2012 AMC 10B Problem 17

Attempt Problem 17 of the 2012 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

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?

18 \dfrac{1}{8}

14 \dfrac{1}{4}

1010 \dfrac{\sqrt{10}}{10}

56 \dfrac{\sqrt{5}}{6}

105 \dfrac{\sqrt{10}}{5}

Answer: C
Concepts:conesectorvolume
Difficulty rating: 1930
Solution:

Each sector forms a cone with slant height 1212. The smaller sector has angle 120120^\circ, so its arc length is 132π12=8π\dfrac13\cdot2\pi\cdot12=8\pi, giving base radius 44. Its cone height is 12242=82\sqrt{12^2-4^2}=8\sqrt2.

The larger sector has arc length 16π16\pi, giving base radius 88. Its cone height is 12282=45\sqrt{12^2-8^2}=4\sqrt5.

The volume ratio is 13π428213π8245=1010.\dfrac{\frac13\pi\cdot4^2\cdot8\sqrt2}{\frac13\pi\cdot8^2\cdot4\sqrt5}=\dfrac{\sqrt{10}}{10}.

Thus, C is the correct answer.

← Problem 16#16
Full Exam

Problem 17 in Other Years