matura.mk› For teachers
Geometry in space

Solid geometry step by step

Solid geometry in one place: 11 modules with video lessons and 3D views, worked problems, a problem generator with new numbers, a quiz and a summary sheet. For class, for homework and for revision before a test.

11modules · ~7 h in total
110worked problems
0/11modules done
Drag to rotate · shortest path A → C₁
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Overview

11 modules · ~7 hyour progress is saved in the browser
Types of problems and how to solve them
Type of problemHow to solve itModule
“Is it true that…”, “Can…”axioms + picture a cube; “No” = find a counterexample01
Prove a statement (S1, S2, parallel lines, skew lines)ready-made templates of 2–4 lines01, 04, 05
Vertices, edges, faces; nets of the cube and the tetrahedron2n · 3n · n+2 and n+1 · 2n · n+1; the cube has 11 nets02
Shortest path over the surfaceunfold → straight line → Pythagoras or the cosine rule → take the smallest03
Count parallel / skew pairsedges × pairs per edge ÷ 204, 05
Distance in a cube (point–line, line–line, point–plane)three perpendiculars or coordinates05
Parallel projection of a segmentc = (na + mb)/(m + n)06
Vectors: simplify, lengthchaining and replacing with an equal vector07
Line–plane angle, three perpendiculars, dihedral angleprojection of the line; right triangle08
Volume of a parallelepiped, melting cubes, a rotated cubeV = abc·sinφ·sinψ; volumes add up09, 10
P and V of prisms and pyramidsright triangles: (H, d, D), (H, r, h), (H, R, s)10, 11
Notation. Problems are numbered by module (e.g. 3.2 is the second problem in module 3). In counting: V = vertices, E = edges, F = faces. In formulas: B = base area, M = lateral surface area, P = surface area, V = volume.

Tip: draw first, then write. A problem about P and V = find the right triangle!

▶

Videos: the basics

6 videos · ~13 minMacedonian voice and subtitles

Watch them before the modules. They cover the tools you need for almost every problem: the right triangle, areas, the prism, the pyramid, the shortest path and lines in space. The videos are in Macedonian with a computer voice and subtitles; the formulas and drawings are easy to follow without the language.

Video 1 · The right trianglePythagoras, sin/cos/tan, values for 30°, 45°, 60°, diagonal of a square, height of an equilateral triangle.2:45 min
Video 2 · Areas of basesRectangle, triangle, equilateral triangle, hexagon, parallelogram, rhombus, trapezoid, Heron’s formula.2:04 min
Video 3 · The prism: P, V and diagonalsP = 2B + M, V = B·H, the diagonals d and D, the angle φ, a worked example, the oblique prism.2:08 min
Video 4 · The pyramid: three trianglesV = ⅓BH, slant height, (H, r, h), (H, R, s), r and R for the bases, a worked example.2:18 min
Video 5 · Shortest path and the cosine ruleUnfolding, √5 on a cube, the cosine rule with 135°, a 5×4×3 cuboid → √74.2:03 min
Video 6 · Lines and planes in spaceParallel, skew, line ∥ plane, line–plane angle, three perpendiculars, dihedral angle.2:02 min
01

Basics: point, line, plane

20 min“Is it true…” and proofs

Picture the classroom. A point is a dot with no size, a line is an endlessly stretched thread, a plane is an endlessly thin sheet (the floor, a wall, the board). Solid geometry = geometry in the whole room, not on a sheet of paper.

Video lesson
Point, line, plane

Axioms (not proved)

  1. A1
    Exactly 1 line passes through 2 points.
  2. A2
    Exactly 1 plane passes through 3 non-collinear points.A table with 3 legs never wobbles.
  3. A3
    If 2 planes have 1 common point, they meet in a whole line.
  4. A4
    There are 4 points that do not lie in one plane.
  5. A5
    Everything from plane geometry holds in every plane.

Two planes

  • 1
    meet in a line
  • 2
    coincide (α ≡ β)
  • 3
    have no common points → parallel

We often count α ≡ β as parallel too. Then the theorem holds: the relation “is parallel to” on the set of planes is an equivalence relation, because it is:

  • reflexive
    α ∥ α (every plane is parallel to itself)
  • symmetric
    α ∥ β ⇒ β ∥ α
  • transitive
    α ∥ β and β ∥ γ ⇒ α ∥ γ
Show:

Statements with proofs

  1. S1
    If the line c has 2 common points A and B with α, then the whole of c lies in α.Proof: exactly one line passes through A and B (A1). There is also a line through A and B in α (A5). The two lines must coincide, so c ⊂ α. ∎
  2. S2
    Exactly one plane passes through a line and a point outside it.Proof: take 2 points of the line and the given point. These are 3 non-collinear points → A2 gives exactly one plane. ∎
  3. S3
    Exactly one plane passes through 2 intersecting lines.Proof: A = a ∩ b, take B ∈ b, B ≠ A. Now we have the line a and the point B outside it → S2. ∎
Problems
1.14 points do not lie in one plane. Can 3 of them lie on one line?No

Proof by contradiction.

  1. Suppose 3 points lie on one line a.
  2. Then a plane passes through the line a and the 4th point (S2).
  3. So all 4 points are in that plane, which contradicts the assumption. ∎
1.2Can 2 different planes have 2 common lines?No

If they have 2 common lines, they also have 3 common non-collinear points. By A2 only one plane passes through them, so the planes coincide. Two different planes meet in only one line (A3).

1.3Is it true that every line that meets each of two given intersecting lines lies in the plane of those lines?No

Counterexample: the line can pass through the intersection point of the two lines and leave the plane (e.g. in a cube AA₁ meets both AB and AD but does not lie in ABCD). If it meets them in 2 different points, then it does lie in the plane (S1).

1.4A line and a point outside it are given. Prove that all lines that meet the given line and pass through the point lie in one plane.from S2 and S1
  1. Let the line be a and the point B ∉ a. By S2 exactly one plane α passes through a and B.
  2. A line b through B meets a at a point A. Then A ∈ α and B ∈ α.
  3. b has 2 common points with α → by S1 the whole of b ⊂ α. This holds for every such line. ∎
1.5A finite number of lines, every 2 of which meet. Prove: either they all lie in one plane, or they all pass through 1 point.S3 + S1
  1. Take 2 of the lines, a and b, which meet at O. A plane α passes through them (S3).
  2. Every other line c meets both a and b. If it meets them in 2 different points, those points are in α, so c ⊂ α (S1).
  3. If c passes through O and is not in α, then every further line must meet c, a and b. That is only possible if it also passes through O. So either they are all in α, or they all pass through O. ∎
Counting

n points, no 3 on a line → lines

n(n−1)/2. Example: 5 points → 10 lines.

n points, no 4 in a plane → planes

n(n−1)(n−2)/6. Example: 5 points → 10, 6 points → 20.

Into at most how many parts do planes divide space?

1 plane → 2 · 2 → 4 · 3 → 8 · 4 → 15.

More problems
1.6Do any a) 3 points b) 4 points always lie in one plane?a) Yes b) No

3 points always do (if they are on a line, infinitely many planes; if not, exactly one). 4 points need not: A4 says there are 4 non-coplanar points (the vertices of a tetrahedron).

1.7A circle has 2 common points with a plane. Does it lie in the plane?No

Picture a hoop standing upright and passing through the floor: 2 points on the floor, the rest outside it. (If it has 3 common points, then it does lie in the plane.)

1.83 vertices of a parallelogram are in α. What about the fourth? / 2 vertices + the intersection of the diagonals are in α?Yes / No

3 vertices determine a plane → the whole parallelogram is in it. But if the 2 vertices are opposite (A and C), the intersection of the diagonals is on the same line AC → 3 collinear points do not determine a plane → B and D need not be in α.

1.9The lines a, b, c meet in pairs. Must they lie in one plane?Not necessarily

The three edges of a cube from one vertex (AB, AD, AA₁) meet in pairs at A but are not in one plane. If they meet in 3 different points, then they are in one plane.

