# Math Help - Pythagoras relationships between perimeter and area

1. ## Pythagoras relationships between perimeter and area

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

2. Hello, SeaN187!

Is there any relationship between the perimeter of a Pythagorean trangle and its area?
No wonder you're stuck . . . There is no "neat" answer.

A Pythagorean Triple is of the form: . $(a,b,c)$ .where $c\:=\:\sqrt{a^2+b^2}$

. . Its perimeter is: . $P\;=\;a + b + \sqrt{a^2+b^2}$

. . Its area is: . $A\;=\;\frac{1}{2}ab$

And there is no "neat" relationship between $P$ and $A.$

Another approach . . .

A Pythagorean Triple can be generated by: $\begin{Bmatrix}a\:=\:m^2-n^2 \\ b\:=\:2mn \\ c\:=\:m^2+n^2\end{Bmatrix}$
. . for any positive integers $m > n$.

So there are (and always will be) two parameters (variables) in the problem.

The perimeter is: . $P \;= \;(m^2-n^2) + 2mn + (m^2+n^2)\:=$ $2m^2 + 2mn \:=\:2m(m+n)$

The area is: . $A\;=\;\frac{1}{2}bh\;=\;\frac{1}{2}(m^2-n^2)(2mn) \;=\;mn(m^2-n^2)$

Again, no "neat" relationship between $P$ and $A.$

The best we can do is a ratio: . $\frac{A}{P}\;=\;\frac{m(m^2-n^2)}{2m(m+n)}\;=\;\frac{n(m-n)}{2}$

Not much better . . .

3. Originally Posted by Soroban
Hello, SeaN187!

No wonder you're stuck . . . There is no "neat" answer.

.
There is alwats Heron's formula, but that is a relation between the area,
perimiter and the triple

RonL

4. 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?

5. Originally Posted by SeaN187
just out of intrest what is heron's formula?
A formula they stopped teaching in American schools because it is too complicated

Given a triangle with sides,
$a,b,c$
Then its area is,
$A=\sqrt{s(s-a)(s-b)(s-c)}$
Where $s=\frac{a+b+c}{2}=\mbox{semiperimeter}$

6. thts not exactly super hard is it and its easy to remember too

7. Originally Posted by SeaN187
thts not exactly super hard is it and its easy to remember too
But remember we are talking about american high schools students