# A triangle has sides with lengths: 1, 8, and 12. How do you find the area of the triangle using Heron's formula?

Jun 8, 2016

You cannot build a triangle with sides $1 , 8 , 12$.

#### Explanation:

The Heron's formula says that the area of a triangle with sides ${s}_{1} , {s}_{2} , {s}_{3}$ is given by

$A = \sqrt{\frac{P}{2} \left(\frac{P}{2} - {s}_{1}\right) \left(\frac{P}{2} - {s}_{2}\right) \left(\frac{P}{2} - {s}_{3}\right)}$

where $P$ is the perimeter $P = {s}_{1} + {s}_{2} + {s}_{3}$.

In your case $P = 1 + 8 + 12 = 21$

then

$A = \sqrt{\frac{21}{2} \left(\frac{21}{2} - 1\right) \left(\frac{21}{2} - 8\right) \left(\frac{21}{2} - 12\right)}$

$= \sqrt{\frac{21}{2} \cdot \frac{19}{2} \cdot \frac{5}{2} \cdot \left(- \frac{3}{2}\right)}$

$= \sqrt{- \frac{5985}{16}}$

$= \sqrt{- 374.0625}$

Of course this tells you that there is something wrong because you obtain the square root of a negative number and the area has to be a positive real number.

What is the problem?

The problem is that with the three numbers you have you cannot obtain a triangle.
The first property of a triangle is the triangular inequality that stats that the sum of two sides must be always greater or equal to the third side. This is clearly not true when you consider the sum of our first two sides $1 + 8 = 9$ that is smaller than $12$.
This tells us that we cannot construct a triangle from those numbers and, consequently, the Heron's formula cannot be applied.