right section of a prism, and AF is a lateral edge. Conclusion. The lateral area = AF(^M]Sr+NO+ OP+etc). Suggestions. Express the area of each lateral face, and form the sum of these area«. Apply § 350 (1). Factor. PRISMS AND CYLINDERS 327 355. Corollary. — The lateral area of a right prism is equal to the product of the altitude and the perimeter of a base. The proof is left to the student. EXERCISES 1. Find the lateral area of a rectangular parallelepiped whose length is 24 in., width 16 in., and height 12 in. Find the total area. 2. Write the formula which expresses the total area of a rectangular parallelepiped whose dimensions are a, b, and c. 3. Find the lateral area of a right hexagonal prism in which each side of a base is 8 in. and the altitude is 20 in. 4. Find the lateral edge of a prism whose lateral area is 714 sq. in. and perimeter of a right section 42 in. 5. A gallon of paint costing $1.50 will cover 300 sq. ft. with one coat. How much will it cost to paint both sides of an 8-foot fence around an athletic field that is 310 ft. wide and 325 ft. long? 6. How many square feet of sheathing are necessary to cover the sides and ends of a box car 34 ft. long, 8 ft. wide, and 7^ ft. high? Add 10 % for waste in cutting and matching. 7. In computing the cost of lathing and plastering a room, the total area of wall space is considered, without any deductions for openings. A room is 9^ ft. high, 15 ft. wide, and 20 ft. long. Laths cost $ 6 per thousand, and are delivered in bundles of 50 each. A bundle will lath 4 sq. yd. How many bundles are required? (Part of a bundle cannot be bought.) What will they cost ? 8. Find the lateral area of a right triangular prism, each side of a base of which is 4 in. and the altitude of which is 9 in. Find also the total area. 9. Find the total area of a right triangular prism whose altitude is 24 in. and whose base is a right triangle in which the sides forming the right angle are 6 in. and 8 in., respectively. 10. Find the lateral area of a right prism whose base is a regular hexagon of which each side is 5 in. and the altitude is 12 in. Find also the total area. 328 SOLID GEOMETRY 11. The lateral area of a rectangular parallelepiped is 1152 sq. in. and the total area is 1632 sq. in. The height is 18 in. Find the length and breadth. Suggestion. Use a system of equations. 12. The total area of a rectangular parallelepiped is 550 sq. ft. The width is twice as great as the depth, and the length is three times as great as the depth. Find the depth, width, and length. 13. The total area of a cube is 96 sq. in. Fii.d the edge. 14. li S represents the total area of the suiface of a cube, what repre- sents the length of an edge ? 15. The lateral area of a right prism whose base is a regular hexagon is iidhVd, where d is the apothem of the base and h is the altitude of the prism. 16. Write the formula which expresses the total area of the right hexagonal prism in Exercise 15. 17. The three face angles at one vertex of a parallelepiped are each 60°, and the three lateral edges of the triedral angle with that vertex are 2 in., 4 in., 6 in., respectively. Find to two decimal places the area of the entire surface of the solid. 356. Volume of a solid. — The volume of a solid is the numerical measure of the solid (See § 115), the unit of measure being a cube whose edge is some linear unit. Thus, if a rectangular parallelepiped contains a cube whose edge is 1 unit ex- actly 24 times, and the cube is taken as the unit of measure, the volume of the paral- lelepiped is 24. If an empty rectangular box is 2 in. deep, 4 in. wide, and 5 in. long, how many cubic blocks each 1 in. long could be packed in it? Then what is the volume of the box? 357. Congruent and equal solids. — Two solids which have the same volume are equal. It is evident that equal solids need not have the same shape. Two solids which are alike in all respects, so that they may be made to coincide, are congruent. PRISMS AND CYLINDERS 329 358. Theorem. — Two prisms are congruent if three faces including a triedral angle of one are congruent respectively to three faces including a triedral angle of the other^ and an similarly placed. H P }0 H3rpothesis. Prisms AQ- and 10 have faces BE^ BG, BB respectively congruent to faces JM^ JO^ JL^ and similarly placed. Conclusion. Prism AG ^ prism 10. Proof. 1. BE^JM,Ba^JO,BD^JL. Hyp. 2. .-. Z.FBA=ANJl ACBF^/.KJN, Z CBA =c ZKJL Def. congruence 3. . •. triedral angles B-A CF and J-IKN^ltq equal. § 346 4. Hence by applying prism ^6r^ to prism /O, angles B-ACF ^nd J-IKN may be made to coincide, face BB coin- ciding with JL, BE with JM, BG with JO, D falling at Z, etc. . § 344 5. The lateral edges are parallel. § 350, (1) 6. .-. DH^V\\\ fall along LP, § 46 7. .-. planes (70" and KP coincide. § 299 8. Since E, F, G coincide with M, iV, 0, respectively, planes EG and MO coincide, and hence H and P coincide. § 297 9. Similarly, the other planes of corresponding lateral faces and the other vertices of ^JTand NP coincide. 10. .-. prism AG ^ prism 10. § 357 330 SOLID GEOMETRY 359. Corollary l. — Two right prisms having congruent bases and equal altitudes are congruent. The proof is left to the student. 360. Corollary 2. — Two truncated prisms are congruent if three faces including a triedral angle of one are congruent respectively to three faces including a triedral angle of the other ^ and are similarly placed. For the steps in the proof of § 358 apply equally to two truncated prisms. Draw two truncated prisms satisfying the hypothesis, and think the steps of the proof. EXERCISES 1. Two rectangular parallelepipeds are congruent if the three edges meeting at a vertex of one are equal respectively to the three edges meet- ing at a vertex of the other. 2. Does the plane determined by two diagonally opposite edges of any parallelopiped divide it into two congruent triangular prisms? Why? 3. Does the plane determined by two diagonally opposite edges of a rectangular parallelopiped divide it into two congruent triangular prisms? Why? 4. Two triangular prisms are congruent if the lateral faces of one are congruent respectively to the lateral faces of the other and are similarly placed. 5. If a square wooden beam of which the ends are at right angles to the longitudinal edges is sawed lengthwise along two diago- nally opposite edges, the two triangular pieces into which it is cut are congruent. 6. For making a right prism of any de- sired kind, it is only necessary to have a pat- tern of a base and a lateral edge. Why ? PRISMS AND CYLINDERS 331 361. Theorem. — An, oblique prism is equal to a right prism having for a base a right section of the oblique prism and for its altitude a lateral edge of the oblique prism. B C Hypothesis. AI is an oblique prism ; KS is a right prism, having for a base KLM "-^ a right section of prism Al^ and its altitude KP equal to AF, a lateral edge of prism AL Conclusion. Oblique prism AI= right prism ^aS'. Suggestions. It can be proved that prism AI= prism KS if it is first proved that truncated prism ^i^ ^ truncated prism FS. How ? The latter may be proved if it is first proved that faces AB^ AL, AG are congruent respectively to faces FI., FQ^ FT. How ? Hence begin by proving faces AD, AL, AG congruent respectively to faces FI, FQ, FT. Write the proof in full. EXERCISES 1. Show that the theorem in § 361 may be proved in two ways, by using truncated prisms AN and FS in two ways. 2. Show that an oblique prism may be changed into an equal right prism by cutting it along a right section into two truncated prisms, then interchanging the positions of the latter. 3. Make a model of wood for illustrating the theorem in £ 361. 332 SOLID GEOMETRY 362. Theorem. — A plane passed through two diagonally op- posite edges of any parallelopiped divides it into two equal triangular prisms. H Hjrpothesis. BH is any parallelopiped; plane AG- passes through diagonally opposite edges AE and (7(r, forming triangular prisms ABQEFG and ADOEHG. Conclusion. Prism AB OEFG = prism AB QEHG, Proof. 1. Plane AG passes through diagonally opposite edges AE and OG of parallelopiped BH^ forming triangular prisms AB OEFG and AD QEHG. Hyp. 2. Let MNOP be a right section of BH, 8. Then MNOP is a parallelogram. § 353, (8) 4. r. AMNO^AMPO. §83 5. Prism AB OEFG = 3i, right prism with A MJSTO for base and AE for altitude, and prism ABOEIIG = aL right prism with AMPO for base and AE for altitude. § 361 6. But these right prisms are equal. § 359 7. .-. prism ABOEFG = prism ABOEHG. Ax. I EXERCISES 1. A plane passed through two diagonally opposite edges of a rec- tangular parallelopiped divides it into two congruent triangular prisms. 2. In the figure of § 362, the plane passed through the diagonally opposite edges AE and CG and the plane passed through the diagonally opposite edges BF and DH divide parallelopiped BH into four equal triangular prisms. PRISMS AND CYLINDERS 333 3. A bin in the form of a rectangular parallelepiped that holds 840 bu. of grain is divided into four compartments by two vertical diagonal planes. How many bushels in each compartment? 363. Volume of a rectangular parallelepiped. The lengths of the three edges of a rectangular parallelopiped which meet at any vertex are called the dimensions of the parallelopiped. If the dimensions of a rectangular parallelopiped are 2, 3, and 5 linear units, respectively, it is evident that the parallelopiped may be divided into 2 x 3 x 5, or 30, cubes, each having its edge equal to the same linear unit. Hence if one of these cubes is taken as the unit of measure of the parallelopiped, the volume of the parallelopiped equals 30. In general, if the dimensions of a rectangular parallelo- piped are a, 5, and c linear units, respectively, the number of unit cubes, or the volume of the parallelopiped, is ahc. This relation also may be proved to be true when the three edges intersecting at any vertex do not have a com- mon unit of measure, or when the dimensions can be ex- pressed only approximately, although the proof is not attempted here. Hence : The volume of any rectangular parallelopiped is equal to the product of its three dimensions. 334 SOLID GEOMETRY 364. Corollary l. — The volume of any rectangular parallelo- piped is equal to the product of its altitude and the area of its base. The proof is left to the student. 365. Corollary 2. — The volumes of two rectangular paral- lelopipeds having equal bases are to each other as their altitudes. For, if M is the volume, b the area of the base, and h the altitude of one parallelopiped, and if N is the volume, b the area of the base, and k the altitude of the other, then by § 364, M=hb Q,nd N=kb. TT 1. J- -J' M hb h Hence, by dividing, —- = — or -• Jy kb k 366. Corollary 3. — The volumes of two rectangular parallel- opipeds having equal altitudes are to each other as their bases. The proof is left to the student. 367. Corollary 4. — The volume of a cube is equal to the cube of its edge. The proof is left to the student. EXERCISES 1. Find, in its lowest terms, without multiplication, the ratio of the volumes of two rectangular parallelopipeds whose dimensions are 6 ft., 8 ft., 5 ft., and 9 ft., 12 ft., 10 ft., respectively. 2. A rectangular water tank is 8 ft. 3 in. long and 5 ft. wide. How many cubic feet of water does it contain when the water is 18 in. deep? 3. One cubic foot of iron weighs 450 lb. If an iron bar is 2\ in. thick, 4 in." wide, and 6 ft. 6 in. long, find its weight. 4. A gallon contains 231 cu. in. How many gallons does a tank hold if it is 8 ft. square and 6 ft. high? 5. 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. Jong, 36 ft. wide, and 7 ft. deep. PRISMS AND CYLINDERS 335 pi i4-' fe: The earth removed is to be used for "filling" the surrounding yard What will be the depth of the filling? 6. In steel construction work, such as the con- struction of bridges and modern city buildings, the weight of steel is computed and not weighed. In a certain building contract the specifications require 100 16-foot beams, a right cross section of which is shown in the margin (8'' = 8 in.). Find the weights of the beams and the cost B.t5fa pound when put in place. Allow 490 lb. per cubic foot. 7. Find the weight of 150 12-foot steel beams, a right cross section of which is shown in the margin. 8. In the construction of a bridge there were T-shaped steel beams 20 ft. long and U-shaped steel beams 16 ft. long, the right cross sections of which are shown in the margin. Compute the weight of a beam of each kind. 9. The total area of a cube is 150 sq. in. Find its volume. 10. The volume of a given cube is v. Find the volume of a cube whose edge is twice as long as the edge of the given cube. 11. The edge of a given cube is e. Find to tenths the edge of a cube containing twice the volume of the given cube. Note. — See § 185. It is discovered in Exercise 11 above that the edge of a cube of which the volume shall be just twice that of a cube of given edge cannot be computed exactly. The corresponding problem of construction, to construct with compasses and unmarked straightedge alone the edge of a cube which shall contain twice the volume of a given cube, is one of the three famous impossible problems of geometry. It dates from the time of the ancient Greeks, and is known as the problem of the Duplication of the Cube. One legend as to the origin of the problem is that the Athenians, who were suffering from a pestilence, consulted the oracle at Delos as to how to stop the plague. Apollo replied through the oracle that in order to stop the pes- tilence the Athenians must double the size of the altar of the god in Athens. This altar was in the form of a cube. A new altar was constructed with an edge twice as long as the edge of the old one. The pestilence then became worse. The Athenians gave the problem to the mathematicians, who in time proved it impossible of solution. 336 SOLID GEOMETRY 368. Theorem. — The volume of any parallelopiped is equal to the product of its altitude and the area of its base. D C^-^ — A / --^^ Hypothesis. P is any parallelopiped of which the altitude is h and the area of the base is a. Conclusion. The volume of P = ha. Proof. 1. P is any parallelopiped of which the altitude is h and the area of the base is a. Hyp. 2. Produce the edge AB and all edges parallel to AB. On AB produced, mark off CD = AB, and through and I) pass planes perpendicular to AB, forming the right parallelo- piped Q. Produce the edge BU of Q and the edges parallel to BB. On BB produced, mark off FGr — BE, and through F and G- pass planes perpendicular to BE, forming the rectangular parallelopiped R. 3. Then P = ^ and $ = i2. § 361 4. .-. P = /2. Ax. I 5. P, Q, and R have a common altitude h. § 323 6. Let the areas of the bases of Q and RhQh and c, re- spectively. 7. Then a = 5 and 5 = and ^, respectively, and that when this is done, C falls upon jP, and hence all points in the perimeter of ABC fall at the same time upon corresponding points in the perimeter of DEF^ etc. EXERCISES 1. The sections of a cylinder made by two parallel planes cutting all of the elements are congruent. 2. Every section of a cylinder made by a plane paral- lel to a base is congruent to the base. 3. Every section of a circular cylinder made by a plane parallel to a base is a circle. 4. The straight line joining the centers of the bases of a circular cylinder passes through the centers of all sections made by planes parallel to the bases. 344 SOLID GEOMETRY f^ Inscribed Prism 374. Inscribed and circumscribed prisms and cylinders. — A prism is said to be inscribed in a circular cylinder when the bases of the prism are inscribed in the bases of the cylinder and the lateral edges of the prism are elements of the cylinder. The cylinder is said to be circumscribed about the prism. A prism is said to be circumscribed about a circular cylinder when the bases of the prism are circumscribed about the bases of the cylinder and the lateral edges are parallel to the elements of the cylinder. The cylinder is said to be inscribed in the prism. It is evident that each lateral face of a cir- cumscribed prism of a cylinder contains one element and no other line nor point of the cylinder. Such a plane is said to be tangent to the cylinder. 375. Limits. — The following principles of limits may be assumed : If a prism whose bases are regular polygons is inscribed iri^ or circumscribed about, a circular cylinder, and if the number of lateral faces is indefinitely increased : (1 ) The perimeter of a right section of the prism approaches the perimeter of a right section of the cylinder as a limit. (2) The lateral area of the prism approaches the lateral area of the cylinder as a limit. (3) The volume of the prism approaches the volume of the cylinder as a limit. Circumscribed Prism PRISMS AND CYLINDERS 345 376. Theorem. — The lateral area of a circular cylinder is equal to the product of an element and the perimeter of a right section of the cylinder. Hyp. Let Hypothesis. /S'= lateral area of a circular cylinder, P = perimeter of a right section, and ^ = an element. Conclusion. S = UP. Proof. 