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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?(6 votes)
- 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.(16 votes)

- What is the thought process behind this? Did a mathematician just sit down and automatically know which steps to take?(7 votes)
- 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. :)(9 votes)

- Does this work for all triangles?(3 votes)
- Yes. As long as you know all three side lengths of any triangle, you can use this formula.(4 votes)

- 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(3 votes)
- 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?(2 votes)
- 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].(3 votes)

- At3:35, Sal says something about (c-x) ^2...=c^2-2cx+x^2... How does that work? Is there a Khan Academy Video about this "trick"?(2 votes)
- (c-x)² = (c-x)(c-x) = c² -2cx + x².

This is called squaring a binomial, Look at this video: https://www.khanacademy.org/math/algebra/polynomials/multiplying_polynomials/v/special-products-of-binomials

This comes up**a lot**in mathematics so it's a key concept to grasp.(2 votes)

- 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:?)(2 votes) - Is it a coincidence that "x" in this derived formula is equal to what seems to closely resemble the law of cosines? At5:09. I thought it was interesting.(2 votes)
- 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)

- at7:35where 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(2 votes)

- It's interesting to see the proof of various formulas, but how useful is it to memorize them? Will you ever need to prove these formulas in high school?(1 vote)
- In high school geometry and geometry in general, you'll probably use this formula a lot. It's useful to see proof of formulas, if you understand them, you'll know when to apply it and how.(2 votes)

