CAT 2018 Question Paper Shift 1 Solved Paper

It is given that radius of the circle $=1\mathit{c}\mathit{m}$
Chord AB subtends an angle of ${60}^{\circ }$ on the centre of the given circle. $\mathit{R}$ be the region bounded by the radii $\mathit{O}\mathit{A},\mathit{O}\mathit{B}$ and the arc $\mathit{A}\mathit{B}$.
Therefore, $\mathit{R}=\frac{{60}^{\circ }}{{360}^{\circ }}\times$ Area of the circle $=\frac{1}{6}\times \mathit{\pi }\times {\left(1\right)}^2=\frac{\mathit{\pi }}{6}$ sq. $\mathit{c}\mathit{m}$It is given that $\overline{\mathit{O}\mathit{C}}=\mathit{O}\mathit{D}$ and area of triangle $\mathit{O}\mathit{C}\mathit{D}$ is half that of $\mathit{R}$. Let $\overline{\mathit{O}\mathit{C}}=\mathit{O}\mathit{D}=\mathit{x}$.
Area of triangle $\mathit{C}\mathit{O}\mathit{D}=\frac{1}{2}\times \mathit{O}\mathit{C}\times \mathit{O}\mathit{D}\times \sin 60°$
$\frac{\mathit{\pi }}{6\times 2}=\frac{1}{2}\times \mathit{x}\times \mathit{x}\times \frac{\sqrt{3}}{2}$
$\Rightarrow {\mathit{x}}^2=\frac{\mathit{\pi }}{3\sqrt{3}}$
$\Rightarrow \mathit{x}={\left(\frac{\mathit{\pi }}{3\sqrt{3}}\right)}^{\frac{1}{2}}\mathit{c}\mathit{m}$