1. AS'=lat. area, P=perim. rt. sec, -^= element. 2. Inscribe a prism whose base is a regular polygon. « = lat. area and p = perim. of right section of prism. 3. Then the lateral edge of the prism = E. .-. s = Ep. But s = S and p= P. Since p = P, then Ep = EP. .'. S=EP. §374 §354 §375 § 275, (2) § 275, (1) 377. Corollary. — The lateral area of a right circular cylinder is equal to the product of the altitude and the circumference of its base. 378. Similar cylinders. — Since a right circular cylinder may be formed by revolving a rectangle about a side as axis, it is called a cylinder of revolution. Two such cylinders formed by revolving similar rectangles about homol- ogous sides as axes are called similar cylinders of revolution. 346 SOLID GEOMETRY 379. Theorem. — The lateral or total areas of two similar cylinders of revolution are to each other as the squares of the radii of their bases or as the squares of their altitudes. ^^^ Hypothesis. Of two similar cylinders of revolution, the lateral areas are S and «, total areas T and f, altitudes S and A, and radii of bases R and r, respectively. Conclusion. —= — = -—=—-. s t r^ h^ Proof. 1. Lateral areas are S and «, total areas ^and t, altitudes IT and A, radii of bases R and r, respectively. Hyp. S_27rRR_RR_R^ff s 2 irrh rh r h §377 But f =:^. h r §127 'S_R ^R_R^_S\ 8 r r r^ P Ax. XII T_2 7rRH+ 2 irR^ _ R^ff-h R) _ t 2 irrh + 2 irr^ r(h + r) r H+R h + r § 377, § 281 ^. H R ir+R R H Since - = — , ; = _ = — . h r h+r r h § 118, (8) , T R R R^ H^ " t r r r2 h^' Ax. XII PRISMS AND CYLINDERS 347 EXERCISES 1. Prove the theorem in § 376 by circumscribing about the cylinder a prism whose base is a regular polygon. 2. The total area of a right circular cylinder is equal to the sum of the altitude and radius of the base, multiplied by the perimeter of the base. 3. The lateral areas of the cylinders formed by revolving a rectangle about each of two adjacent sides are equal. 4. If the total area of the surface of a right circular cylinder is T, and the radius of the base R, what is the altitude ? 5. If the lateral area of the surface of a right circular cylinder is S, and the altitude H, find the radius of the base. 6. Find the lateral area and also the total area of a right circular cylinder whose altitude is 6 in. and diameter of base 4 in. 7. How many square inches of tin are required for making an open cylindrical pail 10 in. in diameter and 12 in. deep ? 8. The altitudes of two similar cylinders of revolution are 3 in. and 4 in., respectively, and the lateral area of the smaller is 45 sq. in. Find the lateral area of the larger. 9. The total areas of two similar cylinders of revolution are 64 sq. in. and 144 sq. in., respectively, and the diameter of the smaller is 6 in. Find the diameter of the larger. 10. The total area of a right circular cylinder is 192 sq. in., and the altitude is 8 in. Find the diameter. 11. In punching round holes through metal plates, the pressure exerted by the punch, in pounds, in the ordinary run of work, must be 60,000 times the area, in square inches, of the cylindrical surface of the hole made. Find the pressure required to punch a hole f in. in diameter through a steel plate | in. thick. 12. Find the pressure required to punch a hole \ in. in diameter through a piece of boiler plate ^^ in. thick. 13. In a steam engine, 72 flues, or cylindrical pipes, each 2 in. in outside diameter and 14 ft. long, convey the heat from the fire box through the water. How much heating surface do they apply to the water ? 14. A condenser contains 800 tubes, each f in. in diameter and 6^ ft. long. Find the total area of their cooling surface. 348 SOLID GEOMETRY 380. Theorem. — The volume of any circular cylinder is equal to the product of the altitude and the area of the lase. Hypothesis. V= volume of circular cylinder, A = area of base, IT = altitude. Conclusion. V = HA. Suggestions. Circumscribe about the cylinder a prism whose base is a regular polygon. Let v = volume of prism, a = area of its base. Then proceed as in the proof of § 376. Write the proof in full. 381. Corollary. — If H denotes the radius of the base and H the altitude of a circular cylinder^ then V = irR^H. The proof is left to the student. 382. Theorem. — The volumes of two similar cylinders of revolution are to each other as the cubes of the radii of their bases., or as the cubes of their altitudes. The proof is left to the student. Proceed as in § 379. EXERCISES 1. Prove the theorem in § 380 by inscribing a prism whose base is a regular polygon. 2. The volume of a right circular cylinder is equal to the product of the lateral area and one half of the radius of the base. 3. The diameter of a well is 8 ft., and the water is 9 ft. deep. Allow- ing 1\ gallons to 1 cubic foot, find the number of gallons of water in the 4. An easy method that may be used for finding M "'H '|| the volume of an irregular solid that may be put into - •'*-'"- '^ ^><^ water is to immerse it in water in a cylindrical vessel, note the diameter of the vessel, and the depth of the water before and after the solid is put in, and then compute the volume of the water displaced by the solid. PRISMS AND CYLINDERS 349 A stone is immersed in a cylindrical jar of water. The diameter of the jar is 8 in., the depths of the water before and after the stone is put in are 5^ in. and 6| in., respectively. Find the volume of the stone. 5. A jagged piece of iron is immersed in a cylindrical jar of water whose diameter is 10 in. The depths of the water before and after the iron is put in are 7 in. and 9^ in., respectively. Find the volume of the iron. 6. A farmer builds a silo 18 ft. in diameter and 32 ft. high. If a cubic foot of silage weighs 25 lb., how many tons of silage does the silo hold? 7. A sheet-metal worker wishes to construct a cylindrical tank of galvanized iron that will hold 63^ gal. and that will fit into a given space 51 in. high. Find the diameter. 8. A wash boiler 12 in. deep, 10 in. wide, and 20 in. long, has round ends, i.e. each end is a half cylinder. How many gallons does it hold? 9. In practical work men use many rules of thumb. It is desirable often to test the accuracy of these rules. For example : To obtain the weight of round iron, multiply the square of the diame- ter in inches by the length in feet, and by 2.63. This gives the weight in pounds approximately. Test the accuracy of this rule when applied to a 2-inch iron rod 12 ft. long, allowing 480 lb. to 1 cu. ft. 10. It often is found necessary in machine-shop work to compute the weight of the rim of a flywheel. The rim of a flywheel is 6 in. wide, the outer diameter 4 ft. 4 in., and the inner diameter 4 ft. Find its weight. Allow 480 lb. to 1 cu. ft. 11. Find the weight of a hollow iron column 12 ft. long, 10 in. in outside diameter, and 1 in. thick, allowing 480 lb. to 1 cu. ft. 12. For irrigating a 5-acre field, water is de- livered through an 8-inch pipe at a speed of 2 ft. a second. How long will it take to deliver 2 in. of water over the entire field? 13. In the manufacture of lard, a cylinder for cooling it is 3^ ft. in diameter and 8 ft. long. As the cylinder revolves, the hot lard adheres 350 SOLID GEOMETRY to the surface to a depth of | in. and is taken off. How many pounds of lard are cooled at each revolution? How many in an hour, if the cylinder makes 5 revolutions a minute? Count 56^ lb. to 1 cu. ft. 14. A lard tank 4 ft. in diameter and 8 ft. deep has a jacket around it, on the bottom and side, 4 in. from the surface. How many gallons of water will the space between the tank and jacket hold? 15. A cylinder of a steam pump, used for pumping city water, is 2 ft. in diameter and 3 ft. long. It is filled and emptied twice at each revo- lution of the piston. Find the number of gallons delivered by the pump in a minute if the piston makes 24 revolutions a minute. 16. The figure represents a view of a rain gauge, used for measuring the amount of rainfall. The open- ing at the top is 12 in. in diameter, and the cylindrical stem is 4 in. in diameter. Suppose that in a rain the stem is filled to a depth of 4 in. What is the precipi- tation, i.e. what is the depth of the rainfall on the level ground ? 17. The drawing shows the plan of a cross section of long. The inner diameter of the brickwork is 8 ft., and the outer diameter 9 ft. 6 in. Allowing 10 % for space taken up by mortar, find how many thousand bricks must be ordered for use in its construction. 18. The Spaulding rule for computing the num- ber of board feet in 16-foot logs is as follows: Diameter in inches 12 16 20 24 28 Board feet 77 161 276 412 569 (A board foot is the quantity of timber in a board 1 ft. square and 1 in. thick. It is equivalent to ^ oi a cubic foot.) Find the amount, in cubic feet, that is wasted in sawing up a 20-inch log, if figured by this rule. Find the per cent wasted. 19. The Doyle rule, which is the rule most generally used through- out this country for computing the number of board feet that a given log will make when sawed, is as follows : PRISMS AND CYLINDERS 351 "Deduct 4 in. from the diameter of the log as an allowance for slab; square one quarter of the remainder; and multiply the result by the length of the log in feet." According to this rule, find the amount, in cubic feet, that is wasted in sawing a 12-foot log that is 25 in. in diameter. Find the per cent wasted 20. A log 10 ft. long and 4 ft. in diameter is de- cayed at the center. The diameter of the decayed part is 10 in. Find the number of cubic feet of good timber. Note. — The amount left or wasted that is required in Exercises 20, 21, and 22 does not refer to the board feet of lumber, but merely to that part of the volume of the log left or wasted. 21. A log 3 ft. in diameter and 12 ft. long has a defect on the surface, due to sun scald, which causes a waste of a slab. The defect extends over one fourth of the surface. Find the number of cubic feet in the slab wasted. Suggestion. — The area of a right section is the difference between a sector of a circle and a triangle. 22. A log 4 ft. in diameter and 16 ft. long has a defect due to decay which causes a waste of a part of the log whose right section is a sector of a circle of which the angle at the center is 60°. Find the num- ber of cubic feet in the part wasted. 23. The " Inscribed Square " rule, a rule of thumb used in lumbering, gives the cubic contents of square pieces of timber that can be cut from cylindrical logs. The width of the square piece is obtained by multiplying the diameter of the log by 17, and dividing the result by 24. Is this accurate? MISCELLANEOUS EXERCISES 1. A section of a tetraedron made by a plane parallel to any face is a triangle. 2. The section of a tetraedron made by a plane which is determined by the middle points of any three edges that do not all meet at one vertex is a parallelogram. 352 SOLID GEOMETRY 3. The line-segments which join the middle points of opposite edges of any tetraedron intersect at one point. 4. Every pair of lateral edges of a prism determines a plane which is parallel to every other lateral edge of the prism. 5. The upper base of a truncated paral- lelopiped is a parallelogram. 6. The sum of two opposite lateral edges of a truncated parallelopiped is equal to the sum of the other two lateral edges. Suggestion. — Compare AE -\- CG with MN, and compare BF + DH with MN. 7. The perpendicular drawn to the lower base of a truncated right triangular prism from the intersection of the medians of the upper base is equal to one third of the sum of the three lateral edges. Suggestion. — Let M be the middle point of DP. Draw MN±ABC. Now express PQ in terms of MN and GO', also express MN in terms of DA and PQ. Eliminate MN. 8. A man contracted to excavate a cellar at 80/* per cubic yard. The location was on sloping ground, so that the depth of the cellar at the upper side was 8 ft. and at the lower side 4 ft. The length of the cellar from the front or lower side to the back or higher side was 32 ft. and the width was 28 ft. What price was the contractor paid for the work ? 9. In the manufacture of many cylindrical articles from sheet metal, such as can lids, shoe-polish boxes, etc., circular blanks are first cut from flat sheet metal. These are then pressed into the required shape in a die. In some large factories one man is kept busy computing the sizes of blanks. The computation is based upon the assumption that the total area of the finished article equals the area of the circular blank. The modification necessary, due to the stretching of the metal, is afterward found by trial. The lid of a 3-pound lard bucket is 5\ in. in diameter, and has a ^-inch flange. Find the diameter of the blank. PRISMS AND CYLINDERS 353 10. A small sample vaseline box is | in. deep and 1| in. in diameter. Find the diameter of the blank from which it is made. 11. A Shinola shoe-polish box lid is 2| in. in diameter, and has a f-inch flange. Find the diameter of the blank from which it is made. 12. A Rumford baking-powder can lid is 3.1 in. in diameter, and has a ^-inch flange. Find the diameter of the blank from which it is made. 13. In the manufacture of door hinges, the blank shown in the middle figure is first cut from flat sheet metal, and then bent into the form shown at the right by forcing it into a die. The die con- sists of a heavy piece of metal into which is cut a vertical slot just wide enough to admit the blank edgewise. This vertical slot terminates in a cylindrical hole. By great pressure against its edge, the blank is forced into the slot, and follows around the wall of the hole, thus being bent into the required form. The size of blank that must be cut is first computed theoretically, and because of the stretching on one side of the metal and compression on the other, modifications to be made in the dimensions are found by trial. The flat part of a steel door hinge is to be 4 in. long and If in. wide, and the outside diameter of the cylindrical part ^ in. Find the size that the blank must be made. 14. In a small brass hinge used in the manufacture of furniture, the flat part is to be 1^ in. by f in., and the outside diameter of the cylin- drical part ^^ in. Find the dimensions of the blank that must be cut. 15. The flat part of a small steel hinge is 1 in. by 4 in., and the outside diameter of the cylindrical part .2 in. Find the dimensions of the blank. 16. Metal tubes are made by a process similar to that used in the manufacture of hinges, except that the die is slightly different, as shown in the drawing. Find the dimensions of the blanks for making brass tubes used as cur- tain poles that are 22 in. long and f in. in diameter. How many square feet of sheet brass will it take to make 1000 ? 354 SOLID GEOMETRY 17. In the manufacture of brass chandeliers, a tube used is 1 in. iD diameter and 3 ft. long. How much sheet brass is required to make 1000 of these tubes? 18. How high must a tomato can that is to hold a quart be made, if its diameter is 4 in. ? Allow 231 cu. in. to a gallon. 19. Find the length of a wire ^ in. in diameter that can be drawn from a cubic foot of brass. 20. A boiler of an engine 4 ft. in diameter and 16 ft. long is traversed by 60 pipes, each 3 in. in diameter, which convey the heat through the water. How many gallons of water does the boiler hold? 21. If the length of a tube is I, and the outer and inner diameters are D and d, respectively, prove that the volume equals ^ttI^D^ — d^). If the thickness of the tube is t, show that the volume equals | '7rlt(D + d). 22. Find the edge of a cube whose volume and area of the entire sur- face contain the same number of units. CHAPTER XIV PYRAMIDS AND CONES Pyramid 383. Pyramids. — A pyramid is a polyedron of which one face, called the base, is a polygon of any number of sides, and of which the other faces are triangles having a common vertex. The triangles are called the lateral faces, and their common ver- tex the vertex of the pyramid. The edges which meet at the vertex of the pyra- mid are called the lateral edges. The sum of the areas of the lateral faces is called the lateral area. The perpendicular distance from the vertex to the plane of the base is called the altitude of the pyramid. A pyramid is called triangular, quadrangu- lar, hexagonal, etc., according as its base is a triangle, quadrilateral, hexagon, etc. A regular pyramid is a pyramid whose base is a regular polygon and whose vertex lies in the perpendicular to the base at its center. 384. A truncated pyra- mid. — A truncated pyra- mid is the part of a pyramid included between its base and a plane cutting all of the lateral edges. 366 Regular Pyramid Truncated Pyramid 356 SOLID GEOMETRY The base of the pyramid and the section of the cutting plane are called the bases of the truncated pyramid. 385. A frustum of a pyramid. — A frustum of a pyramid is a truncated pyramid of which the bases are parallel. The altitude of a frustum of a pyramid is the perpendicular distance between the bases. 