scribed in the cylinder, its surface (B. VIII, Prop. III,) would be less than the cylinder's; much more, then, is it less when the prism does not reach so far as to touch the cylinder. Consequently, also, in the second place, the circumference of a cylinder's base, multiplied by the altitude, cannot measure the surface of a greater cylinder.

The product in question, being, therefore, neither the measure of the convex surface of a less nor greater cylinder, must be the measure of the cylinder itself.

PROPOSITION V.

THEOREM. The solidity of a cone is equal to the product of its base by the third of its altitude.

Let SO be the altitude of

the given cone, AO the radius of its base; the surface of the base being designated by surf. AO, it is to be demonstrated that surf. AO SO is equal to the solidity of the cone.

Suppose, first, that surf.

M

N

AOXSO is the solidity of a greater cone; for example, of the cone whose altitude is also SO, but whose base has OB greater than AO for its radius.

About the circle whose radius is AO, circumscribe a regular polygon MNPT, so as not to meet the circumference whose radius is OB: imagine a pyramid having this polygon for its base, and the point S for its vertex.

The solidity of this pyramid (B. VII, Prop. xvII,) is equal to the area of the polygon MNPT multiplied by a third of the altitude SO; but the polygon is greater than the inscribed circle represented by surf. AO: hence the pyramid is greater than surf. AOX SO, which, by hypothesis, measures the cone having S for its vertex and OB for the radius of its base; whereas, in reality, the pyramid is less than the cone, being contained in it. Hence, first, the base of a cone multiplied by a third of its altitude cannot be the measure of a greater

cone.

Neither can this same product be the measure of a smaller cone. For now let OB be the radius of the given cone's base; and, if possible, let surf. OB SO be the solidity of the cone having SO for its altitude, and for its base the circle whose radius is AO. The same construction being made, the pyramid SMNPT will have for its measure the area MNPT multiplied by SO; but the area MNPT is less than surf. OB: hence the measure of the pyramid must be less than surf. OB × 1 SO, and consequently it must be less than the cone whose altitude is SO, and whose base has AO for its radius; but, on the contrary, the pyramid is greater than the cone, because the cone is contained in it. Hence, in the second place, the base of a cone multiplied by a third of its altitude cannot be the measure of a smaller one.

Consequently the solidity of a cone is equal to the product of its base by the third of its altitude.

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;

241 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.

Schol. Let R be the radius of a cone's base, and H its altitude; the solidity of the cone will be RH, or π R'H.

PROPOSITION VI.

THEOREM.

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

About the smaller circle describe a regular polygon MNPT, the sides of which may not meet the circle whose radius is OB; and let SMNPT be the regular pyramid, having this polygon for its base, and the point S for its vertex. The triangle SMN, one of those which compose the convex surface of the pyramid, has for measure its base MN multiplied by half its altitude SA, or half the side of the given cone; and since this altitude

is the same in all the other triangles SNP, SPQ, &c. the convex surface of the pyramid must be equal to the perimeter MNPTM multiplied by SA. But the contour MNPTM is greater than circ. AO; hence the convex surface of the pyramid is greater than circ. AOX SA, and consequently greater than the convex surface of the cone having the same vertex S, and the circle whose radius is OB for its base. But the surface of this cone is greater than that of the pyramid; because, if two such pyramids are adjusted to each other base to base, and two such cones base to base, the surface of the double cone will envelop on all sides that of the double pyramid, and therefore be greater than it, as is evident; hence the surface of the cone is greater than that of the pyramid, whereas by the hypothesis it is less. Hence, in the first place, the circumference of the cone's base multiplied by half the side cannot measure the surface of a greater cone.

Neither can it measure the surface of a smaller cone : for, let BO be the radius of the base of the given cone; and, if possible, let circ. BOX SB be the surface of a cone having S for its vertex, and AO less than OB for the radius of its base.

14

The same construction being made as above, the surface of the pyramid SMNPT will still be equal to the perimeter MNPT SA. Now, this perimeter MNPT is less than circ. OB; likewise SA is less than SB; consequently, for a double reason, the convex surface of the pyramid is less than circ. OB x SB, which, by hypothesis, is the surface of the cone having SA for the radius of its base: hence the surface of the pyramid must be less than that of the inscribed cone; but it is ob

viously greater; for, adjusting two such pyramids to each other, base to base, and two such cones base to base, the surface of the double pyramid will envelop that of the double cone, and will be greater than it. Hence, in the second place, the circumference of the base of the given cone, multiplied by half the side, cannot be the measure of the surface of a smaller cone.

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

R ; and

Schol. Let L be the side of a cone, R the radius of its base the circumference of this base will be 2 the surface of the cone will be 2 R × L, or

RL.

THEOREM.

PROPOSITION VII.

The convex surface of a truncated cone is equal to its side multiplied by half the sum of the circumferences of its two bases.

In the plane SAB which passes through the axis SO, draw the line AF perpendicular to SA, and equal to the circumference having AO for its radius; join SF, and draw DH parallel to AF.