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

Answer :

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


In \(\triangle\) AOB and \(\triangle\) COD, we have
\(\angle\) 1 = \(\angle\) 2 (Alternate angles)
\(\angle\) 3 = \(\angle\) 4 (Alternate angles)
\(\angle\) 5 = \(\angle\) 6 (Vertically opposite angle)
\(\triangle\) AOB ~ \(\triangle\) COD [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,

\(\frac{Area of ( \triangle AOB)}{Area of ( \triangle COD) } = \frac{AB^2}{CD^2} \)
= \(\frac{(2CD)^2}{CD^2} \) [ AB = CD]
\(\frac{Area of (\triangle AOB)}{Area of (\triangle COD)} \)
= \(\frac{4CD^2}{CD} = \frac{4}{1} \)

Hence, the required ratio of the area of \(\triangle\) AOB and \(\triangle\) COD = 4:1

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