MathsCBSE Class 8 · Chapter 11

Exploring Some Geometric Themes Class 8: Fractals, Nets, Shortest Paths, Projections and Isometric Drawing

Updated 15 October 2026 · Fact-checked

Exploring Some Geometric Themes is Chapter 4 of Ganita Prakash Part II for Class 8 (Chapter 11 overall). It studies fractals such as the Sierpinski carpet and gasket and the Koch snowflake, then visualising solids: faces, edges and vertices, nets, shortest paths on a box, front, top and side views, and isometric drawing.

What Exploring Some Geometric Themes is about

This chapter has two themes.

4.1 Fractals are self-similar shapes: the same pattern repeats at smaller and smaller scales, as in a fern leaf, tree branches, coastlines and lightning. You will build:

the Sierpinski carpet (cut a square into 9 and remove the middle, again and again);
the Sierpinski gasket (join the midpoints of an equilateral triangle and remove the middle one);
the Koch snowflake (replace the middle third of every side by a bump);

and count squares, holes, sides, perimeters and areas at each step. Fractal patterns also appear in art, from Indian temple towers such as the Kandariya Mahadev Temple to Fulani wedding blankets and prints inspired by M.C. Escher.

4.2 Visualising Solids trains your mind's eye:

profiles of solids from different viewpoints;
faces, edges and vertices of prisms and pyramids;
nets that fold into cubes, cuboids, tetrahedrons, prisms, pyramids, cylinders, cones and octahedrons;
the shortest path for an ant on a box, found by unfolding it and using the Baudhayana-Pythagoras theorem;
projections: front, top and side views, and shadows;
isometric projections and drawing on isometric grid paper.

School exams ask fractal counts and patterns, face/edge/vertex counts, net recognition, shortest paths and three views of solids. This page covers each idea with an example and gives exam-level and harder questions with solutions.

Key concepts

Fractals and the Sierpinski carpet

NCERT section 4.1 Fractals

A fractal is a self-similar shape: zoom into a part and you see a copy of the whole. Nature has many (ferns, trees, clouds, coastlines, mountains, lightning).

Sierpinski carpet: start with a square (Step 0). Divide it into 9 equal squares and remove the centre one (Step 1). Repeat on each remaining square, and so on.

Let $R_{n}$ be the number of squares remaining and $H_{n}$ the number of holes at Step $n$ .

Each remaining square produces 8 remaining squares: $R_{n + 1} = 8 R_{n}$ , so $R_{n} = 8^{n}$ .
Each remaining square also produces one new hole, and old holes stay: $H_{n + 1} = H_{n} + R_{n}$ , so $H_{n} = 1 + 8 + 8^{2} + \dots + 8^{n - 1}$ .
Each step keeps $\frac{8}{9}$ of the area, so with a starting area of 1, the remaining area is $(\frac{8}{9})^{n}$ .

Step	0	1	2	3
Remaining squares	1	8	64	512
Holes	0	1	9	73

Worked example

For the Sierpinski carpet, find the number of remaining squares, the number of holes, and the remaining area at Step 3 (starting area 1 square unit).

Try it first, then show the solutionSolution

$R_{3} = 8^{3} = 512$ .
$H_{3} = 1 + 8 + 64 = 73$ .
Area $= (\frac{8}{9})^{3} = \frac{512}{729} \approx 0.70$ square units.

Answer: 512 squares, 73 holes, area $\frac{512}{729}$ .

Sierpinski gasket and Koch snowflake

NCERT section 4.1 Fractals

Sierpinski gasket (triangle): join the midpoints of the sides of an equilateral triangle. This splits it into 4 identical equilateral triangles (each corner triangle is isosceles with a 60° angle, so equilateral). Remove the middle one and repeat on the 3 that remain.

Remaining triangles: $3^{n}$ ; holes: $1 + 3 + 9 + \dots + 3^{n - 1}$ ; remaining area: $(\frac{3}{4})^{n}$ .

Koch snowflake (Von Koch, 1904): start with an equilateral triangle. Divide each side into three equal parts, raise an equilateral triangle on the middle part, and remove that middle part, so every side becomes a 'bump' of 4 pieces, each $\frac{1}{3}$ as long.

Number of sides at Step $n$ : $3 \times 4^{n}$ .
With starting side 1, each side has length $\frac{1}{3 ^{n}}$ , so the perimeter is $3 \times (\frac{4}{3})^{n}$ , which keeps growing without limit.

The snowflake's perimeter grows forever while the shape stays inside a fixed region: a surprising property of fractals.

Worked example

A Koch snowflake starts from an equilateral triangle of side 9 cm. Find the number of sides and the perimeter at Step 2.

