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

(New page: ==Solution== <math>88/5\ \mathrm{(C)}</math>) |
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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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+ | {{image}} | ||

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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== | ==Solution== | ||

<math>88/5\ \mathrm{(C)}</math> | <math>88/5\ \mathrm{(C)}</math> |

## Revision as of 00:24, 4 February 2008

## Problem

Consider the -sided polygon , as shown. Each of its sides has length , and each two consecutive sides form a right angle. Suppose that and meet at . What is the area of quadrilateral ?

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

## Solution