Volume of Frustum of Cone

Question

A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With $\pi = 22/7$, the volume, in cubic ft, of the remaining part of the cone is:

Accepted Answer

198. Total height $H = 12$, base radius $R = 4$. Height of the remaining frustum = 9 ft, so top smaller cone height $h = 12 - 9 = 3$ ft. By similar triangles, smaller radius $r = R \cdot \frac{h}{H} = 4 \cdot \frac{3}{12} = 1$. Volume of remaining frustum = Volume of full cone - Volume of top cone $= \frac{1}{3} \pi R^2 H - \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (4^2 	imes 12 - 1^2 	imes 3)$ $= \frac{1}{3} \cdot \frac{22}{7} (192 - 3) = \frac{1}{3} \cdot \frac{22}{7} \cdot 189 = 22 \cdot 9 = 198$.

Answer

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Volume of Frustum of Cone

Explanation

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