A solid from triangles

The resulting solid is a unit cube with two tetrahedral corners cut off. Here is an image of one of them:

Since the areas of those tetrahedral corners are each $\frac{1}{6}$ then the volume of the resulting cube will be given by:

$V = 1 - \dfrac{1}{6} - \dfrac{1}{6} = \boxed{\dfrac{2}{3}}$

Result is a triangular anti-prism with bases side $a=\sqrt{2}$ . Regular triangular anti-prism is an octahedron with volume (for edge $a=\sqrt{2}$ )

$V=\frac{\sqrt{2}}{3}a^3=\frac{4}{3}$ In top view $AC=\frac{1}{\sqrt{2}}, \angle ACB=90^\circ, \angle CAB=30^\circ$ therefore $AB=\frac{\sqrt{2}}{\sqrt{3}}$

With side edge $a=\sqrt{2}$ the height would be $h_2=\sqrt{2-AB^2}=\sqrt{2-\frac{2}{3}}=\frac{2}{\sqrt{3}}$

With side edge $a=1$ the height is instead $h_1=\sqrt{1-AB^2}=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}$

So we need to take the volume of a regular tetrahedron and multiply it by the ration of the heights.

$V=\frac{4}{3}\times \frac{h_1}{h_2}=\frac{4}{3}\times\frac{1}{2}=\frac{2}{3}$

The isosceles triangles are 45-90-45 with unit sides.