In figure 3.89, line l touches the circle with centre O at point P. Q is the mid point of radius OP. RS is a chord through Q such that chords RS || line l. If RS = 12 find the radius of the circle.

The radius of the circle will bisect the chord RS. Therefore, RQ = QS = 1/2 × 12 = 6
Let the radius of circle be r,
Now, in Δ OQS, we have,
RQ = 6
OR = r
OQ = 1/2 r
Applying Pythagoras theorem, we get,
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3r2 = 4 × 36
r2 = 4 × 12 = 48
r = √48 units
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