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(b) $23 \mathrm{~cm}$

Let the length of the parallel sides be $x \mathrm{~cm}$ and $(x+6) \mathrm{cm}$, respectively.

Then, area of the trapezium $=\left\{\frac{1}{2} \times(x+x+6) \times 9\right\} \mathrm{cm}^{2}$

$=\left\{\frac{1}{2} \times(2 x+6) \times 9\right\} \mathrm{cm}^{2}$

$=4.5(2 x+6) \mathrm{cm}^{2}$

$=(9 x+27) \mathrm{cm}^{2}$

But it is given that the area of the trapezium is $180 \mathrm{~cm}^{2}$.

$\therefore 9 x+27=180$

$\Rightarrow 9 x=(180-27)$

$\Rightarrow 9 x=153$

$\Rightarrow x=\frac{153}{9}$

$\Rightarrow x=17$

Therefore, the length of the parallel sides are $17 \mathrm{~cm}$ and $(17+6) \mathrm{cm}$, which is equal to $23 \mathrm{~cm}$.

Hence, the length of the longer parallel side is $23 \mathrm{~cm}$.