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.

Each side of smaller square WXZY is calculated as $$WX=\sqrt{(WQ)^2+(QX)^2}=\sqrt{(\sqrt{6})^2+(\sqrt{6})^2}=2\sqrt{3}$$
The area common to four circles each having a radius $a$ & center at the vertice of a cube with side $a$ is given by the general expression (obtained by using integration) $$\bbox[4pt, border:1px solid blue;]{A_{common}=a^2\left(\frac{\pi-3(\sqrt{3}-1)}{3}\right)}$$ Hence, by setting $a=2\sqrt{3}$ in the above formula, we get the requited common area ABCD $$=(2\sqrt{3})^2\left(\frac{\pi-3(\sqrt{3}-1)}{3}\right)=4\left(\pi-3(\sqrt{3}-1)\right)=4\pi-12(\sqrt{3}-1)$$
