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 circumradius formula proof

Proof of the formula relating the area of a triangle to its circumradius. Created by Sal Khan.

Want to join the conversation?

  • piceratops ultimate style avatar for user William T. MathGuy
    At around Sal says, "we've proven that an inscribed angle that is subtended by an arc will be half the arc length." When was this proven and how?
    (12 votes)
    Default Khan Academy avatar avatar for user
  • male robot johnny style avatar for user Dandy Cheng
    What does "Subtend" mean?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Sheridan Teasel
      In most maths contexts subtend means CONNECTED TO but normally also on the OPPOSITE side of a figure. e.g. a third of a circle SUBTENDS an angle of 120 degrees at the centre of the circle. If you draw the arc of 1/3 of a circle and CONNECT each end of it to the centre you find the 120 degree angle between the 2 connecting radii. But the angle is kinda OPPOSITE the curve of the arc in that shape. I hope that helps. If so please vote.

      Best regards,
      Sheridan
      (9 votes)
  • blobby green style avatar for user Radha Natarajan
    I am not clear how 2 different quantities are equated here...radius is linear while area is 2 dimensional..
    (4 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Captain Cryptic
      That's an interesting problem. The cool part is that when you multiply and divide everything, it comes out OK. So if we have a triangle with sides 3, 4, and 5 inches, the area would be 6 square inches (since it's a right triangle). So, you multiply it out: abc is 3" times 4" times 5" or 60 cubic inches. Divide 60 cubic inches by 4 to get 15 cubic inches. Divide 15 cubic inches by 6 square inches (the area) to get 2.5 inches! The extra units cancel each other out! Pretty amazing, I think.
      (2 votes)
  • aqualine seed style avatar for user Daniel M
    Is there somewhere I can find a list of all the proofs, postulates, axioms, and theorems, etc in Mathematics: Algebra, Geometry, Trignometry, etc?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • purple pi purple style avatar for user Alex
    At , Sal says that any triangle inscribed in a circle where one of the triangle's sides is the diameter of the circle, it will be a right triangle.
    What video was that proved in?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine sapling style avatar for user Muhammed Atef Abdoullah
    I Dun understand some thing ,, isn't b the circumcenter of triangle ? So how in , You considered the b center of circle.and distance from b to B which is vertex of triangle is a radius ?! Please,I need an instant answer , u used this then in similarity to say c over 2R
    (3 votes)
    Default Khan Academy avatar avatar for user
  • marcimus pink style avatar for user Hannah
    How can you prove that there will always be a "circum circle" for every triangle?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • winston default style avatar for user Yuhang Cui
    Why is this video in the AIME?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Scooley138
    This video seems out of place. I'm trying to go through geometry in a linear way, as material is presented here. Doing this has for the most part felt like new concepts build on old, there are no mysterious unexplained parts of the videos, etc. This video is an exception. It uses words that I don't know in a way that suggests that I've missed something, and talks about proofs that I don't think I've been exposed to. I've never heard the word "subtan" for instance. Am I the only one having this problem? I'm doing a section called "perpendicular bisectors".
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Shinobi
    What does AIME stand for?
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