1.10Find the mistake in the drawing: α and β meet, and A, B, C belong to bothA, B, C must be collinear

Two planes meet in only one line (A3). Every common point must be on that line → A, B, C are on one line. Likewise: if a, b, c meet in pairs and meet α at A, B, C, then A, B, C must be collinear.

1.114 points not in a plane: can 3 of them be on a line? A closed broken line of 3 / 4 segments outside a plane?No · No · Yes

If 3 are on a line, they and the 4th point determine a plane (S2) → all 4 would be in it. A broken line of 3 parts is a triangle → always in a plane. With 4 parts: A→B→C→D→A where D is above the plane ABC → possible.

1.12a ∩ b = C, a ⊂ α, b ⊂ β (α ≠ β). Where does the line of intersection of α and β pass?through C

C is on a (so in α) and on b (so in β) → C is a common point → it lies on the line of intersection.

1.13Proofs: points outside a plane; lines through a point that meet a line; finitely many intersecting linesA4, S1, S2, S3
  1. Every plane has points outside it: by A4 there are 4 non-coplanar points → at least one is not in α.
  2. A line a and a point B ∉ a. All lines through B that meet a lie in one plane: by S2 there is a plane α through a and B. A line b through B that meets a at A has 2 points (A and B) in α → by S1 b ⊂ α.
  3. Finitely many lines, every two meet → either all in one plane, or all through one point: take a and b (plane α). Every other c either passes through the intersection of a and b, or meets them in 2 different points → it lies in α.
02

Polyhedra and nets

25 mineasy marks

A polyhedron is a solid made of polygons (its faces). Their sides are the edges and their corners are the vertices. If the surface is cut along some edges and laid flat so that each face shares an edge with another face, the picture is a net of the polyhedron.

Video lesson
Counting vertices, edges, faces

Formulas

SolidVerticesEdgesFaces
Cube, cuboid, parallelepiped8126
n-gonal prism2n3nn + 2
n-gonal pyramidn + 12nn + 1

Check: vertices − edges + faces = 2

Terms

  • prism
    2 equal parallel bases + lateral faces that are parallelograms. All lateral edges are parallel. Right: the lateral edges are perpendicular to the base. Regular: right + the base is a regular polygon.
  • pyramid
    an n-gon base + triangular lateral faces with a common apex S. Regular: the base is a regular polygon and the lateral edges are equal.
Nets

All 11 nets of a cube

Rule: never 4 squares in a 2×2 block, never more than 4 in a row, and every square must have its own place when you fold it. If you are not sure, mark one face as the “bottom” and imagine how the others fold.

Nets of a regular tetrahedron

The third picture is not a net, because 4 triangles share a vertex. In a tetrahedron exactly 3 meet at each vertex.

Problems
2.1A polyhedron has 12 faces, all pentagons. How many edges does it have?30

Each edge is a side of exactly 2 faces, so each edge is counted twice.

  1. 12 faces · 5 sides = 60.
  2. 60 / 2 = 30 edges.
2.2Is there a polyhedron whose faces are all squares but which is not a cube?Yes, a 3D cross

Glue one cube onto each face of a central cube (a 3D cross). All faces are squares, but the solid is not a cube.

2.3Is there a polyhedron with 11 edges?Yes
  1. Take a square pyramid: 5 vertices, 8 edges, 5 faces.
  2. Cut off a small corner at one vertex of the base (3 edges meet there). This adds 3 new edges and 1 new face (a triangle).
  3. 8 + 3 = 11 edges. Check: 7 − 11 + 6 = 2 ✓. (The only number above 6 for which there is no polyhedron is 7.)
2.4Prove: the number of faces with an odd number of sides is even.contradiction
  1. Each edge is a side of exactly 2 faces → the sum of the numbers of sides of all faces is even.
  2. Let N = the number of faces with an odd number of sides, SN = the sum of their sides, SP = the sum for faces with an even number of sides. SN + SP is even.
  3. SP is always even. If N is odd, SN is a sum of an odd number of odd numbers → odd.
  4. Then SN + SP is odd — a contradiction. So N is even. ∎
More problems
2.5Is there a prism with a) 4 b) 6 c) 12 d) 21 edges?No · No · Yes · Yes

A prism has 3n edges, n ≥ 3.

  1. 4 and 6: 3n = 4 has no whole n; 3n = 6 → n = 2 (there is no 2-gon) → no.
  2. 12 → n = 4 (quadrilateral) → yes. 21 → n = 7 (heptagonal) → yes.
2.6What is the base of a prism with a) 18 edges b) 24 vertices c) 36 faces?6-gon · 12-gon · 34-gon
  1. 3n = 18 → n = 6 → hexagon.
  2. 2n = 24 → n = 12 → 12-gon.
  3. n + 2 = 36 → n = 34 → 34-gon.
2.7A pyramid with 10 / 6 / 24 / 33 edges? The base if it has 8 edges / 22 vertices / 60 faces?Yes Yes Yes No · 4 · 21 · 59
  1. A pyramid has 2n edges → the number must be even: 10 (n=5), 6 (n=3), 24 (n=12) → yes; 33 is odd → no.
  2. 2n = 8 → quadrilateral; n + 1 = 22 → 21-gon; n + 1 = 60 → 59-gon.
2.8A polyhedron: 6 vertices, 4 edges meet at each. / 12 faces, all pentagons. How many edges?12 · 30

Trick: each edge has 2 ends and lies on 2 faces → everything is counted twice → divide by 2.

  1. 6·4/2 = 12.
  2. 12·5/2 = 30.
2.9Is there a polyhedron with 5, 6, 7, 8, 9, 10, 11 edges?No, tetrahedron, No, ...
  1. 5 → no; 6 → tetrahedron; 7 → no; 8 → square pyramid; 9 → triangular prism; 10 → pentagonal pyramid; 11 → yes (problem 2.3).
  2. The tetrahedron has the fewest edges (6). 7 is the only number above 6 that is impossible.
2.10Nets: which are nets of a regular tetrahedron / a prism / a pyramid?see the rule
  1. Regular tetrahedron: 3 triangles meet at each vertex. A figure where 4 triangles share a point is not a net (figure 3).
  2. Net of a prism: 2 equal bases + n rectangles in a row. Net of a pyramid: the base + n triangles around it.
  3. A regular five-pointed star → a regular pentagonal pyramid. 6 rectangles + 2 hexagons → a regular hexagonal prism.
03

Shortest path over the surface

30 mina common type of problem

We draw a net of the polyhedron on whose surface we need the shortest path between 2 points. On the net the shortest path is a straight line. Then Pythagoras, or the cosine rule if the angle is not a right angle.

Video lesson
Unfold and calculate

Recipe

  1. 1
    Draw the faces the path can cross, laid flat in a plane.
  2. 2
    Join the start and the end with a straight line.
  3. 3
    Right angle → Pythagoras. Another angle (105°, 120°, 135°) → the cosine rule: c² = a² + b² − 2ab·cos γ.
  4. 4
    Try all possible nets and take the smallest result.

Cosines you need

cos 120° = −½ · cos 135° = −√2/2 · cos 105° = √2(1 − √3)/4

Problems
3.1A regular unit tetrahedron ABCD: the shortest path between the midpoints of AB and CD.1

The net of 2 neighbouring faces = a rhombus. The midpoints E and F are on opposite sides, EF is parallel to a side → EF = 1.

