Difference between revisions of "Circumradius"
(→Right triangles) |
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draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
I=circumcenter(A,B,C); | I=circumcenter(A,B,C); | ||
+ | draw(I--A,gray); | ||
+ | label("$r$",(I+A)/2,gray,NW); | ||
draw(circumcircle(A,B,C)); | draw(circumcircle(A,B,C)); | ||
label("$C$",I,N); | label("$C$",I,N); | ||
Line 66: | Line 68: | ||
draw(rightanglemark(B,A,C,10)); | draw(rightanglemark(B,A,C,10)); | ||
</asy> | </asy> | ||
+ | |||
+ | This results in a well-known theorem: | ||
+ | ===Theorem=== | ||
+ | The midpoint of the hypotenuse is equidistant from the vertices of the right triangle. | ||
== Equilateral triangles == | == Equilateral triangles == |
Revision as of 22:03, 7 November 2019
The circumradius of a cyclic polygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle.
Contents
Formula for a Triangle
Let and
denote the triangle's three sides and let
denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply
. This can be rewritten as
.
Proof
We let ,
,
,
, and
. We know that
is a right angle because
is the diameter. Also,
because they both subtend arc
. Therefore,
by AA similarity, so we have
or
However, remember that
. Substituting this in gives us
and then bash through the algebra to get
and we are done.
Formula for Circumradius
Where
is the circumradius,
is the inradius, and
,
, and
are the respective sides of the triangle and
is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that
.
But, if you don't know the inradius, you can find the area of the triangle by Heron's Formula:
Euler's Theorem for a Triangle
Let have circumcenter
and incenter
.Then
Proof
Right triangles
The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle.
This results in a well-known theorem:
Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right triangle.
Equilateral triangles
where is the length of a side of the triangle.
If all three sides are known
And this formula comes from the area of Heron and .
If you know just one side and its opposite angle
which is just extension of law of sines
(Extended Law of Sines)