Ex. 1. What is the solidity of a parallelopiped whose alti. tude is 30 feet, breadth 6 feet, and depth 4 feet ? Ans., 720 cubic feet. Ex. 2. What is the solidity of a square prism whose altitude is 8 feet 10 inches, and each side of its base 2 feet 3 inches ? Ans., 44 cubic feet. Ex. 3 What is the solidity of a pentagonal prism whose a.titude is 20 feet 6 inches, and its side 2 feet 7 inches ? Ans., 235.376 cubic feet. PROBLEM III. (108.) To find the surface of a regular pyramid. RULE. Multiply the perimeter of the base by half the slant height for the convex surface. To this add the area of the base when the entire surface is required. See Geometry, Prop. 14, B. VIII. Ex. 1. What is the entire surface of a triangular pyramid whose slant height is 25 feet, and each side of its base 5 feet? Ans., 198.325 square feet. . Ex. 2. What is the entire surface of a square pyramid whose slant height is 30 feet, and each side of the base 4 feet? Ans., 256 square feet. Ex. 3. What is the entire surface of a pentagonal pyramid whose slant height is 20 feet, and each side of the base 3 feet? Ans., 165.484 square feet. PROBLEM IV. (109.) To find the solidity of a pyramid. RULE. Multiply the area of the base by one third of the altitude, See Geometry, Prop. 17, B. VIII. Ex. 1. What is the solidity of a triangular pyramid whose altitude is 25 feet, and each sid; of its base 6 feet? Anm., 129.904 cubic feet. Ex. 2. What is the solidity of a square pyramid whose slant height is 22 feet, and each side of its base 10 feet? Ans., 714.143 cubic feet. Ex. 3. What is the solidity of a pentagonal pyramid whose altitude is 20 feet, and each side of its base 3 feet ? Ans., 103.228 cubic feet. PROBLEM V. (110.) To find the surface of a frustum of a regular pyramid. RULE. Multiply half the slant height by the sum of the perime. ters of the two bases for the convex surface. To this add the areas of the two bases when the entire surface is required. See Geometry, Prop. 14, Cor. 1, B. VIII. Ex. 1. What is the entire surface of a frustum of a square pyramid whose slant height is 15 feet, each side of the greater base being 4 feet 6 inches, and each side of the less base 2 feet 10 inches ? Ans., 248.278 square feet. Ex. 2. What is the entire surface of a frustum of an oc. tagonal pyramid whose slant height is 14 feet, and the sides of the ends 3 feet 9 inches, and 2 feet 3 inches? Ans., 428.344 square feet. PROBLEM VI. (111.) To find the solidity of a frustum of a pyramid. RULE. Add together the areas of the two bases, and a mean proportional between them, and multiply the sum by one third of the altitude. See Geometry, Prop. 18, B. VIII. When the pyramid is regular, it is generally most convenient to find the area of its base by Rule II., Art. 92. If we put a to represent one side of the lower base, and b one side of the upper base, and the tabular number from Art. 92 by T, the area of the lower base will be a’T; that of the upper base will be bạT; and the mean proportional will be abT. Hence, if we represent the height of the frustum by h, its solidity will be hT a’ 3. Ex. 1. What is the solidity of a frustum of an hexagonal pyramid whose altitude is 15 feet, each side of the greater end being 3 feet, and that of the less end 2 feet? Ans., 246.817 cubic feet. Ex. 2. What is the solidity of a frustum of an octagonal pyramid whose altitude is 9 feet, each side of the greater end being 30 inches, and that of the less end 20 inches? Ans., 191.125 cubic feet. E Definition. (112.) A wedge is a solid bounded by five planes, viz., a rectangular base, ABCD, two trapezoids, ABFE, DCFE, meeting in an edge, and two triangular ends, ADE, BCF. The altitude of the wedge is the perpendicular drawn D from any point in the edge to the plane of the base, as EH. G А. I B PROBLEM VII. (113.) To find the solidity of a wedge. RULE. Add the length of the edge to twice the length of the base, and multiply the sum by one sixth of the product of the height of the wedge and the breadth of the base. Demonstration. Put L=AB, the length of the base ; I=EF, the length of the edge; h=EH, the altitude of the wedge. Now, if the length of the base is equal to that of the edge, it is evident that the wedge is half of a prism of the same base and height. If the length of the base is greater than that of the edge, let a plane, EGI, be drawn parallel to BCF. The wedge will be divided into two parts, viz., the pyramid EAIGD, and the triangular prism BCF-G. The solidity of the former is equal to zbh(L-1), and that of the latter is bhl. Their sum is bhl+kbh(L-1)=6h31+16h2L-4h2l= bh(2L+1). If the length of the base is less than that of the edge, the wedge will be equal to the difference between the prism and pyramid, and we shall have bhl-ibh(l-L), which is equal to bhl+{bh(L-1), the same result as before. Ex. 1. What is the solidity of a wedge whose base is 30 inches long and 5 inches broad, its altitude 12 inches, and the length of the edge 2 feet? Ans., 840 cubic inches. Ex. 2. What is the solidity of a wedge whose base is 40 inches long and 7 inches broad, its altitude 18 inches, and the length of the edge 30 inches? Ans., 2310 cubic inches. Definition. (114.) A rectangular prismoid is a solid bounded by six planes, of which the two bases are rectangles having their corresponding sides parallel, and the four upright sides of the sol.. id are trapezoids. PROBLEM VIII. To find the solidity of a rectangular prismoid. RULE. Add together the areas of the two bases, and four times the area of a parallel section equally distant from the bases, and multiply the sum by one sixth of the altitude. Demonstration. Put L and B=length and breadth of one base; M M L Put I and b=length and breadth of the other base; " M " m=length and breadth of middle sec.; h =the altitude of the prismoid. It is evident that if a plane be made to pass through the opposite edges of the upper and lower bases, the prismoid will be divided into B two wedges, whose bases are the bases of the prismoid, and whose edges are L and l. The solidity of these wedges, and, consequently, that of the prismoid, is ;Bh(2L+1)+ bh(21+L)=;h(2BL+Bl+26l+BL). But, since M is equally distant from L and 1, we have 2M=L+1, and 2m=B+b; hence 4Mm=(L+)(B+b)=BL+Bl+bL+bl. Substituting 4Mm for its value in the preceding expression, we obtain for the solidity of the prismoid Th(BL+bl+4Mm). Ex. 1. What are the contents of a log of wood, in the form of a rectangular prismoid, the length and breadth of one end being 16 inches and 12 inches, and of the other 7 inches and 4 inches, the length of the log being 24 feet? Ans., 16} cubic feet. Ex. 2. What is the solidity of a log of hewn timber, whose ends are 18 inches by 15, and 14 inches by 11), its length being 18 feet? Ans., 261 cubic feet. PROBLEM IX. To compute the excavation or embankment for a rail-way. (115.) By the preceding rule may be computed the amount of excavation or embankment required in constructing a railroad or canal. If we divide the line of the road into portions so small that each may be regarded as a straight line, and suppose an equal number of transverse sections to be made, the excavation or embankment between two sections may be regarded as a prismoid, and its contents found by the pre ceding rule. Let ABCD represent the lower surface of the supposed excavation, which we will assume to be parallel to the horizon; and let EFGH represent the upper surface of the excavation |