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AdvancedGrades 9–1125 min57 / 82

Polyhedra: prisms and pyramids

The elements of a polyhedron and Euler’s theorem, the rectangular box and the cube (grade 9 exam), the lateral and total surface area and the volume of prisms and pyramids, where the foot of the height lies, sections, the frustum and similar solids — with DİM-style tasks.

Check yourself
In this lesson you will learn
  • Count the vertices, edges and faces of a polyhedron and check them with Euler’s theorem
  • Find the surface area, volume and diagonal of a rectangular box and a cube, and solve problems with ratios of its dimensions
  • Compute the lateral and total surface area and the volume of prisms and pyramids, and locate the foot of the height (circumcentre or incentre of the base)
  • Solve DİM tasks on sections, frustums and similar solids (k², k³) with full solutions

A matchbox, a brick, an aquarium, the Egyptian pyramids, salt crystals — they are all polyhedra. How much paper does it take to cover such a box, and how many litres of water does it hold? In this lesson we study prisms and pyramids: their elements, lateral and total surface area, volume, sections and the frustum. The tools of the previous lesson «Lines and planes in space» — perpendicular, oblique, projection, dihedral angle — work at every step here. A task on «Çoxüzlülər, onların səthi və həcmi» (polyhedra, their surface area and volume) appeared in 2026 both in stage II of the entrance exam (groups I and II) and in the grade 11 school-leaving exam. In the programme of the grade 9 school-leaving exam the only solid is the rectangular box with the cube — the second section of this lesson is devoted to them.

A polyhedron and its elements. Euler’s theorem

Definition
Polyhedron

A solid whose surface consists of finitely many polygons is a polyhedron. The polygons are its faces, their sides are its edges, and their vertices are the vertices of the polyhedron. A segment joining two vertices that are not on one face is a diagonal. A polyhedron is convex if it lies on one side of the plane of each of its faces; all polyhedra in school problems are convex.

V − E + F = 2
where:
  • Vthe number of vertices
  • Ethe number of edges
  • Fthe number of faces

Euler’s theorem holds for every convex polyhedron — an excellent way to check a count.

PolyhedronVerticesEdgesFaces
n-gonal prism2n3nn + 2
n-gonal pyramidn + 12nn + 1
cube8126
triangular prism695
hexagonal pyramid7127
A prism has n edges on each base and n lateral edges (3n); a pyramid has n base edges and n lateral edges (2n).
Vertices, edges, faces

1) A prism has 30 edges. How many faces and vertices does it have?
2) A pyramid has 14 edges. How many vertices does it have?
3) A convex polyhedron has 12 vertices and 30 edges. How many faces does it have?

Show solution
1) 3n = 30 ⇒ n = 10, the base is a decagon. Faces: n + 2 = 12, vertices: 2n = 20. Check: 20 − 30 + 12 = 2. ✓
2) 2n = 14 ⇒ n = 7, vertices: n + 1 = 8.
3) Euler’s theorem: 12 − 30 + F = 2 ⇒ F = 20. This is the regular icosahedron.

The rectangular box and the cube (also the grade 9 exam)

A prism whose base is a parallelogram is a parallelepiped. Its opposite faces are parallel and equal, and its four diagonals meet at one point and bisect each other there. A parallelepiped whose faces are all rectangles is a rectangular box (cuboid); the lengths a, b, c of the three edges from one vertex are its dimensions (length, width, height). A rectangular box with all edges equal is a cube.

S = 2(ab + bc + ca), V = a · b · c
where:
  • a, b, cthe dimensions of the box
  • Sthe total surface area (all six faces)
  • Vthe volume

The six faces form three equal pairs: ab, bc and ca. The volume is the base area times the height.

d² = a² + b² + c²
where:
  • dthe diagonal of the box

The “Pythagorean theorem in space”: first the base diagonal AC² = a² + b², then in the right triangle ACC₁: AC₁² = AC² + c².

ABCDA₁B₁C₁D₁abcd
d² = AC² + c² = a² + b² + c².
S = 6a², V = a³, d = a√3
where:
  • athe edge of the cube

For a cube a = b = c. A face diagonal is a√2, the space diagonal is a√3.

Surface area, volume and diagonal

1) A rectangular box measures 3 cm, 4 cm and 12 cm. Find its total surface area, volume and diagonal.
2) The diagonal of a cube is 6√3 cm. Find its surface area and volume.

