Difference between revisions of "2012 AMC 12B Problems/Problem 2"
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==Solution== | ==Solution== | ||
If the radius is <math>5</math>, then the width is <math>10</math>, hence the length is <math>20</math>. <math>10\times20=200</math>, <math>\boxed{\text{E}}</math> | If the radius is <math>5</math>, then the width is <math>10</math>, hence the length is <math>20</math>. <math>10\times20=200</math>, <math>\boxed{\text{E}}</math> | ||
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| + | == See Also == | ||
| + | |||
| + | {{AMC12 box|year=2012|ab=B|num-b=2|num-a=2}} | ||
Revision as of 21:47, 12 January 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?
Solution
If the radius is 
, then the width is 
, hence the length is 
. 
, 
See Also
| 2012 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 2 |
| 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 12 Problems and Solutions | |
