Difference between revisions of "2012 AMC 10B Problems/Problem 2"
Math Kirby (talk | contribs) (→Problem) |
RainbowsHurt (talk | contribs) m (→Solution) |
||
| Line 13: | Line 13: | ||
Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is <math>5</math>, the diameter is <math>2\cdot 5 = 10</math>. | Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is <math>5</math>, the diameter is <math>2\cdot 5 = 10</math>. | ||
Since the sides of the rectangle are in a <math>2:1</math> ratio, the longer side has length <math>2\cdot 10 = 20</math>. | Since the sides of the rectangle are in a <math>2:1</math> ratio, the longer side has length <math>2\cdot 10 = 20</math>. | ||
| − | Therefore the area is <math>20\cdot 10 = 200</math> or | + | Therefore the area is <math>20\cdot 10 = 200</math> or \boxed{\textbf{(E)}\ 200}$ |
Revision as of 21:28, 17 February 2013
Problem
A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?
![[asy] draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); draw(circle((10,5),5));[/asy]](http://latex.artofproblemsolving.com/1/c/1/1c19b6dc0eb24a92fe2b043bf705f662d97486a2.png)
Solution
Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is 
, the diameter is 
.
Since the sides of the rectangle are in a 
ratio, the longer side has length 
.
Therefore the area is 
or \boxed{\textbf{(E)}\ 200}$