Show solution
1) S = 2(3 · 4 + 4 · 12 + 12 · 3) = 2(12 + 48 + 36) = 192 cm²; V = 3 · 4 · 12 = 144 cm³; d = √(9 + 16 + 144) = √169 = 13 cm.
2) a√3 = 6√3 ⇒ a = 6 cm; S = 6 · 36 = 216 cm², V = 6³ = 216 cm³.
DİM style: coded task

The dimensions of a rectangular box are in the ratio 2 : 3 : 6, and its volume is 288 cm³. Find the diagonal of the box (in cm).

Show solution
Let the dimensions be 2k, 3k, 6k. V = 2k · 3k · 6k = 36k³ = 288 ⇒ k³ = 8 ⇒ k = 2.
Dimensions: 4 cm, 6 cm, 12 cm.
d = √(16 + 36 + 144) = √196 = 14.
Answer: 14. A curious fact: the surface area of this box is also 2(24 + 72 + 48) = 288 cm².
Volume from the face areas, and litres

1) Three faces of a rectangular box have areas 6 cm², 10 cm² and 15 cm². Find its volume and dimensions.
2) How many litres of water does an aquarium of 50 cm × 30 cm × 40 cm hold?
3) The edge of a cube is doubled. How many times do its surface area and volume increase?

Show solution
1) ab = 6, bc = 15, ca = 10. Multiply all three: (abc)² = 6 · 15 · 10 = 900 ⇒ V = abc = 30 cm³. Then a = 30 : 15 = 2, b = 30 : 10 = 3, c = 30 : 6 = 5 (cm).
2) V = 50 · 30 · 40 = 60 000 cm³ = 60 dm³ = 60 L (1 L = 1 dm³ = 1000 cm³).
3) The surface area is proportional to a²: 2² = 4 times; the volume to a³: 2³ = 8 times.

The prism: lateral and total surface area, volume and sections

Definition
Prism

A polyhedron with two equal n-gons (the bases) in parallel planes and n parallelograms as the other faces (the lateral faces) is a prism. A prism whose lateral edges are perpendicular to the base is a right prism; otherwise it is an oblique prism. A right prism whose base is a regular polygon is a regular prism. The height of a prism is the distance between the bases: in a right prism it equals the lateral edge, in an oblique prism h = l · sin φ (l is the lateral edge, φ the angle between a lateral edge and the base).

Slat = Pbase · h, Stot = Slat + 2S_base, V = Sbase · h
where:
  • Pbasethe perimeter of the base
  • Sbasethe area of the base
  • hthe height

The formula Slat = Pbase · h is for a right prism, whose lateral faces are rectangles of height h. V = Sbase · h holds for every prism.

Regular prisms

1) A regular triangular prism has base side 6 cm and height 10 cm. Find its lateral surface area, total surface area and volume.
2) A regular hexagonal prism has base side 2 cm and height 5 cm. Find its lateral surface area and volume.

Show solution
1) Sbase = (√3/4) · 6² = 9√3 cm²; Slat = Pbase · h = 18 · 10 = 180 cm²; Stot = 180 + 2 · 9√3 = 180 + 18√3 cm²; V = 9√3 · 10 = 90√3 cm³.
2) A regular hexagon splits into 6 equilateral triangles of side 2: Sbase = 6 · (√3/4) · 4 = 6√3 cm². Slat = 12 · 5 = 60 cm², V = 6√3 · 5 = 30√3 cm³.
DİM style: closed task

The base of a right prism is a rhombus with diagonals 10 cm and 24 cm. The lateral surface area of the prism is 520 cm². Find the height of the prism.
A) 8 cm B) 10 cm C) 12 cm D) 13 cm E) 5 cm

Show solution
The diagonals of a rhombus are perpendicular and bisect each other: side a = √(5² + 12²) = 13 cm, Pbase = 4 · 13 = 52 cm.
Slat = Pbase · h ⇒ 520 = 52 · h ⇒ h = 10 cm.
Answer: B. Option D is the side of the rhombus; taking whole diagonals, side √(10² + 24²) = 26, gives h = 5 (E). Bonus: V = ½ · 10 · 24 · 10 = 1200 cm³.

