Question

In the following figure, G is centroid of the triangle ABC.
Prove that : Area $\left(\Delta AGB\right)=\frac{1}{3}\times Area\left(\Delta ABC\right)$

Answer 1

Given that AD and BG are the medians we have to prove that

$ar\left(AGB\right)=\frac{1}{3}ar\left(ABC\right)$

In $△AGB,AGisthemedian\therefore △AGC,AGisthemedianarAGB=arAGC$

Similarly,

BG is the median

$\therefore ar△AGB=ar△BGD⟹ar△AGB=ar△AGC=ar△BGD\therefore ar△ABC=ar△AGB+ar△AGC+ar△BGD\therefore ar△ABC=3ar△\left(AGB\right)⟹ar\left(AGB\right)=\frac{1}{3}ar\left(ABC\right)$