3.2A regular tetrahedron with edge 20 cm, points E and F 7 cm from the vertices D and C (on the heights of two faces).20 cm
  1. Height of a face: h = √(20² − 10²) = √300 = 10√3.
  2. The path over the edge AB: EF = 2(√300 − 7) = 20√3 − 14 ≈ 20.64.
  3. The path over 3 faces (net DBC + its neighbours): E and F are at the same height → EF is parallel to DC → EF = 20.
  4. 20 < 20√3 − 14 → the shortest path is 20.
3.3A regular square pyramid SABCD, all edges 1. The shortest path from A to the midpoint E of SC.√7/2
  1. Net: the faces SAB and SBC (two equilateral triangles).
  2. BE is a height in △SBC → EB² + (½)² = 1² → EB = √3/2, and ∠ABE = 60° + 30° = 90°.
  3. AE² = EB² + AB² = ¾ + 1 = 7/4 → AE = √7/2 ≈ 1.32.
3.4A regular triangular prism ABCA₁B₁C₁, all edges 1. The shortest path from A to the midpoint D of B₁C₁.√(7+2√3)/2
  1. Net: the face ABB₁A₁ + the top base A₁B₁C₁ unfolded next to it.
  2. △ABB₁: AB₁ = √(1² + 1²) = √2, and ABB₁A₁ is a square → ∠A₁B₁A = 45°. The base is an equilateral triangle → ∠A₁B₁C₁ = 60°. B₁D = ½.
  3. ∠DB₁A = 45° + 60° = 105°, cos105° = √2/2·½ − √2/2·√3/2 = √2(1 − √3)/4.
  4. Cosine rule: AD² = 2 + ¼ − 2·½·√2·cos105° = 9/4 − (1 − √3)/2 = (7 + 2√3)/4.
  5. AD = √(7 + 2√3)/2 ≈ 1.62. (Over 2 lateral faces only it would be √13/2 ≈ 1.80, longer.)
3.5A unit cube ABCDA₁B₁C₁D₁: the shortest path over the surface from A to C₁.√5

First method: a net with the top face unfolded.

  1. A₁C₁ is the diagonal of a square → A₁C₁ = √2 and ∠C₁A₁B₁ = 45°.
  2. In the net ∠AA₁C₁ = 90° + 45° = 135°.
  3. Cosine rule in △AA₁C₁: AC₁² = 1² + (√2)² − 2·1·√2·cos135° = 1 + 2 + 2 = 5.
  4. AC₁ = √5. (Faster: the net of 2 neighbouring faces = a 2 × 1 rectangle → √(4 + 1).)
3.6Three edges of a cuboid are 5, 4, 3. The shortest path over the surface from A to C₁.√74

There are 3 cases (which face is the base).

  1. Base 5 × 3 and 5 × 4: legs 5 and 4 + 3 = 7 → √(25 + 49) = √74.
  2. Base 4 × 3: legs 4 and 5 + 3 = 8 → √(64 + 16) = √80.
  3. The third option: √(9² + 3²) = √90.
  4. √74 < √80 < √90 → the shortest is √74 ≈ 8.60. Rule: add the two shortest edges (4 + 3).
3.7A polyhedron with all dihedral angles right angles (an “L shape”, edges 2 and 1). The shortest path from B to C₂.√13
  1. Net I: △BCC₂ with legs 2 and 1 + 2 = 3 → BC₂ = √(9 + 4) = √13 ≈ 3.61.
  2. Net II: BD₁ + D₁C₂ = √((2+1)² + 1) + 1 = √10 + 1 ≈ 4.16.
  3. Net III: √((2 + 2 + 1)² + 2²) = √29 ≈ 5.39.
  4. Shortest: √13.
3.8A regular hexagonal prism, base edge 1, lateral edge 2. The shortest path from A to D₁.√13
  1. Net I (a lateral face + the top base): B₁D₁ = √3 (the short diagonal of the hexagon). AD₁² = AB² + (BB₁ + B₁D₁)² = 1 + (2 + √3)² = 8 + 4√3 → AD₁ = 2√(2 + √3) ≈ 3.86.
  2. Net II (3 lateral faces): a 3 × 2 rectangle → AD₁ = √(3² + 2²) = √13 ≈ 3.61.
  3. The shortest is √13. (If the lateral edge is 1, the net with the base wins: √(5 + 2√3) ≈ 2.91 < √10.)
More problems
3.9The shortest loop that a unit tetrahedron can be passed through2

The cross-section parallel to 2 opposite edges is the rectangle EFGH with perimeter 2 (its sides are parallel to the edges). The tetrahedron goes through starting with the edge AD and moving towards BC. Answer: 2.

3.10A cross-section of a unit tetrahedron that cuts all 4 faces, with the smallest perimeter2

Unfold the 4 faces into a strip; the closed path becomes a straight segment of length 2. Every cross-section parallel to 2 opposite edges has perimeter exactly 2.

3.11A cross-section of a unit cube that cuts all 6 faces, with the smallest perimeter3√2

The regular hexagon through the midpoints of 6 edges. Each side = √2/2, perimeter 6·√2/2 = 3√2 ≈ 4.24. In the net (6 faces “like stairs”) this is a straight segment.

04

Parallel lines and planes

25 mintheorems + proofs

Everything is learnt on a cube. Two lines are parallel if they lie in one plane and do not meet. A line is parallel to a plane if they have no common point.

Video lesson
Parallel, intersecting, skew
Edge: or click an edge
chosenparallelintersectingskew

Theorems

  1. T1
    Through a point outside a line there passes exactly one line parallel to the given line.Proof: there is exactly one plane α through the point A and the line b (S2). Every line through A parallel to b must lie in α, and in a plane the parallel is unique. ∎
  2. T2
    Two lines parallel to a third line are parallel to each other.
  3. T3
    If the plane α passes through a line a ∥ β and meets β in the line b, then b ∥ a.
  4. T4
    If a does not lie in β and a is parallel to a line b that lies in β, then a ∥ β.The criterion used most often in proofs.
  5. T5
    2 intersecting lines in α are parallel to 2 lines in β ⇒ α ∥ β.
Problems
4.1The segments AA₁ and CC₁ are equal in length and parallel. Prove AC ∥ A₁C₁.parallelogram
  1. AA₁ ∥ CC₁ → they lie in one plane, so ACC₁A₁ is a quadrilateral in a plane.
  2. One pair of opposite sides (AA₁ and CC₁) is equal and parallel → ACC₁A₁ is a parallelogram.
  3. Opposite sides of a parallelogram are parallel → AC ∥ A₁C₁. ∎
4.2In the parallelepiped ABCDA₁B₁C₁D₁ prove that AD₁ and BC₁ are parallel.ABC₁D₁ is a parallelogram
  1. AB ∥ DC and DC ∥ D₁C₁ (the faces are parallelograms) → AB ∥ D₁C₁ (T2). Also AB = DC = D₁C₁.
  2. So ABC₁D₁ has a pair of opposite sides that are equal and parallel → a parallelogram.
  3. Then the other pair of sides, AD₁ and BC₁, are parallel. ∎
4.3How many pairs of parallel lines does a regular hexagonal prism have (lines of the edges)?33
  1. Lateral edges (6): all are parallel to each other → each is in 5 pairs → 6·5/2 = 15.
  2. Base edges (12): each is parallel to 3 others (the opposite one in the same base + 2 in the other base) → 12·3/2 = 18.
  3. In total 15 + 18 = 33 pairs.
Proof templates (just change the letters)

Line ∥ plane

AA₁ is parallel to BB₁, which lies in the plane BCC₁. Therefore (T4) AA₁ ∥ (BCC₁). ∎

Plane ∥ plane

AB and BC meet and lie in ABC. They are parallel to A₁B₁ and B₁C₁ in A₁B₁C₁. Therefore (T5) the planes are parallel. ∎

More problems
4.4Must 2 lines that do not meet be parallel? Skew?No · No

They can be parallel or skew. So “they do not meet” is not enough — you also need “in the same plane”.

4.5a is skew to b, and b to c. Are a and c skew?Not necessarily

In a cube: AB and CC₁ are skew, CC₁ and DA are skew, but AB and DA meet.

4.6Two lines parallel to the same plane — are they parallel? A line ∥ α — is it parallel to every line in α?No · No

AB and BC are both parallel to the top face, yet they meet. A₁B₁ ∥ (ABC), but A₁B₁ and BC are skew.

