Angle $E$ marks off a sector of the circle. If we find the area of the sector and subtract the area of the triangle with vertex at $E$, we will have our answer.
The area of a sector can be solved using a proportion: the ratio of angle $E$ to the entire degree measure of the circle is proportional to the ratio of the area of the sector to the area of the circle. The area of the circle is given by $\(\pi r^2\)$, which means we can set up the following:

$$
\(\begin{aligned}
\frac{60}{360} & =\frac{x}{36 \pi} \\
\frac{1}{6} & =\frac{x}{36 \pi} \\
x & =6 \pi
\end{aligned}\)
$$


This is approximately 18.84 .

To find the area of the triangle, it helps to note that this is an equilateral triangle. We know that it is at least isosceles, since all radii of a circle are congruent, and since angle $E$ measures $60^{\circ}$ and the other angles are equal to one another (because their corresponding sides, both radii, are equal.) So, all angles must be $\(60^{\circ}\)$. (If the other two angles are defined as $x$, then $\(2 x+60=180\)$, and $\(x=60\)$.)
This establishes that this triangle is equilateral, which means we can divide it as follows:



The radius 6 corresponds to the hypotenuse of the right triangle. The side length ratio of a 30-60-90 triangle is $\(x: x \sqrt{3}: 2 x\)$, which in this case means $\(3: 3 \sqrt{3}: 6\)$.
The area of a triangle is $\(.5 \times$ base $\times\)$ height. The base is 6 and the height is $\(3 \sqrt{3}\)$, which makes the area of the triangle $\(9 \sqrt{3}\)$, or approximately 15.6.

Subtracting the area of the triangle from the area of the sector gives us:

$$
\(18.84-15.6=3.24\)
$$


This rounds down to 3 .