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Difference between revisions of "2017 AMC 12B Problems/Problem 14"

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<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math>
 
<math>\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}</math>
  
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==Solution==
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==Solution 1==
  
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We add up the volumes separately.
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The top cone has radius 2 and height 4 so it has volume <math> \dfrac{1}{3} \pi (2)^2 \times 4 </math>.  
  
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Top cone = <math> \dfrac{1}{3} \pi (2)^2 \times 4 </math>.  
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The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so it has volume
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<math> \dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4</math>.
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Adding, we get <math>\dfrac{1}{3} \pi (16+32-4) = \boxed{ \textbf{E} \ \dfrac{44\pi}{3}}</math>.  
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Solution by: SilverLion
  
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Frustum  = large inverted cone - small inverted cone =  
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==Solution 2==
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<math> \dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4</math>.
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Find the volume of the cone with the method in Solution 1. The volume of the frustrum is <math>\pi\int_{0}^{4} \left(-\frac{1}{4}x+2\right)^2 dx=\frac{28\pi}{3}</math>
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Adding, we get <math>\dfrac{44\pi}{3}</math>.
  
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Adding we get <math>\dfrac{1}{3} \pi (16+32-4) = \dfrac{44}{3} \pi </math>.
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Solution by IdentityChaos2020
  
 
{{AMC12 box|year=2017|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2017|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 23:31, 24 January 2023

Problem

An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?

$\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}$

Solution 1

The top cone has radius 2 and height 4 so it has volume $\dfrac{1}{3} \pi (2)^2 \times 4$.

The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so it has volume $\dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4$.

Adding, we get $\dfrac{1}{3} \pi (16+32-4) = \boxed{ \textbf{E} \ \dfrac{44\pi}{3}}$.

Solution by: SilverLion

Solution 2

Find the volume of the cone with the method in Solution 1. The volume of the frustrum is $\pi\int_{0}^{4} \left(-\frac{1}{4}x+2\right)^2 dx=\frac{28\pi}{3}$

Adding, we get $\dfrac{44\pi}{3}$.

Solution by IdentityChaos2020

2017 AMC 12B (Problems • Answer Key • Resources)
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Problem 13 		Followed by
Problem 15
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All AMC 12 Problems and Solutions

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