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(A) Let $	heta$ be the angle that the side $AB$ of the rectangle $ABCD$ makes with the side $PQ$ of the rectangle $PQRS$.From the geometry of the fig

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$72$

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(A) Let $	heta$ be the angle that the side $AB$ of the rectangle $ABCD$ makes with the side $PQ$ of the rectangle $PQRS$.From the geometry of the figure,the sides of the rectangle $PQRS$ are $a = 4 \cos 	heta + 2 \sin 	heta$ and $b = 4 \sin 	heta + 2 \cos 	heta$.The area of the rectangle $PQRS$ is $A = a 	imes b = (4 \cos 	heta + 2 \sin 	heta)(4 \sin 	heta + 2 \cos 	heta)$.Expanding this,we get $A = 16 \sin 	heta \cos 	heta + 8 \cos^2 	heta + 8 \sin^2 	heta + 4 \sin 	heta \cos 	heta = 8(\sin^2 	heta + \cos^2 	heta) + 20 \sin 	heta \cos 	heta$.Using trigonometric identities,$A = 8 + 10 \sin 2	heta$.The area is maximum when $\sin 2	heta = 1$,which occurs at $	heta = 45^{\circ}$.At $	heta = 45^{\circ}$,the sides are $a = 4(\frac{1}{\sqrt{2}}) + 2(\frac{1}{\sqrt{2}}) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ and $b = 4(\frac{1}{\sqrt{2}}) + 2(\frac{1}{\sqrt{2}}) = 3\sqrt{2}$.Then $(a+b)^2 = (3\sqrt{2} + 3\sqrt{2})^2 = (6\sqrt{2})^2 = 36 	imes 2 = 72$.

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 :

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