2012 AMC 10B Problems/Problem 17
Problem
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?
\text{(B)} \frac{1}{4} \qquad \text{(C)} \frac{\sqrt{10}}{10} \qquad \text{(D)} \frac{\sqrt{5}}{6} \qquad \text{(E)} \frac{\sqrt{5}}{5}$[[Category: Introductory Geometry Problems]]
==Solution== Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is$ (Error compiling LaTeX. ! Missing $ inserted.)24\pi\dfrac{120}{360} \times 24\pi = 8\pi41212\sqrt{12^2-4^2}=8\sqrt{2}\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi$.
Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is$ (Error compiling LaTeX. ! Missing $ inserted.)812\sqrt{12^2-8^2}=4\sqrt{5}\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi$.
The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find$ (Error compiling LaTeX. ! Missing $ inserted.)\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}\boxed{(C) \frac{\sqrt{10}}{10}}$.
- A side note
We can first simplify the volume ratio:$ (Error compiling LaTeX. ! Missing $ inserted.)\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.rhC.$
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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