1035. COROLLARY II. The surface of a pyramid is less than the surface of a pyramid that envelopes it and has the same base.

[Produce the plane of one of the lateral faces of the inner pyramid until it cuts the surface of the outer pyramid, forming a new polyhedron whose surface is less than that of the outer pyramid. Produce the plane of the next lateral face of the inner pyramid until it cuts the surface of this polyhedron, forming a second polyhedron whose surface is less than that of the first polyhedron.

S'.

S

D

Continue in this way, and the last polyhedron will be the inner pyramid.]

1036. If a pyramid whose base is a regular polygon is inscribed in or circumscribed about a circular cone, and if the number of sides of the base of the pyramid is indefinitely increased,

I. The volume of the cone is the limit of the volume of the pyramid.

II. The lateral area of the cone is the limit of the lateral area of the pyramid.

Let a pyramid whose base is a regular polygon be inscribed in the circular cone, and one whose base is a similar polygon be circumscribed about the circular cone, and let the number of sides of the base be indefinitely increased.

E el

F

N

I. To Prove that the volume of the cone is the limit of the

volume of the pyramid.

[Proof similar to that of § 1013, I.]

II. To Prove that the lateral area of the cone is the limit of the lateral area of the pyramid.

Proof. Let s', s, and s designate the entire surfaces of cone, circumscribed pyramid, and inscribed pyramid, respectively. Let B and b designate the bases of the circumscribed pyramid and inscribed pyramid respectively.

Draw dм in the plane of the ▲ DON parallel to DN. Connect M with the remaining vertices of the inscribed base cde ......, forming a third pyramid M-cde... Designate the entire surface and base of this pyramid by s' and b' respectively.

Show that Mc, Me, etc., are parallel to NC, NE, etc., respectively.

Show that A Med, Mdc, etc., are similar to ▲ NED, NDc, etc., respectively.

Show that S

equal to zero.

s' can be made as small as we please, but not

But s is greater than s' and less than s. (?)

.. S

s can be made as small as we please. But s' is greater than s and less than s. (?)

Therefore S s' or s'-s can be made as small as we please, but not equal to zero.

The entire surface of the cone is consequently the limit of the entire surface of the inscribed or circumscribed pyramid. Since the base of the cone is the limit of the base of the pyramid (?), the lateral area of the cone is the limit of the lateral area of the pyramid.

Q.E.D.

PROPOSITION XII. THEOREM

1037. The lateral area of a cone of revolution is equal to one half the product of the circumference of its base by its slant height.

M

[Circumscribe about the cone a pyramid having for its base a regular polygon, and proceed as in § 1014.]

B

1038. COROLLARY. If S stands for the lateral area, r for the radius of the base, and H for the slant height, then

1039. EXERCISE. The altitude of a cone of revolution is 12 in., and the radius of its base is 9 in. Find its lateral area.

1040. EXERCISE. The slant height of a cone of revolution is equal to the diameter of its base. The lateral area is 25.1328 sq. ft. Find the slant height.

PROPOSITION XIII. THEOREM

1041. The volume of a circular cone is equal to one third the product of its base by its altitude.

[The proof is left to the student.]

B

E

S

1042. COROLLARY. If v stands for the volume, a for the altitude, and r for the radius of the base, then

1043. EXERCISE. The volume of a circular cone is 100 cu. in. Its altitude is 25 ft. What is the area of its base?

1044. EXERCISE. The slant height of a cone of revolution is 25 ft., and the radius of its base is 20 ft. Find its volume.

1045. DEFINITION. Similar cones of revolution are cones generated by similar right-angled triangles revolved about homologous legs as axes.

1046. The lateral or entire areas of two similar cones of revolution are to each other as the squares of their altitudes or as the squares of the radii of their bases; and their volumes are to each other as the cubes of their altitudes or as the cubes of the radii of their bases.

AA

[Proof similar to that of § 1022.]

.1047. EXERCISE. The volume of one of two similar cones of revolution is 125 times the volume of the other. Compare their surfaces.

1048. EXERCISE. A cone of revolution is cut into two portions by a plane parallel to the base. The portion with the vertex is of the remaining part. If the altitude is 8 in., how far from the vertex did the cutting plane pass?

1049. DEFINITIONS. A truncated cone is the portion of a cone included between its base and a plane cutting all of its elements.

The frustum of a cone is the portion of a cone included between its base and a plane parallel to its base.

The base of the cone and the parallel section are called the bases of the frustum.

The altitude of the frustum is the perpendicular distance between the bases. The portion of an element included between the parallel bases of the frustum of a right circular cone is its slant height.

PROPOSITION XV. THEOREM

1050. The lateral area of the frustum of a cone of revolution is equal to one half the sum of the circumferences of its bases multiplied by its slant height.

Let denote the radius of the upper base of the frustum, R the radius of the lower base, and BC the slant height.

To Prove the lateral area of the frustum = (2TR + 2 πr)BC.

(AC — BC) 2 πR=(AB + BC) 2 πr.

· =

AC 2R-AB · 2 πr BC(2πR+2 πr).

But

AC 2TR is the lateral area of the cone A-CM, and AB 2 is the lateral area of the cone A-BN, and their difference is the lateral area of the frustum.