Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Join OA and OC. Let the radius of the circle be r cm and O be the centre. Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P, O and Q are collinear. So, PQ = 6 cm. Let OP = x. Then, OQ = (6 – x) cm. And OA = OC = r. Also, AP = PB = 2.5 cm and CQ = QD = 5.5 cm (Perpendicular from the centre to a chord of the circle bisects the chord). In right triangles OAP and OCQ, we have OA² = OP² + AP² and OC² = OQ² + CQ²
∴r² = x² + (2.5)² (1)
and r² = (6 – x)² + (5.5)² (2)
⇒ x² + (2.5)² = (6 – x)² + (5.5)²
⇒ x² + 6.25 = 36 – 12x + x² + 30.25
12x = 60
∴ x = 5
Putting x = 5 in (1), we get
r² = 5² + (2.5)² = 25 + 6.25 = 31.25
⇒ r² = 31.25
⇒ r = 5.6
Hence, the radius of the circle is 5.6 cm