Question 103834: The city is planning to build a small playground on arectangle section of a park. The perimeter of the playground is 80 feet. The two longer sides of the playground will each be 25 feet.
Now what I did was this..
80/5= 16 25/5 & 25/5=5+5=10*16=160/5=32ft square.
But I have a feeling is wrong like I'm missing something I used the L*W operation
also..
A rectangle yard is surrounded by a fence that is 26 feet long. The area of the yard is 36 ft square. Determine the dimension of the yard.
I did this...
26/2 & 36/2 which gave me
13 * 18 = 234/2= 117 square feet
again I used the L*W operation still I feel is wrong
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The city is planning to build a small playground on a rectangle section of a park. The perimeter of the playground is 80 feet. The two longer sides of the playground will each be 25 feet.
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Perimeter = 2*length + 2*width
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Let the length be 25 ft.; Let the width be "x" ft
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EQUATION:
80 = 2*25 + 2x
30 = 2x
x = 15
The width = 15 ft.
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A rectangle yard is surrounded by a fence that is 26 feet long. The area of the yard is 36 ft square. Determine the dimension of the yard.
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Formula : Area = length*width
Perimeter = 2*length + 2*width
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Let the length by x; let the width by y.
EQUATIONS:
xy = 36
2x+2y=26
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Simplify:
y = 36/x
x+y = 13
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Substitute y=36/x into x+y=13 to get:
x+(36/x)= 13
Multiply thru by x to get:
x^2 + 36 = 13x
Rearrange:
x^2 - 13x + 36 = 0
Factor:
(x-4)(x-9)=0
x = 4 or x=9
If the width is 4 ft, the length = 36/4 = 9 ft
If the length is 9 ft.,the width = 36/9 = 4 ft
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Cheers,
Stan H.
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