In the adjoining figure, ABCD is a trapezium in which AB || DC;

∆ADL is a right angle triangle.

So, $D L=\sqrt{\left(5^{2}-4^{2}\right)}=\sqrt{9}=3 \mathrm{~cm}$

Similarly, in ∆BMC, we have:

$M C=\sqrt{\left(5^{2}-4^{2}\right)}=\sqrt{9}=3 \mathrm{~cm}$

∴ DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm

Now, ar(trapezium. $A B C D)=\frac{1}{2} \times($ sum of parallel sides $) \times($ distance between them $)$

$=\frac{1}{2} \times(7+13) \times 4$

$=40 \mathrm{~cm}^{2}$

Hence, $D C=13 \mathrm{~cm}$ and area of trapezium $=40 \mathrm{~cm}^{2}$