Area of shaded region

In the figure , a square ABCD and a sector OAB of a circle centre O , radius r . Show that the area of square ABCD is $\displaystyle 2{{r}^{2}}(1-\cos \theta )$ and then find the shaded area R , when $\displaystyle \theta =\frac{\pi }{3}$ .

Solution

$\displaystyle \text{Draw}\ \ OE\bot AB\ .$  

$\displaystyle AE=r\sin \frac{\theta }{2}\ \ \ ,\ \ \ OE=r\cos \frac{\theta }{2}$ 

$\displaystyle AB=2AE=2r\sin \frac{\theta }{2}\ $

$\displaystyle \alpha (ABCD)={{(2r\sin \frac{\theta }{2}\ )}^{2}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{r}^{2}}{{\sin }^{2}}\frac{\theta }{2}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ .\ 2{{\sin }^{2}}\frac{\theta }{2}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ (1-\cos \theta )\ \ \ \ [\because \cos 2\theta =1-2{{\sin }^{2}}\theta ]$

$\displaystyle \text{The}\ \text{shaded}\ \text{area}\ R\ =\alpha (ABCD)+\alpha (\Delta OAB)-\text{Area}\ \text{of}\ \text{sector}\ OAB$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ (1-\cos \theta )+\frac{1}{2}(2r\sin \frac{\theta }{2}\ )(r\cos \frac{\theta }{2})-\frac{1}{2}{{r}^{2}}\theta$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{r}^{2}}\ (1-\cos \theta )+\frac{1}{2}{{r}^{2}}\sin \theta -\frac{1}{2}{{r}^{2}}\theta$

$\displaystyle \text{When}\ \theta =\frac{\pi }{3}\ ,$

$\displaystyle R=2{{r}^{2}}(1-\frac{1}{2})+\frac{1}{2}{{r}^{2}}(\frac{{\sqrt{3}}}{2})-\frac{1}{2}{{r}^{2}}(\frac{\pi }{3})$

$\displaystyle \ \ \ =(1+\frac{{\sqrt{3}}}{4}-\frac{\pi }{6})\ {{r}^{2}}\ \text{sq}\ \text{units}$