The hypotenuse of a right-angled triangle is 20 metres.

Let one side of the right-angled triangle be $x \mathrm{~m}$ and the other side be $(x+4) \mathrm{m}$.

On applying Pythagoras theorem, we have:

$20^{2}=(x+4)^{2}+x^{2}$

$\Rightarrow 400=x^{2}+8 x+16+x^{2}$

$\Rightarrow 2 x^{2}+8 x-384=0$

$\Rightarrow x^{2}+4 x-192=0$

$\Rightarrow x^{2}+(16-12) x-192=0$

$\Rightarrow x^{2}+16 x-12 x-192=0$

$\Rightarrow x(x+16)-12(x+16)=0$

$\Rightarrow(x+16)(x-12)=0$

$\Rightarrow x=-16$ or $x=12$

The value of $x$ cannot be negative.

Therefore, the base is $12 \mathrm{~m}$ and the other side is $\{(12+4)=16 \mathrm{~m}\}$.