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?

$8\pi$

$\dfrac{28\pi}{3}$

$12\pi$

$14\pi$

$\dfrac{44\pi}{3}$

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 $4:$ $\tfrac13 \pi (2^2)(8) - \tfrac13 \pi (1^2)(4) = \tfrac{32}{3}\pi - \tfrac{4}{3}\pi = \tfrac{28}{3}\pi.$ 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.

2017 AMC 12B Problem 14

Solution:

Problem 14 in Other Years