Question

In the quadrilateral $MARE$ inscribed in a unit circle $C,$ $AM$ is a diameter of $C,$ and $E$ lies on the angle bisector of $\angle RAM.$ Given that triangles $RAM$ and $REM$ have the same area. Then area (in sq. units) of the quadrilateral $MARE$ is equal to

• A. $\frac{8}{9}$
• B. $\frac{8\sqrt{2}}{9}$
• C. $\frac{8\sqrt{2}}{3}$
• D. $\frac{4\sqrt{2}}{3}$

Answer 1

The correct option is B $\frac{8\sqrt{2}}{9}$



In $△RAM,$
$\sin 2\theta =\frac{RM}{2}$
$\Rightarrow RM=2\sin 2\theta$
In $△REM,$
$\frac{RM}{\sin 2\theta }=\frac{ME}{\sin \theta }\Rightarrow ME=\frac{2\sin 2\theta }{\sin 2\theta }\times \sin \theta \Rightarrow ME=2\sin \theta$

Given, $ar\left(△RAM\right)=ar\left(△REM\right)$
$\Rightarrow \frac{1}{2}\times 2\times RM\sin (\frac{\pi }{2}-2\theta )=\frac{1}{2}\times RM\times ME\times \sin \theta \Rightarrow 2\cos 2\theta =2{\sin }^2\theta \Rightarrow 1-2{\sin }^2\theta ={\sin }^2\theta \Rightarrow {\sin }^2\theta =\frac{1}{3}$
$\Rightarrow \sin \theta =\sqrt{\frac{1}{3}}$ and $\cos \theta =\sqrt{\frac{2}{3}}$

$RM=2\times 2\times \frac{1}{\sqrt{3}}\times \frac{\sqrt{2}}{\sqrt{3}}\Rightarrow RM=\frac{4\sqrt{2}}{3}$
$ME=\frac{2}{\sqrt{3}}$

Area of $MARE$
$=ar\left(△RAM\right)+ar\left(△REM\right)=2ar\left(△RAM\right)=2\times \frac{1}{2}\times 2\times RM\times \sin (\frac{\pi }{2}-2\theta )=2\times \frac{4\sqrt{2}}{3}\cos 2\theta =2\times \frac{4\sqrt{2}}{3}(2{\cos }^2\theta -1)=\frac{8\sqrt{2}}{3}(2\times \frac{2}{3}-1)=\frac{8\sqrt{2}}{9}$