The area of a trapezium is 475 cm2 and the height is 19 cm.

Let the side of a trapezium, $D C=x \mathrm{~cm}$

According to the question,

Other side, $A B=(x+4) \mathrm{cm}$

We know that,

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

[ $\because$ area of a trapezium $=\frac{1}{2}(a+b) \times h$ ]

$\Rightarrow \quad 475=\frac{1}{2}(x+x+4) \times 19$

$\Rightarrow \quad 475=\frac{(2 x+4) \times 19}{2}$

$\Rightarrow \quad \frac{(2 x+4)}{2}=\frac{475}{19}$

$\Rightarrow \quad \frac{2(x+2)}{2}=\frac{475}{19}$

$\Rightarrow \quad x+2=\frac{475}{19}$

$\Rightarrow \quad x+2=25$

$\therefore \quad x=23$

$\therefore$ Other $\quad$ side $=x+4=23+4=27 \mathrm{~cm}$

Hence, the parallel sides are $23 \mathrm{~cm}$ and $27 \mathrm{~cm}$.