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?
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.