5. The maximum value of the orthogonal projection area of a regular tetrahedron with edge length 1 on a horizontal plane is $\qquad$
A regular tetrahedron has 4 vertices, and the distance between any two points is 1. Therefore, when two opposite edges of the regular tetrahedron are parallel to the horizontal plane, the projection area is maximized, and the maximum area is $\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}$.