Heron's formula
Part 2 of the Proof of Heron's Formula Showing that the expression in part 1 is identical to Heron's Formula
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- In
- the last video, I claimed that this result we got for the area
- of a triangle that had sides of length a, b, and c is
- equivalent to Heron's formula.
- And what I want to do in this video is show you that this is
- equivalent to Heron's formula by essentially just doing a
- bunch of algebraic manipulation.
- So the first thing we want to do-- let's just spring this
- 1/2 c under the radical sign.
- So 1/2 c, that's the same thing as the square root
- of c squared over 4.
- You take the square root of that you get 1/2 c.
- So this whole expression is equal to-- instead of drawing
- the radical, I'll just write the square root of this,
- of c squared over 4 times all of this.
- I'll just copy and paste it.
- Copy and paste.
- So times all of that.
- And of course, it has to be distributed.
- So c squared over 4 times all of that.
- And then we have to close the square root.
- Let me just distribute the c squared over 4.
- This is going to be equal to the square root.
- This is going to be hairy, but I think you'll find it
- satisfying to see how this could turn into something as
- simple as Heron's formula.
- The square root of c squared over 4 times a squared is c
- squared a squared over 4, minus c squared over 4.
- I'm just distributing this.
- And I'm going to write it as the numerator squared over
- the denominator squared.
- So times c squared plus a squared minus b
- squared, squared.
- Over-- if I square the denominator that's 4c squared.
- And we immediately see that c squared and that c squared
- are going to cancel out.
- Let me close all of the parentheses just like that.
- And, of course, this 4 times that 4, that's going to
- result in-- well let me write it this way.
- That's the same thing as 4 squared.
- And I'm instead of writing 16, you'll see why
- I'm writing that.
- Now this I can rewrite.
- This is going to be equal to the square root-- I'm
- arbitrarily switching colors-- of ca over 2 squared.
- This is the same thing as that.
- Right?
- I'm just writing it as the whole thing squared.
- If I square that, that's the c squared a squared over 2
- squared over 4, minus-- and I'm going to write this whole thing
- as an expression squared.
- So that's c squared plus a squared minus
- b squared, over 4.
- And we are squaring both the numerator and the denominator.
- Now this might look a little bit interesting to you.
- Let me make the parentheses in a slightly different color.
- You might remember from factoring polynomials that if I
- have something of the form x squared minus y squared, that
- factors into x plus y times x minus y.
- And we're going to be using this over and over again.
- Now if you call ca over 2x, and you call this whole big thing
- y, then we have x squared minus y squared.
- So we can factor it.
- So this whole thing is going to be equal to the square root of
- x plus y, or in this case it's ca over 2 plus the y, which is
- c squared plus a squared minus b squared over 4.
- Times x minus y.
- So this is our x.
- ca over 2, minus all of this business over here.
- Or even better, let me just say plus and then let me
- just write the negative.
- So plus minus c squared minus a squared plus b squared.
- All of that over 4.
- So all I did here is I said this is the same thing as this
- plus this, this plus this, times this minus this, this
- minus-- I just said plus the negative of this.
- So minus c squared minus a squared plus b squared.
- All I did is that right there.
- Now let's see if we can simplify this, or if we
- could add these fractions.
- Well, we can get a common denominator.
- ca over 2, that's the same thing as 2ca over 4.
- ca over 2, that's the same thing as 2ca over 4, just
- multiplying the numerator and the denominator by 2.
- And now we can add the numerators.
- So our whole expression is now going to be equal to the square
- root of this first expression, will become-- and I'm going
- to write it this way.
- I'm going to write c squared plus 2ca plus a squared minus b
- squared, all of that over 4.
- That's our first expression.
- And then our second expression is going to become-- well,
- everything's going to be over 4, so I'll just write
- that right now.
- Everything over 4.
- And then we could write this as b squared, minus c squared
- minus 2ca plus a squared.
- Just to make sure, I have a minus a squared here.
- Plus times a minus, still it's a minus a squared.
