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

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==Problem==
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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>
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==Solution==
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==See also==

Revision as of 09:53, 5 February 2008

Problem

Consider the $12$-sided polygon $ABCDEFGHIJKL$, as shown. Each of its sides has length $4$, and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M$. What is the area of quadrilateral $ABCM$?


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


$\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}$

Solution

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See also

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