Two identical circles, each with radius \(r\), intersect in such a way that each circle passes through the center of the other. The centers of the circles are labeled A and B, and they intersect at points P and Q. The area of triangle \(\triangle APB\) will be:

Q.Two identical circles, each with radius \(r\), intersect in such a way that each circle passes through the center of the other. The centers of the circles are labeled A and B, and they intersect at points P and Q. The area of triangle \(\triangle APB\) will be: (a) \(\cfrac{\sqrt3}{4}r^2\) (b) \(\cfrac{\sqrt3}{2}r^2\) (c) \(\cfrac{\sqrt3}{3}r^2\) (d) \(\sqrt3 r^2\)

Answer: A

AB = AP = PB = radius of the circle = \(r\) ∴ Triangle \(\triangle\)APB is an equilateral triangle with each side of length \(r\) ∴ The area of the equilateral triangle \(\triangle\)APB is \(= \frac{\sqrt{3}}{4}r^2\)