Try it first, then show the solutionSolution

Sides: $3 \times 4^{2} = 48$ .
Each side is $\frac{9}{3 ^{2}} = 1$ cm long.
Perimeter $= 48 \times 1 = 48$ cm (check: $27 \times (\frac{4}{3})^{2} = 27 \times \frac{16}{9} = 48$ ).

Answer: 48 sides; perimeter 48 cm.

Fractals in art and architecture

NCERT section 4.1 Fractals

Fractal patterns have long been used in human-made art.

Indian temples: the tower of the Kandariya Mahadev Temple at Khajuraho (completed around 1025 CE) is built from smaller copies of the whole tower, which carry even smaller copies. Fractal-like patterns also appear in temples at Madurai, Hampi, Rameswaram and Varanasi.
African textiles: some Fulani wedding blankets from Nigeria show diamonds inside diamonds inside diamonds.
Modern art: the Dutch artist M.C. Escher made prints such as 'Smaller and Smaller', where the same lizard pattern repeats at shrinking scales.

What makes these fractal-like is self-similarity: the whole is made of smaller copies of itself, repeated over several levels.

Worked example

A rangoli designer draws a large square, then draws the same design of 8 smaller squares inside it (like Step 1 of the Sierpinski carpet), and repeats it inside each small square once more. How many smallest squares are there, and is the design a true fractal?

Try it first, then show the solutionSolution

Level 1 gives 8 squares; repeating inside each gives $8 \times 8 = 64$ smallest squares.
The design is self-similar over 2 levels.
A true mathematical fractal repeats forever; a drawing can only show a few steps, so it is fractal-like.

Answer: 64 smallest squares; fractal-like (only a mathematical fractal repeats endlessly).

Faces, edges and vertices of solids

NCERT section 4.2 Visualising Solids

For solids with flat faces:

faces are the flat surfaces on the boundary;
edges are the line segments where faces meet;
vertices are the corner points where edges meet.

A cube or cuboid has 6 faces, 12 edges and 8 vertices.

Prism: two congruent polygons as opposite faces, joined by parallelogram faces. For an $n$ -sided base: ** $n + 2$ faces, $3 n$ edges, $2 n$ vertices**. (Triangular prism: 5, 9, 6.)

Pyramid: a polygon base and one apex joined to each base vertex. For an $n$ -sided base: ** $n + 1$ faces, $2 n$ edges, $n + 1$ vertices. A triangular pyramid is a tetrahedron** (4, 6, 4).

Other solids: an octahedron (two square pyramids joined at their bases) has 8 triangular faces; a dodecahedron has 12 pentagonal faces.

Worked example

Find the number of faces, edges and vertices of (a) a prism whose base has 10 sides, (b) a pyramid whose base has 5 sides.

Try it first, then show the solutionSolution

(a) Faces $= 10 + 2 = 12$ ; edges $= 3 \times 10 = 30$ (10 on each base and 10 joining them); vertices $= 2 \times 10 = 20$ .
(b) Faces $= 5 + 1 = 6$ ; edges $= 2 \times 5 = 10$ (5 on the base and 5 to the apex); vertices $= 5 + 1 = 6$ .

Answer: (a) 12, 30, 20 (b) 6, 10, 6

Nets of solids

NCERT section 4.2 Visualising Solids

A net is a flat shape that folds up into a solid, as if the solid were unfolded onto a plane. (Extra flaps used for gluing are not part of the net.)

Cube: 6 squares. There are exactly 11 different nets (counting rotations and flips as the same). A row of 6 squares, or a 'T' made of 5 squares plus one in the wrong place, will not fold into a cube; a cross shape will.
Cuboid: 6 rectangles in three matching pairs.
Regular tetrahedron: 4 equilateral triangles; it has only 2 nets (a large triangle split into 4, and a strip of 4).
Square pyramid: a square with 4 triangles on its sides.
Triangular prism: 2 triangles and 3 rectangles.
Cylinder: a rectangle (its length equals the circumference of the base, its width the height) and 2 circles.
Cone: a sector of a circle (the curved surface) and a circle (the base).
Octahedron: 8 equilateral triangles; it also has 11 nets. A dodecahedron has 43,380 nets.
A sphere has no net: no flat paper can wrap a ball without wrinkles, gaps or overlaps.

Worked example

A cylinder has radius 7 cm and height 10 cm. Describe its net and give the dimensions of the rectangle (use $π = \frac{22}{7}$ ).

Try it first, then show the solutionSolution

Cutting the curved surface along its height and flattening it gives a rectangle; the two ends are circles of radius 7 cm.
The rectangle's length is the circumference of the base: $2 \times \frac{22}{7} \times 7 = 44$ cm.
Its width is the height of the cylinder, 10 cm.

