Area Circumradius Formula Proof

Proof of the formula relating the area of a triangle to its circumradius

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.