Prisms, parallelepipeds and pyramids

Polyhedra are solids bounded by flat polygons (faces): a brick, a box, crystals, pyramids. The elements of a polyhedron are faces (polygons), edges (sides of the faces) and vertices. The key to working with them is to find the right right-angled triangle inside the solid: the height, the slant height (apothem), the edges and elements of the base always form such triangles. This lesson covers prisms, parallelepipeds, pyramids, the regular polyhedra, Euler’s formula and surface areas; volumes are treated in detail in “Volumes”.

§1. The prism

A prism is a polyhedron with two equal parallel bases and lateral faces that are parallelograms. Each side of the base corresponds to one lateral face.

• In a right prism the lateral edges are perpendicular to the base, the lateral faces are rectangles and the lateral edge equals the height.
• A regular prism is a right prism whose base is a regular polygon.
• In an oblique prism the lateral edge is longer than the height.

The lateral area formula is clear from the net: laid side by side, the lateral faces of a right prism form a rectangle whose length is the base perimeter and whose width is the height.

Worked example

A regular triangular prism has base side 4 and height 5. Find its total surface area.

1. Base perimeter 12: S_lat = 12 · 5 = 60.
2. Base area: 4²√3/4 = 4√3.
3. Total area: 60 + 2 · 4√3 = 60 + 8√3.

Worked example

A regular hexagonal prism has base side 2 and height 5. Find its total surface area.

1. Base perimeter 12: S_lat = 12 · 5 = 60.
2. A regular hexagon consists of six equilateral triangles: S_base = 6 · 2²√3/4 = 6√3.
3. S = 60 + 2 · 6√3 = 60 + 12√3.

§2. The parallelepiped and the cube

A parallelepiped is a prism with a parallelogram base; all six faces are parallelograms, and opposite faces are equal and parallel. A cuboid (rectangular parallelepiped) has rectangular faces and dimensions a, b, c:

The diagonal formula is Pythagoras used twice: the base diagonal √(a² + b²), then the space diagonal from it and c. All four diagonals of a parallelepiped meet at one point, which bisects each of them.

Worked example

Find the diagonal and total area of a 2 × 3 × 6 cuboid.

1. d = √(4 + 9 + 36) = √49 = 7.
2. S = 2(2·3 + 3·6 + 2·6) = 2(6 + 18 + 12) = 72.

§3. The pyramid

A pyramid is a polyhedron whose base is a polygon and whose lateral faces are triangles with a common apex. In a regular pyramid the base is a regular polygon and the foot of the height is the centre of the base; hence all lateral edges are equal and all lateral faces are congruent isosceles triangles.

• The apothem (slant height) l is the height of a lateral face of a regular pyramid, drawn from the apex.
• Height h, apothem l and the inradius r of the base: l² = h² + r².
• Height h, lateral edge b and the circumradius R of the base: b² = h² + R².

Why r and R? The foot of the apothem is the midpoint of a base side, at distance r from the centre; the foot of a lateral edge is a base vertex, at distance R.

Worked example

A regular square pyramid has base side 6 and height 4. Find its total surface area.

1. The inradius of the square is r = 3.
2. Apothem: l = √(16 + 9) = 5.
3. S_lat = ½ · 24 · 5 = 60.
4. S = 60 + 36 = 96.

Worked example

A regular square pyramid has base side 12 and lateral edge 10. Find its height and apothem.

1. Half the diagonal of the square: R = 12√2/2 = 6√2.
2. h = √(100 − 72) = √28 = 2√7.
3. Apothem from a lateral face: l = √(10² − 6²) = 8.

§4. Regular polyhedra

A regular polyhedron is a convex polyhedron whose faces are all congruent regular polygons with the same number of edges meeting at every vertex. There are exactly five:

• tetrahedron — 4 triangular faces (4 vertices, 6 edges);
• cube (hexahedron) — 6 squares (8 vertices, 12 edges);
• octahedron — 8 triangles (6 vertices, 12 edges);
• dodecahedron — 12 pentagons (20 vertices, 30 edges);
• icosahedron — 20 triangles (12 vertices, 30 edges).

Why only five? At each vertex of a convex polyhedron the face angles must add up to less than 360°. Triangles (60°) can meet 3, 4 or 5 at a vertex (180°, 240°, 300°), squares (90°) only 3, pentagons (108°) only 3. For hexagons 3 · 120° = 360° — that is already flat. In total 3 + 1 + 1 = 5 possibilities.
The height of a regular tetrahedron with edge a is h = a√(2/3): the distance from the centre of the base to a base vertex is R = a/√3, and h² = a² − a²/3 = 2a²/3.

§5. Euler’s formula and sections

For a cube: 8 − 12 + 6 = 2; for an octahedron: 6 − 12 + 8 = 2. The formula helps find an unknown number of elements.
Rules for constructing sections: a plane cuts parallel faces along parallel segments; the sides of the section polygon lie in the faces.

Worked example

The cube ABCDA₁B₁C₁D₁ with edge 2 is cut by the plane through A, C, C₁. Find the area of the section.

1. The plane contains the diagonal AC and the edge CC₁ ∥ AA₁, so A₁ lies in it too — the section is ACC₁A₁.
2. AA₁ ⊥ AC, so the section is a rectangle with sides AC = 2√2 and AA₁ = 2.
3. Area: 2√2 · 2 = 4√2.

Check yourself

1. The total surface area of a cube with edge 3.

2. The diagonal of a cube with edge 2.

3. How many edges does a hexagonal prism have?

4. How many edges does a hexagonal pyramid have?

5. A regular square pyramid has base side 8 and apothem 5. Its height?

6. The diagonal of a 3 × 4 × 12 cuboid.

7. A polyhedron has 12 vertices and 30 edges. How many faces?

8. The height of a regular tetrahedron with edge 6.

Common mistakes

• Confusing the apothem with a lateral edge.
• Using R in the apothem triangle and r in the lateral-edge triangle — it is the other way round.
• Adding the bases to the lateral area, or forgetting them in the total area.
• Using only two dimensions for the diagonal of a cuboid.
• Mixing up the numbers of vertices, edges and faces of prisms and pyramids.
• Taking the lateral edge of an oblique prism as its height.