Question

The maximum area of the rectangle whose sides pass through the angular points of a given rectangle of sides $a$ and $b$ is

• A. $2\left(ab\right)$
• B. $\frac{1}{2}(a+b{)}^2$
• C. $\frac{1}{2}(a^2+b^2)$
• D. $\frac{a^3}{b}$

Answer 1

The correct option is B $\frac{1}{2}(a+b{)}^2$

A rectangular whose side pass through the angular points of a given rectangle of sides a and b.

So side of ABCD are $DC=DP+PC$

$DC=a\cos \theta +b\sin \theta$ and $AB=AR+RB$ But this side equals to DC. and

$CB=CS+BS$

$CB=b\cos \theta +a\sin \theta$

Now area of rectangle $=DC\times CB$

$=(a\cos \theta +b\sin \theta )(b\cos \theta +a\sin \theta )$

Area = $ab{\cos }^2\theta +ab{\sin }^2\theta +a^2\sin \theta \cos \theta +b^2\sin \theta \cos \theta$

$\because {\cos }^2\theta +{\sin }^2\theta =1$

Area = $ab({\cos }^2\theta +{\sin }^2\theta )+(a^2+b^2)\left(\sin \theta \cos \theta \right)$$

$A=ab+(a^2+b^2)\left(\frac{{\sin }^2\theta }{2}\right)\because sin\theta =\frac{1}{2}\sin 2\theta$

So for maximum Area $\frac{\sin 2\theta }{2}$ will be maximum

So $-1\le \sin 2\theta \le 1$

$\frac{-1}{2}<\frac{\sin \theta }{2}\le \frac{1}{2}$

$A=ab+\frac{1}{2}(a^2+b^2)$

$A=\frac{(a+b{)}^2}{2}$