SOLUTION: Solve. A garden area is 30 feet long and 20 feet wide. A path of uniform width is set around the edge. If the remaining garden area is 400 feet squared, what is the width of the

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Question 121882: Solve.
A garden area is 30 feet long and 20 feet wide. A path of uniform width is set around the edge. If the remaining garden area is 400 feet squared, what is the width of the path?''.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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?
:
Draw diagram of this; label the outside dimensions of the rectangle 30 by 20.
Label the width of the path as x, it will be apparent that the dimensions of
the garden (inside the path), will be (30-2x) by (20-2x)
:
The area of the garden is given as 400 sq/ft
:
A simple area equation:
:
length times width = 400 sq/ft
(30-2x) * (20-2x) = 400
:
FOIL:
600 - 60x - 40x + 4x^2 = 400
4x^2 - 100x + 600 = 400
4x^2 - 100x + 600 - 400 = 0
4x^2 - 100x + 200 = 0; a quadratic equation
:
Simplify, divide by 4 and you have:
x^2 - 25x + 50 = 0
:
We need to use the quadratic formula to solve this: a=1; b=-25;; c=50
x+=+%28-%28-25%29+%2B-+sqrt%28+-25%5E2-4%2A1%2A50+%29%29%2F%282%2A1%29+
:
x+=+%28%2B25+%2B-+sqrt%28625+-+200+%29%29%2F%282%29+
:
x+=+%2825+%2B-+sqrt%28425+%29%29%2F%282%29+
:
Two solutions:
x+=+%2825+%2B+20.62%29%2F%282%29+
x+=+45.62%2F2
x = 22.8, not a possible solution, obviously
and
x+=+%2825+-+20.62%29%2F%282%29+
x+=+4.38%2F2
x = 2.19 ft is the width of the path
:
:
Check our solution by finding the area of the garden
We have to subtract 2x from the outside dimensions: 2*2.19 = 4.38
:
(30-4.38) * (20-4.38) =
25.62 * 15.62 = 400.2 ~ 400 sq/ft
:
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