Area of the region: Area of equilateral triangle and area of a sector

In the figure shown , US and UT are line segments each of length  2 , and $\displaystyle \angle TUS={{60}^{\circ }}$ . Arcs TR and SR are each one - sixth of a circle with radius 2 . What is the area of the region shown ?

Solution

 $\displaystyle \alpha (\Delta UVW)=\frac{{\sqrt{3}}}{4}\times {{4}^{2}}$

$\displaystyle \text{Area}\ \text{of}\ \text{sector}\ VSR\ =\ \text{Area}\ \text{of}\ \text{sector}\ WRT\ =\ \frac{1}{2}\times {{2}^{2}}\times \frac{\pi }{3}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\pi }}{3}$

$\displaystyle \therefore \ The\ area\ of\ the\ region\ =\ \alpha (\Delta UVW)\ -\ [\text{Area}\ \text{of}\ \text{sector}\ VSR\ +\ \text{Area}\ \text{of}\ \text{sector}\ WRT\ ]$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ (4\sqrt{3}-\frac{{4\pi }}{3})\ \ \text{sq}\ \ \text{units}$ 

*Note