Q:

find the area and perimeter of the shaded part.

Accepted Solution

A:
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.[tex]\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}[/tex][tex]\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[/tex]now, about the perimeter, well[tex]\bf \textit{arc's length}\\\\ s=\cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ \cline{1-1} r=8\\ \theta =90 \end{cases}\implies s=\cfrac{(90)\pi (8)}{180}\implies \stackrel{\textit{for one circle}}{s=4\pi } \\\\\\ \stackrel{\textit{for both circles}}{s = 2(4\pi )}\implies s=8\pi \implies \blacktriangleright s\approx 25.13 \blacktriangleleft[/tex]