23. The squares of the volumes of two similar polyedrons are proportional to the cubes of their homologous faces. PROBLEMS. a 1. Find a point in the interior of a tetraedron, such that being joined to the four vertices, the tetraedron is divided into four equivalent tetraedrons. 2. Draw a plane parallel to the base of a pyramid cutting off a small pyramid which shall be of the given pyramid – 24 - 64: 3. In a tetraedron, S-ABC, through E, the middle point of the ed SB, let the plane DEF be passed parallel to the base ABC; the plane EGH, parallel to the face ASC, and the plane EDH; the pyramid, S-ABC, is thus divided into two equivalent triangular prisms, and into two tetraedrons of the same base and altitude. Divide these two pyramids in the same manner, and deduce the volume of the pyramid as the limit of the sum of the series of successive prisms thus obtained. 4. Cut a cube by a plane so as to make the intersection a regular hexagon. 5. Compute the altitude of a prism, knowing the volume, v, and the base, b; the same of a pyramid. 6. Given in a frustum of a pyramid the lower base, B, the height, h, and the volume, v. Compute the other base. 7. Compute the surface and volume of a regular tetraedron, the edge being given. 8. Find the entire surface of a regular pyramid, the slant height being 12 feet, and each side of the hexagonal base being 3 feet. 3 9. Find the convex surface of the frustum of a pentagonal regular pyramid whose slant height is 40 feet, each side of the lower base 8 feet, and each side of the upper base 5 feet. 10. Find the surface and volume of a frustum of a pyramid whose bases are squares, each side of the lower base being 12 feet, each side of the upper base 6 feet, and the height 4 feet. 11. Find the surface and volume of a block of marble in the shape of a rectangular parallelopipedon whose three dimensions are 4 feet 6 inches, 2 feet 3 inches, 3 feet 9 inches. 3 12. Find the whole surface of a triangular prism whose altitude is 15 feet, and its bases equilateral triangles, whose sides are 4 feet. a 13. Find the volume of a regular hexagonal pyramid whose slant height is 18 feet, and each side of the base 5 feet. 14. Find the volume of the frustum of a pyramid, given the altitude 12 feet, and the bases regular dodecagons, the radii of whose circumscribing circles are 3.6 and .8 feet respectively. 15. Compute the three dimensions of a rectangular parallelopipedon, knowing that they are proportional to the numbers, $, and , the volume of the parallelopipedon being 2 cubic yards. 16. Compute the volume of a rectangular parallelopipedon, of which the surface is 5 square yards, and the three dimensions proportional to the numbers 4, 6, 9. 17. The height of a pyramid is 4.5 meters, and its base is a square whose side is 1.2 meters. Compute the corresponding dimensions of a similar pyramid, the volume of which is 7.29 cubic meters. 18. The base of a regular pyramid is a hexagon, each of whose sides is 3 feet in length, and its convex surface is ten times the area of its base. Find its height. 19. Find the volume of a right truncated triangular prism whose base is an equilateral triangle, the side of which is 6 feet, and the three lateral edges of which are 10, 12, and 15 feet. (See Exercises, Theorem 20.) 20. The greatest pyramid of Egypt is 150 yards high, and its base is a square whose side is 250 yards. Find its volume and its convex surface. 21. The edge, SA, of a pyramid, S-ABCD, being four feet six inches long, find the parts into which it is divided by a plane parallel to the base which divides the convex surface, first, into two equivalent parts; second, into two parts proportional to the numbers 3 and 5. 22. A regular pyramid has a hexagon for its base whose side is 15 feet, and its faces make with the base an angle equal to two-thirds of a right angle. Find the volume. 23. A right prism has for its base a regular hexagon. Find its altitude, knowing that its volume is 3 cubic feet and its convex surface 12 square feet. a APPENDIX TO BOOK VI. THE REGULAR POLYEDRONS. PROPOSITION I. THEOREM. a There can only be five regular polyedrons. For regular polyedrons were defined as those all of whose faces are equal regular polygons, and all whose solid angles are equal. These conditions cannot be fulfilled except in a small number of cases. First. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of those triangles, by four, or by five : hence arise three regular bodies, the tetraedron, the octaedron, and the icosaedron. No other can be formed with equilateral triangles, for six angles of these triangles are equal to four right angles, and cannot form a solid angle (Book V., Prop. XXXIII.). Secondly. - If the faces are squares, their angles may be arranged by threes; hence results the hexaedron, or cube. Four angles of a square are equal to four right angles, and cannot form a solid angle. Thirdly.