Trigonometry/Proof: Heron's Formula

Example: Triangle with sides 5-6-7 We know that a triangle with sides 3, 4 and 5 is a right triangle. Two such triangles would make a rectangle with sides 3 and 4, so its area is ${\frac {3\cdot 4}{2}}=6$ . A triangle with sides 5,6,7 is going to have its largest angle smaller than a right angle, and its area will be less than ${\frac {5\cdot 6}{2}}=15$ . Let's see how much by, by calculating its area using Heron's formula. $s={\frac {5+6+7}{2}}=9$ $A={\sqrt {s(s-a)(s-b)(s-c)}}={\sqrt {9(9-5)(9-6)(9-7)}}={\sqrt {9(4)(3)(2)}}={\sqrt {216}}=6{\sqrt {6}}\approx 14.7$ So it's not a lot smaller than the estimate.

Triangle used in proof.

$A={\sqrt {s(s-a)(s-b)(s-c)}}$

$s={\frac {a+b+c}{2}}$

Do you need the proof?

Do you need this proof?

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.

Proof

Video Link

There are videos of this proof which may be easier to follow at the Khan Academy:

• Proof of Heron's formula part I
• Proof of Heron's formula part II

The Proof

Triangle used in proof.

$A={\sqrt {s(s-a)(s-b)(s-c)}}$

$s={\frac {a+b+c}{2}}$

The area A of the triangle is made up of the area of the two smaller right triangles.

$A={\frac {dh+(c-d)h}{2}}={\frac {ch}{2}}$

$A^{2}={\frac {c^{2}h^{2}}{4}}={\frac {c^{2}(b^{2}-d^{2})}{4}}={\frac {c^{2}b^{2}-c^{2}d^{2}}{4}}$

The second step is by Pythagoras Theorem.

To get closer to the result we need to get an expression for $c^{2}d^{2}$ 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:

$(c+d)^{2}-(c-d)^{2}=c^{2}+2cd+d^{2}-(c^{2}-2cd+d^{2})=4cd$

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:

$c^{2}+2cd+d^{2}-(c-d)^{2}=4cd$ next subtract 2cd from both sides

$c^{2}+d^{2}-(c-d)^{2}=2cd$ next use Pythagoras for a

$c^{2}+d^{2}-(a^{2}-h^{2})=2cd$

$c^{2}+d^{2}-a^{2}+h^{2}=2cd$ next use Pythagoras for b

$c^{2}+b^{2}-a^{2}=2cd$

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.

$A^{2}={\frac {4c^{2}b^{2}-(c^{2}+b^{2}-a^{2})^{2}}{16}}$

Which after expanding and simplifying becomes:

$A^{2}={\frac {2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}-a^{4}-b^{4}-c^{4}}{16}}$

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:

$4s^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$

$4s(s-a)=-a^{2}+b^{2}+c^{2}+2bc$

$4(s-a)(s-b)=-a^{2}-b^{2}+c^{2}+2ab$

Having worked those three formulae out the following complete table follows by symmetry:

$4s^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$

$4s(s-a)=-a^{2}+b^{2}+c^{2}+2bc$

$4s(s-b)=a^{2}-b^{2}+c^{2}+2ac$

$4s(s-c)=a^{2}+b^{2}-c^{2}+2ab$

$4(s-a)(s-b)=-a^{2}-b^{2}+c^{2}+2ab$

$4(s-a)(s-c)=-a^{2}+b^{2}-c^{2}+2ac$

$4(s-b)(s-c)=a^{2}-b^{2}-c^{2}+2bc$

Then multiplying two rows from the above table:

$4s(s-a)\times 4(s-b)(s-c)=(-a^{2}+b^{2}+c^{2}+2bc)\times (a^{2}-b^{2}-c^{2}+2bc)$

On the right hand side of the = we have an expression that is like $(-q+p)\times (q+p)$ which is $-(q^{2})+p^{2}$ . 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:

$16s(s-a)(s-b)(s-c)=-(-a^{2}+b^{2}+c^{2})^{2}+(2bc)^{2}$

$=-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}-2b^{2}c^{2}+4b^{2}c^{2}$

$=-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}$

$=16A^{2}$

and so

$A={\sqrt {s(s-a)(s-b)(s-c)}}$

Exercise: What about obtuse angles? Did you notice that just like the proof for the area of a triangle being half the base times the height, this proof for the area also divides the triangle into two right triangles? It has exactly the same problem - what if the triangle has an obtuse angle? Think about these three different ways we could fix the proof: Repeat the proof, this time with an obtuse angle and subtracting rather than adding areas. Choose the position of the triangle so that the largest angle is at the top. Then the problem goes away. Allow lengths and areas to be negative in the above proof. Which of those three choices is the easiest? Would all three approaches be valid ways to fix the proof? The simplest approach that works is the best. It gives you the shortest proof that is easiest to check.