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## Geometry (all content)

### Course: Geometry (all content)>Unit 8

Lesson 9: Heron's formula

# Proof of Heron's formula (1 of 2)

Sal proves Heron's Formula for finding the area of a triangle solely from its side lengths. Created by Sal Khan.

## Want to join the conversation?

• Here is an interesting question: "Why does Heron's Formula yield an area in square units even though four lengths are being multiplied together? • Okay. That's a good question.The important thing to realize is that there is a square root. First of all, the unit of S(semi perimeter) is meter,as S =(a+b+c)/2,and (meter+meter+meter)/2 still gives meter as the unit. Using the same logic, (S-a),(S-b),(S-c),all have meter as their units.Therefore S(S-a)(S-b)(S-c) has m.m.m.m (meter raised to the 4th power) as units. Now,according to the formula, you take the square root of all that, and sq rt(m^4) gives you sq.meters. Note: I just used meter this time.It could be any unit.
• What is the thought process behind this? Did a mathematician just sit down and automatically know which steps to take? • I'm pretty sure that it required some thought, because a long time ago people just expected that what they had was true. I don't think that they automatically knew exactly what steps to do, or exactly what the answer would be, because it's not really something that takes just 1 minute to do. Back then, when new mathematical concepts were being discovered, they didn't have as much evidence as we have today. Well, maybe some could argue that mathematicians knew exactly what to do when they first got the idea, but really, math is something that requires thought. :)
• Does this work for all triangles? • Does Heron's formula simply to (S^2 sqrt(3)) / 4? or does that only work for equilateral triangles? • No, and for the reason, lets consider:

For any triangle, the only way heron's formula = (S^2 sqrt(3)) / 4 is for the trivial case where S = 0.
So no, it cannot simplify in general the way you propose.

If indeed the triangle is equilateral, then a=b=c and then Heron's simplifies to:
A = sqrt[S(S-a)^3] assuming all sides are a.

Then, in order to get your simplifications we must determine specific values for S and a. This is done by solving the equation:

sqrt[S(S-a)^3] = S^2 sqrt/ 4

But, S = 3a/2

So now we solve the following:

sqrt[3a^4]/4 = 3 a^4 sqrt/64
for a
We get real and complex roots, and the trivial case where a = 0.
The real positive root is a = 4/sqrt

So not only does that simplification not work in general, but only works for one equilateral triangle of side a = 4/sqrt. • The problem with the proof seems to be the right triangles areas can not be known because the right triangles have an irrational number in their calculations which can not be calculated by a mantematican in his life time:?( If they allow successors the problem remains.

I find it shocking that mathematicans do not realize the logistaial problems within irrationality. You can not have pi and eat it:?) • But in this non-simplified version of Huron's formula, doesn't it matter which sides you choose as a,b and c as opposed to Huron's formula • I like developing it from the cosine law.

c^2 = a^2 + b^2 - 2ab cosC
2ab cos C = a^2 + b^2 - c^2
( 2ab cos C )^2 = ( a^2 + b^2 - c^2 )^2
4a^2b^2 cos^2 C = ( a^2 + b^2 - c^2 )^2
4a^2b^2 ( 1 - sin^2 C ) = ( a^2 + b^2 - c^2 )^2
4a^2b^2 - 4a^2b^2 sin C = (a^2 b^2 - c^2 )^2
Letting K = ( 1 / 2 )ab sin C ( formula for area of triangle )
4a^2b^2 - 16K^2 = ( a^2 + b^2 - c^2 )^2
4a^2 b^2 - ( a^2 + b^2 - c^2 )^2 = 16K^2
[ 2ab - (a^2 + b^2 - c^2 ) ] [ 2ab + (a^2 + b^2- c^2 ) ] = 16K^2
[ c^2 - a^2 + 2ab - b^2 ] [ a^2 + 2ab + b^2 - c^2 ] = 16K^2
[ c^2 - ( a - b )^2 ] ( a + b - c ) ( a + b + c ) = 16K^2
( c + b - a ) ( c+ a - b ) ( a + b - c ) ( a + b + c ) = 16K^2
( a + b + c - 2a ) ( a + b + c - 2b ) ( a + b + c - 2c ( a + b + c ) = 16K^2
Let 2s = a + b + c

( 2s - 2a ) (2s - 2b ) ( 2s - 2c )2s = 16K^2
16 s( s - a )( s - b ) ( s - c ) = 16k^2
K = sqrt [ s ( s - a ) ( s - b ) ( s - c ) ]   