a rectangular garden measures 80 feet by 60 feet.?
September 4, 2010 by Builder Wong
Filed under Garden & Landscape
a rectilinear grassed area measures 80 feet by 60 feet. A vast trail of unvaried breadth is to be combined along both shorter sides as well as a single longer sides of a garden. The landscape engineer you do a work wants to stand in a garden’s area with a further of this path. How far-reaching should a trail be?"
Current area is 4800 sq feet, double that would be 9600
Let W = width of the new path
New Length becomes 80 + 2W and new width becomes 60 + W
So:
(60 + W)(80 + 2W) = 9600
2W^2 + 200W + 4800 = 9600 or W^2 + 100W – 2400 = 0
The last equation can be factored (W+120)(W-20)
So the width must be 20 feet
80*60=4800m^2
60y+60y+(80+2y)y=4800
120y+80y+2y^2=4800
2y^2+200y=4800
y^2+100y=2400
y^2+100y-2400=0
(y+120)(y-20)=0
y= -120 OR y=20
Length cannot be negative so the width of the path is 20 feet
Draw out a picture of what the garden looks like now, 80×60.
Draw the path that goes up one 60 side, across one 80 side, and down the other 60 side.
You will see that the path makes the 80 side become 80+width of path+width of path.
You will see that the path makes the 60 sides become 60+width of path.
You can write the new area out with x being the width of the path. (80+x+x) x (60 + x). Simplify this and it’s (80+2x)(60+x).
So this new area must equal double the old area. The old area is 80×60=4800. Double that is 4800×2=9600.
So (80+2x)(60+x)=9600.
Multiply this out. 4800+80x+120X+2x^2=9600. Simplify it and you get 2x^2+200x+4800=9600. Subtract the 9600 from both sides 2x^2+200x-4800=0. You can divide by 2 to make it easier x^2+100x-2400=0.
Factor this and you will get (x+120)(x-20)=0. So x+120=0 means x= -120 and x-20=0 means x=20. You know the width can’t be -120 (it’s negative), so x=20.
Then the new area is (80+20+20)(60+20). Which is 120×80=9600!
Therefore the width of the path is 20.