Step-by-step explanation:
[tex]area = \frac{1}{2} (x + 3)(x + 5) = 24 \\ [/tex]
then
[tex](x + 3)(x + 5) = 48[/tex]
[tex] {x}^{2} + 8x + 15 = 48 [/tex]
[tex] {x}^{2} + 8x + 15 - 48 = 0 \\ {x}^{2} + 8x - 33 = 0[/tex]
[tex] {x}^{2} - 3x + 11x - 33 = 0 \\ x(x - 3) + 11(x - 3) = 0 \\ (x - 3)(x + 11) = 0[/tex]
so
[tex]x = 3 \: or \: x = - 11(reject \: because \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x > 0)[/tex]
hence,
[tex]x = 3[/tex]
Thus, one leg is 8 and other leg is 6.
Since it is a right triangle then we have
[tex]hypotenuse = \sqrt{ {8}^{2} + {6}^{2} } = \sqrt{100} = 10[/tex]
Finally,
[tex]perimeter = 8 + 6 + 10 = 24(ft)[/tex]
