Two equal circles of radius 8 cm intersect each other in such a way that each passes through the centre of the other. The length of the common chord is:

$8\sqrt{3}$ cm

According to the diagram,

AD = DB

O1O2 = 8

Again O1A = O2 A = 8 (Radius of the circle)

\(\angle\)ADO1 = \({90}^\circ\)

O1D = O2D = 4

AD = \(\sqrt {64\; - \; 16 }\) = 4\(\sqrt {3}\)

AB = 2 x 4\(\sqrt {3}\) = 8\(\sqrt {3}\)

Therefore, the length of the common chord is 8\(\sqrt {3}\)