Решебник по геометрии 8 класс Мерзляк Задание 163

\[Схематический\ рисунок.\]

\[Дано:\]

\[AD = AB;\]

\[CE = BC;\]

\[DE = 18\ см;\]

\[\angle BDA = 15{^\circ};\]

\[\angle BEC = 36{^\circ}.\]

\[Найти:\]

\[\angle A;\ \angle B;\ \angle C;\]

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

\[Решение.\]

\[1)\ \mathrm{\Delta}DAB - равнобедренный:\]

\[\angle ABD = \angle ADB = 15{^\circ};\]

\[\angle ADB + \angle ABD + \angle BAD = 180{^\circ}\]

\[15{^\circ} + 15{^\circ} + \angle BAD = 180{^\circ}\]

\[\angle BAD = 150{^\circ}.\]

\[2)\ \mathrm{\Delta}ECB - равнобедренный:\]

\[\angle CBE = \angle CEB = 36{^\circ};\]

\[\angle CBE + \angle CEB + \angle ECB = 180{^\circ}\]

\[36{^\circ} + 36{^\circ} + \angle ECB = 180{^\circ}\]

\[\angle ECB = 108{^\circ}.\]

\[3)\ В\ \mathrm{\Delta}ABC:\]

\[\angle A = 180{^\circ} - \angle BAD = 30{^\circ};\]

\[\angle C = 180{^\circ} - \angle ECB = 72{^\circ};\]

\[\angle A + \angle B + \angle C = 180{^\circ}\]

\[30{^\circ} + \angle B + 72{^\circ} = 180{^\circ}\]

\[\angle B = 78{^\circ}.\]

\[P_{\text{ABC}} = AB + BC + AC;\]

\[P_{\text{ABC}} = DA + CE + AC;\]

\[P_{\text{ABC}} = DE = 18\ см.\]

\[Ответ:\ \ 30{^\circ};\ 72{^\circ};\ 78{^\circ};\ 18\ см.\]