MENSURATION Aain frustum = VA B C - VA" B "C" V A I - -A- K/A 3 j - / A 3, a formula which applies to the frusta of all pyramids regular and irregular. The above result may be otherwise expressed. For, let A B = a^ and A" B " = 3 , then, if A"B"C" be a section equidistant from the ends of the frustum, A"B" = a 2 = ^(a 1 + a 3 ). Now K l =pa and A 3 =pa* s (see 70) ; hence A 2 = area of A"B"C"=^=j/^L^V, which gives +pa = A 1 + 2 VAjA.j + A 3 ; 4A 2 =p therefore volume of frustum L 3 + 2 A 3 ) = J/^A! + 4 A 2 + A 3 ) ; or the volume of the frustum of a pyramid is obtained by adding the areas of the ends to four times the area of the middle section, and multiplying the sum by one-sixth of the alti tude. The above result can be ob tained at once from 72, since A B C A "B "C " is a prismatoid with similar bases. 74. Surface of the Frustum of a Regular Pyramid. In fig. 45 let the perimeter of Aj =^ : , that of A 2 =p 2 > an d that of A 3 =j9 3 , and let YD = 7 1; VD " = 1 3 , and therefore D D " = VD - YD " = Zj - 1 3 = I. The lateral surface of the frustum is equal to the difference between the lateral surfaces of the pyramids YA B C and VA" B "C" , Fig. 45. But, since IL_!L __!., we haye Z^ 1 3 3 na 3 p 3 therefore lateral surface of frustum and L Pi- Pa ^Pi-P* or the lateral surface of the frustum of a regular pyramid is equal to the product of the slant height and the perimeter of the section equidistant from the ends. Otherwise. The top and base being regular polygons, the inclined faces are congruent trapeziums. Let I be the height of each trapezium, and let there be n of them, then area of each face = -r r (^ L + 2 n n and therefore the area of lateral surface = Q}] +P 3 ) = 75. If h the altitude of the frustum be given, we deduce the slant height and then proceed as before. Thus let YQ = 7i-!, and using the same notation as in 72, 73, and 74 we have - h 3 , which gives 7 h = a -L^ Again whence I is known since 7 X is known in terms of h. When the pyramid is irregu lar the lateral planes are non- congruent trapeziums, the areas of which can be found separately by 12, and hence the whole surface. 76. Volume of the Frustum of a Triangular Prism. Let A denote the area of ABC (fig. 46), and let 7ij, h. 2 , 7i 3 be the altitudes of A , B , C respectively with reference to the plane ABC. Divide the frustum into three pyramids B A AC, B ABC, and B A CC by the planes B AC and B A C. These three pyramids are respectively equal to BA AC, B ABC, and ABCC ; hence volume of frustum = ^ A + ^7< 2 A + ^h. A . B 77. If the prism be right or oblique, the volume of a frustum is qual to one-third of the area of its right section multiplied by the um of the parallel edges. For divide the frustum AA B C (fig. 7) into two frusta by a plane A"B"C" of area A at right angles o the edges, then AA B C = A A"B"C" + A"A B C = ^A(AA" + BB" + CO") + J A (A A" + B B" + C C") = I A(AA" + BB" + CC" + A A" + B B" + C C") Again, since eveiy prism can be divided into triangular prisms, ve can find by repeated applications of the above proposition the oluine of the frustum of any prism whatever. For example, if A y Fig. 47. Fig. 48. the base of the frustum of a right prism AA B C D (fig. 48) be a rectangle 12 feet by 6 feet, and the parallel edges in order 6, 4, 10, and 12 feet, then A = area of base = 12 x 6 = 72 square feet. Frustum = ABC A B C + ADC A D C = x JA(AA + BB + CC ) + ^ x JA(A A + CC + DD ) 78. Volume of a Wedge. The wedge (fig. 49) being merely the frustum of a triangular prism, we have at once volume = ^A(FE + AD + BC), where A is the area of its right section ; otherwise, the wedge may be con sidered a prismatoid whose upper base is a straight line, and hence its volume = 7t(4A 2 since A 1 = 0. B. Regular Polyhedra. 79. The regular polyhedra are five in number, namely, the tetrahedron, cube, octahedron, dodecahedron, and icosahedron, whose solid angles are formed respectively by three equilateral triangles, three squares, four equilateral triangles, three pentagons, and five equilateral triangles. Since a regular polyhedron admits of having a sphere inscribed within it and described about it, it can easily be shown that the volume of the polyhedron nl X 24 X cos cot 2 m n ( ( ir ir (if TT < -cos + cos > / V m n / m n / ) and from 18, 7, it follows that the surface of the polyhedron In , x cot , 4 n where and 7 = the number of faces, ??z = the number of faces in each solid angle, 7i = the number of edges in each face, a = the lenth of each side. The following table contains the surfaces and volumes for the five regular polyhedra whose edge is 1. Polyhedron. Siirfacc. Volume. 1 7320508 0-1178511 Cube 6-0000000 1 -0000000 Octahedron 3-464101G 0-4714043 Dodecahedron 20-6457788 7-6631189 Icosahedron 8 6602540
2-1816950