Question

$ABCD$ is a trapezium in which $AB\parallel DC$ and $AB=2DC.$ $O$ is the point of intersection of the diagonals. The ratio of the areas of $△AOB$ and $△COD$ is:

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

Answer 1

The correct option is C $4:1$

Given, $AAB=2CD$ ....(1)

In $△$s $AOB$ and $COD$
$\angle AOB=\angle COD$ (vert. opp. $\angle$s)
$\angle OAB=\angle DCO$ ($DC\parallel AB,$ alt. $\angle$s are equal)
$\therefore$ $△AOB\sim △COD$ (AA similarity)
$\Rightarrow$ $\frac{ar\left(△AOB\right)}{ar\left(△COD\right)}=\frac{A{B}^2}{C{D}^2}$ {Ratio of areas of two similar $△$s is equal to the ratio of the squares of the corresponding sides}
$=\frac{{\left(2CD\right)}^2}{C{D}^2}=\frac{4C{D}^2}{C{D}^2}=\frac{4}{1}.$ ....From (1)