JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Properties of Triangle

With $AB=x$, the right triangles give $BE=x(\tan {\theta }_1+\tan {\theta }_2)$. From $\sqrt{3}BE=4AB$ and ${\theta }_2=\frac{\pi }{2}-{\theta }_1$ (so $\tan {\theta }_2=\cot {\theta }_1$), $\tan {\theta }_1+\cot {\theta }_1=\frac{4}{\sqrt{3}}$, giving $\tan {\theta }_1=\sqrt{3}$ or $\frac{1}{\sqrt{3}}$. The ratio $\frac{{\theta }_2}{{\theta }_1}$ is largest when ${\theta }_1=\frac{\pi }{6}$ and ${\theta }_2=\frac{\pi }{3}$. The area $\frac{1}{2}{x}^2\tan {\theta }_1=\frac{x^2}{2\sqrt{3}}=2\sqrt{3}-3$ gives $x^2=12-6\sqrt{3}$, so $x=3-\sqrt{3}$. In the right triangle $CED$, $CD=x$, $DE=x\tan {\theta }_2=\sqrt{3}x$ and $CE=2x$, so its perimeter is $x(3+\sqrt{3})=(3-\sqrt{3})(3+\sqrt{3})=6$. The perimeter of $△CED$ is $6$.