# How do you use Heron's formula to determine the area of a triangle with sides of that are 4, 7, and 8 units in length?

Jan 21, 2016

$A = \sqrt{{S}_{p} \left({S}_{p} - a\right) \left({S}_{p} - b\right) \left({S}_{p} - c\right)}$

$A = \sqrt{19.2 \left(\frac{19}{2} - 4\right) \left(\frac{19}{2} - 7\right) \left(\frac{19}{2} - 8\right)}$

#### Explanation:

Heron formula requires you know only the sides of a triangle to compute the area. Note you can use another approach i.e. determine the height of the triangle and use our familiar:
$A = \frac{1}{2} b h$ but why not use only the sides. Well if you decide to do that then welcome to Heron formula:

$A = \sqrt{{S}_{p} \left({S}_{p} - a\right) \left({S}_{p} - b\right) \left({S}_{p} - c\right)}$

Where ${S}_{p}$ is the semi-perimeter:

${S}_{p} = \frac{a + b + c}{2}$

There is an elegant prove to this leveraging trigonometry and using the following identities:
$\sin \theta = \sqrt{1 - \cos \theta} = \frac{\sqrt{4 {a}^{2} {b}^{2} - {\left({a}^{2} + {b}^{2} - {c}^{2}\right)}^{2}}}{2 a b}$
$A = \frac{1}{2} b h = \left(\frac{1}{2} a b\right) \sin \theta$

Try to complete the derivation...

You can also leverage "Pythagorean Theorem" on the following triangle... see image: The idea is to express d and h only in terms of: $a , b , c , {S}_{p}$