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The solidity of a cone is equal to one third of the product of its base and altitude.

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Let A-BCDF be a cone whose base is the circle BCDEFG, and AH its altitude ; the solidity of the cone will be equal to one thira of the product of the base BCDF by the altitude AH.

In the circle BDF inscribe a regular polygon BCDEFG, and construct a pyramid whose base is the polygon BDF, and having B its vertex in A. The solidity of this pyramid is equal to one third of the product of the polygon BCDEFG by its altitude AH (Prop. XVII., B. VIII.). Let, now, the number of sides of the polygon be indefinitely increased ; its area will become equal to the area of the circle, and the solidity of the pyramid will become equal to the solidity of the cone. But, whatever be the number of faces of the pyramid, its solidity is equal to one third of the product of its base and altitude; hence the solidity of the cone is equal to one third of the product of its base and altitude.

Cor. 1. Since a cone is one third of a cylinder having the same base and altitude, it follows that cones of equal altitudes are to each other as their bases; cones of equal bases are to each other as their altitudes; and similar cones are as the cubes of their altitudes, or as the cubes of the diameters of their bases.

Cor. 2. If A represent the altitude of a cone, and R the radius of its base, the solidity of the cone will be represented by R'XLA, or R’A.

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A frustum of a cone is equivalent to the sum of three cones, having the same altitude with the frustum, and whose bases are the lower base of the frustum, its upper base, and a mean proportional between them.

Let BDF-bdf be any frustum of a cone. Complete the cone to which the frustum belongs, and in the circle BDF in scribe the regular polygon BCDEFG; and upon this poly

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gon let a regular pyramid be construct

A ed having its vertex in A. Then will BCDEFG-bcdefg be a frustum of a regular pyramid, whose solidity is equal to three pyramids having the same altitude

with the frustum, and whose bases are be | the lower base of the frustum, its upper

base, and a mean proportional between them (Prop. XVIII., B. VIII.). Let, now, the number of sides of the polygon be indefinitely increased, its area will become

D equal to the area of the circle, and the frustum of the pyramid will become the frustum of a cone. Hence the frustum of a cone is equivalent to the sum of three cones, having the same altitude with the frustum, and whose bases are the lower base of the frustum, its upper base, and a mean proportional between them.

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The surface of a sphere is equal to the product of its diame. ter by the circumference of a great circle.

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Let ABDF be the semicircle by the revo

А lution of which the sphere is described. In- в. scribe in the semicircle a regular semi-poly

M gon ABCDEF, and from the points B, C, D,

N E let fall the perpendiculars BG, CH, DK, EL upon the diameter AF. If, now, the polygon be revolved about AF, the lines AB, EF will describe the convex surface of two Di cones; and BC, CD, DE will describe the convex surface of frustums of cones.

Froń the center I, draw IM perpendicular to BC; also, draw MN perpendicular to AF,

F and BO perpendicular to CH. Let circ. MN represent the circumference of the circle described by the revolution of MN. Then the surface described by the revolution of BC, will be equal to BC, multiplied by circ. MN (Prop. IV., Cor.).

Now, the triangles IMN, BCO are similar, since their sides are perpendicular to each other (Prop. XXI., B. IV.); whence BC : BO or GH :: ÌM : MN,

:: circ. IM : circ. MN. Hence (Prop. I., B. II.),

BC X circ. MN=GH Xcirc. IM.

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Therefore the surface described by BC, is

A equal to the altitude GH, multiplied by circ.

G IM, or the circumference of the inscribed

M circle.

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C с In like manner, it may be proved that the

H surface described by CD is equal to the alti

I tude HK, multiplied by the circumference of he inscribed circle; and the same may be Di

K к proved of the other sides. Hence the entire surface described by ABCDEF is equal to the circumference of the inscribed circle, multiplied by the sum of the altitudes AG, GH,

F HK, KL, and LF; that is, the axis of the polygon.

Let, now, the arcs AB, BC, &c., be bisected, and the num. ber of sides of the polygon be indefinitely increased, its perimeter will coincide with the circumference of the semicircle, and the perpendicular IM will become equal to the radius of the sphere; that is, the circumference of the inscribed circle will become the circumference of a great circle. Hence the surface of a sphere is equal to the product of its diameter by the circumference of a great circle.

Cor. 1. The area of a zone is equal to the product of its al titude by the circumference of a great circle.

