Step 1: Spot the equilateral triangle hiding inside the segment.
The two straight sides bounding the segment are both radii of length $r$, and the angle between them is $60^\circ$. A triangle with two equal sides and an included angle of $60^\circ$ must actually have all three angles equal to $60^\circ$ (since the two base angles are equal to each other, and all three angles add to $180^\circ$, each one works out to $60^\circ$). So this triangle is equilateral, with every side equal to $r$.
Step 2: Find the sector area using the radian formula rather than the degree-fraction formula.
First convert the central angle to radians:
\[ 60^\circ = \frac{\pi}{3} \text{ radians} \]
The area of a sector in terms of the angle in radians is:
\[ A_{\text{sector}} = \frac{1}{2} r^2 \theta \]
Substituting $\theta = \frac{\pi}{3}$:
\[ A_{\text{sector}} = \frac{1}{2} r^2 \times \frac{\pi}{3} = \frac{\pi r^2}{6} \]
Step 3: Find the area of the equilateral triangle using the standard equilateral-triangle formula.
For an equilateral triangle of side $r$:
\[ A_{\text{triangle}} = \frac{\sqrt{3}}{4} r^2 \]
Step 4: Subtract to get the segment area.
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 \]
Step 5: Final Answer.
The area of the segment is $\frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2$, so option (C) is correct.
\[ \boxed{\dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4} r^2} \]