To find the lateral surface of a frustum of a regular pyramid. RULE. Add the perimeters of the ends or bases together, and multiply the sum by half the slant height; the product will be the surface of the sides. EXAMPLE. Each side of one of the bases of the frustum of a hexagonal pyramid is 4 feet, each side of the other base, 2 feet, and the slant height of the frustum is 24 feet; required its lateral surface. 4X6+2×6×24 432 square feet. = Ans. To find the solidity of the frustum of a pyramid. RULE. To the square root of the product of the areas of the bases, add the areas of the bases, and multiply the sum by the perpendicular height of the frustum ;* the product will be the solidity. EXAMPLE.-The greater base of the frustum of a quadrilateral pyramid is 3 feet on each side, and the less base is 2 feet on each side, the perpendicular height of the frustum is 15 feet; what are its solid contents? 3 X 3 9 feet area of greater base. 2 X 2 = 4 feet = area of less base. 15÷3=5=} height. Then, 9 X = = √36 = 6 +9+4 = 19 X 5 95 feet. Ans. OF PRISMOIDS AND THE WEDGE. To find the solidity of a prismoid. The RULE for finding the solidity of the frustum of a pyramid is equally applicable to the prismoid. Or, RULE. Add the areas of the ends, and four times the area of the mean between the ends, together, and multiply the sum by the perpendicular height; the product will be the capacity or solidity. EXAMPLE. The bottom of a rectangular cistern is 8 feet by 6 feet, the top is 4 by 3 feet, and the perpendicular depth is 12 feet; required its capacity. To find the solidity of a wedge. RULE.-Multiply the sum of the length of the edge and twice the length of the base, by the breadth of the base multiplied by the perpendicular depth or length of the wedge, and divide the product by 6; the quotient will be the solidity. EXAMPLE. The length of the base of a wedge, a c, is 10 inches, the length of the edge, ef, is 8 inches, the breadth of the base, a b, is 4 inches, and the perpendicular depth, d e, is 12 inches; required the contents. 8 +10 × 228 X 4 X 12 = 13446 224 cubic in. Ans. OF CYLINDERS. To find the convex surface of a cylinder. RULE. Multiply the circumference by the length, and the product will be the convex surface. If the surface of the entire cylinder is required, add the areas of the ends to the convex surface. RULE. To find the solidity of a cylinder. Multiply the area of an end by the length of the cylinder, and the product is the solidity. - EXAMPLE. The diameter of a cylinder is 6 feet, and its length is 8 feet; required its solidity or capacity. 6 X 6 = 36 X .7854 = 28.2744 = area of end, and 28.2744 X 8 = 2262-cubic feet. Ans. To find the length of a helix, or spiral, wound round a cylinder. RULE.-Multiply the circumference of the cylinder by the number of revolutions the spiral makes around it, square the product, and thereto add the square of the length of the cylinder; the square root of the sum is the length of the spiral. The rule is applicable in finding the length of the thread of a screw, hand-rail to a winding staircase, &c., &c. OF CONES. To find the convex surface of a cone. RULE. Multiply the circumference of the base by half the slant height, and the product is the surface required. If the surface of the entire figure is required, add the area of the base to the convex surface. EXAMPLE. The diameter of the base of a right cone is 8 feet, and the slant height is 18 feet; required the convex surface. 8 X 3.1416 2262 sq. feet. Ans. To find the solidity of a cone. RULE.-Multiply the area of the base by the perpendicular height, and the product is the solidity. - EXAMPLE. The diameter of the base of a right cone is 8 feet, and the perpendicular height of the cone is 15 feet; required its solidity. 82.7854 X 15 = 251} cubic feet. Ans. To find the solidity or capacity of the frustum of a cone. RULE.1. To the square root of the product of the areas of the ends add the areas of the ends, and multiply the sum by the perpendicular height of the frustum; the product will be the solidity. Or,—2. Divide the difference of the cubes of the diameters of the ends by the difference of the diameters of the ends, and multiply the quotient by the perpendicular height of the frustum; this product multiplied by .7854 gives the solidity. Or, — 3. To the product of the diameters of the ends add the square of the difference of the diameters, multiply the sum by .7854, and the product will be the mean area between the ends, which, multiplied by the perpendicular height of the frustum, gives the solidity. EXAMPLE.—The diameter of the larger end of a stick of round timber is 30 inches, that of the smaller end is 21 inches, and the length of the stick is 36 feet; required its contents. § 2.53 1.753 ÷ 2.5 ✔ 1.75—13.6875 × 36 × .7854 = 129 cubic feet. Ans. By Rule 2. By Rule 3. = 4.375.1875 = 4.5625.7854 X EXAMPLE.The interior diameter of the larger end of a circular cistern is 12 feet, that of the smaller end is 8 feet, and the perpendicular depth of the cistern is 14 feet; required its capacity in cubic feet. NOTE. For an Example under Rule 1, see Mensuration of Pyramids. All rules which are applicable to the measuring of a cone, or frustum thereof, are also applicable to the pyramid, or its frustum; but, inasmuch as the areas of the ends of the two figures are not found with equal readiness, there will usually be a choice in the employment of rules. OF SPHERES. To find the surface of a sphere or globe. RULE.-Multiply the diameter by the circumference; or, multiply the square of the diameter by 3.1416, and the product will be the surface. To find the solidity of a sphere or globe. RULE. Multiply the superfices by of the diameter; or, multiply the square of the diameter by of the circumference; or, multiply the cube of the diameter by .5236, and the product is the solidity. RULE. 9 × 9 × 9 = sq. in., surface, or superfices, and 381.7 cubic inches. Ans. = 729 X.5236 = 381.7. Ans. To find the convex surface of a spherical segment. Multiply the height of the segment by the circumference of the sphere, and the product is the answer. The rule is equally applicable to the finding of the convex surface of a spherical zone. 11 To find the solidity of a spherical segment. RULE. 1. To the square of the height of the segment, add three times the square of half the base, and multiply the sum by the height, multiplied by .5236, and the product is the solidity. Or,-2. From three times the axis of the sphere, e f, subtract twice the height of the segment, and multiply the difference by the square of the height, multiplied by .5236, and the product will be the contents. NOTE.- ad2+bc2bcef, and e f X bcbc2= √ a d2 = a d. EXAMPLE. The base, a d, of the segment a d c, is 18 inches, and the altitude, bc, is 8 inches; required the solidity of the segment. 82 +92 × 3 = 307 8.5236 = 1285.96 cubic inches. Ans RULE. To find the solidity of a spherical zone. Add the square of the radius of one base to the square of the radius of the other, and thereto add the square of the height multiply their sum by the height, multiplied by 1.5708, and the product will be the contents. EXAMPLE.-The base, g h, of the zone g h d a, is 4 feet, the base, a d, is 3 feet, and the height, i b, is 24 feet; the contents of the zone are required. 4÷2= 2, radius of base = g h. 66 66 ad. 2.526.25 ÷ 3 = 2.0833 = square of height. Then, 221.52.083 8.333 X 2.5 X 1.5708 = 32.72+ cub. inches. Ans. To find the side of the greatest cube that can be cut from a given sphere. RULE. Divide the square of the diameter of the sphere by 3, and the square root of the quotient is the side; or, multiply the diameter of the sphere by .57735, and the product is the side. EXAMPLE. The diameter of a globe is 15 inches; required the side of the greatest cube that may be cut from the globe. 15 X 15225 ÷ 3 = √758.66 inches, or |