Difference between revisions of "2017 AMC 10A Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | The region consisting of all points in three-dimensional space within 3 units of line segment <math>\overline{AB}</math> has volume | + | The region consisting of all points in three-dimensional space within <math>3</math> units of line segment <math>\overline{AB}</math> has volume <math>216\pi</math>. What is the length <math>\textit{AB}</math>? |
<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> | ||
− | ==Solution== | + | ==Solution 1== |
− | In order to solve this problem, we must first visualize what the region | + | In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, 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 <math>3</math> units of a segment. It can be seen that our region is a cylinder with two hemispheres/endcaps 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} \cdot 3^3+9 \pi x=216 \pi</math>, where <math>x</math> is equal to the length of our line segment. | <math>\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi</math>, where <math>x</math> is equal to the length of our line segment. | ||
Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>. | Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Because this is just a cylinder and <math>2</math> "half spheres", and the radius is <math>3</math>, the volume of the <math>2</math> half spheres is <math>\frac{4(3^3)\pi}{3} = 36 \pi</math>. Since we also know that the volume of this whole thing is <math>216 \pi</math>, we do <math>216-36</math> to get <math>180 \pi</math> as the volume of the cylinder. Thus the height is <math>180 \pi</math> divided by the area of the base, or <math>\frac{180 \pi}{9\pi}=20</math>, so our answer is <math>\boxed{\textbf{(D)}\ 20}.</math> | ||
+ | |||
+ | ~Minor edit by virjoy2001 | ||
==Diagram for Solution== | ==Diagram for Solution== | ||
http://i.imgur.com/cwNt293.png | http://i.imgur.com/cwNt293.png | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/s4vnGlwwHHw | ||
==See Also== | ==See Also== |
Latest revision as of 23:33, 31 July 2021
Problem
The region consisting of all points in three-dimensional space within units of line segment has volume . What is the length ?
Solution 1
In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, 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 units of a segment. It can be seen that our region is a cylinder with two hemispheres/endcaps 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.
Solving, we find that .
Solution 2
Because this is just a cylinder and "half spheres", and the radius is , the volume of the half spheres is . Since we also know that the volume of this whole thing is , we do to get as the volume of the cylinder. Thus the height is divided by the area of the base, or , so our answer is
~Minor edit by virjoy2001
Diagram for Solution
http://i.imgur.com/cwNt293.png
Video Solution
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
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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