SOLUTION: Here it is copied and pasted exactly out of the book. Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining

Algebra ->  Radicals -> SOLUTION: Here it is copied and pasted exactly out of the book. Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining       Log On


   

Question 42285This question is from textbook
: Here it is copied and pasted exactly out of the book.
Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path?

Paul answered it as
400=(30+2x)(20+20)-(30*20)
400 = 4x^2 + 100x
x^2 + 25x - 100 = 0
Solve that and you get x = 3.5'

You replied with this can't be solved, so I am confused now. This question is from textbook

Found 2 solutions by PaulAllen65270, 303795:
Answer by PaulAllen65270(19) About Me  (Show Source):
You can put this solution on YOUR website!
It can be solved but I am not complete sure how to do it. What is there is almost correct because the complete area of the garden minus the are of the path would be equal to the remainder of the garden. If you were to draw the garden out with path included you would see that the width would be the same for both path and garden however the length would be shorter for the path then the whole garden by 2x since the path is uniform.
400=%2830%2A20%29-%2820%2830-2x%29%29
400=%28600%29-%28600-40x%29
400=600-600%2B40x
400=40x
x=10 so the width of the path is 10 feet.

Answer by 303795(602) About Me  (Show Source):
You can put this solution on YOUR website!
I note that you have two solutions for this problem but to me they are both incorrect.
The area for the garden is 30 m x 20 m. This is a total area of 600 square metres. If you get a final area of the garden as 400 square metres then the other 200 metres of the original garden area must be the path. The path is therefore on the inside of the 30 x 20 metre area. If the width of the path is x then the garden area inside the path must be (30-2x)(20-2x) which must be equal to 400 square metres
%2830-2x%29%2820-2x%29=400
600-100x%2B4x%5E2=400
Subtract 400 from each side of the equation
200-100x%2B4x%5E2-400=0
4x%5E2-100x%2B200=0
Divide through by 4
x%5E2-25x%2B50=0
The use of the quadratic equation then gives answers of x = 22.81 or x = 2.19
Obviously the first is not a reasonable answer so the path width must be 2.19m.

Check (The length - the path is therefore 25.62 m and the width - the path is 15.62 m so the area inside the path is 400.18 square metres (a rounding error accounts for the small difference))