Solving Pool Dimensions - Length = 30 ft, Width = 60 ft

[bacon]

The length of a rectangular swimming pool is twice its width. The pool is surrounded by a cement walk 4 ft wide. If the area of the walk is 748 square feet, determine the dimensions of the pool.

A= the total area bounded by the outer edge of the surrounding walk
W= the area of the walk
P= the area of the pool
x= the length of the short side of the pool

A=P+W
P=x2x=$$2x^{2}$$
W=748
A=(x+8)(2x+8)=$$2x^{2}+24x+64$$=$$2x^{2}+748$$
solving for x gives x=28.5 feet and 2x=57 feet
The answer sheet says the dimensions are 30 and 60 feet.
Where am I going wrong?
Thanks for any replies.

 

[Snazzy]

Well, if you plug 30 and 60 back in, you get 784 so maybe it's a typo.

Or they rounded.

 

[symbolipoint]

Did you draw and label a picture?

 

[bacon]

Did you draw and label a picture?

Yes. I just don't know how to reproduce it in the post.

 

[symbolipoint]

Yes. I just don't know how to reproduce it in the post.

Good. I believe you and see now that some of what you showed indicated that you used a picture. I'm working on the exercise right now. Someone will probably write a response before I finish. So far, I'm looking at A of pool equals A of wholeRectangle minus A of walkway. I obtained 2x^2 = (x+8)(2x+8) - 748, and I'm not finished.

 

[symbolipoint]

bacon, in fact, I found the same results as you did: x=28.5, 2x=57