Question

In this figure, ABCD is a trapezium in which AB || DC and AB = 3DC. Determine the ratio of the areas of $△$ AOB and $△$COD.

• A. 4 : 1
• B. 16 : 1
• C. 3 : 4
• D. 9 : 1

Answer 1

The correct option is D

9 : 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

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

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

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