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:
Answer & explanation
Correct answer: option 1
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\(\sqrt {3}\)