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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 ¬R2× 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
4 x OP (AO2+DP+ 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. A 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 R2.7 (Book V. Prop. XII. Cor. 2.), the expression for the solidities of the frustum will become
for the first pyramid
for the third
AOX PD. is a mean proportional between OA2.7 and PD.n Hence the solidity of the frustum πOP × (OA2+ PD2 + AO × PD).
of the cone is measured bv
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 CO perpen- A dicular 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.)
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.
PROPOSITION IX. LEMMA.
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.
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. From E and D, the extremities 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
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=HI× 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) x circ ON=AFx 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 IPx circ. ON: hence the surface described by ED+DC, is equal to (HI+IP) × circ. ON, or equal to HP circ, ON.
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 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 po
lygon. 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 AE× 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
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 47.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 described by the arc BCD: hence the surface of the zone is equal to EH x circ. OA.
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.
PROPOSITION XI. LEMMA.
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.