Difference between revisions of "2012 AMC 10B Problems/Problem 17"
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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? | 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? | ||
− | <math>\ | + | <math>\text{(A)} \frac{1}{8} \qquad</math> \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}<math> |
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
==Solution== | ==Solution== | ||
− | Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is <math>24\pi< | + | Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is </math>24\pi<math>, so the circumference of the smaller cone would be </math>\dfrac{120}{360} \times 24\pi = 8\pi<math>. This means that the radius of the smaller cone is </math>4<math>. Since the radius of the paper disk is </math>12<math>, the slant height of the smaller cone would be </math>12<math>. By the Pythagorean Theorem, the height of the cone is </math>\sqrt{12^2-4^2}=8\sqrt{2}<math>. Thus, the volume of the smaller cone is </math>\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi<math>. |
− | Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is <math>8< | + | Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is </math>8<math> and the slant height is </math>12<math>. By the Pythagorean Theorem again, the height is </math>\sqrt{12^2-8^2}=4\sqrt{5}<math>. Thus, the volume of the larger cone is </math>\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi<math>. |
− | The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find <math>\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}< | + | The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find </math>\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}<math> after simplifying, or </math>\boxed{(C) \frac{\sqrt{10}}{10}}<math> . |
*A side note | *A side note | ||
− | We can first simplify the volume ratio: <math>\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.< | + | We can first simplify the volume ratio: </math>\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.<math> Now we can find the GENERAL formulas for </math>r<math> and </math>h<math> based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us </math>C.$ |
==See Also== | ==See Also== |
Revision as of 23:51, 16 January 2021
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 | |
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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