Решебник по геометрии 11 класс. Атанасян ФГОС 544

\[\boxed{\mathbf{544.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[конус;\]

\[R_{осн} = 6\ дм;\]

\[R_{сферы} = 3\ дм.\]

\[Найти:\]

\[V_{конуса}.\]

\[Решение.\]

\[1)\ MB - высота\ конуса;\ \ \]

\[OB = 2R_{сферы};\ \ \]

\[AC = 2R_{осн}.\]

\[2)\ \mathrm{\Delta}AOB - прямоугольный:\]

\[tg\ \angle OAB = \frac{\text{OB}}{\text{AB}} = \frac{R_{сферы}}{R_{осн}} = \frac{3}{6} =\]

\[= \frac{1}{2}.\]

\[3)\ В\ \mathrm{\Delta}MAC:\ \ AO - биссектриса\ \]

\[\angle BAM:\]

\[\angle BAM = 2\angle OAB;\ \]

\[\frac{\text{MB}}{\text{AB}} = tg\ (2\angle OAB);\]

\[\text{tg\ }(2\angle OAB) = \frac{2\ tg\ \angle OAB}{1 - tg^{2}\angle OAB} =\]

\[= \frac{2 \bullet \frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3}.\]

\[h = MB = \frac{4}{3} \bullet AB = \frac{4}{3} \bullet 6 = 8\ дм.\]

\[4)\ V_{конуса} = \frac{1}{3}\pi \bullet r^{2} \bullet h =\]

\[= \frac{1}{3} \bullet \pi \bullet 6^{2} \bullet 8 = 96\pi\ дм^{3}.\]

\[\mathbf{Отв}ет:\ \ V = 96\pi\ дм^{3}.\]