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?

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?

Answer: Option D

Solution

Answer: Option D

Solution:

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