RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

(c) Let the radius of the semicircle be R.Now join O to B
OC = OD = R    $\therefore \mathit{O}\mathit{B}=\mathit{R}\sqrt{2}$
The diameter of the smaller circle$=\left(\mathit{R}\sqrt{2}-\mathit{R}\right)=\mathit{R}\left(\sqrt{2}-1\right)$
Area of the semicircle $=\frac{\mathit{\pi }{\mathit{R}}^2}{2};$
Area of the circle = $\frac{\mathit{\pi }{\mathit{R}}^2{\left(\sqrt{2}-1\right)}^2}{2^2}$
Hence the ratio of the area of the smaller circle to that of the semicircle =$\frac{{\left(\sqrt{2}-1\right)}^2}{2}$
or${\left(\sqrt{2}-1\right)}^2:2$