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 $$I_0$$. If part ADE is removed, the moment of inertia of the remaining part about the same axis is $$\frac{NI_0}{16}$$ where N is an integer. Value of N is:

Mass and dimensions of each partitioned triangle: $$m_1 = \frac{M}{4},\ a_1 = \frac{a}{2}$$

Moment of inertia of the central inverted triangle about the common centroid $$G$$: $$I_{\text{central}} = \frac{I_0}{16}$$

By rotational symmetry, the three corner triangles have equal moments of inertia $$I_1$$ about $$G$$:

$$I_0 = I_{\text{central}} + 3I_1$$

$$I_0 = \frac{I_0}{16} + 3I_1 \implies 3I_1 = \frac{15I_0}{16} \implies I_1 = \frac{5I_0}{16}$$

Moment of inertia of the remaining portion after removing one corner piece ($$ADE$$):

$$I_{\text{remaining}} = I_0 - I_1 = I_0 - \frac{5I_0}{16} = \frac{11I_0}{16}$$