$\text{ABCD}$ is a trapezium $\overline{\text{AB}}\text{ //}\overline{\text{CD}}$ . If $\text{AB}=20cm,\text{BC}=8cm,\text{CD}=10cm$ and

$\text{AD}=6cm$. Then the area of $\text{ABCD}$ is

Draw $CE$ parallel to $DA$

$\therefore$$AECD$ is a parallelogram

$AE=CD=10cm$

$AD=CE=6cm$

Then $EB=20-10=10cm$

Area of $\Delta BEC=\sqrt{s(s-a)(s-b)(s-c)}$

$s=\frac{6+8+10}{2}=\frac{24}{2}=12cm$

$=\sqrt{12(12-6)(12-8)(12-10)}$

$=\sqrt{12\times 6\times 4\times 2}$

$=\sqrt{2\times 6\times 6\times 2\times 2\times 2}$

$=\sqrt{(2\times 2)\times (6\times 6)(2\times 2)}$

$=2\times 6\times 2$

$=24c{m}^2$

Area of $\Delta BEC=\frac{1}{2}\left(\text{base}\right)\times \left(\text{height}\right)$=

 $\Rightarrow \frac{1}{2}\times \left(EB\right)\times h=24c{m}^2$

$\Rightarrow \frac{1}{2}\times 10\times h=24c{m}^2$

$h=\frac{24\times 2}{10}=\frac{48}{10}=4.8cm$

$h=4.8cm$

Area of ABCD = $\frac{1}{2}(a+b)\times h$

$=\frac{1}{2}(10+20)\times 4.8$

$=\frac{1}{2}\left(30\right)\times 4.8$

$=15\times 4.8$

$=72sq.cm$