386. Fundamental properties of pyramids. — The following important properties of pyramids Frustum of are easily deduced from the definitions given above. The student should draw figures, and reason out the cor- rectness of each. (1) The lateral edges of a regular pyramid are equal, (2) The lateral faces of a regular pyramid are congruent isosceles triangles. (3) The altitudes of the faces of a regular pyramid drawn from the vertex of the pyramid are equal. (4) The lateral edges of a frustum of a regular pyramid are equal. (5) The lateral faces of a frustum of a regular pyramid are congruent trapezoids. (6) The altitudes of the faces of a frustum of a regular pyramid are equal. 387. Slant height. — The altitude of a lateral face of a regu- lar pyramid, drawn from the vertex of the pyramid, is called the slant height. The altitude of a lateral face of a frustum of a regular pyramid is called the slant height of the frustum. PYRAMIDS AND CONES 357 EXERCISES 1. Draw on cardboard a figure similar to the adjoining figurie, making each side of the hexagon 2 in. Cut olit the pattern, and by folding along the dotted lines and pasting, make a model of a regular hex- agonal pyramid. 2. If E, F, G, and H are the middle points of the edges AB, AD, CD, and BC, respectively, of the triangular pyramid A-BCD, then EFGH is a parallelogram. Suggestion. — What is the relation of EH a^nd FG to ACi Could it be proved by using a different pair of lines ? 3. Find the slant height of a regular quad- rangular pyramid whose altitude is 4 in. and each side of whose base is 6 in. Suggestions. — Let OE be the altitude of the regular quadrangular pyramid 0-ABCD, and let OF be the slant height. Draw FE. Find FE. Then find OF. 4. Find the lateral edge of the pyramid in Ex.3. 5. Find the slant height of a regular hexagonal pyramid whose altitude is 12 in. and each side of whose base is 4 in. 6. Find the lateral edge of the pyra- mid in Ex. 5. 7. The altitude of a frustum of a regular quadrangular pyramid is 8 in. and the sides of the bases are 12 in. and 16 in., respectively. Find the slant height. Suggestions. — Find MO and NP. Then find NK and MK. Then find MN. 8. Find tne lateral edge of the frustum in Ex. 7. 9. Find the slant height of a regular triangular pyramid each side of whose base is 3 in. and whose altitude is 4 in. . - ...... 358 SOLID GEOMETRY Suggestion. — The apothem of the base is a leg of a right triangle whose acute angles are 30° and 60° respectively. What is the relation between the sides of such a triangle ? 10. Find the lateral edge of the pyramid in Ex. 9. 11. Find the slant height of a frustum of a regular triangular pyramid the sides of whose bases are 9 in. and 5 in., respectively, and whose alti- tude is 4 in. Suggestion. — Proceed as in Exercise 7. 12. Find the lateral edge of the frustum in Ex. 11. 13. A work basket is in the form of a frustum of a regular hexagonal pyramid whose slant height is 4| in. and the sides of whose top and bottom bases are 5^ in. and 4 in., respectively. The sides and bottom are covered inside and outside with silk. If one half extra is allowed for fullness in shirring the silk, how much silk is required for the basket? If the silk is 27 in. wide, find to an eighth of a yard how much of a yard is required. 14. The figure shows the plan of a square roof in the form of a frustum of a pyramid, the upper base being a flat deck. CD is 18 ft., AB i^Q ft., and the height of the roof, or altitude of the frustum, is 8 ft. Find the lengths that the rafters A C and AE must be cut in building it. Suggestion. — AE is the hypotenuse of a right triangle whose legs are 6 ft. and 8 ft. respectively. 15. The figure shows the plan of a roof a hexagonal tower. OA, OB, etc., are rafters; CD, EF, etc., are jack rafters. If slope of the roof is 45°, and the length of J B is 8 ft., find the length that each rafter must be cut in building it. 16. Prove that the perimeter of the mid-section of a frustum of any pyramid, made by a plane parallel to the bases, is equal to one half of the sum of the perimeters of the bases. PYRAMIDS AND CONES 359 388. Theorem. — The lateral area of any regular pyramid u equal to one half of the product of the slant height and the perimeter of the bas^. A B H3rpothesis. 0-ABC • • • is a regular pyramid ; aS' = its lateral area ; Z = its slant height ; p = the perimeter of. its base. Conclusion. S = \lp. Proof. The proof is left to the student. Write the proof in full. 389. Theorem. — The lateral area of a frustum of a regular pyramid is equal to one half of the product of the slant height and the sum of the perimeters of the bases. Hypothesis, ^^is a frustum of a regular pyramid ; S = its lateral area ; Z = its slant height ; p = perimeter of one base and q = perimeter of the other base. Conclusion. S = ^ IQp + q) . Proof. The proof is left to the student. Write the proof in full. 360 SOLID GEOMETRY EXERCISES 1. The lateral area of a regular pyramid is equal to the product of the slant height and the perimeter of a mid-section made by a plane parallel to the base. 2. The lateral area of a pyramid is greater than the area of the base. Suggestion. — Draw line-segments from the foot of the altitude to all of the vertices of the base, dividing the base into triangles which may be compared to the corresponding lateral faces. 3. Prove the theorem in § 388 as a corollary of the theorem in § 389, by reducing one base of the frustum to a point. 4. The slant height of a regular pyramid is 24 ft., and the base is a tri- angle each side of which is 10 ft. Find the lateral area. The total area. 5. The base of a regular pyramid is a hexagon each side of which is 16 in., and the slant height is 20 in. Find the lateral area. The total area. 6. A tent is made of canvas stretched tightly over six poles, which are tied together at the top, and has the form of a regular pyra- mid. The distance between the feet of each two adjoining poles is 4| ft., and the slant height of the tent is 14 ft. How many square yards of canvas are there in the cover ? 7. Find the lateral area of a regular quad- rangular pyramid each side of whose base is 6 ft., and whose altitude is 8 ft. 8. The Great Pyramid of Egypt, when completed, was 481 ft. high, and each side of its square base was 764 ft. long. How many acres were there in its surface ? 9. Find the lateral area of a regular hex- agonal pyramid each side of whose base is 4 ft., and whose altitude is 10 ft. 10. Each side of the base of a regular trian- gular pyramid is 4 ft., and its altitude is 6 ft. Find the lateral area. Suggestion. — Let PAB be one face; let be the center of the base; and let OC±AB. First compute OC; then PC. See § 114. PYRAMIDS AND CONES 361 11. Find the lateral area of a triangular pyramid whose lateral faces ire all equilateral triangles and whose altitude is 4 ft. Suggestion. — The apothem of the base equals one third of the slant height. 12. Find the lateral area of a frustum of a regular triangular pyramid the sides of whose bases are 42 in. and 72 in., respectively, and whose slant height is 36 in. 13. The pedestal of a marble column is in the form of a frustum of a pyramid whose bases are regular octagons with sides 3 ft. and 2 ft. 8 in. respectively, and whose slant height is 14 in. How much surface must be polished? 14. A marble monument consists of a frustum of a square pyramid whose bases are 2 ft. and 1 ft. 10 in. square, respectively, and whose slant height is 5 ft., surmounted by a pyramid whose slant height is 2 ft. What is the amount of polished surface ? 15. Find the lateral area of a frustum of a regular quadrangular pyramid the sides of whose bases are 22 in. and 12 in., respectively, and whose altitude is 16 in. Suggestions. — Let A BCD be a lateral face ; let P and Q be the centers of the bases; let PHA.CD and QKjlAB\ and let HE±QK. First compute KE, then HK. 16. Find the lateral area of a frustum of a regular triangular pyramid the sides of whose bases are 16 ft. and 12 ft., respectively, and whose altitude is 12 ft. 17. Find the lateral area of the frustum of a regular hexagonal pyramid the sides of whose bases are 9 ft. and 5 ft., respectively, and whose altitude is 4 ft. 18. A church spire which is in the form of a regular hexagonal pyramid is covered with slate. Each side of the base is 6 ft., and a lateral edge is 38 ft. In figuring on the contract for building the spire, the contractor must estimate the amount of slate required to cover it. How many squares (a square is 100 sq. ft.) of slate are required? "^0/ 362 SOLID GEOMETRY 390. Theorem. — If a pyramid is cut hy a plane parallel to the base : (1) the lateral edges and the altitude are divided 'proportionally ; (2) the section is a polygon similar to the base. B C Hypothesis. In pyramid 0-AB (7 • • •, plane of section FCrH- parallel to the base and meets altitude OilSf at N. ^ , . ... OF Oa ON Conclusion. ( 1 ) = = . . . = ; ^ ^ OA OB OM' (2) FaH ABQ..: Proof. 1. Plane FEW plane AQ. .-. Fa II AB, anw bo, etc., and NH OF^qa^ ^ ON ' ' OA OB '" DM Also Z FaH= ZABO.Z GRJ^ Z BOB, etc. § 317 Z OFa = Z OAB, Z FaO = AABO. § 26 .-. since Z is common, A FOa ^ AAOB, § 128 Similarly, A GOR ^ABOC. 2. 3. 4. 5. 6. 7. Hyp. MO. § 315 § 123, Ax. I 8. 10. 11. FG AB Fa oa OB an and Off BO oa OB AB BO Similarly, |f=g, etc. .-. FGH ABO^'^. Def. sim. A Ax. I Def. sim. poly. PYRAMIDS AND CONES 363 391. Corollary l. — The area of a section of a pyramid par- allel to the base is to the area of the base as the square of its distance from the vertex is to the square of the altitude. The proof is left to the student. 392. Corollary 2. — If tivo pyramids have equal altitudes and equal bases, sections made by planes parallel to the bases at equal distances from the vertices are equal. For, OR EF' CD ON AB oSP and FV' Why Hence ^=-^. Why? AB EF ^ nencQCL^an. Why? EXERCISES 1. What part of the area of the base of a pyramid is the area of a section made by a plane which is parallel to the base and bisects the altitude ? 2. The altitude of a pyramid is 16 in. and its base is a square 12 in. on a side. What is the area of a section parallel to the base whose dis- tance from the vertex is 12 in. ? 3. If two pyramids with equal altitudes are cut by planes parallel to the bases, and at equal distances from their vertices, the sections have the same ratio as the bases. 4. A pyramid 18 ft. high has a base containing 225 sq. ft. How far from the vertex must a plane be passed parallel to the base in order that the section may contain 64 sq. ft. ? 5. At what point of the altitude of a pyramid should a plane be passed parallel to the base so that the section shall equal one half of the base ? 6. The base of a pyramid is 10 yd. square. A plane parallel to the base and 9 yd. from the vertex cuts a section whose area is 36 sq. yd. Find the altitude of the pyramid. 364 SOLID GEOMETRY 393. Theorem. — Two triangular pyramids having equal altitudes and equal bases are equal. Hypothesis. Triangular pyramids V-ABC and W-LMN have equal altitudes and equal bases. Conclusion. V-ABC= W-LMN. Proof. 1. Suppose that W-LMN > V-ABO. 2. Let each of the two equal altitudes be divided into n equal parts of length h. Through the points of division pass planes parallel to the bases, cutting the pyramids in A BEF^ A GrRI, etc., and A OPQ, A EST, etc., respectively. 3. On A BEF, A GUI, etc., as upper bases, and with AB, BG-, etc., as lateral edges, construct prisms a, 5, etc. 4. On A LMN^ A OPQ^ etc., as lower bases, and with iO, OR^ etc., as lateral edges, construct prisms x^ y^ etc. 5. A BBF= AOPQ.A aHI= A EST, etc. § 392 6. a = ^ X A BEF, y = hxA OPQ, etc. § 369 7. .•. a = y, 5 = 2, etc. Ax. IV 8. But V-ABO >a-\-h^- etc., and W-LMN < x-\-y-{-z-{- etc. Ax. X 9. .-. W-LMN-V-ABOv^ 6. On paper or pliable cardboard, draw a figure \ ^ — ^ J similar to the adjoining figure. By cutting out the >v y pattern, folding and pasting, make a model of the ""^ conical surface of a frustum of a right circular cone. 7. The altitude of a frustum of a right circular cone is 8 in., and the radii of the bases are 4 in. and 6 in., respectively. Find the slant height. 376 SOLID GEOMETRY 8. The altitude, slant height, and radius of the larger base of a frus- tum of a right circular cone are 6 in., 8 in., and 10 in., respectively. Find the radius of the smaller base. 9. A plane determined by an element of a circular cone and a tangent to the base con- tains no point of the surface without that ele- ment. Such a plane is said to be tangent to the cone. Suggestion, — Let P be any point of the plane without the element. Draw a plane through P parallel to the base. Prove that this plane intersects the given plane in a tangent to the section of the cone. 406. Inscribed and circumscribed pyramids and cones. — A pyramid is inscribed in a circular cone when the base of the pyramid is inscribed in the base of the cone and the two have the same vertex. / \ / /c-~\ .\ ^-^ Inscribed Pyramid Circumscribed Pyramid A pyramid is circumscribed about a circular cone when the base of the pyramid is circumscribed about the base of the cone and the two have the same vertex. It is evident that when a regular pyramid is circumscribed about a right cir- cular cone the slant height of the pyramid is equal to the slant height of the cone. A frustum of a pyramid is inscribed in a frustum of a cir- cular cone when its bases are inscribed in the bases of the frustum of a cone. PYRAMIDS AND CONES 377 Inscribed Frustum Circumscribed Frustum OF Pyramid op Pyramid A frustum of a pyramid is circumscribed about a frustum of a circular cone when its bases are circumscribed about the bases of a frustum of a cone. It is evident that when a frustum of a regular pyramid is circumscribed about a frustum of a right circular cone the slant height of the frustum of the pyramid is equal to the slant height of the frustum of the cone. 407. Limits. — The following principles of limits are as- sumed: (1) The lateral area of a right circular cone (^frustum of a right circular cone) is the limit approached by the lateral area of an inscribed or circumscribed regular pyramid (^frustum of a regular pyramid) as the number of faces is indefinitely in- creased. (2) The volume of a circular cone (frustum of a circular cone) is the limit approached by the volume of an inscribed or cir- cumscribed pyramid (frustum of a pyramid) whose base is a regular polygon as the number effaces is indefinitely increased, 408. Principles of limits. — The following general principles of limits are assumed, as were similar principles in § 275 : (1) The limit of the sum of two or more variables is the sum of their limits. (2) ITie limit of the product of two or more variables i» the product of their limits. (3) The limit of any principal root of a variable is that root of its limit. 378 SOLID GEOMETRY 409. Theorem. — The lateral area of a right circular cone is equal to one half of the product of its slant height and the cir- cumference of its base. Hypothesis. S = the lateral area, O = the circumference of the base, and L = the slant height of a right circular cone. Conclusion. S = -^ LQ, Proof. 1. /S'==the lateral area, (7= the circumference of the base, and L = the slant height of a right circular cone. Hyp. 2. Let s = the lateral area of a circumscribed regular pyramid, and let p = the perimeter of its base. Then L = the slant height of the pyramid also. § 406 3. .-. s = iLp. §388 4. Now, if the number of lateral faces of the pyramid is increased indefinitely, s = S and p = 0. § 407, § 274 5. .-. lLp = ^La §275,(2) 6. r.S=\LO. §275,(1) Using different letters, write out the proof without refer- ring to the book. 410. Corollary. — If S is the lateral area of a right circular cone, L the slant height, and R the radius of the base, then The proof is left to the student. PYRAMIDS AND CONES 379 EXERCISES 1. Can the theorem in § 409 be proved by inscribing a regular pyramid in the cone ? 2. Compare the lateral areas of a right circular cylinder and a right circular cone with equal bases and equal altitudes when the slant height of the cone is equal to the diameter of the base. 3. Write the formula for the total area of a right circular cone. 4. Since the lateral surface of a cone of revolution may be produced by rolling up a sector of a circle, prove the theorem in § 409 by use of §284. 5. The slant height of a right circular cone is 8 ft. and the radius of the base 6 ft. Find the lateral area. The total area. 6. The altitude of a right circular cone is 12 in., and the diameter of the base 14 in. Find the slant height. The lateral area. The total area. 7. The altitude oi a right circular cone is 18 ft., and the slant height 24 ft. Find the radius of the base. The lateral area. The total area. 8. How many square yards of canvas will be required to make a conical tent whose altitude is 12 ft., and diameter of base 12 ft. ? 9. A conical slate roof of a tower is 20 ft. high, and the diameter of its base is 18 ft. Find the area of the slate. 