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Question
In Fig. O is the centre of the circle of radius 5 cm. OP ⊥ AB, OQ ⊥ CD, AB || CD, AB = 6 cm and CD = 8 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 the right-angled triangle OAP, we have
OA2 = OP2 + AP2
⇒ 52 = OP2 + 32
⇒ OP2 = 52 - 32 = 16 cm
⇒ OP = 4 cm2
In the right angled triangle OCQ we have
OC2 = OQ2 + CQ2
⇒ 52 = OQ2 + 42
⇒ OQ2 = 52 - 42 = 9 cm2
⇒ OQ = 3 cm
∴ PQ = PO - QO
∴ PQ = OP - OQ = (4 - 3) cm = 1 cm
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