Answer: A 44 cm by 10 cm rectangle with two circles of radius 7 cm.

Shortest paths on a box

NCERT section 4.2 Visualising Solids

An ant walking on the surface of a cuboid cannot go through it. To find its shortest path, unfold the box into a net: every path on the surface becomes a path of the same length on the net, and the shortest path on a flat net is a straight line.

But the way you unfold matters. Different unfoldings give different straight lines; some lines leave the net and are not real paths. So list the possible unfoldings and compare the lengths, using the Baudhayana-Pythagoras theorem.

For a box of length $l$ , width $w$ and height $h$ , the shortest surface path from a bottom corner to the opposite top corner is the smallest of
$(l + w)^{2} + h^{2}$ , $(l + h)^{2} + w^{2}$ and $(w + h)^{2} + l^{2}$ .

For a cube of side 5 this gives $1 0^{2} + 5^{2} = 125 \approx 11.2$ , shorter than walking along three edges (15).

Worked example

An ant is at one bottom corner of a closed box 8 cm long, 6 cm wide and 4 cm high. Sugar is at the opposite top corner. Find the length of the shortest path over the surface.

Try it first, then show the solutionSolution

Unfolding 1: $(6 + 4)^{2} + 8^{2} = 164 \approx 12.8$ cm.
Unfolding 2: $(8 + 4)^{2} + 6^{2} = 180 \approx 13.4$ cm.
Unfolding 3: $(8 + 6)^{2} + 4^{2} = 212 \approx 14.6$ cm.
The shortest is $164$ cm.

Answer: $164 \approx 12.8$ cm.

Projections: front, top and side views

NCERT section 4.2 Visualising Solids

The projection of a point P on a plane is the point O where the line through P perpendicular to the plane meets it. The projections of all points of an object form its projection (like the hole a solid leaves when pushed straight through a wall).

Facts about projections:

a line's projection is never longer than the line; it is equal only when the line is parallel to the plane (by the Baudhayana-Pythagoras theorem, a 10 cm rod whose ends differ in height by 6 cm projects to 8 cm on the floor);
parallel lines project to parallel lines, so a parallelogram always projects to a parallelogram (or a segment);
different objects can have the same projection, so one view is not enough.

Engineers use three perpendicular views: front view (on the vertical plane), top view (on the horizontal plane) and side view (on the side plane). A cylinder standing upright has a rectangle as front and side views and a circle as top view; an upright cone has triangles as front and side views and a circle as top view.

Shadows in sunlight falling perpendicular to a wall look just like projections, because the Sun is so far away.

Worked example

Name a solid whose top view is a circle and whose front view is a triangle, and one whose top view is a circle and front view a rectangle.

Try it first, then show the solutionSolution

An upright cone: seen from above, its base circle; from the front, a triangle.
An upright cylinder: from above a circle; from the front a rectangle.

Answer: Cone; cylinder.

Isometric projection and isometric drawing

NCERT section 4.2 Visualising Solids

When a cube is balanced on one corner and projected onto the floor, all 12 edges project to equal lengths. This is the isometric projection ('isometric' means 'equal measure'). The outline is a regular hexagon.

Tiling the plane with such hexagons (or equilateral triangles) gives isometric grid paper, widely used in engineering drawing. On it:

vertical lines show height;
the two slanting directions show length and depth;
one unit along any axis is one grid step, so lengths can be measured in all three directions.

This works because parallel edges project to parallel lines, and in the isometric view unit lengths along the three axes look equal.

Drawing tips: draw edge by edge, counting units along each axis; draw faint lines first and darken visible ones. Shapes made of 4 cubes joined face to face (like Tetris pieces) are good practice. Some drawings, like the 'impossible triangle', look solid but cannot be built.

Worked example

On isometric grid paper, describe how to draw a cuboid 3 units long, 2 units deep and 1 unit high.

Try it first, then show the solutionSolution

From a starting dot, draw 3 steps along the length direction, 2 along the depth direction, and complete the bottom parallelogram.
From each visible bottom corner, draw 1 step straight up (height).
Join the tops to form the top face (3 by 2) and leave hidden edges out (or dotted).

Answer: Count 3, 2 and 1 grid steps along the length, depth and height directions.

Maths in real life: Exploring Some Geometric Themes

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Self-similarity

Suppose you look closely at a fern leaf and notice each small leaflet looks like a tiny copy of the whole leaf.

If each leaf carries 8 leaflets and each leaflet carries 8 smaller leaflets, how many smallest leaflets are on one leaf?

