Three rods, each of mass M and length L, are welded together to form an equilateral triangle as shown in Fig. The moment of inertia of the triangular frame about an axis passing through its centroid O and perpendicular to its plane is

 

Refer fig

AB = BC = AC = L, OD = OE = OF = x (say).

x = OD = BD tan 30°

            =$\frac{L}{2\sqrt{3}}$

Using the parallel axes theorem, the moment of inertia of rod BC about

 $$\begin{gathered} O=\left(\frac{M{L}^2}{12}+M{x}^2\right) \\ =\frac{M{L}^2}{12}+\frac{M{L}^2}{12}=\frac{M{L}^2}{6} \end{gathered}$$

Moment of inertia of the complete frame about

$O=3\times \frac{M{L}^2}{6}=\frac{M{L}^2}{2}$