candygirl200:

An inground rectangular pool has a concrete pathway surrounding the pool. If the pool is 16 feet by 32 feet and the entire area of the pool including the walkway is 924 ft2 , find the width of the walkway.

2 weeks ago
Shadow:

Ok, lets draw a picture

2 weeks ago
Shadow:

|dw:1511233272623:dw|

2 weeks ago
Shadow:

This is the information that we have so far. Length of Pool = 32ft Width of Pool = 16ft Area of Pool = 512ft Area of Pool + Walkway = 924ft Therefore if we subtract the Area of the Pool, we get the Area of the Walkway which is 412ft

2 weeks ago
Shadow:

Now let the variable x stand for the width of our walkway. If 32ft is the length of our pool then 32 + 2x = the length of the entire area. Do you see how this makes sense?

2 weeks ago
candygirl200:

what sis you subtract to get 412?

2 weeks ago
Shadow:

The entire area of the pool + the walkway = 924ft They gave us the dimensions of the pool, it's length and width. I multiplied those dimensions to get the area of the pool. Then I subtracted the area of the pool from the total area to get just the area of the walkway.

2 weeks ago
Shadow:

Do you understand?

2 weeks ago
candygirl200:

yes

2 weeks ago
Shadow:

So with the information we have in front of us. lets write an equation to solve for x. Area = Length times Width (32 + 2x)(16 + 2x) - 512 = 412 So what this is saying, is the length of the pool plus two times the width of the sidewalk is our whole length of the given area. This stands in as one of our dimensions. Next we have 16 + 2x to represent the width of our whole area. This is our second dimension. If we multiply those two numbers, we get the entire area of the pool + walkway. Therefore if we subtract the area of the pool from it, we should get the area of our walkway.

2 weeks ago
Shadow:

Let me know if you understand that so we can solve.

2 weeks ago
candygirl200:

ok

2 weeks ago
Shadow:

Can you FOIL out the problem for me?

2 weeks ago
candygirl200:

512+96x+4x^2

2 weeks ago
candygirl200:

right?

2 weeks ago
Shadow:

4x^2 + 96x - 412 = 0 This is what you should have remaining.

2 weeks ago
candygirl200:

yes

2 weeks ago
candygirl200:

and then u divide by 4 and get x^2+4x-103=0

2 weeks ago
Shadow:

No so at this point, we will be utilizing the quadratic formula. \[x = \frac{ -b \pm \sqrt (b^2 -4ac) }{ 2a }\]

2 weeks ago
Shadow:

a = 4, b = 96, c = -412

2 weeks ago
Shadow:

You were correct on that earlier. We divide by 4 and and use x^2 + 24x - 103> a = 1, b = 24, c = -103 \[\frac{ -24 \pm \sqrt(24^2 -4(1)(-103) }{ 2(1) }\] \[\frac{ 24 \pm \sqrt(576 + 412) }{ 2 }\] \[\frac{ 24 \pm \sqrt 988 }{ 2 }\] I notice that 4 times 247 = 988 \[\frac{ 24 \pm \sqrt (4 \times 247) }{ 2 }\] So we can pull out the 4, as a 2 \[\frac{ 24 \pm 2 \sqrt 247 }{ 2}\] Cancel out, using the denominator two \[12 \pm \sqrt 247\]

2 weeks ago
Shadow:

Since we want a positive number use: \[x = 12 + \sqrt 247\] That is our width.

2 weeks ago
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