Question

Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{\sqrt{10}}{10}$

$\dfrac{\sqrt{5}}{6}$

$\dfrac{\sqrt{10}}{5}$

Solution:

Each sector forms a cone with slant height $12$ . The smaller sector has angle $120^\circ$ , so its arc length is $\dfrac13\cdot2\pi\cdot12=8\pi$ , giving base radius $4$ . Its cone height is $\sqrt{12^2-4^2}=8\sqrt2$ .

The larger sector has arc length $16\pi$ , giving base radius $8$ . Its cone height is $\sqrt{12^2-8^2}=4\sqrt5$ .

The volume ratio is $\dfrac{\frac13\pi\cdot4^2\cdot8\sqrt2}{\frac13\pi\cdot8^2\cdot4\sqrt5}=\dfrac{\sqrt{10}}{10}.$

Thus, C is the correct answer.

2012 AMC 10B Problem 17

Solution:

Problem 17 in Other Years