A TCS problem

In the diagram below, let the circle have radius $r$ :

Consider the triangle formed by regions A, B and C. It has area $r^2$ . Note that $[B] + [C]$ is the desired yellow area (where $[\cdot]$ represents the area of the region). $[A] = r^2 - \frac{\pi r^2}{4}$ since it is a quarter of the square minus a quarter of the circle. Thus $[B]+[C] = ([A] + [B] + [C]) - [A] = r^2 -\left( r^2 - \frac{\pi r^2}{4}\right)=\frac{\pi r^2}{4}$ . Since we are told that $[B]+[C] = \pi$ , $r^2=4$ and $r = 2$ . Thus the area of the square is $(2r)(2r) = 16$