SOLUTION: a footpath of uniform width run round the inside of a rectangular field 38 meter long and 32 meter wide. If path occupies 600 meter square .Find the width ?

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Question 915319: a footpath of uniform width run round the inside of a rectangular field 38 meter long and 32 meter wide. If path occupies 600 meter square .Find the width ?
Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
38%2A32+-+900 = 616 = (38-2w)(32-2w)
4w^2 - 140w + 600 = 0
w^2 - 35w + 150 = 0
(w-30)(x-5) = 0
w = 5m, the width of the path
And...checking
28*22 = 616

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Outer length of the rectangular field = 38 m
Outer breadth of the rectangular field = 32 m
Outer Area:A%5Bo%5D+=+38m%2A+32m+=+1216+m%5E2
Let the width of the path be 'w' meters
Inner length of the rectangular field:L%5BI%5D+=+%2838-+2w%29m
Inner breadth of the rectangular field:W%5BI%5D+=+%2832-2w%29m
Inner area A%5BI%5D=+%2838+-2w%29%2832-2w%29+=1216-76w-64w%2B4w%5E2+=4w%5E2-140w+%2B+1216
Area of the path:A%5Bp%5D+=+A%5Bo%5D+-A%5BI%5D+=+1216-%284w%5E2+-+140w+%2B+1216%29=+600

1216-%284w%5E2-140w+%2B+1216%29=+600
1216-%284w%5E2+-140w+%2B+1216%29-+600=0
616+-4w%5E2+%2B+140w+-1216=0
-4w%5E2+%2B+140w+-600=0
4w%5E2+-140w+%2B600=0
4%28w%5E2+-35w+%2B150%29=0
4%28w%5E2-30w-5w%2B150%29=0
4%28%28w%5E2-30w%29-%285w-150%29%29=0
4%28w%28w-30%29-5%28w-30%29%29=0
4%28w-5%29%28w-30%29=0
solution

if w-5=0=>w=5
if w-30=0=>w=30
Since the outer breadth of the field is 32m, breadth of the rectangular field is 5m.