Strange solid

The final 3D object is actually half of a regular tetrahedron (with edge length two) cut parallel to, and half way between, two opposite edges.

The volume of a regular tetrahedron is given by:

$V_{tetrahedron} = \frac{a^3}{6\sqrt{2}} \approx 0.94$

where $a$ is the edge length, which is $2$ in this case.

So, the volume of our solid will be:

$V \approx \frac{0.94}{2} = \boxed{0.47}$

The shape can be seen as a square pyramid with two tetrahedra attached. Its volume is therefore half the volume of a regular octahedron of side $1$ plus twice the volume of a regular tetrahedron of side $1$ , namely $\frac{\sqrt{2}}{3}$ .

Top of the object is a square ABCD, side 1. The ends are equilateral triangles ADE and BCF, also side 1. The trapezoids are ABFE and DCFE.

The solid can be sliced by two planes, both perpendicular to line EF, one a plane of triangle ADG, the other triangle BCH. The operation results in three solids: a triangular prism in the center and two triangular pyramids on the ends. Triangles ADG and BCH are congruent and one or the other of them functions as the base of every one of these solids.

Value of BH can be calculated from equilateral triangle BFK with sides 1. $BH=\frac{\sqrt{3}}{2}$ . Also $CH=BH=\frac{\sqrt{3}}{2}$ from symmetry or by equivalent calculation.

So the triangle BCH has sides $1, \frac{\sqrt{3}}{2}$ , and $\frac{\sqrt{3}}{2}$ . Height HL of the triangle is from Pythagorean theorem $\frac{1}{\sqrt{2}}$ . This gives us the area of the base as $A=\frac{1}{2\sqrt{2}}$ .

Volume of the prism is $V_p=1\times A=\frac{1}{2\sqrt{2}}$ .

Volume of the end piece is $V_e=\frac{1}{3}\times\frac{1}{2}\times A=\frac{1}{6\sqrt{2}}$ .

Putting it all together $V=V_p+2V_e=\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{2}}=\frac{\sqrt{2}}{3}$ .