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In a trapezium, AB||CD. If the ratio of the sides AB:CD is 2:1, what is the ratio of the area of the $Δ A O B$ : $Δ C O D$ ?

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1:2

2:1

1:4

4:1

answer is D.

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Detailed Solution

Given that, in a trapezium, AB||CD and the ratio of sides AB:CD is 2:1.
Let us consider the below trapezium ABDC.
Question Image

In trapezium ABDC, AB||CD.
So, in

Δ A O B and Δ D O C

∠ A O B = ∠ C O D

[vertically opposite angles]

∠ O A B = ∠ O D C

[alternate interior angles, as DC || AB and AD is transversal]

∠ O B A = ∠ O C D

[alternate interior angles, as DC || AB and BC is transversal]
By AAA similarity criteria,

Δ A O B ∼ Δ D O C

.
We know that the ratio of the areas of similar triangles and the ratio of the squares of their corresponding sides are proportion.
Using this property we get,

a r (Δ A O B) a r (Δ D O C) = A B 2 C D 2 = A O 2 O D 2 = O B 2 O C 2

…………..(1)
Now given the fact that AB:CD = 2:1, we can say that AB = 2CD.
Using this in the equation (1) we get,

⇒ a r (Δ A O B) a r (Δ D O C) = A B 2 C D 2 ⇒ a r (Δ A O B) a r (Δ D O C) = (2 C D) 2 C D 2 ⇒ a r (Δ A O B) a r (Δ D O C) = 2212 ⇒ a r (Δ A O B) a r (Δ D O C) = 41

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