4.7One of 2 parallel lines is ∥ to a plane. Is the other one?No (it may lie in the plane)

A₁B₁ ∥ (ABC), AB ∥ A₁B₁, but AB lies in (ABC).

4.8Through B ∉ α: how many lines ∥ α? How many planes ∥ α?infinitely many · 1

All lines through B in the plane through B parallel to α. And that plane is unique.

4.9A test for ∥ planes: a) 1 line in α ∥ a line in β b) 2 lines in α ∥ 2 lines in β c) 2 intersecting linesNo · No · Yes

The lines in α must meet. If they are parallel to each other, the planes can meet (e.g. the faces ABB₁ and BCC₁ contain AA₁ and CC₁, which are parallel, yet the faces meet in BB₁).

4.10Cube: lines through vertices parallel to AB. Hexagonal prism: lines parallel to AB.A₁B₁, CD, C₁D₁

Cube: A₁B₁; CD; C₁D₁. Regular hexagonal prism: A₁B₁; DE; D₁E₁; CF; C₁F₁ (CF is a long diagonal, parallel to AB).

4.11Cube: planes through vertices parallel to a) AA₁ b) AB₁ c) AC₁BCC₁, CDD₁, BDD₁ ...
  1. AA₁: BCC₁; CDD₁; BDD₁.
  2. AB₁: CDD₁; DA₁C₁; BDC₁.
  3. AC₁ (space diagonal): none.
4.12O is the centre of the face ABCD. How are A₁O and (CB₁D₁) positioned?parallel

Let O₁ be the centre of A₁B₁C₁D₁. A₁O ∥ CO₁ (A₁O₁CO is a parallelogram), and CO₁ lies in (CB₁D₁) ⇒ by T4 A₁O ∥ (CB₁D₁).

4.13Cube: a plane through AC ∥ BC₁; a plane through B ∥ ACD₁; ∥ planes through BA₁ and CB₁ACD₁ · BA₁C₁ · BDA₁ and CB₁D₁
  1. BC₁ ∥ AD₁ ⇒ the plane ACD₁ contains AC and is ∥ to BC₁.
  2. BA₁ ∥ CD₁ and BC₁ ∥ AD₁ ⇒ (BA₁C₁) ∥ (ACD₁).
  3. For the skew lines BA₁ and CB₁: (BDA₁) ∥ (CB₁D₁).
4.14Parallelepiped: prove (ABC) ∥ (A₁B₁C₁) and (AB₁D₁) ∥ (BDC₁)T5
  1. AB and AD (intersecting, in ABC) ∥ A₁B₁ and A₁D₁ (in A₁B₁C₁) ⇒ ∥.
  2. AB₁ and AD₁ (intersecting, in AB₁D₁) ∥ DC₁ and BC₁ (in BDC₁) ⇒ ∥. ∎
4.15Regular hexagonal prism: is ABB₁ ∥ a) CDD₁ b) DEE₁ c) CEE₁ d) CFF₁ e) CFE₁?No Yes No Yes No

AB ∥ ED ∥ CF (in a regular hexagon). The face DEE₁ and the diagonal section CFF₁ are parallel to ABB₁. Proof of b): AB and AA₁ are ∥ to DE and EE₁ ⇒ T5.

4.16Hexagon ABCDEF, the side AF lies in α. Which sides meet α? / A pyramid SABCD with a parallelogram base: the line of intersection of SAB and SCD?CD ∥ α · ∥ to the base
  1. AB, BC, DE, EF meet α; CD ∥ AF ⇒ CD ∥ α.
  2. AB ∥ CD ⇒ AB ∥ (SCD) ⇒ the line of intersection of SAB and SCD is ∥ to AB, so ∥ to the plane ABCD.
05

Skew lines and distances

35 minthe cube-of-edge-6 problem is the hardest

Two lines are skew if they do not lie in one plane. They neither meet nor are parallel, like two bridges at different heights.

Positions of 2 lines

  • in a plane
    intersecting (1 point) or parallel (0 points)
  • not in a plane
    skew
  • test
    a ⊂ α, and b pierces α at a point not on a ⇒ a and b are skew.

Counting

Each edge of the cube is in 4 skew pairs. The cube has 12 edges → 4·12/2 = 24 pairs.

Tetrahedron: 3 pairs · square pyramid: 8 · regular triangular prism: 12 · regular hexagonal prism: 72.

Problems
5.1Name the lines that pass through the vertices of the cube and are skew to AB.12 lines

Not only edges: also the face diagonals and the space diagonals.

  1. Edges: CC₁, DD₁, A₁D₁, B₁C₁.
  2. Face diagonals: CB₁, DA₁, CD₁, DC₁, A₁C₁, B₁D₁.
  3. Space diagonals: CA₁, DB₁.
  4. In total 12. Checked with coordinates.
5.2Prove: if EF and GH do not lie in one plane, then EG and FH are also skew.contradiction
  1. Suppose EG and FH lie in the same plane α.
  2. Then E, G, F, H ∈ α, so the lines EF and GH also lie in α.
  3. That contradicts the condition. So EG and FH are not in one plane → skew. ∎
A problem about distances

The cube ABCDA₁B₁C₁D₁ has edge 6. M and N are the midpoints of AB and B₁C₁. K is on DC with DK = 2KC.

Video lesson
Distances in a cube of edge 6

Coordinates (the safest method)

Put A at the origin: A(0, 0, 0), B(6, 0, 0), D(0, 6, 0), A₁(0, 0, 6).

Then M(3, 0, 0), N(6, 3, 6), K(4, 6, 0).

Distance between points: √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²).

5.3The distance from N to the line AK.6√221/13 ≈ 6.86

The theorem of three perpendiculars: drop N onto the base.

  1. N′ = the projection of N onto ABCD = the midpoint of BC = (6, 3, 0). NN′ = 6.
  2. The distance from N′ to AK in the base: area of △AN′K = ½|6·6 − 3·4| = 12, AK = √(16 + 36) = 2√13 → d = 2·12/(2√13) = 12/√13.
  3. Let H be the foot of the perpendicular from N′ to AK. N′H ⊥ AK and NN′ ⊥ the base → by the theorem of three perpendiculars NH ⊥ AK.
  4. NH² = NN′² + N′H² = 36 + 144/13 = 612/13 → NH = 6√221/13 ≈ 6.86.
5.4The distance between the lines MN and AK.18√53/53 ≈ 2.47

The lines are skew. The distance is the length of the common perpendicular.

  1. Directions: u = MN = (3, 3, 6) ∼ (1, 1, 2), w = AK = (4, 6, 0) ∼ (2, 3, 0).
  2. A vector perpendicular to both: n = u × w = (1·0 − 2·3, 2·2 − 1·0, 1·3 − 1·2) = (−6, 4, 1).
  3. Distance = |AM · n| / |n| = |(3, 0, 0)·(−6, 4, 1)| / √(36 + 16 + 1) = 18/√53.
  4. d = 18√53/53 ≈ 2.47.
5.5The distance from A₁ to the plane MNK.66√173/173 ≈ 5.02
  1. Normal of the plane: MN × MK = (3, 3, 6) × (1, 6, 0) = (−36, 6, 15) ∼ (−12, 2, 5).
  2. Equation: −12x + 2y + 5z = −36 (check with M, N, K).
  3. Distance from A₁(0, 0, 6): |−12·0 + 2·0 + 5·6 + 36| / √(144 + 4 + 25) = 66/√173.
  4. d = 66√173/173 ≈ 5.02.
More problems
5.6Cube: which are skew to AB? Tetrahedron ABCD? Pyramid SABCD?A₁D₁, B₁C₁, DD₁, CC₁
  1. Cube: A₁D₁; B₁C₁; DD₁; CC₁ (4 edges).
  2. Tetrahedron: AB and CD; BC and AD; AC and BD.
  3. SABCD: SA–BC, SA–CD, SB–AD, SB–CD, SC–AB, SC–AD, SD–AB, SD–BC.
06

Parallel projection

20 min

Picture the sun shining at a slant: every point casts a shadow on the floor. All rays are parallel (direction l). The shadow of the point A is its parallel projection A′ onto the plane π.

