\[\boxed{\mathbf{1220.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[Дано:\]
\[конус,\ \]
\[h - высота,\ \]
\[r - радиус\ основания,\ \]
\[V - объем\ конуса.\]
\[Решение.\]
\[\textbf{а)}\ h = 3\ см;r = 1,5\ см:\]
\[V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi \bullet {1,5}^{2} \bullet 3 =\]
\[= \pi \bullet {1,5}^{2} = 2,25\pi\ см^{3}.\]
\[\textbf{б)}\ r = 4\ см;V = 48\pi\ см^{2}:\]
\[V = \frac{1}{3}\pi r^{2}h\]
\[3V = \pi r^{2}h\]
\[\ h = \frac{3V}{\pi r^{2}}\]
\[h = \frac{3 \bullet 48\pi}{\pi \bullet 4^{2}} = \frac{3 \bullet 48}{16} = 3 \bullet 3 =\]
\[= 9\ см.\]
\[\textbf{в)}\ h = m;\ V = p:\]
\[V = \frac{1}{3}\pi R^{2}h\]
\[3V = \pi r^{2}h\]
\[\ r^{2} = \frac{3V}{\text{πh}}\]
\[r = \sqrt{\frac{3p}{\text{πm}}}.\]
\[Ответ:а)\ 2,25\pi;б)\ 9\ см;\]
\[\textbf{в)}\ \sqrt{\frac{3p}{\text{πm}}}.\]