Решебник по геометрии 9 класс Мерзляк Задание 162

\[Схематический\ рисунок.\]

\[Дано:\]

\[O - центр\ впис.\ окружности;\]

\[AB = 17\ см;\]

\[BC = 25\ см;\]

\[AC = 28\ см;\]

\[OH\bot AB.\]

\[Найти:\]

\[S_{\text{AOB}};\ S_{\text{BOC}};\ S_{\text{AOC}}.\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[p = \frac{1}{2}(17 + 25 + 28) =\]

\[= \frac{1}{2} \bullet 70 = 35\ см;\]

\[S = \sqrt{35(35 - 17)(35 - 25)(35 - 28)}\]

\[S = \sqrt{35 \bullet 18 \bullet 10 \bullet 7} =\]

\[= \sqrt{44\ 100} = 210\ см^{2}.\]

\[OH = r = \frac{S}{p} = \frac{210}{35} = 6\ см.\]

\[2)\ В\ \mathrm{\Delta}AOB:\]

\[S = \frac{1}{2}AB \bullet OH = \frac{1}{2} \bullet 17 \bullet 6 = 51\ см^{2}.\]

\[3)\ В\ \mathrm{\Delta}BOC:\]

\[S = \frac{1}{2} \bullet BC \bullet OH =\]

\[= \frac{1}{2} \bullet 25 \bullet 6 = 75\ см^{2}.\]

\[4)\ В\ \mathrm{\Delta}AOC:\]

\[S = \frac{1}{2} \bullet AC \bullet OH =\]

\[= \frac{1}{2} \bullet 28 \bullet 6 = 84\ см^{2}.\]

\[Ответ:\ \ 51\ см^{2};\ 75\ см^{2};\ 84\ см^{2}.\]