Difference between revisions of "2012 AMC 12B Problems/Problem 2"
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− | == Problem== | + | == 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? | 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> | + | <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> | <math>\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200</math> | ||
− | ==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= \boxed{\textbf{(E)}\ 200}.</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= \boxed{\textbf{(E)}\ 200}.</math> | ||
Latest revision as of 01:07, 19 October 2020
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 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 12 Problems and Solutions |
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