# Difference between revisions of "Circumradius"

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== Proof == | == Proof == | ||

Proof: | Proof: | ||

− | + | <asy> | |

pair O, A, B, C, D; | pair O, A, B, C, D; | ||

O=(0,0); | O=(0,0); | ||

Line 19: | Line 19: | ||

D=-B; dot (D); | D=-B; dot (D); | ||

draw(B--D--A); | draw(B--D--A); | ||

− | label(" | + | label("$A$", A, W); |

− | label(" | + | label("$B$", B, N); |

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

− | label(" | + | label("$D$", D, S); |

− | label(" | + | label("$O$", O, W); |

pair E; | pair E; | ||

E=foot(B,A,C); | E=foot(B,A,C); | ||

draw(B--E); | draw(B--E); | ||

dot(E); | dot(E); | ||

− | label(" | + | label("$E$", E, S); |

draw(rightanglemark(B,A,D,20)); | draw(rightanglemark(B,A,D,20)); | ||

draw(rightanglemark(B,E,C,20)); | draw(rightanglemark(B,E,C,20)); | ||

− | + | </asy> | |

We let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>, <math>BE=h</math>, and <math>BO=R</math>. We know that <math>\angle BAD</math> is a right angle because <math>BD</math> is the diameter. Also, <math>\angle ADB = \angle BCA</math> because they both subtend arc <math>AB</math>. Therefore, <math>\triangle BAD \sim \triangle BEC</math> by AA similarity, so we have | We let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>, <math>BE=h</math>, and <math>BO=R</math>. We know that <math>\angle BAD</math> is a right angle because <math>BD</math> is the diameter. Also, <math>\angle ADB = \angle BCA</math> because they both subtend arc <math>AB</math>. Therefore, <math>\triangle BAD \sim \triangle BEC</math> by AA similarity, so we have |

## Revision as of 19:36, 7 December 2014

*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 of the circumradius of the triangle is simply . Also,

## Proof

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.

## Formula for Circumradius

Where is the Circumradius, is the inradius, and , , and are the respective sides of the triangle. Note that this is similar to the previously mentioned formula; the reason being that .

## Euler's Theorem for a Triangle

Let have circumradius and inradius . Let be the distance between the circumcenter and the incenter. Then we have