SOLUTION: Aretangular garden is to be surrounded by a walkway of constant width. the garden's dimensions are 30 ft by 40 ft. the total area, garden plus walkway, is 1800 ft^2. What must be t

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Question 103940: Aretangular garden is to be surrounded by a walkway of constant width. the garden's dimensions are 30 ft by 40 ft. the total area, garden plus walkway, is 1800 ft^2. What must be the width of nthe walkway to the nearest thousandth?
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800ft^2. What must be the width of the walkway to the nearest thousandth?
:
Let x = width of the walkway
:
Draw this out labeling the walkway width as x, and the inner rectangle
representing the garden as 30 by 40. It will be apparent that the overall
dimensions will be (2x+30) by (2x+40), the area of that is given as 1800 sq ft
:
(2x + 30) * (2x + 40) = 1800
:
FOIL
4x^2 + 140x + 1200 = 1800
:
4x^2 + 140x + 1200 - 1800 = 0
:
4x^2 + 140x - 600 = 0
:
Simplify, divide eq by 4 and you have:
x^2 + 35x - 150 = 0
:
Use the quadratic formula to find x: a=1; b=35; c=-150

:

:

:
Do the math and you will get two solutions; -38.86 and +3.86,
obviously it's x = 3.86
:
3.86 ft is the width of the walkway
:
:
Check it:
2x = 2(3.86) = 7.72
So we have:
37.72 * 47.72 = 1799.998 ~ 1800 sq ft
:
How about this? Did it make sense to you? Any questions?

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