How wide a strip must be mowed around a rectangular grass plot 40.0m by 60.0m for one half of the grass to be mowed?
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x = border width
(60-2x)(40-2x) = 60*40/2 = 1200
2400-200x+4x^2=1200
4x^2-200x+1200=0
x^2-50x+300 = 0 use the quadratic formula
x = 6.972 m
(60-2x)(40-2x) = 60*40/2 = 1200
2400-200x+4x^2=1200
4x^2-200x+1200=0
x^2-50x+300 = 0 use the quadratic formula
x = 6.972 m
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Just map out the grass plot on a piece of a paper
Imagine a smaller rectangle with in the grass plot (which is the grass not cut)
to get the length of the sides of the smaller rectangle simply divide 40 and 60 by 2
40/2 = 20
60/2 = 30
But, the stripe around it must have the same width all around.
calculate the distances from the sides of the small rectangle to the big rectangle
40 - 20 = 20
10/2 = 10
60 - 30 = 30
30/2 = 15
Get an average distance between these lengths:
15 + 10 = 25
25/2 = 12.5
Width of strip = 12.5 m
Imagine a smaller rectangle with in the grass plot (which is the grass not cut)
to get the length of the sides of the smaller rectangle simply divide 40 and 60 by 2
40/2 = 20
60/2 = 30
But, the stripe around it must have the same width all around.
calculate the distances from the sides of the small rectangle to the big rectangle
40 - 20 = 20
10/2 = 10
60 - 30 = 30
30/2 = 15
Get an average distance between these lengths:
15 + 10 = 25
25/2 = 12.5
Width of strip = 12.5 m
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let x be the width of the strip
so clearly area
=2(x*60 +(4-2x)*x){used 40-2x as the area of the four squares has already been counted }
other way
2(60*x{area of strip 1}+40*x{area of strip 2}-(4x^2){area of the 4squares on the corners ,as these have been counted twice }
so clearly
we get 4x^2-200x+1200=0
just solve the eq.(m getting about 6.97m)
hope it helps..
so clearly area
=2(x*60 +(4-2x)*x){used 40-2x as the area of the four squares has already been counted }
other way
2(60*x{area of strip 1}+40*x{area of strip 2}-(4x^2){area of the 4squares on the corners ,as these have been counted twice }
so clearly
we get 4x^2-200x+1200=0
just solve the eq.(m getting about 6.97m)
hope it helps..
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The area of the strip is Sₐ = (40.0 m • 60.0 m)/2.
Call the width of the strip s.
The length of the strip is 2((40.0 m - s) + (60.0 - s)) ∴
s = (40 • 60)/2 ÷ 2(40 - s + 60 - s)
s = (2400)/2 ÷ 2(100 - 2s)
s = 300/(50 - s)
(50 - s)s = 300
50s - s² = 300
s² - 50s + 300 = 0
s = (50 ± √(50² - 4 • 300))/2 (quadratic equation)
s = (50 ± √(2500 - 1200))/2
s = (50 ± √1300)/2
s = (50 ± 36.056)/2
s = 86.056/2 or 13.944/2
s = 6.972 meters.
The first zero is ruled out because the strip would be wider than the plot.
Call the width of the strip s.
The length of the strip is 2((40.0 m - s) + (60.0 - s)) ∴
s = (40 • 60)/2 ÷ 2(40 - s + 60 - s)
s = (2400)/2 ÷ 2(100 - 2s)
s = 300/(50 - s)
(50 - s)s = 300
50s - s² = 300
s² - 50s + 300 = 0
s = (50 ± √(50² - 4 • 300))/2 (quadratic equation)
s = (50 ± √(2500 - 1200))/2
s = (50 ± √1300)/2
s = (50 ± 36.056)/2
s = 86.056/2 or 13.944/2
s = 6.972 meters.
The first zero is ruled out because the strip would be wider than the plot.