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

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\begin{aligned} & \sqrt{3} \mathrm{BE}=4 \mathrm{AB} \\ & \mathrm{Ar}(\triangle \mathrm{CAB})=2 \sqrt{3}-3 \\ & \frac{1}{2} \mathrm{x}^2 \tan \theta_1=2 \sqrt{3}-3 \\ & \mathrm{BE}=\mathrm{BD}+\mathrm{DE} \\ & =\mathrm{x}\left(\tan \theta_1+\tan \theta_2\right) \\ & \mathrm{BE}=\mathrm{AB}\left(\tan \theta_1+\cot \theta_1\right) \end{aligned}

\begin{aligned} & \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} \quad \theta_2=\frac{\pi}{3} \\ & \theta_1=\frac{\pi}{3} \quad \theta_2=\frac{\pi}{6} \\ & \text { as } \frac{\theta_2}{\theta_1} \text { is largest } \therefore \theta_1=\frac{\pi}{6} \theta_2=\frac{\pi}{3} \end{aligned}

\begin{aligned} & \therefore \mathrm{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} \end{aligned}\\ Perimeter \, \, of \, \, \triangle C E D\\ \begin{aligned} & =C D+D E+C E \\ & =3 \sqrt{3}+(3-\sqrt{3}) \sqrt{3}+(3-\sqrt{3}) \times 2=6 \end{aligned}

Ans. (6)

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Kshitij

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