2017 AMC 12B Problem 14

Problem 14 of 25IntermediateGeometry

An ice-cream novelty item consists of a cup in the shape of a

4

-inch-tall frustum of a right circular cone, with a

2

-inch-diameter base at the bottom and a

4

-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height

4

inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?

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Solution

Extending the frustum’s sides to a point, similar triangles show the frustum equals a cone of radius

2

and height

8

minus a cone of radius

1

and height

4:

\begin{aligned} &\tfrac13 \pi (2^2)(8) \\ &\quad {}- \tfrac13 \pi (1^2)(4) \\ &\quad {}= \tfrac{32}{3}\pi - \tfrac{4}{3}\pi \\ &\quad {}= \tfrac{28}{3}\pi. \end{aligned}

The top cone of radius

2

and height

4

adds

\tfrac13 \pi (2^2)(4) = \tfrac{16}{3}\pi.

The total is

\tfrac{28}{3}\pi + \tfrac{16}{3}\pi = \tfrac{44}{3}\pi.

Thus, the correct answer is E.