Try it first, then show the solutionSolution

Level 1: 8 leaflets.
Level 2: $8 \times 8 = 64$ .

Answer: 64 smallest leaflets.

Memory hook: Each level multiplies by the same number in a fractal.

Shortest path by unfolding

Suppose an ant on a closed shoebox 30 cm long, 20 cm wide and 10 cm high wants to go from one bottom corner to the opposite top corner.

Which unfolding gives the shortest path and how long is it?

Try it first, then show the solutionSolution

Candidates: $(20 + 10)^{2} + 3 0^{2} = 1800 \approx 42.4$ ; $(30 + 10)^{2} + 2 0^{2} = 2000 \approx 44.7$ ; $(30 + 20)^{2} + 1 0^{2} = 2600 \approx 51.0$ .
The shortest unfolds the width and height together.

Answer: About 42.4 cm.

Memory hook: Add the two smaller edges, pair with the longest.

Nets

Suppose you want to make a cube-shaped gift box from one piece of card without any cuts after folding.

What must the flat piece look like, and how many squares does it have?

Try it first, then show the solutionSolution

It must be a net of a cube: 6 equal squares joined edge to edge.
Any of the 11 cube nets works, for example a cross shape.

Answer: One of the 11 cube nets, with 6 squares (plus gluing flaps).

Memory hook: Six squares, each with an opposite partner.

Views of a solid

Suppose an architect shows you only the top view of a water tank: a circle.

Can you be sure of the tank's shape?

Try it first, then show the solutionSolution

A circle could be the top view of a cylinder, a cone or a sphere.
You need the front or side view as well.

Answer: No; more views are needed.

Memory hook: One view can hide the shape; three views reveal it.

Topics in this chapter

Exploring Some Geometric Themes formula sheet

Sierpinski carpet: $R_{n} = 8^{n}, H_{n + 1} = H_{n} + R_{n}, area = (\frac{8}{9})^{n}$
Sierpinski gasket: $R_{n} = 3^{n}, area = (\frac{3}{4})^{n}$
Koch snowflake: $sides = 3 \times 4^{n}, perimeter = 3 \times (\frac{4}{3})^{n}$; Starting side 1.
Prism with n-sided base: $F = n + 2, E = 3 n, V = 2 n$
Pyramid with n-sided base: $F = n + 1, E = 2 n, V = n + 1$
Cube and cuboid: $F = 6, E = 12, V = 8$
Cylinder net rectangle: $2 π r \times h$
Shortest corner-to-corner path on a box: $min ((l + w)^{2} + h^{2}, (l + h)^{2} + w^{2}, (w + h)^{2} + l^{2})$

How the school exam asks Exploring Some Geometric Themes

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Fractal counting	Number of squares, triangles, holes or sides at Step n; perimeter or area at Step n.	Write how one piece produces the next step (×8, ×3, ×4), make a table, then a formula.
Faces, edges, vertices	Count for prisms and pyramids with given bases; find the base from a given count.	Use $n + 2, 3 n, 2 n$ for prisms and $n + 1, 2 n, n + 1$ for pyramids.
Nets	Which figures are nets of a cube; draw nets of a cuboid, tetrahedron, prism, pyramid, cylinder.	Imagine folding; check that no two faces overlap and each face has a partner opposite it.
Shortest paths	Ant and food on a box; compare unfoldings.	Unfold so both points are on one flat net, draw the straight line, use the Baudhayana-Pythagoras theorem, and compare all possible unfoldings.
Views and projections	Draw or match front, top and side views; identify a solid from its views; projection lengths.	Look straight at each face; remember parallel lines stay parallel and lengths can only shrink.
Isometric drawing	Draw cube arrangements on isometric paper; explain why a cube's isometric outline is a hexagon.	Map height to vertical lines and length/depth to the two slants; count grid steps.

Tricks to remember Exploring Some Geometric Themes

Short memory aids. Say each one aloud once, then use it on a question the same day.

Carpet keeps 8 of 9
Carpet: times 8 squares, area times $\frac{8}{9}$ . Gasket: times 3 triangles, area times $\frac{3}{4}$ .
Use it when: Fractal counts and areas.
Koch: four thirds
Each Koch step makes 4 pieces each one-third long, so the perimeter is multiplied by $\frac{4}{3}$ .
Use it when: Snowflake perimeter.
Prism 2-3-2
Prism: faces $n + 2$ , edges $3 n$ , vertices $2 n$ .
Use it when: Prism counts.
Pyramid plus one
Pyramid: faces and vertices are both $n + 1$ ; edges $2 n$ .
Use it when: Pyramid counts.
Flatten, then straighten
For an ant on a box: unfold first, then draw a straight line.
Use it when: Shortest path questions.

