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In fig., O is the center of the circle of radius 5 cm. OP AB, OQ CD, AB||CD. = 6 cm, = 8 cm. Determine PQ.

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

Since the perpendicular from the center of the circle to a chord bisects the chord.

P and Q are the mid points of AB and CD

In right triangles OAP and OCQ, we have

OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2

52 = OP2 + 32 and 52 = OQ2 + 42

OP2 = 52 - 32 and OQ2 = 52 - 42

OP2 = 16 and OQ2 = 9

OP = 4 and OQ = 3

PQ = OP + OQ = 4 + 3 = 7 cm

Answered by Expert 4th June 2014, 3:23 PM
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