SOLID GEOMETRY 'BY"" '• CLAUDE IRWIN PALMER, A.B. ASSOCIATE PROFESSOR OF MATHEMATICS ARMOUR INSTITUTE OF TECHNOLOGY AND DANIEL POMEROY TAYLOR, A.M. Edited by GEORGE WILLIAM MYERS, Ph.D. PROFESSOR OF THE TEACHING OF MATHEMATICS SCHOOL OF EDUCATION THE UNIVERSITY OF CHICAGO SCOTT, FORESMAN AND COMPANY CHICAGO NEW YORK Q.A4-S7 COPYRIGHT 1918 BY SCOTT, FORESMAN AND COMPANY EDUCATION D£PT* 0tc PREFACE The purpose of this book is to present the subject of solid geometry so as to emphasize some of its applications as well as to give a thorough training in logical reasoning. To accom- plish this the authors have continued the same method of presentation that characterizes their Plane Geometry. The aim of the authors in the preparation of this text has been to present the subject in a logical manner that is com- prehensible to the youthful mind, and to vitalize the subject- matter — making it both interesting and useful through a wide range of practical applications. Some of the features peculiar to the Solid Geometry are out- lined in the following paragraphs: 1. In Chapter VI, the theorems on the relations of lines to planes and of planes to planes are carefully grouped so as to make them more easily . comprehended by the student. 2. Distinct advantage is gained by combining certain related subjects, thus securing brevity and simplicity of treatment. Prisms are combined with cylinders, pyramids with cones, and polyedral angles with spherical polygons. 3. In the theorems regarding the areas and volumes of solids, and requiring a use of limits, great care is taken to make the treatment logical and still keep it within the grasp of the aver- age student. 4. The large number and great variety of exercises are care- fully distributed, and range from some quite elementary to others of considerable difficulty. This enables the teacher to adapt the book to any group of students, whether in a classical or a technical high school. The abstract exercises give mental training and application of the basic theorems, while the prac- tical exercises are used to correlate geometric facts with real life. 54 10*85 i v PREFACE Many of the exercises involve an application of arithmetic and algebra to geometry. In addition to the four points outlined above, attention is called to the following features: While many theorems are proved in full, the proof of others is given only in part. In still others the method of proof is suggested only or the work is left entirely to the student. Thus a middle course is adopted in the use of the suggestive method. The Prismatoid Formula, one of the most powerful theorems of solid geometry, is proved in a simple direct manner, and is followed by various applications. However, this may be omitted and in no way interfere with the proofs of later theo- rems. The more important theorems are printed in heavy type. This is in accordance with the recommendation of the Com- mittee of Fifteen, whose report, as well as various other com- mittee reports, have been given careful consideration in the preparation of the entire work. For convenience, a combination ruler and protractor accom- panies each book. Acknowledgment is due to the McGraw-Hill Book Co., Inc., for permission to use exercises from Palmer's Practical Mathe- matics, in which are gathered numerous exercises from the author's many years of experience in teaching practical students. C. I. Palmer. Chicago, September, 1918. D. P. Taylor. CONTENTS P*GE Introduction 273 Chapter VI. Straight Lines and Planes Determining a Plane 278 Relative Positions of Lines and Planes 280 Intersecting Planes : 281 Lines and Planes, Parallel ■ 283 Lines and Planes, Perpendicular 288 Projections 293 Diedral Angles ' 295 Polyedral Angles 302 Loci 307 General Exercises 309 Chapter VII. Polyedrons, Prisms, Cylinders Definitions 313 Prisms and Cylinders 315 Areas of Prisms and Cylinders 319 Congruent and Equivalent Solids 324 Parallelepipeds 326 Volumes of Prisms and Cylinders 331 Similar Prisms and Cylinders 336 General Exercises 339 Chapter VIII. Pyramids and Cones Definitions and Theorems 343 Areas of Pyramids and Cones 349 Area of Frustum of Pyramid or Cone 353 Volumes of Pyramids and Cones 357 Similar Pyramids and Cones 365 General Exercises 367 Chapter IX. Prism atoids and Polyedrons Prismatoids 369 Polyedrons 373 General Exercises 379 v vi CONTENTS Chapter X. The Sphere and Polyedral Angles Page General Properties 381 Circles of Spheres 384 Distance on a Sphere 386 Relative Positions of Spheres 392 Measurement of the Sphere, Area 393 Volume of a Sphere 400 Spherical Angles 407 Spherical Polygons and Polyedral Angles »; 408 Polar Triangles • 415 Areas of Spherical Polygons 420 General Exercises 424 Formulas for Reference 427 Syllabus of Plane Geometry 429 Index 446 SOLID GEGME3;RX :V INTRODUCTION 528. Solid geometry may be thought of as an extension of plane geometry, with which it has much in common. The figures of plane geometry lie in a plane; while in solid geom- etry we consider the properties of figures that do not lie entirely in one plane. Such figures are sometimes called three- dimensional figures. 529. Space Ideas. We gain ideas concerning space and objects in space through the senses of touch, sight, and hear- ing. Mainly through touch and sight, we determine the size and shape of an object. 530. Difficulties. Most beginners in the study of solid geometry have difficulty in visualizing, or forming a mental image, of a three-dimensional figure from a worded descrip- tion or from a drawing in a plane. Even the same picture may present to the minds of two persons quite different images. For instance, in the figure, one person may see six blocks while another will see seven. 273 274 SOLID GEOMETRY A little practice, however, in, visualizing the blocks enables one at wil; to see six or seven blocks. Frequently a student does not understand a proposition or a proof because he does not visualize the figure properly. Lines may appear to extend in a different direction than was intended, or a plane may seem to lie in front of the page when it was intended that it should appear back of the page. Whatever device may be necessary for a clear image should be used; but care should be taken not to become too dependent upon models. A very important part of the training from the study of solid geometry is that it trains in visualizing three- dimensional figures from a description or a drawing. 531. Devices to Assist in Imaging. Cardboard models may be formed as described in § 753. Planes can be made to intersect by cutting two pieces of cardboard half in two and fitting together along the cuts. Cardboard can be used in connection with sharp pointed wires, such as hatpins. § 568. Hatpins or sharp wires held together by small corks can be used to form open models. § 571. Various figures can be formed with the hatpins and a piece of soft board. 532. Representation of Solid Geometry Figures. In solid geometry, the figures are usually represented on paper or on the blackboard, that is, in a plane. It is well to have some idea of how to make the figures appear solid, and to "stand out" as desired. While it is not thought best to enter into a discussion of perspective, a few general suggestions that are of assistance will be given. The figures may be supposed to be viewed from a point a little above, and either directly in front or a little to the right or left. Parts of the figure that are rectangles, in gen- eral, appear as parallelograms in the drawing. §§ 663-5, 671, 678. Polygons appear as polygons, but reduced in one direc- INTRODUCTION 275 tion. §§ 660, 749. Parts that are circles appear as ellipses, which become narrower as the eye of the observer is nearer to the plane in which the circle lies. §§ 689, 692, 864. Planes are supposed to be opaque, and all lines covered by them should be left out, or, if it is desired to insert them, they should be dashed lines. Two lines that intersect in the drawing do not necessarily intersect in the solid figure. Lines are not usually of the same length as they are drawn. The size of an angle may be quite different from the size in the drawing. In the accompanying drawings representing cubes, the one at the left appears to have the face ABCD in front and can hardly be imaged other- 71 7 H 71 wise. While the one at the right can easily be seen as if viewed from a point a little above and to the right, and so having the face ABCD in front; or it may be visualized as viewed from a point a little below and to the left, and so having the face EFGH in front. 533. The Use of Plane Geometry in Solid Geometry. All of the plane geometry is at the disposal of the student in solid geometry; but, in applying the facts of plane geometry, great care must be taken not to use them except where they will hold. Facts of plane geometry that hold without reference to any plane are the axioms and most definitions, theorems stating that figures are congruent, and all the theorems where it is evident that it does not matter if the figures are not in the same plane. Theorems of plane geometry that hold because the nature of the figure requires that it lie in a plane are those concern- 276 SOLID GEOMETRY ing a triangle, two parallel lines, a parallelogram, a circle, and all other theorems where the conditions stated determine a plane in which the parts must lie. In all other cases, the theorems of plane geometry can be applied only after all the parts concerned are shown to lie in one plane. Under these will fall many theorems concerning parallels, perpendiculars, circles, and polygons. EXERCISES 1. Draw the figure of § 603, omitting the lines DE, EF, and DF. Can you visualize the figure as otherwise than in one plane? 2. Draw the figure of § 572, making the side fines of plane P full. Com- pare with the figure of the text. 3. Draw the figure of § 628, making all lines full and of the same thick- ness. Compare with the figure of the text. 4. Draw the figure of § 671, making all lines of the same thickness. Does your figure appear as if viewed from below? If not, make such lines heavy as will make it appear so. 5. Draw a figure like the one at the right in § 722, but change the fines so that it will appear as if viewed from below. 6. Draw a figure of a sphere as in § 787, but viewed from a point on a level with the center of the sphere. * CHAPTER VI STRAIGHT LINES AND PLANES 534. Surface. A boundary of a portion of space is called a surface. A surface may be limited or unlimited in extent. It has two dimensions. 535. Plane. A surface such that a straight line joining any two of its points, lies wholly in the surface, is called a plane surface or, simply, a plane. Since a straight line may be unlimited in length, this defi- nition implies that a plane may be unlimited in extent, for otherwise the straight line could not lie wholly in the surface. It also follows from the definition that a straight line can intersect a plane in but one point. M 536. Representing a Plane. In drawing a geometric figure a portion of the plane is conveniently represented as a rec- tangle seen obliquely. In order to make the plane appear to "stand out," the front edge is frequently drawn longer than the back edge. Making the edges on the front and one end heavy and shading help one to image a plane as it is intended it should appear. 537. Reading a Plane. A plane may be read by a single letter, by letters at opposite vertices of the rectangle that represents it, or by the letters at all the vertices of the rec- tangle. Thus, in the figure, the planes may be read "plane M," "plane AC," or "plane ABCD." 277 STRAIGHT LINES AND PLANES 279 EXERCISES So that they do D 1. Hold two pencils so that they determine a plane, not determine a plane. 2. In making a kite, two straight sticks AB and CD are fastened together at P. Show that paper stretched over these sticks lies in a plane. 3. Show that a rubber cord fastened at both ends and stretched by grasping it at any other point, lies in one plane. 4. From a given point lines are drawn to points on a given line; show that all these lines lie in the same plane. 5. What is the locus of all lines that pass through a given point and intersect a given line not containing the point? 6. Show that, if a rubber band is stretched by three hooks in any position, all parts of the band he in one plane. 7. Do four points necessarily he in a plane? When do they? Give illustrations by using points in the classroom. 8. Why is a tripod used to support a camera or a surveyor's transit? 9. Why do the four legs of a chair sometimes not all rest upon a floor? 10. How many planes are determined by four points not all lying in the same plane? 11. Given five points four of which are in the same plane. How many planes do they determine? 12. How many planes are determined by three concurrent lines that do not all He in the same plane? 13. How does a carpenter determine whether a floor is a plane? 14. Why does a mason use a trowel with long straight edges when " truing up" a wall? 15. How many planes are determined by four lines all meeting in a point but no three of which he in the same plane? 16. Must a triangle lie in a plane? Must a parallelogram? Must a trapezoid? Must every quadrilateral? 17. Is the following a complete definition for a circle: A curved line every point of which is equally distant from a fixed point called the center is a circle? Why? 18. Through a point in space (1) draw a line parallel to a given line; (2) draw a line perpendicular to a given line. 280 SOLID GEOMETRY RELATIVE POSITIONS OF LINES AND PLANES 545. Definitions. Collinear means lying in the same line. Coplanar means lying in the same plane. 546. Relations of Two Lines. From the study of plane geometry and §§542, 543, it follows: (1) That two coplanar lines may coincide, intersect, or be parallel. (2) That two non-coplanar lines are neither intersecting nor parallel. 547. Relations of a Line and a Plane. A straight line may have three positions relative to a plane: (1) It may lie in the plane. (2) It may intersect the plane. (3) It may be parallel to the plane. A line lies in a plane if all of its points are in common with the plane. A line intersects a plane if it has only one point in com- mon with the plane. The point in common is called the foot of the line. A line is parallel to a plane, and the plane is parallel to the line, if they have no point in common, that is, if they do not meet. 548. Relations of Two Planes. Two planes may have three relative positions: (1) They may coincide. (2) They may intersect. (3) They may be parallel. Two planes are said to coincide if they have the same determining conditions. In plane geometry, the relative positions of points and lines are few and so the treatment of the subject is quite informal. In solid geometry, the study is made of points, lines, and planes, and their relative positions are many. It is advisable then, to make a more formal study of the subject. INTERSECTING PLANES 281 INTERSECTING PLANES 549. Definition. The intersection of two surfaces consists of all points common to the two surfaces, and no other points; that is, it is the locus of all points common to the two sur- faces. 550. Axiom. If two planes have one point in common, they also have another point in common. 551. Theorem. If two planes intersect, their intersection is a straight line. Given two intersecting planes P and Q. To prove that their intersection is a straight line. Proof. Let A and B be two points common to both planes. . § 550 The straight line determined by A and B lies in the plane P and also in Q. § 535 Then AB is common to the two planes. Why? Moreover, no point not in A B is common to the two planes for then they would coincide. § 540 .*. the intersection of planes P and Q is a straight line. EXERCISES 1. Can three planes intersect (1) in one line? (2) in two lines? (3) in ohree lines? (4) in more than three lines? Explain. 2. If a paper is folded and creased, why does it form a straightedge ilong the crease? 3. From a point external to two non-coplanar lines only one line can )e drawn that will cut the two lines. 282 SOLID GEOMETRY 552. Theorem. If three planes, not passing through the sam> line , intersect each other, their three lines of intersection either meet in a point or are parallel each to each. / ' / / ' / Given the three planes AD, CF, and AF not passing througlj the same line, and intersecting in lines AB, CD, and EF. To prove that AB, CD, and EF either meet in a point or ard parallel each to each. Proof. Consider the two lines of intersection lying in anj plane. (1) These lines intersect, or (2) they are parallel. (1) Suppose AB and CD intersect at P. Then P lies in plane AF and in plane CF, and hence in theij intersection EF. § 54fl .-. AB, CD, and EF meet in point P. (2) Suppose AB II CD. Then EF II CD or intersects CD. Suppose EF intersects CD, then AB intersects CD by (1) But this is impossible, for AB II CD. Hence EF II CD. Similarly it can be proved that EF II AB. .'. AB, CD, and EF are parallel each to each. EXERCISES 1. Three planes that do not contain the same straight line can hava but one point in common. 2. Explain the meaning of the expression, "Two planes determine a straight line." Is it always true that two planes determine a straight line? Is the analogous statement, " Two lines determine a point, " of plan^ geometry always true? Explain. 'F PARALLEL LINES AND PLANES 283 LINES AND PLANES, PARALLEL 553. Theorem. Two straight lines that are parallel to a third straight line are parallel to each other. A R Given lines A B and CD, each parallel to EF. To prove AB II CD. Proof. Either AB is parallel to CD or it is not. AB and EF determine plane AF. Why? CD and EF determine plane CF. Why? AB and C determine plane BC. Why? If AB is not parallel to CD, plane BC will intersect plane CF in some line CG other than CD. § 551 Then CG II EF. § 552 (2) But CD II EF. Given Hence CD and CG coincide. § 118 .-. AB II CD. Why? 554. Theorem. A plane containing one, and but one, of two parallel lines is parallel to the other. Given two parallel lines AB and CD, and plane P containing CD but not AB. To prove plane P II AB. Suggestion. AB and CD determine a plane. Suppose AB not parallel to plane P, and show that then AB would not be parallel to CD. EXERCISES 1. As a door is opened, show that all the positions of its outer edge are parallel to each other. 2. Prove that the lines connecting in order the middle points of the sides of a quadrilateral in space form a parallelogram. By a " quadrilateral in space" is meant a quadrilateral not all of whose sides lie in the same plane. It is often called a skew or gauche quadrilateral. D p> A B k Q ■■1 A ~Q — — B t -/ 284 SOLID GEOMETRY 555. Theorem. If a line is parallel to a plane, it is parallel to the intersection of that plane with any plane through the line. 556. Theorem. A line is in a plane if it has a point in the plane, and the line and the plane are both parallel to a second line. The two parallel lines determine a plane intersecting the first plane. 557. Theorem. If a line is parallel to the intersection of two planes, it is parallel to each of the planes. 558. Theorem. A line parallel to each of two intersecting planes is parallel to their intersection. a b A line through a point in the intersection of the two planes and parallel to the given line must lie in both planes by § 556. 559. Problem. Through a given line to pass a plane parallel to any other straight line not intersecting the given line. (1) How many planes can be passed through the given line and parallel to the other line if the given line is parallel to the other line? §§ 541, 554. (2) If the given line is not parallel to the other, construct a fine parallel to the given fine and through a point in the other. Then the intersecting lines determine a plane parallel to the given line. 560. Problem. Through a point to pass a plane parallel to each of two given non-coplanar lines. Given AB and CD, the two lines, and P, the point. Construction. Through P construct lines parallel to both AB and CD. These lines determine the plane parallel to A B and CD. Give proof. EXERCISES 1. If a line in one of two intersecting planes is parallel to the other plane, it is parallel to their intersection. 2. If a line and a plane are parallel to the same line, they are parallel to each other, or the line lies in the plane. A- C \ B l^ G L ^H 1 PARALLEL LINES AND PLANES 285 561. Definition. Two planes are parallel if they have no points in common, that is, if they do not meet however far they may be produced. 562. Theorem. If a plane intersects two parallel planes, the lines of intersection are parallel. Given plane PQ intersecting the two parallel planes MN and RS in AB and CD, respectively. To prove AB II CD. Suggestion. Since AB and CD are in the same plane PQ, they are parallel or intersect. Suppose that they intersect. What follows? 563. Theorem. Parallel lines intercepted between parallel planes are equal. 564. Theorem. A line is in a plane if it has a point in the plane and if the line and plane are both parallel to a second plane. Pass a plane through the line and intersecting the two planes. Then apply § 118. EXERCISES 1. If a line and a plane are parallel to the same plane, they are parallel to each other, or the line lies in the plane. 2. State the converse of the theorem of § 562. Is it true? 3. A board is cut across on a slant. What is the shape of the cut section? 565. Theorem. If two intersecting straight lines are parallel to a plane, the plane determined by these lines is parallel to that plane. Given the intersecting lines AB and CD, each parallel to plane Q, and the plane P determined by AB and CD. To prove P II Q. \^ V Proof. Either P is or is not parallel to Q. If P is not parallel to Q, it will intersect it in a line parallel to both AB and CD. § 555 But this is impossible. § 118 .-. P II Q. 286 SOLID GEOMETRY 566. Theorem. Through a given point one plane, and only one can be passed parallel to any given plane not containing the point. Pass a plane through the given point and intersecting the given plane. Through the point and in this plane draw a line parallel to the intersection of the two planes. This line is parallel to the given plane by"§ 554. In a similar manner draw a second line through the given point, and apply § 565. 567. Theorem. If two angles, not in the same plane, have their sides parallel, right side to right side and left side to left side, the angles are equal and their planes are parallel. Given Zx in plane P and Zz in plane Q, having AB II DE and CB II FE. To prove Zx = Zz, and P II Q. Proof. Take BA=ED, and BC = EF, and draw lines AD, CF, BE, AC, and DF. Both AE and CE are parallelograms. Why? Hence both AD and CF are equal and parallel to BE. Why? Then AD and CF are equal and parallel. Why? Therefore ADFC is a parallelogram.. § 160 And AC = DF. Why? Then AABC^ADEF. § 113 .'.Zx = Zz. Why? Also AB II Q, and CB II Q. § 554 .'. PWQ. § 565 EXERCISES 1. What is the locus of a line that passes through a given point, and i; parallel to a given plane? 2. If two congruent parallelograms lie in parallel planes and hav their sides respectively parallel, how many planes are determined in usinj their sides in all possible combinations? 3. The four lines in which two parallel planes intersect two other par- allel planes are parallel. 4. If two planes are parallel to the same line, their intersections with any plane through the line are parallel to each other. PARALLEL LINES AND PLANES 287 568. Theorem. // two straight lines are cut by three parallel planes, their corresponding segments are proportional. -L \ .--7 C \ / M Wt Given AB and CD two straight lines intersected by three parallel planes M, P, and Q, in the points A, E, and B, and C, F, and D respectively. AE To prove — = CF EB FD Proof. Draw AD intersecting plane P in G. Let the plane determined by AB and AD intersect plane P in EG and plane Q in BD. Then EG II £D. § 562 Similarly, if AD and CD determine a plane intersecting plane M in AC and plane P in ffiF, GF II AC. Why? 4^L4^ and— = — SB GD' an FZ) GD . 4^ = ^ Hence, §411 Why? 569. Theorem. If two straight lines are cut by a series of parallel planes, the corresponding segments are proportional. Exercise. A mine shaft passes for 180 ft. on a slant through two layers of rock. One, a layer of hard rock, is 72 ft. thick; and the other, a layer of soft rock, is 84 ft. thick. If it costs $4 per linear foot in the hard rock and $3 in the soft to make the shaft, find the cost of digging the shaft. 288 SOLID GEOMETRY LINES AND PLANES, PERPENDICULAR 570. Definition. A straight line is perpendicular to a plane if perpendicular to all straight lines in the plane and passing through its foot. The plane is also perpendicular to the line. A straight line is oblique to a plane when it meets the plane but is not perpendicular to it. 571. Theorem. // a line is perpendicular to each of two intersecting lines at their point of intersection, it is perpendicular to the plane determined by these lines. - ■ l f :■* A ? k » / Given AP J_ AB and AD at their point of intersection A, and plane Q determined by AB and AD. To prove AP perpendicular to the plane Q. Proof. In plane Q draw BD, any line not passing through A, and through A draw any line AC meeting BD in C. Prolong PA to E, making AE = AP; and draw BP, CP, DP, BE, CE, and DE. Then AB and AD are perpendicular bisectors of PE. Therefore BP = BE, and DP = DE. Then ABPD^ABED. And ZCBP = ZCBE. Further ACBP^ACBE. And therefore CP = CE. Hence AC _L PE, i.e., AP _L AC, any line in plane Q and passing through its foot. Why? .\ AP is perpendicular to the plane Q. § 570 Why? §205 §113 Why? §247 Why? PERPENDICULAR LINES AND PLANES 289 572. Theorem. All the perpendiculars that can be drawn to a straight line at a given point in the line, lie in a plane perpen- dicular to the line at the given point. Given DC any line J_ AB at C, and plane r— Q±ABatC. \ e \ 1 \ D To prove that all lines perpendicular to \ — , r AB at C lie in plane Q. ■ ' Proof. DC and AB determine a plane P that intersects plane Q in EC. Then EC JL AB. § 570 But DCA.AB. Given Therefore DC coincides with EC. Why? And DC lies in plane Q. .*. all lines J_ AB at C lie in plane Q. 573. Theorem. Through a given point one plane can be passed perpendicular to a given straight line, and only one. If the given point is in the given line, two perpendiculars to the line determine a plane perpendicular to the given line. If the given point is outside the given line, construct a perpendicular from the point to the line, and at its foot construct another perpendicular to the given line. These determine the perpendicular plane. To show that only one plane can be drawn in each case, suppose that there are two planes perpendicular and pass a plane through the given line and intersecting the two planes in two lines. What follows? EXERCISES 1. Show how a perpendicular to a plane can be determined by means of two carpenter's squares. 2. Each of three concurrent lines is perpendicular to each of the other two. Prove that each is perpendicular to the plane of the other two. 3. If a book partly open stands on a table, is the back of the book perpendicular to the table top? 4. One side of a right angle revolved about the other side as an axis generates a plane perpendicular to this other side. 5. Do the hands of a clock revolve in a plane? Why? 290 SOLID GEOMETRY 574. Theorem. From a given point outside perpendicular to the plane, and only one, can be Given plane Q and point A outside Q. To prove one line perpendicular to Q from A, and only one, can be drawn. Proof. In plane Q draw any line CD. Through A pass a plane P J_ CD. § 573 Let BF be the intersection of P with Q. Construct AB 1 BF, and EF II AB. In Q, construct GB II CF. Then CF _L EF. And ZABG = ZEFC. Hence AB J_ G£. Therefore Ai? J_ plane Q. Only one perpendicular can be drawn, for drawn, they would both be perpendicular to of their plane with Q, which is impossible. a plane one line drawn. §§ 312, 117 §117 Why? §567 Why? §571 if two could be the intersection Why? 575. Theorem. At a point in a plane one perpendicular can be drawn to the plane, and, only one. Let be the point in plane Q. Draw AB, any line in Q through 0. Construct plane P ± AB /c through 0. In P construct OE _L CD. Complete and give proof. 576. Theorem. The perpendicular to a plane is the shortest line from a point to a plane. Apply § 185. 577. Definition. The length of the perpendicular from a point to a plane is called the distance from the point to the plane. i, The distance between two parallel planes is the length of the segment of a common perpendicular lying between them. PERPENDICULAR LINES AND PLANES 291 EXERCISES 1. What is the locus of all points equidistant from a circle? 2. What is the locus of all points equidistant from the vertices of a tri- angle? 3. What is the locus of all points equidistant from two given points? 4. Find the length of the locus in a plane of all points 10 in. from a point 6 in. from the plane. 5. With a pair of compasses opened 13 in. show how a circle of 12 in. radius can be drawn. 6. A perpendicular smokestack is braced by three guy wires attached at a point 90 ft. from the ground. If these wires are 150 ft. long arid meet the ground at the vertices of an equilateral triangle, find the length of one side of the triangle. 7. Could a smokestack be fully braced with three guy wires? With two? Why? 8. If from the foot of a perpendicular to a plane a line is drawn perpen- dicular to any given line in the plane, the line joining the point of inter- section to any point in the perpendicular to the plane is perpendicular to the given line in the plane. 578. Theorem. Two planes perpendicular to the same straight line are parallel; and conversely, if one of two parallel planes is perpendicular to a straight line, the other is also. Given planes P and Q _L AB. I To prove P II Q. \ P c\ \ Outline of proof. If the planes were not parallel, they would meet. Two perpendicu- \q d \ \ lars could then be drawn to the line from a 5 point in their intersection. Conversely: Given plane P parallel to plane Q, and P ± AB, To prove Q _L AB. Outline of proof. AB is perpendicular to any line in P through its foot. Draw two such lines. The planes determined by these lines and AB will intersect Q in lines parallel respectively to the lines in P, and, therefore, will be perpendicular to AB, 292 SOLID GEOMETRY 579. Theorem. If two planes are parallel to a third plane, they are parallel to each other. Proof similar to that of § 124. 580. Theorem. If one of two parallel lines is perpendicular to a plane, the other is also; and conversely, two straight lines per- pendicular to the same plane are parallel. Given AB II CD, and AB perpendicular to plane P. To prove CD _L P. Proof. Through D draw any line DF in plane P. Through B draw BE in plane P and parallel to DF. Then ZA BE = ZCDF. § 567 But ZABE is a rt.Z. Why? Therefore ZCDF is a rt.Z and CD ± DF, any line in plane P through D. .'. CD ± P. Why? Conversely: a c e Given AB _L plane P, and CD J_ plane P. To prove AB II CD. Outline of proof. Through D draw ED II . AB. Then ED ± P, and therefore coincides / with CD by § 575. D P/ 581. Theorem. Two parallel planes are everywhere the same distance apart. Choose any two points in one plane and draw perpendiculars to the other plane, § 574. Then these lines are perpendicular to the first plane, § 578. Further these two lines are parallel, § 580, and determine a plane intersecting the two parallel planes in parallel lines, § 562. Then apply § 191. EXERCISES 1. Prove the theorem of § 553 by § 580. 2. A line cannot be perpendicular to each of two intersecting planes. PROJECTIONS 293 A p /Uj \ i B X Y^> \ PROJECTIONS 582. Definitions. The projection of a point upon a plane is the foot of the perpendicular from the point to the plane. The projection of a line upon a plane c is the locus of the projections of all points of the line upon the plane. Thus B is the projection of point A upon plane Q, and EF is the projection of line CD. 583. Theorem. The projection upon a plane of a straight line that is not perpendicular to the plane is a straight line. Given AB, a straight line not _L plane Q. To prove that the projection of AB upon Q is a straight line. Proof. From any point in AB, as E, draw EF ±Q. § 574 Plane P, determined by AB and EF, intersects Q in CD. § 551 From G, any other point in AB, draw GH J_ Q. § 574 Then GH II EF and lies in P. Why? Therefore its foot H lies in CD. Why? And the projections of all points of AB lie in CD. Furthermore, a perpendicular to Q at any point in CD will intersect AB. Why? Then every point in CD is the projection of a point in AB. .'