Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corner is:

 We can easily find R using the Pythagoras theorem,

⇒R=√ a2−(a/2)2 =a√2

We know that for a square plate, the moment of inertia along a perpendicular axis passing through the centre of mass is,

⇒ Iperpendicular = ma2/6

So, using the parallel axis theorem, we get

⇒Iparallel = Iperpendicular + MR2

Here Iparallel is the moment of inertia along the parallel axis, Iperpendicular is the moment of inertia along the axis through the centre of mass, M is the mass of the object and R is the distance between the centre of mass and the parallel axis.

Substituting the value of R and Iperpendicular we get ⇒ Iparallel = Iperpendicular + MR2

∴Iparallel =ma2/6 + ma2/(√2)2 = 2/3ma2