Question

The perimeter of a rectangular playing field is 244 feet. If its length is 2 feet longer than twice its width, what are the dimensions of the field?

L=82 and W=40
Step-by-step explanation:
The answer is W=40 and L=82
If you’re wondering how to solve it, then let “P” represent perimeter “L” represent length and “W” represent width.
P=2L+2W
L=2W+2
244=2(2W+2)+2W
244=4W+4+2W
244=6W+4
6W=244
244/6W=
W=40
Now plug the 40 in for w in: L=2W+2
L=2(40)+2
L=80+2
L=82

2. thanhha
The dimensions of the field are; L=82 and W=40

### What is the perimeter?

It’s the sum of the length of the sides used to make the given figure.
Given that perimeter of a rectangle playing field is 2the 44 feet. If its length is 2 feet longer than twice its width, then;
let “P” represent perimeter “L” represent length and “W” represent width.
Therefore, the  perimeter of a rectangular playing field,
P=2L+2W
Also, L = 2W+2
Thus, P=2L+2W
244=2(2W+2)+2W
244=4W+4+2W
244=6W+4
6W=244
244/6W
W=40
Now plug the 40 in for w;
L=2W+2
L=2(40)+2
L=80+2
L=82
Hence, the dimensions of the field are; L=82 and W=40