Heron or Hero was a Greek Mathematician who discovered a formula
for the area of a triangle, with sides
Before giving this formula, we need a little trigonometry.
We assume that you are familiar with some trigonometry, and know what is
meant by the ``sine'' and the ``cosine'' of an angle.
Consider with a right angle at X and
let
.
We define and
.
Join CB', and note that
Further, note that is a right angle. Observe that
where, as usual, we denote BC by a.Then
This shows that
and so the Sine Rule is obtained.
Consider and draw the perpendicular line to AB through C.
Suppose this line meets AB at D.
Then
so that
and
From this, by squaring the latter two equations, we get:
Adding these together gives
This is the Cosine Rule.
Since we have found that
we can deduce that
We shall use the Sine Rule to get an new formula for the area of a triangle.
The area of is given by
base
height.
Thus
The perimeter of is given by AB+BC+CA.
The semi-perimeter is half of this value!
We denote the semi-perimeter by s, so that
We use equations (1) and (2) to get
A similar argument gives
and so
Using the formulae for the area of a triangle and these, we get
This is HERON's formula.
The great advantage is that the area of the triangle can be calculated
solely from the knowledge of the lengths of the sides. No information about
the angles is required.
Here we give an algebraic derivation of Heron's Formula.
Let CD=h, AB=c, BC=a and CA=b as usual. Then
so that
giving
Squaring gives
or
or
and Heron's formula follows at once.
Here we extend Heron's formula to find the area of CYCLIC quadrilaterals.Consider the cyclic quadrilateral ABCD and denote the sides and a diagonal as follows:
We define .
Let
, so that
.
Then, the cosine rule, applied to and
,
gives that
Eliminating from these equations gives
Thus
yielding
Let represent the area of
and
represent the area of
.Heron's formula then gives us:
Now
Similarly
Together, we now get
so that
From this it is very easy to obtain the following formula for the area of a cyclic quadrilateral: