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## Inscribed shapes problem solving

Current time:0:00Total duration:11:39

# Area of inscribed equilateral triangle

## Video transcript

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