SOLUTION: Would someone help me with this word problem. I think that it is supposed to be solved using the completing the square method.
A garden area is 30ft. long and 20 ft. wide. A path
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-> SOLUTION: Would someone help me with this word problem. I think that it is supposed to be solved using the completing the square method.
A garden area is 30ft. long and 20 ft. wide. A path
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Question 87267: Would someone help me with this word problem. I think that it is supposed to be solved using the completing the square method.
A garden area is 30ft. long and 20 ft. wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft^2, what is the width of the path? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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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Completing the square is not a very good way to solve this
If you us the quadratic formula: 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?
