If $\mathrm{G}$ is the centroid of $∆\mathrm{ABC}$ and $\mathrm{BC}=3, \mathrm{CA}=4, \mathrm{AB}=5$ then $\mathrm{BG}=$

$\mathrm{Given} \mathrm{BC}=3, \mathrm{CA}=4, \mathrm{AB}=5 \mathrm{and} \mathrm{G} \mathrm{is} \mathrm{centriod} \mathrm{of} \mathrm{triangle} \mathrm{ABC}.$

$$\begin{gathered} \mathrm{Now} \mathrm{length} \mathrm{of} \mathrm{median} \mathrm{through} \mathrm{B} \\ BE=\frac{1}{2}\sqrt{2{\mathrm{a}}^2+2{\mathrm{c}}^2-{\mathrm{b}}^2} \\ =\frac{1}{2}\sqrt{2\left(9\right)+2\left(25\right)-16} \\ =\frac{1}{2}\sqrt{52} \\ \mathrm{BE}=\frac{1}{2}\times 2\sqrt{13}=\sqrt{13} \\ \mathrm{Now} \mathrm{BG}=\frac{2}{3}\left(\mathrm{BE}\right) \\ =\frac{2}{3}\sqrt{13} \\ =\frac{\sqrt{52}}{3} \end{gathered}$$