Question

In figure, ABCD is a trapezium in which AB || DC and AB = 3DC. 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, by AA-criterion of similarity, we have

$△$ AOB ~ $△$COD

$\frac{areaof△AOB}{areaof△COD}$ = $\frac{(AB{)}^2}{D{C}^2}$

$\frac{areaof△AOB}{areaof△COD}$ = $\frac{(3DC{)}^2}{D{C}^2}$ = $\frac{9}{1}$

Hence, Area ($△$AOB ) : Area ($△$COD) = 9 :1.