10. Find the area of the surface generated by revolving a right tri- angle about its hypotenuse as axis, the legs of the triangle being 3 in. and 4 in., respectively. 11. If a cone of revolution is formed by revolving an equilateral tri- angle about one of the altitudes as an axis, the lateral area equals twice the area of the base. Suggestion. — Express the lateral area of the cone in terms of a side of the triangle. 12. An equilateral triangle each of whose sides is 6 in. is revolved about a straight line through a vertex and parallel to the opposite side as axis. Find the area of the whole surface traced by the three sides of the triangle. 13. A silo which is 18 ft. in diameter has a conical roof. The rafters are at an angle of 30° with horizontal, and project 1 ft. over the eaves. Find the area of the roof. 380 SOLID GEOMETRY 411. Theorem. — The lateral area of a frusfMm of a right circular cone is equal to one half of the product of its slant height and the sum of the circumferences of its bases. Hypothesis. S = the lateral area, = circumference of larger base, c = circumference of smaller base, and L = slant height of a frustum of a right circular cone. Conclusion. S = ^L(^0 -\- e^. Proof. 1. S = lateral area, and c respectively = circum- ferences of bases, and L = slant height of frustum of right circular cone. Hyp. 2. Let s = lateral area of a circumscribed frustum of a regular pyramid, and let P and p respectively = perimeters of its bases. Then L = its slant height also. § 406 3. .-. 8 = -lL(iF + p). §389 4. Now, if the number of lateral faces is increased indefi- nitely, s = S, F = C, and p = c. § 407 5. .-. (P+i>) = ((7+0. §408,(1) 6. .-. iX(P+j9) = J-Z((7+0- §275,(2) 7. .',S = ^LiO+cy §275,(1) 412. Corollary. If S is the lateral area of a frustum of a right circular cone, L the slant height, and B and r respectively the radii of the bases, then S ^ irLQR + r^. The proof is left to the student. PYRAMIDS AND CONES 381 EXERCISES 1. The lateral area of a frustum of a right circular cone is equal to the perimeter of a mid-section made by a plane parallel to the bases, multiplied by the slant height. 2. Show that the formula for computing the lateral area of a right cir- cular cone is obtained from the formula for the lateral area of a frustum of a right circular cone by making one of the bases of the frustum reduce to a point. 3. Write the formula for the total area of a frustum of a right circu- lar cone. 4. If a trapezoid, one of whose non-parallel sides is perpendicular to the bases, is revolved about this side as axis, the lateral area of a frus- tum of a cone traced equals 2 tt times the product of the other non-parallel side and the line-segment joining the middle points of the non-parallel sides. Suggestion. — See Exercise 1 above. 5. The slant height of a frustum of a right circular cone is 14 in. and the diameters of the bases are 16 in. and 10 in., respectively. Find the lateral area. The total area. 6. The altitude of a frustum of a right circular cone is 8 ft., and the radii of the bases 4 ft. and 9 ft., respectively. Find the slant height. The lateral area. The total area. 7. A funnel is made of tin, in which the diameter AB is 8 in., the diameter CD is 1 in., the diameter EF is I in., the length ^C is 5 in., and the length CE is 4 in. Find the amount of tin required to make it, not allowing for the seams. 8. Find the amount of sheet metal required to make a lot of 1000 pails, each 10 in. deep, 8 in. in diameter at the bottom, and II in. in diameter at the top, not allowing for seams nor waste in cutting. 9. Given the diameters and the altitude of a frustum of a right circular cone, explain how to draw the pattern by which a piece of sheet metal must be cut out in order to be rolled up and made into the re- quired frustum. Suggestions. — Find the slant height. Let x = the radius of the inner arc of the pattern. Then x + the slant height = the radius of the 382 SOLID GEOMETRY outer arc. These radii are proportional to the circumferences of the bases of the frustrum. Solve the proportion for x. 10. A tinner wishes to make a coifee pot which shall be 7 in. deep, 4 in. in diameter at the top, and 6 in. in diameter at the bottom. Al- lowing a quarter of an inch for a seam, cut from paper a pattern for the conical surface. Note. — For explanation of the terms used, and of the process in- volved in the manufacture of articles as indicated in the following prob- lems, see Problem 9, page 352. 11. A stew pan is to be made in the form of a frustum of a right cir- cular cone, 6^ in. wide at the bottom, 8^ in. wide at the top, 3 in. deep, and to have a flange \ in. wide at the top. Find the diameter of the blank from which it must be made. Suggestion. — The area of the pan consists of three parts : the bottom ; the side, which is the lateral area of a frustum of a cone ; and the flange, which is the diif erence between the areas of two circles. The sum of these must equal the area of the blank. Indicate the factors, and avoid as much multiplication as possible. 12. A tin dish pan is to be made 13 in. wide at the bottom, 17 in. wide at the top, 6 in. deep, and have a ^-in. flange at 'the top. Find the diameter of the blank from which it must be made. 13. A tin pie pan is 6f in. in diameter at the bottom, 8 in. in diameter at the top, 1 in. deep, and has a ^-in. flange at the top. Find the diameter of the blank from which it is made. 14. A safety valve of an engine has a valve V in the form of a frus- tum of a cone, which closes the opening D in the boiler AB. The diameter of the opening Z> is 2 in. The conical surface of the valve is inclined at 45°. The " effective area," or area of the opening through which the steam escapes when the valve V is lifted, is evidently the lateral area of a frustum of a cone whose slant height is ss. If the valve V is lifted through a height of \ in., find the effective area. PYRAMIDS AND CONES 383 413. Theorem. — The volume of a circular cone is equal to one third of the product of its altitude and the area of its base. Hypothesis. F"= volume, H= altitude, and B = the area of the base of a circular cone. Conclusion. F'= J HB. Suggestions. Inscribe a pyramid whose base is a regular polygon. What formula expresses the volume of the pyramid ? The formula V=\ HB may be proved from this by § 275, (1), if \ HB is first proved to be the limit of what ? Write the proof in full. 414. Corollary. — If Vis the volume of a circular cone^ H the altitude^ and R the radius of the hase^ then V^iirB^H, The proof is left to the student. EXERCISES 1. The vohime of a circular cone is one third of the volume of a circular cylinder having the same base and altitude. 2. Prove the theorem in § 413 by circumscribing about the cone a pyramid with a regular base. 3. Find the volume of a circular cone of which the altitude is 16 in. and the diameter of the base 12 in. 4. Find the volume of a right circular cone the radius of whose base is 6 in. and whose slant height is 20 in. 384 SOLID GEOMETRY 5. A fruit raiser has a round pile of apples that is 8 ft. across at the bottom, and tapers to a point 5 ft. high at the middle. Allowing 3 bu. to 4 cu. ft., how many bushels are there in the pile? 6. A farmer has a pile of ear corn approximately in the form of a cone whose height is 10 ft. and width at the bottom 20 ft. Allowing 2 bu.to 5 cu. ft., how many bushels does it contain? 7. Crushed stone falling from a stone crusher forms a round pile 30 ft. around at the bottom and 5 ft. high at the center. How many cubic yards of crushed stone are there in the pile ? 8. A farmer wishes to know how many tons of hay there are in a stack in the form of a cylinder 18 ft. in diameter and 8 ft. high, sur- mounted by a cone 10 ft. high. Allowing 512 cu. ft. to the ton, find the amount of hay in the stack. 9. A farmer has a rick of hay 36 ft. long and 16 ft. wide. The lower part is made up of a parallel- opiped with half cylinders at its ends, and is 8 ft. high. The part surmounting this tapers to a line 20 ft. long and 18 ft. above the ground, and is made up of a trian- gular prism with half cones at its ends. Find the number of tons in it, allowing 512 cu. ft. to the ton. 10. A hill, approximately in the form of a cone, is to be removed and the ground leveled down even with its base, and laid off into building lots. The diameter of the base is 280 ft., and the highest point is 12 ft. above the level. How many loads (cubic yards) of earth must be re- moved ? 11. In the center of a building lot is a piece of elevated ground approximately in the form of a cone. The elevated part is to be cut down and used to raise the level of the surrounding ground. The lot is 100 ft. wide and 120 ft. long. The height of the cone is 5 ft. and the diameter of its base 60 ft. Find how much the elevated part must be cut, and how much the rest of the lot must be filled, to bring it to a level. 12. Given the lateral area 5 of a right circular cone, and the radius R of the base. Find the volume V. 13. Given the total area T of a right circular cone, and the lateral area S. Find the volume V, PYRAMIDS AND CONES 385 415. Theorem. — The volume of a fruBtum of a circular cone is equal to one third of the product of its altitude and the sum of the areas of the bases and the mean proportional between them. Hypothesis. V= volume, 11= altitude, and B and b respec- tively = the areas of the bases of a frustum of a circular cone. Conclusion. F = ^ H(iB + 6 + V56) . Suggestions. Inscribe a frustum of a pyramid whose base is a regular polygon. What formula expresses the volume of the frustum of a pyramid ? The formula V=\ H(B + 6 -f- ^Bb) may be proved from this by § 275, (1), if ^H{B + 5 + V^) is first proved to be the limit of what ? Write the proof in full. 416. Corollary. — If Vis the volume of a frustum of a circu- lar cone^ H its altitude^ and R and r respectively the radii of the bases^ then V=\7rH(iB? + r^+ Rry The proof is left to the student. EXERCISES 1. The volume of a frustum of a circular cone is equal to the sum of the volumes of the three circular cones having a common altitude equal to the altitude of the frustum, and for bases the two bases of the frustum and a mean proportional between them, respectively. 2. Find the volume of a frustum of a circular cone of which the alti- tude is 4 in. and the diameters of the bases 6 in. and 5 in., respectively. 386 SOLID GEOMETRY 3. Find the volume of a frustum of a right circular cone of which the slant height is 32 in. and the radii of the bases 16 in. and 28 in. respectively. 4. Does a liquid measure with a diameter of 8 in. at the top, a diameter of 13 1 in. at the bottom, and 12| in. deep, contain 5 gal.? (1 gal. = 231 cu. in.) 5. A Sanford's Ink bottle is in the form of a frus- tum of a cone whose height is If in., and the diame- ters at the top and bottom 1^ in. and 2 in., respectively. How many of these bottles can be filled with a gallon of ink ? 6. The base of a marble column is a frustum of a cone. The height is 1 ft. 6 in., and the diameters of the bases 6 ft. and 5 ft. 6 in., respec- tively. Allowing 170 lb. to the cubic foot, find its weight. 7. A Dutch windmill, in the shape of a frustum of a cone, is 30 ft. high. The outer diameters at the bottom and top are 15 ft. and 12 ft., respectively; and the inner diameters at the bottom and top are 12 ft. and 9 ft., respec- tively. How many cubic yards of stone were required to build it ? 8. The chimney of a factory is to be 100 ft. high ; the outer diameters at the bottom and top are to be 18 ft. and 10 ft., respectively ; and the flue is to be 6 ft. in diameter throughout. How many thousands of bricks, 2 in. by 4 in. by 8 in., must the contractor order for it, allowing 10% for mortar? 9. A churn, in the form of a frustum of a cone, is 28 in. high, 10 in. in diameter at the bottom, and 8 in. in diameter at the top. How many gallons of cream will it hold? 10. A cake pan is 10 in. in diameter at the top and 8 in. in diameter at the bottom, and its depth is 3 in. Rising from the center of the bottom is a stem in the form, of a frustum of a cone 1^ in. in diameter at the bottom and 1 in. in diameter at the top, which makes the hole in the cake. How many cubic inches does the pan hold ? PYRAMIDS AND CONES 387 417. Theorem. — The lateral or the total areas of two similar cones of revolution are to each other as the squares of the slant heights^ the altitudes^ or the radii of the bases ; and their vol- umes are to each other as the cubes of the slant heights^ the alti- tudes^ or the radii of the bases. Hypothesis. S and s are the lateral areas, T and t the total areas, V and v the volumes, L and I the slant heights, IT and h the altitudes, and R and r the radii of the bases, of two similar cones of revolution. Conclusion. - = ^ = _ = _=_; _ = _ = _ = _. Proof. The proof is left to the student. Proceed as in § 379 and § 382. Write the proof in full. EXERCISES 1. The lateral area of a cone of revolution is 144 sq. in. What is the lateral area of a similar cone whose altitude is f that of the given cone ? 2. The altitude of a circular cone is 8 ft. How far from the vertex must a plane be passed parallel to the base to cut off -^^ of the cone from the vertex ? 3. To divide a circular oone into halves by a plane parallel to the base, how far from the vertex must the plane be passed ? 4. From a cone of revolution of which the slant height is 24 ft. and »f which the diameter of the base is 12 ft., a cone of which the slant height is 8 ft. is cut off by a plane parallel to the base. Find the lateral area and the volume of the cone cut off; of the frustum. - •^- 388 SOLID GEOMETRY MISCELLANEOUS EXERCISES 1. A lamp shade, made of a wire frame covered with silk, is in the form of a frustum of a right circular cone whose slant height is 9 in. and the radii of whose bases are 6 in. and 4^ in., respectively. If one half extra is allowed for fullness in shirring the silk, how much silk is required for the lamp? If the silk is 27 in. wide, find to an eighth of a yard how much of a yard is required. 2. The lateral area of a right circular cone is twice the area of the base. Compare the slant height and the radius. At what angle does an element of the cone meet the plane of the base ? 3. If r is the radius of the base of a right circular cone and h the slant height, prove that the ratio of the area of the base to the lateral area is - • h 4. If a right circular cone and a right circular cylinder have equal bases and the slant height of the cone is equal to the altitude of the cylinder, prove that the entire area of the cylinder is equal to twice the entire area of the cone. 5. An ice cream dipper is in the form of a right circular cone whose diameter is 2f in. and altitude 2f in. It is filled level full at each serving. How many servings can be obtained from a gallon (231 cu. in.) of ice cream ? What is received for a gallon at 5 ^ a serving ? 6. Triangle ABC is an isosceles triangle with equal sides AB and AC. Line DE is drawn through A parallel to J5C, and A ABC is revolved about DE as an axis. Find the vol- ume of the figure traced by A ABC. 7. A conical glass beaker, graduated for measuring liquids, holds a quart. At what point of the scale must the mark be placed to measure a pint? measure a gill ? 8. A farmer has two similar piles of corn, each in the form of a right circular cone. The diameters of the two piles are 4 ft. and 18 ft., respectively. He measures the small pile, and finds that it contains PYRAMIDS AND CONES 389 10 bu. How can he estimate, without measuring, the number of bushels in the large pile? How many bushels are there in the large pile ? 9. Two boards, A and B, are placed in parallel vertical positions on a table. A has a square opening cut in it. A lamp is placed at S. Light from this lamp passes through the opening in A, and illuminates part of B. Show, by aid of this diagram, that the intensity of the light falling upon a given surface is inversely proportional to the square of the distance of the surface from the source of light. 10. The altitude of each of two pyramids whose bases are in the same plane is 16 ft. The base of one is a square whose side is 12 ft., and the base of the other is a regular hexagon whose side is 8 ft. The area of the section of the first made by a plane parallel to the base is 81 sq. ft. What is the area of the other pyramid made by the same plane ? 11. A fruit vender has a load of apples in a wagon bed that is 3 ft. wide, 1 ft. deep, and 10 ft. long. The apples fill the bed and are piled up in approximately a pyramid to a point 1 ft. higher than the top of the bed. How many bushels are there ? (Count 3 bu. to 4 cu. ft.) 12. A triangular piece of ground ABC is to be leveled. The side AB is level, and the corner C is 12 ft. higher than AB. Find how much the ground must be cut down at C and how much it must be fiUed along ^^ to make it level. 13. Find the volume of a regular quadrangular pyramid whose altitude is 20 ft. and slant height 28 ft. 14. Find the volume of a triangular pyramid each edge of which is 4 ft. 15. The slant height of a frustum of a regular quadrangular pyramid is 18 in., and the sides of the bases are 20 in. and 12 in., respectively. Find its volume. 