In a trapezium abcd, ab ║ cd, ab = 2 cd. if area of ∆ aob = 84 cm2 , find the area of ∆cod brainly
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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.
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.
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