[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2 \qquad \begin{cases} c=\stackrel{hypotenuse}{84}\\ a=\stackrel{adjacent}{27}\\ b=\stackrel{opposite}{79}\\ \end{cases}\implies 27^2+79^2=84^2 \\\\\\ 729+6241=7056\implies 6970\ne 7056~~\bigotimes~\hfill \boxed{27^2+79^2\ne 84^2} \\\\\\ ~\hspace{34em}[/tex]
c² = a² + b² is true for a right-triangle, if that equation is untrue, namely c² ≠ a² + b², then is not a right-triangle, so, no dice.
