Answer:
[tex]\textsf{C.}\quad \sf 2\sqrt{41}\; m[/tex]
Step-by-step explanation:
To find the value of x, which is the hypotenuse of the given right triangle, we can use the Pythagorean Theorem:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Pythagorean Theorem}}\\\\c^2=a^2+b^2\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a$ and $b$ are the legs of the right triangle.}\\\phantom{ww}\bullet\;\textsf{$c$ is the hypotenuse (longest side) of the right triangle.}\\\end{array}}[/tex]
In this case:
Substitute the values into the formula and solve for x:
[tex]x^2=8^2+10^2\\\\x^2=64+100\\\\x^2=164\\\\x=\sqrt{164}\\\\x=\sqrt{4 \cdot 41}\\\\x=\sqrt{4} \cdot \sqrt{41}\\\\x=2 \cdot \sqrt{41}\\\\x=2\sqrt{41}[/tex]
Therefore, the value of x is:
[tex]\LARGE\boxed{\boxed{\sf 2\sqrt{41}\;m}}[/tex]
