<PRE><B>Question 764060</B>
A cement walk of constant width is built around an 8m x 12m rectangular garden. The total area of the garden and walk is 192m^2. Find the width of the walk. 
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Draw an elongated rectangle inside an elongated rectangle.
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The inside rectangle is 8 by 12 = 96 sq meters
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The outside rectangle is (8+2x)(12+2x) = 96 + 40x + 4x^2 sq. meters
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Solve: 4x^2 + 40x + 96 = 192
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x^2 + 10x + 24 = 48
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x^2 + 10x - 24 = 0
(x+12)(x-2) = 0
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Positive solution:
x = 2 meters
That is the width of the walk.
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Cheers,
Stan H.
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