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Question
AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle.
Solution
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.
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