Number of distinct right triangles formed by connecting vertices of a unit cube.

Suppose we have a unit cube in ${\mathbb{R}}^{3}.$

We want to count the total number of distinct right triangles formed by connecting vertices of the unit cube. I can see that the total number of right triangles simply on a face of the cube will be $4$ and since there are $6$ faces we have $4\times 6=24$ right triangles on the faces alone. Of course there are others across the cube's diagonals. I think the total will be $48$, due a symmetry across the diagonals, but am not entirely sure of my reasoning here, so I may be wrong.

Suppose we have a unit cube in ${\mathbb{R}}^{3}.$

We want to count the total number of distinct right triangles formed by connecting vertices of the unit cube. I can see that the total number of right triangles simply on a face of the cube will be $4$ and since there are $6$ faces we have $4\times 6=24$ right triangles on the faces alone. Of course there are others across the cube's diagonals. I think the total will be $48$, due a symmetry across the diagonals, but am not entirely sure of my reasoning here, so I may be wrong.