Q31

My answer would be sqrt(2) / 4.

First, I draw a diagonal line connecting A and D. Since BE = DE, the triangle BED is an isosceles triangle whose height is 1 (the given length of CE) + half of the square ABCD’s diagonal.

The area of that triangle BED is ½ * sqrt(2) * (1 + sqrt(2)/2) which is (sqrt(2) + 1) / 2.

Now, the area of the two triangles BCE and DCE together is (sqrt(2) + 1) / 2 – ½ = sqrt(2) / 2. The area of the triangle BCE alone is half that which should be sqrt(2)/4.

Apologies for the stupid question, but is the figure supposed to be 3d or 2d, flat on the paper? I can't tell from the problem definition.

EDIT: If it is 2d, here is what I see: A TON of isosceles triangles. Draw a line from B to D and you have

Triangle BCE is isoceles with 2 lenghts 1
Triangle DCE is isoceles with 2 lengths 1
Triangle BCD is isoceles with 2 lengths 1
Triangle BED is isoceles

Redraw the figure (as it is distorted and out of scale) and I believe that you can solve it. Lots of symmetry. I'm not doing all that geometry right now. Its lunch time!

[post lunch]
I get the same result as the original responder.

thank you - second thought I agree with your answer. the answer choices must be wrong.