Class 8 Mathematics · Chapter 11 · Part II

Exploring Some Geometric Themes

Fractals, visualising solids, nets of 3-D shapes and views from different sides.

NCERT Ganita Prakash (Part I and Part II)CBSE Class 8 10 min read

1. Chapter at a Glance

Geometry is not only about measuring angles and lengths. It is also about patterns that repeat and about seeing solid shapes in our mind. This chapter explores two such themes. First, fractals — shapes built by repeating the same step again and again, so that each small part looks like the whole. Second, visualising 3-D solids — drawing their nets, imagining their views from the front, side and top, and counting their faces, edges and vertices.

What you will learn
  • Self-similar patterns: the Sierpinski triangle and the Koch snowflake.
  • Counting patterns in fractals stage by stage.
  • Faces, edges and vertices of polyhedra; Euler's formula.
  • Nets of cubes, cuboids, prisms and pyramids.
  • Front, side and top views of solids.

2. Fractals — Patterns Inside Patterns

A fractal is a shape in which a small part, when enlarged, looks like the whole shape. This property is called self-similarity. Nature is full of near-fractals: the branching of trees, the florets of a cauliflower, fern leaves, river networks and coastlines.

The Sierpinski triangle

  1. Stage 0: Start with a shaded equilateral triangle.
  2. Stage 1: Join the midpoints of its sides. This makes 4 smaller triangles; remove (unshade) the middle one. 3 shaded triangles remain.
  3. Stage 2: Repeat the step in each of the 3 shaded triangles → 9 shaded triangles.
  4. Keep repeating.
Stage 0 Stage 1 Stage 2
Sierpinski triangle: at each stage every shaded triangle is split into four and the middle one is removed.
StageShaded trianglesSide of each (original = 1)Fraction of area shaded
0111
131/23/4
291/49/16
3271/827/64
n3ⁿ(1/2)ⁿ(3/4)ⁿ

The shaded area keeps shrinking (it gets closer and closer to 0), yet the number of triangles keeps growing — a surprising feature of fractals.

The Koch snowflake

  1. Start with an equilateral triangle.
  2. Divide each side into 3 equal parts. On the middle part, build a small equilateral triangle pointing outward and remove its base.
  3. Each side has now become 4 segments, each 1/3 as long. Repeat on every segment.
1 segment 4 segments, each ⅓
One Koch step: each segment is replaced by 4 segments, each one-third as long.
StageNumber of sidesLength of each (start = 1)Perimeter
0313
1121/34
2481/916/3
n3 × 4ⁿ(1/3)ⁿ3 × (4/3)ⁿ

The perimeter is multiplied by 4/3 at every stage, so it grows without limit, while the snowflake always fits inside a fixed circle.

3. Faces, Edges and Vertices

A polyhedron is a solid whose faces are all flat polygons. Every polyhedron has faces (F), edges (E) where two faces meet, and vertices (V) — the corners.

SolidFVEF + V − E
Cube / cuboid68122
Triangular prism5692
Square pyramid5582
Triangular pyramid (tetrahedron)4462
Pentagonal prism710152
Octahedron86122
Euler's formula

For every convex polyhedron: F + V − E = 2.

Patterns for prisms and pyramids (base with n sides):

  • n-sided prism: F = n + 2, V = 2n, E = 3n.
  • n-sided pyramid: F = n + 1, V = n + 1, E = 2n.

Cylinders, cones and spheres have curved surfaces, so they are not polyhedra.

4. Nets of Solids

A net is a flat shape that can be folded to make a solid. Cutting along some edges of a box and opening it out gives a net.

Net of a cube Net of a square pyramid
A cube net (6 squares) and a square-pyramid net (1 square + 4 triangles).
  • A cube has 11 different nets. Not every arrangement of 6 squares folds into a cube — for example, 6 squares in a single row cannot.
  • Cuboid: 6 rectangles in 3 matching pairs.
  • Triangular prism: 2 triangles + 3 rectangles.
  • Cylinder: 2 circles + 1 rectangle (whose length equals the circle's circumference).
  • Cone: 1 circle + 1 sector of a larger circle.
Checking a cube net

In a valid cube net, opposite faces are never next to each other. In a straight run of three squares, the 1st and 3rd are opposite faces. A dice uses this: opposite faces add to 7 (1–6, 2–5, 3–4).

5. Views of Solids

Architects and engineers describe a solid by drawing how it looks from three directions:

  • Top view (plan) — looking straight down.
  • Front view — looking from the front.
  • Side view — looking from the left or right.
SolidTop viewFront viewSide view
CubeSquareSquareSquare
Cylinder (standing)CircleRectangleRectangle
Cone (standing)Circle (with centre dot)TriangleTriangle
Square pyramidSquare with diagonalsTriangleTriangle
SphereCircleCircleCircle

Different solids can share one view (a cylinder and a cone both look like a circle from the top), which is why we need more than one view to identify a solid.

6. Solved Problems (NCERT Style)

Problem 1 — Euler check

A polyhedron has 12 faces and 30 edges. How many vertices does it have?

Solution: F + V − E = 2 → 12 + V − 30 = 2 → V = 20. (This is the dodecahedron.)

Problem 2 — Hexagonal prism

Find F, V and E for a hexagonal prism.

Solution: n = 6: F = 8, V = 12, E = 18. Check: 8 + 12 − 18 = 2 ✓.

Problem 3 — Sierpinski stage 4

How many shaded triangles are in stage 4, and what fraction of the original area do they cover?

Solution: 3⁴ = 81 triangles, covering (3/4)⁴ = 81/256 of the area.

