PROPOSITION III. THEOREM. The convex surface of a cone is equal to the circumference of its base, multiplied by half its side. G Let the circle ABCD be the base of a cone, S the vertex, SO the altitude, and SA the side then will its convex surface be equal to circ. OA× SA. For, inscribe in the base of A the cone any regular polygon ABCD, and on this polygon as a base conceive a pyramid to be constructed having S for its vertex: this pyramid will be a regular pyramid, and will be inscribed in the cone. B From S, draw SG perpendicular to one of the sides of the polygon. The convex surface of the inscribed pyramid is equal to the perimeter of the polygon which forms its base, multiplied by half the slant height SG (Book VII. Prop. IV.). Let now the number of sides of the inscribed polygon be indefinitely increased; the perimeter of the inscribed polygon will then become equal to circ. OA, the slant height SG will become equal to the side SA of the cone, and the convex surface of the pyramid to the convex surface of the cone. But whatever be the number of sides of the polygon which forms the base, the convex surface of the pyramid is equal to the perimeter of the base multiplied by half the slant height: hence the convex surface of a cone is equal to the circumference of the base multiplied by half the side. Scholium. Let L be the side of a cone, R the radius of its base; the circumference of this base will be 27.R, and the surface of the cone will be 2¬R ×¦L, or ¬RL. PROPOSITION IV. THEOREM. S D The convex surface of the frustum of a cone is equal to its side multiplied by half the sum of the circumferences of its two bases. Let BIA-DE be a frustum of a cone: then will its convex surface be circ.OA+circ.CD equal to AD x (circ. circ.CD). 2 For, inscribe in the bases of the frustums two regular polygons of the same number of sides, and having their homologous sides parallel, each to each. The lines joining the vertices of the homologous angles may be regarded as the edges of the frus B tum of a regular pyramid inscribed in the frustum of the cone. The convex surface of the frustum of the pyramid is equal to half the sum of the perimeters of its bases multiplied by the slant height fh (Book VII. Prop. IV. Cor.). I Let now the number of sides of the inscribed polygons be indefinitely increased: the perimeters of the polygons will become equal to the circumferences BIA, EGD; the slant height fh will become equal to the side AD or BE, and the surfaces of the two frustums will coincide and become the same surface. L E G S C K D O d But the convex surface of the frustum of the pyramid will still be equal to half the sum of the perimeters of the upper and lower bases multiplied by the slant height: hence the surface of the frustum of a cone is equal to its side multiplied by half the sum of the circumferences of its two bases. Magage Cor. Through 1, the middle point of AD, draw IKL parallel to AB, and li, Dd, parallel to CO. Then, since Al, ID, are equal, Ai, id, will also be equal (Book IV. Prop. XV. Cor. 2.): hence, K is equal to (OA+ CD). But since the circumferences of circles are to each other as their radii (Book V. Prop. XI.), the circ. Kl=1(circ. OA+ circ. CD); therefore, the convex surface of a frustum of a cone is equal to its side multiplied by the circumference of a section at equal distances from the two bases. Scholium. If a line AD, lying wholly on one side of the line OC, and in the same plane, make a revolution around OC, the surface described by AD will have for its measure AD× circ. AO+ circ. DC DC), or AD × circ. IK; the lines AO, DC, IK, 2 being perpendiculars, let fall from the extremities and from the middle point of AD, on the axis OC. For, if AD and OC are produced till they meet in S, the surface described by AD is evidently the frustum of a coné having AO and DC for the radii of its bases, the vertex of the whole cone being S. Hence this surface will be measured as we have said. This measure will always hold good, even when the point D falls on S, and thus forms a whole cone; and also when the line AD is parallel to the axis, and thus forms a cylinder. In the first case DC would be nothing; in the second, DC would be equal to AO and to IK. PROPOSITION V. THEOREM. The solidity of a cone is equal to its base multiplied by a third of its altitude. Let SO be the altitude of a cone, OA the radius of its base, and let the area of the base be designated by area OA: it is to be proved that the solidity of the cone is equal to area OAXSO. A F B Inscribe in the base of the cone any regular polygon ABDEF, and join the vertices