From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and ∠BCD = 90°,

Since, BFEC is a quarter of a circle.
Hence, BC = EC = 3.5 cm
Now, DC = DE + EC = 2 + 3.5 = 5.5 cm
Area of shaded region = Area of the trapezium ABCD − Area of the quadrant BFEC

$=\frac{1}{2} \times(\mathrm{AB}+\mathrm{DC}) \times \mathrm{BC}-\frac{1}{4} \times \pi(\mathrm{EC})^{2}$

$=\frac{1}{2} \times(3.5+5.5) \times 3.5-\frac{1}{4} \times \frac{22}{7} \times(3.5)^{2}$

$=6.125 \mathrm{~cm}^{2}$

Hence, the area of the shaded region is 6.125 cm2 .