Let mass of triangular lamina =m, and length of side =l, then moment of inertia of lamina about an axis passing through centroid G perpendicular to the plane. I0∝ml2 I0=kml2
Let moment of inertia of DEF=I1 about G So, I1∝(
m
4
)(
l
2
)2∝
ml2
16
or I1=
I0
16
Let IADE=IBDF=IEFC=I2 ∴3I2+I1=I0⇒3I2+
I0
16
=I0⇒I2=
5I0
16
Hence, moment of inertia of DECB i.e., after removal part ADE =2I2+I1=2(