Check the picture below.
so, let's notice, the square of 8x8 is inscribing the overlap of both circles, each circle has a radius of 8.
so we can simply find the area of the square, simple enough, 8x8 = 64, and then subtract the segment of each circle from that, and what's leftover is the shaded area.
![\bf \textit{area of a segment of a circle}\\\\ A=\cfrac{r^2}{2}\left[ \cfrac{\pi \theta }{180}-sin(\theta ) \right]~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ \cline{1-1} r=8\\ \theta =90 \end{cases} \\\\\\ A=\cfrac{8^2}{2}\left[ \cfrac{\pi (90)}{180}-sin(90^o) \right]\implies A=32\left[\cfrac{\pi }{2}-1 \right]\implies \stackrel{\textit{for one segment}}{A=16\pi -32}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6fh4yfjknn41cki1y61phri1ns4srynzgz.png)
![\bf \stackrel{\textit{for both segments}}{A=2(16\pi -32)}\implies \boxed{A=32\pi -64} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large Areas}~~~~~~~~~~}{\stackrel{\textit{square}}{64}~~-~~\stackrel{\textit{circle's segments}}{(32\pi -64)}}\implies 64-32\pi +64 \\\\\\ 128-32\pi ~~\approx~~\blacktriangleright \stackrel{\textit{shaded area}}{27.47} \blacktriangleleft](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1bxqwkrp48pu3ofbwajousfxbghhzhuza9.png)
now, about the perimeter, well