Exploring Some Geometric Themes: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Carpet squares · 1 mark · MCQExam level

How many squares remain at Step 2 of the Sierpinski carpet?
(a) 16
(b) 24
(c) 64
(d) 81

Try it first, then show the solutionSolution

Each remaining square gives 8 at the next step.
$8 \times 8 = 64$ .

Answer: (c) 64

Example 2: Prism edges · 1 mark · Assertion-reasonExam level

Assertion (A): A hexagonal prism has 18 edges.
Reason (R): A prism whose base has n sides has 3n edges.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

n edges on each base and n joining edges: $3 n$ .
With $n = 6$ : 18.

Answer: (a)

Example 3: Hexagonal pyramid · 2 marks · Very short answerExam level

Find the number of faces, edges and vertices of a hexagonal pyramid.

Try it first, then show the solutionSolution

Faces: 6 triangles + 1 base = 7; vertices: 6 + 1 = 7.
Edges: 6 on the base + 6 to the apex = 12.

Answer: 7 faces, 12 edges, 7 vertices

Example 4: Koch at Step 2 · 2 marks · Very short answerHarder exam level

A Koch snowflake starts with an equilateral triangle of side 1 unit. Find the number of sides and the perimeter at Step 2.

Try it first, then show the solutionSolution

Sides $= 3 \times 4^{2} = 48$ , each $\frac{1}{9}$ long.
Perimeter $= \frac{48}{9} = \frac{16}{3}$ units.

Answer: 48 sides; $\frac{16}{3}$ units

Example 5: Gasket at Step 4 · 3 marks · Short answerHarder exam level

For the Sierpinski gasket (start area 1), find the number of holes and the remaining area at Step 4.

Try it first, then show the solutionSolution

Remaining triangles: 1, 3, 9, 27, 81 at Steps 0 to 4.
Holes: $H_{4} = 1 + 3 + 9 + 27 = 40$ .
Area $= (\frac{3}{4})^{4} = \frac{81}{256}$ .

Answer: 40 holes; $\frac{81}{256}$

Example 6: Working back from edges · 3 marks · Short answerHarder exam level

A prism has 30 edges. How many sides does its base have, and how many faces and vertices does it have? Could a pyramid have 30 edges? If so, describe it.

Try it first, then show the solutionSolution

Prism: $3 n = 30$ , so $n = 10$ ; faces 12, vertices 20.
Pyramid: $2 n = 30$ , so $n = 15$ .
Yes: a pyramid on a 15-sided base, with 16 faces and 16 vertices.

Answer: 10-sided base, 12 faces, 20 vertices; a 15-sided pyramid also has 30 edges.

Example 7: Ant on a box · 5 marks · Long answerHarder exam level

A closed box is 10 cm long, 6 cm wide and 3 cm high. An ant at one bottom corner wants to reach the opposite top corner over the surface. (a) Explain why unfolding helps. (b) Find the three candidate lengths. (c) State the shortest.

Try it first, then show the solutionSolution

(a) Unfolding turns the surface into a flat net without changing path lengths, and on a flat sheet the shortest path is a straight line.
Unfolding over the width and height: $(6 + 3)^{2} + 1 0^{2} = 181 \approx 13.5$ cm.
Over the length and height: $(10 + 3)^{2} + 6^{2} = 205 \approx 14.3$ cm.
Over the length and width: $(10 + 6)^{2} + 3^{2} = 265 \approx 16.3$ cm.
(c) The shortest is $181 \approx 13.5$ cm.

Answer: About 13.5 cm ( $181$ ).

Example 8: Cube corner to corner · 3 marks · Short answerHOTS / competency

On a cube of side 5 cm, an ant walks from a bottom corner to the diagonally opposite top corner. Compare three routes: (i) along three edges, (ii) across one face diagonal then up an edge, (iii) the shortest unfolded straight path.

Try it first, then show the solutionSolution

(i) $5 + 5 + 5 = 15$ cm.
(ii) Face diagonal $52 \approx 7.07$ , plus 5: about 12.07 cm.
(iii) Unfold two faces into a 10 by 5 rectangle: $125 \approx 11.18$ cm, the shortest.

Answer: 15 cm, about 12.07 cm and about 11.18 cm; route (iii) is shortest.

Example 9: Identify the solid · 4 marks · Case-basedHOTS / competency

A solid's front view and side view are both rectangles and its top view is a circle. Another solid has a triangle as front view and a circle as top view. Name each solid and give a second solid that has a rectangle as both front and top views.