. the projection of AB upon Q is a straight line. 584. Theorem. Of all oblique lines drawn from a point to a plane: (1) Those that have equal projections are equal. (2) Those that have unequal projections are unequal, and the one having the greater projection is the greater. Another statement of this theorem is: Of all oblique lines, drawn to a plane from a point in a perpendicular to the plane, those that cut off equal distances from the foot of the perpendicular are equal, and of those that cut off unequal distances from the foot of the perpendicular, the more remote is the greater. 294 SOLID GEOMETRY 585. Theorem. The. acute angle that a straight line makes with its projection upon a plane is the least angle that it makes with any line passing through its foot and lying in the plane. Given the straight line AB meeting plane Q at B, its projection CB in Q, and DB, any- other line through B and in Q. To prove ZABC < ZABD. Proof. Make BD = BC, and draw AC and AD. Then ZABC < ZABD. § 259 586. Definition. The acute angle that a straight line makes with its own projection upon a plane is called the inclination of the line to the plane, or the angle that the line makes with the plane. EXERCISES 1. State and prove the converse theorems to § 584. 2. How does the length of the projection of a line upon a plane com- pare with the length of the line when: (1) the line is parallel to the plane; (2) the line is perpendicular to the plane; (3) the line is neither parallel nor perpendicular to the plane? 3. Which is the greatest angle that an oblique line to a plane makes with any line in the plane and through its foot? 4. Parallel lines make equal angles with a plane. 5. If a straight line intersects two parallel planes it makes equal angles with the parallel planes. 6. Find the projection of a line 16 in. long upon a plane if the angle it makes with the plane is 45°. If 30°. If 60°. 7. What is the locus of a point in a given plane and equidistant from two given points not in the given plane? 8. A rectangle 8 in. by 12 in. is intersected along one of its diagonals by a plane. If the other diagonal makes an angle of 45° with the plane, find the distance from the extremities of this diagonal to the plane. Ans. 5.098 in. 9. The projection of a given line on each of two planes is a in length. Are the two planes necessarily parallel? Illustrate. DIEDRAL ANGLES 295 DIEDRAL ANGLES 587. Definitions. An angle formed by two intersecting planes is called a diedral angle. The planes are called the faces of the diedral angle, and their intersection is called the edge. If a plane Q revolves about an axis AB, and if a line CD _L AB is taken in Q, then the plane Q generates a diedral angle, and the line CD generates a plane angle since it revolves in a plane. When the plane angle is acute, right, obtuse, etc., the diedral angle generated in connection with it is acute, right, obtuse, etc. That is, the diedral angle and the plane angle are each formed by the same amount of turning. 588. Plane Angle. The plane angle generated by CD is called the plane angle of the diedral angle. It is formed by two lines, one in each face of the diedral angle, and perpen- dicular to the edge at the same point. A diedral angle is read by naming its faces, as the diedral angle QP; or it may be read C-AB-D; or simply AB. The words complementary, supplementary, adjacent, vertical, etc., when applied to diedral angles, have meanings corresponding to their meanings when applied to plane angles. 589. The following facts are readily deduced from the definitions: a (1) All plane angles of a diedral angle are equal. (2) If two diedral angles are equal, their plane angles are equal; and conversely. (3) Two diedral angles are proportional to their plane angles; that is, the plane angle of a diedral angle can be taken as the measure of the diedral angle. 296 SOLID GEOMETRY EXERCISES 1. By comparison with the corresponding terms in plane geometry, form definitions applying to diedral angles for right, oblique, vertical, adjacent, supplementary, alternate interior. How many of these can be illustrated by the leaves of an open book? 2. With reference to planes, what corresponds to the following axiom in plane geometry: Through a given point .only one straight fine can be drawn parallel to another straight line? 3. If one plane intersects another plane, the vertical diedral angles are equal. 4. If the sum of two adjacent diedral angles is two right diedral angles, the exterior faces lie in the same plane. 5. If two planes are cut by a third plane so as to make the alternate diedral angles equal, the two planes are parallel. 6. Take a piece of heavy paper, rule it as shown in figure (1), and cut along AO. Fold as shown in (2) and show how it may be used in measuring diedral angles. w. U) (2) 590. Perpendicular Planes. If two planes form a right diedral angle, the two planes are said to be perpendicular to each other. 591. Theorem. If a line is perpendicular to a plane, every plane containing this line is perpendicular to the given plane. Given CD perpendicular to plane PQ, and plane MN through CD intersecting PQ in AB. To prove that plane MN is perpendic- ular to plane PQ. Proof. CD _L AB. Why? In plane PQ through D, draw DELAB. Then ZCDE is the plane angle of the -P^ diedral ZM-AB-Q. 9 § 588 But ZCDE is a rt.Z. Why? Hence diedral ZM-AB-Q is a rt. diedral Z. § 589 (3) .*. plane MN is perpendicular to plane PQ. Why? DIEDRAL ANGLES 297 592. Theorem. // a straight line is parallel to a plane, any plane perpendicular to the line is perpendicular to the plane. Given AB II plane M, and plane P _L AB at C. To prove P _L M . Proof. From any point B in AB draw BG _L M. § 574 AB and BG determine a plane Q which intersects plane M in DG, and plane P in CD. Further DG II AB, and CD _L CB. Why? Then CD II BG. Why? Therefore CD 1 M. § 580 .-. P1M. § 591 EXERCISES 1. Through a given point pass a plane perpendicular to a given plane. How many such planes can be drawn? 2. If one line is perpendicular to another line, is every plane containing the first line perpendicular to the second? Prove. 3. A plane determined by a line and its projection upon a given plane is perpendicular to the given plane. 4. If three concurrent lines are each perpendicular to the other two, what can be proved concerning the planes determined by the lines? Illus- trate by the walls of a room. 5. If a plane is perpendicular to a line in another plane it is perpen- dicular to that other plane. 6. State the converse of the theorem of § 592. Is it a true theorem? Prove. 7. A circle is divided into eight equal parts by diameters. A plane containing one of these diameters makes an angle of 45° with the plane of the circle. Find the distance of each division point of the circle from the plane. 298 SOLID GEOMETRY 593. Theorem. If two planes are perpendicular to each other a line drawn in one of them perpendicular to their intersection is perpendicular to the other. Given plane MN J_ plane PQ, AB their intersection, and CD ± AB in plane MN. To prove CD _L plane PQ. Proof. Through D and in plane PQ draw DE _L AB. Then ZCDE is the plane angle of the right diedral ZM-AB-Q. Hence ZCDE is a rt.Z, and CD _L DE. Why? But CD J_ AB. Given .\ CD _L plane PQ. § 571 594. Theorem. If two planes are perpendicular to each other a perpendicular to one of them at any point of their intersection will lie in the other. Given plane MN _L plane PQ, and intersecting it in AB, afso CD JL plane PQ at any point D in AB. To prove that CD must lie in plane MN. Suggestion. In plane MN draw a line _L AB at D. Then this line is J_ plane PQ, and must coincide with CD. Why? 595. Theorem. If two planes are perpendicular to each other a perpendicular to one from any point in the other will lie in the other. Given plane MN J_ plane PQ, and intersecting it in AB, also CD J_ plane PQ from any point C in plane MN. To prove that CD must lie in plane MN. Suggestion. In plane MN draw a line _L AB from C. Then this line is _L plane PQ, and must coincide with CD. DIEDRAL ANGLES 299 596. Theorem. If each of two intersecting planes is perpen- dicular to a third plane their line of intersection is perpendicular to the third plane. ^ H Given planes P and Q, intersecting ^Til^"H^ in AB, and each _L plane M. To prove AB ± M. Proof. Either AB _L M or it is not. If AB is not J_ M, draw a line HB ± M at B. Then HB lies in both P and Q. § 594 Hence HB is the intersection of P and Q. Why? But Ai? is the intersection of P and Q. Given Therefore A B and i7J5 coincide. § 551 .-. AB J_ AT. 597. Theorem. Through any straight line not perpendicular to a plane, one plane, and only one, can be passed perpendicular to the given plane. Given plane P and AB a line not J_ P. To prove that one plane _L P can be passed through AB, and only one. Proof. From any point A in AB, draw AC LP. AB and AC determine a plane JL P. § 591 .*. one plane _L P can be passed through AB. If another plane _1_ P could be drawn through AB, then AB would be ± P by § 596. But AB is not J_ P. Given .*. only one plane J_ P can be passed through AB. EXERCISES 1. Discuss the theorem of § 597 for the case where AB lies in plane P. 2. If a plane is perpendicular to each of two intersecting planes, it is perpendicular to their intersection. 3. Any point in a plane containing the bisector of an angle, and per- pendicular to the plane of the angle, is equally distant from the sides of the angle. 300 SOLID GEOMETRY 598. Theorem. Between two straight lines not in the same plane one common perpendicular, and only one, can be drawn. Given AB and CD two non-coplanar straight lines. To prove that one common perpendicular to AB and CD, and only one, can be drawn. Proof. Through AB pass plane P II CD. § 559 Through CD pass plane EH JL P, and intersecting P in GH. §597 Then GH II CD. Why? And GH intersects AB in some point 0. Why? In plane EH draw NO _L GH at 0. Then NO _L CD and M) _L AB. Why? .*. one common perpendicular to AB and CZ) can be drawn. Suppose that there is another common perpendicular KL. In plane P draw LS II GH, and in plane EH draw 2J72 JL GH. Then LS II CD, and hence KL ± LS. Why? Therefore KL J_ plane P. Why? But this is impossible for KR J_ plane P. Why? Hence KL is not _L AB. ,\ only one common perpendicular to AB and CD can be drawn. 599. Theorem. The common perpendicular between two non- coplanar lines is the shortest line that can be drawn between them. Prove that NO is less than any other line KL, for NO = KR and KR < KL . 600. Definition. The length of the shortest line between two lines is called the distance between them. Compare with §§ 187-190, 577. DIEDRAL ANGLES 301 601. Theorem. Every point in a plane that bisects a diedral angle is equidistant from the faces of the angle. Given the plane R bisecting the diedral angle formed by the planes P and Q, and N any point in plane R. Also NC _L P and ND ± Q. To prove NC = ND. Proof. NC and ND determine plane CD. Why? Plane CD intersects P in CO, Q in DO, and R in NO. Plane CD LP and also 1 Q. § 591 Then plane CD ± AB. Why? And NO, CO, and DO are each _L AB. Why? Therefore ZNOC is the plane angle of diedral ZRP, and /.NOD is the plane angle of diedral ZRQ. Hence ZNOC = ZNOD. Why? Show ANOC^ANOD. .'.NC = ND. Why? 602. Theorem. Every point equidistant from the two faces of a diedral angle lies in the plane bisecting the angle. EXERCISES 1. Are both of the preceding theorems included in the following? The locus of a point equidistant from the faces of a diedral angle is the plane bisecting the angle. Explain. 2. From any point within a diedral angle perpendiculars are drawn to the faces. Prove that the angle formed by these perpendiculars is supplementary to the plane angle of the diedral angle. 3. The plane angle of a diedral angle is 120°. A point in the bisector of the diedral angle is 16 in. from the edge of the angle. Find the distance of this point from the faces of the angle. 302 SOLID GEOMETRY POLYEDRAL ANGLES 603. Definitions. When three or more planes meet at point they form a polyedral angle, or a solid angle. The polyedral angle is formed by a portion of the planes as shown in the figure. The point V in which all the planes meet is the vertex of the polyedral angle; the lines, AV, BV, and CV, of intersection of consecutive planes are the edges; the portions of the planes lying between the edges are the faces; and the angles in the faces between the edges are the face angles. A polyedral angle is read by the letter at the vertex, or by this letter together with a letter on each edge. Thus, the polyedral angle in the figure is read "polyedral angle 7, or V-ABC" 604. Parts of a Polyedral Angle. The face angles are AVB, BVC, and CVA. The diedral angles formed by the faces are the diedral angles of the polyedral angle. The parts of a polyedral angle are its face angles and its diedral angles. For convenience in representing a polyedral angle, a plane is often passed through it cutting all its edges. It should be noted that this plane is not a part of the polyedral angle 605. Convex and Concave Polyedral Angles. If a plane cuts all the edges of a polyedral angle, but not through the vertex, the intersections of the faces with this plane form a polygon. If this polygon is con- vex (§ 169), the polyedral angle is said to be convex. If the polygon is concave, the poly- CONVEX COJSCAVE edral angle is said to be concave. In this text only convex polyedral angles will be considered, unless otherwise stated. POLYEDRAL ANGLES 303 606. Classification. A polyedral angle that has three faces is called a triedral angle. The words tetraedral, pentaedral, hexaedral, etc., may be applied when the polyedral angle has four, five, six, etc., faces. 607. Congruent Polyedral Angles. If the corresponding parts of two polyedral angles are equal and arranged in the same order, the polyedral angles are congruent. Thus, in the figure, the face angles are equal and arranged in the same order, that is, ZAVB = ZA'V'B', ZBVC = ZB'V'C, and ZCVA = ZC'Y'A 1 . Also in the diedral angles, ZAV = ZA'V, ZBV = ZB'V, and ZCV = ZC f V, the arrangement is in the same order. If the corresponding parts of two polyedral angles are equal and arranged in the opposite order, the polyedral angles are said to be symmetric. These will be considered later (§ 836). EXERCISES 1. Does the size of a polyedral angle depend upon the lengths of its edges? How do the number of edges, the number of diedral angles, and the number of face angles of a polyedral angle compare? 2. Construct from stiff paper a triedral angle having face angles equal to 50°, 70°, and 90°. Can you tell the number of degrees in the diedral angles? The paper may be cut as indicated in the figure, folded along the dotted lines and pasted. 3. How many degrees in each of the face angles and the diedral angles of the polyedral angle formed by the walls of a room? 4. The face angles and diedral angles of a convex polyedral angle are each less than 180°. 5. Bearing the plane geometry definitions in mind, define vertical polyedral angles. Are they congruent or symmetric? Explain how, if it is possible to have two vertical triedral angles congruent. 6. Could a triedral angle have one right diedral angle? Could it have two? Three? Give illustrations. 304 SOLID GEOMETRY 608. Theorem. The sum of two face angles of a triedra angle is greater than the third face angle. Given the triedral ZV-ABC. To prove that the sum of two face angles is greater tha the third face angle. y Proof. Three cases arise: (1) all the face //i\ angles equal; (2) two face angles equal; (3) no \ \ \ two face angles equal. A / \\d Cases (1) and (2) are left to the student. j^LZ-^^ In case (3) suppose ZAVB is the greatest. / In the face A VB, construct ZAVD = ZAVC, take VD = VC and draw AB, AC, and BC. Then AA VC £ AA VD. § 24 And AC = AD. Why** AC+CB>AD+DB. § 182J Hence CB>DB. § 174J In ACVB and DVB, BV = BV, and CV = DV, but CB>DB. Therefore ZCVB > ZDVB. § 259 Then ZAVC + ZCVB >ZAVD+ ZDVB. § 173 .-. ZAVC+ZCVB>ZAVB. Or the sum of two face angles is greater than the third. • EXERCISES 1. Any face angle of a triedral angle is greater than the difference the other two. 2. State the theorems in plane geometry that correspond to the theorems of § 608 and Exercise 1. There is a close analogy between the plane triangle and the triedrj angle. Many theorems concerning plane triangles can be changed t theorems concerning triedral angles by replacing the word side by face angle, and the word angle by diedral angle. 3. State theorems for triedral angles analogous to the theorems of plan geometry that have to do with congruent triangles. 4. Referring to the figure of §608, ZAVC = 55° and ZCVB = 65 Make a statement regarding the number of degrees in ZAVB. POLYEDRAL ANGLES 305 609. Theorem. The sum of the face angles of any convex polyedral angle is less than four right angles. Given the convex polyedral ZV-ABCDEF. To prove ZAVB+ZBVC+ • ■ - ZFVA <4 rt.A. Proof. Let a plane intersect the edges of the polyedral angle in A, B, C, etc., and intersect the faces in AB, BC, CD, etc. Join any point in the polygon thus formed to the vertices of the polygon. Then ZVBA + ZVBOZABC. § 608 Similarly ZVCB+ZVCD>ZBCD, etc. Add these and the result is that the sum of all the base angles of the triangles with vertices at V is greater than the sum of all the base angles of the triangles with vertices at 0. Then, since the sum of all the angles of the triangles with vertices at V is equal to the sum of those with vertices at 0, the sum of the angles about V is less than the sum of the angles about 0. § 178 But the sum of the angles about = 4 xtA. Why? .-. ZAVB+ZBVC+ ■ • • ZFVA <4 rt.A EXERCISES 1. Can a polyedral angle have for its face angles three angles from an equilateral triangle? Can it have four? Five? Six? Why? 2. Can a polyedral angle have as face angles the angles of a square? Of a regular pentagon? Of a regular hexagon? Why? 3. Any face angle of a polyedral angle is less than the sum of the others. 4. The sum of the face angles of a polyedral angle is 300°. What is the greatest value that any one of the face angles may have? 5. One of the face angles of a triedral angle is 60°. What are the limitations on the size of each of the other face angles? 306 SOLID GEOMETRY 610. Theorem. Two triedral angles are congruent if the three face angles of one are equal respectively to the three face angles of the other, and arranged in the same order. Given triedral angles V and V, having ZAVB = ZA'V'B', ZBVC = ZB'V'C, ZCVA = ZC'V'A', and ar- ranged in the same order. To prove triedral ZV = triedral ZV. Proof. On the edges of triedral angles V and V lay off the six equal segments VA, VB, VC, V'A', V'B', and VC, and draw AB, BC, CA, A'B', B'C, and C'A'. Then there are three pairs of congruent isosceles triangles as follows: Why? Why? Why? Why? AAVB^AA'VB', ABVC^AB'VC, ACVA^AC'VA'. Hence AB=A'B', BC=B'C, and CA = C'A'. And AABCtSAA'B'C. Also A VAC and VAB are acute. From any point D in VA construct DE in face AVB and DF in face AVC, each J_ VA. These lines meet AB and AC respectively in E and F. Why? Draw EF. Take V'D'=VD and in a similar manner construct, D'E', D'F', and E'F'. Then, since AD=A'D' and ZDAE=ZD'A'E', AADE^AA'D'E' Hence AE =A'E' and DE = D'E'. Similarly AF=A'F' and DF = D'F'. Therefore AAEF^AA'E'F'. And EF=E'F'. Therefore ADEF^AD'E'F'. And ZEDF = ZE'D'F'. Hence diedral ZVA = diedral ZV'A'. Similarly, prove the other two pairs of diedral angles equal. .-. triedral ZFstriedral ZV. § 607 §97 Why? §247 Why? §113 Why? § 589 (3) LOCI 307 LOCI 611. As has been assumed on the previous pages, the idea of a locus here is the same as in plane geometry. Here, too, the proof of a locus theorem must, as in plane geometry, consist of two parts: (1) That all points in the figure satisfy the given conditions. (2) That all points that satisfy the given conditions are in the figure. In the place of (2) one may prove that all points not in the figure do not satisfy the given conditions. 612. In plane geometry it was noticed that usually a locus satisfying one condition is a line or group of lines, and a locus satisfying two conditions is a point or group of points. There, however, a further condition is assumed, namely, that the locus is confined to a plane. In solid geometry usually, one condition will confine the locus to a surface or group of surfaces, two conditions will confine the locus to a line or group of lines, and three conditions will confine the locus to a point or group of points. Here, as in plane geometry, one of the greatest benefits derived from the study of loci is through the imaging and con- structing figures, rather than through the proofs of the loci theorems. This, however, does not mean that the proofs should be neglected. EXERCISES 1. What is the locus of points equally distant from two given points? 2. What is the locus of points equally distant from three given points? 3. What is the locus of points equally distant from four given points? 4. What is the locus of points equally distant from two intersecting lines and in a given plane? 5. What is the locus of the end point of a line-segment of fixed length, that moves so as to remain parallel to a given line and have one end in a given plane? 308 SOLID GEOMETRY 6. What is the locus of points in a given plane equally distant from three given points not in a line? 7. What is the locus of points equally distant from two given parallel lines? 8. What is the locus of the middle point of a line-segment of fixed length, that moves so as to have its end points in two parallel planes? 9. What is the locus of a point equally distant from two parallel planes and equally distant from the faces of a diedral angle? 10. What is the locus of all lines that pass through a given point and are parallel to a given plane not containing the point? 11. What is the locus of the points within a diedral angle and equally distant from its two faces? 12. What is the locus of all points that are twice as far from a line in a plane as from the plane? 13. What is the locus of the points within a triedral angle and equally distant from its three faces? 14. What is the locus of the points equally distant from the edges of a triedral angle? 15. What is the locus of a line making equal angles with each of two intersecting lines? What other exercise in this list has the same locus as this? 16. What is the locus of points in one of two given non-coplanar lines equally distant from two given points in the other? 17. What is the locus of a point that moves so that its distance from one of two given parallel planes is to its distance from the other as 1 : 3? 18. What is the locus of a point whose distances from the faces of a diedral angle are respectively 5 in. and 8 in.? 19. What is the locus of the projections of a given point upon the planes containing a given line? Many questions concerning loci cannot be discussed in elementary geometry as they lead to forms not there considered, for in elementary geometry only a limited number of forms such as lines, circles, planes, cones, cylinders, and spheres are considered. The question: What is the locus of the middle points of all transversals of two non-coplanar lines? leads to none of these. In general, the locus of points equally distant from two different ele- ments, as a point and a line, a line and a plane, or two non-coplanar lines cannot be considered in elementary geometry. QUESTIONS 309 QUESTIONS 1. State the different relations that two lines may have to each other. That a line may have to a plane. That two planes may have to each other. 2. State the different ways of determining that a straight line lies in a plane. 3. What is the intersection of a line and a plane? Of two planes? Of three planes? 4. State the different ways of determining that a line is parallel to a plane. 5. State the different ways of determining that a line is perpendicular to a plane. 6. State the different ways of determining that two planes are parallel. 7. State the different ways of determining that two planes are perpen- dicular. 8. How many intersecting lines can be parallel to the same line? To the same plane? 9. How many intersecting planes can be parallel to the same line? To the same plane? 10. Must planes be parallel if they contain parallel lines? Are lines necessarily parallel if they are in parallel planes? 11. How many lines through a point can be perpendicular to the same line? To the same plane? 12. How many intersecting planes can be perpendicular to the same line? To the same plane? 13. What are the relations of the following to each other: (1) lines parallel to the same line; (2) planes parallel to the same plane; (3) lines parallel to the same plane; (4) planes parallel to the same line; (5) lines perpendicular to the same line; (6) lines perpendicular to the same plane; (7) planes perpendicular to the same line; (8) planes perpendicular to the same plane? 14. If two planes are perpendicular to each other, what lines in one are perpendicular to the other? 15. Can the projection of a straight line upon a plane be a curve? Can the projection of a circle be a straight line? Explain. 16. Can the projection of a square be a square? A rectangle? A parallelogram? Explain. 310 SOLID GEOMETRY GENERAL EXERCISES COMPUTATION 1. Two fixed points, A and B, are 10 in. apart. A point P moves so as to keep 8 in. from A and B. Find the length of the locus of P. 2. Two fixed points, A and B, are 12 in. apart. A point P moves so as to keep 10 in. from A and 4 in. from B. Find the radius of the circle generated. Ans. 2.5 in. 3. A 20-foot pole is placed obliquely in a river. One end of the pole is on the bottom of the river and the other end is 3 ft. above the surface. Find the depth of the river if the length of the part of the pole under water is 15 ft. 4. Plane Q is perpendicular to plane P. Find the x shortest line AED that can be drawn from a point A in Q to a point D in P if AB = 5 ft., CD = 10 ft., and BC = 90. ft. Suggestion. Consider the figure when Q is turned into the same plane as P. 5. The room shown in the figure has a length B < of 40 ft., a width of 20 ft., and a height of 12 ft. Find the length of the shortest line in the walls of A the room from B to H. 6. The sum of the face angles of a tri /\ / X- - F ,'V edral angle is 148°. What is the greatest value a face angle can have? 7. In the figure of a hammock support with the dimensions as given, find the length a"--- , Q^'d* oiABHAB = CB = DE = FE. U 8. The roof of a house makes an angle of 30° with A the plane of the plates. The roof is 18 ft. by 30 ft. Find the area of its projection upon the plane of the plates. 9. In the figure of a cube an edge is 18 in. and BE is 10 in. Find the area of ABCD. 10. Having given the dimensions as shown in the figure, find the area of section S to two decimal ®* places. All the face planes that meet are perpen- dicular to each other. GENERAL EXERCISES 311 11. The figure shows the plan for a half pitch roof. The dimensions are as given and the rafters are to be placed 1 ft. 6 in. from center to center. Find the lengths of the hip rafters, common rafters, and jack rafters. AB is a hip rafter, CD is a common rafter, and EF is a jack rafter. ko' \. 22' / SD Dech 1 a \ A E C THEOREMS AND PROBLEMS 1. If a series of parallel planes intercept equal segments on one straight line, they intercept equal segments on any other straight line that they intersect. 2. If a straight line and a plane are each perpendicular to the same straight line, they are parallel. 3. If two planes are perpendicular to each other, a line perpendicular to one and not in the other is parallel to the other. 4. Prove that the sides of an isosceles triangle make equal angles wfth any plane in which the base lies. 5. From a point within a triedral angle perpendiculars are drawn to the three faces. These perpendiculars are the edges of a second triedral angle. Prove that the face angles of one are supplementary to the diedral angles of the other. 6. Three planes, M, N, and Q, each perpendicular to the other two pass through a common point. Show that a point 5 in. from M, 8 in. from N, and 10 in. from Q may be in any one of eight positions. 7. Given two non-coplanar lines, to construct a plane upon which the projections of the two lines will be parallel. 8. A ray of light from the source L is reflected by the mirror M and enters the eye at E. The path travelled by the light is the shortest possible path from L to the mirror and then to E. Show how to determine where the ray of light strikes the mirror. 9. Three non-coplanar lines meet at a point. Construct a line that will make equal angles with the three lines. 10. Given a polyedral angle with four faces, to construct a plane that will intersect the four faces so that the intersection will be a parallelogram. Suggestion. Pass a plane parallel to one face and intersecting the opposite face. Determine a line in the first face equal to this intersection and parallel to it. These two fines are parallel and hence determine a parallelogram. 312 SOLID GEOMETRY 11. Prove that if a line is parallel to one plane and perpendicular to another, the two planes are perpendicular to each other. 12. If a plane be passed through a diagonal of a parallelogram, the perpendiculars to it from the extremities of the other diagonal are equal. Plane MN passes through diagonal BD of parallelogram ABCD. Prove perpendiculars AP and CQ are equal. 13. Describe how to locate a point that is 4 in. from face of a triedral angle, 5 in. from another face, and 3 in. from the third face. 14. If a series of parallel planes cut all the edges of a triedral angle, the intersections of the planes with the faces form similar triangles. 15. If from P, any point within the diedral angle A-BC-D, PM and PN are drawn perpen- dicular to the faces BD and AC respectively, and MS is drawn perpendicular to AC, then NS is perpendicular to BC. 16. The three planes bisecting the diedral angles of a triedral angle meet in a line. Suggestion. Show that two of them meet in a line, and then show that this line lies in the third plane. Compare this exercise with Exercise 13, page 308. 17. All points within a triedral angle and equally distant from its three faces, lie in the line of inter- section of the planes that bisect the diedral angles. 18. The planes through the bisectors of the face angle of a triedral angle and perpendicular to the planes of the respective faces, meet in a line. CHAPTER VII POLYEDRONS, PRISMS, CYLINDERS 613. In the present and following chapters will be consid- ered some of the solids most commonly observed in nature, and very frequently used in architecture, engineering, and the arts. Besides finding the areas and volumes of these solids it will be necessary to investigate the relations of their parts, and to study the plane figures formed when the solids are cut by planes. 614. Polyedrons. A polyedron is a solid entirely bounded by planes. The intersections of the bounding planes are the edges of the polyedron. The points in which the edges intersect are the vertices. The polygons bounded by the edges are the faces. The faces taken together make up the surface, and the area of this surface is the area of the polyedron. The : amount of space enclosed by the surface is the volume of the • polyedron. 615. Sections. The plane figure formed on a plane passing through a solid, and bounded by the intersections of the plane with the surface of the solid is called a sec- tion of the solid. It is evident that the sec- tion of a solid is a closed figure (§ 146), and that the section of a polyedron is a polygon. 313 314 SOLID GEOMETRY 616. Convex Polyedrons. A polyedron is convex if every section of it is a convex polygon. Otherwise it is concave. Only convex polyedrons are considered unless otherwise stated. 617. Prismatic and Cylindrical Surfaces. A moving straight line that always remains parallel to its original position, and intersects a fixed straight line, generates a plane. Why? N Plane Prismatic Surface Cylindrical Surface A moving straight line that always remains parallel to its original position, and intersects a broken line not coplanar with it, generates a prismatic surface. A moving straight line that always remains parallel to its original position, and intersects a plane curved line not coplanar with it, generates a cylindrical surface. AB is the original position of the moving line, and MN is the fixed line, 618. The moving line is called the generatrix, and the fixed line or curve the directrix. The generatrix in any position is called an element of the surface generated. 