[tex]w-width\\l-length\\\\2w+2l=50\to w+l=25\\w=\frac{2}{3}l\\\\substitute:\\\\\frac{2}{3}l+l=25\\\\\frac{5}{3}l=25\ \ \ \ /\cdot\frac{3}{5}\\\\l=15\ (m)\\\\w=\frac{2}{3}\cdot15=10\ (m)\\\\Area:A_{\fbox{ }}=wl\to A_{\fbox{ }}=10\cdot15=150\ (m^2)[/tex]
