解：$ (1)$由题意，得$∠AOB=\frac {360°}6 = 60°$
∵$OA = OB，$∴$\triangle AOB$是等边三角形
又∵$OA = 1，$∴$OB = AB=OA = 1$
∴$S_{\triangle AOB}=\frac {\sqrt 3}4×1^2=\frac {\sqrt 3}4$
∴$S_{6}=6S_{\triangle AOB}=6×\frac {\sqrt 3}4=\frac {3\sqrt 3}2$
$ (2)$由$(1)，$得$S_{\triangle AOB}=\frac {\sqrt 3}4，$$AB = 1，$$S_{6}=\frac {3\sqrt 3}2$
∵$S_{\triangle AOB}=\frac 12\ \mathrm {A}B·OC$
∴$OC = \frac {2S_{\triangle AOB}}{AB}=\frac {2×\frac {\sqrt 3}4}1=\frac {\sqrt 3}2$
又∵$OD = OA = 1，$∴$CD=OD - OC=1-\frac {\sqrt 3}2$
$ $则$S_{\triangle ABD}=\frac 12\ \mathrm {A}B·CD=\frac 12×1×(1 - \frac {\sqrt 3}2)=\frac 12-\frac {\sqrt 3}4$
∴$S_{12}=S_{6}+6S_{\triangle ABD}=\frac {3\sqrt 3}2+6×(\frac 12-\frac {\sqrt 3}4)=3$
$ (3)$由$ (1) (2)，$得$S_{6}=\frac {3\sqrt 3}2，$$S_{12}=3，$$CD = 1-\frac {\sqrt 3}2，$$AB = 1，$$n = 6$
∴$S_{6}'=6\ \mathrm {A}B·CD+S_{6}=6×1×(1 - \frac {\sqrt 3}2)+\frac {3\sqrt 3}2=6-\frac {3\sqrt 3}2≈3.4$
又∵$S=πr^2=\pi ，$且$S_{12}＜S＜S_{6}'，$∴$3＜\pi ＜3.4$
