Question

$AD$ is a median of triangle $ABC$ and $O$ is a point on $AB$ such that area of $\Delta AOD$ : area of triangle $ADC=3:5$. Then, If $AO:OB=x:2$, find $x$

Answer 1

From the figure, $AD$ is the median of $△ABC$ which divides it into two equal triangles.

i.e, $△ADB=△ADC$

Given that $\frac{ar\left(△AOD\right)}{ar\left(ADC\right)}=\frac{3}{5}$

i.e, $ar\left(ABD\right)=5$

$\Rightarrow ar\left(AOD\right)+ar\left(BOD\right)=ar\left(ABD\right)$

$\Rightarrow ar\left(BOD\right)=2$

$\therefore \frac{ar\left(BOD\right)}{ar\left(ADB\right)}=\frac{2}{5}$

$\therefore AO:BO=3:2$

Hence, $x=3$