# The legs of a right triangle are represented by x+sqrt2, x-sqrt2. What is the length of the hypotenuse?

## List item

The length of hypotenuse is $\sqrt{2 \left({x}^{2} + 2\right)}$
Let hypotenuse is $h$ and legs are ${l}_{1} \mathmr{and} {l}_{2}$
${h}^{2} = {l}_{1}^{2} + {l}_{2}^{2} = {\left(x + \sqrt{2}\right)}^{2} + {\left(x - \sqrt{2}\right)}^{2} = {x}^{2} + \cancel{2 \sqrt{2} x} + 2 + {x}^{2} - \cancel{2 \sqrt{2} x} + 2 = 2 {x}^{2} + 4 = 2 \left({x}^{2} + 2\right) \therefore h = \sqrt{2 \left({x}^{2} + 2\right)}$ [Ans]