Circles

https://gyazo.com/0bb2e632441b774f3d412d086fa55957

I made an equilateral triangle using the centers of the triangle. 
Since an equilateral triangle has angles of 60^o, each sector part is just \(\frac{1}{6}\) of the circle.

Circumference of each circle = 36 units 

Radius = \(\frac{18}{π}\)

Sector areas = \(3 \cdot \frac{1}{6} \cdot (\frac{18}{π})^2 π\)

= \(\frac{1}{2} \cdot \frac{324}{π^2}π\)

= \(\frac{1}{2} \cdot \frac{324}{π}\)

= \(\frac{162}{π}\)

Now the area of the equilateral triangle is: 

\(\frac{\sqrt{3}}{4}\cdot (\frac{36}{π})^2\)

= \(\frac{\sqrt{3}}{4} \cdot \frac{1296}{π^2}\)

= \(\frac{1296\sqrt{3}}{4π^2}\)

Which means the area of the shaded region is: 

\(\frac{1296\sqrt{3}}{4π^2} - \frac{162}{π}\)

= \(\frac{1296\sqrt{3}}{4π^2} - \frac{162π}{π^2}\)

= \(\frac{1296\sqrt{3}}{4π^2} - \frac{648π}{4π^2}\)

= \(\frac{1296\sqrt{3} - 648π}{4π^2}\)

= \(\frac{4(324\sqrt{3} - 162π)}{4π^2}\)

= \(\frac{324\sqrt{3} - 162π}{π^2}\)

please tell me if I did anything wrong...

LOL

I got something complicated as well... I thought I got it incorrect. i actually got the same answer as you, so nothing's wrong... I think.

See image below:

Triangle area - cicle sector area * 3 = area in question        

sqrt(3)/4 ( 36/ pi)²   - 3  * 1/6 *  pi ( 18/pi)²    = shaded                (we use 1/6 because the sector of the circle is 60/360 of the circle)

324 sqrt(3) / pi²    - 162 / pi     <============   Same answer as  LR    Way to go Logarhythm!!