Difference between revisions of "Circumradius"

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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.

Formula for a Triangle

Let $a, b$ and $c$ denote the triangle's three sides, and let $A$ denote the area of the triangle. Then, the measure of the of the circumradius of the triangle is simply $R=\frac{abc}{4A}$ . Also, $A=\frac{abc}{4R}$

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$", A, W); label("$B$", B, N); label("$C$", C, E); label("$D$", D, S); label("$O$", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label("$E$", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy]$

We let $AB=c$ , $BC=a$ , $AC=b$ , $BE=h$ , and $BO=R$ . We know that $\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\angle ADB = \angle BCA$ because they both subtend arc $AB$ . Therefore, $\triangle BAD \sim \triangle BEC$ by AA similarity, so we have $\frac{BD}{BA} = \frac{BC}{BE},$ or $\frac {2R} c = \frac ah.$ However, remember that area $\triangle ABC = \frac {bh} 2$ , so $h=\frac{2 \times \text{Area}}b$ . Substituting this in gives us $\frac {2R} c = \frac a{\frac{2 \times \text{Area}}b},$ and then bash through algebra.

Formula for Circumradius

$R = \frac{abc}{4rs}$ Where $R$ is the Circumradius, $r$ is the inradius, and $a$ , $b$ , and $c$ are the respective sides of the triangle. Note that this is similar to the previously mentioned formula; the reason being that $A = rs$ .

Euler's Theorem for a Triangle

Let $\triangle ABC$ have circumradius $R$ and inradius $r$ . Let $d$ be the distance between the circumcenter and the incenter. Then we have $d^2=R(R-2r)$

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