A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a constant 2 feet wide and has an area of 196 square feet. Find the dimensions of the pool.

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

FWT wrote:A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a constant 2 feet wide and has an area of 196 square feet. Find the dimensions of the pool.

Start by drawing a picture:

+--------------+

| |

| +--------+ |

| | | |

| | | |

| +--------+ |

| |

+--------------+

Since the length of the pool (the inner rectangle) is defined in terms of the width, pick a variable for the width, create an expression for the length, and label:

+--------------+

| |

|--+--------+ |

| | | |

|w | | |

|--+--------+ |

| | 2w | |

+--------------+

What expression will represent the area of the pool? (Hint: Multiply the length and width!)

You are given that the walkway (the outer rectangle) is 2 units wide. If the width of the pool is "w", and if the walkway adds two units to either end of that, what expression then represents the width of the outer rectangle? Use the same reasoning to find an expression for the length of the outer rectangle.

You are given the area of the outer portion, so multiply the "width" and "length" expressions for the whole area, subtract the expression for the inner area, and simplify to find an expression for just the outer portion. Then set this equal to the given "walkway only area" value, and

Remember to interpret your solution(s) within the context of the exercise. For instance, you can't have negative lengths.

my picture is like this

so the outside width is w+2+2=w+4 and th elength is 2w+2+2=2w+4

the pool area is 2w2

the whole area is (w+4)(2w+4)=2w2+4w+8w+16=2w2+12w+16

the outside part is 2w2+12w+16-2w2=12w+16

set this equal 12w+16=196

12w=180 so w=15

is that rite?

+--------------+

| |2

|--+--------+ |

| | | |

|w | | |

|--+--------+ |

| | 2w | |2

+--------------+

2 2

so the outside width is w+2+2=w+4 and th elength is 2w+2+2=2w+4

the pool area is 2w2

the whole area is (w+4)(2w+4)=2w2+4w+8w+16=2w2+12w+16

the outside part is 2w2+12w+16-2w2=12w+16

set this equal 12w+16=196

12w=180 so w=15

is that rite?

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

FWT wrote:...12w=180 so w=15

is that rite?

You can check the answer to any "solving" exercise by plugging your solution back into the original exercise.

In your case, if the width of the pool is 15, then the length, being twice the width, is 30, and the pool area is 15 by 30, or 450 square units. Since the walkway adds another 2 units on either end, the entire area is 34 by 19, or 646 square units. Subtracting the pool area from the entire area gives you the area of just the walkway.

Do the numbers work out right?