Three rods each of mass $m$ and length hare joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of the triangle.

Moment of inertia of rod B C about an axis perpendicular to plane of triangle A B C and passing through the mid-point of rod B C (i.e. D ) is

${\mathrm{I}}_1=\frac{{\mathrm{ml}}^2}{12}$

$\mathrm{r}=\mathrm{BDtan}30°$

$\mathrm{r}=\left(\frac{l}{2}\right)\left(\frac{1}{\sqrt{3}}\right)=\frac{l}{2\sqrt{3}}$

From theorem of parallel axes, moment of inertia of this rod about the asked axis is

${\mathrm{I}}_2={\mathrm{I}}_1+{\mathrm{mr}}^2=\frac{{\mathrm{ml}}^2}{12}+\mathrm{m}{\left(\frac{\mathrm{l}}{2\sqrt{3}}\right)}^2=\frac{{\mathrm{ml}}^2}{6}$

$\therefore$ Moment of inertia of all the three rods is

$\mathrm{I}=3{\mathrm{I}}_2=3\left(\frac{{\mathrm{ml}}^2}{6}\right)=\frac{{\mathrm{ml}}^2}{2}$