What i want to do in this video is to come up with a relationship between the area of a triangle and the triangle's circumscribed circle or circum-circle. So before we think about the circum-circle let's just think about the area of the triangle. So let's say that the triangle looks something like this. Actually I don't want to make it look isosceles. So let me make it a little bit so it doesn't look like any particular type of triangle and let's call this traingle "ABC". That's the vertices and then the length of the side opposite "A" is "a" "b" over here, and then "c" We know how to calculate the area of this triangle if we know its height. If we drop an altitude right here and if this altitude has length "h" we know that the area of [ABC] - and we write [ABC] with the brackets around it means the area of the traingle [ABC] - is equal to 1/2 times the base, which is "b" times the height. Fair enough. We have an expression for the area. Let's see if we can somehow relate some of these things with the area to the radius of the triangle's circumscribed circle. So the circumscribed circle is a circle that passes through all of the vertices of the triangle and every triangle has a circumscribed circle. So let me try to draw it. This is the hard part, right over here so it might look something like this That's fair enough. That's close enough to a circle I think you get the general idea That is the circum-circle for this triangle. or this triangle's circumscribed circle. Let me label it. This is the circum-circle for this triangle. Now let's think about the center of that circum-circle sometimes refer to as the circumcenter. So looks like it would be sitting I don't know, just eyeballing it right on this little "b" here. So that's the circum-circle of the circle Let's draw a diameter through that circumcircle and draw a diameter from vertex "B" through that circumcenter. So then we go there, and we just keep going over here Let's call this point over here "D". Now let's create a triangle with vertices A, B, and D. So we can just draw another line over here and we have triangle ABD Now we proved in the geometry play - and it's not actually a crazy prove at all - that any triangle that's inscribed in a circle where one of the sides of the triangle is a diameter of the circle then that is going to be a right triangle and the angle that is going to be 90 degrees is the angle opposite the diameter So this is the right angle right here. You can derive that, pretty straightforward. You have this arc here that is 180 degrees. because obviously this is a diameter. And it subtends this inscribed angle. We've also proved that an inscribed angle that is subtended by the arc will be half of the arc length This is an 180 degree arc so this is going to be a 90 degree angle. So either way this's going to be 90 degrees over there The other thing we see is that we have this arc right over here that I'm drawing in magenta the arc that goes from "A" to "B" That arc subtends two different angles in our drawing - it subtends this angle right over here, angle ACB it subtends that right over there - but it also subtends angle ADB that's why we construct it this way So it also subtends this So these two angles are going to be congruent. They'll both have half the degree measure of this arc over here because they're both inscribed angles subtended by the same exact arc. Something interesting is popping up. We have two triangles here we have triangle ABD and triangle BEC They have two angles that resemble They have right angle and this magenta angle and their third angle must be the same. We'll do it in yellow The third angle must be congruent to that angle. They have three angles that are the same. They must be similar triangles. or the ratio between the corresponding sides must be the same. So we can use that information now to relate the length of this side which is really the diameter, is two times the radius to the height of this smaller triangle. We know the relationship between the height of the smaller triangle and the area and we essentially are in the home stretch. So let's do that So these are two similar triangles We know that the ratio of C to this diameter right here What's the length of the diameter? The length of the diameter is 2 times the radius This is the radius. We know that the ratio, C to two times the radius is going to be the same exact thing as the ratio of "h" - and we want to make sure we're using the same side - to the hypoteneuse of that triangle to the ratio of "H" to "A". And the way we figured that out we look at corresponding sides. "C" and the hypoteneuse are both the sides adjacent to this angle right over here So you have "H" and "A". So "C" is to "2r "as "H" is to "a". Or, we could do a lot of things. 1, we could solve for h over here and substitute an expression that has the area Actually let's just do that So if we use this first expression for the area. We could multiply both sides by two. And divide both sides by B. That cancels with that. We get that H is equal to 3 times the area over B. We can rewrite this relationship as c/2r is equals to h which is 2 times the area of our triangle over B and then all of that is going to be over A. Or, we could rewrite that second part over here as two times the area over - we're dividing by "b" and then divided by "a", that's the same thing as dividing by ab So we can ignore this right here. So we have c/2r is equals to 2 times the area over ab And now we can cross-multiply ab times c is going to be equal to 2r times 2abc. So that's going to be 4r times the area of our triangle. I just cross multiply this times this is going to be equal to that times that. We know that cross multiplication is just multiplying both sides of the equation by 2r and multiplying both sides of the equation by ab. So we did that on the left hand side we also did that on the right hand side 2r and ab obviously that cancels with that, that cancels with that So we get ABC is equal to 2r times 2abc. Or 4r times the area of our triangle. And now we're in the home stretch. We divide both sides of this by 4 times the area and we're done. This cancels with that, that cancels with that and we have our relationship The radius, or we can call it the circumradius. The radius of this triangle's circumscribed circle is equal to the product of the side of the triangle divided by 4 times the area of the triangle. That's a pretty neat result.