the Given the geometry of the wall as shown below
The wall takes the shape of a trapezoid.
The unpainted door labeled as A takes the shape of a rectangle.
The unpainted window labeled as B takes the shape of a triangle.
The area of the wall painted by Marcus is evaluated as
[tex]\text{Area of the wall - (area of the unpainted door + area of the unpainted window)}[/tex]
Area of the wall:
Recall that the wall takes the shape of a trapezoid. Thus, the area of a trapezoid is evaluated as
[tex]\begin{gathered} A_{wall}=\text{ }\frac{1}{2}(a+b)h \\ \text{where} \\ a\text{ and b are the base lengths of the trapezoid} \\ h\text{ is the height of the trapezoid} \end{gathered}[/tex]
Thus, the area of the wall becomes
[tex]\begin{gathered} \frac{1}{2}\times(9+15)\times13 \\ =\frac{1}{2}\times24\times13\text{ = 12}\times13 \\ \Rightarrow156\text{ square-f}eet \\ \end{gathered}[/tex]
Area of the unpainted door:
Recall that the unpainted door, labeled as A, takes the shape of a rectangle. Hence, the area of a rectangle is evaluated as
[tex]\begin{gathered} A_{\text{rectangle}}=\text{ l }\times b \\ \text{where} \\ l\Rightarrow\text{length of the rectangle } \\ b\Rightarrow\text{width of the rectangle } \end{gathered}[/tex]
The area of the unpainted door becomes
[tex]\begin{gathered} A_{unpainted\text{ door}}=7\times2.5\text{ } \\ \Rightarrow17.5\text{ square-f}eet \end{gathered}[/tex]
Area of the unpainted window:
Recall that the unpainted window, labeled as B, takes the shape of a triangle. Thus, the area of a triangle is evaluated as
[tex]\begin{gathered} A_{triangle}\text{ = }\frac{1}{2}bh \\ \text{where} \\ b\Rightarrow base\text{ length of the triangle} \\ h\Rightarrow height\text{ of the triangle} \end{gathered}[/tex]
Thus, the area of the unpainted window becomes
[tex]\begin{gathered} A_{unpainted\text{ window}}=\frac{1}{2}\times3\times6 \\ =3\times2 \\ \Rightarrow6\text{ square-fe}et \end{gathered}[/tex]
Thus, the area of the painted wall becomes
[tex]\begin{gathered} A_{painted\text{ wall}}\text{ = 156 - (17.5 + 6)} \\ =156\text{ - 23.5} \\ \Rightarrow132.5\text{ square-fe}et \end{gathered}[/tex]
Hence, the area of the wall painted by Marcus is 132.5 square-feet