619. Closed Surface. If the directrix is a closed line, the prismatic or cylindrical surface is closed. 620. Sections of Prismatic and Cylindrical Surfaces. A plane cutting all the elements of a closed prismatic or cylin- drical surface cuts the surface in a closed line. The figure bounded by this closed line is a sec- tion. If the cutting plane is perpendicular to an element, the section is a right section; if not, it is an oblique section. Two sections made by parallel planes are parallel sections. In the figure, ABODE is a right section. PRISMS AND CYLINDERS 315 PRISMS AND CYLINDERS 621. A prism is the solid formed by a closed prismatic surface and two parallel cutting planes. 622. A cylinder is the solid formed by a closed cylindrical surface and two parallel cutting planes. 623* The sections formed by the cut- ting planes are called the bases of the prism or cylinder. It follows from § 620 that the bases of a prism are polygons. 624. The polygons formed on the pris- matic surface between the bases are called the lateral faces. Intersections of the lateral faces are lateral edges, and intersec- tions of the lateral faces with the bases are base edges. 625. The altitude of a prism, or cylinder, is the perpen- dicular distance between its bases. 626. The lateral area of a prism, or cyl- inder, is the prismatic, or cylindrical, surface between the bases. The total area is the lateral area together with the areas of the bases. In the figure read the bases, faces, lateral edges, base edges, altitude. 627. The following facts about prisms and cylinders are readily proved by the definitions and previous theorems: (1) The lateral edges of a prism, or the elements of a cylinder, are equal. (2) The lateral faces of a prism are parallelograms. (3) The right section of a prism, or cylinder, is perpendicular to all the lateral edges, or elements. (4) The section of a prism made by a plane parallel to a lateral edge is a parallelogram. (5) The section of a cylinder made by a plane containing two elements is a parallelogram. 316 SOLID GEOMETRY 628. Theorem. Parallel sections of a prism are congruent Given the prism MN, and the parallel sections AD and A'D'. To prove polygon AD = polygon A'D'. Proof. AB II A'B'. Why? Hence AB = A'B'. § 158 Similarly BC = B'C, CD = C'D', DE = D'E', EA=E'A'. Also ZEAB = E'A'B',ZABC = ZA'B'C, etc. § 567 The two polygons AD and A'D' are then mutually equilateral and mutually equiangular. They then can be proved congruent by superposition. .\ polygon AD = polygon A'D'. 629. Theorem. Parallel sections of a cylinder are congruent. Given the cylinder MN, and the parallel sections P and Q made by the planes. To prove P^Q. Proof. Take two points A and B on the perimeter of P, and let C be any third point of that perimeter. Draw the elements through A, B, and C to the points A', B', and C respectively in the perimeter of Q; and draw AB, BC, CA, A'B', B'C, and CA'. Then AABC^AA'B'C. Why? And, if P is superposed on Q with AB coinciding with A'B', C will fall upon C. Why? But C is any point in the perimeter of P. Hence every point in the perimeter of P will fall upon a point of Q, and conversely. .\P^Q. 630. Theorem. Every section of a prism, or cylinder, parallel to the bases is congruent to them. xzn PRISMS AND CYLINDERS 317 631. A right prism, or cylinder, is one whose bases are right sections. 632. An oblique prism, or cylinder, is one whose bases are oblique sections. 633. A prism is triangular, quadrangular, pentagonal, etc., according as its bases have three, four, five, etc., sides. 634. A regular prism is a right prism whose bases are reg- ular polygons. 635. A circular cylinder is a cylinder whose bases are enclosed by circles. If the bases are also right sections, the cylinder is a right circular cylinder. 636. The line joining the centers of the bases of a circular cylinder is called its axis. The radius of the base of a circu- lar cylinder is called the radius of the cylinder. 637. Theorem. A lateral edge of a right prism, or an ele- ment of a right cylinder, is equal to the altitude, §§ 620, 563. 638. Theorem. The lateral faces of a right prism are rec- tangles. Apply § 627 (2). 639. Theorem. // a rectangle revolves about me of its sides as an axis, it generates a right circular cylinder; and conversely. Show that the side opposite the axis remains parallel to ts original position and moves along a curve. Also show -hat the other two sides generate parallel bases that are circular, and are perpendicular to the elements. 640. Cylinder of Revolution. Since a right circular cylin- ler may be generated by revolving a rectangle about one of ts sides as an axis, it is often called a cylinder of revolution. 318 SOLID GEOMETRY 641. Plane Tangent to a Cylinder. A plane is tangent to a cylinder if it touches the cylindrical surface in one element only, 642. Theorem. If a plane is tangent to a circular cylinder, its intersection with the plane of a base is tangent to that base. Given the cylinder OO' and the tangent _ ^ plane P, meeting the cylinder in the ele- C ment AB, and intersecting the plane of base in DE. To prove that DE is tangent to base 0. Jj| Proof. DE has one point in common / Im, with 00. Why? L m Also DE has only one point in common with OO, for if it had another point as C in common with OO, plane P would have a point outside the element AB in com- mon with the cylinder, which is impossible. Why? DE also lies in the plane of OO. Why? .-. DE is tangent to OO. § 285 643. Theorem. The plane determined by a tangent to a base of a circular cylinder and the element through the point of contact is tangent to the cylinder. , EXERCISES 1. The bases of prisms, or cylinders, are congruent. 2. Would the theorem of § 628 be true if the sections were concave' 3. A line drawn parallel to the elements of a circular cylinder from the center of one base, intersects the other base at its center. 4. The axis of a circular cylinder is parallel to its elements. 5. The axis of a circular cylinder passes through the center of all sec-: tions parallel to the bases. 6. Every lateral edge of a prism is parallel to the plane determined by any other two lateral edges. 7. What is the locus of all points equidistant from a given straight line? 8. Show that the plane determined by an element of a circular cylinder and the center of one base will contain the axis. AREAS OF PRISMS AND CYLINDERS 319 AREAS OF PRISMS AND CYLINDERS 644. Theorem. The lateral area of a prism is equal to the product of the perimeter of a right section and a lateral edge. Given the prism MP', and its right section ABCDE. To prove S = pe, where £ denotes lateral area, p the perimeter of a right section, and e a lateral edge. Proof. Area of OMN' = e-AB. § 355 Similarly area of ON0' = e-BC, etc. Since the lateral area is the sum of the M faces, adding these gives: S = e(AB+BC+--EA), But AB+BC-\ EA=p. r.S = pe. Why? Why? Why? 645. Theorem. The lateral area of a right prism is equal to the product of the perimeter of its base and its altitude. That is, S = ph, where S denotes lateral area, p perimeter of base, and h altitude. 646. Definitions. An inscribed prism is a prism whose bases are respectively inscribed in the bases of a cylinder. The cylinder is then said to be circumscribed about the prism. A circumscribed prism is a prism whose bases are respectively circum- scribed about the bases of a cylinder, said to be inscribed in the prism. The cylinder is then 647. It is evident that the lateral edges of an inscribed prism are elements of the cylinder; and that the lateral faces of the circumscribed prism are tangent to the cylinder by § 641, 320 SOLID GEOMETRY 648. Area of Cylinder. As in §480, with the circle, th general idea of the measurement of a plane area cannot b( used here, for it is impossible to apply a plane unit of area tc a cylindrical surface and thus find its numerical measure. Practically, however, the lateral area of a right circular cylinder can be found by considering the surface as straight- ened out and measured S « 2Tf i as a rectangle having a length equal to the circumference of the; base, and with a width equal to the altitude of the cylinder. It remains to devise a method for measuring the surface o a cylinder, which is accurate and also geometric. 649. In the circular cylinder of base AB, a prism having i regular polygon for base is inscribed, and the num- ber of sides is doubled indefinitely according to § 463. The area of the cylinder is a constant and that of the prism is a variable. It is evident that the area of each prism is greater than the area of the prism of half the number of faces. It is also evident that, at all times, the area of the cylinde is greater than the area of each successive prism. By doub- ling the number of faces of the prism indefinitely, the are; continually increases, and can be made to differ from th* area of the cylinder by less than any assigned value, that is, can be made just as nearly equal to the area of a cylinder as desired^ Similarly, by circumscribing a prism having a regular polygon for base about a circular cylinder and doubling the number of faces indefinitely, the area continually decreases, and can be made to dif- fer from the area of the cylinder by less than any assigned value, that is, it can be made just as nearly equal the area of the cylinder as desired. AREAS OF PRISMS AND CYLINDERS 321 650. Volume. Further, it is evident that the method of the preceding paragraphs can be as readily applied to the volume of a cylinder as to its area. It is thus seen that the volume of an inscribed or circumscribed prism, having a regular polygon for base, as the number of faces is indefin- itely doubled, can be made just as nearly equal to the volume of the cylinder as desired. 651. From the preceding discussion, the following theorem may be considered established: Theorem. If prisms whose bases are regular polygons are inscribed in and circumscribed about a circular cylinder, and if the number of faces of the prisms is indefinitely doubled, (1) The perimeter of a right section of the cylinder is the common limit of the perimeters of right sections of the prisms. (2) The lateral area of the cylinder is the common limit of the lateral areas of the prisms. (3) The volume of the cylinder is the common limit of the volumes of the prisms. 652. Theorem. The lateral area of a circular cylinder is equal to the product of the perimeter of a right section and the length of an element. Given a circular cylinder. To prove S = pe, where 8 denotes lateral area, p the perimeter of a right section, and e the length of an element. Proof. Inscribe a prism, whose base is a reg- ular polygon, in the cylinder, and let S' denote its lateral area and p' the perimeter of a right section. By doubling indefinitely the number of faces of the prism, S'->,S, and p'->p. § 652 (1) and (2) Then p'e-^pe. § 485 (2) But S' = p'e, being variables that are always equal. § 644 ,',S = pe, §485(1) 322 SOLID GEOMETRY 653. Theorem. The lateral area of aright circular cylinder is equal to the product of its altitude and the circumference of its base. That is, S = ch } where S denotes lateral area, c circumference of base, and h altitude. EXERCISES 1. Show that the lateral area and the total area of a right circular cylinder of altitude h and radius r, are given by the formulas: S = 2Wrh, T = 2Trh + 2irr 2 = 2TTr{h+r). In the following use the notation of Exercise 1. 2. Given r = 8 and h = 10; find S and T. 3. Given r = 10 and S = 200; find h and T. 4. Given h = 16 and S = 256; find r and T. 5. Given r = 12 and T = 4800; find h and S. 6. Given ft = 25 and T = 4500; find r and S. 7. Given S - 2000 and T = 2200; find r and ft. 8. Solve the formula T = 27Tr/i+27rr 2 , (1) for r in terms of h and T, (2) for /i in terms of r and T 7 . 9. From the formulas of Exercise 1, find r in terms of S and T. 10. A peck measure made of sheet iron has a diameter of 8 in. and a depth of 10.7 in. Find the number of square inches of sheet iron in it. 11. Find the area (no cover) of a wash boiler , if the bottom is in the form of a rectangle with a semicircle at each end. The rectangle is 10 in. by 14 in., and the depth of the boiler is 16 in. 12. A steam boiler has a diameter of 72 in., is 18 ft. long, and contains 70 tubes each having ' a diameter of 4 in. extending lengthwise of the boiler. Find the heating surface of the boiler, using " one-half" in the rule below. Arts. 1505 sq. ft. Rule. In finding the heating surface of a horizontal boiler, it is cus- tomary to take one-half to two-thirds the lateral area of the shell, the lateral area of the tubes, one-half to two-thirds the area of the ends of the boiler, and subtract the area of both ends of the tubes. 13. Find the heating surface of a horizontal tubular boiler 12 ft. long, 5 ft. in diameter, and having 52 tubes 2\ in. in diameter. Use " two- thirds" in the preceding rule. Ans. 556.7 sq. ft. AREAS OF PRISMS AND CYLINDERS 323 14. The base of a right prism is a triangle whose sides are 12 ft., 15 ft., and 17 ft., and its altitude is 8§ ft. Find its lateral area and total area. 15. Find the lateral area of a regular hexagonal prism each side of whose base is 4 in. and whose altitude is 16 in. Find the total area. 16. Derive the formula S =0.2QlSdl, where 8 is the area of the surface of a cylindrical pipe in square feet, d the diameter in inches, and I the length in feet. 17. Six lines of steam pipes of diameter 2 in. extend along one side and an end of a room 40 ft. by 30 ft. Find the number of square feet of heat- ing surface. 18. Small metal boxes used for various purposes are made from sheet metal as follows: Blanks of the proper shape are first cut from the sheet metal. These are then pressed into the required form in a die. In computing the size of the blank, it is assumed that it has the same area as the finished article. Find the diameter of the blank for the cover of a tin pail for holding lard, if the cover has a diameter of 6§ in. and the flange is -| in. 19. A cylindrical metal box for holding paper fasteners is If in. in diameter and l| in. deep, and the cover has a J-inch flange. Find the diameter of the blanks for box and cover. Ans. 3.68 in.; 2.56 in. 20. What is the locus of points 12 in. distant from two parallel lines 18 in. apart? 21. Holes for rivets are often punched in metal plates. The pressure required for such work, when the material is wrought iron, is 55,000 pounds per square inch of the cut surface in. For example, a hole having a circum- ference of 2 in. punched in a J-in. plate would require a pressure of 2 X J X 55,000 lb., that is, the area of the cylindrical surface sheared off times 55,000 lb. Find the pressure required to punch a J-in. round hole through a piece of sheet iron ^--in. thick. 22. Find the pressure necessary to punch a ^-inch round hole through a boiler plate f in. thick if 60,000 pounds pressure is required per square inch of surface cut. 23. A rectangular box is to be made of sheet steel -^ in. thick. Find the number of pounds the punch press must strike in order to cut out the blank if it is rectangular 5f in. by 2f in., has squares -§- in. on a side cut from each corner, and has two holes y^- in. in diameter punched in the bottom. Use 60,000 pounds per square inch, Ans. 30,297 lb. 324 SOLID GEOMETRY CONGRUENT AND EQUIVALENT SOLIDS 654. Congruent Solids. Two solids are said to be congru- ent if they can be made to coincide completely in all their parts, 655. Equivalent solids are those having the same volume. 656. Theorem. Two prisms are congruent if three face, including a triedral angle of one are congruent respectively to three faces including a triedral angle of the other, and are arranged in the same order. Given prism AI and AT, with face AJ = face A'J', face AG^face A'G', and face AD^face A'D'. To prove prism AI = AT. Proof. ZBAF = ZB'A'F', ZFAE = ZF'A'E', and ZBAE = ZB'A'E'. Why? Then triedral ZA ^triedral ZA'. § 610 Place the prism A' I' upon prism A I so that triedral A A' shall coincide with its congruent triedral ZA. Then face A'J' coincides with its congruent face A J; A'G' with its congruent face AG; and A'D' with its congruent face AD. And points F' $ G', and J' fall upon F, G, and J respectively. Further C'H' coincides with CH. Why? Hence H' falls upon H. Similarly I' falls upon I. .\ prism AI = prism AT. § 654 657. Theorem. Two right prisms, or two right cylinders, are congruent if they have congruent bases and equal altitudes. CONGRUENT AND EQUIVALENT SOLIDS 325 658. Truncated Prism. A portion of a prism included be- tween the base and a section oblique to the base is called a truncated prism. 659. Theorem. Two truncated prisms are con- gruent if three faces including a triedral angle of one are congruent respectively to the three faces including a triedral angle of the other, and are arranged in the same order. 660. Theorem. An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is a lateral edge of the oblique prism. Given the oblique prism AD'; also the right prism GJ' whose base GJ is a right section of the prism AD', and whose altitude is equal to a lateral edge A A' of prism AD'. To prove prism AD' = prism GJ'. Outline of proof. Show that AG = A'G', BH = B'H', etc.; that the angles of face AH equal the angles of face A'H', and that face AH ^f ace A'H'. Likewise show face £I = face B'V , and face AD = face A'D'. Then prism AJ = prism A' J'. § 656 Prism AJ+prism GD' = prism A'J'+prism GD'. Why? .'. prism AD' = prism GJ'. Why? EXERCISES 1. Compare the theorem of § 660 with the theorem of § 354. 2. If a wooden beam has a rectangular right cross section, snow that if it is sawed lengthwise along a diagonal plane the two prisms formed are congruent. 3. In a truncated prism having a parallelogram for base, the sum of two opposite lateral edges is equal to the sum of a the other two opposite lateral edges. Prove a-\-c- 2e, and b +d = 2e. 326 SOLID GEOMETRY PARALLELEPIPEDS 661. A parallelepiped is a prism whose bases are parallel- ograms. 662. A right parallelepiped is a parallelepiped whose lateral edges are perpendicular to its bases. 663. An oblique parallelepiped is a parallelepiped whose lateral edges are oblique to its bases. 664. A rectangular parallelepiped is a right parallelepiped whose bases are rectangles. 665. A cube is a rectangular parallelepiped whose edges are all equal. Oblique Parallelepiped Cube Rectangular Parallelepiped 666. The dimensions of a rectangular parallelepiped are the lengths of the three edges drawn from one vertex. These dimensions are often called length, breadth, and height. 667. A diagonal of a parallelepiped is the line from any vertex to a vertex not in the same face. 668. The volume of any solid is the numerical measure of its magnitude in terms of some unit of measure. The unit of measure usually taken is a cube one linear unit on an edge. 669. Ratio between Solids. By the ratio of one solid to another is meant the ratio of their numerical measures. 670. Prove the following facts about parallelepipeds: ' (1) All the faces (including the bases) of a parallelepiped are parallelograms. (2) A parallelepiped is bounded by three pairs of parallel con- gruent parallelograms. PARALLELEPIPEDS 327 (3) A parallelepiped has three sets of four equal edges. (4) Any two opposite faces of a parallelepiped may be taken as bases. (5) All the faces of a rectangular parallelepiped are rectangles. (6) All the faces of a cube are squares. 671. Volume of Rectangular Parallelepiped. By a consid- eration analogous to that of § 344 the reasonableness of the following statement would be evident; here it is accepted without proof. The volume of a rectangular parallelepiped is equal to the pro- duct of its three dimensions. If V, a, b, and h are the numerical measures of the volume, length, breadth, and height, respectively, of any rectangular parallelepiped, then the above is stated in the formula V = abh. This means that, if the three dimensions are each measured in terms of some unit of length, then the volume is measured in terms of the corresponding unit of volume, which is a cube having an edge one linear unit in length. 672. Theorem. The volume of a rectangular parallelepiped is equal to the product of its base and altitude. 673. Theorem. The volume of a cube is equal to the cube of its edge. 674. Theorem. The volumes of two rectangular parallel- epipeds are to each other as the products of their three dimensions. V : V' = abh'.a'b'h'. 675. Theorem. The volumes of two rectangular parallel- epipeds having equivalent bases are to each other as their altitudes. v abh ; r ■ ;' \ v h bince — =— — — and ab=ab , then — =— . V abh V h 676. Theorem. The volume of two rectangular parallel- epipeds having equal altitudes are to each other as their bases. 677. Theorem. Two rectangular parallelepipeds having equivalent bases and equal altitudes are equal in volume. 328 SOLID GEOMETRY EXERCISES 1. Find the volumes of the following rectangular parallelepipeds: (1) 13 ft. by 27 ft. by 45 ft. (2) 2 ft. 3 in. by 1 ft. 7 in. by 3 ft. 1 in. (3) 30 ft. 6 in. by 41 ft. 6 in. by 12 ft. 2. A common brick is 8 in. by 4 in. by 2 in. Find the number of common brick in a pile 3 ft. by 6 ft. by 12 ft. 3. How many shoe boxes each 3 in. by 4 in. by 9 in. can be put in a packing box 3 ft. by 3 ft. 4 in. by 3 ft. 9 in.? 4. Find the number of cubic yards of earth to be excavated in digging a cellar 40 ft. by 26 ft. by 7 ft. 5. If 180 sq. ft. of zinc are required to line the bottom and sides of a cubical vessel, how many cubic feet of water will it hold? 6. A box car that is 36| ft. long and 8 ft. wide, inside measurements, can be filled with wheat to a height of 4^ ft. Find how many bushels of wheat it will hold if -f- cu. ft. are a bushel. 7. Find the cost at 40 cents a pound for sheet copper to line the bot- tom and sides of a cubical vessel 7 ft. on an edge, if the sheet copper weighs 12 oz. per square foot. Find volume. Ans. $73.50; 81.45 bbl. 8. Find the cost of laying a stone wall 45 ft. long, 6 ft. high, and 2 ft. thick at $2.75 a perch. Use 22 cu. ft. for one perch. 9. The edge of a cube is 10 in. Find the edge of a cube which shall have a volume twice as great. Eight times as great. 10. Show that the edge and diagonal of a cube can be used as the two sides of a right triangle whose acute angles are 30° and 60°. 11. Are the diagonals of a rectangular parallelepiped perpendicular to each other? Are those of a cube? Prove. 12. The diagonals of a rectangular parallelepiped are equal. 13. The diagonals of a rectangular parallelepiped bisect each other. 14. Find the sum of all the face angles of a parallelepiped. 15. Find the edge of a cube if its volume is increased 200 cu. in. when each edge is increased 2 in. 16. The edges of a rectangular parallelepiped are a, b, and c. Find the length of a diagonal, and the entire area of the parallelepiped. 17. Test the following rule if 1 cu. in. of iron weighs 0.28 pounds: The weight of iron bars in pounds per foot of length equals the width in inches times the thickness in inches times ^-. 18. The sum of the squares of the four diagonals of any parallelepiped is equal to the sum of the squares of the twelve edges. Apply Ex. 1 1, p. 181. PARALLELEPIPEDS 329 678. Theorem. The volume of any parallelepiped is equal to the product of its base and altitude. V = Bh. Given X an oblique parallelepiped, with area of base AC — B, and height h. To prove V = Bh, where V denotes volume. Proof. Extend DC and the other edges of X that are parallel to it. On DC extended, take EF = DC. Pass planes EG and FH through E and F and perpendicular to DC, forming a right parallelepiped Y, in which EG is a right section. Then parallelepiped X = parallelepiped Y. § 660 Extend IE and the other edges of Y that are parallel to it. On IE extended, take JK = IE. Pass planes JL and KM through J and K and perpendicular to IE, forming a rectangular parallelepiped Z. Then parallelepiped Y = parallelepiped Z. § 660 Therefore parallelepiped X = parallelepiped Z. § 104 But volume of Z = base JN X NM. § 672 In which NM = h. Why? And UJN = n}IF = OAC = B. Why? /. V = Bh. Why? 679. Theorem. Parallelepipeds having equivalent bases and equal altitudes are equivalent. 330 SOLID GEOMETRY 680. Theorem. Two parallelepipeds are to each other as the products of their bases and altitudes. 681. Theorem. Two parallelepipeds having equal altitudes are to each other as their bases. 682. Theorem. Two parallelepipeds having equivalent bases are to each other as their altitudes. EXERCISES 1. Compare theorems of §§ 678-682 with those of §§ 355, 358-361. 2. Use the notation of § 678, and show that in any parallelepiped V V 3. Find the length of the diagonal of a rectangular parallelepiped whose dimensions are 30 ft., 40 ft., and 12 ft. 4. A parallelepiped has a base in the form of a rhombus whose edges and one diagonal are each 10 in. Find the volume if the altitude is 8 in. 5. A parallelepiped has a rectangular base 8 in. by 15 in., and square ends. Find its volume if one of the sides is a parallelogram having an angle of 60°. 6. Find the diagonal of a cube whose volume is 512 cu. in. 7. Find the dimensions of a rectangular parallelepiped having a vol- ume of 12,960 cu. in., if the dimensions are in the ratio of 3 : 4 : 5. Ans. 18 in., 24 in., 30 in. 8. Find the edge of a cube whose surface and volume have the same numerical value. 9. A parallelepiped of altitude 8 in. has the same volume as a cube that is 12 in. on an edge. Find the area of the base and the dimensions of the base if it is a parallelogram having an angle of 60° and one side equal to 18 in. 10. Find the length of the bar that can be made from 1 cu. ft. of steel, if the bar has a rectangular cross section § in. by 1 \ in. 11. Find the volume to the nearest .001 cu. ft. of a rectangular tank 17 ft. 2 in. by 19 ft. 3 in. by 3 ft. 7 in. Ans. 1184.142 cu. ft. 12. Show geometrically that (a+6) 3 =a 3 +3a 2 6+3a6 2 +6 3 . 13. The total area of a right prism whose base is a rectangle is 118 sq. ft., the volume is 70 cu. ft., and the altitude is 7 ft. Find dimensions of the base. VOLUMES OF PRISMS AND CYLINDERS 331 VOLUMES OF PRISMS AND CYLINDERS 683. Theorem. The plane passed through the two diagonally opposite edges of a parallelepiped divides the parallelepiped into two equivalent triangular prisms. A " jB / Eh / ^^^ / Jp7 G / • r '—Jc Given the parallelepiped AC, divided into two triangular prisms A'-ABD and C'-BCD by the diagonal plane passing through the edges BB' and DD'. To prove prism A'-ABD = prism C'-BCD. Proof. Let EFGH be a right section of the parallelepiped, cutting the diagonal plane in FH. Face AB' II face DC, and face AD' II face EC. Why? Then EF II EG, and EH II FG. § 562 And EFGH is a parallelogram. Why? FHisa diagonal of OEFGH. Why? Hence AEFH^AFGH. § 154 Prism A'-ABD equals a right prism with base EFH and altitude AA'. § 660 Prism C'-BCD equals a right prism with base FGH and altitude AA'. Why? But these two right prisms are congruent and so equivalent. .-. prism A' -ABC = C'-BCD. § 104 EXERCISES 1. Find the edge of a cube that is increased in volume 127 cu. in. when its edge is increased 1 in. 2. A rectangular solid with a square base has a volume of 80 cu. ft., and a surface of 1 12 sq. ft. Find its dimensions. Ans. 4 ft. by 4 ft. by 5 ft. 332 SOLID GEOMETRY 684. Theorem. The volume of a triangular prism is equal to the product of its base and altitude. V = Bh. Given the triangular prism A' -ACT). To prove V = Bh, where V denotes vol- ume, B area of base, and h altitude. Suggestion. Complete the parallel- epiped A' -AGED. Show that prism A'-ACD=i parallelepiped A'-ACED. Apply § 683. 685. Theorem. The volume of any prism is equal to the product of its base and altitude. V = Bh. E < D , Given any prism AD'. J /*CS2^> C ' To prove V = Bh, where V denotes volume, / Ua B area of base, and h altitude. Suggestion. Show that the prism can be divided into triangular prisms by diagonal planes. Add the prisms thus formed. f< 686. Theorem. Prisms having equivalent bases and equal altitudes are equivalent. 687. Theorem. Prisms having equivalent bases are to each other as their altitudes. 688. Theorem. Prisms having equal altitudes are to each other as their bases. EXERCISES 1. Find the volume of a prism whose base is a regular hexagon 6 in. on a side, and whose altitude is 24 in. 2. In a prism given V = 226 and B = 43.6; find h. 3. The sides of a right section of a triangular prism are 4 in., 5 in., and 7 in. Find the volume if a lateral edge is 16 in. 4. The cost of digging a ditch, including all expenses and profits, is estimated at 27 cents a cubic yard. Find the cost of digging a ditch 15 mi. long, 10 ft. wide at the bottom, 20 ft. at the top, and 6 ft. deep. The cross section is a trapezoid Arts. $71,280. VOLUMES OF PRISMS AND CYLINDERS 333 5. Show that an oblique prism of wood may be changed into a right prism by cutting along a right section and interchanging the two parts. 6. One of the concrete pillars to support a floor in a concrete building is 12 ft. high and has as a cross section a regular hexagon 8 in. on a side. Find its weight if the concrete weighs 138 lb. per cubic foot. Arts. 1912 lb. 7. A flow of 300 gallons per second will supply water for a stream of what depth, if the stream is 4 ft. wide and flows 5 miles per hour? Ans. 16-H- in. 8. Find the number of cubic yards of crushed rock to make a road 1 mile in length and of cross section as shown in the figure. 9. One cubic inch of steel weighs 0.29 lb. An I-beam has a cross section as shown in the figure and a length of 22 ft. Find its weight. 10. An iron casting shrinks f in. per linear foot in cool- ing down to 70 degrees Fahrenheit. How many cubic inches is the shrinkage per cubic foot? Ans. 53.44 cu. in. 11. How many cubic yards of soil will it take to fill in a lot 50 ft. by 100 ft. if it is to be raised 3 ft. in the rear and gradually sloped to the front where it is to be 1^ ft. deep? Ans. 416§ cu. yd. 12. In a regular triangular prism, the edge of the base and the altitude are equal. Find these dimensions if the volume is 128\/3 cu. in. 13. In a regular triangular prism, the altitude is 6 in. more than the edge of the base. Find the dimensions if the volume is 224 \/3 cu. in. 14. Find the volume of a regular hexagonal prism whose base has an area of 37.5 sq. ft., and whose altitude equals an edge of the base. 15. The perpendicular drawn to the lower base of a truncated triangular prism from the intersection of the medians of the upper base, equals one third the sum of the three lateral edges. 16. The volume of any oblique prism is equal to the product of the area of a right section by the length of a lateral edge. 17. The volume of a regular hexagonal prism is 30\/3 cu. in. and its lateral area is 180 sq. in. Find its altitude and base edge. 18. The volume of a triangular prism is equal to the area of a lateral face times one-half the perpendicular drawn to that face from the opposite edge. 334 SOLID GEOMETRY 689. Theorem. The volume of a circular cylinder is equal to the product of its base and altitude. V = Bh. Given a circular cylinder. To prove that V = Bh, where V denotes vol- ume, B area of base, and h altitude. Proof. Inscribe a prism, whose base is a regular polygon, in the cylinder, and let V denote its volume and B' the area of its base. By doubling indefinitely the number of faces of the prism, V-+V, and B'->B. §§ 651 (3), 489 Then B'h-^Bh. §485(2) But V'=B'h, being variables that are always equal. § 685 .'.V = Bh. §485(1) EXERCISES 1. Show that the volume of a hollow cylinder with outer diameter D, inner diameter d, and altitude h, is given by the formula, V = lTTh(D*-d*) =lirh(D+d)(D-d). (See Ex. 8, p. 249.) 2. A cylindrical oil tank 3 ft. in diameter and 10 ft. long will contain how many gallons? (1 gal. =231 cu. in.) Arts. 528.8. 3. The outer diameter of a hollow cast iron shaft is 18 in.; and its inner diameter is 10 in. Calculate its weight if the length is 20 ft. and cast iron weighs 0.26 lb. per cubic inch. 4. A peck measure is to have a diameter of 8 in. How deep should it be? (lbu.= 2150.42 cu. in.) 5. Water is flowing at the rate of 10 miles per hour through a pipe of diameter 16 in. into a rectangular reservoir 197 yd. long and 87 yd. wide. Calculate the time in which the surface of the water in the tank will be raised 3 in. Arts. 31.38 minutes. 