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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 767
767
\[\boxed{\mathbf{767.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[MABCD - правильная\ \]
\[четырехугольная\ пирамида;\ \]
\[MA = 8\ см;\]
\[\angle MAH = 60{^\circ};\]
\[MH - высота;\]
\[O - центр\ описанного\ шара.\]
\[Решение.\]
\[\textbf{а)}\ MH = AM \bullet \sin{60{^\circ}} = \frac{8\sqrt{3}}{2} =\]
\[= 4\sqrt{3}\ см.\]
\[\mathrm{\Delta}MAC - правильный\ \]
\[(MA = MC;\ \angle MAC = 60{^\circ}):\]
\[MA = MC = AC = BD.\]
\[AH = \frac{1}{2}AC = 4\ см.\]
\[BC = \frac{\text{AC}}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}\ см.\]
\[Построим\ HP\bot AD:\]
\[MP\bot AD - по\ теореме\ о\ трех\ \]
\[перпендикулярах.\]
\[HP = \frac{1}{2}BC = 2\sqrt{2}\ см.\]
\[В\ \mathrm{\Delta}MPH - прямоугольном:\]
\[MP = \sqrt{48 + 8} = \sqrt{56} =\]
\[= 2\sqrt{14}\ см.\]
\[S_{бок} = 4 \bullet S_{\text{MAD}} =\]
\[= 4 \bullet \frac{1}{2} \bullet 4\sqrt{2} \bullet 2\sqrt{14} = 32\sqrt{7}\ см^{2}.\]
\[\textbf{б)}\ V = \frac{1}{3} \bullet S_{\text{ABCD}} \bullet MH =\]
\[= \frac{1}{3} \bullet 4\sqrt{3} \bullet \left( 4\sqrt{2} \right)^{2} = \frac{128\sqrt{3}}{3}\ см^{3}.\]
\[\textbf{в)}\ Искомый\ угол\ \angle PMT = a.\]
\[\angle PMT = 2\angle PMH.\]
\[В\ \mathrm{\Delta}PMH - прямоугольном:\]
\[\sin{\angle PMH} = \frac{\text{HP}}{\text{MP}} = \frac{\sqrt{7}}{7};\]
\[\cos{\angle PMH} = \sqrt{1 - \sin^{2}{\angle PMH}} =\]
\[= \sqrt{\frac{6}{7}}.\]
\[\sin a = 2\sin{\angle PMH} \bullet \cos{\angle PMH} =\]
\[= \frac{2\sqrt{6}}{7};\]
\[\cos a = \frac{5}{7};\ \]
\[tg\ a = \frac{2\sqrt{6}}{5}\]
\[a = arctg\frac{2\sqrt{6}}{5}.\]
\[\textbf{г)}\ \angle MPH = a - угол\ межу\ \]
\[гранью\ и\ основанием:\]
\[tg\ a = \frac{\text{MH}}{\text{HP}} = \frac{4\sqrt{3}}{2\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{6},\ \]
\[отсюда\ a = arctg\ \sqrt{6}.\]
\[\textbf{д)}\frac{1}{2}\left( \overrightarrow{\text{MB}} + \overrightarrow{\text{MD}} \right) \bullet \overrightarrow{\text{MK}};\]
\[\frac{1}{2}\left( \overrightarrow{\text{MB}} + \overrightarrow{\text{MD}} \right) = \frac{1}{2} \bullet 2\overrightarrow{\text{MP}} = \overrightarrow{\text{MP}};\]
\[\overrightarrow{\text{MP}} \bullet \overrightarrow{\text{MK}} =\]
\[= MP \bullet MK \bullet \cos{\angle PMK} =\]
\[= 4\sqrt{3} \bullet 2\sqrt{14} \bullet \frac{4\sqrt{3}}{2\sqrt{4}} = 48.\]
\[\textbf{е)}\ MO = AO = R.\]
\[В\ \mathrm{\Delta}AOH - прямоугольном:\]
\[R^{2} = OH^{2} + AH^{2} =\]
\[= \left( 4\sqrt{3} - R \right)^{2} + 16\]
\[R^{2} = 48 - 8\sqrt{3}R + R^{2} + 16\]
\[R = \frac{64}{8\sqrt{3}} = \frac{8\sqrt{3}}{3}\ см.\]
