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

The problem doesn't provide a figure, but based on the description, we can infer the following:

1. Triangle ABE is an isosceles triangle, meaning two of its sides are equal. In this case, AB = BE.

2. AB = 2AD. This means that AD is half the length of AB.

3. AE || DC. This means that line AE is parallel to line DC.

4. The area of triangle ABE is 72 cm².

To find the area of trapezium ABCD, we need to find the area of triangle ACD and add it to the area of triangle ABE.

Since AE is parallel to DC and AB = 2AD, triangle ACD is similar to triangle ABE. The ratio of their areas is the square of the ratio of their corresponding sides.

Since AB = 2AD, the ratio of the sides is 2:1. Therefore, the ratio of the areas is 4:1.

This means that the area of triangle ACD is 4 times the area of triangle ABE, which is 4 * 72 cm² = 288 cm².

Finally, the area of trapezium ABCD is the sum of the areas of triangles ABE and ACD, which is 72 cm² + 288 cm² = 360 cm².