$\sqrt{3} BE =4 AB$

$Ar (\triangle CAB )=2 \sqrt{3}-3$

$\frac{1}{2} x ^2 \tan \theta_1=2 \sqrt{3}-3$

$BE = BD + DE$

$= x \left(\tan \theta_1+\tan \theta_2\right)$

$BE = AB \left(\tan \theta_1+\cot \theta_1\right)$

$\frac{4}{\sqrt{3}} \tan \theta_1+\cot \theta_1 \Rightarrow \tan \theta_1=\sqrt{3}, \frac{1}{\sqrt{3}}$

$\theta_1=\frac{\pi}{6}$

$\theta_1=\frac{\pi}{3} \quad \theta_2=\frac{\pi}{3}$

$as \frac{\theta_2}{\theta_1} \text { is } \operatorname{largest} \therefore \theta_1=\frac{\pi}{6} \theta_2=\frac{\pi}{3}$

$\therefore x ^2=\frac{(2 \sqrt{3}-3) \times 2}{\tan \theta_1}=\frac{\sqrt{3}(2-\sqrt{3}) \times 2}{\tan \frac{\pi}{6}}$

$x ^2=12-6 \sqrt{3}=(3-\sqrt{3})^2$

$x =3-\sqrt{3}$

Perimeter of $\triangle C E D$

$= CD + DE + CE$

$=3 \sqrt{3}+(3-\sqrt{3}) \sqrt{3}+(3-\sqrt{3}) \times 2=6$