7. Master Formula Table

ResultFormula
Euler's formulaF + V − E = 2
n-gonal prismF = n + 2, V = 2n, E = 3n
n-gonal pyramidF = n + 1, V = n + 1, E = 2n
Sierpinski stage n3ⁿ triangles, area (3/4)ⁿ
Koch stage n3 × 4ⁿ sides, perimeter 3 × (4/3)ⁿ
Cube nets11 distinct nets

8. Master Summary Table

TopicKey ideaExam focus
FractalsSelf-similar repeated stepsCount triangles/sides at stage n
PolyhedraFlat faces, straight edgesFind F, V, E
Euler's formulaF + V − E = 2Find a missing value; test if a solid can exist
NetsFold flat shapes into solidsIdentify valid nets; opposite faces
ViewsTop, front, sideMatch solids with views

Chart — shaded triangles in the Sierpinski triangle:

Stage 13
Stage 29
Stage 327
Stage 481

9. Mind Map

Geometric Themes
Fractals
  • Sierpinski: 3ⁿ
  • Koch: ×4/3 perimeter
Polyhedra
  • F, V, E
  • F + V − E = 2
Nets
  • 11 cube nets
  • Opposite faces
Views
  • Top / front / side
  • Need more than one

10. Tips and Tricks

Smart ideas
  • For fractals, make a table of the first 3 stages and look for the multiplying pattern.
  • Use Euler's formula to check your counting of F, V and E.
  • To test a cube net, mentally fold it: pick a base square and fold the neighbours up.
Common mistakes
  • Applying Euler's formula to a cylinder, cone or sphere (they are not polyhedra).
  • Counting the hidden edges twice or forgetting the base edges of a pyramid.
  • Thinking any 6 joined squares make a cube net.

11. Important One-liners

  • A fractal is self-similar: parts look like the whole.
  • Sierpinski stage n has 3ⁿ shaded triangles.
  • The Koch snowflake's perimeter grows by a factor 4/3 each stage.
  • For any convex polyhedron, F + V − E = 2.
  • A cube has 6 faces, 8 vertices and 12 edges.
  • A cube has exactly 11 different nets.
  • On a standard dice, opposite faces add to 7.

12. Exam-pattern Questions with Detailed Answers

Q1Easy

How many edges does a triangular prism have?

  1. 6
  2. 8
  3. 9
  4. 12
Show answer

E = 3n = 3 × 3 = 9.

Answer: (C) 9

Q2Easy

Which solid has a circle as its top view and a triangle as its front view?

Show answer

A standing cone.

Answer: Cone

Q3Easy

Which of these is NOT a polyhedron: cube, cylinder, square pyramid, triangular prism?

Show answer

A cylinder has curved surfaces.

Answer: Cylinder

Q4Medium

A polyhedron has 8 faces and 6 vertices. How many edges does it have?

Show answer

8 + 6 − E = 2 → E = 12 (octahedron).

Answer: 12

Q5Medium

How many shaded triangles are there at stage 5 of the Sierpinski triangle?

Show answer

3⁵ = 243.

Answer: 243

Q6Medium

Find F, V and E for an octagonal pyramid and verify Euler's formula.

Show answer

n = 8: F = 9, V = 9, E = 16. 9 + 9 − 16 = 2 ✓.

Answer: F = 9, V = 9, E = 16

Q7Medium

On a dice, the face showing 2 is on top. Which number is at the bottom?

Show answer

Opposite faces add to 7, so 7 − 2 = 5.

Answer: 5

Q8Medium

Starting with an equilateral triangle of side 9 cm, find the perimeter of the Koch snowflake at stage 2.

Show answer

Stage 0 perimeter 27 cm. Each stage multiplies by 4/3: 27 × 4/3 = 36, then 36 × 4/3 = 48 cm.

Answer: 48 cm

Q9Hard

A polyhedron has 20 faces, and every face is a triangle. Each edge is shared by exactly two faces. Find the number of edges and vertices.

Show answer

Counting edges face by face gives 20 × 3 = 60, but each edge is counted twice, so E = 30. Euler: 20 + V − 30 = 2 → V = 12. (This solid is the icosahedron.)

Answer: 30 edges, 12 vertices

Q10Hard

A prism has 24 edges. How many faces and vertices does it have, and what shape is its base?

Show answer

E = 3n = 24 → n = 8. F = 10, V = 16. Base is an octagon. Check: 10 + 16 − 24 = 2 ✓.

Answer: 10 faces, 16 vertices, octagonal base

Q11Hard

In the Sierpinski triangle, after which stage does the shaded area first fall below half of the original?

Show answer

(3/4)¹ = 0.75, (3/4)² = 0.5625, (3/4)³ ≈ 0.42. So after stage 3.

Answer: Stage 3

Q12Hard

Six squares are placed in a row of four with one square attached above the second and one below the third. Does this fold into a cube? Which square is opposite the first one in the row?

Show answer

Yes — a row of four with one flap on each side (on different squares) is a valid cube net. In the row of four, squares 1 and 3 are opposite, and squares 2 and 4 are opposite; the two flaps are opposite each other.

Answer: Yes; the 3rd square in the row is opposite the 1st

13. Quick Revision Summary

Revise in 2 minutes
  • Fractals are self-similar; Sierpinski: 3ⁿ triangles, area (3/4)ⁿ; Koch: perimeter × 4/3 each stage.
  • Polyhedra have flat faces; Euler: F + V − E = 2.
  • Prism: F = n + 2, V = 2n, E = 3n; Pyramid: F = V = n + 1, E = 2n.
  • A cube has 11 nets; opposite faces never touch in a net.
  • Top, front and side views together identify a solid.