Find the expression that represents the length of the hypotenuse of a right triangle whose legs measure

First divide by 2 then excouleate by 1

It A just took the quiz

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(side 1)^2 + (side 2)^2 = hyp^2
 [m^2-n^2]^2  +  [2mn]^ = hyp^2

Then 

  m^4 - 2m^2n^2 + n^4 + 4m^2n^2 = hyp^2

m^4 + 2m^2n^2 + n^4 = hyp^2

Then hyp = +sqrt(m^4 + 2m^2n^2 + n^4) = +(m^2 + n^2)   (answer)

[tex]\bf c^2=a^2+b^2\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
------\\
a=m^2-n^2\\
b=2mn
\end{cases}\\\\\\ c^2=(m^2-n^2)^2+(\underline{2mn})^2
\\\\\\
c^2=m^4-2m^2n^2+n^4+\underline{2^2m^2n^2}
\\\\\\
c^2=m^4-2m^2n^2+n^4+{4m^2n^2}
\\\\\\
c^2=m^4+2m^2n^2+n^4\impliedby \textit{perfect square trinomial}
\\\\\\
c^2=(m^2+n^2)^2\implies c=\sqrt{(m^2+n^2)^2}\implies c=m^2+n^2[/tex]