You need to use the formula of perimeter of rectangle, such that:
`P = 2(L + w)`
`L` represents the length of rectangle
`w` represents the width of rectangle
The problem provides the information that the rectangle athletic field is twice as long as it is wide, hence, translating into equation yields:
`L = 2w`
Replacing 180 for perimeter and `2w` for `L` in equation of perimeter, yields:
`180 = 2(2w + w) => 180 = 2*3w => 180 = 6w => w = 180/6 => w = 30 yards => L = 2*30 => L = 60` yards
Hence, evaluating the dimensions of the rectangle athletic field, under the given conditions, yields `L = 60` yards and `w = 30` yards.
The athletic field is shaped as a rectangle. Its length is twice its width.
Let the length of the field be represented by L and the width is represented by W.
As the length is twice the width, L = 2*W
The perimeter of the field is 2*L + 2*W
2*L + 2*W
= 4*W + 2*W
The perimeter is 180 yards.
6*W = 180
W = 180/6 = 30
The width of the field is 30 yards and the length is 60 yards.