Question

Can u help me with this and six other questions

Answer 1

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

`\begin{gathered} A_1=lw \\ A_1=(9\text{ ft})(6\text{ ft}) \\ A_1=54\text{ ft}^2 \end{gathered}`

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

`\begin{gathered} A_2=(4\text{ ft})(9\text{ ft}) \\ A_2=36\text{ ft}^2 \end{gathered}`

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

`\begin{gathered} A_3=(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}`

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

`\begin{gathered} A_4=(3\text{ ft})(9\text{ ft}) \\ A_4=27\text{ ft}^2 \end{gathered}`

Solving for the surface area

To solve for the surface area, get the sum of the area of all the surfaces that is

`SA=A_1+2A_2+2A_3+A_4`

Note that surface 2 and surface 3 are multiplied by 2 since they have other faces that are congruent

`\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}`

Therefore, the surface area of the figure is 186.3 square feet.