M06-32

Official Solution:


The figure below shows a square inscribed in a circle with a radius of \(\sqrt{6}\). What is the area of the shaded region?



A. \(\pi\)
B. \(6\pi - \sqrt{12}\)
C. \(6\pi - 2\sqrt{6}\)
D. \(3(\frac{\pi}{2} - 1)\)
E. \(3 \frac{\pi}{4} - \frac{3}{2}\)


The area of the shaded region is 1/4th of the area of the circle minus the area of the square.

The area of the circle is \(\pi{r^2}=6\pi\);

Since the diagonal of the square equals to the diameter of the circle then \(\text{diagonal}=2\sqrt{6}\). The area of a square equals \(\frac{\text{diagonal}^2}{2}=12\);

The area of the shaded region equals \(\frac{1}{4}(6\pi - 12)=3(\frac{\pi}{2} - 1)\).


Answer: D

How do we get the area of the square to equal : \(\frac{\text{diagonal}^2}{2}=12\)

I would like to know if we can use the 45-45-90 triangle properties here

what are steps to reduce 1/4 (6 pi - 12) = 3(pi/2 - 1) ? I just can't see it.