PROPOSITION VII. THEOREM. Every section of a sphere, made by a plane, is a circle. Let AMB be a section, made by a plane, in the sphere whose centre is C. From the point C, draw CU perpendicular to the plane AMB; and different lines CM, CM, to different points of the curve AMB, which terminates the section.. 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. PROPOSITION VIII. THEOREM. Every plane perpendicular to a radius at its extremity is tangent to the sphere. Let FAG be a plane perpendicular to the radius OA, at its extremity A. Any point M in this plane being assumed, and OM, AM, being drawn, the angle OAM will be a right angle, and hence the distance OM will be greater than OA. Hence the point M lies without the sphere; and as the same can be shown for every other point of the plane FAG, this plane can have no point but A common to it and the surface of the sphere; hence it is a tangent plane (Def. 12.) PROPOSITION IX. LEMMA. Scholium. In the same way it may be shown, that two spheres have but one point in common, and therefore touch each other, when the distance between their centres is equal to the sum, or the difference of their radii; in which case, the centres and the point of contact lie in the same straight line. Let the regular semi-polygon ABCDEF, be revolved about the line AF as an axis: then will the surface described by its perimeter be equal to AF multiplied by the circumference of the inscribed circle. D From E and D, the extremities of one of the equal sides, let fall the perpendiculars EH, DI, on the axis AF, and from the centre O draw ON perpendicular to the side C DE: ON will be the radius of the inscribed circle (Book V. Prop. II.). Now, the surface described in the revolution by any one side of the regular polygon, as DE, has If a regular semi-polygon be revolved about a line passing through the centre and the vertices of two opposite angles, the surface described by its perimeter will be equal to the axis mul tiplied by the circumference of the inscribed circle. N E F M K B G been shown to be equal to DEx circ. NM (Prop. IV. Sch.). But since the triangles EDK, ONM, are similar (Book IV, Prop. XXI.), ED: EK or HI :: ON: NM, or as circ. ON circ. NM; hence EDx circ. NM-HIx circ. ON; and since the same may be shown for each of the other sides it is plain that the surface described by the entire perimeter is equal to (FH+HI+IP+PQ+QA) × circ ON=AF× circ. ON. Cor. The surface described by any portion of the perimeter, as EDC, is equal to the distance between the two perpendiculars let fall from its extremities on the axis, multiplied by the circumference of the inscribed circle. For, the surface described by DE is equal to HIx circ. ON, and the surface described by DC is equal to IP × circ. ON: hence the surface described by ED+DC, is equal to (HI+IP) × circ. ON, or equal to HP x circ, ON. X PROPOSITION X. THEOREM. The surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Let ABCDE be a semicircle. Inscribe in it any regular semi-polygon, and from the centre O draw OF perpendicular to one of the sides. Let the semicircle and the semi-polygon be revolved about the axis AE: the semicircumference ABCDE will describe the A surface of a sphere (Def. 8.); and the pe- B rimeter of the semi-polygon will describe a surface which has for its measure AE× circ. OF (Prop. IX.), and this will be true whatever be the number of sides of the polygon. But if the number of sides of the polygon be indefinitely increased, its perimeter will coincide with the circumference ABCDE, the perpendicular OF will become equal to OE, and the surface described by the perimeter of the semipolygon will then be the same as that described by the semicircumference ABCDE. Hence the surface of the sphere is equal to AEx circ. OE. Cor. Since the area of a great circle is equal to the product of its circumference by half the radius, or one fourth of the M diameter (Book V. Prop. XII.), it follows that the surface of a sphere is equal to four of its great circles: that is, equal to 4.OA (Book V. Prop. XII. Cor. 2.). Scholium 1. The surface of a zone is equal to its altitude multiplied by the circumference of a great circle. For, the surface described by any portion of the perimeter of the inscribed polygon, as BC+CD, is equal to EH × circ. OF (Prop. IX. Cor.). But when the number of sides of the polygon is indefinitely increased, BC C +CD, becomes the arc BCD, OF becomes equal to OA, and the surface described by BC+CD, becomes the surface of the zone D described by the arc BCD: hence the surface of the zone is equal to EH × circ. OA. B <F H PROPOSITION XI. LEMMA. E Scholium 2. When the zone has but one base, as the zone described by the arc ABCD, its surface will still be equal to the altitude AĚ multiplied by the circumference of a great circle. Scholium 3. Two zones, taken in the same sphere or in equal spheres, are to each other as their altitudes; and any zone is to the surface of the sphere as the altitude of the zone is to the diameter of the sphere. If a triangle and a rectangle, having the same base and the same altitude, turn together about the common base, the solid described by the triangle will be a third of the cylinder described by the rectangle. Let ACB be the triangle, and BE the rectangle. On the axis, let fall the perpen- F dicular AD: the cone described by the triangle ABD is the third part of the cylinder described by the rectangle AFBD (Prop. V Cor.); also the cone described by the triangle ADC is the third par. of the cylinder described by ne rectangle ADCE; hence the sum of the two cones, or the solid described by ABC, is the third part of the two cylinders taken together, or of the cylinder described by the rectangle BCEF. B D E If the perpendicular AD falls without the triangle; the solid described by ABC will, in that case, be the difference of the two cones described by ABD and ACD ; but at the same time, the cylinder described by BCEF will be the difference of the two cylinders described by AFBD and AECD. Hence the solid, described by the revolution of the triangle, will still be a third part of the cylinder described by the revolution of the rectangle having the same base and the same altitude. B C PROPOSITION XII. LEMMA. Scholium. The circle of which AD is radius, has for its neasure л× AD2; hence × AD2 × BC measures the cylinder described by BCEF, and 7× AD2 × BC measures the solid described by the triangle ABC. If a triangle be revolved about a line drawn at pleasure through its vertex, the solid described by the triangle will have for its measure, the area of the triangle multiplied by two thirds of the circumference traced by the middle point of the base. E Let CAB be the triangle, and CD the line about which it revolves. Р A Produce the side AB till it meets the axis CD in D ; from the points A and B, draw AM, BN, perpendicular to the axis, and CP perpendicular to DA produced. MKN The solid described by the tri- C angle CAD is measured by AM2 × CD (Prop. XI. Sch.); the solid described by the triangle CBD is measured by × BN2 × CD; hence the difference of those solids, or the solid described by ABC, will have for its measure (AM2—BN2) × CD. To this expression another form may be given. From I, the middle point of AB, draw IK perpendicular to CD; and through B, draw BO parallel to CD: we shall have AM+BN=2IK (Book IV. Prop. VII.); and AM-BN-AO; hence (AM+ BN) × (AM—NB), or AM2—BN2=2IK × AO (Book IV. Prop X.). Hence the measure of the solid in question is ex pressed by × IK× AO × CD. Ι B |