\[\boxed{\mathbf{344.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Цилиндр\ получен\ от\ вращения\ \]
\[квадрата:\]
\[R = H = a.\]
\[\textbf{а)}\ S_{сеч} = 2RH = 2a^{2}.\]
\[\textbf{б)}\ S_{бок} = 2\pi RH = 2\pi a^{2}.\]
\[\textbf{в)}\ S_{пов} = 2\pi a^{2} + 2\pi a^{2} = 4\pi a^{2}.\ \]