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

Answer: (B)

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

Area of the lune = Area of the smaller semicircle - Area of the segment

Area of the segment = Area of the sector - Area of the triangle formed by intersection points and the center of bigger semicircle

Also, the triangle so formed is an equilateral triangle since all the sides are of 1 unit.

$\therefore$ Area of the lune $=\pi \times \frac{1}{8}-(\frac{60}{360}\times \pi \times 1^2-\frac{\sqrt{3}}{4}\times 1^2)$

Area of the lune $=\frac{\pi }{8}-\frac{\pi }{6}+\frac{\sqrt{3}}{4}$

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