Tick (✓) the correct answer:

(a) 45 cm

Let the length of the parallel sides be $3 \mathrm{x} \mathrm{cm}$ and $4 \mathrm{x} \mathrm{cm}$, respectively.

Then, area of the trapezium $=\left\{\frac{1}{2} \times(3 x+4 x) \times 12\right\} \mathrm{cm}^{2}$

$=\left(\frac{1}{2} \times 7 x \times 12\right) \mathrm{cm}^{2}$

$=42 x \mathrm{~cm}^{2}$

But it is given that the area of the trapezium is $630 \mathrm{~cm}^{2}$.

$\therefore 42 x=630$

$\Rightarrow x=\frac{630}{42}$

$\Rightarrow x=15 \mathrm{~cm}$

$L$ ength of the parallel sides $=(3 \times 15) \mathrm{cm}=45 \mathrm{~cm}$

$(4 \times 15) \mathrm{cm}=60 \mathrm{~cm}$

Hence, the shorter of the parallel sides is $45 \mathrm{~cm}$.