Решебник по геометрии 9 класс Мерзляк Задание 167

\[Схематический\ рисунок.\]

\[Дано:\]

\[O - центр\ окружности,\ \]

\[вписананной\ в\ \angle C;\]

\[OE = OF = r;\]

\[OC - бисс\ \angle C;\]

\[AB = 50\ см;\]

\[BC = 39\ см;\]

\[AC = 41\ см.\]

\[Найти:\]

\[\text{r.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle ACO = \angle BCO = \frac{1}{2}\angle C;\]

\[p = \frac{1}{2}(50 + 39 + 41) =\]

\[= \frac{1}{2} \bullet 130 = 65;\]

\[S_{\text{ABC}} = \sqrt{65(65 - 50)(65 - 39)(65 - 41)};\]

\[S_{\text{ABC}} = \sqrt{65 \bullet 15 \bullet 26 \bullet 24} =\]

\[= \sqrt{608\ 400} = 780\ см^{2}.\]

\[2)\ В\ \mathrm{\Delta}OAC:\]

\[S_{\text{OAC}} = \frac{1}{2}AC \bullet OF = \frac{41}{2}\text{r.}\]

\[3)\ В\ \mathrm{\Delta}OBC:\]

\[S_{\text{OBC}} = \frac{1}{2}BC \bullet OE = \frac{39}{2}r;\]

\[S_{\text{ABC}} = S_{\text{OAC}} + S_{\text{OBC}};\]

\[\frac{41}{2}r + \frac{39}{2}r = 780\]

\[80r = 1560\ \ \ \]

\[r = 19,5\ см.\]

\[Ответ:\ \ 19,5\ см.\]