Am I ever using this thing again?

Question

NO! The thing said I deleted my question when in reality I did no such thing. I've seen it happen on other forums but I though Yahoo would be better than this. PS, even though upon further examantion my conclution was false, the answer was not, in fact 50*25. Good try though.

Answer 1

Here's a copy of the question you posed before it got deleted.

"a 174 ft long fence is placed around the perimeter of a rectangular pool that has a 3 ft sidewalk around it. the actual pool w/o the sidewalk is twice as long as it is wide. what is the dimensions of the pool w/o the sidewalk? I want to see your work."

Before it went away, you had about 10 answers of 50 ft. x 25 ft. Your comments indicated you thought the answer was 52 ft. x 23 ft. but that violates the problem statement where the ratio of the pool length to pool width (without sidewalk) should be 2:1.

If it wasn't 50 ft. x 25 ft., what was the answer?

That's a pool where the length is twice as long as the width.
If you add a 3 foot sidewalk around all the sides, you get an area of 56 x 31. The perimeter of the fence enclosing the pool and the sidewalk is 2(56) + 2(31) = 174 ft., just like the original problem statement.

Solved algebraically:
Let W = width of the pool (w/o sidewalk)
Let L = length of the pool (w/o sidewalk)

From the question we know that the pool is twice as long as it is wide:
#1: L = 2W

The perimeter of the fence can be shown with this formula:
#2: 2 (L + 6) + 2 (W + 6) = 174

Divide both sides by 2:
#3: L + 6 + W + 6 = 87

Simplify:
#4: L + W + 12 = 87

Subtract 12 from both sides:
#5: L + W = 75

Substitute #1 into #5:
#6: 2W + W = 75

Add like terms:
#7: 3W = 75

Divide both sides by 3:
#8: W = 25

Substitute #8 back into #1:
#9: L = 2(25)

Simplify:
#10: L = 50

So the original dimensions of the pool (w/o sidewalk) are:
QED: Length = 50 ft., Width = 25 ft.

Since you respect the opinion of Mathgirl... I invite you to e-mail her and have her post an answer here... I'll agree to live by her answer. If she says the pool is 52 ft x 23 ft , I'll agree to eat my hat. If she says otherwise, you must agree to eat a similar article of clothing...

Answer 2

I got this:

Okay the perimeter of the fence is 174 ft which is equal to twice the size of the length plus twice the size of the width.

2L + 2W = 174ft

This fence is placed around the pool, which has a 3 ft gap from the pool to the fence. The length of the pool is going to be equal to the length of the fence, but will be 3 ft shorter on all sides.

So,

L = l + 6ft
W = w + 6ft

And the length of the pool is twice the width.

l = 2w


Just put the formula in place:

2L + 2W = 174 ft
2 ( l + 6 ft) + 2 (w + 6 ft) = 174 ft
2 (2w + 6 ft) + 2 (w + 6 ft) = 174 ft
4w + 12 ft + 2w + 12 ft = 174 ft
6w + 24 ft = 174 ft
6w = 174 ft - 24 ft
6w = 150 ft
w = 150 ft / 6
w = 25 ft

l = 2 x 25 ft = 50 ft

The length of the pool is 50 ft and the width is 25 ft

Answer 3

I got 25x50 as well.
52x23 can't be correct because 52 is not twice as much as 23.

Answer 4

I'm amazed how quickly users were able to find your question!