Physics - Gravitation Question with Solution | TestHub

$\frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{constant}$ $\frac{{\mathrm{A}}_1}{{\mathrm{t}}_1}=\frac{{\mathrm{A}}_2}{{\mathrm{t}}_2}$ $\frac{\mathrm{A}}{40}=\frac{\frac{\pi \mathrm{ab}}{4}+\text{ Area of }△\mathrm{ESF}}{12}$ $\frac{\mathrm{A}}{40}=\frac{\frac{\mathrm{A}}{4}+\text{ Area of }△\mathrm{ESF}}{12}$ Area of $△\mathrm{ESF}$ $=\frac{12}{40}\mathrm{A}-\frac{\mathrm{A}}{4}=\left(\frac{12-10}{40}\right)\mathrm{A}=\frac{\mathrm{A}}{20}$ $\frac{\mathrm{A}}{40}=\frac{\text{ Area of SFCB }-\text{ Area of }△\mathrm{SCF}}{1}$ $\frac{\mathrm{A}}{40}=$ Area of $\mathrm{SFCB}-$ Area of $△\mathrm{ESF}$

Area of $\mathrm{SFCB}=\frac{\mathrm{A}}{40}+$ Area of $△\mathrm{ESF}$ Total area of shaded region $=$ Area of ESF + Area of SFCB $=\frac{\mathrm{A}}{40}+2$ [Area of ESF] $=\frac{\mathrm{A}}{8}$