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Question
Two chords AB and CD of lengths 6cm and 12cm are drawn parallel inside the circle. If the distance between the chords of the circle is 3cm, find the radius of the circle.
Solution
AP = PB = 3cm
CQ = QD = 6cm (Perpendicular from centre to a chord bisects the chord)
OA = OC = r (say)
Let OP = x, ∴ OQ = 3 - x
In right Δ OQC,
By Pythagoras theorem,
OC2 = OQ2 + CQ2
r2 = (3-x)2 + 62 ----(1)
Similarly, In Δ OPA,
OA2 = AP2 + PO2
r2 = 32 + x2 ----(2}
From (1) and {2}
(3-x)2 + 62 = 32+ x2
-6x + 36 = 0
x = 6
from {2}
r2 = 32 + 62 = 9 + 36 = 45
r = `3 sqrt 5`
Thus , radius of the circle is `3 sqrt 5` cm
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