# Trigonometry/Heron's Formula

Heron's formula states that the area A of a triangle whose sides have lengths a, b, and c is

$A={\sqrt {s(s-a)(s-b)(s-c)}}$ where $\displaystyle s$ is the semiperimeter of the triangle:

$s={\frac {a+b+c}{2}}$ There is a proof here.

 Worked Example todo: add picture Let us try this for the 3-4-5 triangle, which we know is a right triangle. We know its area. It's half that of the rectangle with sides 3x4. So the area $A$ is ${\frac {3\times 4}{2}}=6$ . Let's see if Heron's formula works too. $s={\frac {3+4+5}{2}}=6$ And using Heron's formula: $A={\sqrt {s(s-a)(s-b)(s-c)}}={\sqrt {6(6-3)(6-4)(6-5)}}={\sqrt {6\cdot 3\cdot 2\cdot 1}}={\sqrt {36}}=6$ It worked.

The formula is believed to be due to Hero (or Heron) of Alexandria (10 – 70 AD), a Greek mathematician.

 Exercise todo: add picture Now over to you. We have an equilateral triangle with each side of length 1. The base of the triangle is 1, the height can be worked out by Pythagoras. It's ${\sqrt {1^{2}-{\frac {1}{2^{2}}}}}={\frac {\sqrt {3}}{2}}$ so the area $A$ of our equilateral triangle is ${\frac {\sqrt {3}}{4}}$ . Work out what the semi perimeter S is for this triangle, and put the values for S and the lengths of the sides into Heron's formula and compute the area, A. $s=$ And using Heron's formula: $A={\sqrt {s(s-a)(s-b)(s-c)}}=$ Did the answer work out correctly?