Question

The medians of a triangle ABC intersect at G.
To show: $Area\left(\Delta AGB\right)=\frac{1}{3}Area\left(\Delta ABC\right)$

Answer 1

The medians intersect at G.

We know G divides the median in the ratio 2 : 1

(i)Area $\Delta ABD=\frac{1}{2}area\Delta ABC$

(median divides a triangle into two ${\Delta }^{\prime }s$ of equal area)

$\left(ii\right)\frac{area\Delta AGB}{area\Delta ABD}=\frac{AG}{AD}=\frac{2}{3}$ .........$(△ABDand△ABG$ have a common vertex and their bases are in a straight line)

$\therefore$ area $\Delta AGB=\frac{2}{3}$ area $\Delta ABD$

$=\frac{2}{3}\times \frac{1}{2}$ area $\Delta ABC$ (from (i))

$=\frac{1}{3}$ area $\Delta ABC$