Question

Rectangle are inscribed in a circle of radius r. the dimensions of the rectangle which has maximum area , is

Answer 1

$\Delta$BCD is Right angle

$\frac{CD}{BD}=\cos \theta$

$\Rightarrow CD=BD\cos \theta$

$=2r\cos \theta$

$BC\Rightarrow 2r\sin \theta$

$ar\left(ABCD\right)="s"=BC\times CD=2r\sin \theta \left(2r\cos \theta \right)$

$=4r^2\sin \theta \cos \theta$

Maximize s, $\frac{ds}{d\theta }=0$

$\frac{d}{d\theta }\left(4r^2\sin \theta \cos \theta \right)=0$

$4r^2(\sin \theta \frac{d}{d\theta }\cos \theta +\cos \theta \frac{d}{d\theta }(\sin \theta \left)\right)=0$

$4^2(-{\sin }^2\theta +{\cos }^2\theta )=0$

${\cos }^2\theta -{\sin }^2\theta =0$

$\cos 2\theta =0$ $[\theta <2\theta <\pi ]$

$\therefore 2\theta =\frac{\pi }{2}$

$\theta =\frac{\pi }{4}$ $\left[\begin{array}{c}BC=2r\sin \theta =\sqrt{2}r\\ CD=2r\cos \theta \sqrt{2}r\end{array}\right]$

$\therefore ar\left(ABCD\right)=4r^2\sin \theta \cos \theta$

which is square $=4r^2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=2r^2$

$\therefore$ Dimensions of rectangle $CD=2r\cos \theta =2r\times \frac{1}{\sqrt{2}}=\sqrt{2}r$

$BC=2r\sin \theta =2\times \frac{1}{\sqrt{2}}=\sqrt{2}r$

$\therefore CD=BC=\sqrt{2}r$.