[tex]\bf{La \hspace{0.1cm} diagonal\hspace{0.1cm} de\hspace{0.1cm} un\hspace{0.1cm} cuadrado\hspace{0.1cm} de \hspace{0.1cm}lado\hspace{0.1cm} \bf{n}, de \hspace{0.1cm} acuerdo \hspace{0.1cm} con \hspace{0.1cm} el \hspace{0.1cm} teorema \hspace{0.1cm} de \hspace{0.1cm} Pit\'agoras, mide:[/tex]
[tex]\bf{La \hspace{0.1cm} diagonal\hspace{0.1cm} de\hspace{0.1cm} un\hspace{0.1cm} cuadrado\hspace{0.1cm} de \hspace{0.1cm}lado\hspace{0.1cm} \bf{n}, de \hspace{0.1cm} acuerdo \hspace{0.1cm} con \hspace{0.1cm} el \hspace{0.1cm} teorema \hspace{0.1cm} de \hspace{0.1cm} Pit\'agoras, mide:[/tex]
[tex]\boxed{\bf{d=\sqrt{n^{2}+n^{2}} = \sqrt{2n^{2}}=(\sqrt{2})n = n(\sqrt{2}) }}[/tex]
[tex]\bf{es \hspace{0.1cm} decir,\hspace{0.1cm} n \hspace{0.1cm} veces \hspace{0.1cm}\sqrt{2}[/tex]
[tex]\bf{entonces...}[/tex]
[tex]\bf{Tenemos \hspace{0.1cm} que: \hspace{0.1cm}\\ \hspace{0.1cm}lado = 6 \hspace{0.1cm}cm[/tex]
[tex]\bf{Ahora:}[/tex]
[tex]d= \sqrt{a^{2}+b^{2} }[/tex]
[tex]d= \sqrt{6^{2}+6^{2} }[/tex]
[tex]d= \sqrt{36+36} }[/tex]
[tex]\bf{Respuesta:[/tex]
[tex]\boxed{\bf{d= \sqrt{72} }}}[/tex]
[tex]\boxed{\bf{d=8.48 }}}[/tex]
[tex]\bf{Por \hspace{0.1cm} lo \hspace{0.1cm} tanto,\hspace{0.1cm} la\hspace{0.1cm}respuesta \hspace{0.1cm} es \hspace{0.1cm} el\hspace{0.1cm} inciso\hspace{0.1cm}3)[/tex].