Difference between revisions of "2012 AMC 10B Problems/Problem 2"
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| + | == 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> | ||
| + | |||
| + | <math> \textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200 </math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
| + | |||
| + | ==Solution== | ||
| + | |||
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 <math>\boxed{E}</math> | + | Therefore the area is <math>20\cdot 10 = 200</math> or <math>\boxed{\textbf{(E)}\ 200}</math> |
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2012|ab=B|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:46, 13 August 2014
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 
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
