Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD,

For Competitive Exams

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

Solution:

Given

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

In ΔAOB and ΔCOD, we have

$\angle1 = \angle2$ (Alternate angles)

$\angle3 = \angle4$ (Alternate angles)

$\angle5 = \angle6$ (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides.

Therefore,

$\dfrac{Area \: of \: (ΔAOB)}{Area \: of \: (ΔCOD} = \dfrac{AB^2}{CD^2}$

= $\dfrac{(2CD)^2}{CD^2}$ [∴ AB = 2CD]

∴ $\dfrac{Area \: of \: (ΔAOB)}{Area \: of \: (ΔCOD)}$

= $\dfrac{4CD^2}{CD^2} = \dfrac{4}{1}$

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

Additional Questions

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Diagonals of a trapezium ABCD with AB \|\| DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.	Answer
In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that $\dfrac{ar (ABC)}{ar (DBC)} = \dfrac {AO}{DO}$		Answer

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