Difference between revisions of "2017 AMC 10A Problems/Problem 11"
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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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+ | ==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>4(3^3)/3 \pi = 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 area of the cylinder. Thus the height is <math>180 \pi</math> over the base, or <math>180 \pi/9\pi=20</math>, <math>D</math> | ||
==Diagram for Solution== | ==Diagram for Solution== |
Revision as of 12:39, 12 February 2019
Problem
The region consisting of all points in three-dimensional space within 3 units of line segment has volume 216. What is the length ?
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 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. 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 ):
, 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 area of the cylinder. Thus the height is over the base, or ,
Diagram for Solution
http://i.imgur.com/cwNt293.png
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.