In trapezium ABCD, AB ll CD and AB = 2CD, its diagonals interest at O.  If area of triangle AOB = 84 cm², then area of triangle COD is?

21 sq. cm

DC  ll AB

\(\angle\)DCA = \(\angle\)CAB

\(\angle\)CDB = \(\angle\)DBA

\(\angle\)COD = \(\angle\)AOB

Therefore,

⇒ ΔCOD ∼ ΔAOB

∴ \(\frac{area \;of \;ΔCOD}{area\; of\; ΔAOB}\) = \(\frac{CD^2}{AB^2}\) = \(\frac{CD^2}{4CD^2}\) = \(\frac{1}{4}\)

Area of ΔCOD  = \(\frac{1}{4}\) × area of ΔAOB

                       = \(\frac{1}{4}\) × 84

                       = 21 cm²