Problem 2
Prisms. Take any polygon \(A B C D E\) (Figure 39 ), and through its vertices, draw parallel lines not lying in its plane. Then on one of the lines, take any point \(\left(A^{\prime}\right)\) and draw through it the plane parallel to the plane \(A B C D E\), and also draw a plane through each pair of adjacent parallel lines. All these planes will cut out a polyhedron \(A B C D E A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\) called a prism. The parallel planes \(A B C D E\) and \(A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\) are intersected by the lateral planes along parallel lines (\$13), and therefore the quadrilaterals \(A A^{\prime} B^{\prime} B, B B^{\prime} C^{\prime} C\), etc. are parallelograms. On the other hand, in the polygons \(A B C D E\) and \(A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\), corresponding sides are congruent (as opposite sides of parallelograms), and corresponding angles are congruent (as angles with respectively parallel and similarly directed sides). Therefore these polygons are congruent. Thus, a prism can be defined as a polyhedron two of whose faces are congruent polygons with respectively parallel sides, and all other faces are parallelograms connecting the parallel sides. The faces \(\left(A B C D E\right.\) and \(\left.A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\right)\) lying in parallel planes are called bases of the prism. The perpendicular \(O O^{\prime}\) dropped from any point of one base to the plane of the other is called an altitude of the prism. The parallelograms \(A A^{\prime} B^{\prime} C, B B^{\prime} C^{\prime} C\), etc. are called lateral faces, and their sides \(A A^{\prime}, B B^{\prime}\), etc., connecting corresponding vertices of the bases, are called lateral edges of the prism. The segment \(A^{\prime} C\) shown in Figure 39 is one of the diagonals of the prism. A prism is called right if its lateral edges are perpendicular to the bases (and oblique if they are not). Lateral faces of a right prism are rectangles, and a lateral edge can be considered as the altitude. A right prism is called regular if its bases are regular polygons. Lateral faces of a regular prism are congruent rectangles. Prisms can be triangular, quadrangular, etc. depending on what the bases are: triangles, quadrilaterals, ete.
Problem 8
Parallel cross sections of pyramids. Theorem. If a pyramid (Figure 49) is intersected by a plane parallel to the base, then: (1) lateral edges and the altitude \((S M)\) are divided by this plane into proportional parts; (2) the cross section itself is a polygon \(\left(A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\right)\) similar to the base; (3) the areas of the cross section and the base are proportional to the squares of the distances from them to the vertex.
Problem 9
Lateral surface area of prisms. Theorem. The lateral surface area of a prism is equal to the product of a lateral edge and the perimeter of a perpendicular cross section. By a perpendicular cross section (Figure 51) of a prism, we mean the polygon abcde obtained by intersecting all lateral faces of the prism by a plane perpendicular to the lateral edges. Sides of this polygon are perpendicular to the lateral edges \((\S \S 31,20)\). The lateral surface area of the prism is equal to the sum of areas of parallelograms. In each of them, a lateral edge can be considered as the base, and one of the sides of the perpendicular cross section as the altitude. Therefore the lateral surface area is equal to \(A A^{\prime} \cdot a b+\) \(B B^{\prime}+b c+C C^{\prime} \cdot c d+D D^{\prime} \cdot d e+E E^{\prime} \cdot e a=A A^{\prime} \cdot(a b+b c+c d+d e+e a)\). Corollary. The lateral surface area of a right prism is equal to the product of the perimeter of the base and the altitude, because lateral edges of such a prism are congruent to the altitude, and its base can be considered as the perpendicular cross section.
Problem 11
. Show that in a tetrahedron, or a parallelepiped, each face can be chosen for its base.
Problem 12
Compute the angle between diagonals of two adjacent faces of a cube. (Consider first the diagonals that meet, then skew ones.)
Problem 13
Prove that if every face of a polyhedron has an odd number of sides then the number of the faces is even. there is an edge such that all plane angles adjacent to it are acute. 66.* Prove that in every tetrahedron, there is a vertex all of whose plane angles are acute.
Problem 14
\({ }^{\star}\) Prove that in every polyhedron all of whose faces are triangles there is an edge such that all plane angles adjacent to it are acute.ith legs \(3 \mathrm{~cm}\) and \(4 \mathrm{~cm}\).
Problem 15
\({ }^{\star}\) Prove that in every tetrahedron, there is a vertex all of whose plane angles are acute.
Problem 16
Prove that in a tetrahedron, all faces are congruent if and only if all pairs of opposite edges are congruent.
Problem 18
Prove that a polyhedron is convex if and only if every segment with the endpoints in the interior of the polyhedron lies entirely in the interior.
