Three thin rods each of mass M and length I are welded so as to form an equilateral triangle. What is the moment of inertia about the axis passing through the centre and perpendicular to its plane ?

See fig. Let I_G be the moment of inertia of each rod about an axis passing through its centre of gravity (say D or E or F) and perpendicular to its Plane. So

${\mathrm{I}}_{\mathrm{G}}=\left({\mathrm{ML}}^2/12\right)$
Suppose O be the median of the triangle. It cuts the perpendicular distance in the ratio of 2 : 1' Let OF = h. Then
$\begin{array}{l}\mathrm{h}=\mathrm{OF}=\mathrm{AF}/3=\frac{\mathrm{Lsin}{60}^{\circ }}{3}\\ =\frac{\mathrm{L}}{3}\times \frac{\sqrt{3}}{2}=\frac{\mathrm{L}}{2\sqrt{3}}\end{array}$
Using the theorem of parallel axes, we have
$\begin{array}{l}{\mathrm{I}}_{\text{pop }}=3\left[{\mathrm{I}}_{\mathrm{G}}+{\mathrm{Mh}}^2\right]\\ =3\left[\frac{{\mathrm{ML}}^2}{12}+\frac{{\mathrm{ML}}^2}{12}\right]=\frac{{\mathrm{ML}}^2}{2}\end{array}$