दी गई आकृति में theta_1+theta_2=frac{pi}{2} त | vedclass

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Question

$\sqrt{3} BE =4 AB$

$Ar (	riangle CAB )=2 \sqrt{3}-3$

$\frac{1}{2} x ^2 	an 	heta_1=2 \sqrt{3}-3$

$BE = BD + DE$

$= x \left(	an 	heta

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$\sqrt{3} BE =4 AB$

$Ar (	riangle CAB )=2 \sqrt{3}-3$

$\frac{1}{2} x ^2 	an 	heta_1=2 \sqrt{3}-3$

$BE = BD + DE$

$= x \left(	an 	heta_1+	an 	heta_2ight)$

$BE = AB \left(	an 	heta_1+\cot 	heta_1ight)$

$\frac{4}{\sqrt{3}} 	an 	heta_1+\cot 	heta_1 \Rightarrow 	an 	heta_1=\sqrt{3}, \frac{1}{\sqrt{3}}$

$	heta_1=\frac{\pi}{6}$

$	heta_1=\frac{\pi}{3} \quad 	heta_2=\frac{\pi}{3}$

$as \frac{	heta_2}{	heta_1} 	ext { is } \operatorname{largest} 	herefore 	heta_1=\frac{\pi}{6} 	heta_2=\frac{\pi}{3}$

$	herefore x ^2=\frac{(2 \sqrt{3}-3) 	imes 2}{	an 	heta_1}=\frac{\sqrt{3}(2-\sqrt{3}) 	imes 2}{	an \frac{\pi}{6}}$

$x ^2=12-6 \sqrt{3}=(3-\sqrt{3})^2$

$x =3-\sqrt{3}$

Perimeter of $	riangle C E D$

$= CD + DE + CE$

$=3 \sqrt{3}+(3-\sqrt{3}) \sqrt{3}+(3-\sqrt{3}) 	imes 2=6$

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