Question

A triangle has sides with lengths: 2, 9, 2. How do you find the area of the triangle using Heron's formula?

Answer 1

There is no such triangle, since `2+2 < 9`

Explanation:

If a triangle has sides of length `a`, `b` and `c` then all of these conditions hold:

`a+b > c`

`b+c > a`

`c+a > b`

...unless you count empty triangles, in which case change the `>`'s into `>=`'s.

If you try to apply Heron's formula to lengths `a=2`, `b=9`, `c=2`, then you will find that you end up attempting to take the square root of a negative number, hence no Real area:

The semi-perimeter `sp` is given by:

`sp = (a+b+c)/2 = (2+9+2)/2 = 13/2`

Then Heron's formula for the area `A` is:

`A = sqrt(sp(sp-a)(sp-b)(sp-c))`

`=sqrt(13/2(13/2-2)(13/2-9)(13/2-2))`

`=sqrt((13/2)(9/2)(-5/2)(9/2))`

`=sqrt(-5265/16)`

It is possible to simplify this further, but there's no real point since it's clearly the square root of a negative quantity.