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Question Number 206848 by BaliramKumar last updated on 27/Apr/24

Commented by BaliramKumar last updated on 27/Apr/24

Answered by A5T last updated on 27/Apr/24

$$Letsideofsquare=s$$$${CE}^2={EF}^2+{CF}^2\Rightarrow \mathrm{\angle }EFC=90°\Rightarrow \alpha +\beta =90°$$$$cos\alpha =\frac{s}{CF=8}\Rightarrow s=8cos\alpha ;BF=8sin\alpha$$$$\frac{sin\beta }{AF=s-8sin\alpha }=\frac{1}{EF=6}\Rightarrow s-8sin\alpha =6sin\beta$$$$\Rightarrow 8cos\alpha -8sin\alpha =6cos\alpha \Rightarrow 8sin\alpha =2cos\alpha$$$$\Rightarrow tan\alpha =\frac{1}{4}\Rightarrow \alpha ={tan}^{-1}\left(\frac{1}{4}\right)$$$$\left[ABCD\right]\Rightarrow s^2=64{cos}^2\alpha =64{cos}^2\left({tan}^{-1}\left(\frac{1}{4}\right)\right)=\frac{1024}{17}$$$$$$$$tan\alpha =\frac{1}{4};tan\beta =\frac{sin(90-\alpha )}{cos(90-\alpha )}=\frac{cos\alpha }{sin\alpha }=\frac{1}{tan\alpha }=4$$$$\Rightarrow tan\alpha +tan\beta =\frac{17}{4}$$$$$$$$AE=6cos\beta =6sin\alpha ;DE=s-6sin\alpha$$$$=\sqrt{\frac{1024}{17}}-6sin\left({tan}^{-1}\left(\frac{1}{4}\right)\right)$$$$\Rightarrow \left[CDE\right]=\frac{DE\times DC}{2}$$$$=\frac{(\sqrt{\frac{1024}{17}}-6sin\left({tan}^{-1}\left(\frac{1}{4}\right)\right))\times \sqrt{\frac{1024}{17}}}{2}$$$$=\frac{416}{17}$$

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