Question 229323
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Let *[tex \Large x] represent the measure of one side of the square.  Then the area of the square is *[tex \Large x^2], the measure of one side of the rectangle is *[tex \Large x + 6], the measure of the other side of the rectangle is *[tex \Large x + 8], and the area of the rectangle is *[tex \Large (x+6)(x+8) = x^2 + 14x + 48], but the area of the rectangle is *[tex \Large 188 \text{ft^2}] larger than the square, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 14x + 48 = x^2 + 188]


Solve for *[tex \Large x].


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