Given:
The area of the projection is
[tex]A=(x^2-12x+32)ft^2[/tex]
The width of the projection is w =(x-4) feet.
Let l be the height of the projection.
The shape of the projection is a rectangle.
Consider the formula for the area of the rectangle.
[tex]A=lw[/tex]
[tex]\text{ Sustitute }A=\mleft(x^2-12x+32\mright)\text{ and w=(x-4) in the formula.}[/tex][tex]x^2-12x+32=l(x-4)[/tex][tex]Use\text{ }-12x=-8x-4x.[/tex]
[tex]x^2-8x-4x+32=l(x-4)[/tex]
[tex]x(x-8)-4(x-8)=l(x-4)[/tex]
[tex](x-8)(x-4)=l(x-4)[/tex]
Cancel out the common factor.
[tex](x-8)=l[/tex]
Hence the height of the projection is (x-8).
b)
Given:
Height of the wall x=10 ft.
We know that the width of the projection w =(x-4) and height of the projection l=(x-8).
Substitute x=10 in the equation l and w, we get
[tex]w=10-4=6\text{ ft}[/tex][tex]l=10-8=2ft[/tex]
Consider the perimeter of the rectangle.
[tex]P=2(l+w)[/tex]
Substitute l=2 and w=10 in the formula, we get
[tex]P=2(2+4)=2\times8=16\text{ ft.}[/tex]
Hence the perimeter of the projection is 16 ft when the height of the wall is 10 ft.
