Question 336677
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Let *[tex \Large x] represent the measure of the side of the larger square.  Then the measure of the side of the smaller square is *[tex \Large x\ -\ 5].  Therefore the area of the larger square is *[tex \Large x^2] and the area of the smaller square is *[tex \Large (x\ -\ 5)^2].  And since the difference in the areas is *[tex \Large 95\text{ in^2}] we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ (x\ -\ 5)^2\ =\ 95]


Solve for  *[tex \Large x] and then subtract 7 to get the two length measures.


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