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's arc length is $\tfrac{120}{360}\cdot2\pi\cdot12=8\pi,$ so its base radius is $4$ and its height is $\sqrt{12^2-4^2}=8\sqrt2.$

The larger sector (central angle $240^\circ$ ) has arc length $16\pi,$ base radius $8,$ and height $\sqrt{12^2-8^2}=4\sqrt5.$

The ratio of volumes is $\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}.$

Thus, the correct answer is C.

2012 AMC 12B Problem 15

Solution:

Problem 15 in Other Years