Heron's Formula

Question

Heron's Formula gives the area of a triangle when the length of all three sides are known. There is no need to calculate angles or other distances in the triangle first to determine its area. The formula is given by

$Area=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{2}$, $a, b, c$ are sides of the triangle and $p$ is the semi-perimeter of the triangle. The following is my concern:

If one of the sides of the triangle is greater than $p$, then the Area will not be a real number (which shouldn't be true).

Example. Let the sides of a triangle be 175 metre, 88 metre and 84 metre, then $p=173.5$. Therefore, $Area=\sqrt{-1991498.0625}$, which not a real value.

Therefore, the following is my question:

Why shouldn't the area be expressed as $Area=\sqrt{|p(p-a)(p-b)(p-c)|}$?

Answer 1

Recall that, for a triangle with sides $a, b, c$, we always have $a + b > c, b + c > a, c + a > b$.

This means that $p = \frac {a + b + c} 2 > a$ and similarly $p > b$ and $p > c$.

Thus you don't have to worry about the area being non-real.

Answer 2

If $p$ is less than one of the sides then you don't have a triangle - one of the sides would have to be longer than the other two combined.

So if $p\lt a$ we have $a+b+c\lt 2a$ 0r $b+c\lt a$

Answer 3

In addition to the correct answers' "that's not a triangle!", the imaginary values are still suggestive. Kendig wrote an American Mathematical Monthly article titled "Is a 2000-Year-Old Formula Still Keeping Some Secrets?" exploring ways to make sense of those values. (Currently) you can read the entire article here.