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In the adjoining figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of △AOB and △COD.

In the adjoining figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of △AOB and △COD. Similarity, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

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Answer

Given, 2AB = 3DC

⇒ABDC=32\Rightarrow \dfrac{AB}{DC} = \dfrac{3}{2}.

Considering △AOB and △COD,

∠AOB = ∠COD (Vertically opposite angles are equal)

∠OAB = ∠OCD (Alternate angles are equal)

Hence by AA axiom △AOB ~ △COD.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

∴Area of △AOBArea of △COD=AB2DC2=3222=94\therefore \dfrac{\text{Area of △AOB}}{\text{Area of △COD}} = \dfrac{AB^2}{DC^2} \\[1em] = \dfrac{3^2}{2^2} \\[1em] = \dfrac{9}{4}

Hence, the ratio of area of △AOB : area of △COD is 9 : 4.

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