When the lamina ABCD is at rest, its centre of gravity G must lie vertically below the corner A by which it is hung.
By geometry, the distance between centre of gravity and centre of suspension i.e.,
AG = √2 a = L
M.I. of lamina about horizontal axis passing through C.G. and perpendicular to its plane is given by
Ig = m (2a)2 + (2a)2/12 = 2/3 ma2
Radius of gyration about this axis is given by
K2 = 2/3 a2
Then length of the equivalent simple pendulum