Answer:
2 feet
Step-by-step explanation:
Let 'w' be twice the width of the deck.
The area of the swimming pool is:
[tex]A_p = \frac{\pi d^2}{4}\\A_p = \frac{\pi 28^2}{4} = 196\pi\\[/tex]
The area of the swimming pool plus the deck is:
[tex]A_{p+d} = \frac{\pi (d+w)^2}{4}\\[/tex]
Therefore, the area of the deck is given by:
[tex]A_d = A_{p+d} - A_p\\A_d = 60\pi = \frac{\pi (28+w)^2}{4} - 196\pi\\1,024 = (28+w)^2\\w^2+56w - 240 = 0[/tex]
Solving the quadratic equation:
[tex]w^2+56w - 240 = 0\\w= -56 \pm\frac{\sqrt{56^2-(4*1*-240)} }{2} \\w_1 = -60\\w_2 = 4[/tex]
Since the width cannot be negative, w = 4 feet, ad the width is 2 feet.
