If the perimeter of a right-angled isosceles triangle is ( 4 2 + 4 ) cm, the length of the hypotenuse is :

Answer: Option A

Let the length of each of the sides containing the right angle be x cm
Then,
Hypotenuse :
$$\begin{array}{rl}& =\sqrt{x^2+x^2}cm\\ & =\sqrt{2x^2}cm\\ & =\sqrt{2}xcm\end{array}$$
Perimeter of the triangle :
$$\begin{array}{rl}& =\left(x+x+\sqrt{2}x\right)cm\\ & =\left(2x+\sqrt{2}x\right)cm\\ & =\sqrt{2}x\left(\sqrt{2}+1\right)cm\\ & \therefore \sqrt{2}x\left(\sqrt{2}+1\right)=\left(4\sqrt{2}+4\right)\\ & \Rightarrow \sqrt{2}x\left(\sqrt{2}+1\right)=4\left(\sqrt{2}+1\right)\\ & \Rightarrow \sqrt{2}x=4\\ & \Rightarrow x=2\sqrt{2}\end{array}$$
Hence, hypotenuse :
$$\begin{array}{rl}& =\left(\sqrt{2}\times 2\sqrt{2}\right)cm\\ & =4cm\end{array}$$