Question 1199182
<br>
Also using Edwin's figure....<br>
{{{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)))}}}<br>
OD is the radius, and OF is half the radius; that makes each of the four triangles in the figure 30-60-90 right triangles.  That means all six central angles are 60 degrees, so<br>
(1) The two circular sectors AOD and BOC are each one-sixth of the circle; together their areas are one-third the area of the circle, which is {{{(4/3)pi}}}; and
(2) each of the two triangular regions AOB and COD is a triangle with base {{{2sqrt(3)}}} and height 1; together their areas are {{{2sqrt(3)}}}.<br>
So the area of region ABCD is<br>
ANSWER: {{{(4/3)pi+2sqrt(3)}}}<br>