Given:
Perimeter of a football field = 920 feet.
The width is 60 feet more than one-third of the length.
To find:
The dimensions of a football field.
Solution:
Let the length of the field be x.
Then, width = [tex]\dfrac{1}{3}x+60[/tex]
We know that,
[tex]Perimeter = 2( length + width)[/tex]
[tex]920 = 2(x+\dfrac{1}{3}x+60)[/tex]
[tex]920 = 2(\dfrac{4}{3}x+60)[/tex]
[tex]920 =\dfrac{8}{3}x+120[/tex]
Subtract both sides by 120.
[tex]920-120 =\dfrac{8}{3}x+120-120[/tex]
[tex]800 =\dfrac{8}{3}x[/tex]
Multiply both sides by 3.
[tex]2400 =8x[/tex]
Divide both sides by 8.
[tex]300 =x[/tex]
Now,
[tex]Length = 300\text{ feet}[/tex]
[tex]Width=\dfrac{1}{3}(300)+60[/tex]
[tex]Width=100+60[/tex]
[tex]Width=160\text{ feet}[/tex]
Therefore, the length and width of the football field are 300 feet and 160 feet respectively.
