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Question
In following figure , AB , a chord of the circle is of length 18 cm. It is perpendicularly bisected at M by PQ.
Solution
Given : AB = 18 cm , MQ = 3 cm
To find : PQ
OQ = OA = r cm (say)
∴ OM = OQ = MQ = (r - 3) cm
AM = MB = 9 cm (PQ ⊥ AB)
In right Δ OMA ,
OM2 + MA2 = OA2
⇒ (r - 3)2 + 92 = r2
⇒ r2 - 6r + 9 + 81 = r2
⇒ 6r = 90
⇒ r = 15 cm
PQ = 2r
(Perpendicular bisector of a chord passes through the centre of the circe)
PQ = 2(15)
PQ = 30 cm
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