Originally Posted by

**HallsofIvy** Strictly speaking the term "radius" only applies to circles. And one way to approximate this is to assume that you are talking about a circular pen with circumference 12(12)= 144 feet. For a circle the circumference is given by $\displaystyle 2\pi r$ so we must have $\displaystyle 2\pi r= 144$ so that $\displaystyle r= \frac{72}{\pi}= 22.91$ feet or 22 feet, 11 inches. That is 3 inches (about 1%) shorter than godelproof's answer.

What he did was imagine drawing lines from the center of the pen to the ends of each panel. That divides the 12 sided polygon into 12 isosceles triangles each with a base length of 12 feet and an angle of 360/12= 30 degrees. Now draw a line from the center of the pen to the center of each panel. We now have 24 right triangles with one leg of length 6 and opposite angle 30/2= 15 degrees. Since "sine" is "opposite side/ hypotenuse", taking the distance from the center of the pen to the end of each panel to be x, we have $\displaystyle \frac{6}{x}= sin(15)$ so that $\displaystyle x= \frac{6}{sin(15)}$. Using a calculator to find that sin(15)= 0.25882, we have $\displaystyle x= \frac{6}{0.25882}= 23.18$ feet of 23 feet and 2 inches as godelproof said.

As I said before, strictly speaking the term "radius" only applies to circles. What godelproof calculated was the "ex-radius", the radius of the circle that passes through the vertices of the polygon. One could just as well use the "in-radius", the radius of the circle inscribed in the polygon, that is tangent to each panel at its center. That radius is the other leg of the right triangle so is given by $\displaystyle tan(15)= \frac{6}{x}$ so that $\displaystyle x= \frac{6}{tan(15)}$. tan(15)= 0.26795 so that $\displaystyle x= \frac{6}{0.25795}= 22.4$ feet or 22 feet 4 inches.

For small angles, sine and tangent are almost the same so for polygons with many sides, the "in-radius" and "ex-radius" are almost the same- and my circle approximation is between them.