Решебник по геометрии 10 класс Атанасян ФГОС 551

\[\boxed{\mathbf{551.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[шар\ O;\]

\[вписанный\ конус\ (r;H).\]

\[Найти:\]

\[S_{шара};\]

\[V_{шара}.\]

\[Решение.\]

\[1)\ CH = H - высота\ конуса;\ \ \]

\[OB = OA = OC = R_{шара};\]

\[AH = HB = r - радиус\ \]

\[основания\ конуса.\]

\[2)\ \mathrm{\Delta}ABC - равнобедренный:\]

\[AC = BC.\]

\[3)\ \mathrm{\Delta}ACH - прямоугольный:\]

\[AC^{2} = CH^{2} + AH^{2} = \sqrt{H^{2} + r^{2}}.\]

\[4)\ Радиус\ шара:\]

\[S_{\text{ABC}} = \frac{1}{2} \bullet AB \bullet CH = \frac{1}{2} \bullet 2r \bullet H =\]

\[= r \bullet H;\]

\[R_{шара} = \frac{AB \bullet BC \bullet CA}{4S_{\text{ABC}}} =\]

\[= \frac{\left( \sqrt{H^{2} + r^{2}} \right)^{2} \bullet 2r}{4rH} = \frac{H^{2} + r^{2}}{2H}.\]

\[5)\ Площадь\ поверхности\ шара:\]

\[S_{шара} = 4\pi R^{2} =\]

\[= 4\pi \bullet \left( \frac{H^{2} + r^{2}}{2H} \right)^{2} =\]

\[= \frac{\pi}{H^{2}} \bullet \left( H^{2} + r^{2} \right)^{2}.\]

\[6)\ V_{шара} = \frac{4}{3}\pi R^{3} =\]

\[= \frac{4\pi}{3} \bullet \left( \frac{H^{2} + r^{2}}{2H} \right)^{3} =\]

\[= \frac{\pi}{6H^{3}} \bullet \left( H^{2} + r^{2} \right)^{3}.\]

\[\mathbf{Отв}ет:\ \ S_{шара} = \frac{\pi}{H^{2}} \bullet \left( H^{2} + r^{2} \right)^{2};\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }V_{шара} = \frac{\pi}{6H^{3}} \bullet \left( H^{2} + r^{2} \right)^{3}.\]