Решебник по геометрии 11 класс. Атанасян ФГОС 204

\[\boxed{\mathbf{204.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[Решение.\]

\[\textbf{а)}\ 1)\ Проведем\ высоты\ AD;\]

\[BK;CE.\]

\[O - точка\ их\ пересечения\ в\ \]

\[⊿ABC.\]

\[2)\ ⊿MAO = ⊿MBO = ⊿MCO -\]

\[по\ двум\ катетам:\]

\[OA = OB = OC.\]

\[Отсюда:\]

\[MA = MB = MC = \frac{a}{\sin\varphi}.\]

\[3)\ В\ треугольнике\ MCO:\]

\[\frac{\text{MO}}{\text{MC}} = \sin\varphi;\]

\[MC = \frac{a}{\sin\varphi};\]

\[\frac{\text{MO}}{\text{OC}} = tg\ \varphi;\]

\[OC = \frac{a}{\text{tg\ φ}}.\]

\[4)\ OD = OK = OE = \frac{\text{OC}}{2}:\]

\[OD = \frac{a}{2tg\ \varphi}.\]

\[\mathrm{\Delta}MOD = \mathrm{\Delta}MOK = \mathrm{\Delta}MOE - по\ \]

\[двум\ катетам:\]

\[MK = ME = MD.\]

\[5)\ OD - проекция\ \text{MD\ }на\ \]

\[плоскость\ ABC;\ \ OD\bot BC:\]

\[MD\bot BC - по\ теореме\ о\ трех\ \]

\[перпендикулярах.\]

\[6)\ Из\ треугольника\ MDO:\]

\[MD^{2} = MO^{2} + OD^{2}\]

\[MD^{2} = a^{2} + \frac{a^{2}}{4tg^{2}\varphi} =\]

\[= \frac{4a^{2}tg^{2}\varphi + a^{2}}{4tg^{2}\varphi};\]

\[MD = \sqrt{\frac{a^{2}\left( 4tg^{2}\varphi + 1 \right)}{4tg^{2}\varphi}} =\]

\[= \frac{a}{2tg\ \varphi} \cdot \sqrt{1 + 4tg^{2}\varphi}.\]

\[\textbf{б)}\ l = \pi R;\ \ R = a:\]

\[l = \frac{2\pi a}{\text{tg\ φ}}.\]

\[\textbf{в)}\ S = \frac{AB^{2}\sqrt{3}}{4};\ \ AB = OC \cdot \sqrt{3} =\]

\[= \frac{a\sqrt{3}}{\text{tg\ φ}}:\]

\[S_{\text{ABC}} = \frac{3\sqrt{3}a^{2}}{4tg^{2}\varphi}.\]