have the same height, namely, the perpendicular from F upon ACED, and equal bases, DAC, DEC. Also, FDEC can be considered as having its vertex at C, and base DFE, and therefore is equal to FABC, since the bases DFE, ABC, are equal, and the height is common; hence the three pyramids are equal; and therefore the volume of each of them is equal to one-third of the volume of the whole prism, 2ndly. When the base is a parallelogram. Let F be the vertex, and ACED the base of the pyramid, then it can be divided by a plane through FC, FD, into two pyramids on equal triangular bases ADC, EDC; and, by the previous case, the volume of each of these equal pyramids is one-third of the product of the base and perpendicular height; and therefore the volume of the whole pyramid on ACED 3rdly. When the base is a polygon. The pyramid may be divided into a number of pyramids on triangular bases by planes through the common vertex, and the diagonals of the polygon; and since the volume of each of these = = the whole base × hl. Hence, the volume of any pyramid = 1 basex perp. ht. 3 Ex. A pyramid stands on a base which is an equilateral triangle. Given that a side of the triangle is 4 feet, and the height of the pyramid is 9 feet, find its volume. Vol. = 1 3 ×4 √3×9=12√3 cub. ft., (230). The lateral surface of a pyramid is obviously obtained by taking the sum of all the triangular surfaces which bound it. 291. To measure the volume, and surface, of a cone. A right cone has already been defined in (281). If the line joining the vertex, and the centre of the base, is perpendicular to the base, the cone is a right cone; but if it makes any other angle with it, it is termed oblique. The surface of a right cone has been found in (282), from its being capable of being unwrapped, and measured as a plane surface. Since the cone, whether right or oblique, may be considered as a pyramid upon a polygonal base, with an indefinitely large number of sides, therefore the volume, as in the case of a pyramid, is equal to one-third of the duct of the area of the base and the height. pro COR. Since the volume of the cylinder of the same base and height as the cone Ex. = area of the base × ht. (289), 1 .. volume of a cone = circumscribing cylinder. 3 The height of a conical tent is 12 ft., and the radius of its base is 5 ft., what is its volume? 22 Volume=12x-(1)= 7 =3802 cub. ft. 292. To measure the volume, and surface, of a frus tum of a right cone. E B Let B, and D, be the centres, and AB, CD, radii, of the circular ends of the frustum; and sup pose the slant sides of the section through AB, and CD, to meet in E, the vertex of the complete cone, of which the frustum is a part. Let CF be drawn parallel to DB, the height of the frustum. Then, since AFC, CDE are similar triangles, DE: CD :: FC : AF, :: BD : AB-CD; ...... ..(3). Also, BD-JAC2—(AB-CD)3. From these results the complete cone can be determined, and also the cone cut off. Their difference is the frustum required. Or, the volume of the frustum may be measured approximately, with a rough approximation, by multiplying its height by the area of the section taken midway between the ends. But the correct result obtained by the first method is given by the following Rule*, and is easily remembered: "To the sum of the areas of the ends of the frustum add four times the area of the mid-section, multiply by the height, and take one-sixth of the result." Ex. The radii of the ends of a frustum are 3, and 4 inches, respectively, and the length of its slant side is 3 inches; find the volume of the frustum. Here AB-CD=1, AC2=9; It requires a little knowledge of Algebra to deduce this Rule from (1) and (2). It depends upon the fact, that AB3-CD3 divided by AB-CD is equal to AB2 + CD2+ AB×CD. .. BD=√8=2√2, BE=8√2, and DE=6√2. .. vol. of frustum=22× ×74/2 cub. in. By 2nd method, since radius of mid-section=' .. vol. of frustum=22×(7)*×2 √2, =27×784× √2 cub. in. 22 21 By Rule, vol. of frustum 3+4 or 2 Sometimes the height of the frustum is given, or it can be conveniently measured, as in the following Ex. The height of the frustum is 6 inches, and the radii of the ends are 4, and 24, inches; then BE the height of the complete cone = 16 in., and DE, that of the smaller cone, 10 in.; therefore, By 1st method, vol. of complete cone = 3 =2024 cub. in. By 2nd method, since the radius of the mid-section By the Rule, since the height of the frustum is 6in., To find the curved surface of the frustum, suppose it split down in the line AC, and opened out until it be comes a plane surface. It then forms a portion of a circular ring, which has been measured in p. 244; and surface of frustum = π × (AB + CD) × AC. 293. To measure the surface of a sphere, when its radius, or diameter, is given. DEF. A sphere is a solid generated by the revolution of a semicircle about its diameter. с Let ABC be a quadrant of a circle; and let it revolve completely round AB as an C axis; it will describe a hemisphere. Take a very small portion of the arc, DE; and draw Cc parallel to AB: through D and E draw dDm, and eEn, parallel to CB; and draw DG parallel to AB. Bisect DE in F, join BF, and draw Ff parallel to CB. Fƒ is the average distance of DE from mfn. Then, since |