Two parallel chords AB and CD of lengths 10 cm and 24 cm respectively lie on opposite sides of the center O of a circle. If the distance between the chords AB and CD is 17 cm, find the radius of the circle.

Q.Two parallel chords AB and CD of lengths 10 cm and 24 cm respectively lie on opposite sides of the center O of a circle. If the distance between the chords AB and CD is 17 cm, find the radius of the circle.

Two parallel chords AB = 10 cm and CD = 24 cm lie on opposite sides of the center O of a circle. The distance between the chords AB and CD is PQ = 17 cm. Let the radius of the circle be \(r\) cm, so OA = OC = \(r\). Let OP = \(x\) cm, then OQ = \(17 - x\) cm. Now, AP = \(10/2 = 5\) cm and CQ = \(24/2 = 12\) cm. So, From triangle OAP: \(r^2 = 5^2 + x^2\) — (i) From triangle OCQ: \(r^2 = 12^2 + (17 - x)^2\) — (ii) Equating both expressions for \(r^2\): \(25 + x^2 = 144 + 289 - 34x + x^2\) ⇒ \(25 = 433 - 34x\) ⇒ \(34x = 408\) ⇒ \(x = \frac{408}{34} = 12\) Substituting \(x = 12\) into equation (i): \(r^2 = 5^2 + 12^2 = 25 + 144 = 169\) ⇒ \(r = \sqrt{169} = 13\) ∴ The radius of the circle is 13 cm.