2017 AMC 12B Problem 14

Question

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?

8\pi

\dfrac{28\pi}{3}

12\pi

14\pi

\dfrac{44\pi}{3}

The frustum is a full cone (radius

2,

height

8

) minus a small cone (radius

1,

height

4

)

Add the top solid cone of radius

2

and height

4;

use

V = \tfrac13 \pi r^2 h

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.

Problem 14 in other years