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### 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?
• Yes. As long as you know all three side lengths of any triangle, you can use this formula.
• 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
• Nope. As long as you can draw an altitude from a side (Denoted as C), you can always use this formula, though Obtuse triangles might be a special case.
(1 vote)
• 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[3]/ 4

But, S = 3a/2

So now we solve the following:

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

So not only does that simplification not work in general, but only works for one equilateral triangle of side a = 4/sqrt[3].
• 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:?)
• Is it a coincidence that "x" in this derived formula is equal to what seems to closely resemble the law of cosines? At . I thought it was interesting.
• it is no coincidence. Heron's Formula is an alternative to the law of cosines, which could also be used to solve for the area of a triangle only knowing the sides.
(1 vote)
• at where h = square root of a2 -(( c2+a2-b2)/2c)2 can it be simplified to h = a i ((c2+a2-b2)/2c) cuz the square root and the square signs cancel out
(1 vote)
• A square root of two squares added together cannot cancel out.
Example:
sqrt(a^2 + b^2) is NOT a+b
Think of an example:
sqrt(3^2 + 4^2) is NOT 3+4 or 7
= sqrt(9 + 16)
= sqrt(25)
= 5