2012 AMC 10B Problems/Problem 17

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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?

$\text{(A)} \frac{1}{8} \qquad$ \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$, so the circumference of the smaller cone would be$\dfrac{120}{360} \times 24\pi = 8\pi$. This means that the radius of the smaller cone is$4$. Since the radius of the paper disk is$12$, the slant height of the smaller cone would be$12$. By the Pythagorean Theorem, the height of the cone is$\sqrt{12^2-4^2}=8\sqrt{2}$. Thus, the volume of the smaller cone is$\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi$.

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.)8$and the slant height is$12$. By the Pythagorean Theorem again, the height is$\sqrt{12^2-8^2}=4\sqrt{5}$. Thus, the volume of the larger cone is$\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi$.

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}$after simplifying, or$\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}.$Now we can find the GENERAL formulas for$r$and$h$based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us$C.$

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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