Пусть \(AD = x\), \(BC = y\).
\(p = \left( {AD + BC} \right) \cdot 2 = \left( {x + y} \right) \cdot 2 = 34\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x + y = 17.\)
\(S = x \cdot y = 60.\)
Получаем систему уравнений:
\(\left\{ {\begin{array}{*{20}{c}}{x + y = 17\,\,\,\,\,\,\,}\\{xy = 60\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
Возведём обе части первого уравнения в квадрат:
\({\left( {x + y} \right)^2} = {17^2}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{x^2} + {y^2} + 2xy = 289.\)
Так как из второго уравнения \(xy = 60\), то
\({x^2} + {y^2} + 2 \cdot 60 = 289\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,{x^2} + {y^2} = 169.\)
\(A{C^2} = {x^2} + {y^2} = 169\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,AC = 13.\)
Ответ: 13.