16. The lateral edge of a frustum of a regular hexagonal pyramid is 6 ft., and the sides of the bases are 8 ft. and 4 ft., respectively. Find its volume. CHAPTER XV REGULAR POLYEDRONS. SIMILAR POLYEDRONS. PRISMATOIDS 418. Regular polyedrons. — A regular polyedron is a poly- edron all of whose faces are congruent regular polygons and all of whose polyedral angles are equal. Models of five regular polyedrons are as follows. That these are the only kinds possible is proved in § 419. Tetraedron Hexaedron Octaedron Uodecaedron Icosaedron 419. Theorem. — There cannot he more than jive kinds of regular polyedrons. Proof. 1. Each angle of an equilateral triangle is 60°; of a square, 90°; of a regular pentagon, 108°; etc. § 101 2. .*. not more than three kinds of regular polyedrons are possible with equilateral triangles as faces (tetraedron^ octa- edron^ icosaedron') ; for 3 x 60°, 4 x 60°, or 5 x 60° is less than 360°, while 6 x 60° or any greater multiple of 60° is not less than 360°. § 343 3. Similarly, not more than one kind of regular polyedron is possible with squares as faces (hexaedron or cube); for 3 X 90° is less than 360°, while 4 x 90° or any greater multiple of 90° is not less than 360°. 390 REGULAR POLYEDRONS, ETC. 391 4. Similarl}^ not more than one kind of regular polyedron is possible with regular pentagons as faces (dodecaedron) . 5. No regular polyedron is possible with regular hexagons or regular polygons of more sides as faces, because the sum of the face angles at each vertex would not be less than 360°. 6. .-. only five regular polyedrons are possible. Note. — The regular polyedrons were studied by the ancient Greeks. The members of the Pythagorean school knew that there were five of these solids. It is said that Hippasus (about 470 e.g.), a Pythagorean who dis- covered the regular dodecaedron, was drowned for announcing his discovery, because the Pythagoreans were a secret society, and the members were pledged to refer the honor of any new discovery to Pythagofas, the founder. 420. Construction of models of regular polyedrons. — Models of the five regular polyedrons may be constructed of card- board by first drawing patterns as shown below, then cutting out the patterns, folding along the dotted lines, and pasting. , r 1 j 1/ 1 1 V — 4 r EXERCISES 1. How many faces has a regular tetraedron? Hexaedron? Octa- @dron ? Dodecaedron ? Icosaedron ? 2. Construct cardboard models of one or more of the five regular polyedrons as suggested in § 420. 3. Prove that a regular hexaedron has a center, i.e. a point within equidistant from all vertices. Is this true of all regular polyedrons ? 392 SOLID GEOMETRY 421. Similiar polyedrons. — Two polyedrons are similar when they have the same number of faces, similar each to each and similarly placed, and the homologous polyedral angles are equal. Similar Polyedrons 422. Corollary. — In two similar polyedrons : (1) The ratio of any two homologous edges is equal to the ratio of any other two homologous edges. (2) The areas of any two homologous faces are to each other as the squares of any two homologous edges. (3) The areas of the entire surfaces are to each other as the squares of any two homologous edges. The proof is left to the student. EXERCISES 1. In two similar polyedrons, any two homologous diagonals have the same ratio as any two homologous edges. 2. Prove that any two regular tetraedrons are similar. Is this true of any two regular hexaedrons? Octaedrons? Dodecaedrons ? Icosae- drons ? 3. Two homologous edges of two similar polyedrons are 4 in. and 6 in., respectively. The total area of the smaller is 128 sq. in. Find the total area of the larger. 4. Verify § 422 (3), in the case of two similar rectangular parallelo- pipeds whose concurrent edges are 3 in., 4 in., 6 in., and 6 in., 8 in., 12 in., respectively. REGULAR POLYEDRONS, ETC. 393 423. Theorem. — Two tetraedrons are similar when the faces including a triedral angle of one are similar^ respectively^ to the faces including a triedral angle of the other^ and are similarly 'placed. Hypothesis. Tetraedrons ABOD and EFGRh^iwe AABC'^AEFG, AACD'^AJSaff, AABB'^AEFH. Conclusion. AB CD ~ EFGH. Proof. 1. AABO'^AEFG, AACD^AEaH, A ABB ~ A EFH. Hyp. 3. BO Fa BO Fa 4. Similarly, 5. 6. AO Ea OB an' OB and OB aR' AO Ea §127 Ax. I ^BB aH FE' .\ A BOB ^ A Fan. §129 Now Z BAO = Z FEa, Z. OAB = Z aEH, Z BAB = Z FEE, § 127 7. .-. angle A-BOB = angle E-FaH. § 346 8. Similarly, the other corresponding triedral angles are equal. 9. :.ABOB^EFaH. §421 EXERCISE If a giyen tetraedron is cut by a plane parallel to a face, the tetra- edron cut off is similar to the given one. 394 SOLID GEOMETRY 424. Theorem. — Two tetraedrons are similar when a diedral angle of one is equal to a diedral angle of the other, and the including faces are similar each to each and similarly placed. A Hypothesis. In tetraedrons ABQD and EFGiH, diedral angle C-AB-D = diedral angle G-UF-H, A ABC^AJEFa, AABB^AFFH. Conclusion. AB OB ~ FFaH. Proof. 1. AnglQ C-AB-D = 'dugiQ a-FF-H, QtQ. Hyp. 2. .-. ABCB may be superposed upon FFG-H so that angle 0-AB-B coincides with its equal, angle Gr-FF-H, A falling at F. Test of equality 3. ZBA0=AFFa3indZBAB = ZFFH. §127 4. .-. J. (7 will fall along FG, and AD along Fff. § 13 5. .-. Z CAD^^Z GFR. § 13 a AC AB . AD AB « .07 6. -^^-^=^^7:^ and -—— = ——• § 127 Fa FF FH FF ^ Ax. I 7 . A^^A^ ' ' Fa~ FH 8. ..AACDr^AFaS, §130 9. .'.ABCD^FFaH, §423 EXERCISE Is this a true theorem: Tetraedron ^5CZ> ~ tetraedron EFGH when AABC-^AEFG, angle C-^5-Z> = angle G-EF-H, and angle B-AC-D = angle F-EG-Hl REGULAR POLYEDRONS, ETC. 395 425. Theorem. — Two similar polyedrons may he decomposed into the same number of tetraedrons^ similar each to each and similarly placed. Hypothesis. AM audi UN" are two similar polyedrons. Conclusion. AM and ^iV may be decomposed into the same number of tetraedrons, similar each to each and similarly placed. Proof. 1. Polyedrons J-iHf and JS'iV are similar. Hyp. 2. Let A and ^ be any two homologous vertices. Divide all faces, except those having A and U as vertices, into tri- angles. Pass planes through A and the vertices of the triangles in the faces of AM^ and through U and the vertices of the triangles in the faces of UI^. 3. Let ABQB and EFG-H be two homologous tetra- edrons formed. 4. Now face AB ~ face EF. BO^AO Fa Fa' 6. r.AABC^AFFa, 7. Similarly, AACB'^A EaS. 8. Also angle B-AC-D = angle F-Ea-H. 9. ,'.ABCD^EFaH. 10. It may now be proved that tetraedrou ABDM edron EFHN, etc. Z BOA = Z FaE, and ^ = ^ §421 §127 §130 §421 §424 tetra- 396 SOLID GEOMETRY 426. Theorem. — The volumes of two tetraedrons ha-^xng a pair of equal triedral angles are to each other as the products of the edges including the equal triedral angles. E A Hypothesis. Tetraedrons ABCD and UFG^R hsive angle B-A CD = angle F-UGrff, and their volumes are V and W, respectively. ^ , . r BAxBOxBI) Conclusion. -:fj> = -:prp, ^pr?^ 777^. W FE X Fa X FH Proof. 1. Angle ^-^(7i> = angle J^-^(5^ir. Hyp. 2. .*. ABOD may be superposed upon FFGrH so that angles B-AOD and F-FGrH coincide, ABOD becoming OFPQ, Test of equality 3. If Oilif and ^iV^are J. plane FaH, OMW EK § 321 4. .-. Oil[f and ^i\r determine a plane, which contains FE and intersects plane FGrRin FN, § 299, § 291, § 301 . V_ iOMxAFPQ ^OM AFFQ .004 ' w ^ENxAFas EN A Fas' ^ 6. In AFMO and FNE, Z F is common, Z FMO = Z FNE, Z FOM= FEN. § 26 7. .'.AFMO^AFNE. §128 8. ,,OM^FO ^^ BA^ EN FE FE ^ 9. Also ^^^ = -^^x^g or ^Ox_BB^ 228 A Fas FGxFH FaxFH ^ 10 V ^BA BO xBD ^ BAxBOxBD . ^^j ' "W FE FaxFH FEx FaxFH ^' REGULAR POLYEDRONS, ETC. 397 427. Theorem. — The volumes of two similar tetraedrons are to each other as the cubes of their homologous edges. Hypothesis. Tetraedrons ABOD and EFG-H are similar, with AB and EF homologous edges, and with volumes V and W^ respectively. ^ , . V Aff Concmsion. ^=v=-=t;« W EF^ Itr^^i. 1. Tetraedron^J^CZ^-'tetraedron ^^(5^j?. Hyp. 2. .-. angle A-BCD = angle E-FGH. § 421 3 . V^ ABy^AC^AB ^AB AQ AT) « .^fi ' "W EFxEGxEH EF Ea EH' ^ 4. But — = — = — . §422a) EF Ea EH ^ ^^ c V AB^AB AB Aff . VTT W EF EF EF EF^ 428. Corollary. — The volumes of any two similar polyedrons are to each other as the cubes of their homologous edges. The proof is left to the student. EXERCISES 1. The edge of a cube is 16 in. Find the edge of a cube which shall have a volume twice as great as that of the given cube. 2. If the edge of a given cube is e, find the edge of a cube with n times the volume. 398 SOLID GEOMETRY 3. The edge of a pyramid is x. Find the edge of a similar pyramid with twice the volume. 4. The edge of a given polyedron is E. Find the edge of a similar polyedron with m times the volume. 5. A rectangular tank is 4 ft. by 6 ft. by 8 ft. What are the dimen- sions of a similar tank that will hold 4 times as much? 6. The volumes of two similar polyedrons are 36 cu. in. and 972 cu. in., respectively. The edge of the smaller is 3 in. Find the homologous edge of the larger. 7. A farmer had two rectangular ricks of hay of the same shape. The large rick was 1\ times as long as the small one. He hauled away and weighed the small one, and found that it contained 2\ toL OA. Why ? Write the proof in full. 412 SOLID GEOMETRY 444. Theorem. — A plane tangent to a sphere is perpendiculat to the radius drawn to the point of contact. Suggestions. Let plane MN be tangent at -4 to a sphere with center 0. I^et B be any point of iifiV except A. Show- that OB > OA, etc. Write the proof in full. EXERCISES 1. A perpendicular to a tangent plane at the point of tangency passes through the center of the sphere. 2. A line perpendicular to a tangent plane from the center of a sphere passes through the point of contact. 3. Two planes tangent to a sphere at the extremities of the same diameter are parallel. 4. Show that the diameter of any solid spherical ball may be found by laying the ball upon a table, resting a board upon it in a horizontal position, and measuring the perpendicular dis- tance from the edge of the board to the table. Suggestion. — Draw a diameter perpendicu- lar to the table top. 5. A straight line perpendicular to the radius of a sphere at its extremity is tangent to the sphere. 6. A straight line tangent to a sphere is per- pendicular to the radius drawn to the point of contact. 7. All straight lines tangent to a sphere at the same point lie in the plane which is tangent to the sphere at that point. 445. Circumscribed and inscribed spheres of a polyedron. — A sphere is circumscribed about a polyedron when all of the ver- tices of the polyedron lie in the surface of the sphere. The polyedron is said to be inscribed in the sphere. A sphere is inscribed in a polyedron when all of the faces of the polyedron are tangrent to the sphere. The polyedron is said to be circumscribed about the sphere. THE SPHERE 413 446. Theorem. — A sphere can he circumscribed about any tetraedron. Hypothesis. ABCD is any tetraedron. Conclusion. A sphere can be circumscribed about ABOD. Proof. 1. ABOD is any tetraedron. Hyp. 2. Let M and N be the centers of the circumscribed circles of A ACD and A BQD^ respectively. 3. Then, iHf and iV^lie in perpendicular bisectors ME and NE of CD. § 105 4. .-. plane MEN JL line CD. § 303 .-. plane MEN 1. planes ACD and BCD. § 331 Let MF be A. plane ACD and NG be A. plane BCD. Then MF and iV^^ both lie in plane MEN. § 383 .'. ilO^and NG- intersect at some point 0. § 55 is equidistant from A., (7, and i), and also equidis- tant from B, C, and D. § 310 10. .-. is equidistant from J., ^, (7, and D. Ax. I 11. ,*. the surface of a sphere with center and radius OA will pass through J., B^ (7, and i>. Def . sphere 12. .*. this sphere is circumscribed about ABCD. § 445 EXERCISES 1. A sphere can be circumscribed about any rectangular parallelo- piped. 2. A sphere can be inscribed in any cube. 5. 6. 7. 8. 9. 414 SOLID GEOMETRY 447. Principles of limits. — If one half of a regular polygon having an even number of sides is inscribed in a semicircle^ and the whole figure is revolved about the diameter of the semicircle as axis^ then: (1) the area of the surface of the sphere traced is the limit approached by the area of the surface traced by the perimeter of the half polygon as the number of sides of the polygon is indefinitely increased; (2) the volume of the sphere traced is the limit approached by the volume of the solid traced by the half polygon as the number of sides of the polygon is indefinitely increased. 448. Theorem. — The area of the surface of a sphere is equal to four times the area of a great circle of the sphere. Hypothesis. S is the area of the surface and M the radius of the sphere with center (9, traced by revolving semicircle ACF about diameter AF as axis. Conclusion. S=4:'jrB^. Proof. 1. AS'=area of surface, jB= radius, is center of sphere traced by semicircle AOF revolved about diameter AF as axis. Hyp. 2. Inscribe in the semicircle one half of a regular poly- gon of an even number of sides, ABO-", Draw OM 1. any Def. sim. A THE SPHERE 415 side, as BC. From B, M, and Q draw JiBG; etc., to AF. And draw BK± OH, 3. In revolving trapezoid BCHGi about AF as axis, BO will trace the lateral surface of a frustum of a right circular cone whose area equals ir x BO x {BQ- -\- OH), § 412 4. But il[f is the middle point of ^(7. §156 5. Ba, MN, and (Tff are parallel. § 41 6. .•.MN=^(iBa+OH^. §100 7. .-. lateral area of frustum traced hj B0=2ir x BO x MK Ax. XII 8. It may be proved that A 0KB ~ A MNO. (Proof left to student.) Q . OM^MJSr "BO BK 10. But BK= aH, § 85 11. ,,OM^m. Ax. XII BO an 12. .', OMxaB^BOxMN. Ax. IV 13. .-. area of surface traced by BO =2 ir x OM x GrH. Ax. XII 14. Also it can be proved that area of surface traced by AB = lTr x OM x AG-^ area of surface traced by OD = 2 tt x OM x HL^ etc. 15. .*. if B represents the area of the surface traced by ABO"', « = 27rx 0MxAa+27rx OMx aB:-\-27rx OMx HL + etc. Ax. II 16. .'.s=^2'jr X OMxiAa^ GH^-HL+ Qic.^, Fact. 17. .\8 = 2irx 0Mx2B, Ax. XII 18. If the number of sides of the polygon is indefinitely increased, 8 = aS' and 0M= R. § 447, § 274 19. .'.27rxOMx2B=2'7rxRx2Rov4'7rB^. § 275, (2) 20. .\S=^7rB^. § 275, (1) 416 SOLID GEOMETRY 449. Corollary l. — The areas of the surfaces of two spheres are to each other as the squares of their radii or of their diameters. The proof is left to the student. 450. Zones. — A zone is that portion of the surface of a sphere which lies between two parallel planes. / •-«^f«-.-^.. Zone The sections of the surface made by the planes are the bases of the zone. The perpendicular distance between the planes is the altitude of the zone. If one of the planes is tangent to the sphere, the zone is called a zone of one base. When a semicircle is revolved about the diameter as an axis, the surface traced by any arc of the semicircle is a zone. Thus, in the figure of § 448, BD traces a zone, and GL is the altitude. 451. Corollary 2. — The area of a zone of a sphere is equal to the product of its altitude ayid the circumference of a great circle. For, the method of proof in § 448 applies equally to the zone traced by an arc of the revolving semicircle. The sides of the inscribed polygon which are inscribed in the given arc trace surfaces whose areas are expressed as in step 14 of § 448. By adding these areas and finding the limit of their sum, the equation aS' = 2 irRH is obtained, where S is the area and H the altitude of the zone, and B the radius of the sphere. THE SPHERE 417 EXERCISES 1. Considering the entire surface of a sphere as a zone, deduce the formula for computing the area of the surface of a sphere from § 451. 2. Show that the area of the surface of a sphere is found by the formula S = ttD^ where D is the diameter. 3. What is the area of the surface of a sphere whose diameter is 10 in. ? 4. The area of the surface of a sphere is equal to the product of the diameter and the circumference of a great circle. 5. The diameter of a given sphere is 12 in. If parallel planes divide the diameter into four equal parts, find the area of each of the four zones thus cut off. 6. Zones on the same sphere or on equal spheres are to each other as their altitudes. 7. The diameter of a sphere is 8 in., and a plane is passed. through the sphere 1 in. from the center. Compare the two parts into which the surface of the sphere is divided. 8. Find the diameter of a sphere the area of whose surface is 9C sq. ft. 9. The figure represents a sphere inscribed in a right circular cylinder, the bases of the cylinder being tangent to the sphere and the lateral sur- face of the cylinder touching the surface of the sphere in a circle of contact. Corresponding belts are formed on the two surfaces, included between two planes each parallel to the bases of the cylinder. Prove that the areas of the two belts are equal. Show that the entire area of the surface of the sphere is equal to the lateral area of the cylinder. 