Решебник по геометрии 11 класс. Атанасян ФГОС 842

\[\boxed{\mathbf{842.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\mathrm{\Delta}ABC;\]

\[h_{a} = 4\ см;\]

\[h_{b} = 3\ см;\]

\[h_{c} = 6\ см.\]

\[Найти:\]

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

\[Решение.\]

\[1)\ \ Найдем\ стороны\ \]

\[треугольника\ \text{ABC.}\]

\[Пусть\ BC = a,\ AB = c,\ AC = b\ \ и\ \ \]

\[h_{a},h_{b},h_{c} - высоты,опущенные\]

\[к\ данным\ сторонам:\]

\[S_{\text{ABC}} = \frac{1}{2}h_{a}a = \frac{1}{2}h_{b}b = \frac{1}{2}h_{c}c\]

\[4a = 3b = 6c.\]

\[2)\ c = \frac{2}{3}a\ \ и\ \ b = \frac{4}{3}a;\ \]

\[полупериметр\ \mathrm{\Delta}ABC:\]

\[p = \frac{a + b + c}{2} =\]

\[= \frac{1}{2}\left( a + \frac{4}{3}a + \frac{2}{3}a \right) = \frac{3a}{2}\]

\[(p - a) = \frac{a}{2};\ \]

\[(p - b) = \frac{3a}{2} - \frac{4}{3}a = \frac{a}{6};\]

\[(p - c) = \frac{3}{2}a - \frac{2}{3}a = \frac{5a}{6}.\]

\[3)\ S_{\text{ABC}} =\]

\[= \sqrt{p(p - a)(p - b)(p - c)} =\]

\[= \sqrt{\frac{3a}{2} \bullet \frac{a}{2} \bullet \frac{a}{6} \bullet \frac{5a}{6}} = \frac{a^{2}\sqrt{15}}{12} =\]

\[= \frac{1}{2} \bullet 4 \bullet a\]

\[a = \frac{24}{\sqrt{15}}\ см;\]

\[S_{\text{ABC}} = \frac{1}{2} \bullet 4 \bullet \frac{24}{\sqrt{15}} = \frac{48}{\sqrt{15}} =\]

\[= \frac{48 \bullet \sqrt{15}}{15} = \frac{16\sqrt{15}}{3}\ см^{2}.\]

\[Ответ:\ \ S_{\text{ABC}} = \frac{16\sqrt{15}}{5}\ см^{2}.\]