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What I want to do in this video
is use some of the results from
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the last several videos to
do some pretty neat things.
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So let's say this is a circle,
and I have an inscribed
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equilateral triangle
in this circle.
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So all the vertices of
this triangle sit on the
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circumference of the circle.
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So I'm going to try my best to
draw an equilateral triangle.
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I think that's about as good as
I'm going to be able to do.
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And when I say equilateral
that means all of these
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sides are the same length.
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So if this is side length a,
then this is side length a,
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and that is also a
side of length a.
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And let's say we know that the
radius of this circle is 2.
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I'm just picking a number,
just to do this problem.
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So let's say the radius
of this circle is 2.
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So from the center to the
circumference at any
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point, this distance, the
radius, is equal to 2.
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Now, what I'm going to ask you
is using some of the results of
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the last few videos and a
little bit of basic
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trigonometry-- and if the word
"trigonometry" scares you,
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you'll just need to know maybe
the first two or three videos
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in the trigonometry playlist to
be able to understand
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what I do here.
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What I want to do is figure out
the area of the region inside
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the circle and outside
of the triangle.
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So I want to figure out the
area of that little space, that
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space, and this space combined.
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So the obvious way to do
this is to say, well I can
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figure out the area of
the circle pretty easily.
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Area of the circle.
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And that's going to be
equal to pi r squared.
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Or pi times 2 squared,
which is equal to 4 pi.
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And I could subtract from 4
pi the area of the triangle.
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So we need to figure out
the area of the triangle.
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What is the area
of the triangle?
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Well, from several videos ago I
showed you Heron's formula,
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where if you know the lengths
of the sides of a triangle
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you can figure out the area.
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But we don't know the lengths
of the sides just yet.
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Once we do maybe we can
figure out the area.
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Let me apply Heron's
formula not knowing it.
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So let me just say that the
lengths of this equilateral--
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the lengths of the
sides-- are a.
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Applying Heron's formula, we
first define our variable
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s as being equal to a
plus a plus a, over 2.
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Or that's the same
thing as 3a over 2.
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And then the area of this
triangle, in terms of a.
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So the area is going to be
equal to the square root of
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s, which is 3a over
2, times s minus a.
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So that's 3a over 2 minus a.
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Or I could just
write, 2a over 2.
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Right? a is the same
thing as 2a over 2.
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You could cancel
those out and get a.
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And then I'm going to
do that three times.
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So instead of just multiplying
that out three times for each
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of the sides, by Heron's
formula I could just say
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to the third power.
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So what's this going
to be equal to?
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This is going to be equal to
the square root of 3a over 2.
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And then this right here
is going to be equal
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to 3a minus 2a, is a.
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So a/2 to the third power.
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And so this is going to be
equal to-- I'll arbitrarily
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switch colors.
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We have 3a times a to the
third, which is 3a to the
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fourth, over 2 times
2 to the third.
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Well that's 2 to the
fourth power, or 16.
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Right?
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2 times 2 to the third
is 2 to the fourth.
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That's 16.
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And then if we take the square
root of the numerator and the
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denominator, this is going to
be equal to the square root of
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a to the fourth is a squared.
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a squared times, well I'll just
write the square root of 3,
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over the square root of the
denominator, which is just 4.
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So if we know a, using Heron's
formula we know what the area
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of this equilateral
triangle is.
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So how can we figure out a?
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So what else do we know about
equilateral triangles?
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Well we know that all of
these angles are equal.
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And since they must add
up to 180 degrees, they
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all must be 60 degrees.
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That's 60 degrees,
that's 60 degrees, and
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that is 60 degrees.
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Now let's see if we can use the
last video, where I talked
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about the relationship
between an inscribed angle
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and a central angle.
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So this is an inscribed
angle right here.
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It's vertex is sitting
on the circumference.
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And so it is subtending
this arc right here.
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And the central angle that
is subtending that same arc
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is this one right here.
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The central angles subtending
that same arc is that
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one right there.
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So based on what we saw in the
last video, the central angle
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that subtends the same arc is
going to be double of
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the inscribed angle.
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So this angle right here is
going to be 120 degrees.
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Let me just put an arrow there.
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120 degrees.
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It's double of that one.
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Now, if I were to exactly
bisect this angle right here.
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So I go halfway through the
angle, and I want to just go
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straight down like that.
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What are these two
angles going to be?
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Well, they're going
to be 60 degrees.
