Two identical circles intersect each other such that each passes through the centre of the other. Length of common chord is $16\sqrt{3}$ cm. What is the radius of each circle?

16 cm

Length of common chord BD is 16\(\sqrt {3 }\)

AC = AB = BC = CD = DA = r

So, ABCD is a rhombus

So, \(\Delta \)ABC & \(\Delta \)ACD are equilateral triangle

Then area of equilateral triangle = \(\frac{\sqrt { 3}}{4}\)\( { (side)}^{3 } \)

= \(\frac{\sqrt { 3}}{4}\)\( { r}^{2} \)

So, sum of area of both triangle

ABC & ADC = 2 x \(\frac{\sqrt { 3}}{4}\) x \( { r}^{2} \)

= \(\frac{\sqrt { 3}}{4}\) \( { r}^{2} \)      ..(1)

Now area of rhombus

= \(\frac{1}{2}\) x AC x BD

= \(\frac{1}{2}\) x r x 16\(\sqrt {3 }\)

= 8\(\sqrt {3 }\)r      ..(2)

Now equate 1st and 2nd equation

= \(\frac{\sqrt { 3}}{4}\) \( { r}^{2} \) = 8\(\sqrt {3 }\)r

= r = 16 cm.

Therefore, radius = 16 cm.