Difference between revisions of "2017 AMC 10A Problems/Problem 11"

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(Solution)
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Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>.
 
Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>.
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==Diagram for Solution==
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http://i.imgur.com/cwNt293.png
  
 
==See Also==
 
==See Also==

Revision as of 20:08, 3 January 2019

Problem

The region consisting of all points in three-dimensional space within 3 units of line segment $\overline{AB}$ has volume 216$\pi$. What is the length $\textit{AB}$?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution

In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$):

$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$, where $x$ is equal to the length of our line segment.

Solving, we find that $x = \boxed{\textbf{(D)}\ 20}$.

Diagram for Solution

http://i.imgur.com/cwNt293.png

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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