KASSALMD wrote:

Triangle BCE is congruent to triangle CDE. Also triangle BED is an isosceles triangle, since BE=DE. Now draw a diagonal BD to form triangle BCD. You realize angle CBE IS 45 and so is angle CEB. Therefore, triangle BEC is a rt-angled triangle with base 1 and ht. 1. Hence its area is 1/2 *1*1= 1/2.

Angle CBE is definitely not 45 degrees. We know:

-if you add the three angles around C, you must get 360;

-one of these angles is 90, so the two large angles in triangles BCE and CED must add to 270;

-triangles BCE and CED are congruent (identical sides), so angle BCE and angle DCE must both be 135.

So in BCE, the other two angles must add to 180-135 = 45, and since BCE is isosceles, angle CBE must be 22.5 degrees (not 45 degrees).

The diagram is pretty misleading, since it disguises the symmetry in the picture- in an accurate diagram, triangles BCE and DCE should appear to be the same size, and the square should look like a square. Even with the diagram above, though, it should be clear that BCE is not a right triangle. Regardless, I posted four different solutions to this problem on another forum:

http://www.beatthegmat.com/area-of-tria ... 15073.htmlNone of the answer choices is correct, incidentally- B would be correct if the exponent on 2 in the numerator were 1/2 instead of -2.

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