Difference between revisions of "2017 AMC 10A Problems/Problem 11"
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<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math> | <math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24</math> | ||
| − | 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 <math>3</math> units of a point would be a sphere with radius <math>3</math>. However, we need to find the region containing all points within 3 units of a segment. Therefore, 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: | + | 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 <math>3</math> units of a point would be a sphere with radius <math>3</math>. However, we need to find the region containing all points within 3 units of a segment. Therefore, 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 <math>216 \pi</math>): |
| − | <math>\frac{4\pi}{3}3^3+9\ | + | <math>\frac{4 \pi }{3}3^3+9 \pi x=216</math> |
Where <math>x</math> is equal to the length of our line segment. | Where <math>x</math> is equal to the length of our line segment. | ||
We isolate <math>x</math>. This comes out to be <math>\boxed{\textbf{(D)}\ 20}</math> | We isolate <math>x</math>. This comes out to be <math>\boxed{\textbf{(D)}\ 20}</math> | ||
Revision as of 16:27, 8 February 2017
Problem
The region consisting of all point in three-dimensional space within 3 units of line segment 
has volume 216
. What is the length 
?
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 
units of a point would be a sphere with radius . However, we need to find the region containing all points within 3 units of a segment. Therefore, 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 
):
Where 
is equal to the length of our line segment.
We isolate . This comes out to be 
