Question 1066618
The hexagon is made up of
1 central equilateral triangle of side length 2,
3 squares of side length 2, and
3 isosceles triangles sharing sides of length 2 with the squares.
The surface area of each square is {{{2^2=4}}} .
The surface area of any triangle ABC can be calculated as
{{{(1/2)*AB*BC*sin(B)}}} .
For the central, equilateral triangle ABC,
all sides have equal length, and all angles have the same measure:
{{{AB=BC=AC=2}}} and {{{A=B=C=60^o}}} .,
so {{{area(ABC)=(1/2)*2*2*sin(60^o)=(1/2)*2*2*(sqrt(3)/2)=sqrt(3)}}} .
The angle between the squares measures
{{{120^o=360^o-(90^o+60^o+90^o)}}} ,
because it completes {{{360^o}}} when added to
the right angles of two squares,
plus a {{{60^o}}} angle of the central equilateral triangle.
So each of the three outside isosceles triangles have
two sides of length 2 flanking an angle measuring {{{120^o}}} ,
so the area of each of those 3 isosceles triangles is
{{{(1/2)*2*2*sin(120^o)=(1/2)*2*2*(sqrt(3)/2)=sqrt(3)}}} .
Since the area of the hexagon is the sum of the areas of 
{{{3}}} squares, each with {{{area(square)=4}}} ,
and {{{4}}} triangles, each with {{{area(triangle)=sqrt(3)}}} ,
{{{area(hexagon)=3*4+4*sqrt(3)=highlight(12+4sqrt(3))}}} .