Question

In figure, ABCD is a trapezium in which AB || DC and AB = 2DC. Determine the ratio of the areas of $△$AOB and $△$COD.

Answer 1

In triangle AOB and COD, we have

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

and $\angle$OAB = $\angle$OCD [ Alternate angles ]

So , be AA- criterian of similarity, we have

$△$AOB ~ $△$COD

$\Rightarrow$ $\frac{areaof\left(△AOB\right)}{areaof\left(△COD\right)}=\frac{A{B}^2}{D{C}^2}$

$\Rightarrow$ $\frac{areaof\left(△AOB\right)}{areaof\left(△COD\right)}=\frac{(2DV{)}^2}{(DC{)}^2}$ = $\frac{4}{1}$

Hence , Area ($△$AOB) : Area ($△$ COD) = 4 :1