Задача 8. Периметр прямоугольника равен 42, а площадь 98. Найдите большую сторону прямоугольника.
\(p = \left( {AD + AB} \right) \cdot 2 = 2\left( {x + y} \right) = 42\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x + y = 21.\) Получаем систему уравнений: \(\left\{ {\begin{array}{*{20}{c}}{x + y = 21}\\{x \cdot y = 98}\end{array}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 21 — x}\\{x \cdot y = 98}\end{array}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 21 — x\,\,\,\,\,\,\,\,\,\,}\\{x\left( {21 — x} \right) = 98}\end{array}} \right.} \right.} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 21 — x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} — 21x + 98 = 0}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,{x_1} = 7,\,\,\,\,{x_2} = 14,\,\,\,\,\,\,\,\,{y_1} = 14,\,\,\,\,\,\,\,{y_2} = 7.\) Значит, стороны прямоугольника равны 7 и 14. Большая сторона равна 14. Ответ: 14.
Пусть \(AD = x\), \(AB = y\). \(S = AD \cdot AB = x \cdot y = 98.\)