Question

Diagonals of a trapezium $ABCD$ with $AB||DC$ intersect each other at the point $O$. If $AB=2CD$, find the ratio of the areas of triangles $AOB$ and $COD$.

Answer 1

$ABCD$ is trapezium where $AB\parallel DC$ and diagonals intersect at $O$ and $AB=2CD$

In $△AOB$ and $△COD$

$\Rightarrow$ $\angle AOB=\angle COD$ [ Vertically opposite angles ]

$\Rightarrow$ $\angle OAB=\angle OCD$ [ Since $AB\parallel CD$ with $AC$ as traversal, alternate angles are equal ]

$\therefore$ $△AOB\sim △COD$ [ AA similarity ]

We know that if two triangle are similar, then ratio of areas is equal to squares of ratio of its corresponding sides.

$\Rightarrow$ $\frac{ar\left(△AOB\right)}{ar\left(△COD\right)}={\left(\frac{AB}{CD}\right)}^2$

$\Rightarrow$ $\frac{ar\left(△AOB\right)}{ar\left(△COD\right)}={\left(\frac{2CD}{CD}\right)}^2$ [ Since, $AB=2CD$ ]

$\Rightarrow$ $\frac{ar\left(△AOB\right)}{ar\left(△COD\right)}={\left(\frac{2}{1}\right)}^2$

$\therefore$ $\frac{△AOB}{△COD}=\frac{4}{1}$

$\therefore$ Ratio of the areas of triangles $AOB$ and $COD$ is $4:1$