Herons Formula ...Flaw??

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KHSky

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Hello all...

I have been working with Herons formula and using it to get area of a triangle. I have hit a stumbling block with it and I was hoping to get some help from everyone here.

So in the event where you have a triangle who's legs are AB = 1, BC = 2, CA = 3...using Herons formula seems to yield 0 for area...

S=(AB+BC+CA)/2
S=3
Area=[math]sqrt(S*(S-AB)*(S-BC)*(S-CA))[/math]Area=[math]sqrt(3*(3-1)*(3-2)*(3-3))[/math]
And there is where I find a question. Since 3-3 is 0 everything else in the chain will also be 0

Area=[math]sqrt(3*2*1*0)[/math]
So 0*1 is 0...0*2 is 0...0*3 is 0 and the sqrt of 0 is ...0

But there is area of that triangle...just very small like around .05 or so...so the question that I am left with is why does Herons formula fail here?

Thoughts


Thank you
 
The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. As soon as you know that the sum of any 2 sides is less than or equal to the measure of a third side, then you know that the sides do not make up a triangle.
 
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BigBeachBananas said:
The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. As soon as you know that the sum of any 2 sides is less than or equal to the measure of a third side, then you know that the sides do not make up a triangle.
Click to expand...
The sum of any two sides must be greater than the measurement of the third side....my question here is any?

So 2 + 3 is 5...which is greater than 1. 3 + 1 is 4 which is greater than 2. There is only one condition where that rule does not fit...would this still not be considered a triangle?
 
KHSky said:
The sum of any two sides must be greater than the measurement of the third side....my question here is any?

So 2 + 3 is 5...which is greater than 1. 3 + 1 is 4 which is greater than 2. There is only one condition where that rule does not fit...would this still not be considered a triangle?
Click to expand...
The sum of any 2 sides of a triangle must be greater than the measure of the third side. Since 1+2 =3, the criterion fails. You cannot make a triangle with sides 1,2, and 3. Try to draw it with a ruler & paper and see if you can make such triangle.

Also, imagine you have a rubber band that you laid flat such that it looks like the line BC with a length of 3 units in your picture. You pinched at point A and pulled it upward to make a triangle; the distance BA + AC > 3 units should make sense because you stretched it out. I hope this makes sense.
 
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KHSky said:
The sum of any two sides must be greater than the measurement of the third side....my question here is any?

So 2 + 3 is 5...which is greater than 1. 3 + 1 is 4 which is greater than 2. There is only one condition where that rule does not fit...would this still not be considered a triangle?
Click to expand...
The triangle inequality DEMANDS that

On an Euclidean plane​

the length of EVERY side of a triangle​
​
is SMALLER than​
​
the sum of the lengths of the other two sides​
 
Do it this way.

Put A at 0, 0.

Put B at 0, 1.

That is side AB and has a length of 1. Note AB lies entirely on the x-axis.

Now put C at point p, q so that BC has length 2. That means by the distance formula

[math]2 = \sqrt{(p - 0)^2 + (q - 1)^2} = \sqrt{p^2 + q^2 - 2q + 1} \implies[/math]
[math]4 = p^2 + q^2 - 2q + 1 \implies p^2 = 3 - q^2 + 2q.[/math]
At this point p and q could be lots of different places, but, when we draw a line from A to C, we want that line to have length of 3. So, again by the distance formula

[math]3 = \sqrt{(p - 0)^2 + (q - 0)^2} = \sqrt{p^2 + q^2} \implies 9 = 3 - q^2 + 2q - q^2\implies 2q = 6 \implies q = 3.[/math]
What is p?

[math]p = \pm \sqrt{3 - q^2 + 2q} = \pm \sqrt{3 - 9 + 6} = \pm \sqrt{0} = 0.[/math]
So C is at (0, 3). AC is on the x-axis, AB is on the x-axis. BC is on the x-axis. We have one line, not three. There is no such triangle.
 
The shortest distance between two points is a straight line. Once your distance in your diagram is greater than 3. Done!
 
KHSky said:
Area=\(\displaystyle sqrt(3*(3-1)*(3-2)*(3-3))\)
And there is where I find a question. Since 3-3 is 0 everything else in the chain will also be 0
Area=\(\displaystyle sqrt(3*2*1*0)\)
So 0*1 is 0...0*2 is 0...0*3 is 0 and the sqrt of 0 is ...0

But there is area of that triangle...just very small like around .05 or so...so the question that I am left with is why does Herons formula fail here?
Click to expand...
In computing 3*2*1*0 you do not multiply the 0 by 3, 2 and 1. Fine, you'll get the correct answer but what If you had--
3*2*1*5, this does not equal 15*10*5=750 (I multiply 3, 2 and 1 by 5 to get 15, 10 and 5)
 
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