# 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> \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

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

## Solution

*This problem needs a solution. If you have a solution for it, please help us out by adding it.*