Решебник по геометрии 10 класс Атанасян ФГОС 269

\[\boxed{\mathbf{269.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[AB = BC = AC = 4\ см;\]

\[A_{1}B_{1} = B_{1}C_{1} = A_{1}C_{1} = 2\ см;\]

\[AA_{1} = 2\ см.\]

\[Найти:\]

\[MK;\ \ A_{1}F_{1}.\]

\[Решение.\]

\[O\ и\ O_{1} - центры\ оснований\ \]

\[пирамиды.\]

\[1)\ В\ треугольнике\ ABC:\]

\[AB = R\sqrt{3};\ \ R = AO;\]

\[AO = \frac{4}{\sqrt{3}}\ дм;\]

\[OK = \frac{1}{2}AO = \frac{2}{\sqrt{3}}дм.\]

\[2)\ В\ треугольнике\ A_{1}B_{1}C_{1}:\]

\[A_{1}O = \frac{2}{\sqrt{3}}\ дм;\]

\[O_{1}M = \frac{1}{\sqrt{3}}\ дм.\]

\[3)\ EK = OK - OE;\ \ OE = O_{1}M:\]

\[EK = \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}\ дм.\]

\[4)\ В\ треугольнике\ AA_{1}F:\]

\[FA = AO - FO;\ \ FO = A_{1}O;\]

\[AF = \frac{4}{\sqrt{3}} - \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}}\ дм;\]

\[A_{1}F = \sqrt{AA_{1}^{2} - AF^{2}} = \frac{2\sqrt{6}}{3}\ дм.\]

\[5)\ В\ треугольнике\ MEK:\]

\[MK = \sqrt{ME^{2} + EK^{2}} = \sqrt{3}\ дм.\]

\[Ответ:\ \frac{2\sqrt{6}}{3}\ дм;\sqrt{3}\ дм.\]