In the figure,$\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(BE)=4(AB)$. If the area of $\triangle CAB$ is $2\sqrt{3}-3 \text{ unit}^2$,when $\frac{\theta_2}{\theta_1}$ is the largest,then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.

  • A
    $5$
  • B
    $4$
  • C
    $6$
  • D
    $3$

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