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Question Number 176950 by cortano1 last updated on 28/Sep/22

Answered by mr W last updated on 28/Sep/22

$$say\mathrm{\angle }AEC=\theta$$$$CB=4\sin \theta$$$$BE=4\cos \theta$$$$\frac{AE}{\sin 45°}=\frac{3}{\sin (\theta -45°)}$$$$AE=\frac{3}{\sin \theta -\cos \theta }=4\sin \theta +4\cos \theta$$$${\sin }^2\theta -{\cos }^2\theta =\frac{3}{4}$$$${\sin }^2\theta =\frac{7}{8}\Rightarrow \sin \theta =\frac{\sqrt{14}}{4}$$$$a=4\sin \theta =\sqrt{14}$$$$areaofsquare=a^2=16{\sin }^2\theta =14✓$$

Commented by Tawa11 last updated on 28/Sep/22

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

Commented by cortano1 last updated on 29/Sep/22

$$\mathrm{nice}$$

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