6. A certain handbook gives the following "rules of thumb" for find- ing the volume in gallons of a cylindrical tank: (1) V = (diameter in feet) 2 X 5 J X (height in feet) . (2) V = (diameter in feet) 2 X § (height in inches) less 2% of the product. Find the per cent of error for each rule. 7. Find the height of a 10-gallon wash boiler whose base is 10 in. wide with semicircular ends, the length of the straight part of the sides being 9J in. Arts. 13.5 in VOLUMES OF PRISMS AND CYLINDERS 335 8. A cylinder to cool lard is 4 ft. in diameter and 9 ft. long and makes four revolutions per minute. At each revolution, the hot lard cools upon the surface to a depth of j in. How many pounds of lard will it cool in one hour if 1 cu. ft. of lard weighs 56| pounds? 9. If a tank 5 ft. in diameter and 10 ft. deep holds 10,000 pounds of lard, what will be the depth of a tank of 2000 pounds capacity if its diam- eter is 3 ft.? If this tank has a jacket around it on the bottom and sides 3 in. from the surface of the tank, how many gallons of water will the space between the jacket and the tank hold? 10. A conduit made of concrete has a cross section with dimensions as shown in the figure. How many cubic yards of concrete are used in making one mile of this conduit? 11. The cylindrical water tower shown in the figure is at Long Beach, N.Y. Its diameter is 34 ft., height 150 ft., and it is said to have a capacity of 1,020,000 gallons. Is the capacity given correct? 12. Does a cylindrical water tank 42 in. in diameter and 14 ft. long hold 1000 gallons? 13. The following "rule of thumb" is used for finding the weight of round iron. The weight of round iron in pounds per foot equals the square of the diam- eter in quarter inches divided by 6. Find the per cent of error in using the rule if iron weighs 0.28 lb. per cubic inch. 14. How much water will a horizontal steam boiler 5 ft. in diameter and 16 ft. long with 70 tubes of diameter 3 in. running lengthwise, hold if one third of the volume is for steam? x ,„ 15. Find the height of a cylindrical oil tank with a diameter of 16 in. to hold one barrel. Ans. 36.2 in. 16. A tool steel ring for a steam cylinder is forged from round stock 3 in. in diameter. Find the length of stock to make a ring with the dimensions given in the figure. 17. The segment in the figure is a counter-balance 5 \ in. thick, its weight if made of cast-iron weighing 0.26 lb. per n cubic inch. Ans. 228.5 lb. 18. Find the weight of a hollow hexagonal bar 16 ft. long and weighing 0.28 lb. per cubic inch. The cross section is a regular hexagon l\ in. on a side, with a circle 1^ in. in diameter at the center. . . Ans. 152j lb. Find 336 SOLID GEOMETRY SIMILAR PRISMS AND CYLINDERS 690. Definitions. Two right prisms having similar poly- gons for bases, and whose altitudes are in the same ratio as two corresponding base edges, are similar. Two right circular cylinders are similar if their altitudes are in the same ratio as the radii of their bases. 691. Theorem. The volumes of two similar right prisms are in the same ratio as the cubes of corresponding base edges. Given two similar right prisms P and P', having altitudes h and h', bases B and B\ and two corresponding base edges e and e' ~ V h* e* To prove — = — = — . 17' j,/3 p n V W Proof. Hence But And Also V = Bh, and V '■- = B'h' V _Bh _B . h V B'h' B' ' h'' B _e 2 ¥'~e' 2 ' h e h'~7 V _e 2 e_ " V e' 2 'e'~ e 3 e' 3 ' V _h 3 V h'*' §685 §108 §446 §690 §111 §111 SIMILAR PRISMS AND CYLINDERS 337 692. Theorem. The volumes of two similar right circular cylinders are in the same ratio as the cubes of their altitudes, radii, or diameters. Given two similar right circular cylinders C and C", having volumes V and V, bases B and B', altitudes h and h', radii r and r', and diameters d and d' . V h* r 3 d 3 To prove — = — = — = r Iff L'3 „'Z Y' h f K r' 3 d' 3 The proof is similar to the preceding. r EXERCISES 1. Using the notation of § 691 for similar prisms, also S and S' for lateral areas, and T and T' for total areas, prove the following: ----- (2) ----- 2. Using the notation of § 692 for similar cylinders, also S, S', T, and 7" as in Exercise 1, prove the following: 3. Find the ratios of the volumes, the lateral areas, and the total areas of two similar right prisms having hexagonal bases two of whose corres- ponding edges are 2 in. and 5 in. respectively. 4. Two similar right circular cylinders have radii of 3 in. and 7 in. respectively. Find the ratio of their volumes. Of their lateral areas. Of their total areas. 5. Two similar right circular cylinders have bases whose areas are 20.25 sq. in. and 100 sq. in. respectively. Find the ratio of their volumes. 338 SOLID GEOMETRY 6. The total areas of two similar right circular cylinders are 625 sq. in. and 324 sq. in. respectively. Find the diameter of the second if the diam- eter of the first is 6j in. 7. A rectangle having dimensions of 10 in. and 15 in. is revolved, first about the side that is 10 in., and second, about the side that is 15 in. Find the ratio of the volumes of the two cylinders formed. 8. The number of feet of lumber in a log is often based upon a standard log usually 12 ft. long and 24 in. in diameter inside the bark at the small end. If v, d, and I are the volume, diameter, and length respectively of the standard log; and V, D, and L, the corresponding measurements of vD 2 L the log to be measured, then V= . Derive this formula. QUESTIONS 1. What is a prismatic surface? A cylindrical surface? Is the direc- trix necessarily closed? 2. In § 617, suppose the directrix is coplanar with the generatrix, what is the form of the surface generated? 3. What are the formulas or rules for finding areas of prisms and cylinders? What are the formulas or rules for finding the volumes of each of these solids? 4. Why is there such a close relation between theorems concerning prisms and cylinders? 5. Can you find the area of an oblique circular cylinder? Of an oblique prism? Can you find the volume of each of these? 6. Trace the steps in finding the volume of a parallelepiped. 7. Trace the steps in finding the volume of any prism. Of a cylinder. 8. What effect does it have upon the volume of a prism or cylinder if the base is doubled? If the altitude is doubled? If both base and alti- tude are doubled? 9. What effect does it have upon the lateral area of a right circular cy Under if the circumference of the base is doubled? If the area of the base is doubled? 10. Which do you consider the more common form in nature, the cyl- inder or the prism? Name some examples of each form. 11. Which is the more common of these forms in buildings and archi- tecture? In machinery? Give illustrations of triangular prisms, of quad- rangular prisms, of pentagonal, of hexagonal, etc. GENERAL EXERCISES 339 GENERAL EXERCISES COMPUTATIONS 1. Find the number of barrels each of the following cylindrical tanks will hold: (1) diameter 5 ft. and depth 5 ft., (2) diameter 20 ft. and depth 19 ft. (1 bbl. =3.211 cu. ft.) 2. Find the cost of laying the stonework, at $1.75 per cubic yard, in two abutments for a bridge, each abutment to be 8 ft. high, 3 J ft. thick, 20 ft. V-2&1U Top of Pier 3 ,L Ans. 28y sq. in. long at the bottom, and 15 ft. at the top. 3. Find the cost of common brick in the pier with a cross section as shown in the figure, and a height of 12 ft. 6 in., at $7.00 per thousand. Count 20 brick to 1 cu. ft. Ans. $24.15. 4. A stream flowing 5 miles per hour must be how large in cross section to supply water to a depth of 1 \ in. per week, for 160 acres of land? 5. A water tank in a Pullman car has a vertical section as shown in the figure, and a length of 52 in. Find its capacity in gallons. Ans. 68.3 gal. Consider the arc as a part of a cir-_l cle and apply formula (1) of § 509. 6. The flanges at the joining of two ends of flanged steam pipes 10 in. in inside diameter are bolted together by 14 bolts f in. in diameter. If the pres- sure in the pipes is 200 pounds per square inch, find what each bolt must hold. How much is this per square inch cross section of the bolts? Suppose that the bolts have 10 pitch U.S. standard thread. This makes the root diameter 0.620 in. Ans. 1122 lb.; 3718- lb. 7. Find the diameter of a cylindrical oil tank 40.5 in. high that is to hold 1 barrel. Ans. 15j in. 8. Representing the dimensions of a rectangular solid by x, y, and 2, find their values if when each is increased 2 in. the volume is increased 150 cu. in., the face of dimensions x and y is increased 18 sq. in., and the total surface is increased 110 sq. in. Ans. 3 in., 4 in., and 5 in. 9. Two cubes whose edges differ by 1 in. have volumes that differ by 397 cu. in. Find the edges of each cube. 340 SOLID GEOMETRY 10. A rectangular sheet of tin, 12 in. by 16 in., is made into an open box by cut- .. ting out a square from each corner and 83 turning up the sides. Find the size of the square cut out if the volume of the box is 180 cu. in. 11. A rectangular piece of tin a inches longer than it is wide is made into an open box, containing c cubic inches, by cutting from each corner a square of side b inches. Find the dimensions of the original piece of tin. 12. One edge of a rectangular box is increased 6 in., another 3 in., and the third 4 in., making a cube whose volume is 862 cu. in. greater than that of the original box. Find the dimensions of the box. Arts. 4.14 in. by 7.14 in. by 6.14 in. 13. If in a right prism the altitude is equal to a side of the base, find the volume if the base is an equilateral triangle whose sides are a. 14. Find the capacity in gallons of a water _^~~ 7 — tank for a locomotive tender. The dimensions J are as given in the figure for the length cross section, and the width is 9 ft. 6 in. 15. A certain coal and coke company finds it necessary to construct a wharf wall 150 ft. long, and having a cross section with dimensions as shown in the figure. Find the number of cubic yards in the wall. By " batter 1:12" is meant that the slant is 1 ft. in a vertical rise of 12 ft. 16. In a right prism whose volume is 54, the lateral area is four times the area of the base which is an equilateral triangle. Find the edge of the base. Ans. 6. 17. The total areas of two similar cylinders of revolu- tion are 75 sq. in. and 192 sq. in. respectively. If the volume of the first cylinder is 250 cu. in., what is the volume of the second? 18. In a right circular cylinder, given V —TTr 2 h and T = 2lTr 2 -{-2Trh. Find T in terms of V and h. _ 27 Ans. 19. In a right circular cylinder, find V in terms of the circumference c of the base and the total area T. Ans. V=- 2wcT-c 3 87T 2 20. Find the steam capacity of a horizontal cylindrical boiler 4 ft. in diameter and 16 ft. long, if the height of the segment occupied by the steam is 18 in. (Use (2) of §509). GENERAL EXERCISES 341 21. A cylindrical tank of diameter 30 in. and 34 in. long rests on its side. Find the number of gallons of gasoline in the tank if the depth is 5j in. Ans. 13 gal. nearly. 22. The volume of an irregular shaped body is often found by immersing it in water and determining the amount of water displaced. A cylindrical vessel that has a diameter of 4 in. is partly filled with water. A stone immersed in the water raises its level 3 \ in. Find the vol- ume of the stone. 23. A silo is used to keep fodder in a green and succulent state for feed- ing farm animals. It is usually built in a cylindrical form. Find the capacity in tons of a silo in the form of a right circular cylinder 20 ft. in diameter and 32 ft. high, if a cubic foot of silage weighs 40.7 lb. 24. An 18 acre field yields 11.5 tons of silage per acre. What must be the height of a cylindrical silo 20 ft. in diameter to hold all the silage if 1 cu. ft. of silage weighs 38.4 lb.? 25. A silo is in the form of a right circular cylinder, and is 20 ft. in diameter inside and 32 ft. high. How many cubic yards of concrete did ' it take to build it, if the floor and wall are each 6 in. thick, and the founda- I tion wall is 8 in. thick and 5 ft. deep? ^ 18 '1 > h-sK— 26. The connecting rod with dimen- P ■ — \\» sions as given in the figure is made from Lr stock with the dimensions shown. Find length of stock that it is necessary Connecting Rod to allow for the cylindrical part of the rod. stock 27. Find the volume of the beveled washer with dimensions as shown in the figure. 28. A pine log 2 ft. in diameter and 16 ft. long is floating in water. Find the weight of the log if two-thirds of the volume of the log is under water. (Water weighs 62.5 lb. per cubic foot.) 29. Will a floating pine log lj ft. in diam- eter and 10 ft. long support a man weighing 180 pounds if the specific gravity of the log is 0.72? 30. The following is the record of a test in which a high speed drill removed 70.56 cu. in. of cast-iron per minute. The penetration per minute was 57 J in., the feed y^- in. per turn, and the number of turns per minute was 575. Do these numbers agree? (25" 5 LU 342 SOLID GEOMETRY THEOREMS AND PROBLEMS 1. Derive the following formulas for finding th3 volume, V, of a hollow circular cylinder of length h and cross sectional dimensions as given in the; figure: (1) V = 7Th(R+r)(R-r). (2) V = lTTht(D+d). (3) V = Tht(d+t). (4) V = 7Tht(D-t). 2. Construct a plane through a point and tangent to a given right circular cylinder. 3. If the radius of one cylinder is equal to the alti- tude of a second, and the radius of the second is equal to the altitude of the first, what is the ratio of their volumes? 4. The intersection of two planes each tangent to a circular cylinder is parallel to the elements of the cylinder. 5. Prove that the volume of a right circular cylinder is equal to its lateral area times one-half the radius of its base. 6. The volume of two right circular cylinders are equal. Write a proportion between their lateral areas and their radii. 7. If a straight line has morethan two points common to the curved surface of a right circular cylinder, the fine is an element of the surface. CHAPTER VIII PYRAMIDS AND CONES 693. A moving straight line that always contains a fixed point, and always intersects a given straight line, generates a plane. Why? §542 694. Pyramidal Surface. A moving straight line that always contains a fixed point, and always intersects a broken line not coplanar with it, generates a pyramidal surface. 695. Conical Surface. A moving straight line that always contains a fixed point, and always intersects a plane curved line not coplanar with it, generates a conical surface. 696. The fixed point is called the vertex of the pyramidal, or conical, surface. 697. The two parts of the pyramidal, or conical, surface on opposite sides of the vertex are called nappes. The words generatrix, directrix, element, and closed surface have the same significance here as in §§618, 619. The directrix is not necessarily closed, but in this text only those that are closed are considered. The word " section" has the same significance as before (§ 620), how- ever, the cutting plane must not pass through the vertex. 343 ' 344 SOLID GEOMETRY 698. A pyramid is the solid formed by cutting all the ele- ments of one nappe of a closed pyramidal surface by a plane. 699. A cone is the solid form by cutting all the ele- ments of one nappe of a closed conical surface by a plane. The meanings of the words: base, lateral face, lateral edge, base edge, lateral area, total area, are apparent from the definitions of §§ 623, 624. 700. The altitude of a pyramid, or cone, is the perpendic- ular distance from the vertex to the base. 701. Pyramids Classified According to the Number of Lat- eral Faces. Pyramids, like prisms, are classified as triangular, quadrangular, pentagonal, etc., according as their bases are triangles, quadrilaterals, pentagons, etc. 702. Regular Pyramids. A pyramid whose base is a regu lar polygon, and whose altitude meets the center of its base, is called a regular pyramid. The altitude of a regular pyramid is called its axis. The altitude of one of the lateral faces of a regular pyramid is called the slant height. In the figure, OF is the altitude, BV a lateral edge, B C and NV the slant height. Regular Pyramii 703. Circular Cones. A cone whose base is a circle called a circular cone. The line joining the vertex of a circular cone to the center of its base is called the axis of the cone. 704. A right circular cone is a circular cone whose axis is ,. , ., , Circular Right-Circular perpendicular to its base. Cone Cone PYRAMIDS AND CONES 345 The radius of a circular cone is the radius of its base. Since a right circular cone may be generated by revolving a right triangle about one of its sides as an axis it is often called a cone of revolution. The length of an element of a right circular cone is its slant height. 705. Prove the following facts concerning cones , and pyramids: (1) The lateral faces of a regular pyramid are congruent isos- celes triangles. (2) The lateral edges of a regular pyramid are equal. (3) The elements of a right circular cone are equal. (4) The axis of a right circular cone coincides with its altitude. (5) A straight line drawn from the vertex of a cone to any point in the perimeter of its base is an element. (6) The section of a circular cone made by a plane containing an element is a triangle. (7) The section of a pyramid made by a plane through its vertex is a triangle. EXERCISES 1. Find an element of a right circular cone whose altitude is 15 and radius 7. 2. Find the altitude of a right circular cone whose slant height is s and radius r. 3. Find the altitude of a regular pyramid each face and the base being an equilateral triangle 10 in. on a side. 4. Into how many parts do the nappes of a conical surface divide space? 5. Why is it stated in the definition of a pyramidal surface that the vertex and directrix must not be coplanar? 6. The altitude of a right circular cone is 10 in. and the radius of the base is 6 in. Find the area of a section made by a plane passing through the vertex and 3 in. from the center of the base. In the figure find the area of VAB. 346 SOLID GEOMETRY 706. Theorem. The lateral edges and the altitude of mid, or the elements and the altitude of a cone, are divided tionally by a plane parallel to its base. a pyra propor- Pass a plane through the vertex parallel to the base and apply § 568. 707. Theorem. The section of a pyramid made by a plane parallel to the base is similar to the base. Given pyramid V-AD, cut by plane P parallel to base AD, forming the section FI. To prove polygon Fl^polygon AD. Proof. AFVG^AAVB, AGVH^ABVC, etc. Why? ^ FG VG . GH VG Then = , and — = , etc. AB VB BC VB Hence Also §428 JF FG_GH_HI_IJ _ AB~ BC CD DE EA ZJFG = ZEAB, ZFGH = ZABC, etc. .'. polygon W^polygon AD. Why? §567 §§ 441, 43| EXERCISES 1. The lateral edges of a pyramid are respectively 15 in., 12 in., and 16 in. Find the parts of each made by a plane that is parallel to the base and which divides the altitude into parts that are in the ratio of 1 : [I 2. The area of the base of a pyramid is 125 sq. in. Find the area of a section 8 in. from the base and parallel to it, if two corresponding edges of the base and the section are respectively 10 in. and 8 in. PYRAMIDS AND CONES 347 708. Theorem. The section of a circular cone made by a "plane parallel to its base is a circle, the center of which is the point where the axis intersects it. Given a circular cone V-ABC, with section EH made by the plane P parallel to the base. Also given axis VO piercing plane P in 0'. To prove that EH is a circle having 0' as center. Proof. Take F and G any two points in EH. M Then the planes determined by FO'V B and GO'V intersect the plane of the base in OB and OC, the plane P in O'F and O'G, and the lateral surface in the elements VB and VC, respectively. Why? Prove that AFO'V^ABOV, and AGO'V~ ACOV. FO'_GO' BO~ CO' But BO = CO. Why? Therefore FO' = GO' Why? Since F and G are any two points in the section EH, all points in this section are equally distant from 0' . .*. section EH is a circle with center 0' '. § 263 709. Theorem. The area of any section of a pyramid, or circular cone, parallel 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 of the pyramid, or cone. Then Why? v EXERCISES 1. The area of the base of a pyramid is 64 sq. in. and its altitude is 20 in. Find the area of a section parallel to the base and 4 in. from the vertex. 2. The diameter of the base of a circular cone is 6 in. and the altitude is 10 in. Find the area of a section parallel to the base and 6 in. from the vertex. 348 SOLID GEOMETRY 710. Theorem. 2/ two pyramids, two circular cones, or a pyramid and a circular cone, have equal altitudes and bases of equal area, sections made by planes parallel to the bases and at equal distances from the vertices, have equal areas. Suggestion. Let the equal altitudes be h and the distance from the vertices to the sections h! . Then area of any section is to area of base as h' 2 : h 2 . § 709 But it is given that the bases are equal, and the altitudes equal. Therefore the sections are equal in area. Why? EXERCISES 1. In the figure of §707, VO' = 8, 0'0 = 12. F£ = 25, and VA = find VG and VF. 2. If the base of the pyramid of § 707 has an area of 180 sq. in., using the lengths given in Exercise 1, find the area of the section. 3. The area of the base of a pyramid is 150 sq. cm. and its altitude is 2 dm. Find the area of a section 8 cm. from the vertex. 4. The areas of parallel sections of a pyramid, or of a circular cone, are to each other as the squares of their distances from the vertex. 5. The altitude of a pyramid is h. At what distance from the vertex must a plane be passed parallel to the base so that the section shall be (1) one-half as large as the base? (2) One-third? (3) One-fifth? 6. Each side of the base of a regular hexagonal pyramid is 8 in. the altitude of the pyramid is 20 in., how far from the vertex must a plane parallel to the base be so that the area of the section shall be 24 -\/3 sq. in.? 7. The area of the base of a cone is 216 sq. in. and its altitude is 24 in. Find the distance between two sections parallel to the base, which have areas of 144 sq. in. and 72 sq. in., respectively. If AREAS OF PYRAMIDS AND CONES 349 AREAS OF PYRAMIDS AND CONES 711. Theorem. The lateral area of a regular pyramid is equal to half the product of the perimeter of the base and the slant height. S = \ps. Given the regular pyramid V-ABCDE. To prove S = ^ps, where S denotes lateral area, s slant height, and p perimeter of base. Proof. The lateral faces are congruent tri- angles. Why? The area of each face triangle = |sX its base. Why? The sum of the bases of the triangles = p. Hence the sum of the areas of the faces = ^ps .'.S = %ps. Why? Why? EXERCISES as 2S 1. Using the notation of §711 show that s= — , and p= — . p s 2. Find the lateral area and the total area of a regular hexagonal pyra- mid whose altitude is 8 in. and base edge is 4\/3 in. 3. Find the lateral area and the total area of a regular triangular pyramid having base edges equal to 8 in. and lateral edges equal to 12 in. 712. Plane Tangent to a Cone. A plane is tangent to a cone if it meets the conical surface in one ele- ment only. 713. Theorem. If a plane is tangent to a circular cone, its intersection with the plane of the base is tangent to the base. Proof similar to that of § 642. 714. Theorem. The plane determined by the tangent to the base of a circular cone and the element through the point of contact, is tangent to the cone. Compare with theorem of § 643. 350 SOLID GEOMETRY 715. Inscribed and Circumscribed Pyramids. A pyramid is said to be inscribed in a cone if the pyramid and cone have the same vertex, and if the base of the pyramid is inscribed in the base of the cone. The cone is then said to be circum- scribed about the pyramid. 716. A pyramid is said to be circumscribed about a cone if the pyramid and cone have the same vertex, and if the base of the pyramid is circumscribed about the base of the cone. The cone is then said to be inscribed in the pyramid. It is evident that the lateral edges of an inscribed pyramid are elements of the cone; and that the lateral faces of the circumscribed pyramid are tangent to the cone by §§712, 715. 717. Area of Cone, Practical Consideration. If the lateral surface of a right circular cone is covered with paper, this paper when peeled from the cone can be spread out and will be a sec- tor of a circle, whose radius is the slant height of the cone, and whose arc is equal to the circumference of the base of the cone. Then, by § 508, the lateral area of a right circular cone is equal to half the product of the slant height and the circumference of the base. S = ^sc=irrs. AREAS OF PYRAMIDS AND CONES 351 718. Geometric Treatment. A method for measuring the surface of a cone, which is accurate and geometric follows from a consideration similar to that of § 649. In this manner the truth of the following statement is established. The lateral area of a right circular cone is the common limit of the areas of inscribed and circumscribed regular pyramids, as the number of faces is indefinitely doubled. 719. Theorem. The lateral area of a right circular cone is equal to half the product of the circumference of the base and the slant height. S = ^cs=7rrs. A Given a right circular cone. To prove S = ^cs= wrs, where S denotes lateral J area, s slant height, c circumference of base, Mi and r radius. IMl Proof. Circumscribe a regular pyramid MmA^^^^\ about the cone, and let S' denote its lateral vfiw/ luuOT^ area and p the perimeter of its base. By doubling indefinitely the number of faces of the pyramid, S'->S, and p->c. §§ 718, 488 Then \ps-*\cs. § 485 (2) But S'=^ps, being variables that are always equal. § 711 ,\S=4<». §485(1) Also c = 27rr. §495 .\ S=Trrs. ' §111 Remark. It is to be noted that the formulas of this article apply only to right circular cones. 720. Theorem. The total area T of a right circular cone is given by the formula T =Trrs+irr 2 =7rr(s+r) . EXERCISES 1. Form a cone from a semicircle having a radius of 3 in. Find the lateral area of the cone. The total area. 2. Show how to cut a pattern for a tin cone that is to have the diame- ter of its base 4 in. and its slant height 10 in. 352 SOLID GEOMETRY 2S S 2S % 8 3. In a right circular cone show that s = — = — , c— — , and r= — . c 7Tr s 7Ts 4. Solve T = Wrs+Trr 2 and find r = , and T-7Tr\ T 27r 7Tr 7Tr In a right circular cone, using S for lateral area, T for total area, s for slant height, r for radius, c for circumference, and h for altitude, solve the following : 5. Given r = 4 in., s = 6 in.; find S and T. 6. Given c = 25 in., s = 8 in.; find S and T. 7. Given s = 8 ft. 6 in., c = 3 ft. 7 in. ; find S. 8. Given s = 7.2 in., r = 5.3 in.; find S and T\ 9. Given h = 12 in., r = 5 in.; find S and T. 10. Given c = 30 in., s = 8 in.; find h and &. 11. Given £ = 200 sq. in., r = 6 in.; find s. 12. Given £ = 256 sq. cm., s = 16 cm.; find r. 13. Given T = 500 sq. cm., r = 10 cm.; find s. 14. The circumference of the base of a conical church steeple is 35 ft. and the altitude is 73 ft. Find the lateral area. Ans. 1281— sq.ft. 15. Find the lateral edge and the lateral area of a regular pyramid each side of whose triangular base is 10 ft., and whose altitude is 18 ft. Ans. 18.90+ ft., 273.45- sq. ft. 16. A tower with a regular hexagonal base has dimensions as shown in the figure. The pitch of the pyramdial roof is 1 \> which means that OV = 1^ QR. Find the lateral area of the tower. 17. Cut out of heavy paper a sector of a circle, with a radius of 3 in. and the central angle 120°. Bring the edges together and paste them. Find the lateral area and the total area of the cone thus formed. 18. The cone formed by revolving an equilateral triangle about one of its altitudes has a lateral area equal to twice the U ' area of the base. 19. Find the area of the surface of the solid formed by revolving an equilateral triangle, having a side of 12 in., about one of its sides. 20. Find the area of the surface of the solid formed by revolving a right triangle, having a base of 8 in. and an altitude of 6 in., about its hypotenuse. AREA OF FRUSTUM OF PYRAMID OR CONE 353 AREA OF FRUSTUM OF PYRAMID OR CONE 721. Truncated Pyramid, or Cone. The portion of a pyra- mid, or cone, included between the base and a section not parallel to the base is called a truncated pyramid, or cone. 722. Frustum. The portion of a pyramid, or cone, included between the base and a section parallel to the base is called a frustum of the pyramid, or cone. 723. The base and the parallel section are called the bases of the frustum. The perpendicular distance between the bases is the altitude of the frustum. 724. The slant height of a frustum of a regular pyramid, or of a cone, is that portion of the slant height of the pyra- mid, or cone, included between the bases of the frustum. Thus AB and CD are slant heights, and QP and MN are altitudes. 725. Midsection. If a solid has parallel bases, the section parallel to the bases and half way between them is called the midsection. a ~ — ^f In the figure, NR is the midsection. 726. Prove the following facts concern- ing frustums: (1) .The lateral faces of a frustum of a D ^ regular pyramid are congruent isosceles trapezoids. (2) The lateral edges of a frustum of a regular pyramid are equal; and the slant height is the same for all the faces. S54 SOLID GEOMETRY 727. Theorem. The lateral area of a frustum of a regular pyramid is equal to half the product of the slant height and the sum of the perimeters of the bases. S = ^s(P+p). Given the frustum of a regular pyramid with bases AC and EG. To prove that S = %s(P+p), where 8 denotes lateral area, s slant height, and P and p the perimeter of the lower and upper base respectively. Proof. Hence But Area AF=±s(AB+EF). Avesi BG = ^s(BC+FG). etc. S = %s[(AB+BC+---) + (EF+FG+- AB+BC+ • • • = P, and EF+FG+ • • • .•.s=4*(p+p). =p. 728. Theorem. The lateral area of a frustum of circular cone is equal to half the product of the slant and the sum of the circumferences of the bases. Given the frustum of a right circular cone. To prove S=ir(R+r)s, where S denotes lateral area, s slant height, and R and r the radius of the lower and upper base respec- tively. Proof. Let m be the slant height of the cone and n the slant height of the part above the frustum. Then >S = mirR — rnrr = ir (mR — nr) But R : r = m : n, And therefore mr — nR = . Hence S=ir(mR — nr-\-mr — nR) —ir{R-\-r) (?n — 7i). But m — n = s. .\S**ir{R+ty$. §370 §105 Why? Why? a right height Why? §428 §398 Why? Why? Why? AREA OF FRUSTUM OF PYRAMID , OR GC^T?' ;, | ; J^&x i 729. Theorem. The lateral area of a frustum of a right cir- cular cone is equal to the product of the altitude and the circumfer- ence of a circle whose radius is the perpendicular at the midpoint of an element, and terminated by the axis. ^ r Outline of proof. Let a be the length of EF, \\v\\ the perpendicular at the midpoint of an element / hD \ r>JU and terminated by the axis. Let r' be the / Er ± \\ radius of the midsection, and s and h the slant AllLi ClA->* height and altitude respectively. Show that ADEF^AABC. Then s : h = a : r', and sr f = ah. Show that R+r = 2r', where R and r are the radii of the bases. Then, substituting in the formula of § 728, S = 2irah. EXERCISES 1. Prove the theorem of § 728 by considering the lateral area of a frustum of a right circular cone as the limit of the area of a frustum of pyramid. 2. Prove the theorem of § 727 by using a method similar to that of § 728 for the frustum of a cone. 3. Show that the lateral area of a frustum of a right circular cone is equal to the slant height times the perimeter of the midsection. 4. Find the lateral area of the frustum of a regular quadrangular pyramid, if an edge of the lower base is 16 in., an edge of the upper base 12 in., and the slant height 18 in. 5. Find the total area of the frustum of a right circular cone, if the radii of the bases are 6 ft. and 8 ft. respectively, and the slant height is 7 ft. 6. A portion of the roof of a tower is a frustum of a right circular cone. The radii of the bases are 10 ft. and 6 ft. respectively, and the alti- tude is 8 ft. Find the number of square feet in this part of the roof. 7. The altitude of a right circular cone is h. How far from the vertex must a plane be passed parallel to the base so that the lateral area of the cone cut off shall equal the lateral area of the frustum. Arts. \hy/2. 