\[\boxed{\mathbf{345.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[ABCD - прямоугольник;\]
\[AB = a;\ \ BC = b.\]
\[Решение.\]
\[\textbf{а)}\ S_{1} = 2\pi ba;\ \ S_{2} = 2\pi ab:\]
\[S_{1} = S_{2}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ S_{1} = 2\pi ba + 2\pi b^{2};\ \ \]
\[S_{2} = 2\pi ab + 2\pi a^{2}:\]
\[\frac{S_{1}}{S_{2}} = \frac{2\pi ba + 2\pi b^{2}}{2\pi ab + 2\pi a^{2}} =\]
\[= \frac{2\pi b(a + b)}{2\pi a(a + b)} = \frac{b}{a}.\]
\[Ответ:\ \frac{b}{a}.\]
\[Параграф\ 2.\ Конус\]