Question

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

 

Answer 1

Since AB || CD,

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ ΔAOB ∼ ΔCOD (By AAA similarity criterion)

`:.(ar(ΔAOB))/(ar(COD)) = ((AB)/(CD))^2`

Since AB = 2 CD

`:. (ar(triangleAOB))/(ar(triangleCOD)) =((2CD)/(CD))^2 =4/1=4:1`