Diagonal do quadrado = [tex]l \sqrt{2} [/tex]
[tex]3 \sqrt{3} =l \sqrt{2} [/tex]
[tex]l=\frac{3\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}[/tex]
[tex]Area=l^2+ \frac{l^2}{2} =(\frac{3\sqrt{6}}{2})^2+\frac{(\frac{3\sqrt{6}}{2})^2}{2}[/tex]
[tex]=\frac{9.6}{4}+\frac{\frac{9.6}{4}}{2}=\frac{9.6}{4}+\frac{9.6}{8}[/tex]
[tex]=\frac{54}{4}+\frac{27}{4}=\frac{81}{4}[/tex]
Hugs
