Step 1: Look at the figure and understand the question.

NOTE: In trapezium ABCD,

AB || DC

AB = 2CD.

Step 2: Prove that the triangles (AOB and COD) are similar

NOTE: In ∆AOB and ∆COD

\angle AOB = \angle COD (vertically opposite angles)

\angle OAB = \angle OCD (alternate interior angles)

∆AOB ~ ∆COD (by AA similarity)

Step 3: Recall the relation between the areas of the similar triangles.

THEOREM: The ratio of the areas of two similar triangles is equal to

the ratio of the squares of their corresponding sides

\frac{Area\ \triangle ABC}{Area\ \triangle COD}=(\frac{AB}{DC})^2

Step 4: Find the ratio of areas of triangles

NOTE: \frac{Area\ \triangle ABC}{Area\ \triangle COD}=(\frac{2\cdot DC}{DC})^2 (since AB = 2*DC)

\frac{Area\ \triangle ABC}{Area\ \triangle COD}=\frac{4}{1}\

∴ area (∆AOB) : area (∆COD) = 4 : 1.