2.19 ft
The uniform-width path is of width X. The remaining area will be (20-2x) ft in width, and (30-2x) ft in length. 2x because x is the width decreased by the path, on each side. So the remaining area can be expressed as (20-2x)(30-2x), or 4x^2-100x+600. We're told it is equal to 400. So we have 4x^2-100x+600=400. This becomes 4x^2-100x+200=0, and from there, you can solve it using the quadratic formula. This will yield two possible values of x, but one would make the path longer than the garden. The remaining value is the solution, approx. 2.19 ft.
2006-08-20 19:15:03
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answer #1
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answered by Master Maverick 6
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The path is 2 ft wide if rounded to the nearest whole number.
The easiest way to check this is to subtract 4 feet from each of the given dimensions of lenghth and width. You do that because you must account for the path being on both sides of the space. When you multiply 26x16, you end up with 416, which is as close as you are going to come to your exact area without having to get really technical about it. Your path is actually a tiny bit larger than 2 ft, but that's the closest whole number.
2006-08-21 02:18:45
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answer #2
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answered by Bronwen 7
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yep.. 2ft is exactly right.
The border has to have an area of 200 sq ft. (600 sq ft - 400 sq ft).
If the width of the path is x then to find it's area you would have 4 rectangles - 2 would be x wide and 20 ft long and 2 would be x wide and 30 ft long (the side lengths of the garden).
2*20*x + 2*30 *x = 200
40x+60x=200
100x=200
x=2ft
2006-08-21 02:13:29
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answer #3
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answered by goodlittlegirl11 4
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x is the width of the path.
(30 - 2x)(20 - 2x) = 400
4(15 - x)(10 - x) = 400
(15 - x)(10 - x) = 100
150 - 25x + x^2 = 100
x^2 - 25x + 50 = 0
x = [25 +/- sqrt(625 - 200)] / 2
x = [25 +/- sqrt(425)] / 2
x = (5/2) [5 +/- sqrt(17)]
x = 2.2 feet
Check: the garden, not counting the path, is
(30 - 4.4)(20 - 4.4) = 25.6 * 15.6 = 399.36
With rounding off to the nearest tenth, that's good enough.
[Edit] BTW, 2 feet is right.
(30 - 4)(20 - 4) = 26 * 16 = 416, not 400.
2006-08-21 02:16:11
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answer #4
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answered by bpiguy 7
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2 ft
2006-08-21 02:07:38
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answer #5
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answered by Tony T 4
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2.2 ft (to the nearest tenth)
Doug
2006-08-21 02:24:10
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answer #6
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answered by doug_donaghue 7
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