Try it first, then show the solutionSolution

Rectangles from the front and side with a circle from above: an upright cylinder.
Triangle from the front with a circle from above: an upright cone.
A cuboid has rectangles as front, top and side views.
So the same view can come from different solids; three views together are needed.

Answer: Cylinder; cone; a cuboid (for example).

Example 10: Shrinking projection · 2 marks · Very short answerHOTS / competency

A 17 cm rod is held with one end 8 cm higher than the other. Find the length of its projection on the floor. When is the projection as long as the rod?

Try it first, then show the solutionSolution

Rod, height difference and projection form a right triangle: $289 - 64 = 225 = 15$ cm.
Only when the rod is parallel to the floor (height difference 0).

Answer: 15 cm; when the rod is horizontal.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

A Sierpinski tile

Suppose an artist cuts a Sierpinski carpet design into a square marble tile of side 81 cm, carrying out Steps 1 and 2.

(i) · 1 mark
What is the side of the hole cut at Step 1, and of each new hole at Step 2?
Answer
Answer: 27 cm and 9 cm
Step 1 divides 81 cm into thirds: 27 cm; Step 2 divides 27 cm into thirds: 9 cm.
(ii) · 1 mark
How many holes are there after Step 2, and what total area is cut away?
Answer
Answer: 9 holes; 1,377 cm²
$1 + 8 = 9$ holes; area $= 2 7^{2} + 8 \times 9^{2} = 729 + 648 = 1377$ cm².
(iii) · 2 marks
Check that the remaining area agrees with the fraction $(\frac{8}{9})^{2}$ of the tile.
Answer
Answer: 5,184 cm² both ways
Tile area $8 1^{2} = 6561$ cm²; remaining $6561 - 1377 = 5184$ cm² (1 mark). $6561 \times \frac{64}{81} = 5184$ cm² (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the number of faces, edges and vertices of a prism whose base has 8 sides. (2 marks)

Step	Marks
Faces $= 8 + 2 = 10$	1
Edges $= 3 \times 8 = 24$ ; vertices $= 2 \times 8 = 16$	1

A Koch snowflake starts from an equilateral triangle of side 27 cm. Find its perimeter after 3 steps. (3 marks)

Step	Marks
Each step multiplies the perimeter by $\frac{4}{3}$	1
Starting perimeter $= 81$ cm	1
After 3 steps: $81 \times (\frac{4}{3})^{3} = 81 \times \frac{64}{27} = 192$ cm	1

For the Sierpinski carpet starting from a square of area 1, make a table for Steps 0 to 3 showing remaining squares, holes and remaining area, and write a formula for each. (5 marks)

Step	Marks
Remaining squares: 1, 8, 64, 512; $R_{n} = 8^{n}$	1
Rule for holes: $H_{n + 1} = H_{n} + R_{n}$ (each remaining square makes one new hole)	1
Holes: 0, 1, 9, 73; $H_{n} = 1 + 8 + \dots + 8^{n - 1}$	1
Each step keeps $\frac{8}{9}$ of the area, so area $= (\frac{8}{9})^{n}$	1
Areas: 1, $\frac{8}{9}$ , $\frac{64}{81}$ , $\frac{512}{729}$	1

Where students lose marks in Exploring Some Geometric Themes

The slips examiners see most often, and what to write instead.

Mistake: Counting only new holes in the carpet
Holes from earlier steps are forgotten.
Wrong working
Step 2 of the carpet has 8 holes.
Right working
$H_{2} = H_{1} + R_{1} = 1 + 8 = 9$ holes (the big central hole stays).
Do this: Use $H_{n + 1} = H_{n} + R_{n}$ .
Usually costs: 1 mark.
Mistake: Thinking the Koch perimeter stays the same
Each bump looks like it replaces one piece with one piece.
Wrong working
Each side is still the same length after a step, so the perimeter is unchanged.
Right working
One piece of length $s$ becomes 4 pieces of length $\frac{s}{3}$ : total $\frac{4 s}{3}$ , so the perimeter grows by $\frac{4}{3}$ each step.
Do this: Count pieces and their length separately.
Usually costs: Whole answer.
Mistake: Mixing up prism and pyramid counts
Formulas are memorised without understanding.
Wrong working
A pentagonal pyramid has 15 edges.
Right working
5 base edges and 5 slanting edges: 10 edges. (A pentagonal prism has 15.)
Do this: Picture the solid and count edges by groups.
Usually costs: 1 mark.
Mistake: Using one unfolding for the shortest path
The first straight line found is assumed to be the best.
Wrong working
Box 8 × 6 × 4: path $(8 + 6)^{2} + 4^{2} \approx 14.6$ cm.
Right working
Compare all unfoldings: $164 \approx 12.8$ , $180 \approx 13.4$ , $212 \approx 14.6$ ; the shortest is about 12.8 cm.
Do this: Check every way of unfolding the two faces involved.
Usually costs: 1-2 marks.
Mistake: Assuming a projection keeps true lengths
A projection is treated like a scaled copy.
Wrong working
A 13 cm rod tilted so its ends differ in height by 5 cm has a 13 cm top view.
Right working
Top view $= 1 3^{2} - 5^{2} = 12$ cm; only lines parallel to the plane keep their length.
Do this: Use a right triangle with the height difference.
Usually costs: Answer mark.

