Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, and if $G$ is the centroid of a triangle $ABC$, then $G{A}^2+G{B}^2+G{C}^2=?$

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

Answer 1

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

Centroid = $2/3$ median

${GA}^2$ + ${GB}^2$ + $G{C}^2$ = ${\left(\frac{2}{3}\right)}^2$ $[{AD}^2+{BE}^2+{CF}^2]$

=$\frac{4}{9}[\frac{b^2+c^2}{2}-\frac{a^2}{4}+\frac{a^2+c^2}{2}-\frac{b^2}{4}+\frac{a^2+b^2}{2}-\frac{c^2}{4}]$

= $\frac{4}{9}\left[\frac{2b^2+2c^2-a^2+2a^2+2c^2-b^2+2a^2+2b^2-c^2}{4}\right]$

= $\frac{1}{9}[3a^2+3b^2+3c^2]$

=$\frac{a^2+b^2+c^2}{3}$

Hence, $\frac{a^2+b^2+c^2}{3}$ is the correct answer.