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Question Number 56051 by naka3546 last updated on 09/Mar/19

Answered by mr W last updated on 09/Mar/19

Commented by mr W last updated on 10/Mar/19

$$\cos \frac{\beta }{2}=\frac{a^2+{\left(\frac{a}{\sqrt{2}}\right)}^2-{\left(\frac{a}{2}\right)}^2}{2\times a\times \frac{a}{\sqrt{2}}}=\frac{5\sqrt{2}}{8}$$$$\cos \beta =2\times {\left(\frac{5\sqrt{2}}{8}\right)}^2-1=\frac{9}{16}$$$$\Rightarrow \beta ={\cos }^{-1}\frac{9}{16}$$$$\Rightarrow \sin \beta =\frac{5\sqrt{7}}{16}$$$$l=a\sin \frac{\beta }{2}=a\frac{\sqrt{14}}{8}=\frac{a}{2}\sin \frac{\gamma }{2}$$$$\sin \frac{\gamma }{2}=\frac{\sqrt{14}}{4}$$$$\Rightarrow \gamma =2{\sin }^{-1}\frac{\sqrt{14}}{4}$$$$\cos \gamma =1-2{\left(\frac{\sqrt{14}}{4}\right)}^2=1-\frac{14}{8}=-\frac{3}{4}$$$$\sin \gamma =\frac{\sqrt{7}}{4}$$$$A_{red}={\left(\frac{a}{2}\right)}^2(\gamma -\sin \gamma )-a^2(\beta -\sin \beta )$$$$\frac{A_{red}}{A_{square}}=\frac{1}{4}(\gamma -\sin \gamma )-\beta +\sin \beta$$$$\Rightarrow \frac{A_{red}}{A_{square}}=\frac{\sqrt{7}}{4}+\frac{1}{2}{\sin }^{-1}\frac{\sqrt{14}}{4}-{\cos }^{-1}\frac{9}{16}\approx 0.29276$$

Answered by MJS last updated on 09/Mar/19

$$\mathrm{an}\mathrm{easy}\mathrm{one}:$$$$\mathrm{turn}\mathrm{by}45°$$$$\mathrm{side}\mathrm{of}\mathrm{square}=1$$$$\mathrm{big}\mathrm{circle}:x^2+y^2=1$$$$\mathrm{small}\mathrm{circle}:x^2+{(y-\frac{\sqrt{2}}{2})}^2=\frac{1}{4}$$$$\mathrm{intersection}\mathrm{at}x=\pm \frac{\sqrt{14}}{8}$$$$\mathrm{red}\mathrm{area}:4\underset{0}{\stackrel{\frac{\sqrt{14}}{8}}{\int }}(\frac{\sqrt{2}}{2}+\frac{\sqrt{1-4x^2}}{2}-\sqrt{1-x^2})dx=$$$$=\frac{\sqrt{7}}{4}+\frac{1}{2}\arcsin \frac{\sqrt{14}}{4}-2\arcsin \frac{\sqrt{14}}{8}\approx .292763$$

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