\[\boxed{\text{77.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[S_{1} = \pi R^{2}\]
\[(2R - 1)\ см - новый\ радиус.\]
\[S_{2} = \pi(2R - 1)^{2} =\]
\[= \pi\left( 4R^{2} - 4R + 1 \right).\]
\[Известно,\ что\ S_{2} - S_{1} = \pi.\]
\[Составим\ уравнение:\]
\[\pi\left( 4R^{2} - 4R + 1 \right) - \pi R^{2} = \pi\]
\[\pi\left( 4R^{2} - 4R + 1 - R^{2} \right) = \pi\]
\[3R^{2} - 4R + 1 = 1\]
\[3R^{2} - 4R = 0\]
\[3R\left( R - \frac{4}{3} \right) = 0\]
\[R = 0\ (не\ подходит).\]
\[R = \frac{4}{3} = 1\frac{1}{3}\ (см).\]
\[Ответ:\ \ 1\frac{1}{3}\ см.\]