Two circles of radius of 6 cm each. They intersect each other such that each passes through the centre of the other. What is the length of the common chord?

$6\sqrt{3}$ cm

⇒ Let the centres be at A and B respectively.

Here, AC = CB = BD = DA = radius of circles = 6cm

CD is the common chord

Since they intersect each other such that each passes through the centre of the other, AB = 6 cm,

Hence, triangle ACB is an equilateral triangle whose AC = CB = AB = 6cm,

According to the concept,

CE is the height of triangle ACB

So, CE = \(\frac{√3}{2}\) x 6 = 3\(\sqrt {3 }\) cm,

Similarily, DE = 3\(\sqrt {3 }\) cm,

Now, CD = 3\(\sqrt {3 }\) + 3\(\sqrt {3 }\) = 6\(\sqrt {3 }\) cm

Therefore, the length of common chord is 6\(\sqrt {3 }\).