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3. A new Outdoor Recreation Center is being built in Del Rio. The perimeter of the rectangular playing field is 374 yards. Thelength of the field is 5 yards less than triple the width. What are the dimensions of the playing field?The width isyards.The length isyards.?

Sagot :

The perimeter of a rectangle is given by the formula

[tex]P=2L+2W[/tex]

where

L is the length

W is the width

In this problem

we have that

P=374 yd

L=3W-5

substitute the given values in the formula

[tex]374=2(3W-5)+2W[/tex]

Solve for W

[tex]\begin{gathered} 374=6W-10+2W \\ 374=8W-10 \\ 8W=384 \\ W=48\text{ yd} \end{gathered}[/tex]

Find out the value of L

[tex]\begin{gathered} L=3(48)-5 \\ L=139\text{ yd} \end{gathered}[/tex]

therefore

The length is 139 yards

The width is 48 yards