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In the figure, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4(A B)$. If the area of $\triangle CAB$ is $2 \sqrt{3}-3$ unit $^2$, when $\frac{\theta_2}{\theta_1}$ is the largest, then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.
$5$
$4$
$6$
$3$
Solution
$\sqrt{3} BE =4 AB$
$Ar (\triangle CAB )=2 \sqrt{3}-3$
$\frac{1}{2} x ^2 \tan \theta_1=2 \sqrt{3}-3$
$BE = BD + DE$
$= x \left(\tan \theta_1+\tan \theta_2\right)$
$BE = AB \left(\tan \theta_1+\cot \theta_1\right)$
$\frac{4}{\sqrt{3}} \tan \theta_1+\cot \theta_1 \Rightarrow \tan \theta_1=\sqrt{3}, \frac{1}{\sqrt{3}}$
$\theta_1=\frac{\pi}{6}$
$\theta_1=\frac{\pi}{3} \quad \theta_2=\frac{\pi}{3}$
$as \frac{\theta_2}{\theta_1} \text { is } \operatorname{largest} \therefore \theta_1=\frac{\pi}{6} \theta_2=\frac{\pi}{3}$
$\therefore x ^2=\frac{(2 \sqrt{3}-3) \times 2}{\tan \theta_1}=\frac{\sqrt{3}(2-\sqrt{3}) \times 2}{\tan \frac{\pi}{6}}$
$x ^2=12-6 \sqrt{3}=(3-\sqrt{3})^2$
$x =3-\sqrt{3}$
Perimeter of $\triangle C E D$
$= CD + DE + CE$
$=3 \sqrt{3}+(3-\sqrt{3}) \sqrt{3}+(3-\sqrt{3}) \times 2=6$
