An equilateral triangle ABC is cut from a thin solid sheet of wood (see figure). D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is $I_0$. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is $I$. Then
Answer: (A) $I = \dfrac{15}{16}I_0$
For a uniform equilateral triangular sheet, $I$ about the centroidal perpendicular axis is proportional to $Ma^2$.
Triangle DEF has side $a/2$ and area (so mass) $M/4$, and its centre is also G:
$$I_{DEF} = \frac14\times\left(\frac12\right)^2 I_0 = \frac{I_0}{16}$$
$$I = I_0 - \frac{I_0}{16} = \frac{15}{16}I_0$$
Solution by Sreeraj P, M.Sc Physics