Question 188304
<font face="Goudy Old Style" size="+2">


If the sidewalk is 2 ft wide all the way around, then the rectangle defined by the outer edge of the sidewalk is 4 feet longer and 4 feet wider than the garden.  So let the width of the garden be <i>x</i> and then the length of the garden is <i>x</i> + 5, so the area of the garden is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x + 5) = x^2 + 5x]


And the area of the rectangle defined by the outer edge of the sidewalk is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x + 4)(x + 9) = x^2 + 13x + 36]


And the difference between the two is 76 square feet, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 13x + 36 - (x^2 + 5x) = 76]


Solve for x:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x = 40]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 5]


So the width of the garden is 5 feet, the length is 10 feet.


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9 \times 14 = 126]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 126 - 76 = 50]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 50 = 5 \times 10]


Answer checks.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>