Difference between revisions of "2017 AMC 10A Problems/Problem 11"
(→Solution) |
(→Solution) |
||
| Line 12: | Line 12: | ||
Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>. | Solving, we find that <math>x = \boxed{\textbf{(D)}\ 20}</math>. | ||
| + | |||
| + | ==Diagram for Solution== | ||
| + | http://i.imgur.com/cwNt293.png | ||
==See Also== | ==See Also== | ||
Revision as of 19:08, 3 January 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 
.
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. 
