Two circles of radii 7 cm and 5 cm intersect each other at A and B, and the distance between their centres is 10 cm. The length (in cm) of the common chord AB is:

$\frac{4 \sqrt{66}}{5}$

Let us consider that , O'C = a cm & CO = 10 - a

In a right angled triangle AO'C

O'A² = O'C² + AC²

7² = a² + AC²

AC² = 49 - a² 

In a right angled triangle AOC

OA² = OC² + AC²

5² = (10-a)² + AC²

AC² = 25 - (10-a)² 

Now,

49 - a² = 25 - (10-a)²

49 - a² = 25 - 100 - a² + 20a

a = \(\frac{31}{5}\)

Now, AC² = 49 - a²

= 49 - (\(\frac{31}{5}\))²

= 49 - \(\frac{ 961}{25}\)

= \(\frac{ 264}{25}\)

AC = $\frac{2 \sqrt{66}}{5}$

We know AB = 2AC

= 2 x $\frac{2 \sqrt{66}}{5}$

= $\frac{4 \sqrt{66}}{5}$ cm