# A triangle has sides with lengths: 2, 9, 1. How do you find the area of the triangle using Heron's formula?

Jun 26, 2017

There is no such triangle, but...

#### Explanation:

As noted by other contributors, it is not possible to form a triangle with sides $2$, $9$ and $1$. If the sides of a triangle have lengths $a$, $b$ and $c$ then all of the following must hold:

$\left\{\begin{matrix}a + b > c \\ b + c > a \\ c + a > b\end{matrix}\right.$

and these conditions are sufficient.

For fun, let us try to apply Heron's formula and see what happens.

The semi-perimeter $s$ is given by the formula:

$s = \frac{1}{2} \left(a + b + c\right) = \frac{1}{2} \left(2 + 9 + 1\right) = 6$

Then Heron's formula tells us that the area is:

$\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)} = \sqrt{6 \left(6 - 2\right) \left(6 - 9\right) \left(6 - 1\right)}$

$\textcolor{w h i t e}{\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} = \sqrt{6 \left(4\right) \left(- 3\right) \left(5\right)}$

$\textcolor{w h i t e}{\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} = \sqrt{- 360}$

i.e. not a real area.

Notice that the three conditions we wrote down above hold if and only if all of $\left(s - a\right) > 0$, $\left(s - b\right) > 0$ and $\left(s - c\right) > 0$.

Simply put, Heron's formula applies and gives a positive real value if and only if the conditions for sides of length $a$, $b$ and $c$ to be able to form a triangle hold.