SOLUTION: I am not that good with word problems and I am having problems solving this problem
A garden area is 30 ft long and 20 ft wide. A path of iniform width is set around the edge.
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A garden area is 30 ft long and 20 ft wide. A path of iniform width is set around the edge.
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Question 91763: I am not that good with word problems and I am having problems solving this problem
A garden area is 30 ft long and 20 ft wide. A path of iniform width is set around the edge. If the remaining garden area is 400 ft^2, What is the width of the path?
Thanks a million for the help Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! This problem came up a couple weeks ago, this is what I submitted then.
:
A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path?
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Draw diagram of this; label the outside dimensions of the rectangle 30 by 20.
Label the width of the path as x, it will be apparent that the dimensions of
the garden (inside the path), will be (30-2x) by (20-2x)
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The area of the garden is given as 400 sq/ft
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A simple area equation:
:
length times width = 400 sq/ft
(30-2x) * (20-2x) = 400
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FOIL:
600 - 60x - 40x + 4x^2 = 400
4x^2 - 100x + 600 = 400
4x^2 - 100x + 600 - 400 = 0
4x^2 - 100x + 200 = 0; a quadratic equation
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Simplify, divide by 4 and you have:
x^2 - 25x + 50 = 0
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We need to use the quadratic formula to solve this: a=1; b=-25;; c=50
:
:
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Two solutions:
x = 22.8, not a possible solution, obviously
and
x = 2.19 ft is the width of the path
:
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Check our solution by finding the area of the garden
We have to subtract 2x from the outside dimensions: 2*2.19 = 4.38
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(30-4.38) * (20-4.38) =
25.62 * 15.62 = 400.2 ~ 400 sq/ft
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How about this, did it make sense to you, any questions?
