Question 636224
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Let *[tex \LARGE x] be the "one side" of the rectangle.  Then *[tex \LARGE x\ +\ 1] is the diagonal.


With a tip 'o the hat to Mr. Pythagoras, we can say that the other side of the rectangle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt{(x\ +\ 1)^2\ -\ x^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt{x^2\ +\ 2x\ +\ 1\ -\ x^2}]


So the missing side of the rectangle is *[tex \LARGE \sqrt{2x\ +\ 1}]


The perimeter of a rectangle is 2 times the length plus 2 times the width so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 2\sqrt{2x\ +\ 1}\ =\ 62]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \sqrt{2x\ +\ 1}\ =\ 31]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{2x\ +\ 1}\ =\ 31\ -\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 1\ =\ 961\ -\ 62x\ +\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 64x\ +\ 960\ =\ 0]


Solve the quadratic (it factors).  One of the roots will be extraneous and you can tell by checking them in the original perimeter equation.  Discard the extraneous root.  Then calculate *[tex \LARGE \sqrt{2x\ +\ 1}].  Multiply the length times the width to get the area.


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