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.

If this is an isosceles right triangle, it means that the angles <EAF and <DCG have 45 degrees.

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.

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.

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!