In the diagram above, WZ = XZ, and circular arc

I can bet this is not a real GMAT question. Too much complicated, and I always say GMAT is complicated... but not that much. It is not possible to have a question that will take you more than 3 minutes for sure. And I did GMAT two years ago.

By the way, solution is A because we have to find the shadowed area (1) and divide by the total area (2).

How to find (2): the angle of the arc is 45º (there is a right-angled isosceles triangle), therefore the area of this part must be pi*(radius^2) divided by 8 (because we have 45º and not 360º). In this case, the area is \(\frac{1}{8}*pi*r^2\) --> \(\frac{1}{8}*pi*(L*\sqrt{2})^2\) --> \(\frac{1}{8}*pi*(L^2*2)\).

How to find (1): there is a right-angled isosceles triangle. Therefore the side of the radius must be \(L*\sqrt{2}\), and the equal sides have length L. The right angle is obviously 90º and the other two angles are 45º each. Therefore the area of this triangle is \(\frac{1}{2}*L*L\) --> \(\frac{1}{2}*L^2\). Therefore the shadowed area (1) is \(\frac{1}{8}*pi*(L^2*2)\) minus \(\frac{1}{2}*L^2\).

Finally, we divide (1) by (2) and the solution is A.