10. The area of the surface of a sphere is equal to two thirds of the total area of the surface of the circumscribed right circular cylinder. Note. — The discovery of the relation in Exercise 10 between the areas of the surfaces of a sphere and its circumscribed cylinder, and of a like relation between the volumes of these solids, was first made by Archimedes. He was so pleased with these discoveries that he asked that the figure of a sphere and its circumscribed cylinder be inscribed on his tomb. This request was carried out by his friend Marcellus. 418 SOLID GEOMETRY 11.. Prove that a zone of one base has the same area as a cir«le whose radius is the chord of the generating arc of the zone. Suggestion. — This necessitates showing that the square of the chord is equal to 2 times the product of the radius and the altitude of the zone. Why? 12. Taking the earth as a sphere and the radius as 4000 miles, find the area of the entire surface of the earth. The area of the United States is 3,624,122 sq. mi. What part of the earth's surface is occupied by the United States? 13. The North Frigid Zone is that part of the earth's surface lying north of the Arctic Circle and extending to the North Pole. Its altitude is approximately 330 mi. Find its area. What part of the earth's surface lies in the two frigid zones? 14. Find the diameter of the Arctic Circle. 15. The North Temperate Zone is that part of the earth's surface lying between the Tropic of Cancer and the Arctic Circle. Its altitude is ap- proximately 2165 mi. Find its area. What part of the earth's surface lies in the two temperate zones? 16. Prove that one half of the earth's surface lies within 30° of the equator. Suggestion. — What is the relation between the sides of a right triangle in which one of the acute angles is 30° ? 17. How many miles above the surface of the earth is that point from which one fourth of the earth's surface may be seen? Suggestion. — Apply the same principles as in Exercise 16. 18. An observer stands d feet from the center of a sphere whose diameter is d feet. What part of the surface of the sphere can be seen? 19. Find the total area of a hemispherical bowl 1 in. thick whose external diameter is 12 in. Suggestion. — We are to find the surfaces of two hemispheres and 9 ring (the rim). THE SPHERE 419 20. The surface of a tiled dome, in the form of a hemispherical sur- face whose diameter is 24 ft., is made of colored tiles each 1 in. square. How many tiles are required to make it? 21. In a zone of one base, the diameter of the base is 16 in. and the altitude 4 in. Find the radius of the sphere. Suggestion. — If x is the radius of the sphere, 4(2 a: — 4) = 64. 22. Vessels whose surfaces are in the form of zones of one base some- times are made from sheet metal, by cutting out a cir- cular blank from a flat sheet of the metal, and pressing it into the required form in a die, as described in the exer- ""^^^ilPtes^ cises on the cylinder and the cone. Find the diameter of the blank required to make a hemispherical brass bowl whose diameter is 8 in. 23. The lid to a silver-plated dish is in the form of a zone of one base, with a flange. The depth of the lid is 1^ in., the width of the flange 1 in., and the outer diameter of the flange 9 in. Find the diameter of the blank from which it is made. 24. Hollow metal balls, used as ornaments, casters, anti-friction balls, etc., are made from circular blanks cut from sheet metal, by use of a die. Find the diameter of the blank required to make a hollow metal ball ^ in. in diameter. 25. A method of testing the hardness of metals consists in partly forcing a hardened steel ball into the sample being tested. The liard- ness numeral H is computed by the formula H = —, where K is pressure on ball in kilo- y gi ams, and y is area of surface of depression in sample in square millimeters. Show that y = 2 irr{r — Vr^ - BPi, where r — radius of ball and U — radius of depression in sample. Substitute this value of y, and deduce the formula for finding H in terms of K^ r, and R. 420 SOLID GEOMETRY 452. Theorem. — The volume of a sphere is equal to tht product of the area of its surface and one third of the radius. Hypothesis. F=the volume, AS'=the area of the surface, and R = the radius of a sphere. Conclusion. V= ^ MS. Proof. 1. V, S, and B are the volume, area of surface, and radius, respectively, of sphere. Hyp. 2. Let A0^ be the diameter of the semicircle by revolv- ing which the sphere is traced. Inscribe in the semicircle one half of a regular polygon of an even number of sides, ABOD •... Produce any side, say BQ, to meet ^(7 produced at some point H. Draw OB and 00, Draw OMA.BC. Draw BPl.Aa and OQl-Aa. 3. Vol. traced by Off = vol. traced by OOQ-^ vol. traced by OQH = j7rx Ce'xO^-f Ittx C^xG^ §414 = I TT X (7$^ X ( 0^ -h QH) Factoring = ^7r xOQxOQx Off. Ax. XII 4. But it may be proved that A OMff'^A OQff. (Proof left to student.) -^ = -^. Def. Sim. A OM OH 5. THE SPHERE 421 6. .-. CQ X 0E= OMx OH. Ax. IV 7. .-. vol. traced hy 0011= ^ir x CQ x OMx OR. Ax. XII 8. But TT X OQ X 0H= area of surface traced by OH. §410 9. .-. vol. traced by OOH^ \ OMx area traced by OH. Ax. XII 10. Similarly, vol. traced by OBH=^ OiJf xarea traced hy BH. 11. .-. vol. traced by OBO = I OMx area traced by OH— ^ OMx area traced by BH = ^OMx area traced by BO. Factoring, Ax. XII 12. Similarly, vol. traced by OAB = ^ OMxarea, traced by AB, vol. traced by 0OD = \ Oitfx area traced by CD, etc. 13. .*. if V is volume traced by half polygon ABOD •••, and 8 is the area traced by its perimeter, v = \ OMx (area traced by AB + area traced by 5(7+ etc.) = \ OMx 8. Ax. II, Factoring, Ax. XII 14. If the number of sides of the polygon is indefinitely increased, v=V, 8 = S, and 0M= E. § 447, § 274 15. .'.iOMx8 = i RS. § 275 (2), § 408 (2) 16. .-. V=IRS. §275 (1) 453. Corollary i. — The volume of a 8phere equah ^ 7rM% or J ttZ)^, where R 18 the radiu8 and D the diameter. The proof is left to the student. 454. Corollary 2. — The volumes of any two spheres are to each other as the cubes of their radii., or as the cubes of their diameters. The proof is left to the student. 422 SOLID GEOMETRY 455. A spherical sector. — When a sphere is traced by re- volving a semicircle about its diameter as an axis, that por- tion of the sphere which is traced by a sector of the circle is called a spherical sector. Thus, if the sector A OB of a circle is revolved about the diameter EF as an axis, it traces the spherical sector AB-O-CD. The zone traced by the arc of the sector of a circle is called the base of the spherical sector. 456. Corollary 3. — The volume of a spherical sector is equal to the product of the area of its base and one third of the radius of the sphere. (Apply § 452.) 457. A spherical segment. — A spherical segment is the por- tion of a sphere included between two parallel planes. The sections of the sphere made by the planes are the bases, and the perpendicular distance between the planes is the altitude. When one of the bounding planes is tangent to the sphere, the segment is called a spherical segment of one base. THE SPHERE 423 458. Theorem. — The volume of a spherical segment of which the altitude is h and of which the radii of the bases are a and b, respectively, equals | 'Trh(a?' + ^^) + J tt^^. Hypothesis, the altitude Conclnsion. Proof. 1. The volume of a spherical segment is FJ is A, and the radii of its bases are a and b, V= i irh (a2 + 62) -h 1 TT^s. Vol. sph. seg. = V, alt. = A, etc. Hyp. 2. Let the spherical segment be traced by revolving DNMO, one half of a segment of a circle, about diameter AB as axis. Then OB = R, MN= h, CM= a, and I)N= b. Draw 00 and OD. Let ON = x and 0M= y, 3. Then r= vol. traced by 0(7Z>4-vol. traced by OBliJ - vol. traced h^OCM=\ irB^h 4- J irly^x - \ ira^y. § 456, § 414 4. But A = a: - ?/, 62 = i22 _ 2^^ ^2 ^ 722 _ ^2. § igg 5. .-. r= J TT S2 i^(2: - ?/) 4- (i^2 _ 3,2)^ _ (i22 _ ^2)^1 = ^7r(2:-^) 53i22_(^2^.^^4.^2){ =;| ttAS 3 i22 _ (a^2 + a:i/ + «^2) j . Ax. XII, Fact. But x^-2xy-\-y^=¥. Ax. VI Subtracting this from identity 3 a:^ _|. 3 y2 __ 3 aj2 ^ 3 ^2, 2 a;2 4. 2 2;?/ + 2 ^2 ^ 3 3,2 _^ 3 ^2 _ ^2. .-. ic2 + 2:1/ + ?/2 = 1(2^2 4- 2/2) _ 1 ^2 Ax. V = |(i?2 - 62 4- i22 - a2) _ J ^2 Ax. XII = 3 i22 _ J (^2 _^ 52) _ 1 ^2. .-. F= J rf (a2 4- 52) 4. 1 ^^3. Ax. XII 6. 7. 8. 9. 424 SOLID GEOMETRY EXERCISES 1. Find the volume of a sphere whose diameter is 16 in. 2. Find the volume of a spherical sector if its altitude is 3 in. and the radius of the sphere is 8 in. 3. Find the volume of a spherical segment if its altitude is 4 in. and the radii of its bases are 12 in. and 16 in., respectively. 4. If V is the volume of a spherical sector and H is the altitude of its base, and R is the radius of the sphere, prove that F = f tt R^H. 5. Prove that the volume of a spherical segment of one base equals 7rh^(R — i A), where h is the altitude and R is the radius of the sphere. Suggestion. — In the formula of § 458, let & = 0, and apply § 200. 6. Find the ratio of the volume of a sphere to that of a circumscribed cube. 7. Find the ratio of the volume of a sphere to that of an inscribed cube. 8. Prove that the volume of any sphere is equal to two thirds of the volume of the circumscribed right circular cylinder. 9. A sphere whose diameter is D is transformed into a cylinder of revolution having the same diameter. Find the height of the cylinder. 10. Show that the formula for computing the volume of a sphere may be obtained from the formula for computing the volume of a spherical sector. 11. Show that the formula for computing the volume of a sphere may be obtained from the formula for computing the volume of a spherical segment. 12. Prove that if a sphere and a cube have equal areas, the sphere has the greater volume. Note. — It is interesting to note that of all solids having equal areas, the sphere has the greatest volume. 13. Two spheres of lead, of radii 2 in. and 4 in., respectively, are melted and recast into a solid cylinder of revolution whose altitude is 6 in. Show that the total surface is unchanged in amount. 14. A cubic foot of ivory weighs 114 lb. What is the weight of an ivory billiard ball 2 in. in diameter? THE SPHERE 425 15. A hollow spherical steel shell is 1 in. thick, and its outer diameter is 10 in. Allowing 490 lb. to a cubic foot, find its weight. 16. A washbasin is in the form of a spherical segment of one base. Its depth is 7^ in. and the distance across the top is 16 in. Find how many gallons of water it will hold, allowing 7^ gal. to a cubic foot. 17. Through the center of a 10-in. sphere a hole was bored, taking off one fourth of the surface of the sphere from each end. Find the volume of the ring left. 18. A boiler is made in the form of a 4-ft. cylinder 2 ft. in diameter, with hemispherical ends. How many gallons will it hold? 19. A sphere of lead 5 inches in diameter is formed into a tube 2^ ft. long whose internal diameter is 1 in. Find the thickness of the tube. 20. A hemispherical boiler whose diameter is 4 ft. contains water to a depth of 20 in. How many gallons of water does it contain? 21. Among the rules of thumb used by men in practical work is the following: The weight of a cast-iron ball (in pounds) equals the cube of the diameter times .1377. Allowing 450 lb. to a cubic foot of cast iron, test the accuracy of this rule. 22. Which is the better bargain, oranges 3 in. in diameter at 30^ a dozen, or oranges 3^ in. in diameter at 40 ^ a dozen ? 23. The figure represents a cross section of a ball safety valve. The pressure of the steam from below lifts the ball, and allows the steam to escape. The diameter of the pipe below the ball is | in., and the diameter of the ball is 1 in. If the ball weighs .28 lb. to the cubic inch, what is the pressure in pounds per square inch of the steam in the pipe below the ball when it lifts it and escapes ? 24. A haystack is approximately in the form of a cylinder 16 ft. in diameter and 8 ft. high, surmounted by a hemi- sphere. Allowing 512 cu. ft. to the ton, find the weight of the stack. 25. A building lot which is being leveled off has a mound on it approximately in the form of a spherical segment of one base. It is 40 ft. wide at the base and 10 ft. deep at the highest point. How many cubic yards of earth must be removed in cutting it down to the level of the surrounding ground ? 426 SOLID GEOMETRY 26. When an object floats in water, or is immersed in it, the object is buoyed up by the water with a force equal to the weight of the water displaced by it. A ball floats half submerged in water. Its diameter is 8 in. Find its weight. (Water weighs 62.5 lb. per cubic foot.) 27. A float in a water tank consists of a hollow copper ball which closes the valve that lets the water enter the tank, by means of a lever to which it is attached, when the water buoys it up to a certain point. The valve shuts ofE the entering water when the ball is submerged to three fourths of its depth. If the diameter of the float is 4 in., what is the upward pressure of the water against it ? FIGURES ON THE SURFACE OF A SPHERE 459. Spherical angles. — A spherical angle is the figure formed by two arcs of great circles drawn from the same point on the surface of a sphere. The point from which the arcs are drawn is the vertex of the angle, and the arcs are the sides of the angle. A spherical angle is measured by the plane angle which is formed by the tan- gents to its sides at the vertex. Thus, spherical angle ABC is measured by Z DBE, where BD and BE are tangents to BA and BC respectively. A spherical angle of which the tangents to the sides at the vertex form a right angle is called a right spherical angle. 460. Perpendicular arcs. — The sides of a right spherical angle are perpendicular arcs. Thus, if ^ and ^C in the figure of § 469 form a right spherical angle, they are called perpendicular arcs. FIGURES ON THE SURFACE OF A SPHERE 427 461. Theorem. — A spherical angle has the same measure as the arc of the great circle having the vertex of the angle as pole and included between the sides of the angle^ produced if necessary. A Hypothesis. Z.BAO is a spherical angle on the sphere with center 0\ BCis the arc of a great circle having A as its pole. Conclusion. ^ BAOha^ the same measure as BO. , Proof. 1. Z BAG is spherical angle; BO is arc of great circle having A as pole. Hyp. 2. Draw radii OA, OB, 00, Draw AD tangent to AB and AU tangent to AO. 3. OA ± plane BOO. § 436 4. .-. OB±OA and 001. OA, § 302 5. AB ± OA and AH ± OA. § 164 6. .-. AB II OB and AU W 00. § 41 7. .'.Z.DAE=ZBOO. § 317 8. But Z BAO has the same measure as Z. DAE. § 459 9. And BO has the same measure d^s Z BOO. § 173 10. .*. Z BAOhdi^ the same measure as BO, Ax. I 462. Corollary l. — A spherical angle has the same measure as the diedral angle formed by the planes of its sides. The proof is left to the student. 428 SOLID GEOMETRY 463. Corollary 2. — An are of a great eirele drawn through the pole of another great circle is perpendicular to that circle. The proof is left to the student. 464. Spherical polygons. — A spherical polygon is the figure formed by three or more arcs of great circles which com- pletely inclose a portion of the surface of a sphere. The arcs are the sides, the spherical angles formed by the sides are the angles, and the vertices of the angles are the vertices of the polygon. A diagonal of a spherical polygon is the great circle arc which joins any two non- adjacent vertices. A spherical polygon each of whose angles is less than 180° is convex. Only convex spherical polygons are considered here. A spherical triangle is a spherical polygon of three sides. The terms right triangle, isosceles, equilateral, etc., are applied to spherical triangles with the same meanings as when applied to plane triangles. 465. Relation of spherical polygons to polyedral angles. — The planes of the sides of a spherical polygon form a polyedral angle whose vertex is the center of the sphere. Thus, in the figure of § 464, the planes of the sides of polygon ABCD form the polyedral angle 0-ABCD. It is evident from the preceding sections that: (1) Each face angle of the polyedral angle has the same measure as the corresponding side of the spherical polygon. (2) Uach diedral angle of the polyedral angle has the same measure as the corresponding angle of the spherical polygon. Hence^ from any property of polyedral angles an analogous property of spherical polygons may he inferred. FIGURES ON THE SURFACE OF A SPHERE 429 466. Theorem. — Any side of a spherical triangle is less than the sum of the other two sides. The proof is left to the student. See § 342 and § 425 (1). Write the proof in full. 467. Theorem. — The sum of the sides of any spherical poly- gon is less than 360°. The proof is left to the student. Write the proof in full. EXERCISES 1. The perimeter of any spherical polygon is less than a great circle. 2. Any side of a spherical polygon is less than the sum of the others. 3. No side of a convex spherical polygon is as long as a semicircle. 4. If two sides of a spherical triangle are quadrants, the angles oppo- site them are right angles. 