OPERATION. First, 9x9=81=area of base. 6×6=36=area of upper base. 117 81x36=2916, and 2916=54=sq. root of prod. of 2 areas. Then, 117+54=171 13 of the height. 513 171 57 2280=solidity. Ex. 2. What is the solidity of the frustrum of a regular pentagonal pyramid, whose altitude is 5 feet, each side of one end 18 inches, and each side of the other end 6 inches ? Ans. 9.31925 cubic ft. Ex. 3. If the height of a frustrum of a pyramid be 24, and the areas of the two ends 441 and 121, what is the solidity? Ans. 6344. Ex. 4. What is the solidity of a frustrum of a hexagonal pyramid, whose height is 48, each side of the lower base 26, and each side of the upper base 16 ? Ans. 56034. Ex. 5. What is the solidity of a squared piece of timber, whose length is 18 feet, each side of the lower end 18 inches, and each side of the smaller, 12 inches? PROBLEM VII. Ans. 28.5 cubic ft. To find the Solidity of a Wedge. ART. 44. Rule.-I. To the length of the edge of the wedge add twice the length of the base. II. Then multiply this sum by the height of the wedge, and the breadth of the base; and of the product will be the solid contents. And, 3780×8÷6=5040=solidity of wedge. Ex. 2. How many solid feet are there in a wedge whose base is 5 feet 3 inches long and 9 inches broad, the length of the edge being 3 feet 6 inches, and the perpendicular height 2 feet 3 inches ? Ans. 3.9375 cubic ft. Ex. 3. What is the solidity of a wedge, the length and breadth of whose base are 35 and 15 inches, and the length of the edge is 55 inches, and whose perpendicular height is 18 inches? Ans. 32552 cubic ft. PROBLEM VIII. To find the Solidity of a Rectangular Prismoid. ART. 45. Rule. To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from the parallel ends, and this sum, multiplied by of the height, will give the solidity. Ex. 1. What is the solidity of a rectangular prismoid, the length and breadth of one end being 14 by 12 inches, and the other 6 by 4 inches, and the perpendicular 30 feet 6 inches. First, 14×12=168=area of lower base. 14+6÷2=10_(length and breadth Then, 192 12+4÷2=8 = > of middle section. 320 And, 31232-1728=18.074 cubic ft. Ans. Ex. 2. Required the solidity of a stick of hewn timber whose lower end is 30 inches by 27, and whose upper end is 24 by 18 inches, supposing its height to be 48 feet? Ans. 204 cubic ft. Ex. 3. What weight of water can be put into a prismoidal vessel whose dimensions are as follows: at the top 5 feet by 3 feet, and at the bottom, 7 feet by 6 feet, and the perpendicular depth 4 feet, allowing 62 pounds avoirdupois to the cubic foot? Ans. 3375 pounds. NOTE. The weight of water which a vessel of any given dimensions will contain, may be found by calculating the capacity in cubie feet, and multiplying by 62 lbs. or 1000 ounces avoirdupois, the true weight of a cubic foot of pure water. PROMISCUOUS EXAMPLES. 1. What is the whole area of a regular triangular prism, whose perimeter is 12 feet, and whose height is 22 feet ? Ans. 2. What must be paid for lining a rectangular cistern with lead, a 2d a pound, the thickness of the lead being such as to require 7 lbs. for each square foot of surface, the inner dimensions of the cistern being as follows; viz: the length 3 feet 2 inches, the breadth 2 feet 8 inches, and the depth 2 feet 6 inches? Ans. 3. What is the solidity of a pentagonal pyramid, whose height is 16 feet, and the sides of whose base are each 3 feet 6 inches? Ans. 4. What is the lateral surface of a triangular pyramid whose slant height is 24 feet, and each side of whose base is 4 feet? Ans. 5. If the height of a frustrum of a pyramid be 30 feet, and the areas of the two ends 480 and 136, what is the solidity? Ans. 6. What are the solid contents of a triangular pyramid, whose height is 28 feet, and the sides of whose base are each 4 feet 7 inches? Ans. 7. Required the solidity of a wedge whose base is 4 feet 6 inches long by 2 feet 4 inches wide, the length of the edge being 5 feet, and the perpendicular height 12 feet ? Ans. SECTION IV. MEASURES OF THE FIVE REGULAR BODIES. DEFINITIONS. ART. 46. When a body is contained under a certain number of similar and equal plane figures, it is called a Regular body. Of this description are the following figures: 1. The Tetraedron, which has four equilateral triangular faces. 2. The Hexaedron or Cube, which has six square faces. 3. The Octaedron, whose sides are eight triangles. 4. The Dodecaedron, whose sides are twelve pentagons. 5. The Icosaedron, which has twenty equilateral triangular faces. These comprise all the regular solids. They are formed by triangles, squares, and pentagons. PROBLEM I. To find the Convex Surface of a Regular Solid. ART. 47. Rule. Find the area of one of the sides, then multiply it by the number of sides, and the product will be the area. Or, Multiply the proper tabular area or surface (taken from the following table, Prob. 2,) by the square of the linear edge of the solid, and the product will be the surface. |