Question

A semi-circle of diameter 1 unit sits at the top of a semi-circle of diameter 2 units. The shaded region inside the smaller semi-circle but outside the larger semi-circle is called a lune. The area of the lune is.

• A. $\frac{\pi }{6}-\frac{\sqrt{3}}{4}$
• B. $\frac{\sqrt{3}}{4}-\frac{\pi }{24}$
• C. $\frac{\sqrt{3}}{4}-\frac{\pi }{12}$
• D. $\frac{\sqrt{3}}{4}-\frac{\pi }{8}$

Answer 1

The correct option is B

$\frac{\sqrt{3}}{4}-\frac{\pi }{24}$

area of sector OACB = $\frac{r^2}{2}\theta =\frac{1}{2}.\frac{\pi }{3}=\frac{\pi }{6}$

area of shaded region = $\frac{\pi }{6}$ - area of $\Delta$OAB

$\frac{\pi }{6}-\frac{\sqrt{3}}{4}$

Hence area of line = Area of semi-circle - area of shaded region

$\frac{1}{2}\pi {\left(\frac{1}{2}\right)}^2-(\frac{\pi }{6}-\frac{\sqrt{3}}{4})$

=$\frac{\sqrt{3}}{4}+\frac{\pi }{8}-\frac{\pi }{6}$

=$\frac{\sqrt{3}}{4}-\frac{\pi }{24}$