Two circles having equal radius intersect each other such that each passes through the centre of the other. Sum of diameter of these two circles is 84 cm. What is the length of the common chord ?

$21\sqrt{3}$ cm

Diameter of one circle = \(\frac{84}{2}\) = 42 cm

Radius of the circle = \(\frac{42}{2}\) = 21 cm

According to the diagram,

AD = DB

O1O2 = 21

Again O1A = O2A = 21  [Radius of the circle]

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

O1D = O2D = \(\frac{21}{2}\)

AD = √(441 - \(\frac{441}{4}\))

= \(\frac{21√3}{2}\)

AB = 2 x \(\frac{21√3}{2}\) = 21√3

Therefore, length of common chord is 21√3 cm.