Video lesson
A shadow on the floor
Direction l:

Definition and properties

π is a plane and l is a line that pierces it. Through the point A we draw a line parallel to l. The point where it pierces π is the parallel projection of A in the direction of l.

  1. Prop. 1
    A line parallel to l (or l itself) projects to a point. Every other line projects to a line.
  2. Prop. 2
    The ratio of segments on one line is preserved: AB : BC = A′B′ : B′C′.
  3. Prop. 3
    2 parallel lines (not parallel to l) project to parallel lines (or to the same line).AA′ ∥ BB′. If AB ∥ A′B′, ABB′A′ is a parallelogram; if not, ABB′A′ is a trapezoid.
Problem
6.1The point C divides AB in the ratio m : n. AA′ = a, BB′ = b (parallel projections). Find CC′.c = (na + mb)/(m + n)

Similar triangles.

  1. Through A draw a line parallel to A′B′; it meets CC′ at C″ and BB′ at B″. Then A′C′C″A and A′B′B″A are parallelograms: C″C′ = B″B′ = a.
  2. So B″B = b − a, and let x = C″C.
  3. △ACC″ ∼ △ABB″ (CC″ ∥ BB″): x/(b − a) = AC/AB = m/(m + n) → x = m(b − a)/(m + n).
  4. c = CC′ = a + x = (am + an + mb − ma)/(m + n) = (na + mb)/(m + n).
  5. Check: midpoint (m = n) → c = (a + b)/2 (the midline of a trapezoid).
07

Vectors in space

25 minshort problems

A vector in space is a directed segment (with a start and an end). Every vector has a line of action (the line it lies on), a sense (the order of the points) and a length (magnitude). Two vectors are equal if they have the same line direction, sense and length, so in a cube A₁D₁ = AD = BC.

Video lesson
Adding vectors in a cube

Terms

  • notation
    |AB| = AB — magnitude (length). If A ≡ B, then AB = AA = 0 (the zero vector).
  • free
    The equivalence class (all equal vectors) is called a free vector.
  • operations
    Vector addition and multiplication by a scalar are defined as in the plane. The set of vectors is a vector space over ℝ.
  • collinear
    Vectors on parallel lines (the same direction of line).
  • opposite
    Vectors whose sum is the zero vector: AB + BA = 0.

Rules for adding

  • chain
    AB + BC = AC (the middle letters “cancel”)
  • parallelogram
    AB + AD = AC
  • difference
    AB − AC = CB

Parallelepiped: AC₁ = AB + AD + AA₁

Problems
7.1Parallelepiped: are these true? a) BD₁ = BB₁ + B₁D₁ b) BD₁ = BA + BB₁ + BC c) DB₁ = DA + DC − D₁D d) AC₁ = AB + BC + CC₁ − D₁C₁ + D₁AYes Yes Yes No
  1. a) chaining ✓. b) 3 edges from B → the space diagonal ✓.
  2. c) DA + DC = DB; −D₁D = DD₁ = BB₁; DB + BB₁ = DB₁ ✓.
  3. d) AB + BC + CC₁ = AC₁. The rest: −D₁C₁ + D₁A = C₁D₁ + D₁A = C₁A. In total AC₁ + C₁A = 0 ≠ AC₁ → no.
7.2A vector equal to: a) AA₁ + AB b) AA₁ + BC c) AA₁ + C₁C d) ½CB − ½CAAB₁ · AD₁ · 0 · ½AB
  1. a) parallelogram → AB₁. b) BC = AD → AA₁ + AD = AD₁.
  2. c) C₁C = −AA₁ → the zero vector. d) ½(CB − CA) = ½AB.
7.3Simplify: B₁D₁ + C₁C + C₁B + AC₁ + CA + A₁D₁B₁D
  1. Group: AC₁ + C₁C = AC; AC + CA = 0.
  2. B₁D₁ + C₁B + A₁D₁ remain. Replace A₁D₁ = B₁C₁: B₁C₁ + C₁B = B₁B.
  3. B₁B + B₁D₁ = B₁D (the parallelogram BB₁D₁D).
7.4Unit cube: the length of a) AB + A₁D₁ b) AB₁ + A₁D c) AB₁ + BC₁√2 · √2 · √6
  1. a) A₁D₁ = BC → AB + BC = AC, |AC| = √2.
  2. b) A₁D = B₁C → AB₁ + B₁C = AC → √2.
  3. c) BC₁ = AD₁ → AB₁ + AD₁ = 2·AO₁ (O₁ is the midpoint of B₁D₁). |AO₁| = √(1 + ½) → length 2·√(3/2) = √6.

Check with coordinates: AB₁ = (1,0,1), BC₁ = (0,1,1), sum (1,1,2), length √(1+1+4) = √6.

7.5A regular triangular prism, all edges 1: |AB + A₁C₁|, |AB + CC₁|, |BA + CB₁|√3 · √2 · √2
  1. A₁C₁ = AC → AB + AC = 2·AO (O is the midpoint of BC), |AO| = √3/2 → √3.
  2. CC₁ = BB₁ → AB + BB₁ = AB₁ → √2.
  3. CB₁ = CB + BB₁; BA + CB = CA; CA + BB₁ = CA + AA₁ = CA₁ → √2.
7.6Express: SC in SABCD; SD in a hexagonal pyramid; AC₁ in a hexagonal prism; simplify AB + CD + EE₁ + B₁C₁see below
  1. SC = SA + AC = SA + AB + AD.
  2. A regular hexagon with centre O: AO = AB + AF, AD = 2·AO → SD = SA + 2(AB + AF).
  3. AC = AB + BC, BC = AO = AB + AF → AC₁ = 2AB + AF + AA₁.
  4. CD = BC − AB, B₁C₁ = BC → AB + CD + B₁C₁ = 2BC = AD; plus EE₁ = AA₁ → AD₁.
7.7Tetrahedron ABCD, E and F the midpoints of AB and CD. Prove EF = ½(AD + BC)midlines
  1. Let G be the midpoint of BD.
  2. EG is a midline in △ABD → EG = ½AD. GF is a midline in △BCD → GF = ½BC.
  3. EF = EG + GF = ½(AD + BC). ∎
08

Angles, three perpendiculars, dihedral angle

35 mintheory + proofs

The angle between a line and a plane = the angle between the line and its shadow (orthogonal projection) on the plane. It is the smallest angle the line makes with any line in the plane.

Video lesson
Angles in space
Angle:

Definitions

  1. 2 lines
    For p = AB, q = CD the angle is ∠(AB, CD) or its supplement; always ≤ 90°.
  2. p ⊥ Σ
    p pierces Σ and is perpendicular to every line in Σ. So if x ⊂ Σ and p ⊥ Σ ⇒ p ⊥ x.
  3. line–plane
    p′ = the orthogonal projection of p onto Σ: ∠(p, Σ) = ∠(p, p′), always ≤ 90°.
  4. dihedral
    2 half-planes Σ₁, Σ₂ with a common edge p. The dihedral angle: p₁ ⊂ Σ₁, p₁ ⊥ p and p₂ ⊂ Σ₂, p₂ ⊥ p (from the same point) → ∠(p₁, p₂).
  5. 2 planes
    The angle between 2 intersecting planes = the angle between 2 lines, each perpendicular to one of the planes.
  6. 3 perpendiculars
    n ⊥ Σ, A ∈ n, A′ = n ∩ Σ, B ∈ Σ, q ⊂ Σ through B. If q ⊥ A′B ⇒ q ⊥ AB.
Problems
8.1The line p pierces Σ at O. Prove that the angle between p and the plane is the smallest of the angles between p and the lines in Σ.sin ∠XOY ≤ sin ∠XOZ
  1. It is enough to look at the lines in Σ through O (a parallel shift does not change the angle).
  2. Let X be any point of p, Y its projection onto Σ (it lies on p′), and Z the projection of X onto some line q′ ⊂ Σ through O.
  3. XY is perpendicular to the plane → XY ≤ XZ (the perpendicular is the shortest).
  4. sin ∠XOY = XY/XO ≤ XZ/XO = sin ∠XOZ, and both angles are in [0°, 90°].
  5. So ∠XOY ≤ ∠XOZ, i.e. ∠(p, Σ) is the smallest. ∎
8.2The line p makes equal angles with p₁ and p₂ if and only if p is perpendicular to one of the bisectors of the angles between p₁ and p₂.statement

With vectors (a short explanation).

