Now What Is That Length?

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
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.
 
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.
Click to expand...


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!
 
darth sidious wrote,...
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.
Click to expand...

Hello Darth,

Yes indeed that is an interesting method, but as you say very fiddly and time consuming and very much open to computational error, especially when it involves many iterations. You could of course write a piece of software and incorporate a "do while" loop or similar.

Yes as you say the answer that I am seeking would only require the parameters of P (perimeter) and A (area) to be used as they are provided. A simple equation written with these two variables the only unknowns. No trial and error required. It is not a lengthy process and the equation involved is not complicated.

Ron.
 
I THINK I've got it!

We know the perimeter, and the area.

The perimeter = hypotenuse + adjacent + opposite. adjacent = hypotenuse x cos (theta) and opposite = hypotenuse x sin (theta)

P = h x (1 + cos (theta) + sin (theta))

P^2 = h^2 x [(1 + cos (theta) + sin (theta))]^2 = h^2 x 2 [1 + cos (theta) + sin (theta) + {cos (theta) x sin (theta)}]

P^2 = h x 2 x [h x (1 + cos (theta) + sin (theta)) + {h x cos (theta) x sin (theta)}] = h x 2 x [P + {h x cos (theta) x sin (theta)}]

P^2 = h x 2 x [P + {h x cos (theta) x sin (theta)}] = 2 x [h x P + {h^2 x cos (theta) x sin (theta)}]

Area = adjacent x opposite / 2 = [h x cos (theta) x h x sin (theta)]/2 = h^2 x [cos (theta) x sin (theta)]/2

Therefore,

P^2 = 2 x [h x P + {2 x Area}]

h = [{(P^2) / 2} - {2 x Area}]/P = {P/2} - {2 x Area/P}

hypotenuse = {P/2} - {2 x Area/P}
 
darth sidious wrote,..
hypotenuse = {P/2} - {2 x Area/P}
Click to expand...

Hello Darth,

Yes indeed,...that is the correct answer... :D For simplifcation you can combine the two fractions to form h = (P^2 - 4A) / 2P.

Your approach is quite different compared to the one that I took, but the outcome is identical.

The three sides of the triangle are h, a and b. From Pythagoras h^2 = a^2 + b^2.

The area A is given by 1/2ab and the perimeter P = h + a + b.

Thus our three equations are

h^2 = a^2 + b^2...(1)
A = 1/2ab..........(2)
P = h + a + b.......(3)

So we have 3 equations and 5 unknowns. Not easy to solve like this so....

(a + b)^2 = a^2 + 2ab + b^2.....so now express (a + b)^2 in two ways. From equations (1) and (2)....

(a + b)^2 = (a^2 + b^2) + 2ab = h^2 + 4A

From equation (3)....

(a + b)^2 = (P – h )^2 = P^2 – 2Ph + h^2 thus,….

h^2 + 4A = P^2 – 2Ph + h^2

2Ph = P^2 – 4A therefore

h = (P^2 – 4A) / 2P

It is always good when you can prove a solution by an alternative method.

Ron.
 
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