The diagonal of a rectangular figure is an illustration of Pythagoras theorem
The length is 16 in, and the width is 12 in
The given parameters are:
[tex]\mathbf{Diagonal = 20}[/tex]
[tex]\mathbf{Length = 4 + Width}[/tex]
The diagonal is calculated using, the following Pythagoras theorem
[tex]\mathbf{Diagonal^2 = Length^2 + Width^2}[/tex]
Represent length with L and width with W
So, we have:
[tex]\mathbf{20^2 = L^2 + W^2}[/tex]
[tex]\mathbf{400 = L^2 + W^2}[/tex]
This gives
[tex]\mathbf{400 = (4 + W)^2 + W^2}[/tex]
Expand
[tex]\mathbf{400 = 16 + 8W + W^2 + W^2}[/tex]
[tex]\mathbf{400 = 16 + 8W + 2W^2}[/tex]
Divide through by 2
[tex]\mathbf{200 = 8 + 4W +W^2}[/tex]
Rewrite as:
[tex]\mathbf{W^2 + 4W +8 -200 = 0}[/tex]
[tex]\mathbf{W^2 + 4W - 192 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{W = \{-16,12\}}[/tex]
The width cannot be negative.
So, we have:
[tex]\mathbf{W =12}[/tex]
Recall that:
[tex]\mathbf{Length = 4 + Width}[/tex]
This gives
[tex]\mathbf{L= 4 + 12}[/tex]
[tex]\mathbf{L= 16}[/tex]
Hence, the length is 16 in, and the width is 12 in
Read more about diagonals at:
https://brainly.com/question/18983839
