the greater base, and 1 foot at the leffer, its length being 72 feet. Anf. 89.5356 cubic feet. PROBLEM IX. To find the folidity of the prismoid. RULE I. To the areas of the two ends add four times the area of the middle section. Multiply the sum by the height, and š the product will be the solidity. Rule 2. To the longest side of the leffer base add half the longest side of the greater base, and multiply the sum by the breadth of the lefser bafe ; reserve this product. Again : To the longest of the greater base add half the longest side of the lefser base, and multiply thie fum by the breadth of the greater base; and to the product add the product formerly reserved; multiply this sum by the height, and the product will give the solidity, EXAMPLE I, Required the folidity of a quadrilateral prismoid, of which the shortest and longest sides of the greater base are 20 and 16 feet, and the corresponding sides of the lefler base 12 and 10 feet, the height being 40 feet 32 20+12=32 and is 16 26 10+16=26 and is 13 48 16 208 area of mid. fecte 4 832 320 I 20 1272 40 6)50880 8480 solidity in cubic feet, By RULE II. 12 the longest side of the leffer base. 22 220 reserved number, Again, 20 the longest side of the greater base, 6 half the longest side of the lesser base. 26 416 416 636 3)25440 8480 cubic feet as above. Ex. 2. Required the solid content of a trough, in the form of a prismoid, whose greater base is 24 inches by 30, and lefler base 20 inches by 24, the depth being 18 inches. Anf 10728. Ex. 3. What is the content of the hopper of a mill, 4 feet by s at the greater base, and 12 inches by so at the lesser, its , depth being 4 feet ? Ans: 57408 folid inches, or 33.2 feet, PROBLEM X. To find the folidity of a wedge RULE. Multiply the sum of twice the length of the base, and the length of the edge by the product of the height of the wedge into the breadth of the base, and of the last product will be the folidity Nite. When the length of the base is equal to that of the edge, the wedge is equal to one half a prism of the same base and altitude, EXAMPLE I. How many solid feet are in a wedge whose base is 2 feet 8 inches long and 44 inches broad, its perpendicular height being 14 inches, and the length of the edge i foot 9 inches ? Multiply the circumference by the diameter, and the product will be the surface : Or, Multiply the square of the diameter by 3.1416 for the sur-. face. RULE 2. Multiply the square of the axis by 7854, and four times the product will be the superficies. EXAMPLE I. How many square inches will cover a globe of 6 inches diameter? Note. 4 times the area of a great circle of a sphere is equal to its surface. By Ex. 2. Required the surface of a sphere, whose diameter is 5 feet 6 inches. Anf. 95.0334/9. feet. Es. 3. What is the surface of a ball, whose diameter is inch? Anf. 3.1416 inches. Ex. 4. How many inches will cover a globe of 12 inches diameter ? Anf. 452.3904. Ex. 5. Required the surface of a globe of 18 inches diameter. Ans: 7.0686 sq. feet. Ex. 6. Required the superficies of the terraqueous globe, it: diameter being 7958 miles. And if only one fourth part of its surface be dry land, and two acres sufficient to produce food for one person, how many persons can live on the earth at one time? 198956786.5824 sq. miles. Ans: 349739196.6456 dry land. 15916542927 persons. Note. A square mile contains 640 acreş. PROBLEM |