Question

In the given figure, $ABCD$ is a trapezium in which $AB\parallel DC$. If $2AB=3DC$, find the ratio of the areas of $△AOB$ and $△COD$.

Answer 1

In $\Delta sAOBand\Delta COD.$
$\angle AOB=\angle COD$ (vert. opp . $\angle s$)
$\angle OAB=\angle OCD$
[Alternate $\angle s$ are equal as $AB\parallel CD$ (Given)]
$\therefore \Delta AOB\sim \Delta COD$
[By AA criterion of similarity]
$\Rightarrow \frac{ar\left(\Delta AOB\right)}{ar\left(\Delta COD\right)}=\frac{A{B}^2}{C{D}^2}=\frac{(2CD{)}^2}{C{D}^2}$
[$\therefore$ Ab = 2CD]
$=\frac{4C{D}^2}{C{D}^2}=41$