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Question Number 177910 by HeferH last updated on 11/Oct/22

Answered by mr W last updated on 11/Oct/22

Commented by mr W last updated on 11/Oct/22

$$saysidelengthofsquareiss.$$$$\tan \alpha =\frac{CD}{GD}=2$$$$FD=2s\cos \alpha =\frac{2s}{\sqrt{5}}$$$$ED=\frac{\sqrt{2}s}{2}$$$$\mathrm{\angle }FDE=\mathrm{\angle }EFC=\alpha -\frac{\pi }{4}$$$$S_1=\frac{FD\times ED\times \sin \mathrm{\angle }FDB}{2}$$$$=\frac{1}{2}\times \frac{2s}{\sqrt{5}}\times \frac{\sqrt{2}s}{2}\times \sin (\alpha -\frac{\pi }{4})$$$$=\frac{s^2}{2\sqrt{5}}(\sin \alpha -\cos \alpha )$$$$=\frac{s^2}{2\sqrt{5}}(\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}})$$$$=\frac{s^2}{10}=\frac{S}{10}✓$$

Commented by Tawa11 last updated on 11/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by kapoorshah last updated on 11/Oct/22

Commented by kapoorshah last updated on 11/Oct/22

$$saythesidelengthofthesquareis2$$$$theequationofthecircleis{x}^2+y^2=1$$$$polarlineofDFisx+2y=1$$$$\to x=1-2yissubstitutedtothe$$$$equationofthecircle$$$${(1-2y)}^2+y^2=1$$$$5y^2-4y=0$$$$y(5y-4)=0$$$$y=0y=\frac{4}{5}\to x=-\frac{3}{5}$$$$soD(1,0),E(0,1),F(-\frac{3}{5},\frac{4}{5})$$$$S_1=\frac{1}{2}\left|\begin{array}{cc}1& 0\\ 0& 1\\ -\frac{3}{5}& \frac{4}{5}\end{array}\right|$$$$S_1=\frac{2}{5}$$$$S=2^2$$$$S=4$$$$S_1=\frac{S}{10}$$

Commented by Tawa11 last updated on 11/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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