Question 1153294
<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 is {{{a}}} 

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

So, {{{a/2r = 1/sqrt(3)}}}
Therefore, {{{r = (a*sqrt(3))/2}}}

Area of circle is, 

{{{A =pi*r^2}}}

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

if given that the area of a circle is {{{89.42cm^2}}}, we have

{{{89.42cm^2=(3*pi/4)a^2}}}

{{{89.42cm^2 =2.356a^2}}}

{{{a^2=89.42cm^2/2.356}}}

{{{a^2=37.95cm^2}}}

{{{a=6.16cm}}}

so, 
D. 6.12 cm -> should be {{{6.16cm}}}