  1. Let e₁, e₂ be unit vectors of p₁, p₂, and u of p. Equal angles ⇔ |u·e₁| = |u·e₂|.
  2. That is ⇔ u·(e₁ − e₂) = 0 or u·(e₁ + e₂) = 0.
  3. e₁ + e₂ and e₁ − e₂ are the directions of the two bisectors → p ⊥ one of them. ∎
8.3p makes equal angles with p₁ and p₂, which meet at O and lie in Σ. Prove that the projection of p onto Σ is parallel to the bisector of the angle between p₁ and p₂.congruent △ + 3 perpendiculars
  1. X ∈ p, X′ = the projection of X onto Σ, Xᵢ = the projection of X onto pᵢ (i = 1, 2).
  2. ∠XOX₁ = ∠XOX₂ (given), ∠XX₁O = ∠XX₂O = 90°, XO is common → △XOX₁ ≅ △XOX₂, so OX₁ = OX₂ and XX₁ = XX₂.
  3. Theorem of three perpendiculars: XX₁ ⊥ p₁ ⇒ X′X₁ ⊥ p₁; likewise X′X₂ ⊥ p₂.
  4. △OX₁X′ ≅ △OX₂X′ (right angles, OX₁ = OX₂, OX′ common) → X′X₁ = X′X₂.
  5. X′ is at equal distances from p₁ and p₂ → it lies on the bisector: ∠X₁OX′ = ∠X₂OX′. ∎
09

Volume: Cavalieri and the parallelepiped

25 min5 worked problems

A stack of cards: if you push it so it leans, the volume does not change. That is Cavalieri’s principle: if two solids have cross-sections of equal area at every height, they have the same volume.

Video lesson
An oblique parallelepiped
Slant:
The same base, the same height → V = B · H, however much it leans.

Formulas

  • general cylinder
    Base F, line l ∦ Σ. c = the directrix (the boundary of the base), l = the generator. V = B · H. Prisms and cylinders are general cylinders.
  • parallelepiped
    V = S · h (a face × the height to it)
  • slanted edge
    V = S · c · sin ψ
  • 3 edges
    V = a · b · c · sin φ · sin ψ
Problems
9.1Find the parallelepiped with the largest volume if the sum of the lengths of all its edges is 12.a cube with edge 1
  1. 4a + 4b + 4c = 12 → a + b + c = 3.
  2. V = abc·sinφ·sinψ ≤ abc, with equality when φ = ψ = 90° (a cuboid).
  3. The AM–GM inequality: ∛(abc) ≤ (a + b + c)/3 → abc ≤ ((a+b+c)/3)³ = 1.
  4. Equality when a = b = c → a cube with edge 1, V = 1.
9.22 opposite faces are squares with side 1. The edge joining them is 1 and makes a 60° angle. Volume?√3/2
  1. S = 1, h = 1·sin60° = √3/2.
  2. V = √3/2.
9.3One face is a rhombus with side 1 and angle 60°; an edge of length 1 makes 60° with that face. Volume?3/4
  1. V = abc·sinφ·sinψ = 1·1·1·sin60°·sin60°.
  2. V = √3/2 · √3/2 = 3/4.
9.43 faces with a common vertex O are rhombi with side 1 and acute angle 60°. Volume?√2/2
  1. O is the common vertex and OX is the third edge (length 1). A is the projection of X onto the base, B is the projection of X onto the edge OY. XA ⊥ the base and XB ⊥ OB ⇒ AB ⊥ OB (three perpendiculars).
  2. XB = sin60° = √3/2, OB = cos60° = ½. ∠XOY = ∠XOZ ⇒ A lies on the bisector ⇒ ∠AOB = 30° ⇒ OA = 2AB.
  3. OA² = AB² + ¼ → 4AB² = AB² + ¼ → AB² = 1/12.
  4. XA² = XB² − AB² = ¾ − 1/12 = 2/3 → H = √(2/3) = √6/3.
  5. V = B·H = 1·1·sin60°·√6/3 = √2/2.
9.5Two lead cubes have edges 12 and 5. They are melted and the metal is made into a new cube. What is its edge?∛1853 ≈ 12.28

Melting preserves the volume (not the surface area!).

  1. V₁ = 12³ = 1728, V₂ = 5³ = 125.
  2. V = 1728 + 125 = 1853 = a³.
  3. a = ∛1853 ≈ 12.28. (Compare with the problem about surface areas, where the answer is 13.)
More problems
9.62 neighbouring faces with areas S₁ and S₂, a common edge a, dihedral angle 150°S₁S₂/(2a)
  1. The height of the face S₂ to the edge a: S₂/a.
  2. The height of the solid to the face S₁: h = (S₂/a)·sin150° = S₂/(2a).
  3. V = S₁·h = S₁S₂/(2a).
10

Prism: surface area and volume

40 min

The surface area of a prism is the sum of the areas of its faces, P = 2B + M (B = base, M = lateral surface). V = B · H. Almost every problem comes down to a right triangle with the height H.

Video lesson
The diagonal of a prism

Areas of bases

Squarea²Equilateral △a²√3/4
Rectanglea·bRegular hexagon3a²√3/2
Parallelograma·b·sin αTrapezoid(a+b)/2 · h
Rhombusd₁·d₂/2Heron√(s(s−a)(s−b)(s−c))

Ready formulas

CubeP = 6a², V = a³, D = a√3
CuboidD = √(a²+b²+c²), V = abc
Regular triangularP = a²√3/2 + 3aH
Regular squareP = 2a² + 4aH, V = a²H
Regular hexagonalP = 3a²√3 + 6aH
Oblique prismV = Q · s (Q = the perpendicular section)
Problems
10.1A cuboid with base edges a and b has a diagonal that makes 60° with the plane of the base. Find P and V.c = √3·√(a²+b²)

We have the space diagonal, so we need to find the third edge.

