Answer
length of the rectangular field is $126$ yard
and width of the rectangular field is $ 44 $ yard
its dimensions are $44\; yd \; by\; 126\; yd$.
Work Step by Step
Let $l$ represents the length.
and $w$ represents the width.
The rectangular field is $6$ yards less than triple the width.
In equation form.
$l=3w-6$ ... (1)
the perimeter of the soccer field is $P=2l+2w$.
From the question $P=340$ yards plug into above equation.
$340=2l+2w$
Substitute the value of $l$ from equation (1).
$340=2(3w-6)+2w$
$ 340=6w-12+2w $
$ 340+12=8w $
$ 352 = 8w $
$ \frac{352}{8}=w $
$ 44=w $
Substitute above value into equation (1).
$l=3(44)-6$
$ l=132-6 $
$ l=126 $.
