Question 1153295
<a href="https://www.imageupload.net/image/Zx8k1"><img src="https://imagehost.imageupload.net/2020/03/02/Untitled-143-300x199.th.png" alt="Untitled-143-300x199.png" border="0" /></a>


From the figure, it is clear that, we can divide the regular hexagon into{{{ 6 }}}identical equilateral triangles.

We take one triangle{{{ OAB}}}, with {{{O}}} as the center of the hexagon or circle, & {{{AB}}} as one side of the hexagon.

Let {{{M}}} be mid-point of {{{AB}}}, {{{OM}}} would be the perpendicular bisector of {{{AB}}}, angle {{{AOM = 30}}}° 

Then in right angled triangle {{{OAM}}},side {{{a}}} (in your case is {{{h}}})

{{{tan(x) = tan(30) = 1/sqrt(3)}}}

So, 

{{{h/2r = 1/sqrt(3)}}}

Therefore, 

{{{r = (h*sqrt(3))/2}}}

Area of circle is:
 
{{{A =pi*r^2}}}

{{{A =pi*((h*sqrt(3))/2)^2}}}

{{{A =(3*pi/4)h^2}}}

{{{A =2.356.h^2}}}


answer: A.{{{2.356h^2}}}