Ex. 2. What is the solidity of a regular pentagonal prism, whose altitude is 20, and each side of the base 15 feet? Ans. 7742.1443. Ex. 3. Required the capacity of a cubical vessel which is 3 feet 5 inches deep. OPERATION BY DUODECIMALS. 35' 35' 93' 15' 1" 10 8' 1" 35' 32 0' 3" 45' 4" 5" 36f5' 7" 5"" Ex. 4. A cellar was dug, whose length was 49 feet 6 inches, breadth 23 feet 9 inches, and depth 9 feet 5 inches; how many solid yards of earth were taken out of it? Ans. 410 yards. Ex. 5. Required the solidity of a triangular prism whose altitude is 10 feet, and the three sides of its triangular base 3, 4, and 5 feet. Ans. 60 solid ft. ART. 39. The capacity of a vessel in gallons or bushels of any given dimensions, may be readily ascertained by calculating its contents in inches, and then dividing the contents by the number of cubic inches in one gallon or bushel. Ex. 1. Required the number of ale gallons there are in a cistern which is 9 feet 8 inches deep, and whose base is 5 feet 4 inches square? Ans. 16848 gall. In this operation we simply calculate the capacity of the cistern in cubic inches, and then divide by the number of cubic inches contained in 1 gallon. OPERATION. 1 First, 5 ft. 4 in.=64 inches. --2 And, 64= 4096 Then, 4096x116=475136 And, 475136-282 (cubic inches in gall. ale)=1684.8, Ans. Ex. 2. How many ale gallons are there in a cistern whose base is 5 feet 3 inches square, and which is 9 feet 8 inches deep? Ans. 1632.6. Ex. 3. How many wine gallons are there in a cistern 11 feet 9 inches long, 4 feet 9 inches wide, and 3 feet deep? Ans. 1252.5. PROBLEM III. To find the Lateral Surface of a regular Pyramid. ART. 40. Rule. - Multiply the perimeter of the base by the slant height, and half the product will be the surface. If the whole surface be required, add to this the area of the base. Ex. 1. What is the lateral surface of a regular triangular pyramid whose slant height is 20 feet, and the sides of whose base are each 8 feet? OPERATION.. 8×3=24-perimeter of the base. 2)480 240-lateral surface. Ex. 2. What is the entire surface of a regular pyramid whose slant height is 15 feet, and whose base is a regular pentagon, each side of which is 25 feet? Ans. 2012.798 sq. ft. Ex. 3. Required the convex surface, and also the entire surface, of a regular pentagonal pyramid, whose slant height is 45, and the sides of whose base are each 15 feet. Ans. 1687.5=convex surface. 387.107=area of the base. 2074.607 sq. ft.=whole surface. PROBLEM IV. To find the Lateral Surface of the Frustrum of a regular Pyramid. Авт. 41. Rule. - Multiply the perimeters of the two ends by the slant height of the frustrum, and half the product will be the surface required. To this add the surface of the two ends when the entire surface is required. Ex. 1. What is the lateral surface of the frustrum of a regular octagonal pyramid whose slant height is 42 feet, and the sides of the lower base 5 feet each, and of the upper base, 3 feet each.? OPERATION. First, 5×8=40=perimeter of lower base. 3x8=24= 64-sum of the two ends. 66 F 42-slant height. 128 256 2)2688 1344-area of lateral surface. 69 Ex. 2. How many square feet are there in the lateral surface of the frustrum of a square pyramid, whose slant height is 10 feet, each side of the lower base 3 feet 4 inches, and each side of the upper base 2 feet 2 inches? Ans. 110. Ex. 3. If the slant height of the frustrum of a hexagonal pyramid be 48, each side of the lower base 26, and each side of the upper base 16 feet, what is the lateral surface? Ans. 6048 sq. ft. PROBLEM V. To find the Solidity of a Pyramid. ART. 42. Rule. Find the area of the base, and multiply that area by of the height. This rule follows from that of the prism, because any pyramid is of a prism of the same base and altitude. It is manifest, therefore, that the solidity of a pyramid, whether right or oblique, is equal to the product of the area of the base into of the perpendicular height. Ex. 1. What is the solidity of a square pyr amid the sides of whose base are each 30 feet, and its perpendicular height 25 feet? Ex. 2. How many solid feet in a triangular pyramid the altitude of which is 14 feet 6 inches, and the three sides of its base, 5, 6 and 7 feet? Ans. 71.0352. Ex. 3. What are the solid contents of a triangular pyramid, the sides of whose base are each 3 feet, and height 30 feet? (Find the area of the base by Art. 7.) Ans. 38.97. NOTE. A Triangular Pyramid, whose height, and each side of its triangular base, are each 12 inches, is nearly of the cube whose lin. ear edge is equal to a side of the triangular base, and contains 249.413 cubic inches. This is a short method of obtaining the solidity of the pyramid, and is sufficiently accurate for all practical purposes, as may be seen by applying it to the solution of the above proposition. Thus, 3×3×30÷7=38.571. Ex. 4. What is the solidity of a regular pentagonal pyramid, its altitude being 12 feet, and each side of its base 2 feet? Ans. 27.52 solid ft. PROBLEM VI. To find the Solidity of the Frustrum of a Pyramid. ART. 43. Rule. To the areas of the two ends of the frustrum, add the square root of their product, and this sum, multiplied by of the perpendicular height, will give the solid contents. NOTE. This Rule holds equally true to a pyramid of any form. For the solidities of pyramids are equal when they have equal heights and bases, whatever be the figure of their bases. Ex. 1. In the frustrum of a pyramid, (Art. 40,) one end of which is 9 feet square, the other end 6 feet square, and the height 40 feet, what is the solidity? 1 |