Question 1168086
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*[illustration squareandtriangles.jpg].


BCDF is square, so BF = FD, but BF = BG + GF = 8 = FD.  Triangle GFD is right, and the legs measure 6 and 8, so it is a 3:4:5 right triangle.   Since GFD is right, angles FGD and GDF are complementary.  Since angle GDE is right, angles GDF and FDE are also complementary.  Since triangle FDE is right, angles FDE and FED are complementary.  So by transitive equality angles GDF and FED are congruent.  Therefore triangles GFD and FDE are similar by AAA. So FDE is also a 3:4:5 right triangle and its sides are 3x, 4x, and 5x where 3x = 8, and x = 8/3.  So FE must measure 4 times 8/3.   Now that you have the measures of the two legs of FDE, you should be able to calculate the desired area.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish]


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