L Through the diagonal edges L and ' let a plane be passed, and it will divide the prismoid into two wedges, having for bases, the bases of the prismoid, and for edges the lines L and l'=1. The solidity of these wedges, and consequently of the prismoid, is Bh(2L+1)+bh(2l+L)=÷h(2BL+B1+2b1+bL). But since M is equally distant from L and I, we have 2M=L+1, and 2m B+b; hence, 4Mm=(L+) x (B+b)=BL+Bl+bL+bl. Substituting 4Mm for its value in the preceding equation, and we have for the solidity jh(BL+bl+4Mm). ever. REMARK. This rule may be applied to any prismoid what For, whatever be the form of the bases, there may be inscribed in each the same number of rectangles, and the number of these rectangles may be made so great that their sum in each base will differ from that base, by less than any assignable quantity. Now, if on these rectangles, rectangular prismoids be constructed, their sum will differ from the given prismoid by less than any assignable quantity. Hence the rule is general. 1. One of the bases of a rectangular prismoid is 25 feet by 20, the other 15 feet by 10, and the altitude 12 feet; required the solidity. Ans. 3700. 2. What is the solidity of a stick of hewn timber, whose ends are 30 inches by 27, and 24 inches by 18, its length being 24 feet? Ans. 102 feet. OF THE MEASURES OF THE THREE ROUND BODIES. PROBLEM IX. To find the surface of a cylinder. RULE.-Multiply the circumference of the base by the altitude and the product will be the convex surface (Book VIII. Prop '). To this add the areas of the two bases, when the extir su. face is required. 1. What is the convex surface of a cylinder, the diameter of whose base is 20, and whose altitude is 50? Ans. 3141.6. 2. Required the entire surface of a cylinder, whose altitude is 20 feet, and the diameter of its base 2 feet. Ans. 131.9472. PROBLEM X. To find the convex surface of a cone. RULE.-Multiply the circumference of the base by half the side (Book VIII. Prop. III.): to which add the area of the base, when the entire surface is required. 1. Required the convex surface of a cone, whose side is 50 feet, and the diameter of its base 8 feet. Ans. 667.59. 2. Required the entire surface of a cone, whose side is 36 and the diameter of its base 18 feet. Ans. 1272.348. PROBLEM XI. To find the surface of the frustum of a cone.. RULE.-Multiply the side of the frustum by half the sum of the circumferences of the two bases, for the convex surface (Book VIII. Prop. IV.): to which add the areas of the two bases, when the entire surface is required. 1. To find the convex surface of the frustum of a cone, the side of the frustum being 12 feet, and the circumferences of the bases 8.4 feet and 6 feet. Ans. 90. 2. To find the entire surface of the frustum of a cone, the side being 16 feet, and the radii of the bases 3 feet and 2 feet. Ans. 292.1688. PROBLEM XII. To find the solidity of a cylinder. RULE.-Multiply the area of the base by the altitude (Book VIII. Prop. II.). 1. Required the solidity of a cylinder whose altitude is 12 feet, and the diameter of its base 15 feet. Ans. 2120.58. 2. Required the solidity of a cylinder whose altitude is 20 feet, and the circumference of whose base is 5 feet 6 inches. Ans. 48.144. PROBLEM XIII. To find the solidity of a cone. RULE.-Multiply the area of the base by the altitude, and take one-third of the product (Book VIII. Prop. V.). 1. Required the solidity of a cone whose altitude is 27 feet, and the diameter of the base 10 feet. Ans. 706.86. 2. Required the solidity of a cone whose altitude is 10 feet, and the circumference of its base 9 feet. Ans. 22.56. PROBLEM XIV. To find the solidity of the frustum of a cone. RULE.-Add together the areas of the two bases and a mean proportional between them, and then multiply the sum by one. third of the altitude (Book VIII. Prop. VI.). 1. To find the solidity of the frustum of a cone, the altitude being 18, the diameter of the lower base 8, and that of the upper base 4. Ans. 527.7888. 2. What is the solidity of the frustum of a cone, the altitude being 25, the circumference of the lower base 20, and that of the upper base 10? Ans. 464.216. 