Question

If $G$ is the centroid of a $△ABC,$ then ${GA}^2+{GB}^2+{GC}^2$ is equal to

• A. $(a^2+b^2+c^2)$
• B. $\frac{1}{3}(a^2+b^2+c^2)$
• C. $\frac{1}{2}(a^2+b^2+c^2)$
• D. $\frac{1}{3}{(a+b+c)}^2$

Answer 1

The correct option is B $\frac{1}{3}(a^2+b^2+c^2)$

By Appolloneous theorem,
${GB}^2+{GC}^2=2[{GD}^2+{DC}^2]$
$\Rightarrow {GB}^2+{GC}^2=2[{\left(\frac{1}{2}GA\right)}^2+{\left(\frac{a}{2}\right)}^2]$
$\Rightarrow {GB}^2+{GC}^2=\frac{{GA}^2}{2}+\frac{a^2}{2}$ .............$\left(1\right)$
Similarly, we have
${GC}^2+{GA}^2=\frac{{GB}^2}{2}+\frac{b^2}{2}$ .............$\left(2\right)$
${GA}^2+{GB}^2=\frac{{GC}^2}{2}+\frac{c^2}{2}$ .............$\left(3\right)$
Adding eqns$\left(1\right),\left(2\right)$ and $\left(3\right)$ we get
$2[{GA}^2+{GB}^2+{GC}^2]=\frac{{GA}^2+{GB}^2+{GC}^2}{2}+\frac{a^2+b^2+c^2}{2}$
$\Rightarrow ({GA}^2+{GB}^2+{GC}^2)(2-\frac{1}{2})=\frac{a^2+b^2+c^2}{2}$
$\therefore {GA}^2+{GB}^2+{GC}^2=\frac{a^2+b^2+c^2}{3}$