—In fine, if the faces are regular pentagons, their angles may likewise be arranged in threes; and the regular dodecaedron will result. We can go no farther ; for three angles of a regular hexagon are equal to four right angles ; three of a heptagon are greater. Hence, there can be only five regular polyedrons ; three formed with equilateral triangles, one with squares, and one with pentagons. a SCHOLIUM. It will be proved in the following proposition that these five polyedrons actually exist, and that all their dimensions may be determined when one of their faces is known. PROPOSITION II. PROBLEM. One of the faces of a regular polyedron being given, or only its side, to construct the polyedron. This problem embraces five, which will be solved in succession. 1. CONSTRUCTION OF THE TETRAEDRON. Let ABC be the equilateral triangle which is to be one face of the tetraedron. At the point O, the centre of this triangle, erect S А. OS perpendicular to the plane ABC; terminate this perpendicular in S, so that AS : AB; join SB, SC, and the pyramid S-ABC will be the required tetraedron. For, by reason of the equal distances OA, OB, OC, the oblique lines SA, SB, SC are equal (Prop. IX., Book V.). One of them, SA = AB; hence, the four faces of the pyramid S-ABC are triangles equal to the given triangle ABC. And the solid angles of this pyramid, are all equal, because each of them is formed by three equal plane angles; hence this pyramid is a regular tetraedron. B 2. CONSTRUCTION OF THE HEXAEDRON. Let ABCD be a given square ; on the base, ABCD, construct a right prism whose altitude, AE, shall be equal to the side AB. It is evident that the faces of this prism are equal squares, and that its solid angles are equal, since they are each formed by three right angles ; hence, this prism is a regular hexaedron, or cube. D D A 3. CONSTRUCTION OF THE OCTAEDRON. Let AB be the given edge : on AB describe the square ABCD; at the point O, the centre of this square, erect TS perpendicular to its plane, and terminating on both sides in T and S, so that OT = OS = AO; then join SA, SB, TA, etc. ; you will havé a solid, SABCDT, composed of two quadrangular pyramids, S-ABCD, T-ABCD, united together by their common base, ABCD; this solid will be the required octaedron. For the triangle AOS is right angled at O, as is the triangle AOD; the sides AO, OS, OD are equal ; hence, those triangles are equal ; hence, AS = AD. It may be shown in like manner that all the other right angled triangles, AOT, BOS, COT, etc., are equal to the triangle AOD; hence all the sides, AB, AS, AT, etc., are equal, and consequently the solid SABCDT is bounded by eight equilateral triangles whose sides are equal to AB. Moreover, the solid angles of this polyedron are all equal ; for instance, the angle S is equal to the angle B. For it is evident that the triangle SAC is equal to the triangle DAC, and therefore the triangle ASC is right; hence the figure SATC a square, equal to the square ABCD. But, comparing the pyramid B-ASCT with the pyramid S-ABCD, the base of the first may be placed on the base, ABCD, of the second ; then the point O being a common centre, the altitude OB of the first will coincide with the altitude OS of the second, and the two pyramids will exactly apply to each other in all points ; hence the solid angle S is equal to the solid angle B; hence the solid SABCDT is a regular octaedron. SCHOLIUM. If three equal straight lines, AC, BD, ST, are perpendicular to each other, and bisect each other, the extremities of these straight lines will be the vertices of a regular octaedron. 4. CONSTRUCTION OF THE DODECAEDRON. Let ABCDE be a given regular pentagon ; let ABP, CBP be two plane angles equal to the angle ABC; with these plane angles form the solid angle B. The mutual inclination of two of these planes, * we will call K. In like manner, at the points C, D, E, A, form solid angles equal to solid angle B, and similarly situated : the plane CBP will be the same as the plane BCG, since they are both inclined by the same quantity, K, to the plane ABCD. Hence, the pentagon BCGFP, equal to the pentagon ABCDE, may be described in the plane PBCG. If the same thing is done in each of the other planes, CDI, DEL, etc., we shall have a convex surface, PFGH, etc., composed of six equal regular pentagons, and each inclined to its adjacent plane by the same quantity, K. Let pfgh, etc., be a second surface equal to PFGH, etc., then may these two surfaces be united so as to form a single continuous convex surface. For, the angle opf, for example, may be joined to the two angles OPB, BPF, to make a solid angle, P, equal to the angle B; and in this junction no change will take place in the inclination of the planes BPF, BPO, that inclination being already such as is required to form the solid angle B. But, whilst the , solid angle P is formed, the side pf will fall along its equal PF, and at the point F will be found the three planes, PFG, pfe, efg, united and forming a solid angle equal to each of the solid angles already formed : * K is determined by methods not given in these elements. |