Read the following passage and answer the questions given below
A parking lot in an IT park has an area equal to the smaller part of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ cut-off by the line $\mathrm{\chi }=\frac{\mathrm{a}}{\sqrt{2}}$ .This area is allotted for car owners who
practice carpooling.
On the basis of above information, answer the following questions.
(i) Find the area used for carpooling.
(ii) Find the remaining area.
(iii) Write any two benefits of carpooling.

$$\begin{gathered} \text{ (i) }\therefore \text{ Required area of region }\mathrm{APBCA} \\ =2\left(\text{ Area of region }\mathrm{APCA}\right) \\ \begin{array}{l}=2{\int }_{\mathrm{a}/\sqrt{2}}^{\mathrm{a}} \mathrm{y}\mathrm{dx}=2{\int }_{\mathrm{a}/\sqrt{2}}^{\mathrm{a}} \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\mathrm{dx}\\ =2{\left[\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{x}}{\mathrm{a}}\right]}_{\mathrm{a}/\sqrt{2}}^{\mathrm{a}}\end{array} \end{gathered}$$

$\begin{array}{l}=2\left[\left(0+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{a}}{\mathrm{a}}\right)\right.-\left[\frac{\mathrm{a}}{2\sqrt{2}}\sqrt{{\mathrm{a}}^2-\frac{{\mathrm{a}}^2}{2}}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\left(\frac{\mathrm{a}}{\sqrt{2}}\cdot \frac{1}{\mathrm{a}}\right)\right\}\\ \\ \left.=2\left[\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\left(1\right)\right)-\frac{\mathrm{a}}{2\sqrt{2}}\cdot \frac{\mathrm{a}}{\sqrt{2}}-\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{1}{\sqrt{2}}\right]\\ =2\left[\frac{{\mathrm{\pi a}}^2}{4}-\frac{{\mathrm{a}}^2}{4}-\frac{{\mathrm{a}}^2\mathrm{\pi }}{8}\right]\\ =\frac{{\mathrm{\pi a}}^2}{2}-\frac{{\mathrm{a}}^2}{2}-\frac{{\mathrm{\pi a}}^2}{4}\\ =\frac{2{\mathrm{\pi a}}^2-2{\mathrm{a}}^2-{\mathrm{\pi a}}^2}{4}\\ =\frac{{\mathrm{\pi a}}^2-2{\mathrm{a}}^2}{4}\\ =\frac{{\mathrm{a}}^2(\mathrm{\pi }-2)}{4}\mathrm{sq}\text{ units }\end{array}$
$=\frac{{\mathrm{a}}^2(\mathrm{\pi }-2)}{4}$
(ii) Area of remaining part
= Area of circle-Area of park used for car pooling
$\begin{array}{l}={\mathrm{a}}^2-\frac{{\mathrm{a}}^2(\mathrm{\pi }-2)}{4}\\ ={\mathrm{a}}^2\frac{[4-\mathrm{\pi }+2]}{4}\\ =\frac{{\mathrm{a}}^2(6-\mathrm{\pi })}{4}\text{ sq units }\end{array}$
(iii) (a) Car pooling protects the environment.
(b) Car pooling helps in saving money.