smc

Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is $$2d+d=3d$$ and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we create two different right triangles: - One with base $\frac{38}{2}=19$, height $d$, and hypotenuse $r$ ($△RAO$ on the diagram) - Another with base $\frac{34}{2}=17$, height $3d$, and hypotenuse $r$ ($△LBO$ on the diagram) By the Pythagorean theorem, we can create the following system of equations: $$19^2+d^2=r^2$$ $$17^2+(2d+d{)}^2=r^2$$ Solving, we find $d=3$, so $2d=\boxed{6}$.