Intersection of two tetrahedrons

The side of the regular octahedron is the midsegment of one of the tetrahedron's triangular faces, so it is half the the length of one of the tetrahedron's side, which is $s = \frac{1}{2} \cdot 10 = 5$ .

The volume of a regular octahedron with side $s$ is $V = \frac{\sqrt{2}}{3}s^3$ , so its volume is $V = \frac{\sqrt{2}}{3} \cdot 5^3 \approx \boxed{58.9}$ .

The volume of each of the given tetrahedra is $V=\frac{10^3}{6\sqrt{2}}$ . The volume of the intersection is $\frac{V}{2}\approx\boxed{58.93}$ since we are losing the four ochre tetrahedra each of which has a volume of $\frac{V}{2^3}$ .