2012 AMC 12B Problem 15

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}

Answer: C

Concepts:cone volume Pythagorean Theorem

Difficulty rating: 1800

Small hint:

The arc length of each sector becomes the circumference of that cone’s base, giving its base radius

Big hint:

Each cone has slant height

12,

so find each height with the Pythagorean theorem before comparing

\tfrac13\pi r^2 h

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.