we are asked to determine the surface area of the figure. To do that we need to find the areas of each face and add them together. The front face is a trapezoid, and its area is:
![A_(tz)=(b+B)/(2)h](https://img.qammunity.org/2023/formulas/mathematics/high-school/su7izfqnaemk4q9ia62chfga0mo2y7tnhj.png)
Where "b" is the upper base, "B" is the lower base and "h" is the height. Replacing the values:
![A_(tz)=(9ft+4ft)/(2)*4.3ft](https://img.qammunity.org/2023/formulas/mathematics/high-school/pregfoqmfm7hkwlhr02lqxnn843aqnhksy.png)
Solving the operations:
![A_(tz)=28ft^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/9pf4lnok3xfiviw7ofbl65h52i8mf6oruz.png)
the upper face is a rectangle, its area is the product of its side:
![A_(uf)=(10ft)(9ft)=90ft^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/hpx7v30px820joea4yxbijsu9x2mp0ds4i.png)
The area of the lower face is also a rectangle, therefore its area is:
![A_(lf)=(4ft)(10ft)=40ft^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/jg2rlqgeasrs7drsvxwnpl5kd0bv2eknv9.png)
The side face is also a rectangle, and its area is:
![A_(sf)=(5ft)(10ft)=50ft^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/t5h6r7w7swxls2j9snnbj8kfzhyx6c5sgk.png)
Now we add the areas having into account that the front and side faces repeat themselves. the total surface area is:
![A=2A_(tz)+A_(uf)+A_(lf)+2A_(sf)](https://img.qammunity.org/2023/formulas/mathematics/high-school/oexafhwuhnjt9zymvug72fbc06s27vmci9.png)
replacing the values:
![A=2(28ft^2)+(90ft^2)+(40ft^2)+2(50ft^2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/c5nthsoz8bxioi699eemx3yoc9lssmokro.png)
Solving the operations:
![A=286ft^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/h0ycatqxbkv9nvgheejo3vhmodthco8l3z.png)
Therefore, the surface area is 286 square feet.