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Question Number 70121 by ahmadshahhimat775@gmail.com last updated on 01/Oct/19

Answered by mind is power last updated on 01/Oct/19

$$letaangleleft-,csideofsquar$$$$\Rightarrow c=6sin\left(a\right)$$$$\frac{6sin\left(a\right)+6cos\left(a\right)}{sin\left(45\right)}=\frac{10}{sin(45+a)}$$$$\Rightarrow 6.(sin\left(a\right)+cos\left(a\right)).sin(45+a)=10sin\left(45\right)$$$$\Rightarrow 6.\sqrt{2}sin(a+45).sin(45+a)=5\sqrt{2}$$$$\Rightarrow {sin}^2(45+a)=\frac{5}{6}$$$$a={\sin }^{-1}\left(\frac{5}{6}\right)-45$$$$Areaofsquar=c.c=36.{sin}^2\left(a\right)$$$$sin\left(a\right)=Sin({\sin }^{-1}\left(\frac{5}{6}\right)-45)=\frac{5}{6}.\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}.\frac{\sqrt{11}}{5}$$$$Area=18{(\frac{5}{6}-\frac{\sqrt{11}}{5})}^2$$

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