# How do you use Heron's formula to determine the area of a triangle with sides of that are 6, 4, and 9 units in length?

Mar 19, 2018

$9.56 \setminus 2 \setminus \mathrm{dp}$

#### Explanation:

Heron's Formula tells us that given all three sides of a triangle, $a , b , c$, say, then the area of the triangle is given by:

$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{1}{2} \left(a + b + c\right)$

So, for the given triangle we have:

$s = \frac{1}{2} \left(6 + 4 + 9\right) = \frac{19}{2}$

And so we get:

$A = \sqrt{\frac{19}{2} \left(\frac{19}{2} - 6\right) \left(\frac{19}{2} - 4\right) \left(\frac{19}{2} - 9\right)}$
$\setminus \setminus \setminus = \sqrt{\frac{19}{2} \left(\frac{7}{2}\right) \left(\frac{11}{2}\right) \left(\frac{1}{2}\right)}$
$\setminus \setminus \setminus = \sqrt{\frac{1463}{16}}$
$\setminus \setminus \setminus = \frac{\sqrt{1463}}{4}$
$\setminus \setminus \setminus = 9.562295 \ldots$
$\setminus \setminus \setminus = 9.56 \setminus 2 \setminus \mathrm{dp}$