The given figure shows a circle with centre O. AB and PQ are its chords where AB is at a distance of 4 cm from the centre and OT ⊥ PQ. It is also given that, ∠AOB = ∠POQ = 60° and PT = 3 cm. What is the radius of the circle?

It is given that ∠AOB = ∠POQ = 60°

If two chords subtend equal angles at the centre of the circle, then the chords are equal in length.

∴AB = PQ

It is also given that AB is at a distance of 4 cm from the centre.

Chords of equal lengths are equidistant from the centre of the circle.

∴ OT = 4 cm

Applying Pythagoras Theorem in ΔOTP,

OT² + PT² = OP²

⇒ (4 cm)² + (3 cm)² = OP²

⇒ (16 + 9) cm² = OP²

⇒ OP² = 25 cm²

∴ OP = $\sqrt{25}$ cm = 5 cm

Thus, the radius of the circle is 5 cm.