Решебник по геометрии 9 класс Атанасян ФГОС Задание 1220

\[\boxed{\mathbf{1220.ОК\ ГДЗ - домашка\ н}а\ 5}\]

\[Дано:\]

\[конус,\ \]

\[h - высота,\ \]

\[r - радиус\ основания,\ \]

\[V - объем\ конуса.\]

\[Решение.\]

\[\textbf{а)}\ h = 3\ см;r = 1,5\ см:\]

\[V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi \bullet {1,5}^{2} \bullet 3 =\]

\[= \pi \bullet {1,5}^{2} = 2,25\pi\ см^{3}.\]

\[\textbf{б)}\ r = 4\ см;V = 48\pi\ см^{2}:\]

\[V = \frac{1}{3}\pi r^{2}h\]

\[3V = \pi r^{2}h\]

\[\ h = \frac{3V}{\pi r^{2}}\]

\[h = \frac{3 \bullet 48\pi}{\pi \bullet 4^{2}} = \frac{3 \bullet 48}{16} = 3 \bullet 3 =\]

\[= 9\ см.\]

\[\textbf{в)}\ h = m;\ V = p:\]

\[V = \frac{1}{3}\pi R^{2}h\]

\[3V = \pi r^{2}h\]

\[\ r^{2} = \frac{3V}{\text{πh}}\]

\[r = \sqrt{\frac{3p}{\text{πm}}}.\]

\[Ответ:а)\ 2,25\pi;б)\ 9\ см;\]

\[\textbf{в)}\ \sqrt{\frac{3p}{\text{πm}}}.\]