Question 600786
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Let *[tex \Large x] represent the width of the border.  Since the border goes all the way around the garden, the length and width of the lawn area are both reduced by *[tex \Large 2x] each.  Hence, the area of the lawn part is given by


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (40\ -\ 2x)(70\ -\ 2x)]


The area of the original yard is 40 times 72 which is 2880.  If the flower bed part is 5/12 of the original area, then the remaining lawn part has to be 7/12 of the original value.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (40\ -\ 2x)(70\ -\ 2x)\ =\ 2880\ \times\ \frac{7}{12}]


Simplify and solve the quadratic for *[tex \Large x].  Then calculate *[tex \Large 40\ -\ 2x] and *[tex \Large 72\ -\ 2x] which are the dimensions of the inner lawn.  Using these dimensions, calculate the perimeter of the inner lawn.  There are 5280 feet in a statute mile.  Divide by 4 to get the number of feet in 1/4 mile.  Divide the number of feet in 1/4 mile by the perimeter of the inner lawn area.  Round up to the nearest whole lap.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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