  1. The diagonal of the base d = √(a² + b²).
  2. The right △ (c, d, D) with a 60° angle at the base: c/d = tan60° = √3 → c = √3·√(a² + b²) = √(3a² + 3b²).
  3. V = abc = ab√(3a² + 3b²).
  4. P = 2ab + 2(a + b)c = 2ab + 2(a + b)√(3a² + 3b²).
10.2The base of a right prism is a trapezoid. Prove that V equals the mean of the areas of the two parallel lateral faces times the distance between them.V = (P₁+P₂)/2 · d
  1. V = B·H, B = (a + b)/2 · h → V = (a + b)/2 · h · H.
  2. The parallel lateral faces: P₁ = aH, P₂ = bH → (P₁ + P₂)/2 = (a + b)/2 · H.
  3. The distance between those faces is the height of the trapezoid: d = h.
  4. (P₁ + P₂)/2 · d = (a + b)/2 · H · h = V. ∎
10.3There are 2 equal cubes with edge a. If the first one is turned through 90° about the midline of one of its faces, it fits onto the other. The volume of the common part?a³/4
  1. The midline of the face passes through the midpoints of 2 opposite edges.
  2. After the turn the cube is shifted by half an edge in 2 directions.
  3. The common part is an a × a/2 × a/2 cuboid.
  4. V = a³/4 = V/4.
10.4The base of a prism is an equilateral triangle with side 4 cm. The 6 cm lateral edge makes 60° with the base. The volume and the area of the section perpendicular to the lateral edge?36 cm³ · 6 cm²
  1. H = 6·sin60° = 3√3, B = 4²√3/4 = 4√3.
  2. V = 4√3 · 3√3 = 36 cm³.
  3. V = Q·s (Q = the perpendicular section) → Q = 36/6 = 6 cm².
10.5Cubes with edges 12 and 5. Find the cube whose surface area is the sum of their surface areas.edge 13
  1. P₁ = 6·12², P₂ = 6·5² = 150. 6x² = 6(144 + 25) → x² = 169.
  2. x = 13, V = 13³ = 2197.
10.6A right prism with P = 334 cm², the base a parallelogram with sides 9 and 10 cm and a diagonal of 17 cm. Volume?360 cm³
  1. The diagonal splits the base into triangles 9, 10, 17. s = (9 + 10 + 17)/2 = 18.
  2. Heron: √(18·9·8·1) = 36 → B = 2·36 = 72.
  3. P = 2B + 2(a + b)H → 334 = 144 + 38H → H = 5.
  4. V = 72·5 = 360 cm³.
More problems
10.7A right parallelepiped, the base a parallelogram 1 and 4 cm, angle 60°. The longer diagonal is 5 cm. V?4√3 cm³

The longer diagonal of the base is opposite the obtuse angle of 120°.

  1. d² = 1² + 4² − 2·1·4·cos120° = 17 + 4 = 21.
  2. D² = d² + H² → 25 = 21 + H² → H = 2.
  3. B = 1·4·sin60° = 2√3 → V = 2√3·2 = 4√3 ≈ 6.93 cm³.
10.8The diagonal section of a cuboid is a square of 1 m², a : b = 3 : 4. P and V?3.76 m² · 0.48 m³
  1. The diagonal section = a d × H rectangle. A square → d = H = 1.
  2. a = 3k, b = 4k → d = 5k = 1 → k = 0.2 → a = 0.6, b = 0.8.
  3. P = 2(0.48 + 0.6 + 0.8) = 3.76 m², V = 0.6·0.8·1 = 0.48 m³.
10.9A regular square prism, M = 12√6 cm², the diagonal is inclined to the base at 30°P = 18+12√6, V = 9√6
  1. The triangle (H, d = a√2, D): tan30° = H/(a√2) → H = a√6/3.
  2. M = 4aH = 4a²√6/3 = 12√6 → a² = 9 → a = 3, H = √6.
  3. P = 2·9 + 12√6 = 18 + 12√6 cm², V = 9√6 ≈ 22.05 cm³.
10.10A regular square prism: the diagonal makes 30° with a lateral face, edge a. V?a³√2
  1. The distance from A to the face BCC₁B₁ is AB = a → sin30° = a/D → D = 2a.
  2. D² = 2a² + H² → 4a² = 2a² + H² → H = a√2.
  3. V = a²·a√2 = a³√2.
10.11A right triangular prism H = 5 dm, lateral faces 125, 85, 140 dm². V?1050 dm³
  1. The sides of the base: 125/5 = 25, 85/5 = 17, 140/5 = 28.
  2. Heron: s = 35, B = √(35·10·18·7) = 210.
  3. V = 210·5 = 1050 dm³.
10.12The base is an isosceles triangle with legs 5; H = 10, V = 120. P?204 or 184
  1. B = V/H = 12. Let the base be 2x: B = x·√(25 − x²) = 12.
  2. x⁴ − 25x² + 144 = 0 → x² = 16 or 9 → base 8 or 6.
  3. Base 8: P = 24 + 10·18 = 204. Base 6: P = 24 + 10·16 = 184. (Two solutions!)
10.13A regular triangular prism with edge a; the lateral surface is half of the total surface. V?a³/8
  1. M = P/2 → M = 2B → 3aH = 2·a²√3/4 → H = a√3/6.
  2. V = a²√3/4 · a√3/6 = a³/8.
10.14A right triangular prism: B = 4 cm², lateral faces 9, 10, 17 cm². V?12 cm³
  1. The sides of the base: 9/H, 10/H, 17/H — that is △(9, 10, 17) scaled down H times.
  2. △(9, 10, 17) has area 36 → 36/H² = 4 → H = 3.
  3. V = 4·3 = 12 cm³.
10.15A regular hexagonal prism; the longer diagonal d makes 30° with a lateral edge. V?9d³/64
  1. The projection of the diagonal = 2a (the long diagonal of the hexagon).
  2. H = d·cos30° = d√3/2; 2a = d·sin30° → a = d/4.
  3. V = 3a²√3/2 · H = 9d³/64.
11

Pyramid: surface area and volume

40 min

P = B + M, V = ⅓ · B · H (a pyramid = a third of a prism with the same base and height). The secret: a regular pyramid contains 3 right triangles.

Video lesson
The three triangles in a pyramid
Base:
Triangle:

r and R for the base

BaserR
Equilateral △a√3/6a√3/3
Squarea/2a√2/2
Regular hexagona√3/2a

Angle of a lateral face → (H, r, h). Angle of a lateral edge → (H, R, s).

