Question 1199182
<pre>

{{{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)



)}}}  

Draw the two diameters AC and BD (in red) 

{{{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)))}}}

OA is a radius so OA = 2. And by the marks along the horizontal diameter,
OE is half a radius or half of 2 which is 1.  So OE = 1

So triangle AOE is a 30-60-90 right triangle, so AE={{{sqrt(3)}}}

Triangle AOE has area = {{{expr(1/2)*base*height=expr(1/2)(1)sqrt(3)}}}{{{""=""}}}{{{sqrt(3)/2)}}}

The four triangles AOE, BOE, COE, DOE are congruent. 
So the four of them have area {{{4(expr(sqrt(3)/2))=2sqrt(3)}}} 

Next we find the area of the sector AOD.
   
Angle AOD is 60<sup>o</sup> because angles AOE and DOE are both 60<sup>o</sup>,
so angle AOD = 180<sup>o</sup>-60<sup>o</sup>-60<sup>o</sup> = 60<sup>o</sup>.

The area of a sector is {{{expr(1/2)(theta/360)pi*radius^2}}} 
and substituting the values, the area of sector AOD is

{{{(1/2)(60/360)(pi)(2^2)}}}{{{""=""}}}{{{pi/3}}}

The sector BOC is congruent to the sector AOD

Adding the 4 triangles and the two sectors,

Area of ABCD = {{{(2sqrt(3)+pi/3)}}}{{{cm^2}}}

Edwin</pre>