Question

Diagonals of a trapezium ABCD with AB ll DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD

Answer 1

First show that ∆AOB & ∆COD are similar by AAA similarity , then use property of area of similar triangles to get the required ratio.

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Fig. Is in the attachment.

Given:
ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.


In ΔAOB and ΔCOD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [By AAA similarity criterion]

Now,

ar (ΔAOB)/ ar(ΔCOD)= AB²/CD²

[If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]

= (2CD)²/CD²

[∴ AB = 2CD] given


∴ ar (ΔAOB)/ ar (ΔCOD)= 4CD²/CD²
= 4/1


Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

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Hope this will help you.......

Answer 2

Answer:

Step-by-step explanation:

In triangle AOB and triangle COD

Angle COD is equal to angle AOB(vertically opposite angles)

Angle CDB is equal to angle ABD(alternate interior angles)

Therefore triangle AOB similar to triangle COD(aa similarity criterion)

Now area of

AOB/ area of COD= AB square/CD square

=(2CD)^2/CD^2

=4CD^2/CD^2

=4/1

REQUIRED RATIO IS EQUAL TO 4 :1