Question 196955
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We draw the hexagon:
{{{drawing(300,300,-5,5,-3,3,line(2,2,4,0),line(4,0,2,-2),line(2,-2,-2,-2),line(-2,-2,-4,0),line(-4,0,-2,2),line(-2,2,2,2))}}} --->{{{drawing(300,300,-5,5,-3,3,line(2,2,4,0),line(4,0,2,-2),line(2,-2,-2,-2),line(-2,-2,-4,0),line(-4,0,-2,2),line(-2,2,2,2),line(-2,2,2,-2),line(2,2,-2,-2),line(-4,0,4,0),green(locate(-1,.5,60^o)),green(locate(0,.8,60^o)),green(locate(.8,.6,60^o)),green(locate(.7,-.1,60^o)),green(locate(-.5,-.4,60^o)),green(locate(-1,-.1,60^o)),red(line(-5,0,5,0)),red(line(0,3,0,-3)))}}} 
The interior angles in the center measure 60 deg. => {{{360^o/6=60^o}}}


Being Equilateral, all interior angles are 60 deg.


Let us isolate 1 Triangle:
{{{drawing(300,300,-5,5,-3,3,line(-2,2,-4,0),line(-4,0,0,0),line(0,0,-2,2),red(line(-5,0,5,0)),red(line(0,3,0,-3)),red(line(-2,2,-2,0)),red(locate(-2.3,1.4,h)),red(locate(-3,.3,b)),green(locate(-3.5,1.5,2)),green(locate(-.5,1.5,2)),green(locate(-2,-.3,2)))}}}


As you can see, we make all sides equal to <font color=blue>2 units.</font>


Woking Eqn, {{{A[T]=(1/2)(b)(h)}}},where{{{system(b=(1/2)(2)=red(1))}}}


Solving "h" via Pyth Theorem:
{{{2^2=b^2+h^2}}}
{{{h^2=2^2-b^2=2^2-1^1=4-1}}}
{{{red(h=sqrt(3))}}}


Going Back Working Eqn:
{{{A[T]=(1/2)(1)(sqrt(3))}}}
{{{A[T]=(1/2)(sqrt(3))}}}, sq.units

And there are <font color=blue>2 Right Triangles </font>for 1 Equilateral Triangle. Therefore,
{{{2(A[T])=cross(2)*(1/cross(2))(sqrt(3))}}}
{{{red(sqrt(3))}}} ----> There are <font color=blue>6 Equilateral Triangles</font>:


{{{highlight(red((6)(sqrt(3))))}}} sq.units (Answer)



We can also do it by Trigo function since we know the interior angles.
{{{drawing(300,300,-5,5,-3,3,line(-2,2,-4,0),line(-4,0,0,0),line(0,0,-2,2),red(line(-5,0,5,0)),red(line(0,3,0,-3)),red(line(-2,2,-2,0)),red(locate(-2.3,1.4,h)),green(locate(-3.5,1.5,2)),green(locate(-.5,1.5,2)),green(locate(-2,-.3,2)),green(locate(-3.6,.5,60^o)),red(locate(-2.8,.3,b)),green(locate(-.8,.5,60^o)))}}} ---> {{{sin60^o=opp/hyp=h/2}}}
{{{h=(sin60^o)(2)}}}
{{{red(h=sqrt(3))}}}
Or, {{{cos60^o=adj/hyp=b/2}}}
{{{b=(cos60^o)(2)=0.50(2)}}}
{{{red(b=1)}}}


Just do the same we did for Area of Triangle above.


Thank you,
Jojo</font>