Plane sections of a prism. A section parallel to the base is a polygon equal to the base. The section of a right prism through two non-adjacent lateral edges (a diagonal section) is a rectangle with sides equal to a base diagonal and the height. To find the shape of a section, find the segment along which the cutting plane meets each face — it meets parallel faces along parallel segments.

An oblique prism and sections of a cube

1) The lateral edge of an oblique prism is 10 cm and makes an angle of 30° with the plane of the base; the base area is 12 cm². Find the volume of the prism.
2) The cube ABCDA₁B₁C₁D₁ has edge 4 cm. Find the area of a) the diagonal section AA₁C₁C; b) the section through the points A, B₁ and D₁.

Show solution
1) The height is not the lateral edge: h = 10 · sin 30° = 5 cm. V = Sbase · h = 12 · 5 = 60 cm³.
2a) AA₁C₁C is a rectangle: AC = 4√2, AA₁ = 4 ⇒ S = 16√2 cm².
2b) AB₁, B₁D₁ and AD₁ are face diagonals of the cube, each 4√2. The section is an equilateral triangle: S = (√3/4) · (4√2)² = (√3/4) · 32 = 8√3 cm².

The pyramid: foot of the height, apothem, surface area and volume

Definition
Pyramid

A polyhedron with one face (the base) an n-gon and the other n faces (the lateral faces) triangles with a common vertex S is a pyramid. The perpendicular SO from the apex to the plane of the base is the height. A pyramid whose base is a regular polygon and whose height falls at the centre of that polygon is a regular pyramid: its lateral edges are equal and its lateral faces are equal isosceles triangles. The height of a lateral face of a regular pyramid drawn from the apex is the apothem (m).

SABCDOMhmrφ
The right triangle SOM: m² = h² + r², h = r · tan φ. φ is the linear angle of the dihedral angle between a lateral face and the base.
Slat = ½ · Pbase · m, V = ⅓ · Sbase · h
where:
  • mthe apothem
  • hthe height of the pyramid
  • Sbasethe area of the base

The lateral-area formula is for a regular pyramid (all lateral faces have height m). The volume formula holds for every pyramid: the volume of a pyramid is one third of the volume of a prism with the same base and height. For example, a cube splits into three equal quadrilateral pyramids with a common apex, each with volume a³/3.

m² = h² + r², l² = h² + R²
where:
  • r, Rthe radii of the inscribed and circumscribed circles of the base (OM and OA)
  • lthe lateral edge

Two right triangles in a regular pyramid: SOM (with the apothem) and SOA (with the lateral edge). For a square r = a/2, R = a√2/2; for an equilateral triangle r = a√3/6, R = a√3/3.

Regular pyramids

1) A regular quadrilateral pyramid has base side 6 cm and height 4 cm. Find its apothem, lateral edge, lateral and total surface area, and volume.
2) A regular triangular pyramid has base side 6√3 cm and lateral edge 10 cm. Find its volume.

Show solution
1) r = 6/2 = 3, m = √(4² + 3²) = 5 cm; R = 3√2, l = √(16 + 18) = √34 cm. Slat = ½ · 24 · 5 = 60 cm², Stot = 60 + 36 = 96 cm², V = ⅓ · 36 · 4 = 48 cm³.
2) R = a√3/3 = 6√3 · √3/3 = 6 cm, h = √(10² − 6²) = 8 cm. Sbase = (√3/4) · (6√3)² = 27√3 cm². V = ⅓ · 27√3 · 8 = 72√3 cm³.

When a pyramid is not regular, the key question is where the foot of the height lies. Two cases keep coming back in DİM tasks:

  • If all lateral edges are equal, or all make equal angles with the base, the foot of the height is the circumcentre of the base (equal obliques have equal projections: OA = OB = OC = R). In a right triangle it is the midpoint of the hypotenuse, in a rectangle the intersection of the diagonals.
  • If all lateral faces make equal dihedral angles with the base (or the heights of the lateral faces from the apex are equal), the foot of the height is the incentre of the base. Then h = r · tan φ and the lateral area is Slat = Sbase / cos φ.
  • If one lateral edge is perpendicular to the base, that edge is the height; any face of a triangular pyramid can be taken as the base.
h = r · tan φ, Slat = Sbase / cos φh = r · tan φ, Slat = Sbase / cos φ
where:
  • φthe dihedral angle between each lateral face and the base (all equal)
  • rthe radius of the circle inscribed in the base

The second formula is S′ = S · cos φ from the lesson «Lines and planes in space»: the base is the projection of the lateral surface.

