In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of △AOB and △COD.

Given, AB || DC and AB = 2CD.

$\therefore \dfrac{AB}{CD} = \dfrac{2}{1}$.

Considering △AOB and △COD,

∠ AOB = ∠ COD (Vertically opposite angles)
∠ OCD = ∠ OAB (Alternate angles are equal)

Hence, by AA axiom △AOB ~ △COD.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △AOB}}{\text{Area of △COD}} = \dfrac{AB^2}{CD^2} \\[1em] \Rightarrow \dfrac{\text{Area of △AOB}}{\text{Area of △COD}} = \dfrac{2^2}{1^2} \\[1em] \Rightarrow \dfrac{\text{Area of △AOB}}{\text{Area of △COD}} = \dfrac{4}{1}.$

Hence, the ratio of the area of △AOB : area of △COD = 4 : 1.