In a trapezium AB

Let us consider AOB and COD.

∠AOB = ∠COD (∵ vertically opposite angles)

∠OBA = ∠ODC (∵ alternate interior angles)

∠OAB = ∠OCD (∵ alternate interior angles)

We know that if in two triangles, corresponding angles are equal,

then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar (AAA criteria).

So, ΔAOB ≅ ΔCOD.

Given, AB = 2CD and ar(ΔAOB) = 84 cm²

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

∴ ar(AOB)/ar(COD) = (AB/CD)²

⇒ 84cm²/ar(ΔCOD) = (2CD/CD)²

⇒ 84cm²/ar(ΔCOD) = 4

⇒ ar(ΔCOD) = 84cm²/4

⇒ ar(ΔCOD) = 21cm²

ar(ΔCOD) = 21cm²