Verified Solution • 2025–26 Session moderate

QJEEM23Ma81e1: JEE Main 2023 Mathematics - Class 11 JEEM23Ma81e1 Solution: In the figure, and . If the area of is unit , when is the…

In the figure, θ1+θ2=π2\theta_{1}+\theta_{2}=\frac{\pi}{2} and 3(BE)=4(AB)\sqrt{3}(\mathrm{BE})=4(\mathrm{AB}) . If the area of △CAB\triangle \mathrm{CAB} is 23−32 \sqrt{3}-3 unit 2{ }^{2} , when θ2θ1\frac{\theta_{2}}{\theta_{1}} is the largest, then the perimeter (in unit) of △CED\triangle \mathrm{CED} is equal to _________.

JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Properties of Triangle Question 7 English

Step-by-Step Verified Solution

JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Properties of Triangle Question 7 English Explanation

We have, θ1+θ2=π2\theta_1+\theta_2=\frac{\pi}{2} and 3(BE)=4AB\sqrt{3}(B E)=4 A B

Let AB=xA B=x unit

amp;AC=xtan⁡θ1amp;ED=xtan⁡θ2amp;BE=BD+DE\begin{aligned}& A C=x \tan \theta_1 \\\\& E D=x \tan \theta_2 \\\\& B E=B D+D E\end{aligned}

amp;⇒43x=x(tan⁡θ1+tan⁡θ2)[∵3BE=4AB]amp;⇒43=tan⁡θ1+tan⁡(π2−θ1)[∵θ1+θ2=π2]\begin{array}{rlrl}& \Rightarrow \frac{4}{\sqrt{3}} x =x\left(\tan \theta_1+\tan \theta_2\right) {[\because \sqrt{3} B E=4 A B]} \\\\& \Rightarrow \frac{4}{\sqrt{3}}=\tan \theta_1+\tan \left(\frac{\pi}{2}-\theta_1\right) {\left[\because \theta_1+\theta_2=\frac{\pi}{2}\right]}\end{array}

amp;⇒tan⁡θ1+cot⁡θ1=43=3+13amp;⇒tan⁡θ1=3 or θ1=π3 and θ2=π6amp;or⁡θ1=π6 and θ2=π3\begin{aligned}& \Rightarrow \tan \theta_1+\cot \theta_1=\frac{4}{\sqrt{3}}=\sqrt{3}+\frac{1}{\sqrt{3}} \\\\& \Rightarrow \tan \theta_1=\sqrt{3} \text { or } \theta_1=\frac{\pi}{3} \text { and } \theta_2=\frac{\pi}{6} \\\\& \operatorname{or} \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3}\end{aligned}

∵θ2θ1\because \frac{\theta_2}{\theta_1} is largest

∴θ1=π6 and θ2=π3\therefore \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3}

amp; Area of △CAB=12×x×xtan⁡θ1amp;⇒x2tan⁡θ12=23−3amp;⇒x2=2(23−3)tan⁡π6=12−63&⇒x=3−3\begin{aligned}& \text { Area of } \triangle C A B=\frac{1}{2} \times x \times x \tan \theta_1 \\\\& \Rightarrow \frac{x^2 \tan \theta_1}{2}=2 \sqrt{3}-3 \\\\& \Rightarrow x^2=\frac{2(2 \sqrt{3}-3)}{\tan \frac{\pi}{6}}=12-6 \sqrt{3} \\\\\& \Rightarrow x=3-\sqrt{3}\end{aligned}

 Also, CE=x2+x2tan⁡2π3=(3−3)×2=6−23\text { Also, } C E=\sqrt{x^2+x^2 \tan ^2 \frac{\pi}{3}}=(3-\sqrt{3}) \times 2=6-2 \sqrt{3}

Perimeter of △CED\triangle C E D

amp;=CD+DE+CEamp;=(3−3)+(3−3)tan⁡π3+6−23amp;=3−3+33−3+6−23=6\begin{aligned}& =C D+D E+C E \\\\& =(3-\sqrt{3})+(3-\sqrt{3}) \tan \frac{\pi}{3}+6-2 \sqrt{3} \\\\& =3-\sqrt{3}+3 \sqrt{3}-3+6-2 \sqrt{3}=6\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring