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We may conceive it to be generated by the revolution of a trapezoid ABHG, whose angles A and G are right angles, about the side AG. The immoveable line AG is called the axis or altitude of the frustum, the circles BDC, HEK, are its bases, and BH is its side.

4. Two cylinders, or two cones, are similar, when their axes are to each other as the diameters of their bases.

5. If in the circle ACD, which forms the base of a cylinder, a polygon ABCDE be inscribed, a right prism, constructed on this F base ABCDE, and equal in altitude to the cylinder, is said to be inscribed in the cylinder, or the cylinder to be circumscribed about the prism.

The edges AF, BG, CH, &c. of the prism, being perpendicular to the plane of the base, are evidently included in the convex surface of the cylinder; hence the prism and the cylinder touch one another along these edges.

6. In like manner, if ABCD is a polygon, circumscribed about the base of a cylinder, a right prism, constructed on this base ABCD, and equal in altitude to the cylinder, is said to be circumscribed about the cylinder, or the cylinder to be inscribed in the prism.

Let M, N, &c. be the points of contact in the sides AB, BC, &c.; and through the points M, N, &c. let MX, NY, &c. be drawn A perpendicular to the plane of the base: these perpendiculars will evidently lie both in the surface of the cylinder, and in that

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of the circumscribed prism; hence they will be their lines of

contact.

7. If in the circle ABCDE, which forms the base of a cone, any polygon ABCDE be inscribed, and from the vertices A, B, C, D, E, lines be drawn to S, the vertex of the cone, these lines may be regarded as the sides of a pyramid whose base is the polygon ABCDE and vertex S. The sides of this pyramid are in the convex A surface of the cone, and the pyramid is said to be inscribed in the cone.

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8. The sphere is a solid terminated by a curved surface, all the points of which are equally distant from a point within, called the centre.

The sphere may be conceived to be generated by the revolution of a semicircle DAE about its diameter DE: for the surface described in this movement, by the curve DAE, will have all its points A equally distant from its cen

tre C.

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Whilst the semicircle DAE revolving round its diameter DE, describes the sphere; any circular sector, as DCF or FCH, describes a solid, which is named a spherical sector.

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10. The radius of a sphere is a straight line drawn from the centre to any point of the surface; the diameter or axis is a line passing through this centre, and terminated on both sides by the surface.

All the radii of a sphere are equal; all the diameters are equal, and each double of the radius.

11. It will be shown (Prop. VII.) that every section of the sphere, made by a plane, is a circle: this granted, a great circle is a section which passes through the centre; a small circle, is one which does not pass through the centre.

12. A plane is tangent to a sphere, when their surfaces have but one point in common.

13. A zone is a portion of the surface of the sphere included between two parallel planes, which form its bases. One of these planes may be tangent to the sphere; in which case, the zone has only a single base.

14. A spherical segment is the portion of the solid sphere, included between two parallel planes which form its bases. One of these planes may be tangent to the sphere; in which case, the segment has only a single base.

15. The altitude of a zone or of a segment is the distance between the two parallel planes, which form the bases of the zone or segment.

Note. The Cylinder, the Cone, and the Sphere, are the three round bodies treated of in the Elements of Geometry.

PROPOSITION I. THEOREM.

The convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude.

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whose radius is CA being represented by circ. CA, we are to show that the convex surface of the cylinder is equal to circ. CA × H.

Inscribe in the circle any regular polygon, BDEFGA, and construct on this polygon a right

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prism having its altitude equal to H, the altitude of the cylinder this prism will be inscribed in the cylinder. The convex surface of the prism is equal to the perimeter of the polygon, multiplied by the altitude H (Book VII. Prop. I.). Let now the arcs which subtend the sides of the polygon be continually bisected, and the number of sides of the polygon indefinitely increased: the perimeter of the polygon will then become equal to circ. CA (Book V. Prop. VIII. Cor. 2.), and the convex surface of the prism will coincide with the convex surface of the cylinder. But the convex surface of the prism is equal to the perimeter of its base multiplied by H, whatever be the number of sides: hence, the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude.

PROPOSITION II. THEOREM.

The solidity of a cylinder is equal to the product of its base by its altitude.

Let CA be the radius of the base of the cylinder, and H the altitude. Let the circle whose radius is CA be represented by area CA, it is to be proved that the solidity of the cylinder is equal to area CA × H. Inscribe in the circle any regular polygon BDEFGA, and construct on this polygon a right prism having its altitude equal

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to H, the altitude of the cylinder: this prism will be inscribed in the cylinder. The solidity of the prism will be equal to the area of the polygon multiplied by the altitude H (Book VII. Prop. XIV.). Let now the number of sides of the polygon be indefinitely increased: the solidity of the new prism will still be equal to its base multiplied by its altitude.

But when the number of sides of the polygon is indefinitely increased, its area becomes equal to the area CA, and its perimeter coincides with circ. CA (Book V. Prop. VIII. Cor. 1. & 2.); the inscribed prism then coincides with the cylinder, since their altitudes are equal, and their convex surfaces perpendicular to the common base: hence the two solids will be equal; therefore the solidity of a cylinder is equal to the product of its base by its altitude.

Cor. 1. Cylinders of the same altitude are to each other as their bases; and cylinders of the same base are to each other as their altitudes.

Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases. For the bases are as the squares of their diameters; and the cylinders being similar, the diameters of their bases are to each other as the altitudes (Def. 4.); hence the bases are as the squares of the altitudes; hence the bases, multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes.

Scholium. Let R be the radius of a cylinder's base; H the altitude the surface of the base will be .R2 (Book V. Prop. XII. Cor. 2.); and the solidity of the cylinder will be R2 × H or π.R2.H.

PROPOSITION III. THEOREM.

The convex surface of a cone is equal to the circumference of its base, multiplied by half its side.

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

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regular pyramid, and will be inscribed in the cone.

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.

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

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