NCERT exercise guide: Exploring Some Geometric Themes

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Figure it Out (Sierpinski triangle)

What it tests: Drawing the first steps of the Sierpinski triangle, counting holes and remaining triangles, and the remaining area of both Sierpinski fractals at Step n.

How to approach it: Use the 'one piece gives how many' rule; area keeps the same fraction each step.

Like the counting question on the Sierpinski triangle in Section 4.1

For the Sierpinski gasket, find the remaining triangles, holes and remaining area (start area 1) at Step 3. At which step does the remaining area first drop below half?

Show the solutionSolution

Remaining: $3^{3} = 27$ triangles.
Holes: $1 + 3 + 9 = 13$ .
Area: $(\frac{3}{4})^{3} = \frac{27}{64} \approx 0.42$ ; at Step 2 it was $\frac{9}{16} \approx 0.56$ , so Step 3 is the first below half.

Answer: 27 triangles, 13 holes, area $\frac{27}{64}$ ; first below half at Step 3.

Figure it Out (Koch snowflake)

What it tests: Drawing the first steps, the number of sides at Step n and the perimeter at Step n for a unit starting triangle.

How to approach it: Each side becomes 4 sides, each a third as long.

Like the sides-and-perimeter question in Section 4.1

Find the number of sides and the perimeter at Step 3 when the starting triangle has side 1.

Show the solutionSolution

Sides: $3 \times 4^{3} = 192$ .
Each side: $\frac{1}{27}$ .
Perimeter: $\frac{192}{27} = \frac{64}{9} \approx 7.1$ .

Answer: 192 sides; perimeter $\frac{64}{9}$ .

Figure it Out (nets)

What it tests: Identifying nets of a cube, finding all 11 nets, and drawing nets of cuboids with given dimensions.

How to approach it: Fold in your mind: each face must end up opposite exactly one other face; no two faces may overlap.

Like the cuboid-net question in Section 4.2

Describe a net of a cuboid 4 cm by 3 cm by 2 cm, listing all its rectangles.

Show the solutionSolution

The cuboid has three pairs of opposite faces: two 4 × 3, two 4 × 2 and two 3 × 2 rectangles.
Lay a strip of four rectangles in order 4 × 3, 4 × 2, 4 × 3, 4 × 2 (all 4 cm long), so it can wrap round.
Attach a 3 × 2 rectangle to each side of one of the 4 × 3 rectangles, so they close the ends.

Answer: Six rectangles: 2 of 4 × 3, 2 of 4 × 2, 2 of 3 × 2, arranged as a strip of four with two end flaps.

Figure it Out (projections and isometric drawing)

What it tests: Relations between the lengths of views of lines, views of common solids, matching objects with projections, views of cube arrangements, letters made from cubes, solids from three views, counting cubes in a stack, projections of a cube, more ways to join four cubes, drawing on isometric grids and impossible figures.

How to approach it: Look along each direction separately; use 'parallel stays parallel' and 'lengths never grow'; on isometric paper count units along the three directions.

Like the projection-length questions in Section 4.2

A 13 cm stick is held so that one end is 5 cm higher than the other. How long is its top view (shadow on the floor under overhead light)?

Show the solutionSolution

The stick, its height difference and its top view form a right triangle.
Top view $= 1 3^{2} - 5^{2} = 144$ .
$= 12$ cm, shorter than the stick.

