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Click here to see ALL problems on Geometry Word ProblemsQuestion 503498: a community flower garden measuring 12 meters by 16 meters is to have a pedestrian pathway installed all around it, increasing the total area to 285 square meters. What ill be the width of the pathway?
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Borders are interesting because you have to remember that you have to pave the 4 corners.
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The area of the garden = 12*16 = 192 square meters.
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Then you add the border with width 'x'.
So,
Length is now L + 2x
Width is now W + 2x
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We are told:
(L+2x)(W+2x) = 285
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(12+2x)(16+2x) = 285
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192 +24x +32x +4x^2 = 285
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4x^2 + 56x + 192 = 285
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4x^2 + 56x - 93 = 0
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Solved by pluggable solver: SOLVE quadratic equation with variableQuadratic equation (in our case ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=4624 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: 1.5, -15.5. Here's your graph:
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You cannot have a negative width, so the border is 1.5 m.
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Always check your work.
L = 12 + 2(1.5) = 15
W = 16 + 2(1.5) = 19
15*19 = 285
Correct.
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Answer: The border is 1.5 meters wide.
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A garden measuring 12 meters by 16 meters have 1.5 metes width pathway installed all around it, increasing the total area to 285 square meters.
Given :
A garden measuring 12 meters by 16 meters.
A pathway installed all around it, increasing the total area to 285 square meters.
Let 'x' be the width of the pathway installed
After pathway is installed then the length of the garden becomes [tex]x+12+x=12+2x[/tex]
Width of garden including pathway is
[tex]x+16+x=16+2x[/tex]
Now we find the area using new length and width
[tex]Area = length \cdot width \\Area =(12+2x)(16+2x)\\Total \; area \; 285 \\285=(12+2x)(16+2x)[/tex]
Now solve the equation for x to find out the width of the pathway
[tex]285=(12+2x)(16+2x)\\285=192+56x+4x^2\\192+56x+4x^2-285=285-286\\4x^2+56x-93=0[/tex]
We got an quadratic equation.. To solve for x use quadratic formula
[tex]x=\frac{-56\pm \sqrt{56^2-4\cdot \:4\left(-93\right)}}{2\cdot \:4}\\x=\frac{-56\pm \:68}{2\cdot \:4}\\x=\frac{-56+68}{2\cdot \:4},\:x=\frac{-56-68}{2\cdot \:4}\\x=1.5, x=-15.5[/tex]
Width x cannot be negative
So width of the pathway is 1.5 meters
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