A new youth activity center is being built in Erie. The perimeter of the rectangular playing field is 490 yards. The length of the field is 5 yards less than quadruple the width. What are the dimensions of the playing field?

P = 2(L + W)
P = 490
L = 4W - 5
now we sub and solve
490 = 2(4W - 5 + W)
490 = 2(5W - 5)
490 = 10W - 10
490 + 10 = 10W
500 = 10W
500/10 = W
50 = W <===

L = 4W - 5
L = 4(50) - 5
L = 200 - 5
L = 195 <===

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lisboa lisboa · Answer 2 · 2021-09-09T18:12:21+02:00

Length of the rectangle = 490 yards

Width of the rectangle = 50 yards

Given :

The perimeter of the rectangular playing field is 490 yards.

Let 'w' be the width of the rectangle

Length is 5 yards less than the quadruple the width

length = 4 times width -5

[tex]Length (L)=4w-5[/tex]

Perimeter of rectangle = 2(length + width )

Replace the length and width

[tex]Perimeter =2(L+W)\\P=2(4w-5+w)\\P=2(5w-5)\\P=10w-10[/tex]

Given perimeter is 490 yards. Replace perimeter by 490

[tex]490=10w-10\\490+10=10w\\500=10w\\Divide \; by \; 10\\w=50[/tex]

Length =[tex]4w-5=4(50)-5=195[/tex]

Length of the rectangle = 490 yards

Width of the rectangle = 50 yards

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