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I'm bisecting that angle.
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That is 60 degrees, and that
is 60 degrees right there.
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And we know that I'm
splitting this side in two.
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This is an isosceles triangle.
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This is a radius right here.
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Radius r is equal to 2.
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This is a radius right
here of r is equal to 2.
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So this whole triangle
is symmetric.
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If I go straight down the
middle, this length right
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here is going to be
that side divided by 2.
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That side right there is going
to be that side divided by 2.
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Let me draw that over here.
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If I just take an isosceles
triangle, any isosceles
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triangle, where this side is
equivalent to that side.
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Those are our radiuses
in this example.
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And this angle is going to
be equal to that angle.
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If I were to just go straight
down this angle right
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here, I would split that
opposite side in two.
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So these two lengths
are going to be equal.
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In this case if the whole
thing is a, each of these
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are going to be a/2.
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Now, let's see if we can use
this and a little bit of
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trigonometry to find the
relationship between a and r.
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Because if we're able to solve
for a using r, then we can then
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put that value of a in here and
we'll get the area
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of our triangle.
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And then we could subtract
that from the area of the
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circle, and we're done.
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We will have solved
the problem.
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So let's see if we can do that.
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So we have an angle
here of 60 degrees.
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Half of this whole central
angle right there.
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If this angle is 60 degrees,
we have a/2 that's
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opposite to this angle.
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So we have an opposite
is equal to a/2.
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And we also have
the hypotenuse.
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Right?
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This is a right
triangle right here.
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You're just going straight
down, and you're bisecting
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that opposite side.
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This is a right triangle.
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So we can do a little
trigonometry.
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Our opposite is a/2, the
hypotenuse is equal to r.
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This is the hypotenuse, right
here, of our right triangle.
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So that is equal to 2.
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So what trig ratio is the
ratio of an angle's opposite
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side to hypotenuse?
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So some of you all might get
tired of me doing this all
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the time, but SOH CAH TOA.
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SOH-- sin of an angle is
equal to the opposite
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over the hypotenuse.
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So let me scroll
down a little bit.
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I'm running out of space.
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So the sin of this angle right
here, the sin of 60 degrees, is
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going to be equal to the
opposite side, is going to be
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equal to a/2, over the
hypotenuse, which is
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our radius-- over 2.
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Which is equal to a/2
divided by 2 is a/4.
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And what is sin of 60 degrees?
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And if the word "sin" looks
completely foreign to you,
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watch the first several videos
on the trigonometry playlist.
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It shouldn't be too daunting.
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sin of 60 degrees you
might remember from your
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30-60-90 triangles.
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So let me draw one right there.
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So that is a 30-60-90 triangle.
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If this is 60 degrees, that
is 30 degrees, that is 90.
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You might remember that this is
of length 1, this is going to
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be of length 1/2, and this
is going to be of length
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square root of 3 over 2.
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So the sin of 60 degrees is
opposite over hypotenuse.
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Square root of 3 over 2 over 1.
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sin of 60 degrees.
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If you don't have a calculator,
you could just use this--
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is square root of 3 over 2.
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So this right here is
square root of 3 over 2.
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Now we can solve for a.
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Square root of 3 over
2 is equal to a/4.
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Let's multiply both sides by 4.
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So you get this 4 cancels out.
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You multiply 4 here.
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This becomes a 2.
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This becomes a 1.
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You get a is equal to
2 square roots of 3.
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We're in the home stretch.
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We just figured out the length
of each of these sides.
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We used Heron's formula to
figure out the area of the
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triangle in terms
of those lengths.
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So we just substitute this
value of a into there
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to get our actual area.
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So our triangle's area
is equal to a squared.
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What's a squared?
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That is 2 square roots
of 3 squared, times the
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square root of 3 over 4.
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We just did a squared times
the square root of 3 over 4.
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This is going to be equal
to 4 times 3 times the
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square of 3 over 4.
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These 4's cancel.
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So the area of our triangle
we got is 3 times the
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square root of 3.
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So the area here is 3
square roots of 3.
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That's the area of
this entire triangle.
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Now, to go back to what this
question was all about.
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The area of this orange area
outside of the triangle
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and inside of the circle.
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Well, the area of
our circle is 4 pi.
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And from that we subtract
the area of the triangle,
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3 square roots of 3.
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And we are done.
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This is our answer.
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This is the area of this
orange region right there.
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Anyway, hopefully
you found that fun.
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