ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is  I0. If part ADE is removed, the moment of inertia of the remaining part about the same axis is NI016 where N is an integer. Value of N is __________ .

Divide the larger plate into four equal plates as shown.If  I0 is moment of inertia of bigger plate, we may writeI0=βma2 Where  β is a shape factor.  So each small plate will have moment of inertia about its own centroid as  =I=β m4a22=I016.So their contribution to I0  can be written as  I0=I016+3I016+ x     Using parallel axis theorem​I0−I016−3I016=3 x ⇒   x=I04We require  Iremaining plate=I0−Ismall plate =I0−I016+x=I0−I016+I04=11 I016