Решебник по геометрии 11 класс. Атанасян ФГОС 17

\[\boxed{\mathbf{17.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[DM = MB;\]

\[DN = NC;\]

\[AQ = QC;\]

\[AP = PB;\]

\[AD = 12\ см;\]

\[BC = 14\ см.\]

\[Найти:\]

\[P_{\text{MNQP}}.\]

\[Решение.\]

\[1)\ QP - средняя\ линия\ \]

\[\bigtriangleup ABC:\]

\[AQ = QC;\ \ AP = PB;\]

\[\bigtriangleup ABC \subset \alpha.\]

\[Отсюда:\]

\[QP \subset \alpha;\]

\[QP = \frac{\text{BC}}{2};\]

\[QP \parallel BC;\]

\[MN \subset \beta;\]

\[QP = MN = 7\ см.\]

\[2)\ MN - средняя\ \]

\[линия\ ⊿DBC:\]

\[DM = MB;\]

\[DN = NC;\]

\[⊿DBC \subset \beta.\]

\[Отсюда:\]

\[MN = \frac{\text{BC}}{2};\]

\[MN \parallel BC.\]

\[3)\ Аналогично:\]

\[QN = MP = 6\ см.\]

\[4)\ P_{\text{MNOP}} =\]

\[= 7 + 7 + 6 + 6 = 26\ см.\]

\[Ответ:26\ см.\]