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
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