Помогите решить задачу срочно умоляю!(

Пусть a, b -- стороны прямоугольника, d -- диагональ.
$s = a \times b \\ {d}^{2} = {a}^{2} + {b}^{2} \\ \\ a \times b = 48 \\ {a}^{2} + {b}^{2} = 100 \\ \\ a \times b = 48 \\ {a}^{2} + 2ab + {b}^{2} = 100 + 2ab \\ \\ a \times b = 48 \\ {(a + b)}^{2} = 100 + 2 \times 48 \\ \\ a \times b =48 \\ {(a + b)}^{2} = 196 \\ \\ a + b = 14 \\ a \times b = 48 \\ \\ b = 14 - a \\ a(14 - a) = 48 \\ \\ b = 14 - a \\ 14a - {a}^{2} = 48 \\ \\ b = 14 - a \\ {a}^{2} - 14a + 48 = 0 \\ \\ d = {b}^{2} - 4ac = 196 - 4 \times 48 = 4 \\ a1 = \frac{14 + 2}{2} = \frac{16}{2} = 8 \\ b1 = 14 - 8 = 6 \\ \\ a2 = \frac{14 - 2}{2} = \frac{12}{2} = 6 \\ b2 = 14 - 6 = 8$
=> стороны равны 6 и 8.