Solution:

Put the vertices at angles $30k^\circ$ on a unit circle. The chord joining vertices $i$ and $j$ has direction $15(i+j)^\circ + 90^\circ,$ so chords come in $12$ directions spaced $15^\circ$ apart, and a rectangle uses two chords from each of two perpendicular directions. The six perpendicular direction pairs split into two kinds, three of each, by rotation. When $i + j$ is even, a family of parallel chords has $5$ members, at distances $0, \pm\frac{1}{2}, \pm\frac{\sqrt{3}}{2}$ from the center with half-lengths $1, \frac{\sqrt{3}}{2}, \frac{1}{2}$ respectively; when $i + j$ is odd, a family has $6$ members, at distances $\pm\cos 75^\circ, \pm\cos 45^\circ, \pm\cos 15^\circ$ with half-lengths $\sin 75^\circ, \sin 45^\circ, \sin 15^\circ.$

A corner is the intersection of one chord from each direction, and its offset along a chord equals the other chord's distance from the center. Since half-lengths shrink as distance grows, the four corners lie on all four chord segments exactly when, writing $D_1, D_2$ for the larger distances of the two chosen pairs, each $D$ is at most the half-length of the other pair's farther chord. For the $5$-chord families: pairs with $D = \frac{\sqrt{3}}{2}$ (there are $7$) have half-length bound $\frac{1}{2},$ and pairs with $D = \frac{1}{2}$ (there are $3$) have bound $\frac{\sqrt{3}}{2};$ the valid combinations give $7 \cdot 3 + 3 \cdot 7 + 3 \cdot 3 = 51$ rectangles. For the $6$-chord families: there are $1, 5, 9$ pairs with $D = \cos 75^\circ, \cos 45^\circ, \cos 15^\circ,$ and the valid combinations are $(\cos 75^\circ, \cos 75^\circ),$ both orders of $(\cos 75^\circ, \cos 45^\circ)$ and $(\cos 75^\circ, \cos 15^\circ),$ and $(\cos 45^\circ, \cos 45^\circ),$ giving $1 + 5 + 5 + 9 + 9 + 25 = 54.$

Each kind of direction pair occurs three times, so the total is $3(51 + 54) = 315.$