Question 1199182
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I will borrow the plot from the post by Edwin.


    {{{drawing(400,400,-5,5,-5,5,

circle(0,0,4), line(-2,3.4641016,-2,-3.4641016), line(2,3.4641016,2,-3.4641016),
line(-5,0,5,0), circle(0,0,.2),
line(-3,-.5,-3,.5), line(-1,-.5,-1,.5),line(1,-.5,1,.5), line(3,-.5,3,.5),
locate(-2.2,0,E), locate(1.7,0,F), locate(-2.1,3.9,A),locate(-2.1,-3.5,B)

locate(-2.1+4,3.9,D),locate(-2.1+4,-3.5,C), locate(-.1,.6,O),

red(line(-2,-3.4641016,2,3.4641016),line(2,-3.4641016,-2,3.4641016)))}}}



The area of the figure  ABCD  is the sum of 


      the area of triangle AOB   + the area of triangle DOC + 
    + the area of the sector AOD + the area of the sector BOC.


The area of triangle AOB is the same as the area of equilateral triangle with the side of 2 units,
so it is {{{2^2*(sqrt(3)/4)}}} = {{{sqrt(3)}}} square units.


The area of triangle AOB + the area of triangle DOC = {{{2*sqrt(3)}}} square units.


Next, the area of the sector AOD is  {{{1/6}}}  of the area of the circle with the radius of 2;
so, the area of the sector AOD is  {{{(1/6)*pi*2^2}}} = {{{(4/6)*pi}}} = {{{(2/3)*pi}}} square units.


The area of the sector AOD + the area of the sector BOC is twice that value, i.e.  {{{(4/3)*pi}}}  square units.


So, the  {{{highlight(highlight(ANSWER))}}}  is  {{{2*sqrt(3) + (4/3)*pi}}}  square units = 3.4641 + 4.1867 = 7.651 square units approximately.
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Solved.