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Question

In a trapezium ABCD, O is the point of intersection of AC and BD, AB ∥ CD and AB = 2 × CD. If area of AOB = 84 cm2, find the area of ΔCOD.

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Solution

In Δ AOB and Δ COD, we have:
OAB = OCD [ Alternate interior angles]
OBA = ODC [ Alternate interior angles]
∴ ΔAOB ∼ Δ COD [ By AA similarity]
arAOBarCOD=AB2CD2=2CD2CD2 [ Since AB = 2 × CD]
=4×CD2CD2=4
arCOD=14×arAOB=14×84cm2 = 21cm2
Hence, area of ΔCOD is 21 cm2.

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