SOLUTION: a rectangular pool measuring 46 feet by 78 feet is surrounded by a path of uniform width around the 4 edges. the perimeter of the bigger rectangle formed by the pool and the surrou

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: a rectangular pool measuring 46 feet by 78 feet is surrounded by a path of uniform width around the 4 edges. the perimeter of the bigger rectangle formed by the pool and the surrou      Log On


   

Question 982388: a rectangular pool measuring 46 feet by 78 feet is surrounded by a path of uniform width around the 4 edges. the perimeter of the bigger rectangle formed by the pool and the surrounding path is 320 feet. find the width of the path.
Found 2 solutions by macston, Cromlix:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
x=width of path; P=perimeter; L=length; W=width
.
P=2(L+2x+W+2x)
320 feet=2(78ft+46ft+4x)
160 feet=124feet+4x
36 feet=4x
9 feet=x
.
ANSWER:The width of the path is 9 feet.

Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
Make width of path = 'x'
The width of the pool = 46ft
Width of pool + path = 46 + 2x
The length of the pool = 78ft
The length of the pool + path = 78 + 2x
Perimeter = 2 x Width + 2 x Length
320 = 2(46 + 2x)+ 2(78 + 2x)
320 = 92 + 4x + 156 + 4x
Collect like terms
4x + 4x = 320 -92 - 156
8x = 72
x = 9
The path's width is 9ft
Proof:
Perimeter = 2 x Width + 2 x Length
320 = 2(46 + 2x) + 2(78 + 2x)
320 = 2(46 + 18) + 2(78 + 18)
320 = 2(64) + 2(96)
320 = 128 + 192
320 = 320
Hope this helps:-)