DİM style: closed task

The base of a pyramid is a rectangle with sides 6 cm and 8 cm, and all lateral edges make an angle of 45° with the plane of the base. Find the volume of the pyramid.
A) 240 cm³ B) 80 cm³ C) 160 cm³ D) 80√2 cm³ E) 48 cm³

Show solution
The lateral edges make equal angles with the base ⇒ the foot of the height is the circumcentre of the rectangle, i.e. the intersection O of the diagonals.
Diagonal √(36 + 64) = 10 cm, R = OA = 5 cm.
In triangle SOA, ∠A = 45° ⇒ h = SO = OA · tan 45° = 5 cm.
V = ⅓ · 6 · 8 · 5 = 80 cm³.
Answer: B. A forgets the ⅓; C takes the diagonal (10) as the height; D takes the lateral edge (5√2) as the height.
The foot of the height: circumcentre and incentre

The base of a pyramid is a right triangle with legs 6 cm and 8 cm.
1) All lateral edges are 13 cm. Find the volume of the pyramid.
2) All lateral faces make dihedral angles of 45° with the base. Find the volume and the lateral surface area.
3) In another pyramid the three lateral edges at the apex are mutually perpendicular, with lengths 3 cm, 4 cm and 5 cm. Find the volume.

Show solution
Sbase = ½ · 6 · 8 = 24 cm², the hypotenuse is 10 cm.
1) Equal lateral edges ⇒ O is the midpoint of the hypotenuse, R = 5. h = √(13² − 5²) = 12 cm. V = ⅓ · 24 · 12 = 96 cm³.
2) Equal dihedral angles ⇒ O is the incentre. In a right triangle r = (a + b − c)/2 = (6 + 8 − 10)/2 = 2 cm. h = r · tan 45° = 2 cm, V = ⅓ · 24 · 2 = 16 cm³. Slat = Sbase / cos 45° = 24 : (√2/2) = 24√2 cm².
3) Take two perpendicular edges (3 and 4) as the legs of the base: the third edge (5) is perpendicular to their plane and is the height. V = ⅓ · (½ · 3 · 4) · 5 = 10 cm³.

Parallel sections, the frustum and similar solids

Cut a pyramid with a plane parallel to the base at distance h₁ from the apex. The section is a polygon similar to the base with ratio k = h₁ / h, and the small pyramid cut off is similar to the whole one. The part below the section is a frustum: its bases (areas S₁ and S₂) are similar polygons in parallel planes, its lateral faces are trapezoids, and its height is the distance between the bases. The height of a lateral face of a regular frustum is its apothem.

S₁ / S₂ = k², V₁ / V₂ = k³S₁ / S₂ = k², V₁ / V₂ = k³
where:
  • kthe similarity ratio (ratio of corresponding lengths)
  • S₁, S₂corresponding areas (sections, surfaces)
  • V₁, V₂the volumes of the similar solids

The surface areas of similar solids are in the ratio of the square of the similarity ratio, and their volumes in the ratio of its cube — true for any similar solids (cubes, prisms, pyramids, cylinders, balls).

A parallel section and similar solids

1) A pyramid has height 12 cm and base area 72 cm². Find the area of the section parallel to the base at distance 4 cm from the apex.
2) The same pyramid is cut by a plane parallel to the base through the midpoint of the height. Find the volumes of the small pyramid and of the frustum.
3) Two similar pyramids have volumes 8 cm³ and 27 cm³, and the smaller one has total surface area 20 cm². Find the total surface area of the larger one.

Show solution
1) k = 4/12 = ⅓ ⇒ S = 72 · (⅓)² = 8 cm².
2) The whole pyramid: V = ⅓ · 72 · 12 = 288 cm³. k = ½ ⇒ the small pyramid is 288 · (½)³ = 36 cm³, the frustum 288 − 36 = 252 cm³. The section has a quarter of the base area (18 cm²), but the volume is one eighth!
3) k³ = 27/8 ⇒ k = 3/2, k² = 9/4. The surface of the larger one is 20 · 9/4 = 45 cm².
V = ⅓ · h · (S₁ + S₂ + √(S₁ · S₂)), Slat = ½ · (P₁ + P₂) · m
where:
  • S₁, S₂the areas of the bases of the frustum
  • P₁, P₂the perimeters of the bases
  • h, mthe height and (for a regular frustum) the apothem

The volume formula comes from the difference of the volumes of two pyramids; the lateral-area formula is the sum of the areas of equal trapezoids (regular frustum).

