Tip:
Highlight text to annotate it
X
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.