Question 1071415
.
<pre>
Connect the centers of the circles by straight line segments.
You will obtain an equilateral triangle with the side length of 6 units.

The area of this triangle is 

{{{S[triangle]}}} = {{{(1/2)*6*((6*sqrt(3))/2)}}} = {{{9*sqrt(3)}}}.


We need to add to it the areas of the three segments of the circle.
Each segment of the circle area is 60 degree sector area minus the area of the equilateral triangle with the side length of 6 units.


So, the area of each sector is 

{{{S[sector]}}} = {{{(1/6)*pi*6^2}}} - {{{9*sqrt(3)}}} = {{{6pi - 9*sqrt(3)}}}.


Thus the area under the question is 

S = {{{S[triangle]}}} + {{{3*S[sector]}}} = {{{9*sqrt(3) + 18pi - 27*sqrt(3)}}} = {{{18pi - 18*sqrt(3)}}} = {{{18*(pi-sqrt(3))}}} = 25.343 square units (approximately).


<U>Answer</U>.  The area under the question is  25.343 square units (approximately).
</pre>

Solved.