Question 1190261
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The area of a right triangle is the product of the two short sides divided by 2.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(s)\ =\ \frac{s(s\,+\,1)}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2A\ =\ s^2\ +\ s]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s^2\ +\ s\ -\ 2A\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ \frac{-1\ \pm\ \sqrt{1\ +\ 8A}}{2}]


Since a negative value for the measure of the side of a triangle is absurd, discard the negative value of *[tex \Large s]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s(A)\ =\ \frac{-1\ +\ \sqrt{1\ +\ 8A}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(s)\ =\ s\ +\ s\ +\ 1\ +\ s\ +\ 2\ =\ 3s\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s(P)\ =\ \frac{P\ -\ 3}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(A)\ =\ 3\(\frac{-1\ +\ \sqrt{1\ +\ 8A}}{2}\)\ +\ 3]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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