In the given figure area of isosceles triangle ABE is 72 cm² and BE = AB and AB = 2AD, AE || DC, then what is the area (in cm²) of the trapezium ABCD?

Solution (By Examveda Team)

\[\Delta OAD \sim \Delta ABE\left( \begin{gathered} \because AE\,||\,DC \hfill \\ \therefore \angle E = \angle D \hfill \\ \angle B = \angle A \hfill \\ \end{gathered} \right)\]
$$\eqalign{ & \frac{{{\text{ar}}{\text{. }}\Delta OAD}}{{{\text{ar}}{\text{. }}\Delta ABE}} = \frac{1}{4} \cr & \Delta OBC \sim \Delta ABE \cr & \therefore \frac{{{\text{ar}}{\text{. }}\Delta OBC}}{{{\text{ar}}{\text{. }}\Delta ABE}} = \frac{9}{4} \cr & \because 4 = 72{\text{ c}}{{\text{m}}^2} \cr & \therefore 1 = 18{\text{ c}}{{\text{m}}^2} \cr & 9 = 162{\text{ c}}{{\text{m}}^2} \cr & \therefore {\text{ ar}}{\text{. of }}\square ADEC \cr & \Rightarrow 162 - \left( {72 + 18} \right) = 72{\text{ c}}{{\text{m}}^2} \cr & \therefore {\text{Area of trapezium }}ADCB \cr & = 72 + 72 \cr & = 144{\text{ c}}{{\text{m}}^2} \cr} $$