Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

In trapezium ABCD,

AB is parallel to CD and AB = 2 CD ---------- (1)

Diagonals AC and BD intersect at ‘O’

In ΔAOB and ΔCOD,

∠AOB = ∠COD (vertically opposite angles)

∠ABO = ∠CDO (alternate interior angles)

⇒ ΔAOB ~  ΔCOD (AA criterion)

⇒ Area of ΔAOB / Areaof ΔCOD = (AB)² / (CD)² [Theorem 6.6]

(2CD)² / (CD)² = 4CD² / CD² = 4 / 1 [From equation (1)]

Thus, Area of ΔAOB : Area of ΔCOD = 4:1