**AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the … between them is 17 cm, find the radius…**

###### Answers:

Let O be the centre of the given circle and let it’s radius be r cm. Draw OP ⊥ AB and OQ ⊥ CD. Since, AB ∥ CD. Therefore, points P , O and Q are collinear. So, PQ = 17 cm.

Let OP = x cm, Then, OQ = (17 – x) cm

Join OA and OC. Then, OA = OC = r.

Since, the perpendicular from the centre to a chord of the circle bisects the chord.

∴ AP = PB = 5 cm and CQ = QD = 12 cm

In right triangles OAP and OCQ, we have

OA^{2} = OP^{2} + AP^{2} and OC^{2} = OQ^{2} + CQ^{2}

⇒ r^{2} = x^{2} + 5^{2} **…….(i)**

and r^{2} = (17 – x)^{2} + 12^{2} **…..(ii)**

⇒ x^{2} + 5^{2} = ( 17 – x)^{2} + 12^{2} [On equating the values of r^{2} ]

⇒ x^{2} + 25 = 289 – 34x + x^{2} + 144

⇒ 34x = 408 ⇒ x = 12 cm.

Putting x = 12 cm in (i), we get

r^{2} = 12^{2 }+ 5^{2} = 169 ⇒ r = 13 cm

hence, the radius of the circle is 13 cm.