find the area and perimeter of the shaded part.

Accepted Solution

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]