Find the value of x using the pythagorean theorem.
Do the side lengths form a pythagorean triple?

Answer:
[tex] {x}^{2} = {9}^{2} + {40}^{2} \\ {x}^{2} = 81 + 1600 \\ {x}^{2} = 1681 \\ x = \sqrt{1681} \\ x = 41[/tex]
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