Q:

Find the area of the region $ABCD$.

In the Figure $\square PQRS$ is a square with side $2\sqrt6$.
By joining the midpoints another square $\square WXYZ$ is formed .
Circles are drawn with $4$ vertices as the center and
radius equal to the side of the square $\square WXYZ$.
Find the area common to all the $4$ circles .
$a.)6\pi\left(\dfrac{\sqrt3-1}{2}\right)\\
b.)4\pi-3\sqrt3\\
c.)\dfrac12\left(\pi -3\sqrt3\right)\\
d.)4\pi-12\left(\sqrt3 -1\right)\quad \LARGE \color{green}{\checkmark}\\$
So i have to find the area of region $ABCD$.

I have found that $WX=WY=YZ=XZ=2\sqrt3.$
Let Region $WBA$=Region $XAD$=Region $ZCD$=Region $YBC=a$
and
Region $WAX$=Region $XDZ$=Region $ZCY$=Region $YBW=b$
and 
Region $ABCD=m$
$Area(\square WXYZ)=12\\
m+4a+4b=12$.
Area of sectors.
$Area$(sector $WZY$)=$Area$(sector $XYZ$)=$Area$(sector $WXY$)=$Area$(sector $WZX$)=$m+3a+2b=3\pi$.
Now i m stucked.
I m looking for simple short way.
I have studied maths upto $12th$ grade.

A:

Let $[F]$ be the area of a figure $F$.
Since $\angle{WYA}=30^\circ,\angle{AYZ}=60^\circ$, one has
$$\begin{align}a+b&=[YWA]\\&=[\text{sector}\ YWA]-\left([\text{sector}\ ZYA]-[\triangle{ZYA}]\right)\\&=(2\sqrt 3)^2\pi\times\frac{30}{360}-\left((2\sqrt 3)^2\pi\times\frac{60}{360}-\frac{\sqrt 3}{4}\times(2\sqrt 3)^2\right)\\&=3\sqrt 3-\pi\end{align}$$
Hence, one has$$m=12-4(a+b)=12-4(3\sqrt 3-\pi)=4\pi+12-12\sqrt 3.$$