5. If two sides of a spherical triangle are 80° and 90°, respectively, between what two values does the third side lie ? 430 SOLID GEOMETRY 468. Polar triangles. — If A ABCis any spherical triangle, and if the three great circles are drawn of which A^ B^ and (7, respectively, are poles, these circles evidently form eight new spherical tri- angles. Of these, one triangle BEF is called the polar triangle of A ABC. A is the pole of EF^ B is the pole of DF^ and O is the pole of BE, Of the eight tri- angles, A BEF is so chosen that and F are on the same side of AB^ A and B on the same side of BO^ and B and E on the same side of AC. Note. — The properties of polar triangles were discovered by Girard, a Dutch mathematician, about 1626 a.d. They were discovered independently about the same time by Snell, a prodigy, who at the age of twelve was familiar with the standard works on higher mathematics of that time. 469. Theorem. — If one spherical triangle is the polar of a second, the second spherical triangle is the polar of the first. Hypothesis. A BEF is the polar of A ABC. Conclusion. A ABC is the polar of A BEF. Proof. 1. A BEF is the polar of A ABC. Hyp. 2. Draw BB and CB of great circles. 3. Since B is the pole of BF, BB is a quadrant. § 440 4. Since C is the pole of BE, CB is a quadrant. § 440 5. .-. B is the pole of ^. § 441 6. Similarly, E is the pole of AC and F the pole of AB. 7. .-. A ABCis the polar of A BEF. § 468 FIGURES ON THE SURFACE OF A SPHERE 431 470. Theorem. — In two polar spherical triangles^ each angle of one has the same measure as the supplement of the side of the other of which its vertex is the pole. Hjrpothesis. A ABC and aDEF are polar spherical tri- angles. Conclusion. Z C has the same measure as the supplement of DE^ Z F has the same measure as the supplement of ABy etc. Proof. 1. A AjBC and A BEF are polar triangles. Hyp. 2. Prolong CA and (S, if necessary, to meet BE at M and iV, respectively. 3. Then Z (7 has the same measure as MN, § 461 4. ^=90°andMfe=90°. §440 5. .-. BN^-ME= 180°. Ax. II 6. ByxtBN+ME=MN+BE. ^ Ax. XII 7. .'. MN+ BE^ 180°, i.e. ilEV'and BE are supp. Ax. I 8. .'. Z (7 has the same measure as the supplement of BE. Ax. XII 9. Similarly, Z F has the same measure as the supplement of AB., etc. 471. Corollary l. — If a spherical triangle is equilateral, its polar triangle is equiangular ; and conversely. The proof is left to the student. 432 SOLID GEOMETRY 472. Definitions. — Two mutually equiangular triangles (See § 127) are triangles such that to each angle of one there corresponds an equal angle of the other. Two mutually equilateral triangles are triangles such that to each side of one there corresponds an equal side of the other. 473. Corollary 2. — If two spherical triangles on the same sphere^ or on equal spheres^ are mutually equiangular^ their polar triangles are mutually equilateral ; and conversely. The proof is left to the student. 474. Theorem. — The sum of the angles of a spherical tri- angle is greater than 180° and less than 540°. Hypothesis. A ABO i^ any spherical triangle. Conclusion. Z^ + Z5-fZ(7> 180° and < 540°. Proof. 1. Let d^ e, and / be the sides of the polar of AABQoi which ^, B, and (7, respectively, are the poles. 2. Then Z A has the same measure as 180° — d, Z. B has the same measure as 180°— e, Z (7 has the same measure as 180° — /. § 470 3. .'./.A-^Z.B-\-/.0 has the same measure as 540° -((^ + g+/). Ax. Ill 4. Butc^ + e+/<360°and >0°. §467 5. .-. 540°-((^4-e-f/)> 180° and < 540°. Subtraction 6. .-. Z^-|-Z^ + Z(7>180°and <540°. Ax. XII FIGURES ON THE SURFACE OF A SPHERE 433 475. Corollaxy. — A spherical triangle may have one^ two^ or three right angles. And it may have one^ two^ or three obtuse angles. The proof is left to the student. 476. Definitions. — A spherical triangle having two right angles is a birectangular spherical triangle. One having three right angles is a trirectangpilar spherical triangle. The difference between the sura of the angles of any spheri- cal triangle and 180° is the spherical excess of the triangle. EXERCISES 1. The angles of a spherical triangle are 80°, 65°, and 135°. Find the sides of its polar triangle. 2. The sides of a spherical triangle are 120°, 84°, and 66°, respectively. How many degrees are there in each angle of its polar triangle ? 3. A spherical triangle has two of its sides quadrants and the third side equal to 60°. Determine the angles of its polar triangle. 4. If two sides of a spherical triangle are quadrants, the triangle is birectangular. 5. If a spherical triangle is birectangular, the sides opposite the right angles are quadrants. 6. Three planes passed through the center of a sphere, each perpen- dicular to the other two, form on the spherical surface eight trirectangular triangles. 7. Find the sides of a spherical triangle if the angles of its polar triangle are 96° 37' 36", 87° 17' 57", 72° 46' 32". 8. What is the polar triangle of a spherical triangle all of whose sides are quadrants? 9. In any birectangular spherical triangle the side opposite the angle that is not a right angle has the same measure as that angle. , 10. In any trirectangular spherical triangle each of the sides is a quadrant. 11. The angles of a spherical triangle are 80°, 90°, and 120°. What is the spherical excess ? 434 SOLID GEOMETRY 12. Any trirectangular spherical triangle coincides with its polai triangle. 13. Any exterior angle of a spherical triangle is less than the sum of the two opposite interior angles. 14. The arcs of the great circles which bisect the angles of a spherical triangle are concurrent. 15. Any arc of a great circle which is perpendicular to a second great circle passes through the pole of the latter. 477. Symmetrical spherical triangles. — Two triangles on the same sphere, or on equal spheres, are symmetrical when to each part of one there corresponds an equal part of the other, but the equal parts are arranged in reverse order. It is easily seen that if three planes pass through the center of a sphere and do not intersect in one straight line, they cut the surface of the sphere in a pair of symmetrical spherical triangles, as A ABQ and A DEF in the figure. Also it may be assumed that two symmetrical triangles of a sphere may be moved into such positions as are occupied by A ABO and A DEF. It is evident that, because of the curvature of the spherical surface, in general two symmetrical spherical triangles cannot be made to coincide, and hence are not congruent. See § 344. If, however, two symmetrical spheri- j} ]g £) cal triangles are isosceles^ as A ABC and A DEF in the figure, it is possible to make them coincide throughout. Hence it is inferred that : If two symmetrical spherical triangles are isosceles, they are congruent. C F FIGURES ON THE SURFACE OF A SPHERE 435 478. Theorem. — Two symmetrical spherical triangles have equal areas. Hjrpothesis. A ABC and ADEF are symmetrical spheri- cal triangles. Conc^asion. AABC^aBEF, Proof. 1. A ABO and A BEF are symmetrical spherical triangles. Hyp. 2. .'.A ABC and A BEF may be placed, as in the figure, so that AB, BE, and CF are diameters. § 477 3. Let P be the pole of the circle determined by points A, jB, (7, and let P^ be a diameter. Draw great circle arcs AP, BP, CP, BQ, EQ, FQ. 4. Then AP=^BP=CP. § 438 5. AP = BQ,BP = fQ,6P=f^. §20(V), §173 6. .'.BQ=EQ = FQ. Ax.I 7. .-. the symmetrical A APB and BQE are isosceles. §464 8. .'.AAPB^aBQE. §477 9. Similarly, A BPC ^ A EQF, A CPA ^ A FQB. 10. .'.AABC^A BEF. Ax. II EXERCISE Draw a figure for § 478 in which the pole P falls outside of A ABC and Q falls outside of A DEF, and give the proof of the theorem. Suggestion. — Each of the given triangles will be equivalent to the sum of two isosceles triangles minus a third one. 436 SOLID GEOMETRY 479. Theorem. — If two spherical triangles on the same sphere or equal spheres have two sides and the included angle of one equal respectively to two sides and the included angle of the other, they are either congruent or symmetrical. Hypothesis. A ABO and A DUF are spherical triangles on the same sphere or equal spheres ; AB = BE, A0= BF, ZA^ZB. Conclusion. A ABO and A BFJF are either congruent or symmetrical. Suggestions. If the equal parts of the triangles occur in the same order, superpose A BFF upon A ABO so that Z B coincides with Z.A, and prove that the triangles coincide throughout. If the equal parts occur in reverse order, construct A GrHK symmetrical to A BEF. Prove that A ABO ^. A GHK, and hence that A ABO i^ symmetrical to A BEF. Write the proof in full. 480. Theorem. — If two spherical triangles on the same sphere or equal spheres have a side and two adjacent angles of one equal respectively to a side and two adjacent angles of the other, they are either congruent or symmetrical. The proof is left to the student. Proceed as in the proof of § 479. Write the proof in full. FIGURES ON THE SURFACE OF A SPHERE 437 481. Theorem. — If two spherical triangles on the same sphere or equal spheres are mutually equilateral^ they are either congruent or symmetrical. C F F D'^" y^^^-^D Left to the student. Prove diedral angles with edges A and DP equal, and that Z ^ = Z D. Apply § 479. 482. Theorem. — If two spherical triangles on the same sphere or equal spheres are mutually equiangular^ they are either congruent or symmetrical. K N N H M '^ Hypothesis. A ABQ 2^vA A DEF are triangles on the same sphere or equal spheres ; /.A = Z.D^Z.B = Z.I]^/.0=/.F. Conclusion. A ABO and A DEF are either congruent or symmetrical. Suggestions. Draw A GrRK ^nd A LMN" polars of A ABO and A DEF. Prove A GrRKsmd A LMN mutually equilat- eral. Then they are congruent or symmetrical. Why ? Hence they are mutually equiangular. But A ABO and A DEF are polars of A GrIIKsind A LMJY. Now, from these facts, by similar steps, show that A ABO and A DEF 2iTe congruent or symmetrical. 438 SOLID GEOMETRY 483. Theorem. — In any isosceles spherical triangle the angles opposite the equal sides are equal. Draw a great circle arc from (7 to Z>, the middle point of Ah, Compare A ^1>(7 and A i>J?(7 by § 481. Write the proof in full. 484. Theorem. — If two angles of a spherical triangle are equals the sides opposite these angles are equals and the tri- angle is isosceles,- .-'% Hypothesis. In spherical A ABO^ Z A = Z B. Conclusion. AO=BO. Proof. 1. In spherical AABC,ZA = ZB. Hyp. 2. Let A BUF be the polar of A ABO, 3. Then JDF and JEF have the same measures as the sup- plements oi Z B and Z A^ respectively. § 470 4. ,',]5f=EF, Supp. of equals 5. .-. ZD=ZE, §483 6. But A A^CMs the polar of A DJ^^. §469 FIGURES ON THE SURFACE OF A SPHERE 439 7. .*. AO and ^(7 have the same measures as the supple- ments of Z ^ and Z i>, respectively. § 470 8. .-. JX/= Bu. Supp. of equals EXERCISES 1. The great circle arc from the vertex of an isosceles spherical tri- angle to the middle point of the base bisects the vertical angle, is per- pendicular to the base, and divides the triangle into two triangles with equal areas. 2. Every point in the arc of a great circle perpendicular to another arc of a great circle at its middle point is equidistant from the ends of that arc. 3. Every point on the surface of a sphere that is equidistant from the ends of an arc of a great circle is on the arc perpendicular to that arc at its middle point. 4. What is the locus of points on the surface of a sphere that are equidistant from the ends of a given great circle arc ? 5. If an arc of one great circle is perpendicular to an arc of another great circle at its middle point, any point without the former is unequally distant from the ends of the latter. 6. Any point on the bisector of a spherical angle is equally distant from the sides of the angle. Suggestion. — Let BD bisect ZABC, and let MP be JL AB. Mark ofE jBQ = BP, and draw MQ. Prove that MQ±BC and MP = MQ. 7. In any spherical triangle, if two angles are unequal, the opposite sides are unequal, the greater side being opposite the greater angle. Hypothesis. — In spherical A ABC, ZBAOZCBA. Conclusion. — sd > AC. Suggestions. — Draw AD making Z BAD = Z CBA. Prove AD = BD. Then AD + DC' > AC, etc. 8. State and prove the converse of Exercise 7. 440 SOLID GEOMETRY 485. Area of a trirectangular triangle. — If three planes are passed through the center of a sphere, each perpendicular to the other two planes, they form on the surface of the sphere eight trirectangular triangles^ which are equal in area by § 482. Hence: The area of a trirectangular triangle is equal to one eighth of the area of the surface of the sphere. Such a triangle is sometimes used as a unit of measure of spherical surfaces. 486. A spherical degree. — One of the 720 equal parts of the surface of a sphere is called a spherical degree. A spherical degree is used as a unit of measure of spherical surfaces. 487. A lime. — A lune is that figure on the surface of a sphere which is formed by two great semicircles whose end points coincide. A lune, then, has two equal angles. Thus, ABCD is a lune whose equal angles are Z A and /. C. It is evident that lunes on the surface of the same sphere which have equal angles may be made to coincide, and hence are congru- ent. Consequently, it is inferred that: The area of a lune whose angle is n degrees is — - of the area of the surface of the sphere. 488. Corollary. — The area of a lune., in spherical degrees^ is equal to twice the number of degrees in the angle of the lune. For, by § 486 and § 487, the area of a lune whose angle is n degrees, is equal to -^ x 720, or 2 w, spherical degrees. FIGURES ON THE SURFACE OF A SPHERE 441 489. Theorem. — The area of a spherical triangle^ in spheri- cal degrees^ is equal to the spherical excess of the triangle. Hypothesis. Spherical excess of A ABC — e. Conclusion. The number of sph. deg. in A ABQ = e. Proof. 1. Spherical excess of A ABQ = e. Hyp. 2. Produce each side of A ABO to form the complete circle, and let AF^ BU, and OB be the diameters in which their planes intersect. 3. The vertical angrles at are equal. § 20, V 4. .-. Bd= DE. BF = AF, OF^ lb. § 173 5. .'. A ABE and A BOF are either congruent or sym- metrical. § 481 6. .-. A ABE and A BOF are equal. Def. cong., § 478 7. ^o\Y\unQABFO=AABO+ABOF, lune BOEA = A ABO+ A AOE, Ax. X luae OBBA = A ABO+ A ABD. 8. .'.ABFO-\-BOEA+OBBA = 2AAB0+ (A ABO-^ A BOF-}- aAOE-\- A ABB). Ax. II 9. Bnt ABFO+BOEA+ 0BBA = 2iA-\-B+ 0) sph, deg. § 488 10. And A J.5(7+A^(7^+A^(7^+A^5i)=a hemi- spherical surface, or 360 sph. deg. § 486 11. .'.2(A-\-B+ 0) sph. deg. = 2aABO-{-SQ0 sph. deg. Ax. XII 12. .-. A^^(7=^ + 5+6'-180sph. deg. =e. Solving 442 SOLID GEOMETRY 490. Spherical excess of a spherical polygon. — The sphericav excess of a spherical polygon is the difference between the sum of its angles and the sum of the angles of a plane polygon of the same number of sides. 491. Theorem. — The area of a spherical polygon^ in spheri- cal degrees^ is equal to the spherical excess of the polygon. The proof is left to the student. Draw diagonals from any vertex, dividing the polygon into triangles. How many triangles are there ? Write the proof in fall. EXERCISES 1. Find the area of a trirectangiilar triangle on a sphere whose diameter is 28 in. 2. Find the area of a lune whose angle is 30° on a sphere whose diameter is 20 in. 3. What part of the whole sphere is a triangle whose angles are 70°, 110°, and 120° ? 4. What is the spherical excess of a triangle whose angles are 84° 27' 20", 96° 42' 36", and 116° 12' 24" ? 5. What is the area of a triangle whose angles are 80°, 92°, and 112°, on a sphere whose diameter is 36 in. ? 6. What is the spherical excess of a hexagon whose angles are 84°, 167°, 140°, 106°, 98°, and 157°? 7. Find the area of a spherical pentagon whose angles are 120°, 140°, 155°, 158°, and 163°, on a sphere whose diameter is 10 in. . 8. Find the area of that part of the earth's surface lying between the 75th and 90th meridians. Call the radius 4000 mi. 9. The areas of two lunes on the same sphere or on equal spheres are to each other as the angles of the lunes. 10. The areas of lunes having equal angles, but situated on unequal spheres, are to each other as the squares of the radii of the spheres on which they are situated. FIGURES ON THE SURFACE OF A SPHERE 443 MISCELLANEOUS EXERCISES 1. The radius of a sphere is 12 in. Find the area of a section made by a plane 8 in. from the center. 2. Find the locus of all points in space which are equidistant from two given points and at a given distance d from a third given point. 3. Any two vertical spherical angles are equal. 4. A right circular cylinder whose altitude is 8 in. is inscribed in a sphere whose radius is 6 in. Find the volume of the cylinder. 5. Any side of a spherical polygon is less than 180°. 6. If two adjacent sides of a spherical quadrilateral are greater, respectively, than the other two sides, the angle included by the two shorter sides is greater than the angle included by the two greater sides, 7. The radii of two concentric spheres are 8 in. and 12 in., respec- tively. A plane is tangent to the inner sphere. Find the area of the section of the outer sphere made by the plane. Find the area of the smaller zone of the outer sphere cut o£E by the plane. 