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

Therefore, according to the question:

Hypotenuse $=(2 x-1) m$

Third side $=(x+1) m$

On applying Pythagoras theorem, we get:

$\begin{array}{l}
(2 x-1)^{2}=(x+1)^{2}+x^{2} \\
\Rightarrow 4 x^{2}-4 x+1=x^{2}+2 x+1+x^{2} \\
\Rightarrow 2 x^{2}-6 x=0 \\
\Rightarrow 2 x(x-3)=0 \\
\Rightarrow x=0 \text { or } x=3
\end{array}$

The length of the side cannot be $0 ;$ therefore, the shortest side is $3 \mathrm{~m}$.

Therefore,

Hypotenuse $=(2 \times 3-1)=5 m$

Third side $=(3+1)=4 m$