Question reads: Square DEPG is inscribed in isosceles right triangle ABC. If the area of triangular region ABC is r, what is the area of triangular region AFE?
It’s a tough problem.
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If this is an isosceles right triangle, it means that the angles <EAF and <DCG have 45 degrees.
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The other angles inside all triangles must also be 45 degrees, because one of the angles inside each triangle has 90 degrees. So we can deduce that AE = EF = GD = ED = DC = FG.
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If we take 3 as a side of AB and BC, then AC must equal 3√2. Given that the AC consists of 3 lengths of equal measure, we can divide 3√2 into 3. Which means that we can say that AE, EF etc. all equal √2.
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Having that, we can now calculate the area of the smaller triangle √2 x √2 / 2 = 1. The bigger triangle is going to have an area of 3x3 / 2 = 4.5
Now, the area of the small triangle to the big triangle is 1 divided by 4.5 = 2/9
The correct answer is B.
Hope that helps!