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$=\frac{r^2}{2}\theta =\frac{1}{2}.\frac{\pi }{3}=\frac{\pi }{5}$
area of shaded region =$\frac{\pi }{6}-$ area of $△OAB$
=$\frac{\pi }{6}-\frac{\sqrt{3}}{4}$
Hence area of line = Area of semi circle - area of shaded region
=$\frac{1}{2}\pi {\left(\begin{array}{c}\frac{1}{2}\end{array}\right)}^2-\left(\begin{array}{c}\frac{\pi }{6}-\frac{\sqrt{3}}{4}\end{array}\right)$
=$\frac{\sqrt{3}}{4}+\frac{\pi }{8}-\frac{\pi }{6}$
=$\frac{\sqrt{3}}{4}-\frac{\pi }{24}$