Fig. 259. THEOREM. 524. The solidity of a cone is equal to the product of its base by a third part of its altitude. Demonstration. Let SO (fig. 259) be the altitude of the given cone, AO the radius of the base; representing by surf. AO the surface of the base, we say that the solidity of the cone is equal to surf. AO SO. 1. Let surf. AO × SO be supposed to be the solidity of a greater cone, of a cone, for example, whose altitude is always SO, but of which BO, greater than AO, is the radius of the base. About the circle, whose radius is AO, circumscribe a regular polygon MNPT, which shall not meet the circumference of which OB is the radius (285); suppose then a pyramid having this polygon for its base and the point S for its vertex. The solidity of this pyramid is equal to the area of the polygon MNPT multiplied by a third of the altitude SO (416). But the polygon is greater than the inscribed circle represented by surf. OA; consequently, the pyramid is greater than surf. AOSO, which, by hypothesis, is the measure of the cone of which Sis surf. OB SO, and accordingly it would be less than the cone, of which AO is the radius of the base and SO the altitude. But on the contrary the pyramid is greater than the cone, since it contains it; therefore it is impossible that the base of a cone multiplied by a third of its altitude should be the measure of a less cone. We conclude then, that the solidity of a cone is equal to the product of its base by a third of its altitude. 525. Corollary. A cone is a third of a cylinder of the same base and same altitude; whence it follows, 1. That cones of equal altitudes are to each other as their bases; 2. That cones of equal bases are to each other as their altitudes; 3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes. 526. Scholium. Let R be the radius of the base of a cone, H its altitude; the solidity of the cone will be л R2 × ¦ Н, or π R2 H. THEOREM. 527. The frustum of a cone ADEB (fig. 260) of which OA, DP, Fig. 260. are the radii of the bases, and PO the altitude, has for its measure × OP × (AO+DP+ AO × DP). Demonstration. Let TFGH be a triangular pyramid of the same altitude as the cone SAB, and of which the base FGH is equivalent to the base of the cone. The two bases may be supposed to be placed upon the same plane; then the vertices S, T, will be at equal distances from the plane of the bases; and the plane EPD produced will be in the pyramid the section IKL. We say now, that this section IKL is equivalent to the base DE; for the bases AB, DE, are to each other as the squares of the radii AO, DP (287), or as the squares of the altitudes SO, SP; the triangles FGH, IKL, are to each other as the squares of these same altitudes (407); consequently the circles AB, DE, are to each other as the triangles FGH, IKL. But, by hypothesis, the triangle FGH, is equivalent to the circle AB; therefore the triangle IKL is equivalent to the circle DE. Now the base AB multiplied by SO is the solidity of the cone SAB, and the base FGH multipled by SO is that of the pyramid TFGH; the bases therefore being equivalent, the solidity of the pyramid is equal to that of the cone. For a similar reason the pyramid TIKL is equivalent to the cone SDE; therefore the frustum of the cone ADEB is equivalent to the frustum of the pyramid FGHIKL. But the base FGH, equivalent to the circle of which the radius is AO, has for its measure л× AO; likewise the base IKLл × DP, and the mean pro AO portional between л× АО and л× DP is л AО × DP; therefore the solidity of the frustum of a pyramid or that of the frustum of a cone has for its measure or ¦ OP × (ñ × AO +л × DP +л × AO × DP】 (422), 528. The convex surface of a cone is equal to the circumference of its base multiplied by half its side. Demonstration. Let AO (fig. 259), be the radius of the base of the given cone, S its vertex, and SA its side; we say that the surface will be circ. AOSA. For, if it be possible, let circ. AO × SA be the surface of a cone which has S for its vertex, and for its base the circle described with a radius OB greater than AO. Circumscribe about the small circle a regular polygon MNPT, the sides of which shall not meet the circumference of which OB is the radius; and let SMNPT be a regular pyramid, which has for its base the polygon, and for its vertex the point S. The triangle SMN, one of those which compose the convex surface of the pyramid, has for its measure the base MN multiplied by half of the altitude SA, which is at the same time the side of the given cone; this altitude being equal in all the triangles SNP, SPQ, &c., it follows that the convex surface of the pyramid is equal to the perimeter MNPTM multiplied by ¦ SA. But the perimeter MNPTM is greater than circ. AO; therefore the convex surface of the pyramid is greater than circ. AO × §SA, and consequently greater than the convex surface of the cone, which, with the same vertex S, has for its base the circle described with the radius OB. But on the contrary the convex surface of the cone is greater than that of the pyramid; for, if we apply the base of the pyramid to the base of an equal pyramid, and the base of the cone to that of an equal cone; the surface of the two cones will enclose on all sides the surface of the two pyramids; consequently the first surface will be greater than the second (514), and therefore the surface of the cone is greater than that of the pyramid, which is comprehended within it. The contrary would be the consequence of our hypothesis; accordingly this hypothesis cannot be maintained; therefore the circumference of the base of a cone multiplied by the half of its side cannot be the measure of the surface of a greater cone. 2. We say also, that this same product cannot be thc measure of the surface of a less cone. For let BO be the radius of the base of the given cone, and, if it be possible, let circ. BOX SB be the surface of a cone of which S is the vertex, and AO, less than OB, the radius of the base. The same construction being supposed as above, the surface of the pyramid SMNPT will always be equal to the perimeter MNPT multiplied by SA. Now the perimeter MNPT is less than circ. BO, and SA is less than SB; therefore for this double reason the convex surface of the pyramid is less than circ. BOX SB. which, by hypothesis, is the surface of a cone of which AO is the radius of the base; consequently the surface of the pyramid would be less than that of the inscribed cone. But on the contrary it is greater; for by applying the base of the pyramid to that of an equal pyramid, and the base of the cone to that of an equal cone, the surface of the two pyramids will enclose that of the two cones, and consequently will be greater. Therefore it is impossible that the circumference of the base of a given cone multiplied by the half of its side should be the measure of the surface of a less cone. We conclude then, that the convex surface of a cone is equal to the circumference of the base multiplied by half of its side. 529. Scholium. Let L be the side of a cone, and R the radius of the base, the circumference of this base will be 2 R, and the surface of the cone will have for its measure 2 R × 1⁄2 L, ог л RL. THEOREM. 530. The convex surface of the frustum of a cone ADEB (fig. 261) is equal to its side AD multiplied by the half sum of the Fig. 261. circumferences of the two bases AB, DE. Demonstration. In the plane SAB, which passes through the axis SO, draw perpendicularly to SA the line AF, equal to the circumference which has for its radius AO; join SF, and draw DH parallel to AF. On account of the similar triangles SAO, SDC, AO: DC: SA: SD; and, on account of the similar triangles SAF, SDH, AF: DH:: SA: SD; whence AF: DH:: AO: DC :: circ. AO: circ. DC (287). But, by construction, AF = circ. AO; consequently This being premised, the triangle SAF, which has for its meas ure AFSA, is equal to the surface of a cone SAB, which has for its measure circ. AO × 1 SA. For a similar reason the triangle SDH is equal to the surface of the cone SDE. Whence the surface of the frustum ADEB is equal to that of the trapezoid ADHF: This has for its measure AD × (AF+DH) (178) Therefore the surface of the frustum of a cone ADEB is equal to its side AD multiplied by the half sum of the circumferences of the two bases. 531. Corollary. Through the point I, the middle of AD, draw IKL parallel to AB, and IM parallel to AF; it may be shown as above that IM= circ. IK. But the trapezoid (179) ADHF = AD× IM = AD ×x circ. IK. Hence we conclude further that the surface of the frustum of a cone is equal to its side multiplied by the circumference of a section made at equal distances from the two bases. 532. Scholium. If a line AD, situated entirely on the same side of the line OC and in the same plane, make a revolution about OC, the surface described by AD will have for its measure circ. AO+ circ. DC AD × (circ. 1+ circ. DC), 2 or AD x circ. IK; the lines AO, DC, IK, being perpendiculars let fall from the extremities and from the middle of the line AD upon the axis OC. For, if we produce AD and OC till they meet in S, it is evident that the surface described by AD is that of the frustum of a cone, of which OA and DC are the radii of the bases, the entire cone having for its vertex the point S. Therefore this surface will have the measure stated. This measure would always be correct, although the point D should fall upon S, which would give an entire cone, and also when the line AD is parallel to the axis, which would give a cylinder. In the first case DC would be nothing, in the second DC would be equal to AO and to IK. |