# Difference between revisions of "Circumradius"

m (→Equilateral triangles) |
m (→Equilateral triangles) |
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draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||

I=incenter(A,B,C); | I=incenter(A,B,C); | ||

− | draw( | + | draw(circumcircle(A,B,C)); |

− | label("$ | + | label("$C$",I,E); |

dot(I); | dot(I); | ||

</asy> | </asy> |

## Revision as of 18:57, 1 August 2017

*This article is a stub. Help us out by expanding it.*

The **circumradius** of a cyclic polygon is the radius of the cirumscribed 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 . Also,

## 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 area , so . Substituting this in gives us and then bash through algebra to get and we are done.

--Nosaj 19:39, 7 December 2014 (EST)

## 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 .

## 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.

## 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