AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and AB||CD. If the distance between AB and CD is 3 cm, find the radius
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Since, AB ∥ CD and OP ⊥ AB, OQ ⊥ CD. Therefore points O, Q and P are collinear. Clearly, PQ = 3 cm.
Let OQ = x cm. Then, OP = (x + 3) cm
In right triangles OAP and OCQ, we have
OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2
⇒ r2 = (x + 3)2 + 32 and r2 = x2 + 62
[∵ AP = ½ AB = 3 cm and CQ = ½ CD = 6 cm
⇒ (x + 3)2 + 32 = x2 + 62 (on equating the value of r2)
⇒ x2 + 6x + 9 + 9 = x2 + 36
⇒ 6x = 18 ⇒ x = 3 cm
Putting the values of x in r2 = x2 + 62, we get
r2 = 32 + 62 = 45
⇒ r = √45 cm = 6.7 cm
Hence, the radius of the circle is 6.7 cm.
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