Question

$the \: area \: of \: a \: rectangular \: carpet \: is \: 120 {m}^{2} and \: its \: perimeter \: is \: 46m \: the \: length \: of \: its \: diagonal \: is$

Answer 1

$let \: \: carpet \: \: length \: \: be \: \: x \: \: and \\ breadth \: \: be \: \: y \\ area \: \: xy = 120 - - - (1) \\ perimeter \: \: 2(x + y) = 46 \\ x + y = 23 \: \: \: substituting \: \: from \: \: (1) \\ x + \frac{120}{x} = 23 \\ {x}^{2} - 23x + 120 = 0 \\ (x - 8)(x - 15) = 0 \\ x = 8 \: \: or \: \: 15 \\ y = 23 - x = 15 \: \: or \: \: 8 \\ since \: \: x \: \: is \: \: length \: \: it \: \: should \: be \: \: larger \\ so \: \: x = 15 \: \: and \: \: y = 8 \\ diagonal = \sqrt{ {x}^{2} + {y}^{2} } \\ = \sqrt{ {15}^{2} + \sqrt{ {8}^{2} } } = 17 \: \: m$