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

8 cm $4\sqrt{3}$ cm $8\sqrt{3}$ cm $8\sqrt{2}$ cm

$8\sqrt{3}$ cm

We know that, AD = DB Again O1A = O2A = 8 [Radius] ∠ADO1 = 90° O1D = O2D = 4 AD = √(64 - 16) = √48 = 4√3 AB = 2 × 4√3 = 8√3