Question

If the vertices of a triangle are $A(2,1),B(-2,-2),C(1,0),$ then sum of squares of the lengths of the medians of the triangle is

• A. $25$
• B. $40$
• C. $30$
• D. $45$

Answer 1

The correct option is C $30$



Here $D,E,F$ are the mid-points of $BC,AC,AB$ respectively.
From the figure,
$A{D}^2=\frac{41}{4}$
$B{E}^2=\frac{74}{4}$
$C{F}^2=\frac{5}{4}$
Thus $A{D}^2+B{E}^2+C{F}^2=30$

Alternative method:
We know that in a triangle, $3$ times sum of the squares of side lengths $=4$ times sum of the squares of the median lengths.
i.e., $A{D}^2+B{E}^2+C{F}^2=\frac{3}{4}(A{B}^2+B{C}^2+C{A}^2)$
$=\frac{3}{4}(25+13+2)=30$