Given chords AB = 6 cm, CD = 12 cm and AB ∥ CD Draw OP ⊥ AB. Let it intersect CD at Q and AB at P ∴ AP = PB = 3 cm and CQ = DQ = 6 cm [Since perpendicular drawn from the centre of the chord bisects the chord] Let OD = OB = r In right ΔOQD,r2=x2+62 [By Pythagoras theorem ] r2=x2+36 → (1) In right ΔOPB, r2=(x+3)2+32 [By Pythagoras theorem ] r2=x2+6x+9+9=x2+6x+18 → (2) From (1) and (2) we get x2+36=x2+6x+18 ⇒ 6x=18 x = 3 Put x = 3 in (1), we get r2=32+36=9+36=45 ∴ r=√45=3√5 cm