Question 1155309


area:

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

{{{a}}} is equal to distance between two consecutive vertices

given:consecutive vertices at ({{{0}}},{{{8}}}) and ({{{0}}},{{{4}}})

since x coordinate equal to zero,  distance between {{{y=8}}} and{{{y= 4}}} is
{{{a=4}}} units

{{{A=(3sqrt(3)/2)4^2}}}

{{{A=(3sqrt(3)/2)16}}}

{{{A=3sqrt(3)*8}}}

{{{A=24sqrt(3) }}}-> exact area

What are the possible coordinates of the center?


<a href="https://www.imageupload.net/image/T5qTg"><img src="https://imagehost.imageupload.net/2020/04/05/hexaon.th.png" alt="hexaon.png" border="0" /></a>

since the triangle formed by each side and the lines joining the end points of the side to the circumcentre are equilateral,{{{ r=a}}}

{{{r=4}}}

if center is at ({{{x}}},{{{y}}})

{{{y}}} coordinate of the center is midpoint  consecutive vertices at ({{{0}}},{{{8}}}) and ({{{0}}},{{{4}}})

{{{y=(8+4)/2=6}}}

calculate {{{b}}} using the right triangle with one side being {{{6-4=2}}} units and hypotenuse {{{r=4}}}

{{{b^2=4^2-2^2}}}

{{{b^2=16-4}}}

{{{b^2=12}}}

{{{b=sqrt(12)}}}

{{{b=sqrt(4*3)}}}

{{{b=2sqrt(3)}}}

so, the possible coordinates of the center are:

C({{{2sqrt(3)}}},{{{6}}}) or C({{{-2sqrt(3)}}},{{{6}}})