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.