Geometry (all content)
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.
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- In3:32how did he get 2cx?(16 votes)
- Have a look at "multiplying binomials" which explains this step:
This is how it works out:
(c-x) * (c-x)
c(c-x) - x(c-x) [here I take each c & x from the left and multiply them against the (c-x) on the right]
(c^2 - cx) - (-x^2 + cx)
(c^2 - cx) + (x^2 - cx)
c^2 - 2cx + x^2
Another way of doing the third step is called FOIL (multiply First,Outside,Inside,Last terms):
(c * c) + (c * -x) + (-x * c) + (-x * -x)(26 votes)
- Here is an interesting question: "Why does Heron's Formula yield an area in square units even though four lengths are being multiplied together?(5 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.(15 votes)
- What is the thought process behind this? Did a mathematician just sit down and automatically know which steps to take?(6 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)
- 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/ 4
But, S = 3a/2
So now we solve the following:
sqrt[3a^4]/4 = 3 a^4 sqrt/64
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.(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)
- 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(2 votes)
- 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 ) ](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)
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.