Difference between revisions of "Circumradius"
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<cmath>\frac{BD}{BA} = \frac{BC}{BE},</cmath> or <cmath> \frac {2R} c = \frac ah.</cmath> | <cmath>\frac{BD}{BA} = \frac{BC}{BE},</cmath> or <cmath> \frac {2R} c = \frac ah.</cmath> | ||
However, remember that <math>[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b</math>. Substituting this in gives us | However, remember that <math>[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b</math>. Substituting this in gives us | ||
− | <cmath> \frac {2R} c = \frac a{\frac{2 \times [ABC]}b},</cmath> and then | + | <cmath> \frac {2R} c = \frac a{\frac{2 \times [ABC]}b},</cmath> and then simplifying to get |
<cmath> R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}</cmath> | <cmath> R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}</cmath> | ||
and we are done. | and we are done. |
Revision as of 17:55, 20 July 2020
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
[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 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 simplifying 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 the law of sines.
(Extended Law of Sines)