Two equal circles intersect each other at point A and B, whose centers are O and O’. OO’ = 24 mm and AB = 10 mm, then find the area (in sq. mm) of the circle.

AB is chord to each of the circle and AB = 10 mm and OO' = 24 mm

Let radius of each circle = $$r$$ mm

A line drawn from the centre of the circle perpendicular to the chord bisects it in two parts.

=> AC = $$\frac{10}{2}=5$$ mm

Similarly, OC = $$\frac{24}{2}=12$$ mm

Now, in $$\triangle$$ OAC

=> $$(OA)^2 = (OC)^2 + (AC)^2$$

=> $$OA=\sqrt{(12)^2+(5)^2}$$

=> $$OA = \sqrt{144+25} = \sqrt{169}$$

=> $$OA = 13$$ mm

$$\therefore$$ Area of circle = $$\pi r^2$$

= $$3.14\times(13)^2=530.66$$ $$mm^2$$

=> Ans - (A)