What is the derivation of Heron's Formula?

Answer 1

The following proof is trigonometric, and basically uses the cosine rule. First we compute the cosine squared in terms of the sides, and then the sine squared which we use in the formula A=1/2bc·sinA to derive the area of the triangle in terms of its sides, and thus prove Heron's formula.

We use the relationship x2−y2=(x+y)(x−y) [difference between two squares] [1.2]

Finding the cosine squared in terms of the sidesFrom the cosine rule:

We have:

[1.3]

Rearranging:

[1.4]

Because we want the sine, we first square the cosine:

[1.5]

Finding the SineTo use in:

[1.6]

Using Equation 1.5 in 1.6, we have:

[1.7]

Bringing all under the same denominator:

[1.8]

Using the difference between two squares (Equation 1.2)

[1.9]

Putting the above into a form where we can use the difference between two squares again we have:

[1.10]

Actually using the difference between two squares in both brackets, we find:

[1.11]

Substituting (a+b+c) for 2s, (b+c-a) for 2s-2a, etc:

[1.12]

Taking the square root:

[1.13]

Finding the Area

Recalling:

[1.14]

We have:

[1.15]

And simplified: