Пусть радиус окружности равен R.
Тогда: \(AB = 2R\) и \({S_{ABCD}} = {\left( {2R} \right)^2} = 4{R^2}.\)
\({A_1}{C_1} = 2R.\)
По теореме Пифагора из треугольника \({A_1}{B_1}{C_1}\):
\({A_1}{C_1}^2 = {A_1}{B_1}^2 + {B_1}{C_1}^2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{\left( {2R} \right)^2} = 2{A_1}{B_1}^2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{A_1}{B_1} = \sqrt 2 R.\)
\({S_{{A_1}{B_1}{C_1}{D_1}}} = {\left( {\sqrt 2 R} \right)^2} = 2{R^2}.\)
\(\frac{{{S_{ABCD}}}}{{{S_{{A_1}{B_1}{C_1}{D_1}}}}} = \frac{{4{R^2}}}{{2{R^2}}} = 2.\)
Ответ: 2.