Solve for the area of each faces, there are 6 faces, 2 of which are congruent.
Solving for area of surface 1
The first surface has a dimension of 9 ft by 6 ft, and is a rectangle. The volume is therefore is
[tex]\begin{gathered} A_1=lw \\ A_1=(9\text{ ft})(6\text{ ft}) \\ A_1=54\text{ ft}^2 \end{gathered}[/tex]
Solving for area of surface 2
The second surface has a dimension of 4 ft by 9 ft, and is also a rectangle. The volume is
[tex]\begin{gathered} A_2=(4\text{ ft})(9\text{ ft}) \\ A_2=36\text{ ft}^2 \end{gathered}[/tex]
Solving for area of surface 3
The third surface has a dimension of the following
first base = 3 ft
second base = 6 ft
height = 3.7 ft
and is in the shape of trapezoid
[tex]\begin{gathered} A_3=\frac{a+b}{2}h\frac{}{} \\ A_3=\frac{(3\text{ ft}+6\text{ ft})}{2}(3.7\text{ ft}) \\ A_3=\frac{9\text{ ft}}{2}(3.7\text{ ft}) \\ A_3=(4.5\text{ ft})(3.7\text{ ft}) \\ A_3=16.65\text{ ft}^2 \end{gathered}[/tex]
Solving for area of surface 4
The fourth surface has a dimension of 3 ft by 9 ft, and is in the shape of a rectangle
[tex]\begin{gathered} A_4=(3\text{ ft})(9\text{ ft}) \\ A_4=27\text{ ft}^2 \end{gathered}[/tex]
Solving for the surface area
To solve for the surface area, get the sum of the area of all the surfaces that is
[tex]SA=A_1+2A_2+2A_3+A_4[/tex]
Note that surface 2 and surface 3 are multiplied by 2 since they have other faces that are congruent
[tex]\begin{gathered} SA=A_1+2A_2+2A_3+A_4 \\ SA=54\text{ ft}^2+2(36\text{ ft}^2)+2(16.65\text{ ft}^2)+27\text{ ft}^2 \\ SA=54\text{ ft}^2+72\text{ ft}^2+33.3\text{ ft}^2+27\text{ ft}^2 \\ SA=186.3\text{ ft}^2 \end{gathered}[/tex]
Therefore, the surface area of the figure is 186.3 square feet.
