a rectangle athletic field is twice as long as it is wide. If the perimeter of the athletic field is 180 yards, what are it's dimensions?

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.

tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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

= 6*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.

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