Three identical rods each of length L and mass m are joined to form a U-shape as shown in figure .The distance of center of mass of the combined system from the corner  'O' is

${\mathrm{m}}_1=\mathrm{m}, {\mathrm{m}}_2=\mathrm{m}, {\mathrm{m}}_3=\mathrm{m}$

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}$

$=\frac{0\times \mathrm{m}\times {\displaystyle \frac{\mathrm{L}}{2}}\times \mathrm{m}\times \mathrm{L}}{\mathrm{m}+\mathrm{m}+\mathrm{m}}$

$=\frac{\mathrm{m}\left(\frac{3\mathrm{L}}{2}\right)}{3\mathrm{m}}=\frac{3\mathrm{L}}{6}=\frac{\mathrm{L}}{2}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{my}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}$

$=\left(\frac{\mathrm{m}\times \frac{\mathrm{L}}{2}+\mathrm{m}\times 0+\mathrm{m}\times \frac{\mathrm{L}}{2}}{3\mathrm{m}}\right)$

$=\frac{{\displaystyle \frac{\mathrm{mL}}{2}}+\frac{\mathrm{mL}}{2}}{3\mathrm{m}}=\frac{\frac{2\mathrm{mL}}{2}}{3\mathrm{m}}=\frac{\mathrm{L}}{3}$

${\mathrm{r}}_{\mathrm{cm}}={\mathrm{x}}_{\mathrm{cm}}\mathrm{i}+{\mathrm{y}}_{\mathrm{cm}}\mathrm{j}$

$=\frac{\mathrm{L}}{2}\mathrm{i}+\frac{\mathrm{L}}{3}\mathrm{j}$

$=\sqrt{{\left(\frac{\mathrm{L}}{2}\right)}^2+{\left(\frac{\mathrm{L}}{3}\right)}^2}$

$=\sqrt{\frac{{\mathrm{L}}^2}{4}+\frac{{\mathrm{L}}^2}{9}}=\sqrt{\frac{9{\mathrm{L}}^2+4{\mathrm{L}}^2}{36}}$

$=\sqrt{\frac{13{\mathrm{L}}^2}{36}}=\frac{\sqrt{13}\mathrm{L}}{6}$