Ready formulas

Regular squareP = a² + 2ah, V = ⅓a²H
TetrahedronH = a√6/3, V = a³√2/12
FrustumV = H/3·(B₁ + √(B₁B₂) + B₂)
Section ∥ baseS/B = (x/H)²
Square pyramid
11.1A regular square pyramid: M = 135 cm², the base edge is 50% longer than the height. V?162 cm³
  1. a = 1.5H. h² = H² + (a/2)² = H² + 0.5625H² → h = 1.25H.
  2. M = 4·a·h/2 = 2ah = 3.75H² = 135 → H = 6, a = 9.
  3. V = ⅓·81·6 = 162 cm³.
11.2The base edge is 1 cm longer than the height, the lateral edge is 9 cm. V?448/3 cm³
  1. The triangle (H, R, s): s² = H² + a²/2 (because R² = a²/2).
  2. 81 = H² + (H+1)²/2 → 3H² + 2H − 161 = 0 → H = 7, a = 8.
  3. V = ⅓·64·7 = 448/3 ≈ 149.3 cm³.
11.3Height H; one lateral face has the same area as the base. P and V?4H²/3 · 4H³/45
  1. a·h/2 = a² → h = 2a. h² = H² + a²/4 → a² = 4H²/15.
  2. P = a² + 4a² = 5a² = 4H²/3.
  3. V = ⅓a²H = 4H³/45.
11.4The base is an isosceles trapezoid 3 and 5 cm with legs 7; the foot of the height = the intersection of the diagonals; the longer lateral edge is 10. V?80 cm³
  1. The height of the trapezoid √(7² − 1²) = 4√3; the diagonal √(4² + 48) = 8.
  2. The intersection divides the diagonal 5 : 3 → the longer part = 5.
  3. H = √(10² − 5²) = 5√3, B = (3+5)/2·4√3 = 16√3 → V = ⅓·16√3·5√3 = 80 cm³.
11.5Edge a; a lateral face makes 60° with the base. P and V?3a² · a³√3/6
  1. (H, r = a/2, h): h = r/cos60° = a, H = r·tan60° = a√3/2.
  2. P = a² + 2a·a = 3a².
  3. V = ⅓a²·a√3/2 = a³√3/6.
11.6Height H, diagonal section Q. V? (H = 8, Q = 60)2Q²/(3H) = 300
  1. Q = ½·a√2·H → a² = 2Q²/H².
  2. V = ⅓a²H = 2Q²/(3H). For H = 8, Q = 60: V = 7200/24 = 300.
11.7A cube of edge a: the apex = the centre of a face, the base = the midpoints of the sides of the opposite face. P and V?2a² · a³/6
  1. Base: a square with side a√2/2 → B = a²/2. H = a.
  2. Slant height h = √(a² + (a√2/4)²) = 3a√2/4 → M = 4·½·(a√2/2)(3a√2/4) = 3a²/2.
  3. P = a²/2 + 3a²/2 = 2a², V = ⅓·(a²/2)·a = a³/6.
11.8a = 12, s = 2√34. An inscribed cube (its top vertices on the lateral edges). The edge of the cube and V above it?4.8 cm · 24.576 cm³
  1. s² = H² + a²/2 → 136 = H² + 72 → H = 8.
  2. At height x the section is a square a·(H − x)/H. Cube: x = 12(8 − x)/8 → x = 4.8 cm.
  3. Above the cube: a pyramid with base 4.8² and height 3.2 → V = ⅓·23.04·3.2 = 24.576 cm³.
11.9The lateral edge s is inclined to the base at an angle α. V?⅔s³cos²α·sinα
  1. (H, R, s): H = s·sinα, R = s·cosα, a = R√2 → a² = 2s²cos²α.
  2. V = ⅓·2s²cos²α·s·sinα = ⅔s³cos²α·sinα.
11.10Diagonal section Q, lateral edges at 45°. V?2Q√Q/3
  1. 45° → H = R = d/2. Q = ½·d·H = d²/4 → d = 2√Q, H = √Q.
  2. a² = d²/2 = 2Q → V = ⅓·2Q·√Q = 2Q√Q/3.
11.11Edge a; the base and the diagonal section have equal areas. M and V?3a² · a³√2/3
  1. a² = ½·a√2·H → H = a√2. h = √(2a² + a²/4) = 3a/2.
  2. M = 2ah = 3a², V = ⅓a²·a√2 = a³√2/3.
11.12The base is a parallelogram a = 10, b = 18, B = 90; the foot of the height = the intersection of the diagonals, H = 6. M?192 cm²
  1. The heights of the parallelogram: 90/10 = 9 and 90/18 = 5 → from the centre to the sides: 4.5 and 2.5.
  2. Slant heights: √(36 + 4.5²) = 7.5 and √(36 + 2.5²) = 6.5.
  3. M = 2·½·10·7.5 + 2·½·18·6.5 = 75 + 117 = 192 cm².
Triangular, hexagonal and truncated pyramids
11.13A regular triangular pyramid: P = 8√3, the height is twice the base edge. V?4√3/3
  1. H = 2a, r = a√3/6 → h = √(4a² + a²/12) = 7a√3/6.
  2. P = a²√3/4 + 3·½·a·h = a²√3/4 + 7a²√3/4 = 2a²√3 = 8√3 → a = 2, H = 4.
  3. V = ⅓·√3·4 = 4√3/3 ≈ 2.31.
11.14Derive V of a regular tetrahedron. The ratio cube : tetrahedron AB₁CD₁a³√2/12 · 3 : 1
  1. H = a√6/3, B = a²√3/4 → V = ⅓·(a²√3/4)·(a√6/3) = a³√2/12.
  2. AB₁CD₁ = the cube without 4 “corners”, each with V = ⅓·(a²/2)·a = a³/6 → a³ − 4a³/6 = a³/3 → ratio 3 : 1.
11.15The surface area of a regular tetrahedron with height H3√3H²/2
  1. H = a√6/3 → a² = 3H²/2.
  2. P = a²√3 = 3√3H²/2.
11.16A regular triangular pyramid, edge a, lateral faces at 60°. P?3a²√3/4
  1. r = a√3/6, h = r/cos60° = a√3/3.
  2. P = a²√3/4 + 3·½·a·a√3/3 = 3a²√3/4.
11.17The base is an isosceles triangle with base 6 and height 9; all lateral edges are 13. V?108 cm³

Equal lateral edges → the foot of the height is the centre of the circumscribed circle (R).

  1. Leg = √(3² + 9²) = 3√10, B = 27, R = abc/(4B) = 6·90/108 = 5.
  2. H = √(13² − 5²) = 12 → V = ⅓·27·12 = 108 cm³.
11.18A regular hexagonal pyramid, a = 2; the lateral surface is 3 times the base. P and V?24√3 · 12√2
  1. B = 3a²√3/2 = 6√3, M = 18√3 = 6·½·2·h → h = 3√3.
  2. r = a√3/2 = √3 → H = √(27 − 3) = 2√6.
  3. P = 24√3, V = ⅓·6√3·2√6 = 12√2 ≈ 16.97.
11.19The interior angles of the base add up to 720°; the lateral edge s makes 30° with the height. V?3s³/16
  1. (n−2)·180° = 720° → n = 6 (a hexagon, R = a).
  2. H = s·cos30° = s√3/2, a = R = s·sin30° = s/2.
  3. V = ⅓·(3a²√3/2)·H = 3s³/16.
11.20A regular hexagonal pyramid, H = 12 dm; a plane ∥ to the base, 9 dm from it, gives a section of 450 cm². V?288 dm³
  1. The section is 12 − 9 = 3 dm from the apex → ratio 3 : 12 = 1 : 4.
  2. The areas are in the ratio (1/4)² = 1/16 → B = 16·450 = 7200 cm² = 72 dm².
  3. V = ⅓·72·12 = 288 dm³.
11.21A regular square frustum: a = 13, b = 1, s = 10. P and V?394 cm² · 122√7 cm³
  1. A lateral face = an isosceles trapezoid: h′ = √(10² − 6²) = 8 → M = 4·(13+1)/2·8 = 224.
  2. P = 169 + 1 + 224 = 394 cm².
  3. H = √(8² − 6²) = 2√7 → V = (2√7/3)(169 + 13 + 1) = 122√7 ≈ 322.8 cm³.
11.22A regular square frustum: a = 12, b = 6, the lateral surface is half of the total surface. V?336 cm³
  1. M = B₁ + B₂ = 144 + 36 = 180 = 4·(12+6)/2·h′ → h′ = 5.
  2. H = √(5² − 3²) = 4.
  3. V = (4/3)(144 + 72 + 36) = 336 cm³.
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Summary sheet

everything in one place

Axioms and statements

2 points
1 line
3 non-collinear
1 plane
S1
2 points in α ⇒ the line ⊂ α
S2
line + point ⇒ 1 plane
S3
2 intersecting lines ⇒ 1 plane
“∥” planes
equivalence

Counting

prism
2n · 3n · n+2
pyramid
n+1 · 2n · n+1
Euler
V − E + F = 2
nets of a cube
11
n points→lines
n(n−1)/2
parts
2, 4, 8, 15

Pairs in a cube

parallel
18
skew
24
lines through vertices skew to AB
12
hexagonal prism ∥
33
tetrahedron skew
3

Shortest path

cube a
a√5
cuboid
√((small+mid)² + large²)
triangular prism, 1
√(7+2√3)/2
hexagonal, 1 and 2
√13
pyramid, 1
√7/2

Projection and angles

ratio
AB : BC = A′B′ : B′C′
segment
c = (na + mb)/(m+n)
line–plane
∠(p, p′), the smallest angle
3 perpendiculars
q ⊥ A′B ⇒ q ⊥ AB
dihedral
p₁ ⊥ p, p₂ ⊥ p

Vectors

chain
AB + BC = AC
parallelogram
AB + AD = AC
parallelepiped
AC₁ = AB + AD + AA₁
zero
AA = 0

Volume

parallelepiped
V = abc·sinφ·sinψ
melting
V = V₁ + V₂
max V, edges 12
cube 1
rotated cube
a³/4

Prism

all
P = 2B + M, V = B·H
diag.
D² = d² + H², tanφ = H/d
△
B = a²√3/4
hexagon
B = 3a²√3/2

Pyramid

all
P = B + M, V = ⅓BH
△ 1
h² = H² + r²
△ 2
s² = H² + R²
△ 3
s² = h² + (a/2)²
tetrahedron
V = a³√2/12