Difference between revisions of "2006 AMC 12A Problems/Problem 13"

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The vertices of a <math>3-4-5</math> right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?
 
The vertices of a <math>3-4-5</math> right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?
  
<math> \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}</math><math>\mathrm{(E) \ }  14\pi</math>
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<math> \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ }  14\pi</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 19:06, 4 November 2006

Problem

2006 AMC 12A Problem 13.gif

The vertices of a $3-4-5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?

$\mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ }  14\pi$

Solution

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$\rm{(E)}\,14\pi$

See also

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