In a trapezium AB

Given: AB ∥ CD

AB = 2CD ……….(i)

ar (∆ AOB) = 84 cm²

To find: ar (∆ COD)

In ∆ AOB and ∆ COD,

∠ AOB = ∠ COD [Vertically Opposite angles]

∠ OAB = ∠ OCD [Alternate interior angles (AB ∥ CD)]

∠ OBA = ∠ ODC [Alternate interior angles (AB ∥ CD)]

⇒ ∆ AOB ∼ ∆ COD [By AAA criterion]

Now,

∵ The ratios of the areas of two similar triangles are equal to the ratio of squares of any two corresponding sides.

∴ We have

Also, from (i), we have