Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the…

Question

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b) (^2\] is equal to :

Accepted Answer

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

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

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

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

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

Solutions · 1