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In a circle with center $O$ and radius $1$ cm, an arc $AB$ makes an angle $60$ degrees at $O$. Let $R$ be the region bounded by the radii $OA$, $OB$ and the arc $AB$. If $C$ and $D$ are two points on $OA$ and $OB$, respectively, such that $OC = OD$ and the area of triangle $OCD$ is half that of $R$, then the length of $OC$, in cm, is

  1. $\bigg(\dfrac{\pi}{3\sqrt 3} \bigg)^\frac{1}{2} \\$
  2. $\bigg(\dfrac{\pi}{4} \bigg)^\frac{1}{2} \\$
  3. $\bigg(\dfrac{\pi}{6} \bigg)^\frac{1}{2} \\$
  4. $\bigg(\dfrac{\pi}{4\sqrt 3} \bigg)^\frac{1}{2}$
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Given that, radius $ = \text{OA} = \text{OB} = 1 \; \text{cm},$ and $ \boxed{\text{OC} = \text{OD}} $

So, the $\triangle \text{OCD},$ is isosceles triangle.

An isosceles triangle is a triangle that :

  • Have two sides equal 
  • The base angles are also equal
  • The perpendicular from the apex angle bisects the base

We can draw the diagram,

We know that, sum of all the angles of a triangle $ = 180^ {\circ}$

Now, the sum of all the angles of a $ \triangle \text{OCD} = 180^{\circ}$

$ \Rightarrow 60^{\circ} + x + x =  180^{\circ} $

$ \Rightarrow 2x = 120^{\circ} $

$ \Rightarrow x = 60^{\circ} $

So, $ \triangle \text{OCD}$ is a equilateral triangle.

Area of $\triangle \text{OCD} = \frac{1}{2} \text{(area of R)} \quad \longrightarrow (1) $

Area of sector $ = \frac{Q}{360^{\circ}} \times \pi \times (\text{radius})^{2};$  where $Q$ is the angle subtended at the center.

Area of $ R = \frac{60^{\circ}}{360^{\circ}} \times \pi \times (1)^{2} = \frac{\pi}{6} \; \text{cm}^{2} $

Now, the area of $\triangle \text{OCD}  = \frac{\sqrt{3}}{4} \; \text{(side)}^{2}  = \frac{\sqrt{3}}{4} \; \text{OC}^{2} \; \text{cm}^{2} $

From the equation $(1),$ we get 

$ \frac{\sqrt{3}}{4} \; \text{OC}^{2} = \frac{1}{2} \times \frac{\pi}{6} $

$ \Rightarrow \text{(OC)}^{2} = \frac{\pi}{3 \sqrt{3}} $

$ \Rightarrow \text{OC} = \sqrt{\frac{\pi}{3 \sqrt{3}}} = \left( \frac{\pi}{3 \sqrt{3}} \right)^{\frac{1}{2}} \; \text{cm}.$

Correct Answer $: \text{A}$

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