- I have a plus 2ca over here.
- Minus times a minus, that's a plus 2ca.
- I have a minus c squared here.
- I have a minus c squared here.
- So these two things are equivalent.
- Now the next thing we need to recognize, or hopefully we can
- recognize, is that this over here-- this might get a little
- bit messy-- that's the same thing as c plus a squared.
- Let me write this.
- This is equal to the square root, open parentheses, of this
- over here is c plus a squared minus b squared, over 4.
- That's that first term.
- And then the second term.
- This over here is the same thing as c minus a squared.
- So that whole thing will simplify to b squared
- minus c minus a squared, all of that over 4.
- So we're making some headway.
- As I told you, this is a hairy problem.
- But we're seeing some neat applications of factoring
- polynomials, and we're seeing how a fairly bizarre looking
- equation can be transformed into a simpler one.
- Now we can use this exact same property-- we have that
- pattern-- something squared minus something else squared.
- So we can factor it out.
- And I'll do it in the same line.
- So this is going to be equal to-- I'm going to write a
- little bit small, just so I don't run out of space--
- the square root.
- This will factor into this plus this.
- So c plus a plus b times c plus a minus b.
- Right?
- It's the exact same pattern that I did over here.
- This is x squared, this is y squared.
- So times c plus a minus b, all of that over 4.
- And then we have this one.
- This is going to be b plus c minus a.
- Let me scroll down to the right a little bit.
- Times b plus c minus a-- that's x plus y-- times
- b minus c minus a.
- Or that's the same thing as b minus c plus a.
- This is the same thing as b minus c minus a.
- Right?
- All right.
- And all of that over 4.
- Now, I can rewrite this whole expression.
- I don't want to run out of space.
- I can rewrite this whole expression as, well 4 is
- the product of 2 times 2.
- So our whole area expression has been, arguably, simplified
- to it equals the square root-- and this is really the home
- stretch-- of this right here, which I can just write as
- a plus b plus c over 2.
- That's that term right there.
- Times this term.
- Times that term.
- And let me write a simplification here. c plus a
- minus b, that's the same thing as a plus b plus c minus 2b.
- These two things are equivalent.
- Right?
- You have an a, you have a c, and then b minus 2b is going
- to be equal to minus b.
- Right? b minus 2b, that's minus b.
- So this next term is going to be a plus b plus
- c minus 2b, over 2.
- Or instead of writing it like that, let me write this
- over 2 minus this over 2.
- And then our next term right here.
- Same exact logic.
- That's the same thing as a plus b plus c minus 2a,
- all of that over 2.
- Right?
- If we add the minus 2a to the a we get minus a.
- So we get b plus c minus a.
- These are identical things.
- So all this over 2, or we can split the denominators
- just like that over 2.
- And then one last term.
- And you might already recognize the rule of
- Heron's formula popping up.
- I was thinking not the rule of Heron-- Heron's formula.
- That term right there is the exact same thing as a
- plus b plus c minus 2c.
- Right?
- You take 2c away from the c you get a minus c, and then you
- still have the a and the b.
- And then all of that over 2.
- You could write that over 2 minus that over 2.
- And, of course, we're taking the square root
- of all of this stuff.
- Now, if we define an S to be equal to a plus b plus c over
- 2, then this equation simplifies a good bit.
- This right here is S.
- That right there is S.
- That right there is S.
- And that right there is S.
- And these simplify a good bit too.
- Minus 2b over 2, that's just the same thing as minus b.
- Minus 2a over 2, that's the same thing as minus a.
- Minus 2c over 2, that's the same thing as minus c.
- So this whole equation for our area now is equal to-- I'll
- rewrite the square root.
- The radical, the square root, of S-- that's that right there.
- I'll do it in the same colors.
- Times S minus b, times this is S minus a, times-- and we're
- at the last one-- S minus c.
- And we have proved Heron's formula is the exact same thing
- as what we proved at the end of the last video.
- So this was pretty neat.
- And we just had to do a little bit of hairy algebra
- to actually prove it.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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