# How do you use Heron's formula to determine the area of a triangle with sides of that are 12, 16, and 19 units in length?

$95.503 \setminus \setminus {\textrm{u n i t}}^{2}$

#### Explanation:

The semi-perimeter $s$ of the triangle having sides $a = 12$, $b = 16$ & $c = 19$ is given as

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

$= \setminus \frac{12 + 16 + 19}{2}$

$= 23.5$

Now, using Hero's formula, the area $\setminus \Delta$ of triangle is given as follows

$\setminus \Delta = \setminus \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

$= \setminus \sqrt{23.5 \left(23.5 - 12\right) \left(23.5 - 16\right) \left(23.5 - 19\right)}$

$= 95.503 \setminus \setminus {\textrm{u n i t}}^{2}$

Jul 16, 2018

$\text{The area of the triangle is } \Delta \approx 95.5036 s q . u n i t s$

#### Explanation:

We have,

$\text{Heron's formula : the area of the triangle is}$

$\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)} \text{ , where s is semi perimeter}$

Let ,

$a = 12 , b = 16 , \mathmr{and} c = 19$

$\implies s = \frac{a + b + c}{2} = \frac{12 + 16 + 19}{2} = 23.5$

$\implies s - a = 23.5 - 12 = 11.5$

$s - b = 23.5 - 16 = 7.5$

$s - c = 23.5 - 19 = 4.5$
So ,
$\Delta = \sqrt{23.5 \left(11.5\right) \left(7.5\right) \left(4.5\right)} \approx 95.5036$ square units