Find another method of solving the shaded area. The diagram shows a square of side length $10\;\rm cm$. A quarter circle, of radius $10\;\rm cm$, is drawn from each vertex of the square. Find the exact area of the shaded region.


And This is my answer.



The answer is right, but I am searching for other ways.
Thanks.


Area of closed figure contains square $EFGH$ and 4 segments congruent to $EF$
Area of segment can be found as difference of sector and triangle:
$$S_{seg}=\frac12 AF^2 \frac{\pi}{6}-\frac12 AF^2 \sin \frac{\pi}{6}=\frac{\pi-3}{12}AF^2$$
Square side can be found in many ways, for example with cosine rule
$$AB^2=AG^2+BG^2-2AG\cdot BG\cos \frac{5\pi}{6}=(2+\sqrt{3})AG^2$$
$$FG^2=\frac{1}{2+\sqrt{3}}AB^2=(2-\sqrt{3})AF^2$$
$$S=4S_{seg}+FG^2=\frac{\pi-3}{3}AF^2+(2-\sqrt{3})AF^2=\left(1-\sqrt{3}+\frac{\pi}{3}\right)AF^2$$
