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PROPOSITION XXXII. THEOREM

664. The volume of a circular cone is equal to onethird the product of its base by its altitude.

Hyp. V is the volume, B the base, and H the altitude of the cone; V' the volume, B' the regular polygon forming the base of an inscribed pyramid.

V= {B x H.

To prove

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665. Cor. If the cone is a cone of revolution, with R as radius of B, then

V= } R H.

Ex. 1131. Find the lateral area of a cone of revolution whose radius is 4 and whose altitude is 3.

Ex. 1132. Find the volume of a cone of revolution with radius 6 and altitude 2.

Ex. 1133. Find the volume of a cone of revolution whose radius is 5 and whose slant height is 13.

Ex. 1134. Find the lateral surface of a cone of revolution if its volume is 314 and its altitude is 2.

PROPOSITION XXXIII. THEOREM 666. The lateral areas, or the total areas, of two similar cones of revolution are to each other as the squares of their altitudes, as the squares of their radii, or as the squares of their slant heights; and their volumes are to each other as the cubes of their altitudes, as the cubes of their radii, or as the cubes of their slant heights.

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Нур. . S, S' are the lateral areas, T, T' the total areas, V, V' the volumes, H, H' the altitudes, R, R' the radii, L, L,' the slant heights of two similar cones of revolution.

S:S=T:T::H: H=R2: R=L: L2

To prove

and

V:V' = H3: H13

R3: R'3 = L : L'3

The proof is similar to that of Prop. XXVIII.

Ex. 1135. Find the ratio of the total areas of two similar cones of revolution whose altitudes are

(a) 12 inches and 24 inches respectively.

(6) 5 inches and d inches respectively. Ex. 1136. The volume of a cone of revolution is 343 cubic inches and its altitude is 5 inches. Find volume of a similar cone of altitude

(a) 7 inches.
(6) 15 inches.

PROPOSITION XXXIV. THEOREM

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

Hyp. S is the lateral area, C and C" the circumferences of the bases, R and R' their radii, and L the slant height of the frustum ; S' the lateral area, P and Pl the perimeters of the bases, of a frustum of a regular pyramid circumscribed about the cone.

S= }(C + C") * L.

To prove

Proof. The slant height of the frustum of the circumscribed pyramid = L,

(Why ?) then S' = 1(P + P') x L.

(Why ?) [To be completed by the student. Hint. — Use Theorem of Limits.]

Note. - It can be shown that the limit of the sum of a finite number of variables is equal to the sum of their respective limits.

668. COR. The lateral area of a frustuin of a cone of revolution is equal to the circumference of a section equidistant from its bases multiplied by its slant height. For

(R+R C=2 #R, and C" 2 #R', and 27

2

)

is the circumference of a section equidistant from the bases.

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669. The volume of a frustum of a circular cone is equivalent to the sum of its bases, and the mean proportional between its bases, multiplied by one-third its altitude.

To prove

Hyp. V is the volume, B the lower base, b the upper base, H the altitude of the frustum, V' the volume, B' the lower base, b' the upper base, of an inscribed frustum of a pyramid with regular polygons as bases.

V=JH(B + 6 + Bxb).
Proof. The altitude of the frustum of the pyramid = H.
Then

V'= {H(B'+b' + VB' xb'). (601) By increasing the number of the lateral faces of the inscribed frustum indefinitely,

V' becomes a variable with V as a limit,

B' approaches B as a limit, b' approaches b as a limit, and B' x b', B x b as a limit.

.. B' +! + B' x b' approaches

B +b + VB x b as a limit.
:: V= {H(B+b+VB x b).

Q.E.D.

Note. — See note, p. 49. It can also be shown that the limit of the product of two or more variables is the product of their respective limits when no one of these limits is zero.

670. Cor. If the frustum is that of a cone of revolution, with R and R' as radii of its bases,

V=H(R' + R" + RR'). Note. — The preceding proposition may be stated :

The volume of a frustum of a circular cone is equivalent to the sum of the volumes of three cones whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and the mean proportional between the bases of the frustum. For, the value of V may be written,

{ H x B + | H xb + {H X VB xb.

MISCELLANEOUS EXERCISES Ex. Find the lateral edge, lateral area, and volume of a regular triangular pyramid each side of whose base is 9 and whose altitude is 3.

D

B

AK=(

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2

Let K be the center of the base ABC of the regular triangular pyramid D-ABC. Draw the median BE and the altitude DK. Join A and K, and D and E.

2
alt. of equil. A of side 9); = 3V3.
DA=VAR? + DK= V (373)2 + 9 = 6.

DE = V DA – Ē A’ = V36 – $1 = {V7.
Lat. Area = 3 x 9 x

DE_8177

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