Given: \[ \text{Radius of circle} = 1 \ \text{cm} \] Chord \( AB \) subtends \( 60^\circ \) at the centre \( O \). The sector \( R \) is bounded by \( OA, OB \) and the arc \( AB \). Area of sector \( R \): \[ R = \frac{60^\circ}{360^\circ} \times \pi \times (1)^2 \] \[ R = \frac{1}{6} \pi \ \text{sq.cm} \]
It is given: \[ OC = OD = x \] and area of \( \triangle COD \) is **half** the area of \( R \). So: \[ \text{Area}(\triangle COD) = \frac{1}{2} \times R = \frac{\pi}{12} \ \text{sq.cm} \] Also: \[ \text{Area}(\triangle COD) = \frac{1}{2} \times OC \times OD \times \sin 60^\circ \] \[ \frac{\pi}{12} = \frac{1}{2} \times x \times x \times \frac{\sqrt{3}}{2} \] \[ \frac{\pi}{12} = \frac{\sqrt{3}}{4} x^2 \]
\[ x^2 = \frac{\pi}{3\sqrt{3}} \] \[ x = \sqrt{\frac{\pi}{3\sqrt{3}}} \ \text{cm} \]
✅ Final Answer: \( x = \sqrt{\frac{\pi}{3\sqrt{3}}} \ \text{cm} \)