Let AB be a line segment of length 2. Constru | vedclass

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(B) Let $O$ be the center of the semicircle $S$ with diameter $AB = 2$. Thus,the radius of $S$ is $r_S = 1$.Since $C$ is the mid-point of the arc $AB$

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$\frac{1}{2}$

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(B) Let $O$ be the center of the semicircle $S$ with diameter $AB = 2$. Thus,the radius of $S$ is $r_S = 1$.Since $C$ is the mid-point of the arc $AB$,$	riangle AOC$ is a right-angled isosceles triangle with $OA = OC = 1$ and $\angle AOC = 90^\circ$.The length of the chord $AC$ is $\sqrt{OA^2 + OC^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.Semicircle $T$ is constructed with $AC$ as diameter,so its radius is $r_T = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.The area of the region inside $T$ but outside $S$ is the area of the semicircle $T$ minus the area of the segment of $S$ cut off by chord $AC$.Area of semicircle $T = \frac{1}{2} \pi r_T^2 = \frac{1}{2} \pi \left(\frac{1}{\sqrt{2}}ight)^2 = \frac{\pi}{4}$.The area of the segment of $S$ cut off by chord $AC$ is (Area of sector $OAC$) - (Area of $	riangle OAC$).Area of sector $OAC = \frac{90^\circ}{360^\circ} \pi (1)^2 = \frac{\pi}{4}$.Area of $	riangle OAC = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$.Area of segment = $\frac{\pi}{4} - \frac{1}{2}$.Required Area = (Area of semicircle $T$) - (Area of segment) = $\frac{\pi}{4} - (\frac{\pi}{4} - \frac{1}{2}) = \frac{1}{2}$.

Let $AB$ be a line segment of length $2$. Construct a semicircle $S$ with $AB$ as diameter. Let $C$ be the mid-point of the arc $AB$. Construct another semicircle $T$ external to the $\triangle ABC$ with chord $AC$ as diameter. The area of the region inside the semicircle $T$ but outside $S$ is

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