2. Let $A B C D$ be a regular tetrahedron with side length 2. The plane parallel to edges $A B$ and $C D$ and lying halfway between them cuts $A B C D$ into two pieces. Find the surface area of one of these pieces.
Solution: $1+2 \sqrt{3}$
The plane intersects each face of the tetrahedron in a midline of the face; by symmetry it follows that the intersection of the plane with the tetrahedron is a square of side length 1. The surface area of each piece is half the total surface area of the tetrahedron plus the area of the square, that is, $\frac{1}{2} \cdot 4 \cdot \frac{2^{2} \sqrt{3}}{4}+1=1+2 \sqrt{3}$.