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a cement walk of constant width is built around a 20-ft by 40-ft rectangular pool. The total area of the pool and the walk is 1500ft sq. Find the width of walk

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3 Answers

I drew this because pictures help me visualize the problem.
I named the rectangles 1, 2, 3, and 4. The width of the walk is x.
________
|  |__3__|  | x
|  |         |  |          
|1|         |2 |            
|  |         |  |
|  |_____|  |
|_|__4__|_|
x

Let x be the width of the walk. The area of the pool would be 800 square feet (20x40). So the area of the walk must be 1500 - 800, or 700 square feet. Now we have to find ways to represent the area of the walk using only these values: 20, 40, x, and the total area, 700.
First, break up the walk into 4 rectangles. In the picture above, I broke the walk up into two long rectangles (#1 and 2) whose lengths are: the length of the pool (40) plus an x for the width of the walk at the top of the pool, plus another x to account for the width of the walk at the bottom of the pool. So their lengths are 40 + 2x. And their widths are x, the width of the walk.
I also have two small rectanges (#3 and 4) whose lengths are the width of the pool (20) and whose widths are x, the width of the walk.
Now that we now the lengths and widths of all four rectangles, we can find their areas (LxW):

Area of rectangle 1: (40 + 2x)(x) = 40x + 2x2
Area of rectangle 2: (40 + 2x)(x) = 40x + 2x2
Area of rectangle 3: (20)(x) = 20x
Area of rectangle 4: (20)(x) = 20x

As we already found, we know that the area of the entire walk must be 700 sq ft. So we add the areas of all 4 rectangles together and set it equal to 700. Then we can solve for x, the width of the walk.
40x + 2x2 + 40x + 2x2 + 20x + 20x = 700 *add all areas together, set = to 700
4x2 + 120x = 700 *combine like terms
4x2 + 120x - 700 = 0 *subtract 700 from both sides

*Note: Now that the right hand side of the equation is 0, we can try to factor this quadratic equation.
*Since every number is divisible by 4, we can simplify the problem to:

x2 + 30x - 175 = 0

*At this point you can factor or use the quadratic formula. This does factor easily:

(x - 5)(x + 35) = 0

*Split into two problems:
x - 5 = 0 OR x + 35 = 0
x = 5 OR x = -35

**Since the width of anything cannot be negative, we know the answer is x = 5. So the width of the walk is 5 feet all the way around the pool.**

Comments

The width cannot be 5 feet.
40 x 20 = 800
 
(40 + 5) x (20 + 5) = 1500
(45)(25) ? 1500
 
However...
 
(40 + 10) x (20 + 10) = 1500
(50)(30) = 1500
 
Yes, precisely.. The reason why 10 works is because the walk is 5 feet on BOTH sides of the pool. The first equation should be written:
 
(40 + 2x)(20 + 2x) = 1500
 
Now you can substitute 5 in for x.
You cannot double the x, without doubling the 40 and 20.
 
The pool and walkway is 50 x 30=1500
 
The pool is 40 x 20=800
 
The walkway is 10 on all four sides.
Thank you for the answer, I had originally gotten that after going over the question, but was not sure of it. 
Good for you! Glad to help assure you :)
Hi Josh;
Thank you for asking this question.
 
(20 + x) (40 + x) = 1500
 
x represents the additional width you are looking for.
 
I do not know your educational background.  I am therefore going to assume you have not yet learned FOIL.
 
FIRST...20 x 40 = 800
OUTER...20x
INNER...40x
LAST...(x)(x)=x2
 
 
The equation is now...
800 + 20x + 40x + x2=1500
 
Let's simplify and subtract 800 from both sides...
 
-800 + 800 + 60x+ x2=1500-800
 
Simplify...
 
60x + x2=700
 
The answer can only be obtained through visual inspection.  I tried everything to isolate the x and calculate.
 
x=10 feet.
!!!!!!!!!!!!!!!!!!!!
 
 
 
 
Let x be the constant width of the walk that we need to find. Then the total area of the pool and the walk is given by
 
A = (40+2x)(20+2x) = 4x²+120x+800
 
Set this equal to 1500 and simplify:
 
4x²+120x+800=1500
4x²+120x-700=0
x²+30x-175=0
 
Use the quadratic formula or factoring to find the two solutions of this quadratic equation:
 
x = -35 and 5
 
Since x has to be positive, the solution is x=5 ft.