Question

Question 2
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Answer 1

ABCD is a trapezium with AB || DC.
Diagonals AC and BD intersect each other at point O.


In $\Delta AOBand\Delta COD$, we have
$\angle 1=\angle 2$ (Alternate angles)
$\angle 3=\angle 4$ (Alternate angles)
$\angle 5=\angle 6$ (Vertically opposite angle)
$\therefore \Delta AOB\sim \Delta COD$ [By AAA similarity criterion]

Now,
$\frac{Areaof\Delta AOB}{Areaof\Delta COD}$ =$\frac{A{B}^2}{C{D}^2}$
[If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides.]
$\frac{Areaof\Delta AOB}{Areaof\Delta COD}$ =$\frac{A{B}^2}{C{D}^2}=\frac{(2CD{)}^2}{C{D}^2}[Giventhat,AB=2CD]$
$\therefore \frac{Areaof\Delta AOB}{Areaof\Delta COD}$ = $\frac{4C{D}^2}{C{D}^2}=\frac{4}{1}$

Hence, the required ratio of the areas of $\Delta AOBand\Delta COD=4:1$