# How do you use Heron's formula to find the area of a triangle with sides of lengths 12 , 15 , and 18 ?

Feb 15, 2016

${\text{Area}}_{\triangle} = \frac{135}{4} \sqrt{7} \approx 89.3$ sq.units

#### Explanation:

Heron's formula says that a triangle with sides of length, $a , b , c$ has an area:
color(white)("XXX")"Area"_triangle=sqrt(s(s-a)(s-b)(s-c))
where $s$ is the semi-perimeter $= \frac{a + b + c}{2}$

Given sides $12 , 15 , 18$
$\textcolor{w h i t e}{\text{XXX}} s = \frac{45}{2}$
and
color(white)("XXX")"Area"_triangle=sqrt(45/2 * (45/2-12) * (45/2-15) * (45/2-18)

color(white)("XXXXXXX")=sqrt(45/2) * (21/2) * (15/2) * (9/2))

$\textcolor{w h i t e}{\text{XXXXXXX}} = \sqrt{\frac{\left(3 \cdot 3 \cdot 5\right) \cdot \left(3 \cdot 7\right) \cdot \left(3 \cdot 5\right) \cdot \cdot \left(3 \cdot 3\right)}{2 \cdot 2 \cdot 2 \cdot 2}}$

$\textcolor{w h i t e}{\text{XXXXXXX}} = \sqrt{\frac{{\left({3}^{3}\right)}^{2} \cdot {5}^{2} \cdot 7}{{\left({2}^{2}\right)}^{2}}}$

$\textcolor{w h i t e}{\text{XXXXXXX}} = \frac{{3}^{3} \cdot 5}{2} \sqrt{7}$

$\textcolor{w h i t e}{\text{XXXXXXX}} = \frac{135}{4} \sqrt{7}$

$\textcolor{w h i t e}{\text{XXXXXXX}} \approx 89.3$