Question 478619: 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 of inner surface area of the walkadjacent to the poolside.but i thinkthis is wrong too.
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 scott8148(6628)   (Show Source):
You can put this solution on YOUR website! you are kind of on the right track
the area of the pool is 600 (30 * 20), this is also the area of the walk
___ so the area of the walk plus the pool is 1200
if w is the width of the walk ___ (30 + 2w)(20 + 2w) = 1200
(15 + w)(10 + w) = 300 ___ w^2 + 25w + 150 = 300 ___ w^2 + 25w - 150 = 0
(w + 30)(w - 5) = 0
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