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Question Number 82659 by naka3546 last updated on 23/Feb/20

Commented by naka3546 last updated on 23/Feb/20

$$areaofredregionis...$$

Commented by john santu last updated on 23/Feb/20

$$area=16(0.4-0.{5\tan }^{-1}\left(\frac{3}{4}\right))$$

Commented by naka3546 last updated on 23/Feb/20

$$showyourworkings,please$$

Answered by mr W last updated on 24/Feb/20

Commented by mr W last updated on 24/Feb/20

$$r=4$$$$\tan \alpha =\frac{4}{8}=\frac{1}{2}$$$$\beta =\pi -2\alpha =\pi -2{\tan }^{-1}\frac{1}{2}$$$$\sin \beta =\sin 2\alpha =2\times \frac{1\times 2}{5}=\frac{4}{5}$$$$A_{grayshaded}=\frac{r^2}{2}(\beta -\sin \beta )=\frac{r^2}{2}(\pi -2{\tan }^{-1}\frac{1}{2}-\frac{4}{5})$$$$A_{blueshaded}=r^2-\frac{\pi r^2}{4}=r^2(1-\frac{\pi }{4})$$$$$$$$A_{red}=A_{\mathrm{\Delta }ABC}-A_{grayshaded}-A_{blueshaded}$$$$=\frac{4\times 8}{2}-\frac{r^2}{2}(\pi -2{\tan }^{-1}\frac{1}{2}-\frac{4}{5})-r^2(1-\frac{\pi }{4})$$$$=16-8(\frac{\pi }{2}-2{\tan }^{-1}\frac{1}{2}+\frac{6}{5})$$$$=\frac{32}{5}-4\pi +16{\tan }^{-1}\frac{1}{2}$$$$=1.251991$$

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