Answer:
[tex] \sqrt{41} [/tex]
Step-by-step explanation:
[tex]let \: the \: hypothenuse =\: x \\ by \: pythagoras \: theorem \\ x {}^{2} = 4 {}^{2} + 5 {}^{2} \\ x {}^{2} = 16 + 25 \\ x {}^{2} = 41 \\ sqare \: root \: bothsides \\( \sqrt{x}) {}^{2} = \sqrt{41} \\ x = \sqrt{41} \\ lenght \: of \: the \: hypothenus = \sqrt{41} [/tex]
