Question

In a trapezium ABCD, AB is parallel to DC and AB = 2 DC. If AC and BD meet at O, then area of $\Delta$AOB is equal to:

• A. the area of $\Delta$COD
• B. twice the area of $\Delta$COD
• C. thrice the area of $\Delta$COD
• D. four times the area of $\Delta$COD

Answer 1

The correct option is D four times the area of $\Delta$COD

$ABCD$ is a trapezium $AB||CD$ and $AB=2CD$

In $\Delta AOB$ $COD$

$\angle AOB=\angle COD$ ( vertical angle )

$\angle DAB=\angle OCD$ (alternate angle )

$\therefore \Delta AOB\sim \Delta COD$ ($AA$ simiilar)

So,$\frac{area\left(\Delta AOB\right)}{area\left(\Delta COD\right)}=\frac{(AB{)}^2}{(CD{)}^2}$

$\Rightarrow \frac{area\left(\Delta AOB\right)}{area\left(\Delta COD\right)}=\frac{(2CD{)}^2}{(CD{)}^2}=4$

$\Rightarrow areaAOB=4\times area\Delta COD$

Then four times the area of triangle COD