What is the length of a chord which is at a distance of 6 cm from the center of a circle with radius 10 cm?
Explanation:
A perpendicular drawn from the center of the circle to the chord bisects the chord.
In the above figure, OD bisects chord AB i.e. AD = DB
In ∆ODB, ∠ODB = 90°
Using Pythagoras theorem, we get,
OB2 = DB2 + OD2
∴ 102 = DB2 + 62
∴ DB2 = 100 – 36 = 64
∴ DB = 8 cm
Length of chord AB = 2 × DB = 2 × 8 = 16 cm
Hence, option (b).
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