Question

ABCD is a trapezium in which AB parallel to CD and AB= CD and diagonals cut each other at point 'O' .Find the ratio of area of triangle AOB and area of triangle COD.​

Answer 1

Answer:

given: ABCD is a trapezium with AB||CD.......(1)

and AB = 2 CD......(2)

In the triangles AOB and COD;

∠DOC = ∠ BOA [vertically opposite angles are equal]

∠ CDO = ∠ ABO [alternate interior angles ]

∠ DCO = ∠ BAO

thus Δ AOB ≈ Δ COD.

by the similarity rule, the ratio of the areas of the similar triangles is the ratio of the square of corresponding sides.

therefore

area(Δ AOB) : area(Δ COD)=AB^2:CD^2

area(Δ AOB) : area(Δ COD)=(2CD)^2:CD^2

area(Δ AOB) : area(Δ COD)=4CD^2:CD^2

area(Δ AOB) : area(Δ COD) = 4 : 1.

Step-by-step explanation:

dear user use this example to solve and instead of AB=2CD which I have mentioned you use AB=CD.

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