Question 1103080
{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
circle(0,0,0.866),green(triangle(0,0,0,-0.866,0.5,-0.866)),
green(rectangle(0,-0.866,0.1,-0.766)),triangle(-0.5,-0.866,0.5,-0.866),
triangle(-0.5,0.866,0.5,0.866),triangle(-1,0,-0.5,-0.866,0,0),
triangle(-1,0,0,0,-0.5,0.866),triangle(1,0,0,0,0.5,0.866),
locate(0.02,-0.4,R),triangle(0,0,1,0,0.5,-0.866)
)}}}
{{{R}}}= radius of the circle = height of each of the 6 triangles forming the hexagon.
{{{pi*R^2}}}= area of the circle
{{{2sqrt(3)R^2}}}= area of the hexagon
{{{pi*R^2/(2sqrt(3)R^2)=highlight(pi/(2sqrt(3))=sqrt(3)pi/6)}}} is the ratio asked for.
 
You may have a formula to calculate the area of a regular hexagon
as a function of its apothem (which in this case is R),
but I calculated it by figuring the sides and area of the triangles.
The angles of those triangles measure {{{60^o}}} .
That makes {{{sin(60^o)=sqrt(3)/2=R/side}}} for the triangles,
so {{{side=2R/sqrt(3)=2sqrt(3)R/3}}} .
Area for 1 triangle would be
{{{(1/2)(2sqrt(3)R/3)R=sqrt(3)R^2/3}}} ,
and the area of the hexagon is 6 times that
{{{6(sqrt(3)R^2/3)=2sqrt(3)R^2}}} .