SydneyRoverP6B said:
We all know that thanks to Pythagoras, the famous Greek Philosopher and Mathematician (579 - 495 BC) we can with great precision determine the length of the hypotenuse within a right angle triangle. This of course hinges upon the reality that we know the lengths of the other two sides.
What happens though if we know the length of the perimeter along with the total area. Are we still in a position to calculate the length of the hypotenuse? If so, what might it be? Using P for perimeter and A for area,..what will be the length of the hypotenuse?
Ron.
There IS one method I know would solve this, but it's very trial and error, and I doubt it's the answer you seek...
Nevertheless...
The rule of heron states
A^2 = p x (p-a) x (p-o) x (p-h), or of course [(A^2) / p] = (p-a) x (p-o) x (p-h)
Where A - Area of the triangle,
p - half the perimeter (i.e. P/2),
a being the adjacent side length,
o being the opposite side length and,
h being the hypotenuse length.
We know the area, we know the perimeter (so we also know half the perimeter!)
We could compute what [(A^2) / p] is, then try a value for a, divide [(A^2) / p] by (p-a), choose a value for b, divide [(A^2) / p (p-a)] by (p-b), calculate what c is, then hope/test that a^2 +b^2 =c^2
Example..
The perimeter is 12 cm, Area is 6 sq cm
p = 12/2 = 6, A =6
So, A^2 /p = 36/6 = 6
6 = (6-a) x (6-o) x (6-h)
Say a = 4, 6 = (6-4) x (6-b) x (6-c) = 2 x (6-b) x (6-c), or 3 = (6-b) x (6-c)
Try o = 5
3 =(6-5) x (6-c) = (6-c)
c = 3...
However, we note that c is less than a and b, so lets interchange, say, a and c...
a = 3, b = 4, c = 5.
a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25. c^2 = 5^2 = 25
They agree!
That's A method of possibly finding the hypotenuse (and the adjacent and opposite sides too!), but very fiddly and 'trial-and-error'. Not the answer sought, I'm sure!