\[\boxed{\mathbf{467.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - квадрат;\]
\[\text{EFMN} - ромб;\]
\[P_{\text{ABCD}} = P_{\text{EFNM}}.\]
\[\mathbf{Сравнить:}\]
\[S_{\text{ABCD}}\ и\ S_{\text{EFNM}}.\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABCD}} = AB^{2};\]
\[S_{\text{EFNM}} = \text{MN} \bullet \text{EH}.\]
\[2)\ P_{\text{ABCD}} =\]
\[= \text{AB} + \text{BC} + \text{CD} + \text{AD} = 4 \bullet \text{AB};\]
\[P_{\text{EFNM}} =\]
\[= \text{FE} + \text{FN} + \text{NM} + \text{EM} =\]
\[= 4 \bullet \text{EM};\]
\[Отсюда:\]
\[\text{AB} = \text{EM}.\]
\[3)\ \text{EH} < \text{EM}\ \]
\[\left( так\ как\ \text{EM} - гипотенуза\ \mathrm{\Delta}\text{EHM} \right).\]
\[4)\ EM^{2} > \text{EM} \bullet \text{EH};\]
\[S_{\text{ABCD}} > S_{\text{EFNM}}.\]
\[Ответ:S_{\text{ABCD}} > S_{\text{EFNM}}.\]
\[\boxed{\mathbf{467.еуроки - ответы\ на\ пятёрку}}\]
\[Пусть\ BC = x,\ тогда\ AB = x + 8;\]
\[CD = x - 8;\]
\[AD = 3(x - 8).\]
\[Периметр\ равен\ 66\ см.\]
\[Составим\ уравнение:\]
\[x + x + 8 + x - 8 + 3(x - 8) =\]
\[= 66\ \]
\[3x + 3x - 24 = 66\]
\[6x = 66 + 24\]
\[6x = 90\ см\text{\ \ \ }\]
\[x = 15\ (см) - сторона\ \text{BC.}\]
\[CD = x - 8 = 15 - 8 = 7\ см.\]
\[AB = x + 8 = 15 + 8 = 23\ см\text{.\ \ \ \ }\]
\[AD = 3 \cdot (x - 8) = 3 \cdot 7 =\]
\[= 21\ см.\]
\[Ответ:7\ см;\ 15\ см;21\ см;23\ см.\]