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In a circle centred at O, AB is a chord and C is any point on AB, such that OC is perpendicular to AB. If the length of the chord is 16 cm and OC = 6 cm, the radius of circle is:
Let the radius of the circle = r
Given, AB = 16 cm, OC = 6 cm
OC is perpendicular to the chord AB.
In a circle, perpendicular from centre of the circle to the chord bisects the chord.
$$\Rightarrow$$ AC = BC = 8 cm
In $$\triangle$$OAC,
AC$$^2$$ + OC$$^2$$ = OA$$^2$$
$$\Rightarrow$$ 8$$^2$$ + 6$$^2$$ = r$$^2$$
$$\Rightarrow$$ 64 + 36 = r$$^2$$
$$\Rightarrow$$ r$$^2$$ = 100
$$\Rightarrow$$ r = 10 cm
$$\therefore\ $$Radius of the circle = 10 cm
Hence, the correct answer is Option C
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