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.
 A: 
Consider circle center on D it intesect circl center on C at point G. The area under first circle and  axis can be fd as follows:
Equation of corcle :
$x^2+(y-10))^2=10^2$
Or:
$y=10-\sqrt{100-x^2}$
If $y=\sqrt{a^2-x^2}$ then:
$S=\int\sqrt{a^2-x^2}dx=\frac x2\sqrt{a^2-x^2}+\frac {a^2}2 \sin^{-1}\frac xa$
$$S=\int^{5}_0(10-\sqrt{100-x^2})dx$$
WE finally get:
$S_{AGI}=50\big(1-\frac{\sqrt 3}4-\frac{\pi}6\big)$
The  area between this circle lines AB and BC which include S is:
$$S_1=10^2-\frac 14 10^2\pi=100(1-\frac {\pi}4)$$
And the area of required region is:
$$S_2=10^2-[4\times 100(1-\frac{\pi}4)-8\times 50(1-\frac{\sqrt3}4-\frac{\pi}6)]=100\big(1-\sqrt 3+\frac {\pi}3\big)$$
A: 
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
$$AB^2=AG^2+BG^2-2AG\cdot BG\cos \frac{5\pi}{6}=(2+\sqrt{3})AG^2$$
$$FG^2=\frac{1}{2+\sqrt{3}}AB^2=(2-\sqrt{3})AF^2$$
$$S=4S_{seg}+FG^2=\frac{\pi-3}{3}AF^2+(2-\sqrt{3})AF^2=\left(1-\sqrt{3}+\frac{\pi}{3}\right)AF^2$$
