Question.
Diagonals of trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at the point $\mathrm{O}$. If $\mathrm{AB}=2 \mathrm{CD}$, find the ratio of the areas of triangles $\mathrm{AOB}$ and COD.
Diagonals of trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at the point $\mathrm{O}$. If $\mathrm{AB}=2 \mathrm{CD}$, find the ratio of the areas of triangles $\mathrm{AOB}$ and COD.
Solution:
In $\Delta \mathrm{AOB}$ and $\Delta \mathrm{COD}$,
$\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate interior angles)
$\angle \mathrm{OBA}=\angle \mathrm{ODC}$ (Alternate interior angles)
$\therefore$ By AA, similarity
$\Delta \mathrm{AOB} \sim \Delta \mathrm{COD}$
So, $\frac{\operatorname{ar} . \Delta \mathrm{AOB}}{\operatorname{ar} . \Delta \mathrm{COD}}=\left(\frac{\mathrm{AB}}{\mathrm{CD}}\right)^{2}$
$=\left(\frac{2}{1}\right)^{2} \quad\{\because \mathrm{AB}=2 \mathrm{CD}\}$
$=4: 1$
In $\Delta \mathrm{AOB}$ and $\Delta \mathrm{COD}$,
$\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate interior angles)
$\angle \mathrm{OBA}=\angle \mathrm{ODC}$ (Alternate interior angles)
$\therefore$ By AA, similarity
$\Delta \mathrm{AOB} \sim \Delta \mathrm{COD}$
So, $\frac{\operatorname{ar} . \Delta \mathrm{AOB}}{\operatorname{ar} . \Delta \mathrm{COD}}=\left(\frac{\mathrm{AB}}{\mathrm{CD}}\right)^{2}$
$=\left(\frac{2}{1}\right)^{2} \quad\{\because \mathrm{AB}=2 \mathrm{CD}\}$
$=4: 1$