Question

AB and CD are two parallel chords of a circle such that

AB=10cm and CD=24 cm. If the chords are on the same

sides of the centre and the distance between them is 17 cm,

find the radius of the circle.
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Answer 1

Step-by-step explanation:

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

OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2

⇒ r2 = x2 + 52 …….(i)

and r2 = (17 – x)2 + 122 …..(ii)

⇒ x2 + 52 = ( 17 – x)2 + 122 [On equating the values of r2 ]

⇒ x2 + 25 = 289 – 34x + x2 + 144

⇒ 34x = 408 ⇒ x = 12 cm.

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

r2 = 122 + 52 = 169 ⇒ r = 13 cm

hence, the radius of the circle is 13 cm.

Hope its help..