A uniform metre rod is bent into L shape at mid point with the bent arms at 90° to each other. The distance of the centre of mass from the bent point is

center of mass of one m/2 is at (0.25,0) and other part is at (0,0.25)
So center of mass after bending ${\mathrm{r}}_{\mathrm{can}}=\frac{(\mathrm{m}/2)(0.25)\widehat{\mathrm{i}}+(\mathrm{m}/2)(0.25)\widehat{\mathrm{j}}}{\mathrm{m}}=\frac{\widehat{\mathrm{i}}+\widehat{\mathrm{j}}}{4}$
$\begin{array}{l}{\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}=\frac{\widehat{\mathrm{i}}+\widehat{\mathrm{j}}}{8}\\ {\mathrm{r}}_{\mathrm{cm}}=\sqrt{\frac{1}{8^2}+\frac{1}{8^2}}=\frac{\sqrt{2}}{8}=\frac{1}{4\sqrt{2}}\end{array}$