Geometry (all content)
Area of inscribed equilateral triangle
A worked example of finding the area of an equilateral triangle inscribed within a circle who's area is known. This video uses Heron's formula and some trigonometry. Created by Sal Khan.
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- What is Heron's formula, and why didn't they show it to us before? I'm so confused...(29 votes)
- I don't know why, but the Heron's formula is in the Volume unit of the All Geometry section on Khan Academy (https://www.khanacademy.org/math/geometry-home/geometry-volume-surface-area#heron-formula-tutorial). So if you're studying trough "Basic Geometry" and "High School Geometry" sections, you won't find it there.(21 votes)
- I don't think you even need trig to solve the problem.
All you have to do is follow the video until around6:06and from there, use the rules of the side of a 30:60:90 right triangle. You can find the side a/2 and the line from the center of the circle to where it reaches side a perpendicularly. With these you can find the eight of the triangle and the base. Then you can just subtract the area of the triangle to the circle.(17 votes)
- Yes but you find out about this 30:60:90 rule in trig(6 votes)
- at8:50, I thought a/2/2 was 2a/2 not a/4, am I mistaken?(8 votes)
- Think about this it is a/2 / 2.
Then you take the inverse of two and get a/2 * 1/2.
Lastly, you multiply a and i and two and two and get a/4.(18 votes)
- Where did he get 3a minus 2a from.(6 votes)
- It's the Heron's Formula. There's a helpful video on that topic.(6 votes)
- why not use area of a triangle = 1/2 * base * height ?
would be easier than Heron's Formula(4 votes)
- There may be multiple ways to tackle math problems. In fact I had to use the triangle area formula, law of cosines and knowledge of inscribed angles because this is the first time I hear about Heron's Formula.
I can only suggest you to try to solve any problem using everything you know so far and afterwards watch the rest of the video. Maybe there will be a better and easier way that you can learn and use next time.
(Sorry for my bad English)(5 votes)
- Do I have to read Euclid's book on geometry to understand geometry better?(1 vote)
- It really isn't necessary. I made 2 documents with some geometry theorems, definitions, postulates, etc. :
- isn't there an easier way? area of triangle = 3(1/2 abSinC) (draw radii from the origin) = 3(1/2(2*2 Sin 120)) = 3(2Sin120) = 5.196, then the area of the circle minus this = 4pi-5.196 = 7.370(1 vote)
- You are correct. But, here's how I would have done it:
Area outside the triangle = πr² - ¼ a²√3
Because the area of an equilateral triangle is ¼ a²√3
Since a = r√3 also stated as a² = 3r²
Substituting, πr² - ¾r²√3
Since r = 2, we get 4π - 3√3 = 7.370
Of course my way does require knowing that a² = 3r² for an inscribed equilateral triangle (though it isn't too hard to derive if you didn't know that)(5 votes)
- how would i be able to find the area if the circle was in a triangle(2 votes)
- Your question is probably about finding the area of an equilateral triangle with an inscribed circle given the circle's radius. This turns out to be very similar to Sal's question!
You can draw an equilateral triangle inside the circle, with vertices where the circle touches the outer triangle. Now, you know how to calculate the area of that inner triangle from Sal's video. Specifically, this is 3/4 * r^2 * sqrt(3). (When r=2 like in the video, this is 3 * sqrt(3).)
The outer triangle is simply 4 of these triangles (ASA postulate). So, the big triangle's area is 3 * r^2 * sqrt(3). (When r=2, this is 12 * sqrt(3).)(2 votes)
- Does anybody know what to do if it's the opposite (circle transcribed in an equilateral triangle)? thanks!(0 votes)
- Yes you can. For example, if you know either the length of the (outside) equilateral triangle... or the radius of the (inside) circle, you can figure out the other. Let's take the example that you know the radius of the (inside) circle. This is a bit more challenging to do without video, but if you understood the video, you could probably follow:
R = radius of circle (known)
x = length of equilateral triangle
Draw the figure (outside equilateral triangle with inscribed circle).
Put a dot in the center of the circle.
Draw a line from the center of the circle to a point where the circle barely touches the triangle. This forms a right angle with that line, and that length = R (radius).
Also draw a line from the center to a nearby apex of the triangle (on the same triangle line edge). You have just formed an inside triangle. You know one of the angles (90 degrees). The other two are 60 and 30 degrees. You know this because the equilateral triangle angle was 60 degrees, and you just bisected it.
Just like the example in the video, the distance from the center to the outside triangle apex is 2*R, and the distance of the distance of the base of the inside triangle is SQRT(3)R, and by definition that distance is x/2 (half the length of the outside triangle edge).
Thus x/2 = SQRT(3)*R
x = 2*SQRT(3)*R
You can now use the formulas shown in the video for the area of the triangle and the circle.
If I did my math correctly...
Area = (SQRT(3)*x^2) / 4 - pi * R^2 ... then enter formula for 'x' above...
= (SQRT(3) * (2*SQRT(3)*R)^2) / 4 - pi * R^2
= SQRT(3) * 3 * R^2 - pi * R^2
or = (3 SQRT(3) - pi) * R^2(5 votes)
- Is this a general result? Like, if you take a circle of any radius and inscribe an equilateral triangle, would the area outside the triangle (but inside the circle) be
4π - 3*sqrt(3)?
I asked because my class 10 finals are coming up and this could be asked as a 1-mark question and it would have been really helpful to just memorize this. Thank you in advance.(2 votes)
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.