Question 478617: A swimming pool 30 ft. long and 20 ft. wide has a cement walk of uniform width around it. If the area of the walk is the same as the area of the surface of the pool, how wide is the Walk?
I am having problems setting up the equation(s) to solve this as I am somewhat confused re. the information provided in the last sentence of the problem.
How can the the area of the walk be the same as the area of the the surface of the pool?
The mental image I have of this problem is the outer area of the walk is the 20 ft. by 30 ft. dimensions,then the inner dimensions(which are not given) are for the pool itself,separated by the width of the walk?
I have no idea how to set it up.
I tried using (30 - 2x)(20 - 2y)= 600
and got 2x^2 -100 xy + 2y^2= 3600
i don't think this is right.
so i tried:
2x + 2y =100 (perimeter)
xy = 600 (Area of outer area of the walk which I think,but am not sure is the 20 ft. by 30 ft.dimensions.
Please help me.
I appreciate any help given.
Despi
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A swimming pool 30 ft. long and 20 ft. wide has a cement walk of uniform width around it.
If the area of the walk is the same as the area of the surface of the pool, how wide is the Walk?
:
Let x = the width of the walk
If the area of the pool and the walkway is the same, the overall area = 1200
:
the equation should be, (we don't need y)
(2x + 30)(2x + 20) = 1200
FOIL
4x^2 + 100x + 600 - 1200 = 0
4x^2 + 100x - 600 = 0
Simplify divide by 4
x^2 + 25x - 150 = 0
Factors to
(x + 30)(x - 5) = 0
the positive solution is all we want here
x = 5 ft is the width of the walkway
:
:
Check this by finding the overall area with 2x=10
40 * 30 = 1200, half is pool and half is walkway
:
You seemed to have the right idea, but were confused about the details. C
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