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Question Number 85756 by TawaTawa1 last updated on 24/Mar/20

Commented by TawaTawa1 last updated on 24/Mar/20

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{red}.$$

Commented by Tony Lin last updated on 24/Mar/20

Commented by Tony Lin last updated on 24/Mar/20

$$red=orange-purple-yellow$$$$orange=\frac{1}{2}\times 8\times 4=16$$$$purple=4\times 4-4^2\pi \times \frac{1}{4}=16-4\pi$$$$yellow=\frac{1}{2}\times 4^2(\theta -sin\theta )$$$$tan\left(\frac{\pi -\theta }{2}\right)=\frac{1}{2}\Rightarrow \theta =\pi -2{tan}^{-1}\frac{1}{2}$$$$sin\left(\frac{\pi -\theta }{2}\right)=\frac{1}{\sqrt{5}}$$$$\Rightarrow \sqrt{\frac{1-cos(\pi -\theta )}{2}}=\frac{1}{\sqrt{5}}$$$$\Rightarrow cos\theta =-\frac{3}{5},sin\theta =\frac{4}{5}$$$$yellow$$$$=\frac{1}{2}\times 4^2(\pi -2{tan}^{-1}\frac{1}{2}-\frac{4}{5})$$$$=8\pi -16{tan}^{-1}\frac{1}{2}-\frac{32}{5}$$$$red=16-(16-4\pi )-(8\pi -16{tan}^{-1}\frac{1}{2}-\frac{32}{5})$$$$=16{tan}^{-1}\frac{1}{2}+\frac{32}{5}-4\pi$$

Commented by TawaTawa1 last updated on 24/Mar/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by TawaTawa1 last updated on 26/Mar/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{it}.$$

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