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Question
In Fig. O is the centre of the circle with radius 5 cm. OP⊥ AB, OQ ⊥ CD, AB || CD, AB = 8 cm and CD = 6 cm. Determine PQ.
Solution
Join OA and OC.
Since the perpendicular from the centre of the circle to a chord bisects the chord. Therefore, P and Q are midpoints of AB and CD respectively.
Consequently, AP = PB = `1/2"AB"` = 3 cm.
and CQ = QD = `1/2`CD = 4 cm
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) cm = 7 cm.
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