8. The diameter of the bottom of a pail is 10 in. and that of the top 12 in. Find the number of square inches of tin in the pail if the slant height is 11 in. Arts. 458.67. : 856 1 -.SOLID GEOMETRY 9. Determine the diameter of the blank to make a pressed basin of the form shown in the figure. The depth is 3 in., the bottom has a diameter of 6 in., the top an inside diameter of 7 in., and the rim is J in. wide. Ans. 11.4+ in. Suggestion. Consider the rim as a ring between two concentric circles. The blank must have an area equal to the total area of the basin. 10. A pie tin has a diameter of 7 \ in. at the bottom and 9 in. at the top inside. It is 1 in. deep and has a flange J- in. wide around the edge. Find the diameter of the blank required to make the tin. 11. Find the area generated by revolving a square a inches on a side about a diagonal. 12. A tower whose cross section is a regular octagon 6 ft. on a side has a roof that is full pitch. Find the area of the roof. By full pitch is meant that the vertex of the roof is the same distance above the plates as the width of the tower from face to face. 13. The conical roof over a water tank is half pitch. Find its area if the diameter of the tank is 12 ft. and the roof projects 1^ ft. at the eaves. Ans. 221.5- sq.ft. By half pitch is meant that the vertex of the roof is one-half the diameter of the tower above the top of the tower. 14. Find the total area of the solid generated by revolving an isosceles trapezoid about a line connecting the middle points of its bases, if the bases are 8 in. and 12 in. respectively, and the angles at one base are each 60°. 15. Determine how to draw a IV c C pattern for the upper part of the v funnel with dimensions as shown, the diameter at A being J in. and at B, 1 in. The allowance for the seam can be added. Suggestion. The length OP = x may be determined as follows: x :z+3 = l : 4. .-. x = l. The radius to use is therefore 4 in. To determine the central angle u, QnR _ u ~87r~~360° But QnR = 47T. .\u = 180°. 16. Determine how to draw a pattern for the lower part of the funnel shown in the figure of the previous exercise. VOLUMES OF PYRAMIDS ANfr CONES VOLUMES OF PYRAMIDS AND CONES 3o7: 730. Inscribed Prisms. If the altitude of a triangular pyramid is divided into equal parts by a series of planes par- allel to the base, the prisms within the pyramid and having the sections formed by the parallel planes as bases, and each having one of the segments of one of the lateral edges of the pyramid as one of its lateral edges, are called a set of inscribed prisms. In the figure M , N, and R are a set of inscribed prisms, each being entirely within the pyramid. 731. Circumscribed Prisms. In a similar manner, the prisms having the base of the pyramid and the sections as bases are called a set of circumscribed prisms. In the figure, S, T, U, and V are a set of circumscribed prisms, each being partly outside of the pyramid. 732. The two sets of such prisms, formed by using the same lateral edge and the same parallel planes, are called corre- sponding sets of inscribed and circumscribed prisms. 733. The following statement may be considered as evident from the preceding: The volume of a triangular pyramid is definite, and is greater than that of any set of inscribed prisms, and less than that of any set of circumscribed prisms. ; ggg ! ; : / ' •, ' ! ; \ : $OLID GEOMETRY 734. Theorem. The difference in volume between a set of inscribed prisms and the corresponding set of circumscribed prisms is the volume of the prism upon the base of the pyramid. 735. Theorem. The volume of a triangular pyramid is the common limit of the volumes of the set of inscribed prisms and the corresponding set of circumscribed prisms, as the number of these prisms is increased indefinitely. Given the triangular pyramid D-ABC, a set of inscribed prisms, and corresponding set of circumscribed prisms. To prove that volume of D-ABC is the common limit of these two corresponding sets of prisms. t^p^d Outline of proof. The volume of the triangular pyramid is greater than the sum of the inscribed prisms and less than the sum of the circumscribed prisms. § 733 Therefore, the volume of the pyramid differs from either by an amount less than the volume of the prism upon the base of the pyramid. § Why? But the volume of this prism can be made as small as desired, that is, it has zero for a limit. § 485 (2) Hence, the volume of the set of inscribed prisms and the set of circumscribed prisms, each have the volume of the triangular pyramid for a limit. § 483 EXERCISES 1. Corresponding sets of inscribed and circumscribed prisms are formed in a triangular pyramid whose base has an area of 25 sq. in. Compute the series of differences between the corresponding sets when their alti- tudes are successively 0.1 in., 0.01 in., 0.001 in., and 0.0001 in. 2. The base of a pyramid is an equilateral triangle 3 in. on a side and its altitude is 7 in. Find the difference between the corresponding sets of inscribed and circumscribed prisms when the altitude of each prism is TWTT5 °f the altitude of the pyramid. 3. Could a quadrangular pyramid be used instead of a triangular pyramid in the discussion of §§ 730-733? VOLUMES OF PYRAMIDS AN1> CONES 359 736. Theorem. The volumes of two triangular pyramids, having equivalent bases and equal altitudes, are equivalent. Given the triangular pyramids O-ABC and O'-A'B'C ', having equivalent bases and equal altitudes; and, for convenience, having their bases in the same plane. To prove that V=V, where V denotes the volume of O-ABC and V the volume of O'-A'B'C. Proof. Let the common altitude ON be divided into any number of equal parts, and pass planes through the points of division parallel to the plane of the bases. Construct the set of circumscribed prisms for each pyramid upon the sections formed by these planes. Then the sections of the pyramids made by each plane are equivalent. § 710 Let DEF and D'E'F' be the sections made by any one of these planes, and P and P' the corresponding prisms. Then volume of P = volume of P'. § 686 Let S denote the volume of the sum of the prisms in the set of which P is one; and S' that of which P' is one. Then S = S'. § 105 But S->7, and S'->V. § 735 And S and S' are variables that are equal for all their succes- sive values as the number of the divisions of ON is increased indefinitely. .\V=V. §485(1) S0OCJ: ' %SOLID GEOMETRY 737. Theorem. The volume of a triangular pyramid is equal to one-third the product of its base and altitude. V = \Bh. Given the triangular pyramid O-PQR. To prove that V = \Bh, where V denotes the volume of the pyramid, B the area of its base, and h its altitude Proof. Upon the base PQR construct a triangular prism NR of altitude h, and having its lateral edges parallel to OQ The prism NR is divided into three triangular pyramids by the sections OPR and ONR. Pyramid R-NOM = pyramid O-PQR. § 736 Pyramid R-OPQ = pyramid R-ONP. § 736 But pyramid R-OPQ is the same as pyramid O-PQR. Therefore the three triangular pyramids are equal, and O-PQR is one-third the volume of prism NR. But volume of prism NR = Bh. § 68 /. V = \Bh. Why 738. Theorem. The volume of any pyramid is equal to one-third the product of its base and altitude. V = \Bh. Suggestion. From one vertex of the base draw all the diagonals of the base. Pass b { planes through the vertex of the pyramid and each of these diagonals. All the pyramids have m the same altitude. Apply § 737 to each triangular pyramic formed and add the results. VOLUMES OF PYRAMIDS AND CONES 361 739. By a like treatment to that referred to in § 650, the truth of the following statement may be established: The volume of a circular cone is the common limit of the volumes of inscribed and circumscribed pyramids with regular polygons for bases, as the number of faces is indefinitely doubled. 740. Theorem. The volume of a circular cone is equal to one-third the product of its base and altitude. V = | Bh = \irr 2 h. Given a circular cone. To prove V =\Bh = \irr 2 h, where V denotes volume, B area of base, h altitude, and r radius. Proof. Inscribe a pyramid with a regular polygon for base, and let V denote its volume and B' the area of its base. By doubling indefinitely the number of faces of the pyramid, 7'->7, and B'->B. § 739, § 489 \B'h->\Bh. § 485 (2) But V = \B'h, being variables that are always equal. § 738 .\7-jBft. §485(1) Also B=>irr\ §498 .-. F=fnr 2 fc. § 111 Remark. It is to be noted that the formulas of this article apply to any circular cone. Compare with the remark of § 719. EXERCISES 37 3V 1. Show that for any pyramid or circular cone Z?=— , and h = ——. h B 2. Find the volume of a square pyramid whose base is 16 in. on a side, and whose altitude is 10 in. 3. Find the volume of a pyramid whose base is a regular hexagon 8 in. on a side, and whose altitude is 14 in. 4. Find the volume of a pyramid whose base is a square 3\/2 in. on a side, and whose altitude is 6f in. 362 SOLID GEOMETRY 5. The base edges of a triangular pyramid are 6 in., 8 in., and 10 in. Find its volume if the altitude is 15 in. 6. Find the volume of a circular cone whose radius is 5 in. and alti- tude 10 in. 7. Find the volume of a circular cone whose altitude is 70 ft. and the circumference of whose base is 31 ft. 8. Derive a formula for finding the radius of a circular cone in terms of the volume and altitude. 9. Derive a formula for finding the altitude of a circular cone in terms of the volume and radius. 10. A right circular cone has a slant height s and a radius r; find its volume V. Ans. V =\lVrWs 11. The Pyramid of Cheops has a square base 720 ft. on a side, and an altitude of 480 ft. Find the number of cubic yards in it. 12. What is the locus of the vertices of all pyramids having the same base and equal volumes? 13. Find the total area and the volume of the solid generated by revolv- ing an equilateral traingle of side a about one side. Find the values if a = 2 in. Ans. Area = a 2 7T\/3; 21.765 sq. in, Volume = Ja 3 7T; 6.283 cu. in. 14. Find the total area and the volume of the solid generated by revolv- ing a parallelogram with sides 22 in. and 16 in., and larger angle 120°, about one of the longer sides. Ans. 3308.3— sq. in.; 13270+ cu. in. 15. A pyramid whose base is a square 4 ft. on a side, and whose alti- tude is 12 ft. is bisected by a plane parallel to the base. How far from the vertex is the plane? 16. A circular sheet of copper 3 ft. in diameter is cut in half, and each half formed into a cone. These are placed base to base as shown in the figure. This is to be used as a float. Find the number of pounds it will support if the sheet copper weighs 0.92 lb. per square foot and water weighs 62.5 per cubic foot. 17. Hard coal dumped in a pile lies at an angle of 30° with the horizontal. Estimate the number of tons in a pile in the shape of a right circular cone having an altitude of 10 ft. Large egg size weighs 38 lb. per cubic foot. Ans. About 60 tons. 18. Show how to construct a triangular pyramid equivalent to any given pyramid having a polygon as base VOLUMES OF PYRAMIDS AND CONES 363 741. Theorem. The volume V of a frustum of a pyramid, or circular cone, of bases B x a nd B 2 and altitude h, is given by the formula V = lh(B 1 +B 2 +VB^B' 2 ). Given the frustum of a pyramid, or of a circular cone. To prove V = \h{Bi+B 2 + y/B x B 2 ), where V denotes volume, h altitude, and Bi and B 2 the areas of the lower base and upper base, respectively. Proof. Complete the pyramid, or cone, and let m denote the altitude of the portion above the frustum in each case, and V 2 the volume of that portion. Also let V\ denote the volume of the entire pyramid, or cone. Then V=V 1 -V 2 . = iBi(/*+m)-^£ 2 m. §740 = |[5 1 / ? ,+ (5 1 -5 2 )m]. It remains to eliminate m. To do this one more equation is needed. This is given by B l _ {h+mY B 2 ~ m 2 *\fW\ h+m 709 Vb 2 m Solving for m, Substituting this for m, m = h\/W 2 Vb[-Vb~2 V^h(B 1 +B 2 +VB^ 2 ). 364 SOLID GEOMETRY 742. Theorem. The volume V of a frustum of a circular cone having bases with diameters d\ and a\, and radii 7*1 and r%, and an altitude h y is given by the formulas: (1) V^frrhW+rf+nrt), (2) VmfahW+df+ddti: Substitute irn 2 and 7Tr 2 2 for B\ and B 2 respectively of § 741 to find (1). Use i7Tdi 2 and lir~ — ! — i"~">\ J turn of a cone. The slant height is 14 ft. and makes N • — ^ ^S\ an angle of 60° with the horizontal, and the lower base \ 8" H is a circle 20 ft. in diameter. Find the volume of the tank in barrels. Use 1 bbl. =4.211 cu. ft. Ans. 1685.4 bbl. 9. A tank is in the form of an inverted cone with its vertex angle 60° and its axis vertical. Find the depth of water in the cone if it has been running in for 10 minutes at the rate of 8 cu. ft. per minute. SIMILAR PYRAMIDS AND CONES 365 SIMILAR PYRAMIDS AND CONES 743. Similar Pyramids. Pyramids that are formed from the same pyramidal surface by parallel planes are similar. 744. Similar Cones. Cones that are formed from the same conical surface by parallel planes are similar. 745. Theorem. The lateral areas, or the total areas, of two similar regular pyramids, or right circular cones, are to each other as the squares of two corresponding lines, where the corresponding lines may be edges, elements, altitudes, slant heights or radii. One case only will be proved., The other cases are left as exercises. Given two similar regular pyramids O-ACDEF and O-A'C'D'E'F'. S h 2 T h 2 To prove — = — , and — = — , where S and S! h' 2 r h' 2 S' denote the lateral areas of O-ACDEF and O-A'C'B'E'F' respectively, h and h! altitudes, and T and T' total areas. Proof. If p and p' are the perimeters of the respective bases, and s and s' the slant heights, S = ^ps, and S' = ^p's'. § 711 Then jS=l^i=Z.i. wh ? S' %p's' p' s' S h 2 . p h , s h w , 9 .*. — = — since — = — , and— =— . Why: S' h' 2 p' h' s' h' Further, if B and B' are the areas of the respective bases, B h 2 _ 8 B ■ 8 S' w , ? — = — and so — = — , and— = — . Why: B' h' 2 S' B' 9 B B f Then ^J? = ^±^, ail d &+*-=*. §§ 406,405 B B' S'+B' B' .-. — =— since S+B = T, and S'+B' = T'. Why? T' h' 2 366 SOLID GEOMETRY 746. Theorem. The volumes of two similar pyramids, o circular cones , are to each other as the cubes of two correspondim lines; where the corresponding lines may J)e edges, elements, alii tudes, slant heights, or radii. Using V and V for the respective volumes, B and B' for bases and h and h' for altitudes, Y _ \Bh _B h V \B'h' B' ' ft 7 ' B h 2 V ft 3 But— =— . Why? .-.— = —. Why' B' ft' 2 Y' ft* 747. Definition. The vertex angle of a right circular com is twice the angle between an element and the axis. 748. Theorem. All right circular cones with equal vertex angles are similar. EXERCISES 1. Two cones generated by similar right triangles revolving abou corresponding sides are similar. 2. Two cones are generated by revolving two similar right triangle about corresponding sides of length 7 in. and 10 in. respectively. Fine the ratio of the lateral areas of the cones. The ratio of the total areas The ratio of the volumes. 3. The total area of a right circular cone is 300 sq. in. and its altitudi is 8 in. Find the total area of the cone cut off by a plane parallel to th4 base and 6 in. from the vertex. 4. How far from the vertex of a right circular cone, or a regular pyra mid, of altitude h must a plane be passed parallel to the base so that thel lateral area of the part cut off shall be one-half that of the original? 5. From a given pyramid, cut off a frustum whose volume shall W fy that of the given pyramid. Must the pyramid be regular? 6. The volumes of two pyramids are 343 cu. in. and 1000 cu. in. respec- tively. The lateral area of the smaller is 100 sq. in. Find the latera area of the larger. 7. The lateral areas of two similar right circular cones are in the ratio of 9 : 64. Find the volume of the smaller if the volume of the larger h 1024 cu. in. QUESTIONS 367 QUESTIONS 1. What is a pyramidal surface? A conical surface? Is the direc- trix necessarily closed? 2. What are the formulas or rules for finding the areas of pyramids, cones, and frustums of pyramids and cones? 3. What are the formulas or rules for finding the volumes of each of these solids? 4. Can you find the area of an oblique circular cone? Of an oblique pyramid? Can you find the volume of each of these? 5. If the altitude of a cone is divided into 10 equal parts by planes passed parallel to the base, how does the area of each section of the cone L formed by these planes compare with the base of the cone? 6. Trace the steps in finding the volume of a triangular pyramid. Of any pyramid. Of a cone. 7. What is the relation between two parallel sections of a pyramid or cone? 8. State theorems concerning the sections of a cone. Of a pyramid. 9. What effect does it have upon the volume of a pyramid or cone if the area of the base is doubled? If the altitude is doubled? If both base and altitude are doubled? 10. What effect does it have upon the lateral area of a right circular cone if the circumference of the base is doubled? If the area of the base is doubled? 11. Point out forms of this chapter that occur in nature. That occur in the arts. GENERAL EXERCISES 1. Given the following formulas for the right circular cone: (1) V = fe-7Tr 2 /i, (2) S = T rs, (3) S = TTrVh*+r* t (4) T = A+WrVh 2 +r 2 , (5) T = fTr 2 +7Tr V h 2 -\-r 2 ) solve each for each letter appearing. 2. In a right circular cone, A=507T, and F = 2007T. Findr, h, and s. 3. If R and r are the radii of the two bases of a frustum of a cone, Isv |;how that R+r=\—+Rr. \7T/i 4. Find the volume and the total area of a right circular cone whose ladius of base is 6 in. and altitude is 5.3 in. Arts. 199.8 cu. in. ; 263.9 sq. in. 368 SOLID GEOMETRY 5. Find the weight of a tapered brick smoke stack 175 ft. high in the form of a frustum of a right circular cone enclosing a cylinder. The inner diameter is 10 ft., the wall is 4 ft. thick at the base and 1 ft. 6 in. at the top. One cubic foot of brick weighs 112 pounds. Ans. 1095.5 tons. 6. Find the lateral edge, lateral area, and volume of a regular pyra- mid each side of whose triangular base is 5 ft., and whose altitude is 9 ft; Ans. 9.45 ft.; 68.36 sq. ft.; 32.5 cu. ft. 7. A bin in a warehouse is 12 ft. square. A hopper is constructed on the base, which has a slope of 1 : 1. The distance from the vertex of the hopper to the top of the bin is 18 ft. Find the capacity of the bin if t bushel equals f cu. ft. 8. The largest possible cylinder of diameter 6 in. is cut from a right circular cone having a diameter and altitude of 10 in. and 26 in. resp< tively. Find the volume of the cylinder. 9. Find the area and volume of the solid generated hy revolving isosceles triangle of base 10 in. and equal sides 16 in., about its base. Fin the area and the volume of the solid when this triangle is revolved about one of its equal sides. 10. In the frustum of a pyramid whose base is 50 sq. ft. and altitud 6 ft., the basal edge is to the corresponding top edge as 5 to 3. Find th volume of the frustum. Ans. 196 cu. f 11. The base of a pyramid is a rectangle 10 in. by 8 in. Find the vo ume if each lateral edge makes with the base an angle of 45°. 12. The total surface of a regular quadrangular pyramid is T, and eight edges are equal. Find the length of an edge. 13. Find the volume of a cone of revolution whose slant height is equal to the diameter of its base, and whose total area is T. 14. Show how to cut a pattern for the frustum of a right circular cone, if the upper and lower bases have diameters of 4 in. and 6 in., respectively, and the altitude is 8 in. 15. A right circular cone whose altitude is 8 in. and radius 6 in. rolls on a floor without slipping, making one complete revolution. What is the shape of the surface covered? Find its area. 16. A pyramid whose altitude is 12 in. weighs 30 pounds. At wl distance from the vertex must it be cut by a plane parallel to its base that the frustum cut off shall weigh 20 pounds? 17. A cube is cut by a plane passed through the other extremities the three edges meeting at a vertex. What part of the volume of the cul is thus cut off? CHAPTER IX PRISMATOIDS AND POLYEDRONS PRISMATOIDS 749. A polyedron all of whose vertices lie in two parallel planes is called a prismatoid. In either plane, the vertices may lie in any polygon, as Fig. (1); or, in one plane, the ver- tices may lie in a line, as in Fig. (2) ; or there may be a point only, as in a pyramid. The faces whose vertices are in the parallel planes are evidently triangles, or quadrilaterals of types that have at least two parallel sides. Why? 750. The altitude of a prismatoid is the distance between the two parallel planes. EXERCISES 1. Show that a prism is a prismatoid whose bases are congruent poly- gons. 2. Show that a frustum of a pyramid is a prismatoid whose bases are similar polygons. 3. Draw or describe a prismatoid that has a face that is a trapezoid. A parallelogram. A rectangle. 4. Compute the area of the midsection of a pyramid whose base is a regular hexagon having a side of 6 in. 6. Compute the areas of the midsection of the prismatoid in Fig. (2) if ABCD is a rectangle having A£ = 18 in. and 2?C = 11 in., and EF = 12 in. 369 370 SOLID GEOMETRY N, 751. Theorem. If Bi and B 2 are the areas of the two bases, M the area of the midsection, and h the altitude, then the volume V of a prismatoid is given by the formula V = ^h(Bi+B 2 +4:M). Given a prismatoid of volume V, alti- tude h, bases Bi and B 2 , and midsection M. To prove V = ih(B 1 +B 2 +4M). Proof. If any lateral face is not a tri- angle divide the face into triangles by a E diagonal. Take any point in the midsection and^- connect it with the vertices of the prisma- toid and the midsection. These lines form triangles with the edges of the prismatoid, which separate the prismatoid into pyramids that have the bases and faces of the prismatoid as bases. (1) Pyramid 0-ABCD=lhB v §737 (2) Pyramid 0-NPQ = \hB 2 . Each of the pyramids like O-CDP, which may be called a lateral pyramid, can have its volume ex- pressed in terms of the part of the midsection com- mon to it. To prove this consider O-CDP apart from the prismatoid, and draw HD. Altitude of each of pyramids P-OHI and D-OHI is \h Pyramid P-OHI = \h times area OHI. § 737 Pyramid D-OHI - \h times area OHI. Why? Pyramid O-CDH = 2 times pyramid 0-HDL Why? Then pyramid O-CDH = %h times area OHI. Hence pyramid O-CDP = %h times area OHI. Why? Similarly each lateral pyramid is equal to -f h times the area of its part of the midsection. (3) Therefore the sum of the lateral pyramids = j^hM = \h-±M. Adding (1), (2), Why? and (3), 7 = |fc(B 1 +B 2 +4M). PRISM ATOIDS 371 752. Other Applications of the Prismatoid Formula. The formula for the volume of a prismatoid can be applied also to various other solids such as cylinders, cones, spheres, combi- nations of these, and various other forms that cannot well be described here. The prismatoid formula, because of its wide application, is used often by engineers and others in determining the volumes of embankments, abutments, and other irregular solids of vari- ous kinds. EXERCISES 1. Derive the following formulas from the formula for the volume of a prismatoid : (1) Prism and cylinder, V = Bh. (2) Pyramid and cone, V = ^Bh. (3) Frustum of pyramid or cone, V = %h {Bi +B 2 + \ZBiB 2 ) . Derivation for frustum. Given V=*§h(Bi+B s +4M). To derive V = \h{B l +B 2 + y/BiB 2 ). It is then necessary to express M in terms of B\ and B 2 . Using the notation of § 741, — =- 7T - — , and — = 711 T~- § 709 M (J/i+ra) 2 M $h+m) 2 _ V#i h+m . \/B 2 m Or , — =y~ , and- / M \h-\-m \/M f/i+ra' Solving each of these for m, ra = j= — 2 ; — — from the first: Vb[-Vm and m — — j== y= from the second. Vm-Vb, h(VM-§y/Bi) _ hhVW 2 Vb[-Vm Vm-VK Dividing by to^Sr^ Then VBi-VM Vm-Vb 2 a/M a/M Then 2VM-V^ ^Vg' ^ ^ M = VW 1 + VB, § 406 Or 4M = £ 1 +£ 2 +2V£ 1 £ 2 . .'. V =ih(2B l +2B 2 +2\ / BA) =^(Bi+B, + Vb,B,). 372 SOLID GEOMETRY 2. Show that, in general, the midsection of a prismatoid has as many sides as the two bases combined. When will this not be so? 3. Find the volume of the frustum of a square pyramid, the lower base being 8 ft. square, the upper base 6 ft. square, and the altitude 12 ft. Solve both by the frustum formula and the prismatoid formula, and compare. 4. Both bases of a prismatoid of altitude h are squares, and the lateral faces isosceles triangles. The sides of the upper base are parallel to the diagonals of the lower base and half the length of these diagonals. If b is a side of the lower base, find the volume. Ans. %b 2 h. 5. Find the volume of the prismatoid with dimen- sions as shown in the figure. The bases are right tri- angles having their corresponding sides parallel. Ans. 3420 cu. in. 6. Use the prismatoid rule to find the volume of a frustum of a pyramid whose bases are regular hexagons 10 in. and 6 in. on a side respectively, and whose altitude is 18 in. 7. Find the volume of the frustum of the preceding exercise by § 741 and compare the result with that of the preceding. n ^ 8. The volume of a truncated triangular prism is readily determined by the prismatoid formula. To do this consider one face and the opposite edges as the bases. Find the volume of the truncated triangular prism shown in the figure. The base ABC, which has a right angle at C, is a right section. 9. Show that the volume of any truncated triangular prism is equal to the product of the area of a right section and D |" one-third the sum of the three edges. Suggestion. Let the edges be c, d, and e, and the altitude and base of a right section a and b respectively. Consider the prism as a prismatoid with bases ABED and CF, and midsection M. A Then 7 = £a[i&(c+d)+0+4M]. §751 But M = j6-i[i(c+e)+|(d+c)]. •*• V — \ab' J(c+d-j-e)=area of right section X \ (c-j-d+e). 10. A truncated right triangular prism has for base an isosceles triangle whose sides are 10 in., 10 in., and 8 in. Find its volume if the three lateral edges are 5 in., 7 in., and 11 in. respectively. POLYEDRONS 373 POLYEDRONS 753. Regular Polyedrons. A polyedron whose faces are congruent regular polygons v and whose polyedral angles are congruent is called a regular polyedron. A polyedron of four faces is a tetraedron; one of six faces, a hexaedron; one of eight faces, an octaedron; one of twelve faces, a dodecaedron; and one of twenty faces, an icosaedron. It is instructive to construct the regular polyedrons as shown in the following figures. Draw on cardboard the diagrams shown below, cut along the full lines, and fold along the dotted lines. The edges that meet may be fastened with gummed paper. Durable models may be constructed from tin and soldered at the edges. There are many interesting facts connected with the five regular poly- edrons. The so-called fourteenth, fifteenth, and sixteenth books of Euclid give many theorems and problems concerning these solids. 374 SOLID GEOMETRY 754. Theorem. There can be no more than five regular poly- edrons. The proof of this theorem depends upon the facts: (a) At least three planes must meet to form a polyedral angle. (§ 603). (b) The sum of the face angles of a polyedral angle is less than 360°. (§ 609). (c) The faces of a regular polyedron are congruent regular polygons. (1) Faces equilateral triangles. A polyedral angle can be formed of three, four, or five equilateral triangles, and no more. Why? Therefore no more than three regular polyedrons can be formed having equilateral triangles as faces. (2) Faces squares. A polyedral angle can be formed of three squares, and no more. Why? Therefore no more than one regular polyedron can be formed having squares for faces. (3) Faces regular pentagons. A polyedral angle can be formed of three regular pentagons, and no more. Why? Therefore no more than one regular polyedron can be formed having regular pentagons for faces. Regular polygons with a greater number of sides than five cannot be used to form polyedral angles because the sum of three or more angles of such polygons is equal to or greater than 360°. Hence there can be no more than five regular polyedrons. 755. Relation between the Number of Faces, Edges, and Vertices of a Regular Polyedron. It is interesting to note the relation between the number of faces, edges, and vertices of a regular polyedron. Complete the following table, and com- pare with the table on page 67. Name Number of Faces Form of Faces Number of Edges Number of Vertices Number of Faces at Each Vertex Sum of Face Angles Tetraedron 4 Equi. A 6 4 3 720° Hexaedron Octaedron Dodecaedron Icosaedron POLYEDRONS 375 756. Euler's Theorem. This theorem is stated because of its interest. The proof is too difficult to be given here. As an exercise it may be tested for the regular polyedrons. Use the preceding table. Theorem. In any polyedron the number of edges plus two is equal to the sum of the number of faces and the number of vertices. Note. Leonhard Euler was born in Switzerland in 1707 and died in Petrograd in 1783. He was one of the greatest physicists, astronomers, and mathematicians of the eighteenth century. 757. Theorem. The sum of the face angles of any polyedron is equal to 360° multiplied by two less than the number of ver- tices. The proof of this theorem is based upon the preceding theorem. Test it for the regular polyedrons. EXERCISES 1. Find the area of the surface of a regular tetraedron 3 in. on an edge. Of a regular octaedron 4 in. on an edge. Of a regular icosaedron 6 in. on an edge. Ans. 15.588+ sq. in.; 55.426— sq. in.; 311.769+ sq. in. 2. Find the area of the surface of a regular dodecaedron 2 in. on an edge. Ans. 82.583— sq. in. 3. If the edge of a regular tetraedron is e, show that its volume is given by the formula V = Y^e 3 \/ r 2. Solution. V = %ON X area of AABC. J* DC = %e\/3. See p. 166, Ex. 7 (3). //I\ Area of AABC = \AB X ZX7 = Je 2 y/l, / \\ \ OC = JD C = \eVz. ^_ _ §211. P Z /|___A C ON = VcN 2 -OC 2 = V> - (p V3) 2 - e Vf . dQ3^ ... 7=JeV / fxie 2 V3=A-^V / 2. J 4. If the edge of a regular octaedron is e, show that its volume is given by the formula V =\e z \f2. 5. Compare the volumes of a regular tetraedron, a cube, and a regular octaedron, each 2 in. on an edge. 6. Find the area and volume of a regular tetraedron having an altitude of 8 in. 376 SOLID GEOMETRY 758. Theorem. The volume of two tetraedrons that have a triedral angle of one congruent to a triedral angle of the other are to each other as the products of the three edges of these triedral angles. c t L >JS' Given the tetraedrons O-ABC and O'-A'B'C, with triedral ZO = triedral ZO' . „ V OA-OB-OC u T7 . ry , , To prove— =___-____, where V and V denote the volumes of the tetraedrons. Proof. Place tetraedron O-ABC so that triedral ZO will coincide with triedral ZO'. Draw CN and CM perpendicular to O'A'B'. CN and CM determine a plane which intersects O'A'B' in O'M. Why? AT V O'AB-CN O'AB CN Now — = — = . Why? V O'A'B' -CM O'A'B' CM But VA^^O'A-0'B^ O'A'B' O'A'-O'B' Further AO'NC and AO'MC are similar rt.A. Why? Then M-^.. 1428 CM O'C V O'A-0'B-O'C OA-OB-OC §111 V O'A'-O'B' -O'C O'A'-O'B' -O'C Compare this theorem and the proof with § 375. 759. Theorem. Two parallelepipeds thai have a triedral angle of one congruent to a triedral angle of the other are to each other as the products of the three edges of these triedral angles. POLYEDRONS 377 760. Similar Polyedrons. Two polyedrons are similar if they have the same number of faces similar each to each and similarly placed, and have their corresponding poly- edral angles congruent. It is evident that the corre- sponding diedral angles are also equal. 761. Theroem. The cor- responding edges of similar polyedrons are proportional. Corresponding edges belonging to corresponding faces are proportional by § 444. Use the fact that each edge belongs to two faces and prove the theorem. 762. Theorem. The surfaces of two similar polyedrons are to each other as the squares of any two corresponding edges. Given two similar polyedrons P and P\ with corresponding faces A and A', B and B f , C and C, • • • . Also given a and a', b and &', c and c', • • • , corresponding sides of these faces respec- tively. _ A+B+C+--- To prove A'+B' + C'+- corresponding edges. A a 2 B = — , where n and n' are any two n ' Proof. But And hence Therefore = ¥ C A' a n -' B' bf ' C' a b c a'~b'~7'~ a 2 _ ¥ _ & _ A' B' C'~ ■ A+B+C +--- " A'+B'+C'+-- §446 Why? Why? §403 378 SOLID GEOMETRY 763. Theorem. The volumes of two similar tetraedrons are to each other as the cubes of any two corresponding edges. Use §§ 746, 761. 764. Theorem. The volumes of two similar polyedrons are to each other as the cubes of any two corresponding edges. This theorem will not be proved. It is accepted as true since it has useful applications. EXERCISES 1. What is the ratio of the volumes of two cubes that are 5 in. and 4 in. on an edge? What is the ratio of their areas? 