How to extract a part of a side of a regular polygon?

Is there a better way of drawing the following picture without the ab segment? I used the following code

\documentclass[tikz,border=5]{standalone}\usetikzlibrary{shapes.geometric}\begin{document}\begin{tikzpicture}[scale=5.5]    \node[regular polygon, regular polygon sides=6, minimum size=10cm, rounded corners, draw] at (0,0) {};%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Erase ab segment%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%    \draw[ultra thick, white](-.377,.787025) -- (.377,.787025);%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   \draw(.377,.787025)arc(276:90:.06);  \draw(-.377,.787025)arc(-96:90:.06);   \def\mypath{(-.06,-.98) -- (-.06,-.95) arc (180:0:.06cm) -- (.06,-.98)}    \foreach \t in {0,120,240} {\draw   [rotate=\t] \mypath;}\def\mypath{(0,.98) -- (0,.98) arc (90:55.5:.98cm)}% -- (0,0)}\draw   [rotate=56.5] \mypath;\def\mypath{(0,.98) -- (0,.98) arc (90:55.5:.98cm)}% -- (0,0)}\draw   [rotate=-22] \mypath;\def\mypath{(0,.98) -- (0,.98) arc (90:-23:.98cm)}% -- (0,0)}\draw   [rotate=176.5] \mypath;\def\mypath{(0,.98) -- (0,.98) arc (90:-23:.98cm)}% -- (0,0)}\draw   [rotate=296.5] \mypath;\end{tikzpicture}\end{document}