# How do you use Heron's formula to find the area of a triangle with sides of lengths 6 , 4 , and 9 ?

Mar 10, 2016

$\frac{\sqrt{1463}}{4} \approx 9.562$

#### Explanation:

For a triangle with side $a$, $b$ and $c$, we first calculate the semi-perimeter, which is half of the perimeter of the triangle.

$s = \frac{a + b + c}{2}$

Heron's formula states that the area of the triangle is given by

$\text{Area} = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

In this question, we have

• $a = 6$
• $b = 4$
• $c = 9$

The semi-perimeter, $s$, is

$s = \frac{6 + 4 + 9}{2} = \frac{19}{2}$

So, the area of the triangle is

$\text{Area} = \sqrt{\frac{19}{2} \left(\frac{19}{2} - 6\right) \left(\frac{19}{2} - 4\right) \left(\frac{19}{2} - 9\right)}$

$= \frac{\sqrt{1463}}{4}$

$\approx 9.562$