Answer: Option A
Let the length of each of the sides containing the right angle be x cm
Then,
Hypotenuse :
$$\eqalign{
& = \sqrt {{x^2} + {x^2}} \,cm \cr
& = \sqrt {2{x^2}} \,cm \cr
& = \sqrt 2 x\,cm \cr} $$
Perimeter of the triangle :
$$\eqalign{
& = \left( {x + x + \sqrt 2 x} \right)cm \cr
& = \left( {2x + \sqrt 2 x} \right)cm \cr
& = \sqrt 2 x\left( {\sqrt 2 + 1} \right)cm \cr
& \therefore \sqrt 2 x\left( {\sqrt 2 + 1} \right) = \left( {4\sqrt 2 + 4} \right) \cr
& \Rightarrow \sqrt 2 x\left( {\sqrt 2 + 1} \right) = 4\left( {\sqrt 2 + 1} \right) \cr
& \Rightarrow \sqrt 2 x = 4 \cr
& \Rightarrow x = 2\sqrt 2 \cr} $$
Hence, hypotenuse :
$$\eqalign{
& = \left( {\sqrt 2 \times 2\sqrt 2 } \right)cm \cr
& = 4\,cm \cr} $$