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### Course: Geometry (all content) > Unit 14

Lesson 10: Area of inscribed triangle# 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.

## Want to join the conversation?

- 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)

- 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. :

https://docs.google.com/document/d/1s_uhlU1Fpg8xwhKfsv3hEBR7DjFlrW0K-m2GuIA0Jc4/edit

https://docs.google.com/document/d/1fY1RPGGYEKdp4_XObMANtUa5KcTD6ukyja6OKI7BDYk/edit(5 votes)

- 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).**SQRT(3) - pi) * R^2

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(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) - Why at9:59is srt of 3 over 2 equal to a/4?(2 votes)
- HOW Can we do the opposite..

i.e. an equilateral triangle of side 24 cm has a circle inscribed ..find the remaining are after circle is cut from the triangle(2 votes)- 1. Draw the figure and label the sides of the equilateral triangle in this case 24.

2. Connect a radii to any part of the circumference of the circle that is touching the triangle.

3. Connect another line segment to an apex of the same side. (For simplicity draw the line segments to the base, so connect the radii you are drawing to the base of the triangle where it is being touched by the circumference of the circle)

4. A right triangle has been formed

5. This a 30 - 60 - 90 triangle because you bisected one of the original angles of the equilateral triangle

6. The radii you drew is the short side of the triangle you created.

7. So you know the longer leg is r * SQRT(3)

8. Now draw the height of the equilateral triangle. This cuts the equilateral triangle in half.

9. If you drew the 30-60-90 triangle on the base of the equilateral triangle you can see the longer leg of the 30-60-90 triangle is the same as half the base of the equilateral triangle. Half the base of the equilateral triangle is 12. So r * SQRT(3) = 12

10. So r = 12 / SQRT(3)

11. If you rationalize the fraction you get r = 12 * SQRT(3) / 3

12. So r = 4 * SQRT(3)

13. The area of the circle is (4 * SQRT(3)^2 * pi which is also 48 pi

14. So the area of the circle is 48 pi you need to find the area of the equilateral triangle.

15. Using Heron's formula the area is 24^2 * SQRT(3)/4

16. So the area is 144 * SQRT(3)

17. So when you take off the circle the area is 144 * SQRT(3) - 48 pi(1 vote)

## 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.