Wasn't sure where to put this so I put it here
Is there any relationship between the perimeter of a pythagorean triple and the area of one. I am really stuck on this help? thanks
Hello, SeaN187!
No wonder you're stuck . . . There is no "neat" answer.Is there any relationship between the perimeter of a Pythagorean trangle and its area?
A Pythagorean Triple is of the form: .$\displaystyle (a,b,c)$ .where $\displaystyle c\:=\:\sqrt{a^2+b^2}$
. . Its perimeter is: .$\displaystyle P\;=\;a + b + \sqrt{a^2+b^2}$
. . Its area is: .$\displaystyle A\;=\;\frac{1}{2}ab$
And there is no "neat" relationship between $\displaystyle P$ and $\displaystyle A.$
Another approach . . .
A Pythagorean Triple can be generated by: $\displaystyle \begin{Bmatrix}a\:=\:m^2-n^2 \\ b\:=\:2mn \\ c\:=\:m^2+n^2\end{Bmatrix}$
. . for any positive integers $\displaystyle m > n$.
So there are (and always will be) two parameters (variables) in the problem.
The perimeter is: .$\displaystyle P \;= \;(m^2-n^2) + 2mn + (m^2+n^2)\:=$ $\displaystyle 2m^2 + 2mn \:=\:2m(m+n)$
The area is: .$\displaystyle A\;=\;\frac{1}{2}bh\;=\;\frac{1}{2}(m^2-n^2)(2mn) \;=\;mn(m^2-n^2)$
Again, no "neat" relationship between $\displaystyle P$ and $\displaystyle A.$
The best we can do is a ratio: .$\displaystyle \frac{A}{P}\;=\;\frac{m(m^2-n^2)}{2m(m+n)}\;=\;\frac{n(m-n)}{2}$
Not much better . . .
yeh i searched on wikipedia and got the stuff on m>n and all the formulas but really i need sum1 to talk me through it all and explain it
although it was an extention for my maths coursework so im just gunna leave it
just out of intrest what is heron's formula?
A formula they stopped teaching in American schools because it is too complicatedOriginally Posted by SeaN187
Given a triangle with sides,
$\displaystyle a,b,c$
Then its area is,
$\displaystyle A=\sqrt{s(s-a)(s-b)(s-c)}$
Where $\displaystyle s=\frac{a+b+c}{2}=\mbox{semiperimeter}$