Question

Diagonals of a trapezium ABCD with $AB\parallel DC$ interest each other at the point O. If $AB=2CD$, find the ratio of areas triangles AOB and COD.

Answer 1

$AB=CD$

$find:\frac{area\text{of}\Delta AOB}{area\text{of}\Delta \text{COD}}$

$\mathrm{Proof}:In\Delta \text{AOB and}\Delta \text{COD,}$

$\angle \text{A =}\angle \text{C}\left(alternateangle\right)$

$\angle o=\angle D\left(vertically\text{opp}\text{.}\right)$

$\therefore \frac{areaof\Delta AOB}{area\text{of}\Delta \text{COD}}={\left(\frac{AB}{CD}\right)}^2$

$\text{=}\frac{A{B}^2}{C{D}^2}$

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

$\text{=}\text{4:1}$