In the given figure ABCD is a trapezium in which AB = 40 m,

(a) 306 m2

In the given figure, AECD is a rectangle.

Length AE = Length CD = 28 m

Now

$B E=A B-A E=40-28=12 \mathrm{~m}$

Also,
AD = CE = 9 m

Area of trapezium $=\frac{1}{2} \times$ Sum of parallel sides $\times$ Distance between them

$=\frac{1}{2} \times(D C+A B) \times C E$

$=\frac{1}{2} \times(28+40) \times 9$

$=\frac{1}{2} \times 68 \times 9$

$=306 \mathrm{~m}^{2}$

In the given figure, if DA is perpendicular to AE, then it can be solved, otherwise it cannot be solved.