Example.-What is the lateral surface of the frustrum of a regular octagonal pyramid, A B C D, whose slant height, a A, is 42 feet, and the sides of the lower base, D C, 5 feet each, and of the upper base, a b, 3 feet each? First, 5×8=40=perimeter of lower base. 66 upper 3×8=24= 66 64 sum of the two ends. A D Then, 64 × 42÷2=1344=area of lateral surface. PROBLEM V. To find the Solidity of a Pyramid. Rule. Find the area of the base, and multiply that area by of the height. NOTE. 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. Example.-What is the solidity of a square pyramid, a b c d, the sides of whose base are each 30 feet, and its perpendicular height, e ƒ, 25 feet? First, 30×30=900=area of the base. 25-3= 7200 300 7500=solidity. f PROBLEM VI. To find the Solidity of the Frustrum of a Pyramid. 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. Example.-What is the cubic or solid contents of the frustrum of a marble pyramid, whose lower base, a b c d, is 20 inches square, and upper base, e f, 14 inches, and whose height, h g, is 8 feet 4 inches? And what will be its weight, reckoning 169 lbs. to the cubic foot? h 2 20=400 area of lower base. 14=196= 66 8 ft. 4=100 100÷3=33}=} of height. Then, 400 × 196=280. 66 upper 596 sum of areas. And, 596+280 × 331=29200. 29200 1728 16.9 cubic feet. Ans. To find the weight, 16:9 × 169=2856 lbs. Ans. NOTE. By this rule, marble cutters can easily determine the solidity and weight of any piece of marble, such as shafts of monuments, slabs, &c., by reference to the Table of Specific Gravities, for a multiplier for the weight of a cubic foot or inch. OF WEDGES AND PRISMOIDS. PROBLEM VII. To find the Solidity of a Wedge. Rule. To the length of the edge of the wedge add twice the length of the base. 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. Example.-Required the solidity of a wedge whose base, a b, is 27 feet, bd, 8 feet, and whose edge, c b, is 36 feet, and the perpendicular height 22 feet? First, 36 length of edge. 54 twice the Igth. of the base. 90X22X8÷6=2660 cubic ft. PROBLEM VIII. To find the Solidity of a Rectangular Prismoid. Rule. To the sum of the areas of the two ends, a b c, d e f, add four times the area of a section, g h, parallel to and equally distant from the parallel ends, and this sum, multiplied by of the height, will give the solidity. Example.-What is the solidity of a rectangular prismoid, a b c d, 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 x 12=168=area of lower base. 320 area of 4 times middle section. And 31232-1728-18 074 cubic ft. Ans. SECTION IV. OF THE CYLINDER, CONE, AND SPHERE. DEFINITIONS. 1. A cylinder is a solid, having equal and parallel circles for its ends, and is described by the revolution of a rectangle about one of its sides. 2. A cone is a solid body, of a true taper from the base to a point, which is called the vertex, and has a circle for its base. 3. A frustrum of a cone is what remains after a portion is cut off by a plane, parallel to the base. 4. A conoid is a solid, generated by the revolving of a parabola or hyperbola around its axis. 5. A spheroid is a solid, generated by the revolution of an ellipse about either of its axes. 6. A sphere is a solid, terminated by a curved surface, all the points of which are equally distant from a point within, called the center. A sphere may be described by the revolution of a semicircle about a diameter. 7. A radius of a sphere is a line drawn from the center to anv part of the surface; as, 8. The diameter of a sphere is a line drawn through the center, and terminated at both ends by the surface. All diameters of a sphere are equal to each other, and each is double the radius. 9. A segment of a sphere is a portion of the sphere cut off by any plane. This plane is called the base of the segment. The height of a segment is the distance from the middle of its base to the convex surface. 10. A zone is a portion of the surface of a sphere, included between two parallel planes, which form its bases. If the bases are equally distant from the center, it is called the middle zone. The height of a zone is the perpendicular distance between the two planes which form its bases. 11. A cylindrical ring is a solid, formed by bending a cylinder, as a cylindrical bar of iron, until the two ends meet each other. 12. A parabola is a section of a cone when cut by a plane parallel to its sides. 13. A hyperbola is the section of a cone when cut by a plane, making a greater angle with the base than the side of a cone makes. 14. The transverse axis is the longest straight line that can be drawn in an ellipse. 15. The conjugate axis is a line drawn through the center, at right angles to the transverse axis. 16. An abscissa is a part of any diameter contained between its vertex and an ordinate. 17. The focus is the point in the axis where the ordinate is equal to half the perimeter. PROBLEM I. To find the Convex Surface of a Cylinder. Rule. Multiply the circumference of the base by the length of the cylinder, and the product will be the convex surface required: To this add the areas of the two ends when the entire surface is required. Example.-What is the convex surface of a right cylinder, whose length is 23 feet, and the diameter of its base 3 feet? 3×3 14159=9.42477 Then, 9.42477 × 23=216.76971 surface. PBOBLEM II. To find the Solidity of a Cylinder. Rule. Multiply the area of the base by the height, and the pro duct will give the solid contents. Examples-1. What is the solidity of a cylinder, the diameter, a b, of whose base is 16 feet, and its height, e f, 28 feet? First, find the area of the base by 162=256. base. Then, 201 0624 × 28=5629 7472= solid con tents. 2. The Winchester bushel is a hollow cylinder, 18 inches in diameter and 8 inches deep: what is its capacity? First, the area of the base 18.5 × 7854 268.8025. NOTE. By this rule, every sealer of weights and measures may determine the exact capacity of any measure submitted to his inspection. And so any one may test the accuracy of any measure, whether dry or liquid, by reducing its capacity to cubic inches, and dividing by the number of cubic inches contained in such measure. The divisor for any measure may be found in the Table of Weights and Measures, page 13. 3. How many gallons of oil will a can of a cylindrical form hold, whose diameter is 28 inches, and whose height is 4 feet 3 inches. Area of the base by Prob. iii., Sec. ii., or by the Tables of Areas of Circles, 643 54; and 643 54 ×51÷221 184 148:39 gallons. 1 gallon =221 184 cubic inches. PROBLEM III. To find the Convex Surface of a Cone. Rule. Multiply the perimeter of the base by the slant height, and the product will be the surface; to which add the area of the base when the entire surface is required. Example. The diameter of the base of a right cone, à b, is 3 feet, and the slant height, c a, is 15 feet what is the convex surface? First, 3×3 14159=9.42477= circum. of base. PROBLEM IV To find the Solidity of a Cone. Rule.-Multiply the area of the base by of the height, and the product will be the solidity Example.-What is the solidity of a right cone, whose perpendicular height, cd, is 10 feet, and the circumference of the base is 9 feet? We here multiply the area of the base by of the height, and the product is the solidity. First, 92-81, and 101÷3=31= height. Now, 81 × 7854-63 6174, area of base. PROBLEM V. To find the Surface of a Frustrum of a Cone. Rule. Add together the circumferences of the two ends, and multiply the sum by the slant of the frustrum; the product will be the convex surface: to which add the areas of the two bases when the entire surface is required. NOTE. This rule is precisely the same as that for a frustrum of a pyramid, and f a cone be considered as a pyramid of an infinite number of sides, it is equally applicable to the measurement of the frustrum of a cone. Example.-What is the convex surface of the frustrum of a cone, the circumference of the greater base, a b, being 30 feet, and of the smaller, e f, 10 feet, the slant height, c a, being 20 feet? 30+10=40= circum. of two ends. 10 slant height. PROBLEM VI. To find the Solidity of the Frustrum of a Cone. Rule. Add to the areas of the two ends of the frustrum the square |