find the area of the shaded region - 2

The diagonal of square $ABCD$ is $8\sqrt{2}$ . The diagonal of square $EBFG$ is $2\sqrt{2}$ . Let $d$ be the diameter of the smaller circle, then

$d=8\sqrt{2}-2(2)-2(2\sqrt{2})=8\sqrt{2}-4-4\sqrt{2}=4\sqrt{2}-4$

The area of the shaded part at the center including the small circle (consider diagram 2) is $4^2-\pi (2^2)=16-4\pi$ .

So the area of the shaded part at the center is $16-4\pi-\left[\dfrac{\pi}{4}(4\sqrt{2}-4)^2\right]=16-4\pi-\left[\dfrac{\pi}{4}(32-32\sqrt{2}+16)\right]=\boxed{16-16\pi+8\pi \sqrt{2}}$ .

The area of the shaded part excluding the shaded part at the center is $[4^2-\pi(2^2)](3)=\boxed{48-12\pi}$ .

So the desired area is $48-12\pi + 16 - 16\pi + 8\pi \sqrt{2}=\color{#D61F06}\boxed{64-28\pi + 8\pi \sqrt{2}}$