Question

In the figure, $ABCD$ is a trapezium in which $AB\parallel DC$ and $AB=2DC$. Determine the ratio of the areas of $△AOB$ and $△COD$.

• A. $4:1$
• B. $4:3$
• C. $8:1$
• D. $5:1$

Answer 1

The correct option is A $4:1$

Given: $AB\parallel CD$ and $AB=2CD$
In $△OAB$ and $△OCD$
$\angle AOB=\angle COD$ (Vertically opposite angles)
$\angle ODC=\angle OBA$ (Alternate angles)
$\angle OCD=\angle OAB$ (Alternate angles)
thus, $△OAB\cong △OCD$ (AAA rule)
$\frac{A\left(△OAB\right)}{A\left(△OCD\right)}=\frac{A{B}^2}{C{D}^2}$ (Similar triangle Property)
$\frac{A\left(△OAB\right)}{A\left(△OCD\right)}=\frac{(2CD{)}^2}{C{D}^2}$
$\frac{A\left(△OAB\right)}{A\left(△OCD\right)}=4:1$