Moment of Inertia & Radius of Gyration question from JEE Main 2020 (4 Sep, Shift 1)

Answer: 11

1. For a triangular lamina of side $a$ about its centroid, $I_0 = \frac{Ma^2}{12}$.
2. ADE is a similar triangle of side $\frac{a}{2}$ and mass $\frac{M}{4}$, so about its own centroid $I' = \frac{M}{4}\cdot\frac{(a/2)^2}{12} = \frac{I_0}{16}$.
3. With height $h = \frac{\sqrt{3}a}{2}$: G is $\frac{2h}{3}$ below A and the centroid of ADE is $\frac{h}{3}$ below A, so $d = \frac{h}{3} = \frac{a}{2\sqrt{3}}$, $d^2 = \frac{a^2}{12}$.
4. Parallel axes: $I_{ADE} = \frac{I_0}{16} + \frac{M}{4}\cdot\frac{a^2}{12} = \frac{I_0}{16} + \frac{I_0}{4} = \frac{5I_0}{16}$.
5. Remaining part: $I_0 - \frac{5I_0}{16} = \frac{11I_0}{16}$, so N = 11.