Area circumradius formula proof Proof of the formula relating the area of a triangle to its circumradius
Area circumradius formula proof
- 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.
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