WEBVTT

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We're told the triangle ABC
has perimeter P and inradius r

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and then they want us
to find the area of ABC

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in terms of P and r.

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So we know that the
perimeter is just

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the sum of the sides
of the triangle,

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or how long a fence
would have to be

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if you wanted to go
around the triangle.

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And let's just remind
ourselves what the inradius is.

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If we take the angle
bisectors of each of these

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vertices-- each of these
angles right over here.

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So bisect that right over
there and then bisect

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that right over there.

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This angle is going to
be equal to that angle.

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This angle is going to
be equal to that angle

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and then this angle is going to
be equal to that angle there.

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And the point where those angle
bisectors intersect, that right

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over there, is our
incenter and it

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is equidistant from
all of the three sides.

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And the distance from those
sides, that's the inradius.

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So let me draw the inradius.

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So when you find the distance
between a point and a line,

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you want to drop
a perpendicular.

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So this length right over
here is the inradius.

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This length right over
here is the inradius

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and this length right
over here is the inradius.

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And if you want, you could
draw an incircle here

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with the center at the
incenter and with the radius r

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and that circle would
look something like this.

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We don't have to necessarily
draw it for this problem.

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So you could draw a circle
that looks something like that.

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And then we'd call
that the incircle.

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So let's think about how we can
find the area here, especially

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in terms of this inradius.

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Well, the cool thing
about the inradius

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is it looks like the
altitude-- or this looks

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like the altitude for this
triangle right over here,

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triangle A. Let's
label the center.

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Let's call it I for incenter.

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This r right over here is
the altitude of triangle AIC.

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This r is the altitude
of triangle BIC.

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And this r, which we didn't
label, that r right over there

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is the altitude of triangle AIB.

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And we know-- and
so we could find

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the area of each
of those triangles

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in terms of both
r and their bases

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and maybe if we sum up the
area of all the triangles,

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we can get something in terms of
our perimeter and our inradius.

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So let's just try to do this.

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So the area of the entire
triangle, the area of ABC,

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is going to be
equal to-- and I'll

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color code this-- is going to
be equal to the area of AIC.

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So that's what I'm
shading here in magenta.

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It's going to be equal to
the area of AIC plus the area

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of BIC, which is this
triangle right over here.

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Actually let me do that
in a different color.

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I've already used the blue.

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So let me do that in orange.

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Plus the area of BIC.

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So that's this area
right over here.

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And then finally plus the area--
I'll do this in a, let's see,

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I'll use this pink color--
plus the area of AIB.

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That is the area AIB.

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Take the sum of the areas
of these two triangles,

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you got the area of
the larger triangle.

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Now AIC, the area
of AIC, is going

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to be equal to 1/2
base times height.

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So this is going to be 1/2.

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The base is the length
of AC, 1/2 AC times

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the height-- times this
altitude right over here,

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which is just going
to be r-- times r.

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That's the area of AIC.

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And then the area of BIC is
going to be 1/2 the base,

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which is BC, times the
height, which is r.

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And then plus the area of
AIB, this one over here,

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is going to be 1/2
the base, which

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is the length of this side
AB, times the height, which

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is once again r.

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And over here, we can factor out
a 1/2 r from all of these terms

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and you get 1/2 r times
AC plus BC plus AB.

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And I think you see
where this is going.

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Plus-- now that's a different
shade of pink-- plus AB.

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Now what is AC plus BC plus AB?

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Well that's going to
be the perimeter, P,

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if you just take the
sum of the sides.

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So that is the perimeter of P
and it looks like we're done.

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The area of our triangle ABC
is equal to 1/2 times r times

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the perimeter, which is
kind of a neat result.

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1/2 times the inradius times
the perimeter of the triangle.

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Or sometimes you'll see
it written like this.

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It's equal to r times P
over s-- sorry, P over 2.

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And this term right over here,
the perimeter divided by 2,

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is sometimes called
the semiperimeter.

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And sometimes it's denoted
by s so sometimes you'll

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see the area is
equal to r times s,

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where s is the semiperimeter.

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It's the perimeter divided by 2.

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I personally like it
this way a little bit

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more because I remember
that P is perimeter.

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This is useful because obviously
now if someone gives you

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an inradius and a perimeter,
you can figure out

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the area of a triangle.

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Or if someone gives you
the area of the triangle

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and the perimeter, you
can get the inradius.

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So if they give you
two of these variables,

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you can always get the third.

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So for example, if this was
a triangle right over here,

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this is maybe the most famous
of the right triangles.

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If I have a triangle that
has lengths 3, 4, and 5,

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we know this is
a right triangle.

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You can verify this from
the Pythagorean theorem.

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And if someone
were to say what is

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the inradius of this
triangle right over here?

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Well we can figure out
the area pretty easily.

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We know this is
a right triangle.

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3 squared plus 4 squared
is equal to 5 squared.

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So the area is going to be
equal to 3 times 4 times 1/2.

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So 3 times 4 times
1/2 is 6 and then

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the perimeter here
is going to be

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equal to 3 plus 4, which
is 7, plus 5 is 12.

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And so we have the area.

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So let's write this area is
equal to 1/2 times the inradius

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times the perimeter.

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So here we have 12 is equal
to 1/2 times the inradius

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times the perimeter.

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So we have-- oh
sorry, we have 6.

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Let me write this in.

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The area is 6.

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We have 6 is equal to 1/2
times the inradius times 12.

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And so in this situation,
1/2 times 12 is just 6.

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We have 6 is equal to 6r.

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Divide both sides by 6,
you get r is equal to 1.

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So if you were to draw
the inradius for this one,

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which is kind of a neat result.

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So let me draw some
angle bisectors here.

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This 3, 4, 5 right triangle
has an inradius of 1.

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So this distance
equals this distance,

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which is equal to
this distance, which

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is equal to 1, which is
kind of a neat result.