• # question_answer The centre of mass of triangle shown in figure has coordinates A)  $x=\frac{h}{2},y=\frac{b}{2}$           B)  $x=\frac{b}{2},y=\frac{h}{2}$C)  $x=\frac{b}{3},y=\frac{h}{3}$         D)         $x=\frac{h}{3},y=\frac{b}{3}$

We can assume that three particles of equal mass m placed at the corners of triangle $\overset{\to }{\mathop{{{r}_{1}}}}\,=0\widehat{i}+0\widehat{j,}\overrightarrow{{{r}_{2}}}=b\widehat{i}+0\widehat{j}$and $\overset{\to }{\mathop{{{r}_{3}}}}\,=0\widehat{i}+h\widehat{j}$ $\therefore {{\widehat{r}}_{cm}}=\frac{{{m}_{1}}{{\overrightarrow{r}}_{1}}+{{m}_{2}}{{\overrightarrow{r}}_{2}}+{{m}_{3}}{{\overrightarrow{r}}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}=\frac{b}{3}\widehat{i}+\frac{h}{3}\widehat{j}$ i.e. coordinates of centre of mass is $\left( \frac{b}{3},\frac{h}{3} \right)$