SOLUTION: Joannie wants to enclose a rectangular area with 408 meters of fencing. If the width is 24 metes less than the length, what will the length and width of the enclosed area be?

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Question 65465This question is from textbook Essential Algebra
: Joannie wants to enclose a rectangular area with 408 meters of fencing. If the width is 24 metes less than the length, what will the length and width of the enclosed area be? This question is from textbook Essential Algebra

Found 2 solutions by uma, stanbon:
Answer by uma(370) (Show Source):
You can put this solution on YOUR website!
Let the length of the field be x m
Then the width = (x - 20)m
The perimeter = 2(l + w)
= 2(x + x - 20)
= 2(2x - 20)
Given the perimeter = 408.
THus the equation is 2(2x - 20) = 408
==> 2x - 20 = 204 [dividing by 2 throughout]
==> x - 10 = 102 [dividing by 2 again]
==> x = 102 + 10 [adding 10 to both the sides]
==> x = 112
Thus the length of the field = 112 m
The width = x - 20
= 112 - 20
= 92 m

Good Luck !!!

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Joannie wants to enclose a rectangular area with 408 meters of fencing. If the width is 24 meters less than the length, what will the length and width of the enclosed area be?
----------------
Perimeter = 2(length + width)=408
length + width = 204 meters
Let length be "x" ft.
Then width is "x-24" ft.
x+x-24=204
2x=228
x=114 meters
x-24=90 meters
Area= length * width
=90*114
=10260 sq meters
Cheers,
Stan H.