8. A sphere is inscribed in a right circular cylin- der, touching the lateral surface of the cylinder all the way around and tangent to both bases. Two cones have the bases of the cylinder as their bases and the center of the sphere as their common vertex. Any plane is passed through the figure parallel to the bases of the cylinder. Prove that the ring between the sec- tions of the cylinder and cone is equal to the section of the sphere. 9. If two circles on a sphere have the same poles, their planes are parallel. 10. The line of centers of two intersecting spheres meets the surfaces of the spheres in the poles of their common circle. 11. Find the locus of the centers of all spheres tangent to a given plane at a given point. 12. Three spheres each of radius R are placed on a horizontal plane so that each is tangent to the other two. A fourth sphere of the same radius is placed on top of them so that it touches each. Find the dis- tance from the highest point of the top sphere to the plane. 444 SOLID GEOMETRY 13. If a spherical triangle is isosceles, its polar triangle is also isosceles. 14. Prove that the volume of a hollow spherical shell whose outer and inner radii are R and r, respectively, is ^^i^ - r){R'^ -[- Rr + r^) o 15. If a surveyor desires that the sum of the angles in any spherical triangle which he uses on the earth's surface shall be within one minute of 180°, what is the largest area that the triangle may inclose? Suggestion. — See § 489. 16. The portion of a sphere bounded by the planes of two great semicircles and the lune which they form on the surface is called a spherical wedge. The angle between the planes of a spherical wedge is 15°, and the diameter of the sphere is 12 in. Find the volume of the wedge. 17. How many straight lines can be tangent to a sphere from a point outside of the sphere ? Compare their lengths between the given point and the points of contact. 18. The points of contact of all lines tangent to a sphere from an external point lie in a circle. 19. The lines in Exercise 18 are the elements of a cone. 20. If two spheres are tangent to the same plane at the same point, the straight line joining their centers passes through the point of contact. - 21. What is the locus of the centers of all spheres tangent to a given plane at a given point ? 22. What is the locus of the centers of all spheres tangent to the faces of a diedral angle ? 23. Find the edge of a cube inscribed in a sphere with radius 10 in. Find its area. Find its volume. 24. The volume of a polyedron circumscribed about a sphere is equal to the product of the area of its surface and one third of the radius of the sphere. 25. The sum of the arcs of great circles drawn from any point within a spherical triangle to the extremities of a side is less than the sum of the other two sides of the triangle. 26. If from a point within a spherical triangle arcs of great circles are drawn to the three vertices, their sum is less than the perimeter. refere;nces to plane geometry The following axioms, theorems, etc. , which are given in the Plane Geom- etry are referred to in the proofs in the Solid Geometry. They are placed here for the convenience of the student in looking up the references. AXIOMS I. Things which are equal to the same thing, or to equal things, are equal to each other. II. If equals are added to equals, the sums are equal. III. If equals are subtracted from equals, the remainders are equal. IV. If equals are multiplied by equals, the products are equal. V. If equals are divided by equals, the quotients are equal. "VI. Like powers, or like positive roots, of equals are equal. VII. If equals are added to or subtracted from unequals, or if unequals are multiplied or divided by the same positive number, the results are unequal in the same order. VIII. If unequals are subtracted from equals, the remainders are unequal in the reverse order. IX. If unequals are added to unequals in the same order, the sums are unequal in that order. X. The whole of a thing is equal to the sum of all of its parts, and is greater than any one of its parts. XL If the first of three things is greater than the second, and the second greater than the third, then the first is greater than the third. XII. A quantity may be substituted for its equal in any expression. DEFINITIONS, THEOREMS, ETC. § 6. A straight line is represented by placing a ruler, or straightedge, upon a plane surface and marking along the edge of the ruler. § 13. Two angles are equal if, and only if, they may be made to coincide throughout. If an angle is placed upon an equal angle so that the vertices and a pair of sides coincide, the other sides must coincide. § 17. A right angle is one half of a straight angle. § 20. I. Any two straight angles are equal. 11. Any two right angles are equal. III. The complements of equal angles are equal. IV. The supplements of equal angles are equal. V. Vertical angles are equal. 445 446 REFERENCES TO PLANE GEOMETRY § 26. If two parallel lines are cut by a transversal, the corresponding angles are equal. § 34. If a line is perpendicular to one of two parallel lines, it is perpendic- ular to the other also. § 41. Two straight lines perpendicular to the same straight line are parallel. § 46. Only one straight line can be drawn through a given point parallel to a given straight line. § 48. The sum of the angles of any triangle is equal to a straight angle. § 53. At a given point on a given line only one perpendicular can be drawn to the line. § 54. Only one perpendicular to a given line can be drawn through a given point not on the line. § 55. Two straight lines perpendicular respectively to two intersecting straight lines must meet. § 60. If one of two figures may be placed upon the other so that they coincide throughout, the figures are called congruent. § 63. If two sides and the included angle of one triangle are equal respec- tively to two sides and the included angle of another, the triangles are con- gruent. § 64. Two right triangles which have the legs of one equal respectively to the legs of the other are congruent. § 67. Two right triangles are congruent if a leg and an acute angle of one are equal respectively to a leg and an acute angle of the other. § 68. Two right triangles are congruent if the hypotenuse and an acute angle of one are equal respectively to the hypotenuse and an acute angle of the other. § 76. If the three sides of one triangle are equal respectively to the three sides of another, the triangles are congruent. § 82. The opposite sides of a parallelogram are equal. § 83. A diagonal divides a parallelogram into two congruent triangles. § 85. Two parallel lines are everywhere equidistant. § 90. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. § 100. If a line-segment is parallel to the bases of a trapezoid and bisects one of the non-parallel sides, then it bisects the other also and is equal to one half of the sum of the bases. § 101. The sum of the interior angles of a convex polygon of n sides is (n — 2) straight angles. § 105. The locus 6f points equidistant from the ends of a line-segment is the perpendicular bisector of it. § 107. Two points each equidistant from the ends of a line-segment de- termine the perpendicular bisector of it. REFERENCES TO PLANE GEOMETRY 447 § 114. The medians of any triangle are concurrent at a point of trisection of each. § 118. (1) If four numbers are in proportion, the product of the extremes equals the product of the means. (2) If the product of two numbers equals the product of two other num- bers, the four numbers are in proportion, one pair of factors being extremes and the other pair means. (3) If four numbers are in proportion, they are in proportion by inversion. (4) In any proportion, the means may be interchanged, or the extremes interchanged, without destroying the proportion. (6) The terms of any proportion are in proportion by addition. (6) The terms of any proportion are in proportion by subtraction. (7) Like powers or like roots of the terms of a proportion are in proportion. (8) If two or more ratios are equal, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. § 123. If DE is parallel to side AB of triangle ABC and meets AC a.t D and BC at E^ then ^ = ^and^=^. AD BE DC EC § 127. Two polygons which have the angles of one equal respectively to the angles of the other, taken in order, are called mutually equiangular. Similar polygons are those which (1) are mutually equiangular and (2) have their corresponding sides proportional. § 128. If two triangles are mutually equiangular, they are similar. § 129. If two triangles have their corresponding sides proportional, the triangles are similar. § 130. If two triangles have an angle of one equal to an angle of the other, and the including sides proportional, they are similar. § 146. The sum of any two sides of a triangle is greater than the third side. § 148. If two triangles have two sides of one equal respectively to two sides of the other, but the third side of the first greater than the third side of the second, the angles opposite these sides are unequal, the greater angle being opposite the greater side. § 151. (1) A diameter of a circle equals two times a radius. (2) Radii of the same circle or equal circles are equal. (3) If the radii of two circles are equal, the circles are equal. (4) In the same or equal circles, equal central angles intercept equal arcs. (5) In the same or equal circles, equal arcs subtend equal central angles. (6) In the same or equal circles, equal arcs are subtended by equal chords. (7) In the same or equal circles, equal chords subtend equal arcs. (8) A point is at a distance from the center of a circle equal to, greater 448 REFERENCES TO PLANE GEOMETRY tlian, or less than the radius, according as it is on, outside of, or within the circle ; and conversely. § 156. A line through the center of a circle perpendicular to a chord bisects the chord and the subtended arcs. § 164. A tangent to a circle is perpendicular to the radius drawn to the point of contact. § 173. A central angle has the same numerical measure as its intercepted arc. § 185. If, in geometric constructions, the instruments used are limited to the ungraduated straightedge and compasses alone, there are some construc- tions which cannot be made. There are three such impossible constructions which have interested mathematicians since the time of the ancient Greeks : (1) To trisect (cut into three equal parts) any given angle. (2) To construct a square which shall have the same area as a given circle. (3) To construct a cube which shall have twice the volume of a given cube. § 196. In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. § 200. If two chords intersect, the product of the segments of one is equal to the product of the segments of the other. § 215. Parallelograms having equal altitudes and equal bases are equal. § 228. Two triangles having an angle of one equal to an angle of the other are to each other as the products of the sides including those angles. § 274. (1) If the number of sides of a regular inscribed polygon is indefi- nitely increased, the apothem approaches the radius of the circle as a limit. (2) If the number of sides of a regular inscribed polygon is indefinitely increased, the perimeter approaches the circumference of the circle as a limit. (3) If the number of sides of a regular circumscribed polygon is indefi- nitely increased, the perimeter of the polygon approaches the circumference of the circle as a limit. (4) If the number of sides of a regular circumscribed polygon is indefi- nitely increased, the area of the polygon approaches the area of the circle as a limit. § 275. (1) If two variables are equal and each approaches a limit, the limits are equal. (2) If the limit of a variable x is a, then the limit of Tex is Tca^ where Tc is any constant. (3) If the limit of a variable x is a, then the limit of - is -, where k is Tc k any constant. § 281. The area of a circle equals vB^. § 284. The area of a sector of a circle is ecjual to one half of the product of its radius and its arc. SOLID GEOMETRY INDEX OF DEFINITIONS Altitude of a cone, § 398. of a cylinder, § 371. of a frustum of a cone, § 405. of a frustum of a pyramid, § 385. of a prism, § 349. of a prismatoid, § 429. of a pyramid, § 383. of a spherical segment, § 457. of a zone, § 450. Angle, diedral, § 325. of a spherical polygon, § 464. polyedral, § 341. right spherical, § 459. spherical, § 459. triedral, § 341. Arcs, perpendicular, § 460. Axis of a circle, § 436. Base of a cone, § 398. of a pyramid, § 383. of a spherical sector, § 455. Bases of a cylinder, § 371. of a frustum of a cone, § 405. of a prism, § 349. of a prismatoid, § 429. of a spherical segment, § 457. of a truncated pyramid, § 384. of a zone, § 450. Center of a sphere, § 432. Cone, § 398. circular, § 398. of revolution, § 398. right circular, § 398. Cones, similar, § 398, Cube, § 352. Cylinder, § 371. circular, § 371. circumscribed, § 374, inscribed, § 374. oblique, § 371. of revolution, § 378. right, § 371. right circular, § 371. Cylinders, similar, § 378. Degree, spherical, § 486. Determining a plane, § 295. Diagonal of a polyedron, § 348. of a spherical polygon, § 464. Diameter of a sphere, § 432. Diedral angle, § 325. plane angle of, § 326. right, acute, obtuse, § 329. Diedral angles, adjacent, § 329. complementary, § 329. supplementary, § 329. vertical, § 329. Dimensions of a parallelopiped, § 363. Distance between parallel planes, § 322. from a point to a plane, § 309. on the surface of a sphere, § 437. polar, § 439. Dodecaedron, § 348. Edges of a polyedral angle, § 341. of a polyedron, § 348. Element of a conical surface, § 398. of a cylindrical surface, § 371. Ellipse, § 404. Excess, spherical, of spherical polygon, § 49Q spherical, of spherical triangle, § 476. Faces of a polyedral angle, § 341. of a polyedron, § 348. Frustum of a pyramid, § 385. circumscribed, § 406. inscribed, § 406. of a cone, § 405. of a cone, inscribed, § 406. Geometry, solid, § 288. Great circle of a sphere, § 435. Hexaedron, § 348. Hyperbola, § 404. Icosaedron, § 348. Inclination of line to plane, § 340. 449 450 INDEX OF DEFINITIONS Lateral area of a prism, § 349, of a pyramid, § 383. Lateral edges of a prism, § 349. of a pyramid, § 383. Lateral faces of a prism, § 349. of a pyramid, § 383. Lateral surface of a cone, § 398. of a cylinder, § 371. Line oblique to a plane, § 302. perpendicular to a plane, § 302. Lune, § 487. Mid-section of a prismatoid, § 429. Nappes of a conical surface, § 398. Octaedron, § 348. Parabola, § 404. Parallel line and plane, § 312. planes, § 312. Parallelopiped, § 352. rectangular, § 352. right, § 352. Perpendicular planes, § 330. Plane angle of diedral angle, § 326. Plane perpendicular to a line, § 302. tangent to a cylinder, § 374. Planes, parallel, § 312. perpendicular, § 330. Poles of a circle, § 436. Polyedral angles, § 341. convex, § 341. equal and symmetrical, § 344. Polyedrons, § 348. regular, § 418. similar, § 421. Polygon, spherical, § 464. convex, § 464. Projection of line upon plane, § 338. of point upon plane, § 338. Prism, § 349. circumscribed, § 374. inscribed, § 374. oblique, § 349. quadrangular, § 349. right, § 349. triangular, § 349. Prismatoid, § 429. Pyramid, § 383. circumscribed, § 406. hexagonal, § 383. inscribed, § 406. quadrangular, § 383. regular, § 383. triangular, § 383. Radius of a sphere, § 432. Right section of a cylinder, § 371. of a prism, § 349. Section, conic, § 404. of a cylinder, § 371. of a polyedron, § 348. Sector, spherical, § 455. Segment, spherical, § 457. of one base, § 457. Slant height of a cone, § 400. of a frustum of a cone, § 405. of a frustum of a pyramid, § 387< of a pyramid, § 387. Small circle of a sphere, § 435. Solids, congruent, § 357. equal, § 357. geometric, § 347. Sphere, § 432. circumscribed, § 445. inscribed, § 445. Surface, conical, § 398. cylindrical, § 371. spherical, § 432. Tangent to a sphere, § 442. Tetraedron, § 348. Triangle, spherical, § 464. equilateral, § 464. isosceles, § 464. right, § 464. Triangles, birectangular, § 476. mutually equiangular, § 472. mutually equilateral, § 472. polar, § 468. symmetrical, § 477. trirectangular, § 476. Triedral angle, § 341. Truncated prism, § 351. pyramid, § 384. Vertex of a conical surface, § 398. of a polyedral angle, § 341. of a pyramid, § 383. of a spherical angle, § 459. Vertices of a polyedron, § 348. of a spherical polygon, § 464. Volume of a solid, § 356. Zone, § 450. of one base, § 450. 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. RECD LD \ SEP 21^^ 1 8Afre'fii<^H - r ! IN STACKS till ? 1QR1 - UUL ^"^ ^ loOl - 1 i 7 >tBl4 1966 . pWfci) AUG 1 1980 RECCIR. JUL 1 '80 LD 21A-50m-4,'59 (A1724sl0)476B General Library University of California Berkeley LD21-100m-7,'40 (6936s) THE UNIVERSITY OF CAUFORNIA LIBRARY