Cut Tetrahedron

The distance between any two midpoints of an equilateral triangle of side length $1$ is $\dfrac{1}{2}$ , so the answer is the square of that.

The graphic below is the orthographic view of a regular tetrahedron with the vertices located as indicated. The blue line $BC$ here is at "the other side." $\;$ The midpoints of $AB$ , $AC$ , and $CD$ are shown as indicated, connected with red lines that form the four-sided cross-section. For reasons of symmetry, the four-sided cross-section is a perfect square, of sides equal to $\dfrac{1}{2}$ of the sides of the tetrahedron, which is $1$ as given.