#### Answer

the measurements of the legs are $(1+\sqrt{2})$ $m$ each and the hypotenuse is $(2+\sqrt{2})$ $m.$

#### Work Step by Step

Let $x$ be the legs of an isosceles right triangle. Then $x+1$ is the length of the hypotenuse.
Using $a^2+b^2=c^2$ or the Pythagorean Theorem, then
\begin{array}{l}\require{cancel}
x^2+x^2=(x+1)^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x^2+x^2=(x)^2+2(x)(1)+(1)^2
\\
x^2+x^2=x^2+2x+1
\\
(x^2+x^2-x^2)-2x-1=0
\\
x^2-2x-1=0
.\end{array}
In the equation above, $a=
1
,$ $b=
-2
,$ and $c=
-1
.$ Using the Quadratic Formula which is given by $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
,$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}}{2(1)}
\\
x=\dfrac{2\pm\sqrt{4+4}}{2}
\\
x=\dfrac{2\pm\sqrt{8}}{2}
\\
x=\dfrac{2\pm\sqrt{4\cdot2}}{2}
\\
x=\dfrac{2\pm\sqrt{(2)^2\cdot2}}{2}
\\
x=\dfrac{2\pm2\sqrt{2}}{2}
\\
x=\dfrac{\cancel2\pm\cancel2\sqrt{2}}{\cancel2}
\\
x=1\pm\sqrt{2}
.\end{array}
Since $x$ is a measurement (and hence should be nonnegative), then $
x=1+\sqrt{2}
.$
Hence, the measurements of the legs, $x,$ are $(1+\sqrt{2})$ $m$ each and the hypotenuse, $x+1,$ is $(2+\sqrt{2})$ $m.$