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Question Number 72930 by TawaTawa last updated on 05/Nov/19

Answered by ajfour last updated on 05/Nov/19

Commented by ajfour last updated on 05/Nov/19

$$R\sin \alpha =R-2r$$$$EF=\frac{r}{\tan (\frac{\pi }{4}-\frac{\alpha }{2})}+r=2R\sin \alpha$$$$\Rightarrow let\frac{r}{R}=x$$$$\Rightarrow \sin \alpha =1-2x$$$$let\frac{\alpha }{2}=\theta ,\tan \theta =m$$$$\Rightarrow \frac{2m}{1+m^2}=1-2x$$$$\Rightarrow m^2-\frac{2m}{(1-2x)}+1=0$$$$andx(\frac{1+m}{1-m}+1)=2(1-2x)$$$$\Rightarrow x=(1-2x)(1-m)$$$$orm=1-\frac{x}{1-2x}=\frac{1-3x}{1-2x}$$$$\Rightarrow {\left(\frac{1-3x}{1-2x}\right)}^2-\frac{2(1-3x)}{{(1-2x)}^2}+1=0$$$$\Rightarrow {(1-3x)}^2-2(1-3x)+{(1-2x)}^2=0$$$$13x^2-4x=0$$$$\Rightarrow x=\frac{4}{13}$$$$\frac{shadedarea}{areaofsemicircle}=\frac{2\pi r^2}{\pi R^2/2}$$$$=4x^2=\frac{64}{169}\cdot$$

Commented by TawaTawa last updated on 05/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}$$

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