SOLUTION: A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width if the area of the pool and path combined is 416 square meters what is the width of the path

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Question 895844: A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width if the area of the pool and path combined is 416 square meters what is the width of the path
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Total dimensions: Using w for width of the path,
2w+10 and 2w+20.
%282w%2B10%29%282w%2B20%29=416 to account for combined pool area and path area.

Simplify.
2%28w%2B5%292%28w%2B10%29=2%2A2%2A104
%28w%2B5%29%28w%2B10%29=104
w%5E2%2B15w%2B50=104
w%5E2%2B15w-54=0
Discriminant, 15%5E2%2B4%2A54=441=21%5E2
w=%28-15%2B21%29%2F2
highlight%28w=3%29

That quadratic IS factorable as (w-3)(w+18) and the equation could be solved through factoring.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width if the area of the pool and path combined is 416 square meters what is the width of the path

Let UNIFORM width of path, be W
Then width on either end of pool = W + W, or 2W
Area of pool and path = (2W + 10)(2W + 20), and area of pool and path = 416.
Thus, (2W + 10)(2W + 20) = 416
4W%5E2+%2B+60W+%2B+200+=+416
4W%5E2+%2B+60W+%2B+200+-+416+=+0
4W%5E2+%2B+60W+-+216+=+0
4%28W%5E2+%2B+15W+-+54%29+=+4%280%29
W%5E2+%2B+15W+-+54+=+0
(W - 3)(W + 18) = 0
W, or width of path = highlight_green%28highlight_green%283%29%29 m
You can do the check!!
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