Difference between revisions of "Circumradius"
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==Formula for a Triangle== | ==Formula for a Triangle== | ||
Let <math>a, b</math> and <math>c</math> denote the triangle's three sides, and let <math>A</math> denote the area of the triangle. Then, the measure of the of the circumradius of the triangle is simply <math>R=\frac{abc}{4A}</math>. Also, <math>A=\frac{abc}{4R}</math> | Let <math>a, b</math> and <math>c</math> denote the triangle's three sides, and let <math>A</math> denote the area of the triangle. Then, the measure of the of the circumradius of the triangle is simply <math>R=\frac{abc}{4A}</math>. Also, <math>A=\frac{abc}{4R}</math> | ||
+ | |||
+ | == Proof == | ||
+ | Proof: | ||
+ | [asy] | ||
+ | pair O, A, B, C, D; | ||
+ | O=(0,0); | ||
+ | A=(-5,1); | ||
+ | B=(1,5); | ||
+ | C=(5,1); | ||
+ | dot(O); dot (A); dot (B); dot (C); | ||
+ | draw(circle(O, sqrt(26))); | ||
+ | draw(A--B--C--cycle); | ||
+ | D=-B; dot (D); | ||
+ | draw(B--D--A); | ||
+ | label("<math>A</math>", A, W); | ||
+ | label("<math>B</math>", B, N); | ||
+ | label("<math>C</math>", C, E); | ||
+ | label("<math>D</math>", D, S); | ||
+ | label("<math>O</math>", O, W); | ||
+ | pair E; | ||
+ | E=foot(B,A,C); | ||
+ | draw(B--E); | ||
+ | dot(E); | ||
+ | label("<math>E</math>", E, S); | ||
+ | draw(rightanglemark(B,A,D,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 | ||
+ | <cmath>\frac{BD}{BA} = \frac{BC}{BE},</cmath> or <cmath> \frac {2R} c = \frac ah.</cmath> | ||
+ | However, remember that area <math>\triangle ABC = \frac {bh} 2</math>, so <math>h=\frac{2 \times \text{Area}}b</math>. Substituting this in gives us | ||
+ | <cmath> \frac {2R} c = \frac a{\frac{2 \times \text{Area}}b},</cmath> and then bash through algebra. | ||
==Formula for Circumradius== | ==Formula for Circumradius== |
Revision as of 19:35, 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
[hide]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: [asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label("", A, W); label("", B, N); label("", C, E); label("", D, S); label("", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label("", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy]
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