Question

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

• A. (π/3√3)1/2
• B. (π/4)1/2
• C. (π/6)1/2
• D. (π/4√3)1/2

Answer 1

The Correct Option is A

Step 1: Area of sector \( R \) 

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 \( R \): \[ R = \frac{60^\circ}{360^\circ} \times \pi \times (1)^2 \] \[ R = \frac{1}{6} \pi \ \text{sq.cm} \]

Step 2: Area of triangle \( OCD \)

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: \[ \text{Area}(\triangle COD) = \frac{1}{2} \times OC \times OD \times \sin 60^\circ \] \[ \frac{\pi}{12} = \frac{1}{2} \times x \times x \times \frac{\sqrt{3}}{2} \] \[ \frac{\pi}{12} = \frac{\sqrt{3}}{4} x^2 \]

Step 3: Solving for \( x \)

\[ x^2 = \frac{\pi}{3\sqrt{3}} \] \[ x = \sqrt{\frac{\pi}{3\sqrt{3}}} \ \text{cm} \]

✅ Final Answer: \( x = \sqrt{\frac{\pi}{3\sqrt{3}}} \ \text{cm} \)