Question

In the figure, a semi-circle with centre $O$ is drawn on $AB$. The ratio of the larger shaded area to the smaller shaded area is ? $\Delta PBO={60}^o$.

• A. $\frac{4\pi -2\sqrt{3}}{2\pi -2\sqrt{3}}$
• B. $\frac{4\pi -3\sqrt{3}}{3\pi -3\sqrt{3}}$
• C. $\frac{4\pi -3\sqrt{3}}{2\pi -3\sqrt{3}}$
• D. $\frac{4\pi -2\sqrt{3}}{3\pi -2\sqrt{3}}$

Answer 1

The correct option is C $\frac{4\pi -3\sqrt{3}}{2\pi -3\sqrt{3}}$

Let radius $=r$
Area of sector $APO=\frac{120}{360}\times \pi r^2=\frac{1}{3}\pi r^2$
Area of sector $PBO=\frac{60}{360}\times \pi r^2=\frac{1}{6}\pi r^2$
Area of $\Delta AOP=\frac{1}{2}\times \frac{\sqrt{3}r}{2}\times r=\frac{\sqrt{3}{r}^2}{4}$
Area of $\Delta BOP=\frac{\sqrt{3}{r}^2}{4}$
Area of major shaded area : Area of minor shaded area
$=(\frac{1}{3}\pi r^2-\frac{\sqrt{3}}{4}{r}^2):(\frac{1}{6}\pi r^2-\frac{\sqrt{3}}{4}{r}^2)$
$=\frac{4\pi -3\sqrt{3}}{2\pi -3\sqrt{3}}$.

Hence, option C is correct.