If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle (in cm) is:
Given : AB = 16 cm and OC = 15 cm
To find : OB = $$r$$ = ?
Solution : The line from the centre of the circle to the chord bisects it at right angle.
=> AC = BC = $$\frac{1}{2}$$ AB
=> BC = $$\frac{16}{2}=8$$ cm
In $$\triangle$$ OBC,
=> $$(OB)^2=(BC)^2+(OC)^2$$
=> $$(OB)^2=(8)^2+(15)^2$$
=> $$(OB)^2=64+225=289$$
=> $$OB=\sqrt{289}=17$$ cm
=> Ans - (C)
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