ABCD is a trapezium in which AB parallel CD a | vedclass

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(D) In a trapezium $ABCD$ with $AB \parallel CD$,triangles $\Delta AOB$ and $\Delta COD$ are similar by $AA$ similarity criterion because $\angle OAB

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(D) In a trapezium $ABCD$ with $AB \parallel CD$,triangles $\Delta AOB$ and $\Delta COD$ are similar by $AA$ similarity criterion because $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (alternate interior angles).The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Therefore,$\frac{	ext{Area}(\Delta AOB)}{	ext{Area}(\Delta COD)} = \left( \frac{AB}{CD} ight)^2$.Given $AB = 2CD$,we have $\frac{AB}{CD} = \frac{2}{1}$.Thus,$\frac{	ext{Area}(\Delta AOB)}{	ext{Area}(\Delta COD)} = \left( \frac{2}{1} ight)^2 = \frac{4}{1}$.The ratio is $4:1$.

$ABCD$ is a trapezium in which $AB \parallel CD$ and $AB = 2CD$. If its diagonals intersect each other at $O$,the ratio of the area of triangle $AOB$ and $COD$ is

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