2. The edges of a parallelepiped are 10, 12, and 14. Find the edges of a similar parallelepiped having ^ as great an area. Find the ratio of their volumes. 3. Two pyramids are cut from the same pyramidal surface. The lateral edges of one are 9 in., 12 in., and 14 in. and of the other 18 in., 21 in., and 20 in. Find the ratio of their volumes. 4. Find the ratio of the volume of a regular octaedron 6 in. on an edge to the volume of a regular octaedron having half as great an area. 5. The center of gravity of a tetraedron is J its altitude above the base. Find the center of gravity of a regular tetraedron 8 in. on an edge. 6. The edge of a regular tetraedron is e. Find the edge of a regular tetraedron that has a volume n times as great as the volume of the given tetraedron. 7. Find the edge of a regular tetraedron such that its volume multi- plied by \/2 is 288. 8. Find the volume of a regular tetraedron that has an altitude of 17 in. 9. A section of a tetraedron by a plane parallel to two opposite edges is a parallelogram. 10. The midpoints of the edges of a regular tetraedron are the vertices of a regular octaedron. 11. Two similar polyedrons have volumes of 121.5 cu. in. and 4.5 cu. in., respectively. An edge of the smaller is 1 J in. Find the corresponding edge of the larger. Ans. 4.5 in. 12. The area of the entire surface of a polyedron is 108 sq. in., and its volume is 432 cu. in. If the area of the entire surface of a similar polyedron is 75 sq. in., find its volume. Ans. 250 cu. in. GENERAL EXERCISES 379 GENERAL EXERCISES U—iot—i 1. If from any point within a regular tetraedron perpendiculars are drawn to its faces, their sum equals an altitude of the tetraedron. (Compare this with Ex. 2, p. 96.) 2. A wedge whose altitude is 10 in. and edge 4 in., has a base that is a square having a perimeter of 36 in. Find the volume of the wedge. 3. A concrete pier for a railway bridge has dimen- sions as shown in the figure, the bases being rec- tangles with semicircles. Find the number of cubic yards of concrete in the pier. Ans. 37.8— cu. yd. 4. A railroad cut has the dimensions given in the figure, which shows the vertical sec- tion and three cross sections, one at each end and one in the middle. Find the number of cubic yards of earth removed in digging the cut. Ans. 7972f cu. yd. Suggestion. The part on each side of the center is a prismatoid, and can be computed separately. 5. A tank of reinforced concrete is 160 ft. long, 100 ft. wide, and 10 ft. 6 in. deep, outside dimensions. The side walls are 8 in. thick at the top and 18 in. thick at the bottom, with the slope on the inside. The bottom is 6 in. thick. Find the number of cubic yards of cement used in building the tank, and the capacity of the tank in gallons. 6. The vertices of one regular tetraedron are the centers of the faces of another regular tetraedron. Find the ratio of the volumes of the two tetraedrons. 7. The three planes passing through the lateral edges of a triangular pyramid, and bisecting the base edges, meet in a common straight line. 8. The six planes passing through the edges of a tetraedron, and bisecting the opposite edges, meet in a common point. 9. The point of intersection of the planes in the preceding exercise is the center of gravity of the tetraedron. Prove that the center of gravity of a tetraedron divides the line from a vertex to the center of gravity of the opposite face in parts that are in the ratio of 3 : 1. The center of gravity of a /ace is at the intersection of the medians of that face. 380 SOLID GEOMETRY 10. The corresponding edges of three similar tctraedrons are 3 in 4 in., and 5 in., respectively. Find the corresponding edge of a simila tetraedron equal in volume to their sum. 11. By means of the prismatoid formula show that the volume of truncated quadrangular prism whose opposite faces are parallel, is equal to the area of a right section times one-fourth the sum of the lateral edges. Suggestion. Consider two of the parallel faces, as Bi and B 2 , as bases. Then M is the midsection between these. Let A be the area of a right section having a side in Bi equal to n, and altitude h, the distance between Bi and B 2 . Then £i=4 (a-f-d) n, -'&«■'$. (6+c) n, and M ~§ { \ (a+b)+%(c+d) }n '• V-\h \\ (a+d) n+J (6+c) *+2{| (a+&)+| {c+d)}n] = \hn (a+&+c+d)=jA (a+b+c+d). 12. The bottom of a bin of wheat is a rectangle 5 ft. by 12 ft. The top of the wheat is in a plane such that the depths at the corners are 6 ft., 5 ft., 3 ft., and 4 ft. respectively. Find the number of bushels in the bin if 1 bu. = l| cu. ft. x" 1 — i i 7 I M 1 i i Bi b ,'--- '" CHAPTER X THE SPHERE AND POLYEDRAL ANGLES GENERAL PROPERTIES 765. Sphere. A sphere is a closed surface all points of which are equally distant from a point within, called the center of the sphere. A sphere is usually designated by a single letter at its center. The radius of' a sphere is the straight line connecting the center to a point in the surface. A straight line connecting two points in the surface and also passing through the center is a diameter. A sphere separates space into two parts so that any point that does not lie in the surface lies within the sphere or outside it. The distance from the center to a point within a sphere is less than the length of the radius; while a point outside a sphere is at a greater distance from the center than the length of the radius. 766. Great Circle of a Sphere. It is evident from the definitions of a sphere and of a circle that a plane passing through the center of sphere intersects the sphere in a circle ivhose radius is equal to the radius of the sphere. Such a rircle i-s called a great circle of the sphere. 381 382 SOLID GEOMETRY 767. The following facts follow readily from the definitions given: (1) Radii of the same sphere, or equal spheres, are equal. (2) Diameters of the same sphere, or equal spheres, are equal (3) Spheres having equal radii, or equal diameters, are equal (4) All great circles of a sphere are equal. (5) The points of intersection of any two great circles of a sphen are in a diameter. (6) A great circle of a sphere bisects the sphere. (7) A sphere may be generated by the revolution of a semicircU about Us diameter. 768. The Straight Line and the Sphere. In order to stud^ the relative positions of a straight line and a sphere, pass i plane through the line and the center of the S* "\ sphere. The line will then have the same /p="^-~~3m 2f relations to the sphere that it bears to the ^jl^t- -^j\ great circle in this plane. ^^_^^ (1) If the line intersects the great circle it cuts the circU in two points, and likewise the sphere. (2) If the line is tangent to the great circle it has only om point in common with the sphere, and is perpendicular to the radius at this point. 769. Definition. If a line has but one point in commoi with the sphere it is said to be tangent to the sphere. EXERCISES 1. A line perpendicular to a radius at its extremity is tangent to th sphere. 2. The locus of all lines tangent to a sphere at a point is a plane per pendicular to a radius at its outer extremity. 3. Find the locus of the centers of all spheres tangent to a given lin at a given point. 4. Find the locus oi the centers of all spheres of a given radius tangen to a fixed line. GENERAL PROPERTIES OF SPHERE 383 5. The plane perpendicular to a tangent at its point of contact passes through the center of the sphere. 6. What is the locus of the vertices of the right angles of the right triangles having a given hypotenuse? 770. The Plane and the Sphere. If a plane has but one point in common with a sphere, it is said to be tangent to the sphere. 771. Theorem. A plane perpendicular to a radius of a sphere at its outer extremity is tangent to the sphere. ^^_^^ Given plane P perpendicular to the / >v radius OC at its outer extremity C. r~~- -A To prove plane P is tangent to the fc" ~a ~~^y\ sphere 0. r\ / ■ J~\ Give a proof similar to that of § 284. / \j Given the tetraedron ABCD. / \ \. To prove that one sphere and only one / E A \. can be inscribed in ABCD. a<~-— U --^>c Outline of proof. Construct the planes ^\^^^^ bisecting the diedral angles AD and BD. B These planes will intersect in a line DS. Why? Then DS is the locus of all points equally distant from faces ADC, ABD, and BDC. Why? Construct the plane bisecting the diedral angle AB. This plane will intersect DS in some point 0. Why? Then is equally distant from the four faces of the tetrae- dron. Why? If a sphere is constructed with as center and the perpen- dicular from to any face, as OE, for radius, this sphere will be tangent to each face of the tetraedron, and therefore is an inscribed sphere. § 789 To show that only one inscribed sphere can be constructed show that there is only one point 0. Compare this theorem and proof with the corresponding problem in plane geometry. (§ 325). Exercise. Find the locus of the centers of all spheres tangent to the faces of a triedral angle. SPHERES AND TETRAEDRONS 389 791. Theorem. One, and only one, sphere can be circum- scribed about a given tetraedron. Given the tetraedron ABCD. To prove that one sphere and only one can be circumscribed about ABCD. Outline of proof. Circumscribe a circle about AABC, and erect a perpendicular O'Q to its plane at its center 0'. Then O'Q is the locus of all points equally distant from A, B, and C. Why? At the midpoint E of the edge CD construct a plane P per- pendicular to CD. Then plane P is the locus of all points equally distant from C and D. Why? But plane P intersects O'Q in a point 0. Why? Then is equally distant from the four vertices of the tet- raedron, and a sphere having as center and OA as radius will pass through the four vertices A, B, C, and D. Why? Therefore a sphere can be circumscribed about the tetraedron. To show that only one sphere can be circumscribed about ABCD, show that there is only point 0. Compare this theorem and proof with the corresponding problem in plane geometry. (■§§ 316, 317). EXERCISES 1. What is the locus of points equidistant from four points not all in the same plane? 2. How many points are necessary to determine a sphere? How many points on a sphere are necessary to determine it if its center is given? 3. Prove that the planes that are the perpendicular bisectors of the edges of a tetraedron pass through a common point. 4. Prove that, if a circle is circumscribed about each face of a tetra- edron, the perpendiculars to the respective faces at the centers of the circumscribing circles are concurrent. 5. One, and only one, sphere can be inscribed in or circumscribed about a cube. 390 SOLID GEOMETRY 792. Problem. To find the radius of a given material sphere. Given a material sphere 0. To find its radius. Construction. Choose any point N on the sphere as a pole, and describe a convenient circle ABC. Choose any three points on this circle as A, B, and C. Construct in a plane AA'B'C congruent to AABC. § 25 Circumscribe a circle with center Q' about AA'B'C. Draw Q"B" equal to the radius Q'B', and then construct XY± Q"B" and through Q". With the compasses lay off B"N' equal to BN, to meet XY at N', and construct B"S' _L B"N' at B" and extend it to meet XY at S'. Determine the bisecting point 0' of S'N'. Then O'N' is the radius of the sphere 0. The proof is left to the student. EXERCISES 1. If NB = 13 in. and QB = 12 in., compute ON. 2. By means of an instrument called a sphero- meter, the distances QN and QB can be measured. Show how the radius ON can then be computed. Such an instrument is used practically in deter- mining the radius of curvature of a spherical lens. 3. Find the radius of curvature of a spherical lens, that is, the radius of the sphere of which the surface of the lens is a part, if QN =0.125 cm. and QB m 2 cm. If QN - 0.643 cm, and QB m 2.5 cm, Arts. 16.9 in. SPHERES, CYLINDERS, CONES 391 793. Sphere Inscribed in Cylinder or Cone. A sphere is said to be inscribed in a cylinder, or the cylinder circum- scribed about the sphere, when the bases of the cylinder and all its elements are tangent to the sphere. A sphere is said to be inscribed in a cone, or the cone cir- cumscribed about the sphere, when the base of the cone and all its elements are tangent to the sphere. 794. Cylinder or Cone Inscribed in a Sphere. A cylinder is said to be inscribed in a sphere, or the sphere circum- scribed about the cylinder, when its bases are circles of the sphere. A cone is said to be inscribed in a sphere, or the sphere circumscribed about the cone, when its base is a circle of the sphere and its vertex lies on the sphere. 795. Theorem. All the tangents to a sphere from a given point are equal, and their points of contact are in a circle of the sphere. Given sphere 0, and point P outside the sphere. To prove that the tangents to the sphere from the point P are equal, and that their points of contact lie in a circle. Outline of proof. Let PB be one tangent. Then PB is tangent to a great circle determined by P, B, and 0. Why? Therefore ZPBO is a rt.Z. § 288 Pass a plane through B _L PO. This will intersect the sphere in a circle with center 0' '. Further, aline drawn from P to any point, as C, in the QO f is tangent to the sphere 0, for AOCP = AOBP, and therefore PC J_ OC. Why? Show that no tangent from P could have its point of contact outside of OO'. That the tangents are all equal follows from § 584. 392 SOLID GEOMETRY have a common RELATIVE POSITIONS OF SPHERES 796. The relative positions of two spheres are analogous to the relative positions of two circles. (See §§ 291-296.) 797. Line of Centers. The line joining the centers of two spheres is the line of centers. 798. Tangent Spheres. If two spheres are tangent to the same plane at the same point, they are tangent spheres. They may be tangent internally or externally. Spheres A and B are tangent internally; and A and C externally. 799. Concentric Spheres. Spheres that center are concentric spheres. 800. Theorem. The intersection of two spheres is a circle, whose center is in a straight line joining the centers of the spheres and whose plane is perpendicular to that line. Given two intersecting spheres and 0'. To prove that the intersection of the spheres is a circle whose center is in 00' and whose plane is perpendicular to 00' . Outline of proof. Through 0, 0' , and any point A of the intersection, pass a plane. This plane intersects the two spheres in two great circles intersecting in two points. Why? If A and B are these points, then 00' is the perpendicular bisector of AB. § 295 Show that if the entire figure is revolved about 00', the great circles generate the spheres, and point A generates a circle with center C and radius CA, which is common to the spheres. Also show that the plane of the circle generated by A is per- pendicular to 00'. Further, show that if any point of the intersection was out- side of this circle, then the spheres would coincide. (§ 791). MEASUREMENT OF THE SPHERE, AREA 393 801. Theorem. // two spheres are tangent to each other , the line of centers passes through the point of contact. Proof similar to that of § 296. EXERCISES 1. State and prove exercises concerning the sphere analogous to Exer- cises 1 to 5, page 115. 2. What is the locus of points at a distance r from a given point O, and at a distance r' from a given point 0"? 3. Find the center of a sphere which contains a given circle and also contains a given point not lying in the plane of the circle. 4. Two spheres with radii 34 and 50, respectively, intersect. If the distance between their centers is 56, find the radius of their circle of intersection, and the distance from the center of each sphere to the center of the circle of intersection. 5. The line of contact of a sphere inscribed in a circular cone with the conical surface of the cone, is a small circle of the sphere. 6. The line of contact of a sphere inscribed in a circular cylinder with the cylindrical surface, is a great circle of the sphere. MEASUREMENT OF THE SPHERE, AREA 802. Area of a Sphere. By the area of a sphere is meant the numerical measure of the surface, that is, the number of units of area in it. As with the cylinder and cone, a sphere is a curved surface, and a plane unit of area cannot be applied directly to it to find its numerical measure. Its area can be determined as a limit. It is based upon (7) of § 767, which states that a sphere is generated by revolving a semicircle about its diameter, and the following statement which is accepted without proof: If in a semicircle generating a sphere, one-half of a regular polygon be inscribed so that a vertex lies at each end of the diameter, then, as the number of the sides of the inscribed semipolygon is indefinitely doubled, the area generated by the semipolygon approaches the area of the sphere as a limit. 394 SOLID GEOMETRY 803. Theorem. The area of the surface generated by a seg- ment of a straight line revolving about an axis, in its plane but not perpendicular to it and not intersecting it, is equal to the projection of the segment upon the axis multiplied by the circumference of the circle whose radius is the perpendicular erected at the midvoint of the segment and terminated by the axis. Given the segment AB revolving about the axis XY in its plane. To prove that S = 00'X2wXEF, where S denotes the area of the surface generated, 00' is the projection of AB upon XY, and FE is the perpendicular at the midpoint of AB. Proof. Three cases arise: Case I, in which AB is oblique to and does not meet XY. The surface generated is the lateral surface of a frustum of a right circular cone. The proof of this is given in § 729. Case II, in which A B is oblique to and meets XY. The sur- face generated is the lateral surface of a right circular cone. S = irXOAxAB. §719 Show that ADEF^AOAB. Then AB : OB = EF : DF, and ABXDF = OBX EF. Since OB = 00' and OA = 2DF, ABxOA= 200' X EF. .'.S = 00'X2ttXEF. § 111 Case III, in which AB is parallel to XY. The area generated is the lateral surface of a right circular cylinder. Give proof. MEASUREMENT OF THE SPHERE, AREA 395 804. Theorem. The area of a sphere is equal to four times the area of a great circle of the sphere. S = 47rr 2 . Given the area S of a sphere generated by the revolution of the semicircle of radius r, about the diameter AE. To prove that £=47rr 2 . Proof. Inscribe in the semicircle one-half of a regular poly- gon of an even number of sides, as ABCDE. Let A denote the apothem of the polygon, and let F, 0, and G be the projections of B, C, and D respectively upon the diameter AE. For each side, the apothem is the perpendicular bisector. Why? Then area generated by AB = AF X 2tcl. § 803 Similarly area generated by BC = FOX2ira. Etc. Adding these equations and denoting by S' the area generated by the revolution of the semipolygon ABCDE, S f =(AF+FO+'-)2ira. But (AF+FO+---)=AE = 2r. Then £' = 47rra, which is always true as the number of the sides of the polygon is continuously doubled. Further S'->S and since a->r, ^wra^wr 2 . §§ 802, 490, 485 (2) /. S=47rr 2 . §485(1) Show that other forms of the formula for the area of a sphere are S = 2irrd J and S=7rd 2 . 805. Theorem. The areas of two spheres are to each other as the squares of their radii; or, as the squares of their diameters. For, if S, and S' are the areas of two spheres, _^ = j^l! = l!_ and— = — = — S' 4:wr' 2 r ,2> ^ S'~ ird' 2 ~ d' 2 396 SOLID GEOMETRY EXERCISES 1. Given the area of a sphere, to find its radius and its diameter. 2. Find the area of a sphere whose radius is 5. Of a sphere whose diameter is 16. Of a sphere whose circumference is 24 in. 3. Find the radius of a sphere whose area is 100 sq. in. 4. How many square feet of tin will it take to roof a hemispherical dome 40 ft. in diameter? 5. Find the diameter of a sphere that has the same area as a cube that is 8 in. on an edge. 6. Find the area of a sphere that is circumscribed about a cube that is 12 V3 in. on an edge. Ans. 4071.5 sq. in. 7. The number of square inches in the area of a sphere is equal to the number of linear inches in the circumference of a great circle of the sphere. Find the radius of the sphere. 8. Show that, if a cylinder is circumscribed about a sphere, the lateral area of the cylinder is equal to the area of the sphere. Also show that the area of the sphere equals two-thirds the total area of cylinder. 9. The radius of the earth is 3960 miles and the radius of the moon is 1080 miles. Find the ratio of their areas. 10. Show how to determine the distance apart the points of a compass must be placed to draw a great circle upon a material sphere that has a diameter of 14 in. So as to draw a circle having a radius of 5 in. 11. What is the locus of the centers of all spheres tangent to a given plane at a given point? 12. What is the locus of the centers of all spheres passing through three given points? 13. What is the locus of points at a distance n from a point Oi and at a distance r 2 from another point O2? Discuss. 14. Find the area of a sphere circumscribed about a tetraedron whose edge is 4 in. 15. The radii of two intersecting spheres are 10 in. and 12 in., respec- tively, and the distance between their centers is 16 in. Find the radius of the circle of intersection of the two spheres. 16. Three equal spheres each tangent to the other two rest on a plane. A fourth sphere rests on the first three spheres. Find the distance from the center of the fourth sphere to the plane, if each sphere has a radius of 8 in. Ans. 21.064 in. MEASUREMENT OF THE SPHERE, AREA 397 806. Zones. The portion of the surface of a sphere in- cluded between two parallel planes is called . a zone. /^SS^\\ It is evident that a zone can be consid- 4/x^_ 1Z^>\ ered as generated by an arc of a great y 5 T 1 circle. A, >A The circles made by the two parallel L — ^ — A planes are called the bases of the zone,, and the 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. The zones on the surface of the earth are included between certain parallels of latitude. The torrid and temperate zones are zones of two bases, and the frigid zones are zones of one base. 807. Theorem. The area of a zone of a sphere is equal to the product of the altitude of the zone and the circumference of a great circle of the sphere; that is, area = 2wrh, where r denotes the radius of a great circle and h the altitude of the zone. The proof is similar to that of § 804. 808. Theorem. The areas of zones of the same sphere or of equal spheres are to each other as their altitudes. EXERCISES 1. Find the area of a zone of one base and having an altitude of 5 ft. on a sphere 8 ft. in diameter. 2. Find the radius of a sphere, on which a zone of altitude 5 in. has an area of 100 sq. in. 3. The diameter of a sphere is 12 in., and is divided into four equal parts by parallel planes. Find the area of each zone. 4. Find what part of the earth's surface is north of the parallel 45° north. North of the parallel 60° north. 5. Show that one-half of the earth's surface lies within 30° of the equator. 6. What part of the earth's surface could be seen from a point whose distance from the surface is equal to the radius? Twice the radius? 398 SOLID GEOMETRY 809. Lunes. That portion of a sphere between the halves of two great circles of the sphere is called a ^jf^ lune. The diedral angle between the planes of the c {^_|_|q_|!^2> great circles bounding the lune is the angle of the lune. From § 589 it follows that the angle of a lune is measured by the plane angle of the diedral angle formed by the planes of the great circles bounding the lune. 810. Comparison of Lunes. If, on the same sphere, one lune is placed on another with one bounding semicircle in common, the other bounding semicircles either coincide, or one lies wholly within the lune bounded by the other. It is evident then that: (1) On the same sphere or equal spheres , two lunes having equal angles are equal; and conversely. (2) On the same sphere or equal spheres, two lunes are in the same ratio as their angles. 811. Theorem. The area of a lune is given by the formula u L = (47rr 2 ), where L denotes the area of the lune, u the anqle of 360 . the lune in degrees, and r the radius of the sphere. To prove this, consider the sphere as a lune whose angle is 360°, then by § 810 (2), — = — , or L = — (47rr 2 ). 4?rr 2 360 360 EXERCISES 1. Find the area of a lune whose angle is 30° on a sphere 8 ft. in diameter. 2. Find the part of the earth's surface between the equator and the parallel 30° north, and between the meridians 120° west and 150° west. 3. The chord of the polar distance of a circle is 8 in. If the radius of the sphere is 10 in., what is the radius of the circle? What is the area of the zone cut off by the circle? MEASUREMENT OF THE SPHERE, AREA 399 4. Find the altitude of a zone whose area equals the area of a great circle of the sphere. 5. A plane passes through a sphere bisecting at right angles a radius of the sphere. Compare the areas of the two parts of the sphere. 6. If h is the height of P above the surface of the earth and r the radius, then S, the extent of the surface of the earth visible from P, is given by 2lTr 2 h the formula S = - r+h S = 2TrXMQ. But MQ=r-OM, and OM : OA=OA : OP. §807 Then OM = -, and MQ=r- r+h ~r+h rh 'r+h' :.S = 2TTrX- rh 2lTr 2 h r-\-h r-\-h 7. How many miles above the surface of the earth is a point from which one-fourth of the surface can be seen? Arts. r. 8. , From a point 6 ft. from a sphere one-fourth of its area is visible. Find the radius of the sphere. 9. An ash tray is in the form of the zone of a sphere. If the tray is 1 in. deep and 3^ in. in diameter at the top, find the diameter of the blank from which it is pressed. C 10. A zone of one base has the same area as a circle whose radius is the chord of the gener- Ai ating arc of the zone, that is, the area of the D zone generated by arc BC revolving about DC is equal in area to a circle having chord BC for a radius. 11. Show that the area of a zone of one base is given by the formula S = lT(w 2 + W) =7T[(|w) 2 +/i 2 ], where S denotes the area, w the diameter of the base of the zone and h the altitude. 12. In a practical handbook, the following rule is given as nearly correct. Show that it is accurate. The area of a flanged spherical segment is equal to a circle of radius equal in length to the fine drawn from the top of the segment to the edge of the flange, that is equal to a circle of radius AB. Suggestion. Area of zone = 7r[(|^) 2 +/i 2 ]. Area of flange = 7T[(^) 2 -(^) 2 ]._ Area of zone and flange = 7T [ (^d) 2 +h 2 ]=TT A B 2 . 400 SOLID GEOMETRY VOLUME OF A SPHERE 812. By the volume of a sphere is meant the volume within the sphere. 813. If a regular polyedron, for instance a cube, is circum- scribed about a sphere, and lines drawn from each vertex to the center of the sphere, these lines may be taken as the lateral edges of pyramids having their vertices at the center of the sphere and having the faces of the polyedron as bases. The volume of the polyedron can then be expressed as the volume of these pyramids. Further, if tangent planes be drawn where each lateral edge intersects the sphere, a circumscribed polyedron of a greater number of faces will be formed. It is evident that this process can be continued indefinitely, thus forming polyedrons whose areas and volumes steadily decrease, and can be made to differ as little as desired from the area and volume, respectively, of the sphere. The following statement can then be accepted. 814. If each vertex of a circumscribed polyedron is joined to the center of the sphere, and tangent planes are drawn at the points where these lines intersect the sphere, then, as the number of faces is increased indefinitely, the area and the volume of the polyedron approach as limits the area and the volume, respectively, of the sphere. VOLUME OF A SPHERE 401 815. Theorem. The volume of a sphere is equal to the product of its area by one-third of its radius. H N jot Given a sphere of radius r and area S. To prove that the volume V = SX^r. Proof. Circumscribe about the sphere any polyedron, for instance a cube, and denote the surface of the polyedron by S'. Since the polyedron may be considered as composed of pyra- mids having the faces of the polyedron as bases and the center of the sphere as common vertex, the volume V of the poly- edron is V' = S'xb If the number of faces of the polyedron is indefinitely in- creased as described in § 813, f->V, S'->S, and S'x\r->Sx\r. §§ 814, 485 (2) But V f = S f X^r is always true as the number of faces is increased. v=sxh r. §485 816. Theorem. The volume V of a sphere of radius r and diameter d is given by the formulas: V = j^Trr z and V = \ird z . For V = \rS by § 815. But >S = 4?rr 2 or ird? by § 804. V = 4irr 3 and V = \ird\ 817. Theorem. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. Let Vi, r h di and Fi, r 2 , a\ be the volumes, radii and, diam- eters, respectively. Then Ii=i^l = r Z 0Y Yi=b^l^ fjrr 2 * r 2 * V 2 Indf a\* 402 SOLID GEOMETRY 818. Theorem. The prismoid formula holds for a sphere. In the formula, V = %h(Bi+B 2 +4M) of § 751, /i = 2r, Bi = Q, 5 2 = 0, 4M = 47rr 2 . ,\ 7 = |7rr 3 . 819. Theorem. The volume of a sphere is equal to two-thirds the volume of the circumscribed right cylinder. EXERCISES 1. Consider the surface of a sphere divided as shown in the figure. Can you conclude that the volume is as given in § 815? 2. Show that the volumes of the right circular cylinder, the sphere, and the right circular cone with dimensions as given in the figure, are in the ratio of 3:2:1. Note. Archimedes (287- 212 b.c.) who was the great- est mathematician of antiq- uity, proved the theorem of Exercise 2. This theorem may well be regarded as one of the most beautiful in elementary mathe- matics. Archimedes proved many of the most important theorems in regard to the sphere. 3. State a formula for finding the radius of a sphere when the volume is given. One for finding the diameter when the volume is given. 4. A ball of diameter 2 in. is inside a right circular cylinder of diameter and altitude each 2 in. How many cubic inches of water can be poured into the cylinder? 5. Find the weight of a spherical cast iron ball of diameter 6 in., if cast iron weighs 0.26 lb. per cubic inch. 6. Find the volume of a hollow spherical shell, the outer diameter being 8 in. and the thickness 2 in. Ans. 234.6 cu. in. 7. How many spherical bullets -f of an inch in diameter can be made from 5 pounds of lead, if lead weighs 0.412 pounds per cubic inch? 8. The volume of two spheres are in the ratio of 512 : 27. Find the radius of each if the sum of their radii is 44 in. 9. A water tank, having a total length of 6 ft. and a diameter of 18 in., is in the form of a right circular cylinder with two hemispherical ends. Find its capacity in gallons. Ans. 72.7+ gal. VOLUME OF A SPHERE 403 820. Spherical Sector. If a sphere is generated by the revolution of a semicircle, the solid generated by a sector whose arc is a part of this semicircle is called a spherical sector. A spherical sector generated by a circular sector revolving about one of its bounding radii is called a spherical cone. The zone generated by the arc of the generating sector is called the base of the spherical sector. 821. Theorem. The volume of a spherical sector is equal to one-third the product of the area of Us base, and its radius. EXERCISES 1. A right circular cone having a vertex angle of 60° has its vertex at the center of a sphere 4 ft. in diameter. What is the volume cut from the sphere by the cone? 2. Find the volume of a spherical sector cut from a sphere 4 ft. in diameter, and whose base is a zone having an altitude of 8 in. 822. Spherical Segment. The portion of the volume of a sphere contained between two parallel planes that intersect the sphere or are tangent to it, is called a spherical segment. The sections of the sphere made by the two parallel planes are the bases of the J spherical segment. If one of the parallel planes is tangent to the sphere, the segment is called a spherical segment of one base. The distance between the parallel planes is the altitude of the spherical segment. 