A new Youth Sports Center is being built in Hadleyville. The perimeter of the rectangular playing field is 282 yards. The length of the field is 3 yards less than double the width. What are the dimensions of the playing field?

Answer 1

Let's assume that the width of the playing field is "w" yards.

According to the problem, the length of the field is 3 yards less than double the width. So, the length can be expressed as:

L = 2w - 3

The perimeter of a rectangle is given by the formula:

P = 2(L + w)

Substituting the values of L and P, we get:

282 = 2[(2w - 3) + w]

Simplifying the equation, we get:

282 = 6w - 6

288 = 6w

w = 48

So, the width of the playing field is 48 yards.

Using the equation for the length, we get:

L = 2w - 3 = 2(48) - 3 = 93

So, the length of the playing field is 93 yards.

Therefore, the dimensions of the playing field are 48 yards by 93 yards.