The diagonals of a trapezium ABCD with AB || DC

Given that the diagonals of trapezium ABCD withintersect each other at point O and if, then we have to find the ratio of areas of triangle AOB and triangle COD.

We have the following diagram.

From the above figure, in triangles AOB and COD, we have

$\angle A O B=\angle C O D$

$\angle O A B=\angle O C D$

$\Rightarrow \triangle A O B \sim \triangle C O D$

Therefore, we can apply the area property of similar triangles.

$\frac{\text { Area }(\Delta A O B)}{\text { Area }(\Delta C O D)}=\frac{A B^{2}}{D C^{2}}$

$=\frac{(2 D C)^{2}}{D C^{2}}$    $($ Since $A B=2 C D)$

$=4 \frac{D C^{2}}{D C^{2}}$

$=4$

Hence, $\frac{\text { Area }(\triangle A O B)}{\text { Area }(\Delta C O D)}=\frac{4}{1}$