In the figure, 1 + 2 = 2 and 3 ? ( B E ) = 4 ( A B ) . If the area of C A B is 2 3 - 3 unit 2 , when 2 1 is the largest, then the perimeter (in unit) of C E D is equal to ______ . [2023]

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Solution

Q.

In the figure, $θ_{1} + θ_{2} = \frac{π}{2}$ and $\sqrt{3} (B E) = 4 (A B) .$ If the area of $∆ C A B$ is $2 \sqrt{3} - 3$ ${unit}^{2}$ , when $\frac{θ_{2}}{θ_{1}}$ is the largest, then the perimeter (in unit) of $∆ C E D$ is equal to ______ . [2023]

Ans.
(6)

$Given, \sqrt{3} (B E) = 4 (A B)$

$ar (∆ C A B) = 2 \sqrt{3} - 3$

$So, \frac{1}{2} x^{2} \tan θ_{1} = 2 \sqrt{3} - 3$

$B E = B D + D E$

$= x (\tan θ_{1} + \tan θ_{2})$

$B E = A B (\tan θ_{1} + \cot θ_{1})$

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

$θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3};$ $θ_{1} = \frac{π}{3}, θ_{2} = \frac{π}{6}$

$as \frac{θ_{2}}{θ_{1}} is largest$

$∴$ $θ_{1} = \frac{π}{6}, θ_{2} = \frac{π}{3}$

$∴$ $x^{2} = \frac{(2 \sqrt{3} - 3) \times 2}{\tan θ_{1}} = \frac{\sqrt{3} (2 - \sqrt{3}) \times 2}{\tan \frac{π}{6}}$

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

$Perimeter of ∆ C E D = C D + D E + C E$

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

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