Question

A new youth center is being built in erie . The perimeter of the rectangular playing field is 296 yards. The length of the field is 2 yards less than double the width. What are the dimensions of the playing​ field?

Answer 1

`\Large \boxed{\mathrm{98 \ yards \cdot 50 \ yards}}`

Explanation:

Let the length be l.

Let the width be w.

`l = 2w - 2`

The perimeter of the rectangle is 296 yards.

`P = 2l + 2w`

Plug in the value for l and P.

`296 = 2(2w-2) + 2w`

Evaluate and solve for w.

`296 = 4w - 4 + 2w \\\\296 = 6w - 4 \\\\300 = 6w \\\\50 = w`

The width is 50 yards.

Let w = 50 for l.

Evaluate and solve for l.

`l = 2(50) - 2 \\ \\ l = 100 - 2 \\\\l = 98`

The length is 98 yards.

Answer 2

The dimensions are 50 yards by 98 yards

Explanation:

If the width is w, then the length is 2w - 2. Perimeter can be expressed by 2 * (length + width), therefore:

2 * (2w - 2 + w) = 296

2 * (3w - 2) = 296

3w - 2 = 148

3w = 150

w = 50 so 2w - 2 = 2 * 50 - 2 = 98