Area of Inscribed Equilateral Triangle (some basic trig used) Problem that requires us to figure out the area of an equilateral triangle inscribed in a circle (A little trigonometry used)
Area of Inscribed Equilateral Triangle (some basic trig used)
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- 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.
- 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.
- 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.
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