The frustum

1) A frustum has base areas 4 cm² and 16 cm² and height 6 cm. Find its volume.
2) A regular quadrilateral frustum has base sides 2 cm and 8 cm and height 4 cm. Find its apothem, lateral and total surface area, and volume.

Show solution
1) V = ⅓ · 6 · (4 + 16 + √64) = 2 · 28 = 56 cm³.
2) The distances from the centres of the bases to their sides are 1 and 4, a difference of 3. Apothem m = √(4² + 3²) = 5 cm. Slat = ½ · (8 + 32) · 5 = 100 cm², Stot = 100 + 4 + 64 = 168 cm². V = ⅓ · 4 · (4 + 64 + √256) = ⅓ · 4 · 84 = 112 cm³.
Check: the frustum is a pyramid of height 16/3 cm minus a pyramid of height 4/3 cm: ⅓ · 64 · 16/3 − ⅓ · 4 · 4/3 = 1024/9 − 16/9 = 112. ✓
DİM style: written task (full solution)

The base side of a regular quadrilateral pyramid is 8 cm, and its lateral faces make an angle of 60° with the plane of the base. Find the height, the apothem, the total surface area and the volume of the pyramid. Write out every step of the solution.

Show solution
Solution.
1) Let O be the centre of the square ABCD and M the midpoint of AB. SO ⊥ (ABC) and OM ⊥ AB, so by the three-perpendiculars theorem SM ⊥ AB. Hence SM is the apothem and ∠SMO is the linear angle of the dihedral angle between a lateral face and the base: ∠SMO = 60°.
2) OM = 8/2 = 4 cm (the radius of the circle inscribed in the square).
3) In triangle SOM: h = SO = OM · tan 60° = 4√3 cm; m = SM = OM / cos 60° = 4 : ½ = 8 cm.
4) Sbase = 8² = 64 cm²; Slat = ½ · Pbase · m = ½ · 32 · 8 = 128 cm² (check: Sbase / cos 60° = 64 : ½ = 128 ✓); Stot = 128 + 64 = 192 cm².
5) V = ⅓ · Sbase · h = ⅓ · 64 · 4√3 = 256√3/3 cm³ ≈ 147.8 cm³.
Answer: h = 4√3 cm; m = 8 cm; Stot = 192 cm²; V = 256√3/3 cm³.

The next lesson is «Solids of revolution: cylinder, cone and sphere». There the same ideas pass to round solids: a cylinder is like a prism and a cone like a pyramid (V = Sbase · h and V = ⅓ · Sbase · h), and combinations of polyhedra with a ball turn into plane problems in an axial section.

Check yourself: fill in the gap
  1. 1.The diagonal of a box measuring 2, 3 and 6 is .
  2. 2.A prism has 18 edges ⇒ its base is a -gon.
  3. 3.Pyramid: S_base = 27, h = 10 ⇒ V =
  4. 4.Similar solids with volumes 8 : 27 ⇒ surfaces 4 :
  5. 5.The edge of a cube is tripled ⇒ its volume grows times.

Key points

  • Euler’s theorem: V − E + F = 2; an n-gonal prism has 3n edges, a pyramid 2n.
  • Rectangular box: S = 2(ab + bc + ca), V = abc, d² = a² + b² + c²; cube: S = 6a², V = a³, d = a√3.
  • Prism: V = Sbase · h (any), Slat = Pbase · h (right); pyramid: V = ⅓ · Sbase · h, regular pyramid Slat = ½ · Pbase · m.
  • Equal lateral edges ⇒ the foot of the height is the circumcentre; equal dihedral angles ⇒ the incentre, Slat = Sbase / cos φ.
  • Parallel sections and similar solids: areas as k², volumes as k³; frustum: V = ⅓ · h · (S₁ + S₂ + √(S₁ · S₂)).

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
How is the lateral surface area of a right prism computed?