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

- (π/3√3)
^{1/2} - (π/4)
^{1/2} - (π/6)
^{1/2} - (π/4√3)
^{1/2}

Radius of the circle = 1 cm

Chord AB subtends an angle of 60° on the centre of the circle. R is the region bounded by the radii OA, OB and the arc AB.

R = 60°/360° × Area of the circle

= 1/6 × π × (1)^{2}

= π/6 sq. cm

Given that OC = OD and area of triangle OCD is half that of R.

Area of triangle OCD = 1/2 × OC × OD × sin 60°

π/6 × 1/2 = 1/2 × OC × OC × √3/2

⇒ OC^{2} = π/3√3

**The correct option is A.**

- Let ABCD be a rectangle inscribed in a circle of radius 13 cm
- ABCD is a square. X is the mid-point of AB
- ABC is an isosceles triangle such that AB = BC = 8 cm
- A circle is inscribed in a given square and another circle is circumscribed about the square
- Two parallel chords of a circle whose diameter is 13 cm are respectively 5 cm and 12 cm