#### Answer

the measurements of the legs are $(2+2\sqrt{2})$ $cm$ each and the hypotenuse is $(4+2\sqrt{2})$ $cm$

#### Work Step by Step

Let $x$ be the legs of an isosceles right triangle. Then $x+2$ 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+2)^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)(2)+(2)^2
\\
x^2+x^2=x^2+4x+4
\\
(x^2+x^2-x^2)-4x-4=0
\\
x^2-4x-4=0
.\end{array}
In the equation above, $a=
1
,$ $b=
-4
,$ and $c=
-4
.$ 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{-(-4)\pm\sqrt{(-4)^2-4(1)(-4)}}{2(1)}
\\
x=\dfrac{4\pm\sqrt{16+16}}{2}
\\
x=\dfrac{4\pm\sqrt{32}}{2}
\\
x=\dfrac{4\pm\sqrt{16\cdot2}}{2}
\\
x=\dfrac{4\pm\sqrt{(4)^2\cdot2}}{2}
\\
x=\dfrac{4\pm4\sqrt{2}}{2}
\\
x=\dfrac{\cancel2(2)\pm\cancel2(2)\sqrt{2}}{\cancel2}
\\
x=2\pm2\sqrt{2}
.\end{array}
Since $x$ is a measurement (and hence should be nonnegtive), then $
x=2+2\sqrt{2}
.$
Hence, the measurements of the legs, $x,$ are $(2+2\sqrt{2})$ $cm$ each and the hypotenuse, $x+2,$ is $(4+2\sqrt{2})$ $cm.$