ГДЗ по геометрии 7 класс Атанасян Задание 467

\[\boxed{\mathbf{467.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\text{ABCD} - квадрат;\]

\[\text{EFMN} - ромб;\]

\[P_{\text{ABCD}} = P_{\text{EFNM}}.\]

\[\mathbf{Сравнить:}\]

\[S_{\text{ABCD}}\ и\ S_{\text{EFNM}}.\]

\[\mathbf{Решение.}\]

\[1)\ S_{\text{ABCD}} = AB^{2};\]

\[S_{\text{EFNM}} = \text{MN} \bullet \text{EH}.\]

\[2)\ P_{\text{ABCD}} =\]

\[= \text{AB} + \text{BC} + \text{CD} + \text{AD} = 4 \bullet \text{AB};\]

\[P_{\text{EFNM}} =\]

\[= \text{FE} + \text{FN} + \text{NM} + \text{EM} =\]

\[= 4 \bullet \text{EM};\]

\[Отсюда:\]

\[\text{AB} = \text{EM}.\]

\[3)\ \text{EH} < \text{EM}\ \]

\[\left( так\ как\ \text{EM} - гипотенуза\ \mathrm{\Delta}\text{EHM} \right).\]

\[4)\ EM^{2} > \text{EM} \bullet \text{EH};\]

\[S_{\text{ABCD}} > S_{\text{EFNM}}.\]

\[Ответ:S_{\text{ABCD}} > S_{\text{EFNM}}.\]