Difference between revisions of "2017 AMC 12A Problems/Problem 8"
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<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 14 </math> | <math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 14 </math> | ||
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| + | ==Solution== | ||
| + | |||
| + | Let the length <math>AB</math> be <math>L</math>. Then, we see that the region is just the union of the cylinder with central axis <math>\overline{AB}</math> and radius <math>3</math> and the two hemispheres connected to each face of the cylinder (also with radius <math>3</math>). Thus the volume is | ||
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| + | <math>9\pi L + \frac{4}{3}\pi(3)^3 = 9\pi L + 36\pi ( = 216\pi)</math> | ||
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| + | <math>9\pi L = 180\pi</math> | ||
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| + | <math>L = \boxed{(D)=\ 20}</math> | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2017|ab=A|num-b=7|num-a=9}} | ||
Revision as of 16:57, 8 February 2017
Problem
The region consisting of all points in three-dimensional space within 
units of line segment 
has volume 
. What is the length 
?
Solution
Let the length be 
. Then, we see that the region is just the union of the cylinder with central axis and radius and the two hemispheres connected to each face of the cylinder (also with radius ). Thus the volume is
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 | |
