Question

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

Answer 1

Given,

$ABCD$ is a trapezium with $AB||CD$ $.......\left(1\right)$

And

$AB=2CD$ $......\left(2\right)$

In the triangles $AOB$ and $COD$,

$\angle DOC=\angle BOA$ [vertically opposite angles are equal]

$\angle CDO=\angle ABO$ [alternate interior angles ]

$\angle DCO=\angle BAO$

Thus,

$△AOB\approx △COD$

By the similarity rule, the ratio of the areas of similar triangles is the ratio of the square of corresponding sides.

therefore

$Area\left(△AOB\right):Area\left(△COD\right)=A{B}^2:C{D}^2$

$Area\left(△AOB\right):Area\left(△COD\right)=(2CD{)}^2:C{D}^2$

$Area\left(△AOB\right):Area\left(△COD\right)=4C{D}^2:C{D}^2$

$Area\left(△AOB\right):Area\left(△COD\right)=4:1$

Hence, this is the answer.