Question 797298
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Length of a rectangular area:  *[tex \Large l]


Width of a rectangular area:  *[tex \Large w]


Width of a uniform-width enclosing area:  *[tex \Large w_p]


Overall length is then:  *[tex \Large l\ +\ 2w_p]


Overall width is then:  *[tex \Large w\ +\ 2w_p]


(Note factor of 2 because the width of the enclosing area is on all four sides)


Total Area is then *[tex \Large \left(l\ +\ 2w_p\right)\left(w\ +\ 2w_p\right)\ =\ A_T]


Plug in your numbers for *[tex \Large l,\ \ ]*[tex \Large w,\ \ ]and *[tex \Large A_T], FOIL the two binomials, and then solve the resulting quadratic for *[tex \Large w_p].  If you happen to get a negative root (and you will given correct arithmetic), discard it.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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