3. If a cask, which is composed of two equal conic frustums joined together at their larger bases, have its bung diameter 28 inches, the head diameter 20 inches, and the length 40 inches how many gallons of wine will it contain, there being 231 cubic inches in a gallon? Ans. 79.0613. PROBLEM XV. To find the surface of a sphere. RULE I.-Multiply the circumference of a great circle by the diameter (Book VIII. Prop. X.). RULE II.-Multiply the square of the diameter, or four times the square of the radius, by 3.1416 (Book VIII. Prop. X. Cor.). 1. Required the surface of a sphere whose diameter is 7. Ans. 153.9384. 2. Required the surface of a sphere whose diameter is 24 inches. Ans. 1809.5616 in. 3. Required the area of the surface of the earth, its diameter being 7921 miles. Ans. 197111024 sq. miles. 4. What is the surface of a sphere, the circumference of its great circle being 78.54? Ans. 1963.5. PROBLEM XVI. 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, and the product will be the surface (Book VIII. Prop. X. Sch. 1). 1. The diameter of a sphere being 42 inches, what is the convex surface of a zone whose altitude is 9 inches? Ans. 1187.5248 sq. in. 2. If the diameter of a sphere is 12 feet, what will be the surface of a zone whose altitude is 2 feet? Ans. 78.54 sq. ft. PROBLEM XVII. To find the solidity of a sphere. RULE I.-Multiply the surface by one-third of the radius (Book VIII. Prop. XIV.). RULE II.-Cube the diameter, and multiply the number thus found by : that is, by 0.5236 (Book VIII. Prop. XIV. Sch. 3). 1. What is the solidity of a sphere whose diameter is 12? Ans. 904.7808. 2. What is the solidity of the earth, if the mean diameter be taken equal to 7918.7 miles? Ans. 259992792083. PROBLEM XVIII. To find the solidity of a spherical segment. RULE. Find the areas of the two bases, and multiply their sum by half the height of the segment; to this product add the solidity of a sphere whose diameter is equal to the height of the segment (Book VIII. Prop. XVII.). REMARK.-When the segment has but one base, the other is to be considered equal to 0 (Book VIII. Def. 14). 1. What is the solidity of a spherical seginent, the diameter of the sphere being 40, and the distances from the centre to the bases, 16 and 10. Ans. 4297.7088. 2. What is the solidity of a spherical segment with one base the diameter of the sphere being 8, and the altitude of the segment 2 feet? Ans. 41.888. 3. What is the solidity of a spherical segment with one base, the diameter of the sphere being 20, and the altitude of the segment 9 feet? Ans. 1781.2872. PROBLEM XIX. To find the surface of a spherical triangle. RULE.-1. Compute the surface of the sphere on which the triangle is formed, and divide it by 8; the quotient will be the surface of the tri-rectangular triangle. 2. Add the three angles together; from their sum subtract 180°, and divide the remainder by 90°: then multiply the trirectangular triangle by this quotient, and the product will be the surface of the triangle (Book IX. Prop. XX.). 1. Required the surface of a triangle described on a sphere whose diameter is 30 feet, the angles being 140°, 92°, and 68°. Ans. 471.24 sq. ft. 2. Required the surface of a triangle described on a sphere of 20 feet diameter, the angles being 120° each. Ans. 314.16 sq. ft. PROBLEM XX. To find the surface of a spherical polygon. RULE.-1. Find the tri-rectangular triangle, as before. 2. From the sum of all the angles take the product of two right angles by the number of sides less two. Divide the remainder by 90°, and multiply the tri-rectangular triangle by the quotient: the product will be the surface of the polygon (Book IX. Prop. XXI.). 1. What is the surface of a polygon of seven sides, described on a sphere whose diameter is 17 feet, the sum of the angles being 1080° ? Ans. 226.98. 2. What is the surface of a regular polygon of eight sides. described on a sphere whose diameter is 30, each angle of the polygon being 140° ? Ans. 157.08 OF THE REGULAR POLYEDRONS. In determining the solidities of the regular polyedrons, it Decomes necessary to know, for each of them, the angle contained between any two of the adjacent faces. The determi nation of this angle involves the following property of a regu lar polygon, viz. |