Pythagoras relationships between perimeter and area

**SeaN187** · June 10th 2006, 02:45 AM

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

**Soroban** · June 10th 2006, 04:20 AM

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 . . .

**CaptainBlack** · June 10th 2006, 06:03 AM

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

**SeaN187** · June 10th 2006, 01:43 PM

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?

**ThePerfectHacker** · June 10th 2006, 06:11 PM

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 $s=\frac{a+b+c}{2}=\mbox{semiperimeter}$

**SeaN187** · June 11th 2006, 04:02 AM

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

**ThePerfectHacker** · June 11th 2006, 07:14 AM

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

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