If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Area of a regular hexagon

Using what we know about triangles to find the area of a regular hexagon. Created by Sal Khan.

Want to join the conversation?

  • piceratops ultimate style avatar for user Jai
    What is the formula of a hexagon?
    (20 votes)
    Default Khan Academy avatar avatar for user
    • leaf grey style avatar for user Zhemin Shao
      The formula to calculate the area of a regular hexagon with side length s:
      (3 √3 s^2)/2
      Remember, this only works for REGULAR hexagons. For irregular hexagons, you can break the parts up and find the sum of the areas, depending on the shape.
      Hope that helped!
      (34 votes)
  • leaf red style avatar for user freyawolf
    instead of dividing the hexagon into 6 triangles wouldn't it be slightly easier to draw a hypothetical line from point f to point b and again from point e to point c turning it into 2 triangles and a rectangle?
    (10 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Zygmunt
    At , isn't the area of an equilateral triangle (sqrt(3)*s^2)/4? I still get 3*sqrt(3), so I guess it's not as important as I thought...
    (10 votes)
    Default Khan Academy avatar avatar for user
    • leafers seedling style avatar for user Fieso Duck
      Well, you are actually right. That would be the special formula that gives you the area of equilateral triangles.
      However the general area formula for triangles used in the video (A = 1/2*h*b), works for all triangles, including equilateral ones.
      Using the special formula as suggested by you would have been quicker though, as you only need to know the side measurement of the equilateral, while the general formula requires the height and the base measurement.
      (10 votes)
  • aqualine ultimate style avatar for user TheEpicOne
    Couldn't you just divide it into separate triangles and add up the area of those?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user fleoul15
    What is the length of a side of a regular six sided polygon with radius of 8cm?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user loumast17
      Radius is the distance from the center to a corner. it is also important to know the apothem This works for any regular polygon.

      Choose a side and form a triangle with the two radii that are at either corner of said side. You know both radii are 8 cm, which means you have an isosceles triangle.

      You want to count how many of these triangles you can make. Basically each side will have one of these. this means each triangle will have an angle of measure 360/n, where n is the number of sides. In your case that is 360/6 =60. Since it is a scalene triangle you know the measure of the other two angles are the same. Also, you should know the angles of a triangle add up to 180. so in other words use some algebra to find the two other angles. Here that works out like this.

      one angle is 60 and the other two are some other angle x where all three equal 180. So that works out to 60 + x + x = 180.
      60 + 2x = 180
      2x = 120
      x = 60.

      So this shows al four angles are 60 degrees, which means not only is it a scalene triangle, but an equilateral triangle. This means all sides are the same. And since we know the radii that means the remaining side is the sme measure at 8 cm.

      If these were not equilateral you would have to use the apothem and the Pythagorean theorem.
      (5 votes)
  • aqualine ultimate style avatar for user Kollen Platt
    i dont get this. whats going on?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • male robot donald style avatar for user Daniel Chernov
    Is there a video or tutorial on how to find the area of the hexagon in the case when it is inscribed into a circle?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • mr pants teal style avatar for user leah.bieniek
    what about a polygon? I'm confused.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Connor Miles
    At you failed to mention that all exterior angles are congruent and have the same measure as well as the interior angles. Of course, even if the hexagon isn't regular and all sides aren't congruent, the exterior angles could still be congruent provided they are attached the right kind of polygon. Anyways, I just felt like pointing that out because it really itched my brain. Side note: Thanks for the great math videos, they really help!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user Unpotato
      I feel like defending Khan here, and I don't want to be a jerk, but:

      He doesn't need to point out that the exterior angles are congruent because it's not relevant to what he's trying to solve: the area of the hexagon. Why mention it if it could be confusing the audience of why it's important?

      He also told us that the angles all have the same measure at , which also means the interior angles are congruent, as by the Definition of Congruent Angles.

      Your second argument was confusing, yet I get what you mean. Imagine that AB and DE were 4 units long, which would keep the interior angles at 120 degrees and thus the exterior angles congruent. Yet, again, the argument is about exterior angles, and exterior angles are not needed to find the area.
      (5 votes)
  • leaf blue style avatar for user Smit
    OK, so each triangle has 180°. 6x180=1080°, not 360°. I don't see why this doesn't work out.
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

