Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.

Area enclosed by the given figure $=$ (Area of trapezium FEDC) $+$ (Area of square $\mathrm{ABCF}$ )

$=\left[\left\{\frac{1}{2} \times(6+20) \times 8\right\}+(20 \times 20)\right] \mathrm{cm}^{2}$

$=(104+400) \mathrm{cm}^{2}$

$=504 \mathrm{~cm}^{2}$

Hence, the area enclosed by the figure is $504 \mathrm{~cm}^{2}$.