Question

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

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is D $4:1$

Given : ABCD is trapezium with $AB\parallel CD$ ......(i)
and $AB=2CD$ .....(ii)
In the $△AOB$ and $△COD$,
$\angle DOC=\angle BOA$ ...[ Vertically opposite angles are equal]
$\angle CDO=\angle ABO$ ...[ Alternate interior angles]
$\angle DCO=\angle BAO$
Thus, $△AOB\sim △COD$ ....AAA test
By the similarity rule, the ratio of the areas of the similar triangles is the ratio of the square of corresponding sides.
Therefore,
$A\left(△AOB\right)$ : $A\left(△COD\right)=A{B}^2:C{D}^2$
$A\left(△AOB\right)$ : $A\left(△COD\right)$ $={\left(2CD\right)}^2:C{D}^2$
$A\left(△AOB\right)$ : $A\left(△COD\right)$ $=4C{D}^2:C{D}^2$
$A\left(△AOB\right)$ : $A\left(△COD\right)$ $=4:1$