We're told that ABCDEF is a regular hexagon. And this regular part-- hexagon obviously tells us that we're dealing with six sides. And you could just count that. You didn't have to be told it's a hexagon. But the regular part lets us know that all of the sides, all six sides, have the same length and all of the interior angles have the same measure. Fair enough. And then they give us the length of one of the sides. And since this is a regular hexagon, they're actually giving us the length of all the sides. They say it's 2 square roots of 3. So this side right over here is 2 square roots of 3. This side over here is 2 square roots of 3. And I could just go around the hexagon. Every one of their sides is 2 square roots of 3. They want us to find the area of this hexagon. Find the area of ABCDEF. And the best way to find the area, especially of regular polygons, is try to split it up into triangles. And hexagons are a bit of a special case. Maybe in future videos, we'll think about the more general case of any polygon. But with a hexagon, what you could think about is if we take this point right over here. And let's call this point G. And let's say it's the center of the hexagon. And when I'm talking about a center of a hexagon, I'm talking about a point. It can't be equidistant from everything over here, because this isn't a circle. But we could say it's equidistant from all of the vertices, so that GD is the same thing as GC is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE. So let me draw some of those that I just talked about. So that is GE. There's GD. There's GC. All of these lengths are going to be the same. So there's a point G which we can call the center of this polygon. And we know that this length is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length. We also know that if we go all the way around the circle like that, we've gone 360 degrees. And we know that these triangles are all going to be congruent to each other. And there's multiple ways that we could show it. But the easiest way is, look, they have two sides. All of them have this side and this side be congruent to each other because G is in the center. And they all have this third common side of 2 square roots of 3. So all of them, by side-side-side, they are all congruent. What that tells us is, if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all six of these triangles over here. And let me call that x. That's angle x. That's x. That's x. That's x. That's x. And if you add them all up, we've gone around the circle. We've gone 360 degrees. And we have six of these x's. So you get 6x is equal to 360 degrees. You divide both sides by 6. You get x is equal to 60 degrees. All of these are equal to 60 degrees. Now there's something interesting. We know that these triangles-- for example, triangle GBC-- and we could do that for any of these six triangles. It looks kind of like a Trivial Pursuit piece. We know that they're definitely isosceles triangles, that this distance is equal to this distance. So we can use that information to figure out what the other angles are. Because these two base angles-- it's an isosceles triangle. The two legs are the same. So our two base angles, this angle is going to be congruent to that angle. If we could call that y right over there. So you have y plus y, which is 2y, plus 60 degrees is going to be equal to 180. Because the interior angles of any triangle-- they add up to 180. And so subtract 60 from both sides. You get 2y is equal to 120. Divide both sides by 2. You get y is equal to 60 degrees. Now, this is interesting. I could have done this with any of these triangles. All of these triangles are 60-60-60 triangles, which tells us-- and we've proven this earlier on when we first started studying equilateral triangles-- we know that all of the angles of a triangle are 60 degrees, then we're dealing with an equilateral triangle, which means that all the sides have the same length. So if this is 2 square roots of 3, then so is this. This is also 2 square roots of 3. And this is also 2 square roots of 3. So pretty much all of these green lines are 2 square roots of 3. And we already knew, because it's a regular hexagon, that each side of the hexagon itself is also 2 square roots of 3. So now we can essentially use that information to figure out-- actually, we don't even have to figure this part out. I'll show you in a second-- to figure out the area of any one of these triangles. And then we can just multiply by 6. So let's focus on this triangle right over here and think about how we can find its area. We know that length of DC is 2 square roots of 3. We can drop an altitude over here. We can drop an altitude just like that. And then if we drop an altitude, we know that this is an equilateral triangle. And we can show very easily that these two triangles are symmetric. These are both 90-degree angles. We know that these two are 60-degree angles already. And then if you look at each of these two independent triangles, you'd have to just say, well, they have to add up to 180. So this has to be 30 degrees. This has to be 30 degrees. All the angles are the same. They also share a side in common. So these two are congruent triangles. So if we want to find the area of this little slice of the pie right over here, we can just find the area of this slice, or this sub-slice, and then multiply by 2. Or we could just find this area and multiply by 12 for the entire hexagon. So how do we figure out the area of this thing? Well, this is going to be half of this base length, so this length right over here. Let me call this point H. DH is going to be the square root of 3. And hopefully we've already recognized that this is a 30-60-90 triangle. Let me draw it over here. So this is a 30-60-90 triangle. We know that this length over here is square root of 3. And we already actually did calculate that this is 2 square roots of 3. Although we don't really need it. What we really need to figure out is this altitude height. And from 30-60-90 triangles, we know that the side opposite the 60-degree side is the square root of 3 times the side opposite the 30-degree side. So this is going to be square root of 3 times the square root of 3. Square root of 3 times the square root of 3 is obviously just 3. So this altitude right over here is just going to be 3. So if we want the area of this triangle right over here, which is this triangle right over here, it's just 1/2 base times height. So the area of this little sub-slice is just 1/2 times our base, just the base over here. Actually, let's take a step back. We don't even have to worry about this thing. Let's just go straight to the larger triangle, GDC. So let me rewind this a little bit. Because now we have the base and the height of the whole thing. If we care about the area of triangle GDC-- so now I'm looking at this entire triangle right over here. This is equal to 1/2 times base times height, which is equal to 1/2-- what's our base? Our base we already know. It's one of the sides of our hexagon. It's 2 square roots of 3. It's this whole thing right over here. So times 2 square roots of 3. And then we want to multiply that times our height. And that's what we just figured out using 30-60-90 triangles. Our height is 3. So times 3. 1/2 and 2 cancel out. We're left with 3 square roots of 3. That's just the area of one of these little wedges right over here. If we want to find the area of the entire hexagon, we just have to multiply that by 6, because there are six of these triangles there. So this is going to be equal to 6 times 3 square roots of 3, which is 18 square roots of 3. And we're done.