Question 351281
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One question per post.  I'm going to guess you want the first one solved.


Let *[tex \Large x] represent the measure of the side of one of the squares.  Then the perimeter of that square AND the measure of the piece of wire that created that square is *[tex \Large 4x].  The other piece of wire measures *[tex \Large 44\ -\ 4x], hence the measure of the side of the other square must be *[tex \Large 11\ -\ x].


The area of the first square, *[tex \Large x^2], plus the area of the second square, *[tex \Large (11\ -\ x)^2], equals 65, so:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x^2\ +\ (11\ -\ x)^2\ =\ 65]


Expand the binomial, collect like terms, put the equation into standard quadratic form, and solve for *[tex \Large x].  Multiply either root by 4 to find the perimeter of one of the squares.  Multiply the other root by 4 to find the perimeter of the other square.


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