Question 1068016
<b><pre><font size=4><b>
{{{drawing(400,2800/13,-4,9,-1,6,
locate(1.875,5.85,6),
line(-1,5.196152423,-2,3.464191615),locate(-1.05,5.8,A),locate(-1.8,4.7,2),
line(-2,3.464191615,0,0),locate(-2.3,3.6,B),locate(-1.21,1.8,4),
line(0,0,6,0),locate(0,0,C),locate(3,0,6),locate(6,3.95,4),
line(6,0,7,1.732050808),locate(6,0,D),locate(6.62,1.05,2),
line(7,1.732050808,5,5.196152423),locate(7.05,2.1,E),
line(5,5.196152423,-1,5.196152423), locate(4.8,5.8,F) )}}} 

It's easy to show by the formula for the sum of the angles that
each interior angle is 120°, whose supplement is 60°.

Now we draw in some green lines to enclose the hexagon in a rectangle.
We have added four 30°-60°-90° right triangles to the hexagon.

{{{drawing(400,2800/13,-4,9,-1,6,
locate(1.875,5.85,6),

green(line(-2,5.196152423,-2,0),line(-1,5.196152423,-2,5.196152423),
line(-2,0,0,0), line(5,5.196152423,7,5.196152423), line(7,5.196152423,7,0),
line(7,0,6,0)

),line(-1,5.196152423,-2,3.464191615),locate(-1.05,5.8,A),locate(-1.8,4.7,2),
line(-2,3.464191615,0,0),locate(-2.3,3.6,B),locate(-1.21,1.8,4),
line(0,0,6,0),locate(0,0,C),locate(3,0,6),locate(6,3.95,4),
line(6,0,7,1.732050808),locate(6,0,D),locate(6.62,1.05,2),
line(7,1.732050808,5,5.196152423),locate(7.05,2.1,E),
line(5,5.196152423,-1,5.196152423), locate(4.8,5.8,F) )}}} 

In a 30°-60°-90° right triangle the shorter leg
is one-half of the hypotenuse.  And the longer leg
is the shorter leg times &#8730;3.

Since we know all 4 hypotenuses, we can find the dimensions
of the enclosing rectangle. 

{{{drawing(400,2800/13,-4,9,-1,6,
locate(1.875,5.85,6), 

green(line(-2,5.196152423,-2,0),line(-1,5.196152423,-2,5.196152423),
locate(-2.8,1.8,2sqrt(3)),locate(-2.6,4.7,sqrt(3)),
locate(6,5.85,2), locate(-1.21,0,2),locate(7.1,3.95,2sqrt(3)),
line(-2,0,0,0), line(5,5.196152423,7,5.196152423), line(7,5.196152423,7,0),
line(7,0,6,0),locate(-1.8,5.85,1),locate(6.62,0,1),locate(7.07,1.05,sqrt(3))

),line(-1,5.196152423,-2,3.464191615),locate(-1.8,4.7,2),
line(-2,3.464191615,0,0),locate(-1.21,1.8,4),
line(0,0,6,0),locate(3,0,6),locate(6,3.95,4),
line(6,0,7,1.732050808),locate(6.62,1.05,2),
line(7,1.732050808,5,5.196152423),
line(5,5.196152423,-1,5.196152423) )}}} 

The area of the big rectangle is 

(length)(width)= (9)(3&#8730;3) = 27&#8730;3

The area of each of the two larger 30°-60°-90° right triangles
is  

(base)(height)/2 = (2)(2&#8730;3)/2 = 2&#8730;3

The area of both of them is 4&#8730;3

The area of each of the two smaller 30°-60°-90° right triangles
is  

(base)(height)/2 = 1(&#8730;3)/2 = &#8730;3/2

The area of both of them is &#8730;3

We get the area of the hexagon by subtracting the
areas of the 4 right triangles:

27&#8730;3 - 4&#8730;3 - &#8730;3 = 22&#8730;3

Edwin</pre></b></font>