Something wierd happens when herons formula is used with a=47 b=38 c=85, why?

Question

Herons formula: sqare root of s(s-a)(s-b)(s-c)
s= semiperimeter

Answer 1

semiparameter
S = (a + b + c) / 2
S = ( 47 + 38 + 85) / 2
S = 170 / 2
S = 85

A = √s (s-a)(s-b)(s-c)
A = √85 (85 - 47) (85 -38)(85 - 85)
where (85 - 85) = 0
then
A = 0
then it's not triangle
because
a + b = c

in any trinagle must be
a + b > c
a + c > b
c + b > a

in words the sum of any 2 sides must be greater than the third side

I hope this helps..

Answer 2

This triangle is impossible. The sum of two sides of triangle must ALWAYS be greater than the measure of the third side. But here this is not satisfied.
(a + b) > c
This must be the case for the triangle to exist.

But here, a + b = c
47 + 38 = 85

I bet you got area = 0
That's because s = 85 = c
s - c = 0
So the whole calculation under the square root is 0.

This triangle cannot exist. The area of such triangle is either 0 or imaginary.

Answer 3

because 47 + 38 = 85, thus the "triangle" in question is flat (i.e. a line segment) and has zero area. (in a true triangle, a + b < c, as opposed to a + b = c)

85 = s, so the factor s-85 is zero, supporting this point.