![\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}](/tpl/images/0171/3126/3fde9.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](/tpl/images/0171/3126/19a63.png)
Find the area and perimeter of the shaded part.
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Mathematics, 05.08.2019 00:10 nostalgiafaith
Find the area and perimeter of the shaded part.
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