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(c) 8 cm

Let the distance between the parallel sides be $\mathrm{x} \mathrm{cm}$.

Then, area of the trapezium $=\left\{\frac{1}{2} \times(19+13) \times \mathrm{x}\right\} \mathrm{cm}^{2}$

$=\left(\frac{1}{2} \times 32 \times x\right) \mathrm{cm}^{2}$

$=16 x \mathrm{~cm}^{2}$

But it is given that the area of the trapezium is $128 \mathrm{~cm}^{2} .$

$\therefore 16 x=128$

$\Rightarrow x=\frac{128}{16}$

$\Rightarrow x=8 \mathrm{~cm}$