Difference between revisions of "2007 AMC 10A Problems/Problem 18"

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== Problem==
 
Consider the <math>12</math>-sided polygon <math>ABCDEFGHIJKL</math>, as shown. Each of its sides has length <math>4</math>, and each two consecutive sides form a right angle. Suppose that <math>\overline{AG}</math> and <math>\overline{CH}</math> meet at <math>M</math>. What is the area of quadrilateral <math>ABCM</math>?
 
  
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<math>\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}</math>
 
 
==Solution==
 
<math>88/5\ \mathrm{(C)}</math>
 

Revision as of 00:25, 4 February 2008

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