In a trapezium ABCD, it is given that AB∥CD and AB = 2 CD. Its diagonals AC and BD intersect at a point O such that ar(△AOB) = 84 cm²
Find ar(△COD)

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Solution

In △AOB and COD
∠ABO = ∠CDO (Alternte angles in AB∥CD)
∠AOB = ∠COD (Vertically opposite angles)
By AA similarity criterion, △AOB ∼ △COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ AOB)}{area (△ COD)} = {(\frac{AB}{CD})}^{2} \Rightarrow \frac{84}{area (△ COD)} = {(\frac{2 CD}{CD})}^{2} \Rightarrow area (△ COD) = 12 {cm}^{2}$

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In a trapezium ABCD, it is given that AB∥CD and AB = 2 CD. Its diagonals AC and BD intersect at a point O such that ar(△AOB) = 84 cm2 Find ar(△COD)

In a trapezium ABCD, it is given that AB∥CD and AB = 2 CD. Its diagonals AC and BD intersect at a point O such that ar(△AOB) = 84 cm²
Find ar(△COD)