Question

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.

Answer 1

Ok, lets draw a picture

Answer 2

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

Answer 3

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?

Answer 4

what sis you subtract to get 412?

Answer 5

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.

Answer 6

Do you understand?

Answer 7

yes

Answer 8

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.

Answer 9

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

Answer 10

ok

Answer 11

Can you FOIL out the problem for me?

Answer 12

512+96x+4x^2

Answer 13

right?

Answer 14

4x^2 + 96x - 412 = 0

This is what you should have remaining.

Answer 15

yes

Answer 16

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

Answer 17

No so at this point, we will be utilizing the quadratic formula.

\[x = \frac{ -b \pm \sqrt (b^2 -4ac) }{ 2a }\]

Answer 18

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

Answer 19

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

\[12 \pm \sqrt 247\]

Answer 20

Since we want a positive number use:

\[x = 12 + \sqrt 247\]

That is our width.