Trigonometry/Proof: Heron's Formula
Example: Triangle with sides 567
So it's not a lot smaller than the estimate. 
Do you need the proof?Edit
For most exams you do not need to know this proof. You can skip over it on a first reading of this book.
It is here for two reasons.
 It is good practice in rather more involved algebra than you would normally do in a trigonometry course. Keep a cool head when following the steps. Most courses at this level don't prove it because they think it is too hard.
 The proof shows that Heron's formula is not some new and special property of triangles. It has to be that way because of the Pythagorean theorem.
ProofEdit
This proof needs more steps and better explanation to be understandable by people new to algebra. 
Video LinkEdit
There are videos of this proof which may be easier to follow at the Khan Academy:
The ProofEdit
The area A of the triangle is made up of the area of the two smaller right triangles.
The second step is by Pythagoras Theorem.
To get closer to the result we need to get an expression for somehow, that does not involve d or h. There is a useful trick in algebra for getting the product of two values from a difference of squares. We can get cd like this:
It's however not quite what we need. On the left we need to 'get rid' of the d, and to do that we need to get the left hand side into a form where we can use one of the Pythagorean identities for a^2 or b^2. Some experimentation gives:
 next subtract 2cd from both sides
 next use Pythagoras for a
 next use Pythagoras for b
We have made good progress. We have a formula for cd that does not involve d or h. We now can put that into the formula for A so that that does not involve d or h.
Which after expanding and simplifying becomes:
This is very encouraging because the formula is so symmetrical. We want a formula that treats a, b and c equally.
We've still some way to go. This formula is in terms of a, b and c and we need a formula in terms of s.
One way to get there is via experimenting with these formulae:
Having worked those three formulae out the following complete table follows by symmetry:
Then multiplying two rows from the above table:
On the right hand side of the = we have an expression that is like which is . That's a shortcut to calculating it. We could just multiply it all out, getting 16 terms and then cancel and collect them to get:
and so
Exercise: What about obtuse angles?
