and the solid angles are all equal to each other, it is evident that the faces are all equally distant from a point in the solid called the center. If planes be made to pass through the center and the several edges of the solid, they will divide it into as many equal pyramids as it has faces. The base of each pyramid will be one of the faces of the polyedron; and since their altitude is the perpendicular from the center upon one of the faces, the solidity of the polyedron must be equal to the areas of all the faces, multiplied by one third of this perpendicular. Also, similar pyramids are to each other as the cubes of their homologous edges (Geom., Prop. 17, Cor. 3, B. VIII.). And since two regular polyedrons of the same name may be divided into the same number of similar pyramids, they must be to each other as the cubes of their edges. (118.) The solidity of a tetraedron whose edge is unity, may be computed in the following manner: Let C-ABD be a tetraedron. From one angle, C, let fall a perpendicular, CE, on the opposite face; draw EF perpendicular to AD; and join CF, AE. Then AEF is a right-angled triangle, in which EF, being the sine of 30°, is one half of AE or BE; and therefore FE is one third of BF or CF. Hence the cosine of the angle CFE is equal to ; that A 1 F B is, the angle of inclination of the faces of the polyedron is 70° 31′ 44′′. Also, in the triangle CAF, CF is the sine of 60°, which is 0.866025. Hence, in the right-angled triangle CEF, knowing one side and the angles, we can compute CE, which is found to be 0.8164966. Whence, knowing the base ABD (Art. 92), we obtain the solidity of the tetraedron =0.1178513. In a somewhat similar manner may the solidities of the other regular polyedrons, given in Art. 116, be obtained. Ex. 1. What is the solidity of a regular tetraedron whose edges are each 24 inches? Ans., 0.9428 feet. Ex. 2. What is the solidity of a regular icosaedron whose edges are each 20 feet? Ans., 17453.56 feet. THE THREE ROUND BODIES. PROBLEM I. (119.) To find the surface of a cylinder. RULE. Multiply the circumference of the base by the altitude for the convex surface. To this add the areas of the two ends when the entire surface is required. See Geometry, Prop. 1, B. X. Ex. 1. What is the convex surface of a cylinder whose altitude is 23 feet, and the diameter of its base 3 feet? Ans., 216.77 square feet. Ex. 2. What is the entire surface of a cylinder whose altitude is 18 feet, and the diameter of its base 5 feet? Ans. PROBLEM II. (120.) To find the solidity of a cylinder. RULE. Multiply the area of the base by the altitude. See Geometry, Prop. 2, B. X. Ex. 1. What is the solidity of a cylinder whose altitude is 18 feet 4 inches, and the diameter of its base 2 feet 10 inches? Ans., 115.5917 cubic feet. Ex. 2. What is the solidity of a cylinder whose altitude is 12 feet 11 inches, and the circumference of its base 5 feet 3 inches? Ans., 28.3308 cubic feet. PROBLEM III. (121.) To find the surface of a cone. RULE. Multiply the circumference of the base by half the side for the convex surface; to which add the area of the base when the entire surface is required. See Geometry, Prop. 3, B. X. Ex. 1. What is the entire surface of a cone whose side is 10 feet, and the diameter of its base 2 feet 3 inches? Ans., 39.319 square feet. Ex. 2. What is the entire surface of a cone whose side is 15 feet, and the circumference of its base 8 feet? Ans., 65.093 square feet. PROBLEM IV. (122.) To find the solidity of a cone. RULE. Multiply the area of the base by one third of the altitude. See Geometry, Prop. 5, B. X. Ex. 1. What is the solidity of a cone whose altitude is 12 feet, and the diameter of its base 2 feet? Ans., 19.635 cubic feet. Ex. 2. What is the solidity of a cone whose altitude is 25 feet, and the circumference of its base 6 feet 9 inches? PROBLEM V. Ans. (123.) To find the surface of a frustum of a cone. RULE. Multiply half the side by the sum of the circumferences 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. 4, B. X. Ex. 1. What is the entire surface of a frustum of a cone, the diameters of whose bases are 9 feet and 5 feet, and whose side is 16 feet 9 inches? Ans., 451.6036 square feet. Ex. 2. What is the convex surface of a frustum of a cone whose side is 10 feet, and the circumferences of its bases 6 feet and 4 feet? Ans., 50 square feet. PROBLEM VI. (124.) To find the solidity of a frustum of a cone. 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. 6, B. X. If we put R and r for the radii of the two bases, then πR2 will represent the area of one base, πr the area of the other, and Rr the mean proportional between them. Hence, if we represent the height of the frustum by h, its solidity will be Ex. 1. What is the solidity of a frustum of a cone whose altitude is 20 feet, the diameter of the greater end 5 feet, and that of the less end 2 feet 6 inches? Ans., 229.074 cubic feet Ex. 2. The length of a mast is 60 feet, its diameter at the greater end is 20 inches, and at the less end 12 inches: what is its solidity? Ans., 85.521 cubic feet PROBLEM VII. (125.) To find the surface of a sphere. RULE. Multiply the diameter by the circumference of a great circle; or, Multiply the square of the diameter by 3.14159. See Geometry, Prop. 7, B. X. Ex. 1. Required the surface of the earth, its diameter being 7912 miles. Ans., 196,662,896 square miles. Ex. 2. Required the surface of the moon, its circumference being 6786 miles. Ans. PROBLEM VIII. (126.) To find the solidity of a sphere. RULE. Multiply the surface by one third of the radius; or, Multiply the cube of the diameter by π; that is, by 0.5236. See Geometry, Prop. 8, B. X. Where great accuracy is required, the value of π must be · taken to more than four decimal places. Its value, correct to ten decimal places, is .52359,87756. Ex. 1. What is the solidity of the earth, if it be a sphere 7912 miles in diameter? Ans., 259,332,805,350 cubic miles. Ex. 2. If the diameter of the moon be 2160 miles, what is Ans. its solidity? PROBLEM IX. (127.) To find the surface of a spherical zone. RULE. Multiply the altitude of the zone by the circumference of a great circle of the sphere. See Geometry, Prop. 7, Cor. 1, B. X. Ex. 1. If the diameter of the earth be 7912 miles, what is the surface of the torrid zone, extending 23° 27′ 36′′ on each side of the equator? Ans., 78,293,218 square miles. G H K D B Let PEP'Q represent a meridian of the earth; EQ the equator; P, P' the poles; AB one of the tropics, and GH one of the polar circles. Then PK will represent the height of one of the frigid zones, KD the height of one of the temperate zones, and CD half the height of the torrid zone. Each of the angles ACE, CAD, and GCK is equal to 23° 27′ 36′′. In the right-angled triangle ACD, A E IC P' Where great accuracy is required, the sine and cosine of 23° 27′ 36′′ must be taken to more than six decimal places, The following values are correct to ten decimal places : Natural sine of 23° 27' 36"-.39810,87431. 66 cosine of 23° 27' 36"-.91733,82302. |