478. A sphere is a body bounded by a uniformly curved face, all the points of which are equally distant from a point within called the center. The diameter of a sphere is a straight line passing through the center of the sphere, and terminated at both ends by its surface. The radius of a sphere is a straight line drawn from the center to any point in the surface. PROBLEMS. 479. To find the convex surf. of a prism or cylinder. 1. Find the area of the convex sur face of a prism whose altitude is 7 ft., and its base a pentagon, each side of which is 4 feet. SOLUTION.-4 ft. x 5 = 20 ft., perimeter. 20 ft. x7 = 140 sq. ft., convex surface. 2. Find the area of the convex surface of a triangular prism, whose altitude is 8 feet, and the sides of its base 4, 5, and 6 feet, respectively. SOLUTION.-4 ft.+5 ft. +6 ft. = 15 ft., pe 3. Find the area of the convex surface of a cylinder whose altitude is 2 ft. 5 in., and the circum ference of its base 4 ft. 9 in. SOLUTION.-2 ft. 5 in. = 29 in.; 4 ft. 9 in. = 57 in. 57 in. x 29=1653 sq. in. = 11 sq. ft. 69 sq. inches, convex surface. FORMULA: Perimeter of base x altitude = convex surface. To find the entire surface, add the area of the bases or ends. 4. If a gate 8 ft. high and 6 ft. wide revolve upon a point in its center, what is the entire surface of the cylinder described by it? 5. Find the superficial contents, or entire surface of a prism 8 ft. 9 in. long, 4 ft. 8 in. wide, and 3 ft. 3 in. high. 6. Find the entire surface of a cylinder formed by the revolution about one of its sides of a rectangle that is 6 ft. 6 in. long and 4 ft. wide? 7. Find the entire surface of a prism whose base is an equilateral triangle, the perimeter being 18 ft., and the altitude 15 feet? 480. To find the volume of any prism or cylinder. 1. Find the volume of a triangular prism, whose altitude is 20 ft., and each side of the base 4 ft. SOLUTION. The area of the base is 6.928 sq. ft. (465). 6.928 sq. ft. x 20 == 138.56 cu. ft., volume. 2. Find the volume of a cylinder whose altitude is 8 ft. 6 in., and the diameter of its base 3 ft. SOLUTION.--32.7854 = 7.0686 sq. ft., area of base (472). FORMULA: Area of base × altitude = volume. 3. What is the volume of a log 18 ft. long and 1 ft, in diameter ? 4. Find the solid contents of a cube whose edges are 6 ft. 6 in.? 5. Find the cost of a piece of timber 18 in. square and 40 ft. long, at $.30 a cubic foot. 6. What is the value of a log 24 ft. long, of the average circumference of 7.9 ft., at $.45 a cubic foot? 481. To find the convex surface of a pyramid or cone. 1. Find the convex surface of a triangular pyramid, the slant height being 16 ft., and each side of the base 5 feet. SOLUTION. (5 ft. +5 ft. +5 ft.) × 16÷2 = 120 sq. ft., convex surface. 2. Find the convex surface of a cone whose diameter is 17 ft. 6 in., and the slant height 30 ft. SOLUTION.-17.5 ft. x 3.1416 = 54.978 ft., circumference To find the entire surface, add to this product the area of the base. 3. Find the entire surface of a pyramid whose base is 8 ft. 6 in. square, and its slant height 21 feet. 4. Find the entire surface of a cone the diameter of whose base is 6 ft. 9 in. and the slant height 45 feet. 482. To find the volume of any pyramid or cone. 1. What is the volume, or solid contents, of a square pyramid whose base is 6 ft. on each side, and its altitude 12 ft. SOLUTION.-6 × 6 × 12÷3 = 144 cu. ft., volume. 2. Find the volume of a cone, the diameter of whose base is 5 ft. and its altitude 10 ft. SOLUTION.-52 ft. ×.7854 × 101÷3 = 68.721 cu. ft., volume. FORMULA: Area of base × } altitude = volume. 3. Find the solid contents of a cone whose altitude is 24 ft., and the diameter of its base 30 inches. 4. What is the cost of a triangular pyramid of marble, whose altitude is 9 feet, each side of the base being 3 feet, at $2 per cubic foot? 5. Find the volume and the entire surface of a pyramid whose base is a rectangle 80 ft. by 60 ft., and the edges which meet at the vertex are 130 feet. 483. To find the convex surface of a frustum of a pyramid or of a cone. 1. What is the convex surface of a frustum of a square pyramid, whose slant height is 7 ft., each side of the greater base 4 ft., and of the less base 18 inches? SOLUTION.-The perimeter of the greater base is 16 ft., of the less, 6 ft. 16 ft.+6 ft. x 7÷2 = 77 sq. ft., convex surface. 2. Find the convex surface of a frustum of a cone whose slant height is 15 ft., the circumference of the lower base 30 ft., and of the upper base 16 ft. FORMULA: Sum of perimeters slant height = conv. surf To find the entire surface, add to this product the areas of both ends. 3. How many square yards in the convex surface of a frustum of a pyramid, whose bases are heptagons, each side of the lower base being 8 ft., and of the upper base 4 ft., and the slant height 55 ft.? 484. To find the volume of a frustum of a pyramid or cone. 1. Find the volume of the frustum of a sq. pyramid, whose altitude is 10 ft., each side of the lower base 12 ft., and of the upper base 9 ft. SOLUTION.—122+92 = 225; (225+ √/144 × 81) × 10÷3 = 1110; hence, 1110 cu. ft. is the volume. 2. How many cubic feet in the frustum of a cone whose altitude is 6 ft. and the diameters of its bases 4 ft. and 3 ft. ? RULE.—To the sum of the areas of both bases add the square root of the product, and multiply this sum by one-third of the altitude. 3. How many cu. ft. in a piece of timber 30 ft. long, the greater end being 15 in. square, and that of the less 12 in. ? 4. How many cu. ft. in the mast of a ship, its height being 50 ft., the circumference at one end 5 ft. and at the other 3 ft.? 485. To find the surface of a sphere. 1. Find the surface of a sphere whose diameter is 9 in. SOLUTION.-92 x 3.1416 254.4696; hence, 246.4696 sq. in., surface. FORMULA: Diameter x3.1416 surface of sphere. = 2. What is the surface of a globe 3 feet in diameter? 3. Find the surface of a globe whose radius is 1 foot. 486. To find the volume of a sphere. 1. Find the volume of a globe whose diameter is 30 in. 3053.6352; hence, 3053.6352 cu. ft., volume. SOLUTION.-183 × .5236 FORMULA: Diameters x.5236 = volume of sphere. 2. Find the volume of a globe whose diameter is 30 in. 3. Find the solid contents of a globe whose radius is 5 yd. 487. Gauging is the process of finding the capacity or volume of casks and other vessels. For ordinary purposes the diagonal rod is used, which gives only approximate results. A cask is equivalent to a cylinder having the same length and a diameter equal to the mean diameter of the cask. To find the mean diameter of a cask (nearly). RULE. Add to the head diameter 3, or, if the staves are but little curved, .6, of the difference between the head and bung diameters. To find the volume of a cask in gallons: RULE.-Multiply the square of the mean diameter by the length (both in inches) and this product by .0034. 1. How many gallons in a cask whose head diameter is 24 in., bung diameter 30 in., and its length 34 inches? SOLUTION.—24+ (30—24 × 3) = 28 in., mean diameter. 282 x 34 x .0034 = 90.63 gal., capacity. |