404 SOLID GEOMETRY 823. Theorem. The volume of a spherical segment is given by the formula V =%Trh(r l 2 +r 2 2 )+^7rh z , where h denotes the alti- tude, and ri and r 2 the radii of the bases of the segment. Given the spherical segment with altitude h, and bases ADB and EHF having radii n and r 2 respectively. To prove V = %irh(r l 2 +r 2 2 )+%7rh\ Proof. Let OG = x, V = volume of spherical sector formed by revolving OBF+ volume of cone O-ADB -volume of cone O-EHF. Why? (1) V = %irr 2 h+lwr 1 2 (h+x) -%7rr 2 2 x. Why? To reduce to the form desired it is necessary to eliminate ? and x, and this requires two other equations. They are: (2) r 2 = r l 2 +(h+x) 2 , and (3) r 2 = r 2 2 +x 2 . Why? From (2) and (3) x = r _lzll^E^ 2h Substituting this value of x in (3), 9 h*+r 1 *+r 2 *+2r 1 2 h 2 +2r 2 2 h 2 -2r 1 2 r 2 2 r z = . \h 2 Substituting these values of x and r 2 in (1) and simplifying, V = ±irh(n 2 +r 2 2 )+lTh\ Remark. The same formula is derived if the center of the sphere is within the spherical segment. If the spherical segment has but one base, n = 0, and the formula becomes V = ^irhr 2 2 +\irh z . The student should carry out the work in each of these cases. EXERCISES 1. What part of the volume of the earth is between the planes cutting its surface in the parallels of 30° and 45° north latitude? What part is north of a plane cutting the surface in the parallel of 60° north latitude? 2. In the formula for the volume of a spherical segment, solve for each letter so far as possible. VOLUME OF A SPHERE 405 824. Spherical Wedge. That portion of the volume of a sphere contained between the planes of two great semicircles is called a spherical wedge. Its curved surface is its base, and is, evidently, a lune. The solid A-NS-B is a wedge. The lune between the semicircles NAS and NBS is the base of the wedge. The angle AOB of the wedge is the angle of its lune. 825. Theorem. The volume of a spherical wedge is equal to one-third the product of the area of its base and its radius. That is, V = (f^rr 3 ), where r denotes the radius of the sphere and u 360 the angle of the lune in degrees. By a discussion similar to that for the volume of a sphere, the volume of a wedge is given by the formula V = \Lr, where L is the area of the lune. u But L = — (4 tit 2 ) by §811. 360 y=- 360 (iirr*). EXERCISES 1. Find the volume of a spherical wedge whose angle is 20°, cut from a sphere whose radius is 12 in. 2. Find the angle of a spherical wedge if its volume is 418.88 cu. in., and the radius of the sphere is 10 in. 3. A circular flower bed is in the form of a spherical segment that is 25 ft. in diameter and 2j ft. in altitude. How many loads of dirt did it take to build it up if one load is \\ cu. yd.? 4. The figure is a vertical cross section of a V~ 16 /-tf'T casting, the inner and outer "skins" being \ V / / i spherical zones. Find the weight of metal at x^ Ss "^*-— **£— . JL 1 ** 0.35 lb. per cubic inch necessary to make the \^ ^ ^ ^/ _J casting. Ans. 176 lb. 406 SOLID GEOMETRY 6. Find the volume of the segment between two parallel planes 6 in. apart that cut a sphere having a radius of 12 in., if one plane passes 2 in. from the center. There are two cases: (1) when the center of the sphere lies outside the segment, and (2) when the center lies in the seg- ment. Ans. (1) 2186.6 cu. in.; (2) 2638.9 cu. in. 6. The number of square inches in the area of a certain sphere is equal to four times the number of cubic inches in its volume. Find the radius of the sphere. 7. A copper ball to be used as a float must contain 22,449.4 cu. in. What must be its diameter? Ans. 35 in. 8. Derive the formula for the volume of a sphere from the formula for the volume of a spherical sector. 9. Derive the formula for the volume of a sphere from the formula for the volume of a spherical segment. Suggestion. Let n = and r 2 =r and obtain the volume of a hemisphere. 10. If oranges 2 \ in. in diameter are selling for 30 cents a dozen, what should be the price for oranges 3 in. in diameter? 11. The water tank shown in the figure is in the form of a right circular cylinder with a hemisphere at the bottom. Find the capacity in gallons if the diameter is 22 ft. and the altitude of the cylinder is 24 ft. 6 in. 12. Find how many square feet of sheet steel in the tank if it is open at the top. 13. The roof over the tank is third pitch and pro- jects 2 ft. at the eaves. Find the number of square feet in it. 14. A shop received an order for 100 twelve-inch steel balls to be used in crushing cement in a cement mill. They were made from 8^-inch round steel shafting. What length of shaft was taken for each ball if 5% was allowed for scale? 15. A round or button-head machine screw has a head in the form of a spherical segment of one base. Show that the volume of the head is given by the formula: V —Trh{\w 2 -\-\h 2 ) y where w is the diameter of the head and h its height. 16. Find the volume of the head of a round head machine screw if the diameter of the head is 0.731 in. and the height is 0.279 in. Ans. 0.070 cu. in. SPHERICAL ANGLES 407 SPHERICAL ANGLES 826. Angles Formed by Arcs. The angle formed by two arcs of circles is the plane angle formed by the lines tangent to the arcs at the point of intersection. Thus the angle formed by the arcs AB and CB is the #* angle formed by the tangent lines DB and EB. "\ \c \A According to this definition, any two intersect- ing circles on a sphere form angles. In this text, however, only those angles formed by great circles are considered. 827. Spherical Angle. The angle between two great circle arcs is called a spherical angle. The point where the two arcs meet is the vertex of the angle, and the arcs are the sides of the angle. The terms right, acute, and obtuse are applied to spherical angles and have the same meaning as with plane angles. 828. Since the planes of two great circles intersect in a diameter (§ 767), and the tangents at the points of intersec- tion are perpendicular to this diameter (§ 288), the angle formed by the tangents is the plane angle of the diedral angle formed by the planes of the great circles § 588. It follows then from §826 that: A spherical angle equals in degrees the diedral angle between the planes of its sides. s Spherical ZANB = ZCND = °diedral ZA-NS-B. EXERCISES 1. Form definitions for the following when applied to spherical angles: adjacent, complementary, supplementary, vertical. 2. Prove that the sum of all the spherical angles about a point on a sphere is equal to 360°. 3. At what angle does a meridian of the earth intersect the equator? 4. Prove that vertical spherical angles are equal. 408 SOLID GEOMETRY 829. Theorem. A sperical angle is equal in degrees to the arc of the great circle described from its vertex as a pole, and included between its sides, produced if necessary. P Given spherical ZAPB, AB an arc of a great jt J |Y\ circle whose pole is P, and which is included L ■*- — AA between the sides AP and BP. \ \ ni To prove that spherical ZAPB = °AB. \" J Proof. Draw PO, AO, and BO. ^""1^ PO is perpendicular to plane of AB. § 775 Then OA _L PO and BO JL PO. Why? And ZAO£ is the plane angle of diedral ZA-QP-B. Why? Hence diedral ZA-OP-B = °ZAOB. § 589 (3) But ZAOB = °AB. § 305 .-. spherical ZAPB = °AB. § 111 SPHERICAL POLYGONS AND POLYEDRAL ANGLES 830. Spherical Polygons. A portion of a sphere bounded by three or more arcs of great circles is called a spherical polygon. The bounding arcs are the sides of the polygon, the spherical angles formed by the sides are the angles of the polygon, and their vertices the vertices of the poly- gon. 831. A diagonal of a spherical polygon is the arc of a great circle joining any two vertices not adjacent. It is evident that a side of a spherical polygon can be greater than a semicircle, but in this text only those will be considered whose sides are each less than a semicircle. 832. A spherical triangle is a spherical polygon of three sides. The terms right, acute, obtuse, isosceles, and equilateral are applied to spherical triangles and have the same meaning as with plane triangles. It will be found, however, that a spherical triangle may have one, two, or three right or obtuse angles. SPHERICAL POLYGONS AND POLYEDRAL ANGLES 409 833. Central Polyedral Angles. It follows from the defini- tion of a spherical polygon that the planes of the sides of the polygon pass through the center of the sphere. These planes form a polyedral angle, called the corresponding central polyedral angle of the spherical polygon. This polyedral angle and the spherical polygon are so related that it is convenient to consider them in connection with each other. It follows directly from §§ 305 and 829 that: (1) The sides of a spherical polygon are equal in degrees to the face angles of the polyedral angle. (2) The angles of a spherical polygon are equal in degrees to the diedral angles of the polyedral angle. Because of these relations it follows that to each property of a polyedral angle there is a corresponding property of the spher- ical polygon, and vice versa. One of these properties can be stated when the other is given by making the substitution indicated in the following: Spherical Polygon Sides Angles of polygon Vertices of polygon Center of sphere Polyedral Angle Face angles Diedral angles Edges Vertex In this manner a theorem relating to spherical polygons can readily be worded so as to apply to polyedral angles, and vice 834. A convex spherical polygon is a spherical polygon no pide of which, when produced, will enter the polygon. It is evident that the corresponding central polyedral angle of a convex spherical polygon is a convex polyedral angle. (See § 605.) 410 SOLID GEOMETRY 835. Two spherical polygons on the same sphere or equal spheres are congruent if their corresponding central polyedral angles are congruent. (See § 607.) The spheres and 0' are equal and the polyedral angles having their ver- tices at the centers of the spheres are congruent. Then the corresponding spherical polygons ABCD and A'B'C'D' are congruent. 836. Symmetric Polyedral Angles and Spherical Polygons. Two polyedral angles are symmetric if the face angles and diedral angles of one are respectively equal to those of the other but arranged in reverse order. The triedral angles V-ABC and V'-A'B'C are symmetric, for diedral angles AV, BV f and CV are equal respec- tively to diedral angles A'V, B'V, and C'F'j and the face angles AVB, BVC, and CVA are equal respectively to face angles A'V'B, B'V'C, and C'V'A'. It is to be noticed that if the triedral angles are viewed from the vertices V and V, then the parts of the triedral angle V-ABC are arranged in the reverse order to the parts of V'-A'B'C'. If the two triedral angles were made of pieces of cloth sewed together, then, if one were turned "inside out," they could be made to coincide. They would then be congruent polyedral angles. A pair of gloves are analogous to two symmetric polyedral angles. All parts of one are equal to corresponding parts of the other, but are arranged in reverse order. Right and left parts of animals are symmetric to each other in this sense. Numerous such illustrations of symmetry can be found in nature and in the arts. 837. Two spherical polygons on the same sphere or an equal spheres are symmetric if their corre- sponding central polyedral angles are sym- metric. In the figure, spherical triangles ABC and A'B'C are symmetric for their corresponding central triedral angles are symmetric. SPHERICAL POLYGONS AND POLYEDRAL ANGLES 411 838. The two polyedral angles formed by a pyramidal surface are symmetric; and, if a sphere is drawn with its center at the vertex of the pyramidal surface, then the pyramidal surface will intersect the sphere in two symmetric spherical polygons. It is evident that the parts of the two nappes of the pyramidal surface with vertex are arranged in reverse j order, then the corresponding spherical polygons ABCD \ and A'B'C'D' are symmetric. Of course, symmetric spherical polygons can He in any position on the sphere, for a figure on a sphere can be moved about without changing its shape. 839. Theorem. In any spherical triangle, the sum of two sides is greater than the third side. Given spherical AABC on sphere 0. To prove that AB+BC>AC. Proof. ZAOB+ZBOOZAOC. §608 But ZAOB = °AB, ZBOC = °BC, and ZAOC = °AC. .\AB+BC>AC. 840. Theorem. In any convex spherical poly- gon, the sum of the sides is less than a great circle of the sphere, or less than 360°. ZAOB+ZBOC+ZCOD+ZDOA<3W°. § 609 .*. AB+BC+CD+DA180°. § 178 Also a'+b f +c f >0°. Why? .\A+B+C<540°. Why? 859. Theorem. The sum of the diedral angles of a triedral angle is greater than 180° and less than 51fi>°. EXERCISES 1. How would you draw a triangle on a sphere, the sum of the angles of which would be only a little greater than 180°? Only a little less than 540°? Could you draw such triangles on a small sphere or would it be necessary to have a large sphere? 2. What is the locus of points on a sphere and equally distant from two intersecting arcs of great circles of the sphere? 3. What is the locus of the points on a sphere that are equally distant from the ends of an arc of a great circle? 4. If the sides of a spherical triangle each have less than 3°, what can be said of the angles of its polar triangle? Could the angles of a spher- ical triangle be respectively 170°, (S°, and 12°? 420 SOLID GEOMETRY AREAS OF SPHERICAL POLYGONS 860. Theorem. On the same sphere or on equal spheres two symmetric spherical triangles are equal in area. Given the two symmetric spherical triangles ABC &nd A' B'C. To prove that AABC = AA'B'C Proof. Place the two triangles so that they are formed by the same pyramidal surface with its vertex at 0. § 838 Let P be the pole of a small circle passing through A, B, and C. Draw the diameter POP' and the great circle arcs PA, PB, PC, P'A', P'B', and P'C. Then PA=PB = PC. §783 But P^A^PA, P'B'^PB, fiV^PC. Why? Then P 7 A , = i% , = P 7 C^ Why? And APAB and P'A'B' are isosceles as well as symmetric. Hence APAB^AP'A'B'. §843 Similarly APBC^AP'B'C, and APCA^AP'C'A'. Then APAB+ APBC+ APCA = AP'A'B'+ AP'B'C'+AP'C'A'. § 105 .-. AABC = AA , B , C. §§ 109, 104 EXERCISES 1. Draw a figure in which the pole P, § 860, falls outside the triangle ABC. What difference would this make in the proof? 2. A sphere can be divided into four, eight, or twenty equal equilateral spherical triangles. AREAS OF SPHERICAL POLYGONS 421 861. Spherical Excess. The excess of the sum of the angles of a spherical triangle over 180° is called the spherical excess of the triangle. The spherical excess of a spherical polygon of n sides is the excess of the sum of its angles over (n — 2)180°, that is, it is the excess over the sum of the angles of a plane polygon of the same number of sides. 862. Area of a Spherical Triangle. The sides as well as the angles of a spherical triangle are usually given in degrees. It is evident then that the number of square units in the area of a spherical triangle will depend upon the radius of the sphere on which the triangle is. When the radius of the sphere is not given it is convenient to have a unit of meas- ure that is some fractional part of the surface of a sphere. 863. Spherical Degree. One half the ar,ea of a lune of 1°, or the area of a birect angular triangle having a vertex angle of 1°, or y^-Q of the surface of a sphere, is called a spherical degree. It is the unit of measure of spherical polygons. A polyedral angle whose vertex is at the center of a sphere cuts a spherical polygon out of the sphere. The area of this polygon in spherical degrees is called the measure of the polyedral angle. In this manner the magnitudes of polyedral angles may be compared. EXERCISES 1. How many spherical degrees in a lune of 20°? Of 46° 30'? 2. How many spherical degrees in a trirectangular spherical triangle? What is the area in square inches of a trirectangular triangle on a sphere having a radius of 10 in.? 3. How many square feet in the area of a birectangular triangle with vertex angle 18° on a sphere 6 ft. in. diameter? 4. Find the spherical excess in spherical triangles having the following angles: (1) 100°, 117°, 136°. (2) 141°, 97° 30', 67° 45'. (3) 116° 23' 14", 127° 43' 18", 69° 45' 15". 6. Find the spherical excess in a spherical pentagon having angles of 113°, 97°, 119° 17', 141° 23', and 172° 27', 422 SOLID GEOMETRY 864. Theorem. A spherical triangle is equivalent to a lune on the same sphere, whose angle is half the spherical excess of the tri- angle. Given the spherical AABC on a sphere of surface S and center 0, with A, B, and C denoting the value in degrees of the angles at the respective vertices. To prove that AABC = a lune of angle ^(A+B+C -180°). Proof. Complete the great circles of the sides of AABC, and let one of them as BC from the boundary of the hemisphere which is divided into four triangles, ABC, or X, ABC or Y, AB'C or Z, and AB'C or U. ABA'C on the reverse hemisphere = AZ. § 860 Lune of ZA = AX+AZ. Why? Lune of ZB = AX+ A*7. Why? Lune of AC = AX+AY. Why? Then lune of Z(A+B+C) = ZAX+AZ+AU+AY. § 105 But AX+AZ+AU+AY = ±S = lune of Z180°. Why? Then lune of Z(A+B+C) =2AX+lune of Z180°. § 111 Or 2AX = lune of Z(A+B+C -180°). Why? .-. AABC = lune of Z\(A+B+C -180°). Why? 865. Theorem. On the same sphere or on equal spheres, the areas of two spherical triangles are to each other as their spherical excesses. 866. Theorem. // A denotes the area of a spherical triangle, E the number of degrees in its spherical excess, and r the radius of irr 2 E the sphere, then the area is given by the formula: A = -7^r« ? AREAS OF SPHERICAL POLYGONS 423 867. Theorem. A spherical polygon is equivalent to a lune on the same sphere, whose angle is half the spherical excess of the polygon. Divide the spherical polygon into spherical triangles by drawing the diagonals from one vertex. Then the area of the polygon is equivalent to the sum of the areas of the triangles, and the spherical excess of the polygon is equal to the sum of the excesses of the triangles. 868. Theorem. The area of a spherical polygon is given by 7rr 2 E the formula A — , where r is the radius of the sphere and E the number of degrees in the spherical excess. Note. This theorem was stated by A. Girard in 1629 and was first completely proved by Cavalieri a few years later. 869. Spherical Pyramid. The portion of the volume of a sphere bounded by a spherical polygon and the planes of its sides is called a spherical pyramid. The polygon is the base of the spherical pyramid, and the center of the sphere is its vertex. Thus, O-ABCD is a spherical pyramid. 870. Theorem. The volume of a spherical pyramid is equal to one-third the product of the area of its base by the radius of the sphere. EXERCISES 1. Find the areas of triangles on spheres of the given radii, and having angles as follows: (1) 80°, 140°, 120°, r = 10 in. (3) 125°, 135°, 145°, r = 50 ft. (2) 150°, 75°, 110°, r = 16 in. (4) 179°, 179°, 179°, r = 10 ft. (5) 112° 30' 15", 127° 43' 30", 175° 27' 15", r = 20 ft. (6) 37° 43' 27", 15° 45' 34", 170° 45' 43", r = 4000 mi. 2. Find the areas of polygons on spheres of the given radii, and having angles as follows: (1) 70°, 170°, 100°, 125°, r = 12in. (2) 100°, 98°, 175°, 160°, 128°, 96°, r = 8 ft. (3) 75° 45' 17", 127° 14' 16", 102° 43' 16", 150° 43' 41", r = 20 in. 424 SOLID GEOMETRY 3. Find the area of the triangle on the earth's surface, determined by the meridians at 48° and 75° west longitude, and the equator. Use 4000 miles for the radius of the earth. 4. Find the volume of a spherical pyramid, given that its base is an equiangular triangle having each angle 120°, on a sphere of radius 9 in. 5. Find the volume of a spherical pyramid, given that its base is a spherical hexagon each angle of which is 130°, on a sphere of 12 in. radius. 6. The volume of a spherical pyramid whose base is an equiangular spherical triangle with angles of 105° is 2567T cu. in. Find the diameter of the sphere. 7. What is the greatest area that a triangle on the surface of the earth may have if the sum of its angles is not to differ by more than 1° from 180°? If the sum is not to differ by more than 1' from 180°? 8. A triangle having an area of 62 sq. in. is on a sphere that has an area of 500 sq. in. Two of its angles are 92° and 135° respectively. Find the third angle. 9. The sides of a spherical triangle are 79°, 88°, and 115°, and it is on a sphere whose radius is 18 in. Find the area of its polar triangle. QUESTIONS 1. Define a circle. A sphere. Is a circle an area? Is a sphere a solid? Define a sphere as a locus. Define a circle as a locus. 2. Is it in agreement with the definition of a sphere to speak of the area of a sphere? What is the volume of a sphere? Compare the defini- tion of the area of a circle with that of the volume of a sphere. 3. What is a spherical angle? A spherical triangle? Is a triangle an area? 4. How is a plane angle measured? A spherical angle? A diedral angle? A polyedral angle? 5. What is a lune? A wedge? How may lunes be compared in size? How may wedges be compared in size? 6. Define the sector of a sphere, and compare with the sector of a circle. 7. Define the segment of a sphere, and compare with the segment of a circle. Do we speak of the segment of a circle of one base? Is there any reason for not defining the segment of a circle so that we can speak of a segment of two bases? 8. Is the distance between two points on a sphere defined so that it is analogous to the distance between two points in a plane? GENERAL EXERCISES 425 9. Explain why it is well to treat spherical polygons in connection with polyedral angles. State the relations between the parts of a spherical triangle and the parts of the corresponding polyedral angle. 10. State several theorems about spherical triangles that are analogous to theorems about plane triangles. Is a plane triangle determined when its angles are known? Is a spherical triangle determined when its angles are known? 11. What is a polar triangle? What are some relations between a spherical triangle and its polar triangle? State a striking difference between the angles of a spherical triangle and those of a plane triangle. 12. How many points determine a circle? A sphere? 13. How many points on a sphere determine a small circle? A great circle? 14. Define axis, pole, polar distance. 15. If a great circle is perpendicular to a small circle, what must it contain? Is the converse of this true? 16. Give an outline of the steps in determining the formulas for the area of a sphere. State these formulas, and state when each is the best one to use. 17. Do the same for the volume of a sphere. 18. What is the formula for the area of a zone? A lune? A spherical triangle? Can you give a practical application of the use of each of these areas? 19. What is the formula for the volume of a spherical segment? A wedge? A spherical pyramid? A spherical sector? Can you give a practical application of the use of each of these volumes? 20. What remarkable relation is there between tlie volumes of a cylinder, sphere, and cone? Is there a similar relation between their areas? GENERAL EXERCISES 1. The volume of any polyedron circumscribed about a sphere is equal to the area of its surface times one-third the radius of the sphere. 2. The volumes of polyedrons circumscribed about the same sphere are to each other as their surfaces. 3. What is the locus of all points in space equally distant from two given points and at a distance r from a third given point? 4. Show how to find the pole of a given circle on a sphere. 5. Construct a circle through three given points on a sphere. 426 SOLID GEOMETRY 6. Four balls of equal radii r are placed on a plane and touching each other in such a way that their centers form a square. A fifth ball of the same radius is placed upon the four first balls. Find the distance of its center from the plane. 7. A cylinder is inscribed in a sphere of radius r and has a lateral area equal to one-half the area of the sphere. Find the radius of the cylinder. Ans. %r\/~2. 8. A zone of one base is a mean proportional between the remaining part of the area of the sphere and the total area of the sphere. How far is its base from the center of the sphere? 9. A spherical segment of one base is half as large as the spherical sector to which it belongs. If the radius of the sphere is 4 in., find the altitude of the segment. 10. The inside of a glass is in the form of a right circular cone whose vertex angle is 60° and the diameter of whose base is 4 in. The glass is filled with water and the largest sphere that can be immersed in the water is placed in the glass. Find the volume of water remaining in the glass. 11. An irregular portion, but less than half, of a material sphere, is given. Show how to construct the diameter of the sphere. 12. In the figure of § 813, the plane MNP is tangent to the sphere inscribed in the cube. If the edge of the cube is a find DN. 13. Two spheres whose radii are respectively 6 in. and 8 in. have their centers 10 in. apart. Find the volume of the portion common to the two spheres. This is the form of a spherical lens. 14. A hole of diameter 4 in. was bored through the center of a sphere of diameter 12 in. Find the volume of the part cut away. 15. A sphere is inscribed in a right circular cylinder having an altitude equal to the diameter of the sphere. Two planes parallel to the base of the cylinder cut the cylinder and the sphere. Prove that the surface of the cylinder lying between the planes equals the area of the zone between the planes. 16. A circle is circumscribed about an equilateral triangle. Find the ratios of the areas and the volumes of the solids formed by revolving the triangle and the circle about an altitude of the triangle as axis. 17. A hollow copper sphere used as a float in water weighs 10 oz., and has a diameter of 5 in. How heavy a weight will it support? Arts. Less than 27.9 oz. FORMULAS FOR REFERENCE 427 FORMULAS FOR REFERENCE Lateral area of prism. S = pe, S = ph. Lateral area of circular cylinder. S — pe> S = ch. Volume of rectangular parallelepiped. V = abh. Volume of parallelepiped. V = Bh. Volume of prism or cylinder. V = Bh. Volume of hollow circular cylinder. V = \irh(D+d)(D — d). Lateral area of regular pyramid. S = ^ps. Lateral area of right circular cone. S = ^cs = wrs. Total area of right circular cone. T = Trs+wr 2 = irr(s+r). Lateral area of frustum of regular pyramid. S = ^s(P+p). Lateral area of frustum of right circular cone. S = ir(R+r)s. Volume of pyramid or cone. V = ^Bh. Volume of circular cone. V — \irr 2 h. Volume of frustum of pyramid or cone. Vm%h(Bi+Bt+\fihBl). Volume of frustum of circular cone. 7 = |7r/i(r 1 2 +r 2 2 +r 1 r 2 ) =-&7rh(d 1 *+d 2 2 +d l d 2 ). Volume of prismatoid. 7 = fft(J?i+ J B 2 +4Af). Area of a sphere. >S = 47rr 2 . Area of a zone. Z = 2irrh. u Area of a lune. L — — (47rr 2 ). 360 Volume of a sphere. V = jfirr' 6 = f 7rd 3 = Jr& Volume of spherical segment. V = ^irh(ri 2 +r2 2 )+^wh 3 . 11 Volume of spherical wedge. V = Of Trr 3 ) . Area of spherical polygon. A = . 428 SOLID GEOMETRY USEFUL NUMBERS 1 cu. ft. of water weighs 62.5 lb. (approx.) = 1000 oz. 1 gal. of water weighs 8 J lb. (approx.). 1 atmosphere pressure = 14.7 lb. per sq. in. = 2116. lb. per sq. ft 1 atmosphere pressure = 760 mm. of mercury. A column of water 2.3 ft. high = a pressure of 1 lb. per sq. in. 1 Kg. = 2.2 lb. (approx.). 1 gal. = 231 cu. in. (by law of Congress). 1 cu. ft. = 7^ gal. (approx.) or better 7.48 gal. 1 cu. ft. = 4 bu. (approx.). 1 bbl. =4.211- cu. ft. (approx.). 1 bu. = 2150.42 cu. in. (by law of Congress) = 1.24446- cu. ft. 1 bu.=f cu. ft. (approx.). 1 perch = 24f cu. ft. but usually taken 25 cu. ft. 1 in. = 25.4 mm. (approx.). 1 m. = 39.37 in. (by law of Congress). 1 lb. (avoirdupois) = 7000 grains (by law of Congress). 1 lb. (troy or apothecaries) = 5760 grains. 7r = 3.14159265358979+=3.1416 = fff = 3+ (all approx.). V2= 1.4142136. \/3 = 1.7320508. V5 = 2.2360680. \/6 = 2.4494897. V^ 1.2599210. V^ = 1.4422496. SYLLABUS OF PLANE GEOMETRY (FOR REFERENCE OR REVIEW) Chapter I. Straight-Line Figures PRELIMINARY PROPOSITIONS AND AXIOMS § 19. Experiment. If two triangles have two angles and the included side of one equal respectively to two angles and the included side of the other, the triangles are equal. § 21. In equal triangles, the corresponding sides are equal and the corresponding angles are equal. § 23. Experiment. If two triangles have two sides and the included angle of one equal respectively to two sides and the included angle of the other, the triangles are equal. § 25. Construction. To construct a triangle when the three sides are given. § 27. Experiment. If two triangles have the three sides of one equal respectively to the three sides of the other, the triangles are equaL § 30. Experiment. The sum of the angles of a triangle is equal to 180°. § 32. If two triangles have two angles of one equal respectively to two angles of the other, the third angles are equal. § 35. Experiment. If one straight line intersects another straight line, the vertical angles are equal. § 49. If two adjacent angles have their exterior sides in the same straight line, they are supplementary. § 51. If two adjacent angles are supplementary, their exterior sides are in the same straight line. § 53. Complements of the same angle or of equal angles are equal. § 55. Supplements of the same angle or of equal angles are equal. § 58. If one of the angles formed by two intersecting lines is a right angle, the other three angles also are right angles. § 59. The sum of all the angles about a point is equal to 360°, or four right angles. § 60. The sum of all the angles about a point, on the same side of a straight line passing through the point, is equal to 180°. § 61. All right angles are equal. § 62. All straight angles are equal. 429 430 SOLID GEOMETRY § 63. A right angle is half a straight angle. § 65. One straight line, and only one, can be drawn from one point to another. § 66. Two points determine a straight line. § 67. A straight line may be produced indefinitely in both directions. § 68. Any number of straight lines can be drawn through a given point. § 70. Two straight lines can intersect in only one point. § 71. Two intersecting lines determine a point. § 74. At a given point in a given straight line, only one perpendicular can be drawn to that line. § 76. Only one perpendicular can be drawn to a given straight line from a given external point . § 77. Through a given point, one straight line, and only one, can be drawn perpendicular to a given straight line. TRIANGLES § 84. Problem. To construct an equilateral triangle when a side is given. § 85. Problem. To construct an isosceles triangle when the base and one of the equal sides are given. § 86. Theorem. In an isosceles triangle, the angles opposite the equal sides are equal. § 87. Theorem. An equilateral triangle is also equiangular. § 88. Theorem. Each angle of an equilateral triangle is equal to 60°. § 89. Theorem. If two angles of a triangle are equal the sides oppo- site the equal angles are equal and the triangle is isosceles. § 90. Theorem. An equiangular triangle is also equilateral. § 91. Theorem. The sum of the acute angles of a right triangle is equal to 90°, or one right angle. § 92. Theorem. If two right triangles have one acute angle of one equal to one acute angle of the other, the other acute angles, also, are equal. § 93. Theorem. Each acute angle of an isosceles right triangle is equal to 45°. § 97. If two right triangles have a side and an acute angle of one equal respectively to the corresponding side and the corresponding acute angle of the other, the triangles are equal. § 98. Theorem. If two triangles have a side and any two angles of one equal respectively to the corresponding side and the two correspond- ing angles of the other, the triangles are equal. § 101. If two right triangles have two sides of one equal respectively to two corresponding sides of the other, the triangles are equal. SYLLABUS OF PLANE GEOMETRY 431 AXIOMS OF EQUALITY § 104. Axiom. Quantities which are equal to the same quantity or to equal quantities are equal to each other. § 105. Axiom. If equals are added to equals, the sums are equal. § 106. Axiom. If equals are subtracted from equals, the remainders are equal. § 107. Axiom. If equals are multiplied by equals, the products are equal. § 108. Axiom. If equals are divided by equals, the quotients are equal. The divisor must not be zero. § 109. Axiom. The whole is equal to the sum of all its parts. § 110. Axiom. The whole is greater than any of its parts. § 111. Axiom. A quantity may be substituted for its equal in any operation. § 113. Theorem. If two triangles have the three sides of one equal respectively to the three sides of the other, the triangles are equal. PARALLEL LINES § 115. Theorem. Straight lines in the same plane which are per- pendicular to the same straight line can never meet. § 117. Theorem. Straight lines in the same plane which are per- pendicular to the same straight line are parallel. § 118. Axiom. Through a given point only one straight line can be drawn parallel to another straight line. § 119. Axiom. Straight lines in the same plane which are not par- allel will meet if sufficiently prolonged. § 122. Experiment. If two parallel lines are cut by a transversal: (1) The alternate interior angles are equal. (2) The alternate exterior angles are equal. (3) The exterior interior angles are equal. (4) The consecutive interior angles are supplementary. (5) The consecutive exterior angles are supplementary. § 123. Theorem. If a straight line is perpendicular to one of two parallel lines, it is perpendicular to the other also. § 124. Theorem. Two straight lines parallel to a third straight line are parallel to each other. § 127. Theorem. If two straight lines in the same plane are cut by a transversal, making the alternate interior angles equal, the lines are parallel. § 128. Theorem. If two straight lines in the same plane are cut by a transversal, making the exterior interior angles equal, the lines are parallel. 