For the surface described by the lines BC, CD is equal to the altitude GK, multiplied by the circumference of the inscribed circle. But when the number of sides of the polygon is indefinitely increased, the perimeter BC+CD becomes the aic BCD, and the inscribed circle becomes a great circle. Hence the area of the zone produced by the revolution of BCD, is equal to the product of its altitude GK by the cir cumference of a great circle.

Cor. 2. The area of a great circle is equal to the product of its circumference by half the radius (Prop. XII., B. VI.), or one fourth of the diameter; hence the surface of a sphere is equivalent to four of its great circles.

Cor. 3. The surface of a sphere is equal to the convex sur face of the circumscribed cylinder.

For the latter is equal to the product of its altitude by the circumference of its base. But its base is equal to a great circle of the sphere, and its altitude to the diameter; hence the convex surface of the cylinder, is equal to the product of its diameter by the circumference of a great circle, which is also the measure of the surface of a sphere.

Cor. 4. Two zones upon equal spheres, are to each other as their altitudes; and any zone is to the surface of its sphere, as the altitude of the zone is to the diameter of the sphere.

Cor. 5. Let R denote the radius of a sphere, D its diameter, C the circumference of a great circle, and S the surface of the sphere, then we shall have

C=2R, or 7D (Prop. XIII., Cor. 2, B. VI.).
Also, S=27RX2R=47R', or D’.
If A represents the altitude of a zone, its area will be

27RA.

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The solidity of a sphere is equal to one third the product of its surface by the radius.

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Let ACEG be the semicircle by the revolution of which the sphere is described. Inscribe in the semicircle a regular semi-polygon ABCDEFG, and draw the radii BO, CO, DO, &c.

The solid described by the revolution of the polygon ABCDEFG about AG, is com

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O posed of the solids formed by the revolution of the triangles ABO, BCO, CDO, &c., about AG.

First. To find the value of the solid formed by the revolution of the triangle ABO. From O draw OH perpendicular to AB,

G and from B draw BK perpendicular to AO. The two triangles ABK, BKO, in their revolution about AO, will describe two cones having a common base, viz., the circle whose radius is BK. Let area BK represent the area of the circle described by the revolution of BK. Then the solid described by the triangle ABO will be represented by

Area BK XAO (Prop. V.). Now the convex surface of a cone is expressed by TRS (Prop. III., Cor.); and the base of the cone by R. Hence the convex surface : base :: TRS : TR',

::S:R (Prop. VIII., B. II.). But AB describes the convex surface of a cone, of which BK describes the base; hence the surface described by AB : area BK :: AB BK

::A0: OH, because the triangles ABK, AHO are similar. Hence

Area BKX AO= OH x surface described by AB,
Area BK XŞAO= OH x surface described by AB.

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But we have proved that the solid described by the triangle ABO, is equal to area BKX;A0; it is, therefore, equal to OH surface described by AB.

Secondly. To find the value of the solid M
formed by the revolution of the triangle C
BCO.
Produce BC until it meets AG produced

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in L. It is evident, from the preceding
demonstration, that the solid described by
the triangle LCO is equal to

ES OM x surface described by LC; and the solid described by the triangle LBO

G is equal to

JOMX surface described by LB; hence the solid described by the triangle BCO is equal to

JOMX surface described by BC. In the same manner, it may be proved that the solid described by the triangle CDO is equal to

ZONX surface described by CD; and so on for the other triangles. But the perpendiculars OH, OM, ON, &c., are all equal; hence the solid described by the polygon ABCDEFG, is equal to the surface described by the perimeter of the polygon, multiplied by JOH.

Let, now, the number of sides of the polygon be indefinitely increased, the perpendicular OH will become the radius ÓA, the perimeter ACEG will become the semi-circumference ADG, and the solid described by the polygon becomes a sphere; hence the solidity of a sphere is equal to one third of the product of its surface by the radius.

Cor. 1. The solidity of a spherical sector is equal to the product of the zone which forms its base, by one third of its radius.

For the solid described by the revolution of BCDO is equal to the surface described by BC+CD, multiplied by OM. But when the number of sides of the polygon is in definitely increased, the perpendicular OM becomes the radius OB, the quadrilateral BCDO becomes the sector BDO, and the solid described by the revolution of BCDO becomes a spherical sector. Hence the solidity of a spherical sector is equal to the product of the zone which forms its base, by one third of its radius.

Cor. 2. Let R represent the radius of a sphere, D its diameter, S its surface, and V its solidity, then we shall have

S=4nR or Do (Prop. VII., Cor. 5). Also, VaRxS=TR or 1 D'; hence the solidities of spheres are to each other as the cubes of their radii

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