2007 AMC 10A Problems/Problem 18
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 ?
We can obtain the solution by calculating the area of rectangle minus the combined area of triangles and .
We know that triangles and are similar because . Also, since , the ratio of the distance from to to the distance from to is also . Solving with the fact that the distance from to is 4, we see that the distance from to is .
The area of is simply , the area of is , and the area of rectangle is .
Taking the area of rectangle and subtracting the combined area of and yields .
Extend and and call their intersection .
The triangles and are clearly similar with ratio , hence and thus . The area of the triangle is .
The triangles and are similar as well, and we now know that the ratio of their dimensions is .
Draw altitudes from onto and , let their feet be and . We get that . Hence . (An alternate way is by seeing that the set-up AHGCM is similar the 2 pole problem(http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, must be , by the harmonic mean. Thus, must be .)
Then the area of is , and the area of can be obtained by subtracting the area of , which is . Hence the answer is .
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