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