Answer:
510.4 ft²
Explanation:
We have:
two trapezoids with bases 15ft and 7ft and height 5ft.
four rectangles 5ft × 11ft, 15ft × 11ft, 9.4ft × 11ft and 7ft × 11ft.
The formula of an area of a trapezoid:

b₁, b₂ - bases
h - height
Substitute:

The formula of an area of a rectangle:

l - length
w - width
The dimensions of rectangle l × w
Subtitute:

The surface area of the figure:
[tex]S.A.=2A_t+A_1+A_2+A_3+A_4\\\\S.A.=2(55)+55+165+103.4+77=510.4\ ft^2[/text]