MENSURATION OF VOLUMES. To find the volume of a prism. 116. From the principle demonstrated in Book VII., Prop. XIV., we may write the following RULE. Multiply the area of the base by the altitude; the product will be the volume required. EXAMPLES. 1. What is the volume of a cube, whose side is 24 inches? Ans. 13824 cu. in. 2. How many cubic feet in a block of marble, of which the length is 3 feet 2 inches, breadth 2 feet 8 inches, and height or thickness 2 feet 6 inches? Ans. 214 cu. ft. 3. Required the volume of a triangular prism, whose height is 10 feet, and the three sides of its triangular base 3, 4, and 5 feet. Ans. 60. To find the volume of a pyramid. 117. From the principle demonstrated in Book VII., Prop. XVII., we may write the following RULE. Multiply the area of the base by one-third of the altitude; the product will be the volume required. EXAMPLES. 1. Required the volume of a square pyramid, each side of its base being 30, and the altitude 25. Ans. 7500. 2. Find the volume of a triangular pyramid, whose altitude is 30, and each side of the base 3 feet. 38.9711 cu. ft. 3. What is the volume of a pentagonal pyramid, its altitude being 12 feet, and cach side of its base 2 feet. Ans. 27.5276 cu. ft. 4. What is the volume of an hexagonal pyramid, whose altitude is 6.4 feet, and each side of its base 6 inches? Ans. 1.38564 cu. ft. To find the volume of a frustum of a pyramid. 118. From the principle demonstrated in Book VII., Prop., XVIII., C., we may write the following RULE. Find the sum of the upper base, the lower base, and a mean proportional between them; multiply the result by onethird of the altitude; the product will be the volume required. EXAMPLES. 1. Find the number of cubic feet in a piece of timber, whose bases are squares, each side of the lower base being 15 inches, and each side of the upper base 6 inches, the altitude being 24 feet. Ans. 19.5. 2. Required the volume of a pentagonal frustum, whose altitude is 5 feet, each side of the lower base 18 inches, and each side of the upper base 6 inches. Ans. 9.31925 cu. ft. 119. Since cylinders and cones are limiting cases of prisms and pyramids, the three preceding rules are equally applicable to them. EXAMPLES. 1. Required the volume of a cylinder whose altitude is 12 feet, and the diameter of its base 15 feet. Ans. 2120.58 cu. ft. 2. Required the volume of a cylinder whose altitude is 20 feet, and the circumference of whose base is 5 feet 6 inches. Ans. 48.144 cu. ft. 3. Required the volume of a cone 27 feet, and the diameter of the base 10 feet. whose altitude ᎥᎦᎸ 4. Required the volume of a cone whose altitude is 10 feet, and the circumference of its base 9 feet. Ans. 22.56 cu. ft. 5. Find the volume of the frustum of a cone, the altitude being 18, the diameter of the lower base 8, and that of the upper base 4. 6. What is the volume of the frustum altitude being 25, the circumference of the and that of the upper base 10? Ans. 527.7888. of a cone, the lower base 20, An's. 464.216. 7. If a cask, which is composed of two equal conic frustums joined together at their larger bases, have its bung diameter 28 inches, the head diameter 20 inches, and the length 40 inches, how many gallons of wine will it contain, there being 231 cubic inches in a gallon? Ans. 79.0613. To find the volume of a sphere. 120. From the principle principle demonstrated in Book VIII., Prop. XIV., we may write the following RULE. Cube the diameter of the sphere, and multiply the result by 17, that is, by 0.5236; the product will be the volume required. 12 ? EXAMPLES. 1. What is the volume of a sphere, whose diameter is Ans. 904.7808. 2. What is the volume of the earth, if the mean diameter be taken equal to 7918.7 miles. Ans. 259992792083 cu. miles. To find the volume of a wedge.. 121. A WEDGE is a volume bounded by a rectangle ABCD, called the back, two trapezoids ABHG, DCHG, called faces, and two triangles ADG, CBH, called ends. The line GH, in which the faces meet, is called the edge. کچھ D A G H B It may happen that the edge is either greater or less than the length of the back. In deducing an expression for the volume, we shall take the latter case, premising that the result, so far as the volume is concerned, is the same in either parallel to HCB, dividing the wedge into the triangular prism GNM-B, and the pyramid AMND-G. The altitude of the prism GH, is equal to 7, and the area of its base CDH, is equal to 1) bh; hence, its volume is equal to bhl. The altitude GP, of the pyra mid, is equal to h, and its base AMND is equal to b(L-l); hence, the volume of the pyramid is equal to ļbh(L — l). The volume of the wedge, denoted by V, is equal to the sum of the volumes of the prism and pyramid ; hence, V = {bhl + {bh(L — 1) Factoring and reducing, we have, 1⁄2bhl + дbhL — 1bhl. hence, the following RULE. Add twice the length of the back to the length of the edge; multiply the sum by the breadth of the back, and that result by one-sixth of the altitude; the final product will be the area required. EXAMPLES. 1. If the back of a wedge is 40 by 20 feet, the edge 35 feet, and the altitude 10 feet, what is the volume ? Ans. 2. What is the volume of a wedge, whose feet y 9, edge 20 feet, and altitude 6 feet? To find the volume of a prismoid. 122. A PRISMOID is a frustum of a wedge. Let L and B denote the length and breadth of the lower base, 7 and b the length and breadth of the upper base, M and m the length and breadth of the section equidistant from the bases, and h the altitude of the prismoid. Through the edges L and l'′, let a plane be passed, and it will B m divide the prismoid into two wedges, having for bases, the bases of the prismoid, and for edges the lines L and 7, The volume of the prismoid, denoted by V, will be equal to the sum of the volumes of the two wedges; hence, or, V = ↓ Bh(l + 2Z) + дbh(I +27); V = jh(2BL+ 2b1 + Bl + bL); |