The length of a new rectangular playing field is 6 yards longer than quadruple the width. If the perimeter of the rectangular playing field is 552 ​yards, what are its​ dimensions?

Answer 1

L= 4x +6....................................given information. (length )

W=x . .........................................given information. ( width )

P=perimeter = 2x+2y...................P= perimeter of the rectangle . x=w, y=L

552 yards= 2L +2w..................... using the formula for the perimeter

552-2L = 2W. ............................subtracting 2L in both sides

552- 2(4w)=2W........................... substitute. L= 4W

552-8W=2W ...........................multiplying 2(4W)= 8W

552= 8W+2W .......................grouping like terms

552= 10W

(552)/(10) = W ...........................dividing both sides by 10 I got 55,2=W

55.2 yds =w

L=4W =4(55.2)=220.8 yds .........................substitute w in L =4W. I got 220,8 yds.

The dimensions are 55.2 yds by 220.8 yds.