The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side.

Let the shortest side be $x \mathrm{~m}$.

Therefore, according to the question:

Hypotenuse $=(2 x-1) \mathrm{m}$

Third side $=(x+1) \mathrm{m}$

On applying Pythagoras theorem, we get:

$(2 x-1)^{2}=(x+1)^{2}+x^{2}$

$\Rightarrow 4 x^{2}-4 x+1=x^{2}+2 x+1+x^{2}$

$\Rightarrow 2 x^{2}-6 x=0$

$\Rightarrow 2 x(x-3)=0$

$\Rightarrow x=0$ or $x=3$

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,

Hypotenuse $=(2 \times 3-1)=5 \mathrm{~m}$

Third side $=(3+1)=4 \mathrm{~m}$