A, B, C, &c. with the vertex S of the cone : then will there be inscribed in the cone a regular pyramid having the same vertex as the cone, and having for its base the polygon ABDEF. The solidity of this pyramid is equal to its base multiplied by one third of its altitude (Book VII. Prop. XVII.). Let now the number of sides of the polygon be indefinitely increased: the polygon will then become equal to the circle, and the pyramid and cone will coincide and become equal. But the solidity of the pyramid is equal to its base multiplied by one third of its altitude, whatever be the number of sides of the polygon which forms its base: hence the solidity of the cone is equal to its base multiplied by a third of its altitude. D E Cor. A cone is the third of a cylinder having the same base and the 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. P* Cor. 2. The solidity of a cone is equivalent to the solidity of a pyramid having an equivalent base and the same altitude (Book VII. Prop. XVII.). Scholium. Let R be the radius of a cone's base, H its altitude; the solidity of the cone will be R2x H, or R2H. PROPOSITION VI. THEOREM The solidity of the frustum of a cone is equal to the sum of the solidities of three cones whose common altitude is the altitude of the frustum, and whose bases are, the upper base of the frustum, the lower base of the frustum, and a mean proportional between them. Let AEB-CD be the frustum of a cone, and OP its altitude; then will its solidity be equal to × OP × (AO2+DP2+AO × DP). For, inscribe in the lower and upper bases two regular polygons having the same number of sides, and having their homologous sides parallel, each to each. Join the vertices of the homologous angles and there will then be inscribed in the frustum of the cone, the frustum of a regular pyramid. The solidity of the frustum of the pyramid is equivalent to three pyramids having the common altitude of the frustum, and for bases, the lower base of the frustum, the upper base of the frustum, and a mean proportional between them (Book VII. Prop. XVIII.). Let now, the number of sides of the inscribed polygons be indefinitely increased: the bases of the frustum of the pyramid will then coincide with the bases of the frustum of the cone, and the two frustums will coincide and become the same solid. Since the area of a circle is equal to R2. (Book V. Prop. XII. Cor. 2.), the expression for the solidities of the frustum will become π for the first pyramid A D P E }OPxOA”. OP x AOX PD.; since B AOX PD. is a mean proportional between OA. and PD2.7. Hence the solidity of the frustum of the cone is measured by πOP × (OA2+PD2+AO × PD). PROPOSITION VII. THEOREM. Every section of a sphere, made by a plane, is a circle. D Let AMB be a section, made by a plane, in the sphere whose centre is C. From the point C, draw CU perpen- A dicular to the plane AMB; and different lines CM, CM, to different points of the curve AMB, which terminates the section. M M B The oblique lines CM, CM, CA, are equal, being radii of the sphere; hence they are equally distant from the perpendicular CO (Book VI. Prop. V. Cor.); therefore all the lines OM, OM, OB, are equal; consequently the section AMB is a circle, whose centre is O. Cor 1. If the section passes through the centre of the sphere, its radius will be the radius of the sphere; hence all great circles are equal. Cor. 2. Two great circles always bisect each other; for their common intersection, passing through the centre, is a diameter. Cor. 3. Every great circle divides the sphere and its surface into two equal parts: for, if the two hemispheres were separated and afterwards placed on the common base, with their convexities turned the same way, the two surfaces would exactly coincide, no point of the one being nearer the centre than any point of the other. Cor. 4. The centre of a small circle, and that of the sphere, are in the same straight line, perpendicular to the plane of the small circle. Cor. 5. Small circles are the less the further they lie from the centre of the sphere; for the greater CO is, the less is the chord AB, the diameter of the small circle AMB. Cor. 6. An arc of a great circle may always be made to pass through any two given points of the surface of the sphere; for the two given points, and the centre of the sphere make three points which determine the position of a plane. But if the two given points were at the extremities of a diameter, these two points and the centre would then lie in one straight line, and an infinite number of great circles might be made to pass through the two given points. |