Question 68
The length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm is
Solution
Given : OB is the radius of circle = 13 cm and OC = 12 cm
To find : AB = ?
Solution : The line from the centre of the circle to the chord bisects it at right angle.
=> AC = BC = $$\frac{1}{2}$$ AB
In $$\triangle$$ OBC,
=> $$(BC)^2=(OB)^2-(OC)^2$$
=> $$(BC)^2=(13)^2-(12)^2$$
=> $$(BC)^2=169-144=25$$
=> $$BC=\sqrt{25}=5$$ cm
$$\therefore$$ AB = $$2 \times$$ BC
= $$2 \times 5=10$$ cm
=> Ans - (A)
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