A hexagon inscribed in a circle has three consecutive sides each of length $3$ and three consecutive sides each of length $5$. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length $3$ and the other with three sides each of length $5$, has length equal to $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

$\text{(A)}\ 309 \qquad \text{(B)}\ 349 \qquad \text{(C)}\ 369 \qquad \text{(D)}\ 389\qquad \text{(E)}\ 409$
1. **Define Variables and Angles:**
   Let the radius of the circle be \( r \). Let \( 2\alpha \) be the measure of each of the angles subtended by segments of length \( 3 \), and let \( 2\beta \) be the measure for segments of length \( 5 \). Since the hexagon is inscribed in a circle, the sum of the central angles is \( 360^\circ \). Therefore, we have:
   \[
   6\alpha + 6\beta = 360^\circ \implies \alpha + \beta = 60^\circ
   \]

2. **Express Sides in Terms of \( r \):**
   Using the sine rule in the circle, we can express the sides in terms of \( r \):
   \[
   2r \sin \alpha = 3 \quad \text{and} \quad 2r \sin \beta = 5
   \]
   Solving for \( \sin \alpha \) and \( \sin \beta \):
   \[
   \sin \alpha = \frac{3}{2r} \quad \text{and} \quad \sin \beta = \frac{5}{2r}
   \]

3. **Find the Length of the Chord:**
   The chord that divides the hexagon into two trapezoids is the chord subtending the angle \( 3\alpha \) or \( 3\beta \). Using the sine rule for the chord length:
   \[
   \text{Chord length} = 2r \sin 3\alpha = 2r \sin 3\beta
   \]
   Using the triple angle formula for sine:
   \[
   \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta
   \]
   For \( \alpha \):
   \[
   \sin 3\alpha = 3 \sin \alpha - 4 \sin^3 \alpha
   \]
   Substituting \( \sin \alpha = \frac{3}{2r} \):
   \[
   \sin 3\alpha = 3 \left(\frac{3}{2r}\right) - 4 \left(\frac{3}{2r}\right)^3 = \frac{9}{2r} - \frac{108}{8r^3} = \frac{9}{2r} - \frac{27}{2r} = \frac{9 - 27}{2r} = -\frac{18}{2r} = -\frac{9}{r}
   \]
   For \( \beta \):
   \[
   \sin 3\beta = 3 \sin \beta - 4 \sin^3 \beta
   \]
   Substituting \( \sin \beta = \frac{5}{2r} \):
   \[
   \sin 3\beta = 3 \left(\frac{5}{2r}\right) - 4 \left(\frac{5}{2r}\right)^3 = \frac{15}{2r} - \frac{500}{8r^3} = \frac{15}{2r} - \frac{125}{2r} = \frac{15 - 125}{2r} = -\frac{110}{2r} = -\frac{55}{r}
   \]

4. **Equate and Solve for \( r \):**
   Since the chord length is the same for both angles:
   \[
   2r \sin 3\alpha = 2r \sin 3\beta
   \]
   Substituting the values:
   \[
   2r \left(-\frac{9}{r}\right) = 2r \left(-\frac{55}{r}\right)
   \]
   Simplifying:
   \[
   -18 = -110
   \]
   This indicates an error in the calculation. Let's correct the approach by solving for \( \cos \beta \) using the given relationship.

5. **Correct Calculation for \( \cos \beta \):**
   Using the relationship:
   \[
   \frac{3}{5} = \frac{\sin \alpha}{\sin \beta} = \frac{\sqrt{3}}{2} \cot \beta - \frac{1}{2}
   \]
   Solving for \( \cot \beta \):
   \[
   \frac{3}{5} = \frac{\sqrt{3}}{2} \cot \beta - \frac{1}{2}
   \]
   \[
   \frac{3}{5} + \frac{1}{2} = \frac{\sqrt{3}}{2} \cot \beta
   \]
   \[
   \frac{6 + 5}{10} = \frac{\sqrt{3}}{2} \cot \beta
   \]
   \[
   \frac{11}{10} = \frac{\sqrt{3}}{2} \cot \beta
   \]
   \[
   \cot \beta = \frac{11}{10} \cdot \frac{2}{\sqrt{3}} = \frac{22}{10\sqrt{3}} = \frac{22\sqrt{3}}{30} = \frac{11\sqrt{3}}{15}
   \]

6. **Find \( \cos \beta \):**
   \[
   \cot^2 \beta = \frac{121}{75} = \frac{1}{\sin^2 \beta} - 1
   \]
   \[
   \sin^2 \beta = \frac{75}{196}
   \]
   \[
   \cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{75}{196} = \frac{121}{196}
   \]
   \[
   \cos \beta = \frac{11}{14}
   \]

7. **Final Chord Length Calculation:**
   \[
   5(4 \cos^2 \beta - 1) = 5 \left(4 \left(\frac{11}{14}\right)^2 - 1\right) = 5 \left(4 \cdot \frac{121}{196} - 1\right) = 5 \left(\frac{484}{196} - 1\right) = 5 \left(\frac{484 - 196}{196}\right) = 5 \left(\frac{288}{196}\right) = \frac{1440}{196} = \frac{360}{49}
   \]

The final answer is \( \boxed{409} \)