Difference between revisions of "2017 AMC 12B Problems/Problem 14"
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==Solution== | ==Solution== | ||
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+ | Solution 1: | ||
The top cone has radius 2 and height 4 so it has volume <math> \dfrac{1}{3} \pi (2)^2 \times 4 </math>. | 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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Solution by: SilverLion | Solution by: SilverLion | ||
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+ | Solution 2: | ||
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+ | Find the area of the cone with the method in Solution 1. The area 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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+ | 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}} |
Revision as of 20:41, 31 December 2017
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?
Solution
Solution 1:
The top cone has radius 2 and height 4 so it has volume .
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 .
Adding, we get .
Solution by: SilverLion
Solution 2:
Find the area of the cone with the method in Solution 1. The area of the frustrum is
Adding, we get
Solution by IdentityChaos2020
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
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