2. A hexagonal prisim measures 28 inches across the centre of the end, from corner to corner, and is 134 inches in length; required the solidity and surface. Ans. S The solidity is 39.48835 feet. Surface 85.2392 feet. 3. A decagonal prism, or pillar, measures 50 inches in circumference, and is 30 feet high; required the solidity and surface. Ans. The solidity 40.074 feet. 4. The gallery of a church is supported by 10 octagonal prisms of wood, which measure each 48 inches in circumference, and are each 12 feet high; what will be the expense of painting them at 10 pence per square yard? Ans. 1. 2: 7s91 5. A trapezoidal prism of earth, or part of a canal, is to be dug, whose perpendicular depth is 10 yards, the width at the top 20 yards, at the bottom 16, and the length 50 yards, the two ends being cut perpendieularly down; how many solid yards of earth are contained in this part ? Ans. 9000 yards. SIV. Of a PYRAMID. A pyramid is a solid figure, the base of which is a polygon, and its sides plain triangles, their several vertical angles meeting together in one point. To find the Solidity. RULE. Multiply the area of the base by a third part of the altitude, or length; and the product is the solid content of the pyramid. EXAMPLE 1. Let ABD be a square pyramid, having each side of its base 18.5 inches and the perpendicular height CD 15 feet, what is the solidity ? Multiply 18.5 by 18.5, and the product is 342.25, the area of the base, in inches, which multiply by 5, a third part of the height, and the product is 1711.25; this divided by 144 gives 11.88 feet, the solid content. To find the Superficial Content. RULE. Multiply the slant height by half the circumference of the base, and the product will be the upright surface. To which the area of the base may be added, for the whole surface. NOTE. This rule will serve for the surface of all pyramids. Perhaps it may be proper here to acquaint the learner, that the slant height of any pyramid is not the height from one of the corners of the base to the vertex or top, but from the middle of one side of the base. And the perpendicular height of a pyramid, is a line drawn from the vertex, to the middle or centre of the base; hence it will be necessary to find the distance between the centre of the base of a pyramid, and the middle of one of the sides.This distance may always be found by multiplying the tabular perpendicular in section V. p. 84, by one of the sides of the base; then, if to the square of this number, you add the square of the perpendicular height of the pyramid, the square root of the sum will give the slant height. EXAMPLES. 1. Required the surface of the foregoing pyramid. To the square of the perpendicular height De 15 feet or 180 inches, add the square of de 9.25, the distance from the centre e, of the base, to the middle d of one of the sides; which, in a square base, is always equal to half the side. The square root of the sum, viz. 32485.5625 is 180.24 nearly, the slant height Dd. Now half the circumference of the base is 37, which multiplied by 180.24, gives 6668.88 inches for the upright surface; to which add 342.25 the area of the base, the sum is 7011.13 inches, the whole surface equal to 48.69 feet nearly. DEMONSTRATION OF THE RULE. Every pyramid is a third part of a prism, that has the same base and height (by Euclid, XII. 7.) That is, the solid content of the pyramid ABD (in the last figure) is one third part of its circumscribing prism ABEF. Now the solidity of a prism is found by multiplying the area of the base into the height; therefore the solidity of a pyramid will be found by multiplying the area of the base by the height, and taking one third of the product. You may very easily prove a trianglular pyramid to be a third part of a prism of equal base and altitude, mechanically, by making a triangular prism of cork, and then cutting that prism into three Pyramids, in a diagonal direc tion. 2. Let ABCD be a triangular pyramid, each side of the base being 21.5 inches, and its perpendicular height 16 feet; required the solidity and sur face. First the area of the base, by sect. V. is 200.19896 inches; which multiplied by 64, the third part of the height in inches, gives 12812.73344 inches, the solidity, equal to 7.41477 feet. Again, the distance from the centre of the triangle ABC to the middle of one of the sides, as d in the side AC, is 6.20652; to the square of this number, which is 38.5208905104, add the square of the height 36864 inches, the square root the sum 36902.5208905104, is 192.10028 inches, the slant of height dD. Hence the upright surface is 6195.23403 inches, and the whole surface 6395.43299 inches, or 44.41272 feet. 0 3. Let ABCDEFGH be a pyramid, whose base is a heptagon each side of it being 15 inches, and the perpendicular height of the pyramid, HI, 13.5 feet; required the solidity and superficies. First, 15 multiplied by 1.0382617, the tabular perpendicular for a heptagon (sect. VIII.) gives 15.5739255 inches, the distance from the centre I to the middle of the side AB; this multiplied by 52.5, half the sum of the sides of the base, gives 817.63108875 inches, the area of the base. This last number multiplied by 4.5, one third of the height, and divided by 144. will give 25.55 feet, the solidity.-Again, if to the square of 15.5739255, you add the square of the height in inches, the square root of the B sum 26486.54715548 will be 162.74688, the slant height of the pyramid; this multiplied by 52.5 gives 8544.2112 inches, the upright surface, to which add the area of the base, and the whole surface will be 9361.84228875 inches, or 65.0128 feet. By Scale and Compasses. First, find a mean proportional between 15.57 and 52.5, by dividing the space between them into two equal parts, and you will find the middle point to be 28.6, the side of a square equal in area to the base of the pyramid; then extend the compasses from 12 to 28.6, that extent will reach from 4.5 (twice turned over) to 25.55 feet, the solid content. Or, extend the compasses from 1 to the area of the base, that extent |