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?

$\mathbf{(A)}$ $\dfrac{1}{8}$ $\space$ $\mathbf{(B)}$ $\dfrac{1}{4}$ $\space$ $\mathbf{(C)}$ $\dfrac{\sqrt{10}}{10}$ $\space$ $\mathbf{(D)}$ $\dfrac{\sqrt{5}}{6}$ $\space$ $\mathbf{(E)}$ $\dfrac{\sqrt{5}}{5}$

Solution

Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is $24\pi$, 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 $8$ 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 $\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}$ after simplifying, or $\boxed{(C)  \frac{\sqrt{10}}{10}}$ .

  • A side note

We can first simplify the volume ratio: $\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.$ 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
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All AMC 10 Problems and Solutions

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