Which would you rather have? Q.  Which would you rather have, a piece of an 8-inch pie that's been cut into sixths or a piece of a 10-inch pie that's been cut into eights?
A.  This is a problem involving sectors. One-sixth of a pie is $\frac {1}{6}$ of $ 2\pi$ radians. The measure of the central angle is $\frac{1}{6} \cdot 2\pi = \frac{\pi}{3}$. An 8-inch pie has a 4-inch radius. Putting the angle and radius into the formula, for the area of a sector, $A= \frac{1}{2} \cdot \frac{\pi}{3} \cdot 4^2 = \frac{16\pi}{6} = \frac{8\pi}{3} \approx 8.38$ square inches. One-eighth of a pie is $\frac{1}{8}$ of $2\pi$ radians. The measure of the angle is $\frac{1}{8} \cdot 2\pi = \frac{2\pi}{8} = \frac{\pi}{4}$. A 10-inch pie has a 5-inch radius. Putting the angle and radius into the formula, for the area of a sector, $A= \frac{1}{2} \cdot \frac{\pi}{4} \cdot 5^2 = \frac{25\pi}{8} \approx 9.82$ square inches. There isn't too much difference, but it looks like the smaller part of the bigger pie has the larger piece, in terms of area.
My question: 
Is how do one, get, $2\pi$ radians?

Area per piece of 8 inch pie = ${1\over 6} \times \pi (4)^2 = 8.38$
Area per piece of 10 inch pie = ${1\over 8} \times \pi (5)^2 = 9.82$
So the we will take the piece of 10 inch pie. $\ddot \smile$.