## Video transcript

Let's say I've got a triangle. There is my triangle
right there. And I only know the lengths of
the sides of the triangle. This side has length a, this
side has length b, and that side has length c. And I'm asked to find the
area of that triangle. So far all I'm equipped with is
the idea that the area, the area of a triangle is equal
to 1/2 times the base of the triangle times the
height of the triangle. So the way I've drawn this
triangle, the base of this triangle, would be side c, but
the height we don't know. The height would be that h
right there and we don't even know what that h is. So this would be the h. So the question is how do
we figure out the area of this triangle? If you watched the last
video you know that you use Heron's formula. But the idea here is to try
to prove Heron's formula. So let's just try to figure
out h from just using the Pythagorean theorem. And from there, once we know h,
we can apply this formula and figure out the area
of this triangle. So we already
labeled this as h. Let me define another
variable here. This is a trick you'll see
pretty often in geometry. Let me define this is x, and if
this is x in magenta, then in this bluish-purplish color,
that would be c minus x, right? This whole length is c
-- the whole base is c. So if this part is x, then
this part is c minus x. What I could do now, since
these are both right angles, and I know that because this is
the height, I can set up two Pythagorean theorem equations. First, I could do this left
hand side and I can write that x squared plus h squared
is equal to a squared. That's what I get from
this left hand triangle. Then from this right hand
triangle, I get c minus x squared plus h squared
is equal to b squared. So I'm assuming I know a, b and
c, so I have two equations with two unknowns. The unknowns are x and h. And remember, h is what
we're trying to figure out because we already know c. If we know h, we can
apply the area formula. So how can we do that? Well, let's substitute
for h to figure out x. When I say that I mean let's
solve for h squared here. If we solve for h squared
here we just subtract x squared from both sides. We can write that x squared --
sorry, we could write that h squared is equal to a
squared minus x squared. Then we could take this
information and substitute it over here for h squared. So this bottom equation
becomes c minus x squared plus h squared. h squared we know from this
left hand side equation. h squared is going to be equal
to -- so plus, I'll do it in that color -- a squared minus x
squared is equal to b squared. I just substituted the
value of that in here, the value of that in there. Now let's expand this
expression out. c minus x squared, that
is c squared minus 2cx plus x squared. Then we have the minus --
sorry, we have the plus a squared minus x squared
equals b squared. We have an x squared and
a minus x squared there, so those cancel out. Let's add the 2cx to both
sides of this equation. So now our equation
would become c squared plus a squared. I'm adding 2cx to both sides. So you add 2cx to this,
you get 0 is equal to b squared plus 2cx. All I did here is I canceled
out the x squared and then I added 2cx to both sides
of this equation. My goal here is to solve for x. Once I solve for x, then
I can solve for h and apply that formula. Now to solve for x, let's
subtract b squared from both sides. So we'll get c squared
plus a squared minus b squared is equal to 2cx. Then if we divide both sides by
2c, we get c squared plus a squared minus b squared
over 2c is equal to x. We've just solved for x here. Now, our goal is to solve
for the height, so that we can apply 1/2 times
base times height. So to do that, we go back to
this equation right here and solve for our height. Let me scroll down
a little bit. We know that our height
squared is equal to a squared minus x squared. Instead of just writing x
squared let's substitute here. So it's minus x squared -- x
is this thing right here. So c squared plus a
squared minus b squared over 2c, squared. This is the same
thing as x squared. We just solved for that. So h is going to be equal to
the square root of all this business in there -- I'll
switch the colors -- of a squared minus c squared plus
a squared minus b squared -- all of that squared. Let me make it a little bit
neater than that because I don't want to--. The square root -- make sure I
have enough space -- of a squared minus all of this stuff
squared -- we have c squared plus a squared minus b
squared, all of that over 2c. That is the height
of our triangle. The triangle that we
started off with up here. Let me copy and paste that
just so that we can remember what we're dealing with. Copy it and then let me
paste it down here. So I've pasted it down here. So we know what the height
is -- it's this big convoluted formula. The height in terms of a, b
and c is this right here. So if we wanted to figure out
the area -- the area of our triangle -- let me
do it in pink. The area of our triangle is
going to be 1/2 times our base -- our base is this entire
length, c -- times c times our height, which is this
expression right here. Let me just copy and
paste this instead of--. So let me copy and paste. So times the height. So this now is our
expression for the area. Now you're immediately saying
gee, that doesn't look a lot like Heron's formula,
and you're right. It does not look a lot like
Heron's formula, but what I'm going to show you in the next
video is that this essentially is Heron's formula. This is a harder to remember
version of Heron's formula. I'm going to apply a lot of
algebra to essentially simplify this to Heron's formula. But this will work. If you could memorize this,
I think Heron's a lot easier to memorize. But if you can memorize this
and you just know a, b and c, you apply this formula
right here and you will get the area of a triangle. Well, actually let's just apply
this just to show that this at least gives the same
number as Heron's. So in the last video we had a
triangle that had sides 9, 11 and 16, and its area using
Heron's was equal to 18 times the square root of 7. Let's see what we get when we
applied this formula here. So we get the area is equal
to 1/2 times 16 times the square root of a squared. That is 81 minus -- let's see,
c squared is 16, so that's 256. 256 plus a squared, that's
at 81 minus b squared, so minus 121. All of this stuff is squared. All of that over 2 times c
-- all of that over 32. So let's see if we can
simplify this a little bit. 81 minus 121, that is minus 40. So this becomes 216 over 32. So area is equal to
1/2 times 8 is 8. Let me switch colors. 1/2 times 16 is 8 times the
square root of 81 minus 256. 81 minus 121, that's minus 40. 256 minus 40 is 216. 216 over 32 squared. Now, this is a lot of
math to do so let me get out a calculator. I'm really just trying to show
you that these two numbers should give us our same number. So if we turn on
our calculator--. First of all, let's just
figure out what 18 square root of 7 are. 18 times the square root
of 7 -- this is what we got using Heron's. We got 47.62. Let's see if this is 47.62. So we have 8 times the square
root of 81 minus 216 divided by 32 squared, and then we
close our square roots. And we get the
exact same number. I was worried -- I actually
didn't do this calculation ahead of time so I might have
made a careless mistake. But there you go, you get
the exact same number. So our formula just now gave
us the exact same value as Heron's formula. But what I'm going to do in the
next video is prove to you that this can actually be reduced
algebraically to Heron's.