The two sides of a right triangle are 10 square root 2

\(?\,\,\,:\,\,\,3{\rm{rd}}\,\,{\rm{side}}\,\,\,\,\left( {\,{\rm{length}}\,} \right)\)

\(\left. \matrix{\\
{\left( {10\sqrt 2 } \right)^2} = {\left( {5\sqrt 2 } \right)^2} + {x^2}\,\,\, \Rightarrow \,\,\,\,200 = 50 + {x^2}\,\,\,\,\mathop \Rightarrow \limits^{x\, > \,0} \,\,\,\,x = 5\sqrt 6 \,\,\,\,\, \hfill \cr \\
{y^2} = {\left( {5\sqrt 2 } \right)^2} + {\left( {10\sqrt 2 } \right)^2}\,\,\, \Rightarrow \,\,\,\,{y^2} = 250\,\,\,\,\mathop \Rightarrow \limits^{y\, > \,0} \,\,\,\,y = 5\sqrt {10} \,\, \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\,:\,\,\,5\sqrt 6 \,\,{\rm{or}}\,\,5\sqrt {10} \,\,\,\, \Rightarrow \,\,\,\,\left( D \right)\)

Important: in the case of right triangles, the \(a^2=b^2+c^2\) equality guarantees that the (positive) numbers obtained will be lengths of a viable triangle!
(In short: there is no need to check whether each side is less than the sum of the other two, for instance.)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.