QJEEM24Maf9ae: JEE Main 2024 Mathematics - Class 12 JEEM24Maf9ae Solution: Let a rectangle ABCD of sides 2 and 4 be inscribed in another…

$\begin{aligned}& P D=4 \cos \theta \Rightarrow P S=4 \cos \theta+2 \sin \theta \\& D S=2 \sin \theta \\& A P=4 \sin \theta \\& Q A=2 \cos \theta \Rightarrow P Q=2 \cos \theta+4 \sin \theta \\& \Rightarrow \text { Area of } P Q R S=4(2 \cos \theta+\sin \theta)(\cos \theta+2 \sin \theta) \\&=4\left[2 \cos ^2 \theta+2 \sin ^2 \theta+5 \sin \theta \cos \theta\right] \\&=8+10 \sin 2 \theta\end{aligned}$

Area is maximum when $\sin 2 \theta=1 \Rightarrow \theta=45^{\circ}$

$\Rightarrow$ Maximum area $=8+10=18$

$\therefore P S=4 \times \frac{1}{\sqrt{2}}+2 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}}$

$\begin{aligned}& P Q=2 \times \frac{1}{\sqrt{2}}+4 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}} \\\therefore \quad & (a+b)^2=\left(\frac{6}{\sqrt{2}}+\frac{6}{\sqrt{2}}\right)^2=\left(\frac{12}{\sqrt{2}}\right)=(6 \sqrt{2})^2=72\end{aligned}$