Two circles having radii $$r$$ units intersect each other in such a way that each of them passes through the centre of the other. Then the length of their common chord is

Solution (By Examveda Team)

According to question

Let the radius of the circle be = $$r$$
∴ DO = OC = $$\frac{r}{2}$$
In right angle ΔAOD
By using Pythagoras theorem
$$\eqalign{ & A{D^2} = O{D^2} + A{O^2} \cr & {r^2} = \frac{{{r^2}}}{4} + A{O^2} \cr & A{O^2} = {r^2} - \frac{{{r^2}}}{4} \cr & A{O^2} = \frac{{3{r^2}}}{4} \cr & AO = \frac{{\sqrt 3 r}}{2}\,\,\,\left( {\because AB = 2 \times AO} \right) \cr & AB = \frac{{\sqrt 3 r}}{2} \times 2 \cr & AB = \sqrt 3 r{\text{ units}} \cr} $$