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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>\mathbf{(A)}</math> <math>\dfrac{1}{8}</math> <math>\qquad</math> <math>\mathbf{(B)}</math> <math>\dfrac{1}{4}</math> <math>\qquad</math> <math>\mathbf{(C)}</math> <math>\dfrac{\sqrt{10}}{10}</math> <math>\qquad</math> <math>\mathbf{(D)}</math> <math>\dfrac{\sqrt{5}}{6}</math> <math>\qquad</math> <math>\mathbf{(E)}</math> <math>\dfrac{\sqrt{5}}{5}</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</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>.
+
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</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>.
+
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}</math> after simplifying, or <math>\boxed{(C)  \frac{\sqrt{10}}{10}}</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}<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}.</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.</math>
+
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 00:51, 17 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?

$\text{(A)} \frac{1}{8} \qquad$ \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. Unknown error_msg)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. Unknown error_msg)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. Unknown error_msg)\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. Unknown error_msg)\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 (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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