\[\boxed{\mathbf{467.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - квадрат;\]
\[\text{EFMN} - ромб;\]
\[P_{\text{ABCD}} = P_{\text{EFNM}}.\]
\[\mathbf{Сравнить:}\]
\[S_{\text{ABCD}}\ и\ S_{\text{EFNM}}.\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABCD}} = AB^{2};\]
\[S_{\text{EFNM}} = \text{MN} \bullet \text{EH}.\]
\[2)\ P_{\text{ABCD}} =\]
\[= \text{AB} + \text{BC} + \text{CD} + \text{AD} = 4 \bullet \text{AB};\]
\[P_{\text{EFNM}} =\]
\[= \text{FE} + \text{FN} + \text{NM} + \text{EM} =\]
\[= 4 \bullet \text{EM};\]
\[Отсюда:\]
\[\text{AB} = \text{EM}.\]
\[3)\ \text{EH} < \text{EM}\ \]
\[\left( так\ как\ \text{EM} - гипотенуза\ \mathrm{\Delta}\text{EHM} \right).\]
\[4)\ EM^{2} > \text{EM} \bullet \text{EH};\]
\[S_{\text{ABCD}} > S_{\text{EFNM}}.\]
\[Ответ:S_{\text{ABCD}} > S_{\text{EFNM}}.\]