Question

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

Answer 1

In $△AOB$ and $△COD$

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

$\angle OAB=\angle OCD$ [Since AB||CD with AC as traversal alternate angle are equal]

Hence, $△AOB\sim △COD$ [AA similarity]

We know that if two triangle re similar, then

Ration of areas is equal to square of ratio of its corresponding sides

Hence, $\frac{Area△AOB}{Area△COD}={\left(\frac{AB}{CD}\right)}^2$

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

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

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

$\therefore Area△AOB:Area△COD=4:1$

Hence ratio of areas is 4:1.