If the perimeter of an isosceles right-angle triangle is 8($$\sqrt 2 $$ + 1) cm, then the length of the hypotenuse of the triangle is:

$$\eqalign{ & \left( {2 + \sqrt 2 } \right){\text{unit}} \to 8\left( {\sqrt 2 + 1} \right) \cr & 1\,{\text{unit}} \to \frac{{8\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}} = 4\sqrt 2 \cr & {\text{AC}} = 4\sqrt 2 \times \sqrt 2 = 8{\text{ cm}} \cr} $$