432 SOLID GEOMETRY § 129. Theorem. If two straight lines in the same plane are cut by a transversal, making the alternate exterior angles equal, the lines are parallel. § 130. Theorem. If two straight lines in the same plane are cut by a transversal, making the consecutive exterior angles supplementary, the lines are parallel. § 131. Theorem. If two straight lines in the same plane are cut by a transversal, making the consecutive interior angles supplementary, the lines are parallel. § 132. Theorem. If two parallel lines are cut by a transversal, the alternate interior angles are equal. § 133. Theorem. If two parallel lines are cut by a transversal, the exterior interior angles are equal. § 134. Theorem. If two parallel lines are cut by a transversal, the alternate exterior angles are equal. § 135. Theorem. If two parallel lines are cut by a transversal, the consecutive interior angles are supplementary. § 136. Theorem. If two parallel lines are cut by a transversal, the consecutive exterior angles are supplementary. § 138. Theorem. If two angles have their sides parallel, right side to right side, and left side to left side, the angles are equal. § 139. Theorem. If two angles have their sides parallel, right side to left side, and left side to right side, the angles are supplementary. § 140. Theorem. If two angles have their sides perpendicular, right side to right side, and left side to left side, the angles are equal. § 141. Theorem. If two angles have their sides perpendicular, right side to left side and left side to right side, the angles are supplementary. § 143. Theorem. The sum of the angles of a triangle is equal to 180°, or two right angles. § 144. Theorem. An exterior angle of a triangle is equal to the sum of the opposite interior angles, and is therefore greater than either of them. § 145. Theorem. In a triangle, there can be only one right angle, or one obtuse angle. QUADRILATERALS § 154. Theorem. A diagonal divides a parallelogram into two equal triangles. § 155. Theorem. In any parallelogram, the opposite sides are equal, and the opposite angles are equal. § 156. Theorem. The diagonals of a parallelogram bisect each other. § 157. Theorem. Any two consecutive angles of a parallelogram are supplementary. SYLLABUS OF PLANE GEOMETRY 433 § 168. Theorem. Parallels cut off by two parallel lines are equal. § 169. Theorem. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. § 160. Theorem. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. § 161. Theorem. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. § 162. Theorem. The line joining the middle points of two opposite sides of a parallelogram is parallel to and equal to each of the other two sides. § 163. Theorem. The diagonals of a rectangle are equal. § 164. Theorem. If a parallelogram has equal diagonals, it is a rectangle. § 165. Theorem. The diagonals of a square are perpendicular to each other, and bisect the angles of the square. POLYGONS § 170. Theorem. The sum of the interior angles of any polygon is equal to 180° multiplied by two less than the number of sides. § 171. Theorem. In an equiangular polygon, each angle is equal to (tt-2)180° (n-2)2rt.^ , or . n n § 172. Theorem. The sum of the exterior angles of a polygon, formed by prolonging one side at each vertex, is 360°. INEQUALITIES § 173. Axiom. If equals are added to unequals, the sums are unequal in the same order. § 174. Axiom. If equals are subtracted from unequals, the remainders are unequal in the same order. § 175. Axiom. If unequals are multiplied by positive equals, the products are unequal in the same order. § 176. Axiom. If unequals are divided by positive equals, the quo- tients are unequal in the same order. § 177. Axiom. If unequals are added to unequals in the same order, the sums are unequal in that order. § 178. Axiom. If unequals are subtracted from equals, the remain- ders are unequal in the reverse order. § 179. Axiom. If the first of three quantities is greater than the second and the second is greater than the third, then the first is greater than the third, 434 SOLID GEOMETRY § 181. Axiom. A straight line is the shortest line that can be drawn from one point to another. § 182. Theorem. The sum of two sides of a triangle is greater than the third side. The difference between two sides of a triangle is less than the third side. § 183. Theorem. If two sides of a triangle are unequal, the angles opposite these sides are unequal, and the greater angle is opposite the greater side. § 184. Theorem. If two angles of a triangle are unequal, the sides opposite these angles are unequal, and the greater side is opposite the greater angle. § 185. Theorem. A perpendicular is the shortest line that can be drawn from a point to a straight line. § 186. Theorem. If two straight lines are drawn from a point in a perpendicular to a given line, cutting off on the line unequal lengths from the foot of the perpendicular, the more remote is the greater. DISTANCE § 191. Theorem. Two parallel lines are everywhere equally distant. § 192. Theorem. If a line bisects one side of a triangle and is par- allel to a second side, it bisects the third side. § 193. Theorem. The middle point of the hypotenuse of a right triangle is equidistant from the three vertices. § 194. Theorem. If one acute angle of a right triangle is twice the other, the hypotenuse is twice the shorter side. §. 196. Theorem. The line joining the middle points of two sides of a triangle is parallel to the third side and is equal to one-half the third side. § 197. Theorem. If three or more parallel lines intercept equal parts on one transversal, they intercept equal parts on every transversal. § 198. Theorem. The median of a trapezoid is parallel, to the bases and is equal to half the sum of the bases. § 199. Theorem. All points in the bisector of an angle are equi- distant from the sides of the angle. § 200. Theorem. All points equidistant from the sides of aji angle are in the bisector of the angle. § 205. Theorem. All points in the perpendicular bisector of a straight line are equidistant from the extremities of the line. § 206. Theorem. All points equidistant from the extremities of a straight line lie in the perpendicular bisector of the line. § 207. Theorem. Two points, each equidistant from the extremities of a straight line, determine the perpendicular bisector of the line. SYLLABUS OF PLANE GEOMETRY 435 OTHER THEOREMS § 209. Theorem. The bisector of the angles of a triangle pass through a common point, which is equidistant from the sides of the triangle. § 210. Theorem. The perpendicular bisectors of the three sides of a triangle pass through a common point, which is equidistant from the three vertices of the triangle. § 211. Theorem. The medians of a triangle pass through a common point, which is two-thirds of the length of each median from its vertex. § 213. Theorem. The three altitudes of a triangle pass through a common point. § 247. Theorem. Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. § 248. Theorem. Two triangles are congruent if two angles and the included side of one are equal respectively to two angles and the included side of the other. § 249. Theorem. Two parallelograms are congruent if they have two adjacent sides and the included angle of one equal respectively to two adjacent sides and the included angle of the other. § 251. Theorem. If two angles of a triangle are unequal, the sides opposite these angles are unequal, the greater side being opposite the greater angle. § 252. Theorem. Only one perpendicular can be drawn to a given line at a given point in the line. § 253. Theorem. Only one perpendicular can be drawn to a given line from a given external point. § 254. Theorem. If two parallel lines are cut by a transversal, the alternate interior angles are equal. § 255. Theorem. If two parallel lines are cut by a transversal, the exterior interior angles are equal. § 256. Theorem. If two parallel lines are cut by a transversal, the consecutive interior angles are supplementary. § 257. Theorem. If two parallel lines are cut by a transversal, the alternate exterior angles are equal. § 258. Theorem. If two triangles have two sides of one equal respectively to two sides of the other but the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second. § 259. Theorem. If two triangles have two sides of one equal to two sides of the other but the third side of the first greater than the third side of the second, then the angle opposite the third side of the first is greater than the angle opposite the third side of the second. 436 SOLID GEOMETRY Chapter II. The Circle § 272. Experiment. Equal central angles in the same circle or in equal circles intercept equal arcs. Conversely: Equal arcs are inter- cepted by equal central angles. § 273. Of two unequal central angles, in the same circle or in equal circles, the greater central angle intercepts the greater arc. Conversely: Of two unequal arcs, the greater arc is intercepted by the greater central angle. ARCS AND CHORDS § 275. Theorem. In the same circle or in equal circles, equal chords subtend equal arcs. Conversely: Equal arcs are subtended by equal chords. § 276. Theorem. A diameter of a circle is greater than any other chord. § 277. Theorem. A diameter perpendicular to a chord bisects the chord and the arc subtended by it. § 278. Theorem. A diameter bisects the circle. § 279. Theorem. A diameter which bisects a chord is perpendicu- lar to the chord. § 280. Theorem. The perpendicular bisector of a chord passes through the center of the circle. § 281. Theorem. In the same circle or in equal circles, equal chords are equidistant from the center. Conversely: Chords equidistant from the center are equal. § 282. The distance from the center of a circle to any point without the circle is greater than the radius, and the distance from the center to any point within the circle is less than the radius. § 283. A point is without a circle when its distance from the center is greater than the radius. Also, a point is within a circle when its dis- tance from the center is less than the radius. TANGENTS AND SECANTS § 284. Theorem. A perpendicular to a radius at its outer extremity can touch the circle at only one point. § 287. Theorem. A straight line perpendicular to a radius at the point where the radius meets the circle is tangent to the circle. § 288. Theorem. The radius drawn to the point of tangency is per- pendicular to the tangent. § 289. Theorem. Parallel lines intercept equal arcs on a circle. SYLLABUS OF PLANE GEOMETRY 437 J 290. Theorem. Tangents to a circle from an external point are equal, and make equal angles with the line joining that point with the center. § 295. Theorem. The line of centers of two intersecting circles is the perpendicular bisector of their common chord. § 296. Theorem. If two circles are tangent to each other, the line of centers passes through the point of contact. ANGLE MEASUREMENT § 303. Theorem. The arcs formed by two perpendicular diameters are equal. § 305. Experiment. A central angle equals in degrees its intercepted arc. § 306. Theorem. An inscribed angle equals in degrees one-half its intercepted arc. § 307. Theorem. Angles inscribed in the same arc are equal. § 308. Theorem. An angle inscribed in a semicircle is a right angle. § 309. Theorem. An angle formed by two intersecting chords equals in degrees one-half the sum of the intercepted arcs. § 310. Theorem. An angle formed by a tangent and a chord equals in degrees one-half the intercepted arc. § 311. Theorem. An angle formed by two secants, two tangents, or a tangent and a secant, equals in degrees one-half the difference of the intercepted arcs. PROBLEMS OF CONSTRUCTION § 312. Problem. To construct a perpendicular to a given straight line from a given external point. § 313. Problem. To construct a perpendicular to a given straight line at a given point in the line. § 314. Problem. To bisect a given straight line. § 315. Problem. To bisect a given arc. § 316. Problem. To circumscribe a circle about a given triangle. § 317. Problem. To construct a circle through three points which are not in the same straight line. § 318. Theorem. Three points not in the same straight line deter- mine a circle. § 319. Problem. To bisect a given angle. § 320. Problem. From a given point in a given line, to draw a line making an angle equal to a given angle. § 321. Problem. Lines may be added, subtracted, and multiplied. § 322. Problem'. A line may be divided into equal parts. 438 SOLID GEOMETRY § 323. Problem. To construct a triangle when two angles and the included side are given. § 324. Problem. To construct a triangle when two sides and the included angle are given. § 325. Problem. To inscribe a circle in a given triangle. § 326. Problem. To construct a tangent to a given circle. § 327. Problem. With a given straight line as a chord, to construct an arc of a circle in which a given angle may be inscribed. § 329. Problem. To construct a triangle when two sides and the angle opposite one of them are given. Chapter III. Area of Polygons MEASURING § 333. Axiom. Every geometric magnitude has a numerical measure. § 346. The area of a rectangle is equal to the product of its base and altitude. § 347. Theorem. The area of a square equals the square of its side. § 348. Theorem. The areas of two rectangles are to each other as the products of their bases and altitudes. § 349. Theorem. The areas of two rectangles having equal bases are to each other as their altitudes. § 350. Theorem. The areas of two rectangles having equal altitudes are to each other as their bases. § 351. Theorem. Two rectangles having equal bases and equal altitudes are equal in area. § 354. Theorem. Any parallelogram is equivalent to a rectangle that has its base and altitude equal respectively to the base and altitude of the parallelogram. § 355. Theorem. The area of a parallelogram is equal to the prod- uct of its base and altitude. § 356. Theorem. The altitude of a parallelogram is equal to its area divided by its base. § 357. Theorem. The base of a parallelogram is equal to its area divided by its altitude. § 358. Theorem. Parallelograms having equal bases and equal alti- tudes are equivalent. § 359. Theorem. Two parallelograms having equal bases are to each other as their altitudes. § 360. Theorem. Two parallelograms having equal altitudes are to each other as their bases. SYLLABUS OF PLANE GEOMETRY 439 § 361. Theorem. Any two parallelograms are to each other as the products of their bases and their altitudes. § 362. Theorem. The area of a triangle is equal to half the product of its base and its altitude. § 363. Theorem. The area of a triangle equals half the*area of a parallelogram having the same base and altitude as the triangle. § 364. Theorem. The base of a triangle equals twice its area divided by its altitude. § 365. Theorem. The altitude of a triangle equals twice its area divided by its base. § 366. Theorem. Triangles having equal bases and equal altitudes are equivalent. § 367. Theorem. Two triangles having equal bases are to each other as their altitudes. § 368. Theorem. Two triangles having equal altitudes are to each other as their bases. § 369. Theorem. Any two triangles are to each other as the prod- ucts of their bases and their altitudes. § 370. Theorem. The area of a trapezoid is equal to half the prod- uct of its altitude and the sum of its bases. § 371. Theorem. The area of a trapezoid equals the product of its altitude and its median. § 375. Theorem. Two triangles that have an angle of one equal to an angle of the other are to each other as the products of the sides includ- ing the equal angles. § 376. Theorem. The square constructed on the hypotenuse of a right triangle is equivalent to the sum of the squares constructed on the other two sides. § 377. Theorem. The square on either side of a right triangle is equivalent to the square on the hypotenuse minus the square on the other side. PROJECTION § 380. Theorem. In any obtuse triangle, the square of the side opposite the obtuse angle is equivalent to the sum of the squares of the other two sides, plus twice the product of one of those sides and the pro- jection of the other side upon it. § 381. Theorem. In any triangle, the square of a side opposite an acute angle is equivalent to the sum of the squares of the other two sides, minus twice the product of one of these sides and the projection of the other side upon it. 440 SOLID GEOMETRY § 382. Theorem. An angle of a triangle is acute, right, or obtuse according as the square of the opposite side is less than, equal to, or greater than the sum of the squares of the other two sides. § 383. Theorem. If any median of a triangle be drawn to a side: (1) The sum of the squares of the other two sides is equal to twice the square of the median plus twice the square of half the bisected side. (2) The difference of the squares of the other two sides is equal to twice the product of the bisected side and the projection of the median upon that side. TRANSFORMATIONS AND CONSTRUCTIONS § 385. Problem. To construct a square that is equivalent to the sum of two given squares. § 386. Problem. To construct a square that is equivalent to the difference of two given squares. § 388. Problem. To transform a given quadrilateral into an equiv- alent triangle. § 389. Problem. To transform a given polygon into an equivalent triangle. § 390. Problem. To construct a square equivalent to a given rec- tangle or parallelogram. § 391. Problem. To construct a square equivalent to a given triangle. § 392. Problem. To construct a square equivalent to a given polygon. Chapter IV. Proportion and Similarity THEOREMS OF PROPORTION § 398. Theorem. If four numbers form a proportion, the product of the extremes equals the product of the means. § 399. Theorem. The mean proportional between two numbers equals the square root of their product. § 400. Theorem. If the two antecedents of a proportion are equal, the two consequents are equal. § 401. Theorem. If three terms of one proportion are equal respec- tively to three corresponding terms of another proportion, the fourth terms are equal. § 402. Theorem. If the product of two numbers equals the product of two other numbers, either two may be made the means and the other two the extremes of a proportion. § 403. Theorem. In a series of equal ratios, the sum of the anteced- ents is to the sum of the consequents as any antecedent is to its consequent. SYLLABUS OF PLANE GEOMETRY 441 § 404. Theorem. If four numbers form a proportion, they are in proportion by inversion; that is, the second term is US the first as the fourth is to the third. § 405. Theorem. If four numbers form a proportion, they are in proportion by alternation; that is, the first term is to the third as the second is to the fourth. § 406. Theorem. If four numbers form a proportion, they are in proportion by addition; that is, the sum of the first two terms is to the second as the sum of the last two terms is to the fourth. § 407. Theorem. If four numbers form a proportion, they are in proportion by subtraction; that is, the difference of the first two terms is to the second as the difference of the last two terms is to the fourth. § 408. Theorem. If four numbers form a proportion, they arc in proportion by addition and subtraction; that is, the sum of the first two is to their difference as the sum of the last two is to their difference. PROPORTIONAL LINES § 411. Theorem. A straight line parallel to one side of a triangle divides the other two sides in the same ratio. § 412. Theorem. If two straight lines are cut by a series of parallel straight lines, the corresponding segments are proportional. § 413. Theorem. A line which divides two sides of a triangle in the same ratio is parallel to the third side. § 414. Problem. To construct the fourth proportional to three given lines. § 415. Problem. To divide a given straight line into parts that are in a given ratio. § 416. Problem. .To divide a given straight line into parts propor- tional to any number of given lines. § 417. Theorem. The bisector of an angle of a triangle divides the opposite side into two parts which are proportional to the adjacent sides. SIMILAR TRIANGLES § 421. Theorem. Two triangles similar to the same triangle are similar to each other. i § 422. Theorem. Two triangles are similar if two angles of one are equal respectively to two angles of the other. § 423. Theorem. Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. § 424. Theorem. Two isosceles triangles are similar if the angle at the vertex, or a base angle, of one equals the corresponding angle of the other. 442 SOLID GEOMETRY § 425. Theorem. Two equilateral triangles are similar. § 426. Theorem. If the corresponding angles of two triangles have their sides respectively parallel, right side to right side and left side to left side, the triangles are similar. § 427. Theorem. If the corresponding angles of two triangles have their sides respectively perpendicular, right side to right side and left side to left side, the triangles are similar. § 428. Theorem. The corresponding sides of similar triangles are proportional. § 430. Theorem. If two triangles have their corresponding sides proportional, they are similar. § 431. Theorem. If two triangles have an angle of one equal to an angle of the other and the including sides proportional, they are similar. . § 432. Theorem. The areas of two similar triangles are in the same ratio as the squares of any two corresponding sides. § 434. . Theorem. In any right triangle, if the altitude is drawn from the vertex of the right angle to the hypotenuse: (1) The two triangles thus formed are similar to the given triangle and to each other. (2) The altitude is the mean proportional between the segments of the hypotenuse. § 435. Theorem. Each side about the right angle is the mean pro- portional between the hypotenuse and its adjacent segment. § 436. Theorem. The two segments of the hypotenuse and the squares of the two sides are proportionals. § 437. Theorem. The square of the hypotenuse is to the square of either side as the hypotenuse is to the segment of the hypotenuse adjacent to that side. § 438. Theorem. The square of the hypotenuse equals the sum of the squares of the two sides. § 439. Theorem. A perpendicular drawn to a diameter from any point on the circle is a mean proportional between the segments of the diameter. § 440. Problem. To construct the mean proportional between two given lines. SIMILAR POLYGONS § 442. Problem. Construct a polygon similar to a given polygon having given a side corresponding to a side of the given polygon. § 443. Theorem. The corresponding angles of two similar polygons are equal. SYLLABUS OF PLANE GEOMETRY 443 § 444. Theorem. The corresponding sides of two similar polygons are proportional. § 445. Theorem. The perimeters of two similar polygons are in the same ratio as any two corresponding sides. § 446. Theorem. The areas of two similar polygons are in the same ratio as the squares of any two corresponding sides. PROPORTIONAL LINES CONNECTED WITH CIRCLES § 447. Theorem. If two chords intersect within a circle, the prod- uct of the segments of one is equal to the product of the segments of the other. § 450. Theorem. If a tangent and a secant are drawn to a circle from an external point, the tangent is the mean proportional between the secant and its external segment. § 451. Theorem. If a tangent and a secant are drawn to a circle from an external point, the square of the tangent is equal to the product of the secant and its external segment. § 452. Theorem. If two secants are drawn to a circle from an exter- nal point, the product of one secant and its external segment is equal to the product of the other secant and its external segment.. Chapter V. Measurement of Circles § 458. Problem. To inscribe a regular hexagon in a circle. § 459. Theorem. Any equilateral polygon inscribed in a circle is a regular polygon. § 460. Problem. To inscribe a square in a circle. § 461. Theorem. If a circle is divided into any number of equal parts, the chords joining the successive points of division form a regular inscribed polygon; and the tangents drawn at the successive points of division form a regular circumscribed polygon. § 462. Theorem. Tangents drawn to a circle at the vertices of a regular inscribed polygon form a regular circumscribed polygon of the same number of sides. § 463. Theorem. Chords drawn from the middle points of the arcs subtended by the sides of a regular inscribed polygon to the adjacent ver- tices of the polygon form a regular inscribed potygon of double the num- ber of sides. § 464. Theorem. The perimeter of a regular inscribed polygon is less than the perimeter of a regular inscribed polygon of double the num- ber of sides, and less than the circumscribed circle. 444 SOLID GEOMETRY § 465. Theorem. The area of a regular inscribed polygon is less than the area of a regular inscribed polygon of double the number of sides. § 466. Theorem. The area of a regular polygon is less than the area within the circumscribed circle. POLYGONS AND CIRCLES § 469. Problem. To divide a line in extreme and mean ratio. § 470. Problem. To construct a regular decagon in a given circle. § 471. Problem. To inscribe a regular pentagon in a circle. § 472. Problem. To inscribe in a given circle a regular pentadecagon, or polygon of fifteen sides. § 473. Problem. To circumscribe a circle about any regular polygon. § 474. Problem. To inscribe a circle in any regular polygon. § 476. Theorem. Two regular polygons of the same number of sides are similar. § 477. Theorem. The perimeters of two similar regular polygons are to each other as their radii, and also as their apothems. § 478. Theorem. The area of a regular polygon is equal to half the product of the perimeter by the apothem. § 479. Theorem. The areas of two regular polygons of the same number of sides are to each other as the squares of their radii, and also as the squares of their apothems. . MEASUREMENT OF THE CIRCLE § 486. Theorems on Limits. (1) If two variables that approach limits are equal for all their succes- sive values, their limits are equal. (2) If x is a variable having the limit a, and c is a constant not zero, . x a then cx—>ca and — > - as x— >a. c c § 488. The length of a circle is the common limit of the perimeters of regular inscribed and regular circumscribed polygons as the number of sides is indefinitely doubled. § 489. The area of a circle is the common limit of the areas of regular inscribed and regular circumscribed polygons as the number of sides is indefinitely doubled. § 490. Theorem. The apothem of a regular inscribed polygon approaches as a limit the radius of the circle, as the number of sides is indefinitely doubled. Also, the radius of a regular circumscribed polygon approaches as a limit the radius of the circle, as the number of sides is indefinitely doubled. SYLLABUS OF PLANE GEOMETRY 445 § 492. Theorem. Two circumferences have the same ratio as their radii, or as their diameters. § 493. Theorem. The ratio of a circumference to its diameter is constant. § 495. Theorem. The circumference of any circle is equal to 7Td, or to 27Tr. § 496. Theorem. If an arc s is intercepted by a central angle of u u degrees the length of the arc is TTd. 360 § 497. Theorem. The area of a circle is equal to one-half the product of the circumference by the radius. § 498. Theorem. The area of any circle is equal to 7Tr 2 . § 499. Theorem. The area of any circle is equal to \ird 2 . § 500. Theorem. The areas of two circles are in the same ratio as the squares of their radii or as the squares of their diameters. § 508. Theorem. The area of a sector is equal to half the product of its radius by its arc. INDEX Page Angles, between line and plane 294 diedral 295 face . . 302 polyedral 302 solid 302 spherical 407 triedral 303,413 Circles, of sphere 384 axis of . 385 great 384 poles of 385 small 385 Coilinear . . . 280 Concentric spheres 392 Cones, area of ■. 350 axis of . 344 circumscribed .... 350, 391 definition of 344 frustum of 353 inscribed 350,391 of revolution 345 similar . 365 spherical 403 truncated 353 vertex angle of 366 volume of 357 Conical surface 343 Cube 326 Cylinders 315 area of 315,319 bases of 315 circular 317 oblique 317 of revolution 317 right 317 Page Cylinders, similar 336 volume of 321,331 Cylindrical surface 314 Degree, spherical . . . . . .421 Devices in imaging 274 Diedral angle 295 plane angle of 295 Dimensions 273 Directrix 314 Distance, between parallel planes 290 between two lines 300 from point to plane .... 290 on spheres 386 polar 386 Dodecaedron 373 Euler's theorem 375 Excess, spherical . . . . . .. 421 Face angles 302 Figures, drawing . . • 274 Formulas 427 Frustum, area of 354 of cone 353 of pyramid . . . . . . . . 353 volume of 363 Generatrix 314 Hexaedron 373 Icosaedron 373 Inclination 294 447 448 INDEX Page Lines, inclination of 294 parallel to planes 283 perpendicular to planes . . 288 projection of 293 relations of 280 tangent to sphere 382 Loci 307 Lime ...... 398 angle of 398 area of 398 Nappes 343 Numbers 428 Octaedron 373 Parallelpiped 326 oblique 326 rectangular 326 right 326 volume of 327 Planes, coincide 280 definition of ....... 277 determined 278 how represented 277 intersecting 281 parallel 285 parallel to lines 283 perpendicular 296 perpendicular to lines . . . 280 relations of 280 tangent to cone . . . * . . . 349 tangent to cylinder . . . .318 tangent to sphere 383 Point, projection of 293 Polar distance 386 Polar triangles 415 Polyedral angles 302 central 409 concave 302 congruent ■ . . 303 convex 302 Page Polyedral angles, parts of . . 302 symmetric 303, 410 Polyedrons, circumscribed . . 388 concave 314 convex 314 definition of ....... 313 inscribed 388 regular 373 sections of 313 similar 377 Prismatic surface 314 Prismatoid formula 371 Prismatoids, definition of . . 369 volume of . 370 Prisms 315 area of 315,319 basis of . . : 315 circumscribed . . . . 319, 357 definition of 315 inscribed ...... 319,357 oblique 317 regular 317 right 317 similar 336 truncated 325 volume of 331 Projection 293 Pyramidal surface 343 Pyramids, areas of 344 circumscribed 350 definition of 344 frustum of 353 inscribed 350 regular 344 similar 365 spherical 423 truncated 353 volume of . 357 Quadrant 386 Quadrilateral, gauche or skew . 283 INDEX 449 Page Sector, spherical 403 Segment, spherical . . . . . 403 Solids, congruent ...... 324 equivalent 324 ratio between 326 Space ideas 273 Sphere, and straight line . . .381 area of 393 circles of 384 circumscribed 391 concentric 392 definition of 381 distance on 386 great circle of 381 inscribed in cone 391 inscribed in cylinder . . .391 inscribed in polyedron . . . 388 small circle of 384 tangent to 392 volume of 400 Spherical angle 407 Spherical degree 421 Spherical cone 403 Spherical excess 421 Spherical polygon 408 area of 420 congruent 410 convex 409 Page Spherical polygon, symmetric . 410 Spherical pyramid 423 Spherical sector 403 Spherical segment 403 Spherical triangle 408 area of 421 Spherical wedge 405 Surface, conical 343 cylindrical .314 definition of 277 . prismatic 314 pyramidal 343 Tangent planes, to cone . . . 349 to cylinder 318 to sphere 383 Tetraedron 373 Triangles, polar 415 spherical 408 Triedral angles ..... 303, 414 isosceles 413 Truncated prisms 325 Wedge, spherical 405 Zones 397 area of 397 definition of 397 THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW EDUCATION LIBRARY AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. j J l I: o Ioto 1348 LD 21-100w-12,'43 (8796s) ^B 35956 ^o