The given area of the [tex]52\frac{1}{2}[/tex] ft. by 33 ft. wall less the areas of the Windows and the Door give the area of the painted part of the wall as [tex]\underline{1642\frac{9}{16}}[/tex] ft.²
The area of the painted part of the wall is [tex]\underline{1642\frac{9}{16}}[/tex] ft.²
The given dimensions of the wall are;
The width of the wall = [tex]52\frac{1}{2}[/tex] ft.
The height of the wall = 33 ft.
The painted area = Area of the wall - (Area of the widows + The door)
Which gives;
[tex]52\frac{1}{2} \times 33 - \left(6\frac{1}{4} \times 5\frac{3}{4} + 3\frac{1}{8} \times 4 + 9\frac{1}{2} + 4 \times 8 \right) = \mathbf{ 1642 \frac{9}{16}}[/tex]
The area of the painted part of the wall is [tex]\underline{1642\frac{9}{16}}[/tex] ft.²
Learn more about the area of a rectangle here:
https://brainly.com/question/18101587
