In a trapezium ABCD, it is given that AB || CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar (ΔA0B) = 84 cm2. Find ar(ΔCOD).

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
Couldn't generate an explanation.
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