Решебник по геометрии 9 класс Мерзляк Задание 168

\[Схематический\ рисунок.\]

\[Дано:\]

\[O - центр\ впис.\ окружности;\]

\[S_{\text{AOB}} = 26\ см^{2};\]

\[S_{\text{BOC}} = 28\ см^{2};\]

\[S_{\text{AOC}} = 30\ см^{2};\]

\[OH\bot AB.\]

\[Найти:\]

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

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}AOB:\]

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

\[AB \bullet OH = 52\ \ \ \]

\[AB = \frac{52}{\text{OH}}.\]

\[2)\ В\ \mathrm{\Delta}BOC:\]

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

\[BC \bullet OH = 56\ \ \ \]

\[BC = \frac{56}{\text{OH}}.\]

\[3)\ В\ \mathrm{\Delta}AOC:\]

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

\[AC \bullet OH = 60\ \ \]

\[AC = \frac{60}{\text{OH}}.\]

\[4)\ В\ \mathrm{\Delta}ABC:\]

\[p = \frac{1}{2} \bullet \frac{52 + 56 + 60}{\text{OH}} = \frac{84}{\text{OH}};\]

\[S_{\text{ABC}} = \sqrt{\frac{84}{\text{OH}} \bullet \frac{84 - 52}{\text{OH}} \bullet \frac{84 - 56}{\text{OH}} \bullet \frac{84 - 60}{\text{OH}}};\]

\[S_{\text{ABC}} = \sqrt{\frac{84 \bullet 32 \bullet 28 \bullet 24}{OH^{4}}} = \frac{1344}{OH^{2}}.\]

\[S_{\text{ABC}} = S_{\text{AOB}} + S_{\text{BOC}} + S_{\text{AOC}}:\]

\[\frac{1344}{OH^{2}} = 26 + 28 + 30\]

\[OH^{2} = \frac{1344}{84} = 16\ \ \ \]

\[OH = 4\ см.\]

\[AB = \frac{52}{4} = 13\ см.\ \ \]

\[BC = \frac{56}{4} = 14\ см.\ \ \ \]

\[AC = \frac{60}{4} = 15\ см.\]

\[Ответ:\ \ 13\ см;\ 14\ см;\ 15\ см.\]