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(B) The smaller semi-circle has a diameter of $1$ unit,so its radius $r = \frac{1}{2}$ unit. Its area is $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (\frac

Vedclass · Accepted Answer

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

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(B) The smaller semi-circle has a diameter of $1$ unit,so its radius $r = \frac{1}{2}$ unit. Its area is $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (\frac{1}{2})^2 = \frac{\pi}{8}$.The larger semi-circle has a diameter of $2$ units,so its radius $R = 1$ unit. The chord $AB$ of length $1$ unit forms an equilateral triangle $OAB$ with the center $O$ of the larger semi-circle,where $OA = OB = AB = 1$. Thus,$\angle AOB = 60^{\circ}$.The area of the segment $AEB$ of the larger circle is the area of the sector $OAB$ minus the area of the equilateral triangle $OAB$.Area of sector $OAB = \frac{60^{\circ}}{360^{\circ}} 	imes \pi R^2 = \frac{1}{6} \pi (1)^2 = \frac{\pi}{6}$.Area of equilateral triangle $OAB = \frac{\sqrt{3}}{4} (side)^2 = \frac{\sqrt{3}}{4} (1)^2 = \frac{\sqrt{3}}{4}$.Area of segment $AEB = \frac{\pi}{6} - \frac{\sqrt{3}}{4}$.The area of the lune is the area of the smaller semi-circle minus the area of the segment $AEB$ of the larger semi-circle:Area of lune $= \frac{\pi}{8} - (\frac{\pi}{6} - \frac{\sqrt{3}}{4}) = \frac{\pi}{8} - \frac{\pi}{6} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} - \frac{\pi}{24}$.

$A$ 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

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