Question 782788
{{{drawing(300,300,-20,20,-15,25,
circle(0,0,10),triangle(0,20,17.3,-10,-17.3,-10),
green(line(0,0,0,-10)),green(line(0,0,17.3,-10)),
green(rectangle(0,-10,1,-9)),locate(0,22,B),
locate(0,-4,green(10cm)),
locate(17.3,-10,A),locate(-17.3,-10,C),
locate(0,2,O),locate(0,-10,D)
)}}} Side AC is tangent to the circle, perpendicular to radius OD at the point of tangency (point D).
We can draw {{{6}}} angles congruent with AOD connecting the center with points of tangency and vertices, and their measures add to {{{360^o}}},
so AOD measures {{{360^o/6=60^o}}}
Using trigonometry, we know that
{{{DA/OD=tan(60^o)}}} --> {{{DA/10cm=sqrt(3)}}} --> {{{DA=(10cm)*sqrt(3)}}}
and since {{{CD = DA}}},
{{{CA = CD+DA= 2DA =2*(10cm)*sqrt(3)=20*sqrt(3)}}}{{{cm}}}
Rounding, we get an approximmate value of {{{highlight(34.64cm)}}}.