Answer: 12 cm

Practice questions on Exploring Some Geometric Themes

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
A cube has
(a) 6 faces, 8 edges, 12 vertices
(b) 6 faces, 12 edges, 8 vertices
(c) 8 faces, 12 edges, 6 vertices
(d) 12 faces, 6 edges, 8 vertices
Hint
Count corners.
Answer
(b)
Q2 · 1 mark · MCQExam level
The isometric projection of a cube has the outline of a
(a) square
(b) rectangle
(c) regular hexagon
(d) rhombus
Hint
Balanced on a corner.
Answer
(c) regular hexagon
Q3 · 1 mark · Very short answerExam level
How many holes are there at Step 2 of the Sierpinski carpet?
Hint
$H_{2} = H_{1} + R_{1}$ .
Answer
9
Q4 · 1 mark · Very short answerExam level
Write the faces, edges and vertices of a tetrahedron.
Hint
Triangular pyramid.
Answer
4 faces, 6 edges, 4 vertices
Q5 · 2 marks · Very short answerHarder exam level
Find the number of sides and the perimeter at Step 4 of a Koch snowflake with starting side 1.
Hint
$3 \times 4^{n}$ and $3 \times (\frac{4}{3})^{n}$ .
Answer
768 sides; perimeter $3 \times (\frac{4}{3})^{4} = \frac{256}{27} \approx 9.5$
Q6 · 2 marks · Very short answerHarder exam level
A pyramid has 16 edges. What shape is its base? How many faces and vertices does it have?
Hint
$2 n = 16$ .
Answer
An octagon; 9 faces and 9 vertices.
Q7 · 2 marks · Very short answerHarder exam level
What fraction of the original area remains at Step 4 of the Sierpinski carpet? Is it more or less than half?
Hint
$(\frac{8}{9})^{4}$ .
Answer
$\frac{4096}{6561} \approx 0.62$ , more than half.
Q8 · 2 marks · Very short answerHarder exam level
A cylinder has radius 14 cm and height 20 cm. What are the dimensions of the rectangle in its net? (Use $π = \frac{22}{7}$ .)
Hint
Circumference by height.
Answer
88 cm by 20 cm
Q9 · 3 marks · Short answerHarder exam level
Which of these can fold into a cube: (i) a straight row of 6 squares; (ii) a cross made of a column of 4 squares with one square on each side of the second square? Explain.
Hint
Fold in your mind.
Answer
(i) No: the row wraps round as 4 sides, but the other two squares also go round, overlapping, and the ends stay open. (ii) Yes: the column of 4 forms the front, top, back and bottom, and the two side squares close the left and right.
Q10 · 5 marks · Long answerHarder exam level
An ant is at a bottom corner of a cube of side 4 cm and food is at the opposite top corner. (a) Find the length along three edges. (b) Find the length of the shortest path over the surface. (c) Find how much shorter it is, to one decimal place.
Hint
Unfold two faces into an 8 by 4 rectangle.
Answer
(a) 12 cm. (b) $8^{2} + 4^{2} = 80 \approx 8.9$ cm. (c) About $12 - 8.9 = 3.1$ cm shorter.
Q11 · 3 marks · Short answerHOTS / competency
For the Sierpinski carpet starting with area 1, at which step does the remaining area first fall below one half? Show your working.
Hint
Multiply by $\frac{8}{9}$ each step.
Answer
Areas: Step 4 $\approx 0.624$ , Step 5 $\approx 0.555$ , Step 6 $= (\frac{8}{9})^{6} \approx 0.493$ . First below half at Step 6.
Q12 · 3 marks · Short answerHOTS / competency
A 25 cm rod is tilted so its ends differ in height by 7 cm. Find the length of its top view. Can the top view of a line ever be longer than the line? Explain.
Hint
Right triangle.
Answer
$625 - 49 = 576 = 24$ cm. No: the line is the hypotenuse of a right triangle whose other sides are the projection and the height difference, so the projection is never longer.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a fractal?Show answer
A self-similar shape whose parts are smaller copies of the whole, repeating at smaller and smaller scales.
2.How many squares remain at Step n of the Sierpinski carpet?Show answer
$8^{n}$ .
3.What fraction of area remains at each step of the Sierpinski gasket?Show answer
$\frac{3}{4}$ .
4.How many sides does the Koch snowflake have at Step n?Show answer
$3 \times 4^{n}$ .
5.Give the faces, edges and vertices of a cube.Show answer
6 faces, 12 edges, 8 vertices.
6.How many edges does a prism with an n-sided base have?Show answer
$3 n$ .
7.How many nets does a cube have?Show answer
11.
8.What shape is the curved surface of a cylinder when unrolled?Show answer
A rectangle whose length is the base circumference and width is the height.
9.Name the three standard views of a solid.Show answer
Front view, top view and side view.
10.What is the isometric projection of a cube?Show answer
A regular hexagon, with all edges projected to equal lengths.

Best order to study it

Read about self-similarity in nature, then draw Steps 0 to 2 of the Sierpinski carpet.
Make tables for remaining pieces, holes and area for the carpet and the gasket (first Figure it Out set).
Draw the Koch snowflake and find its sides and perimeter (second set).
Practise the 'build it in your imagination' prompts and profiles of solids.
Learn faces, edges and vertices of prisms and pyramids; practise nets of a cube and other solids (third set).
Solve shortest-path problems by unfolding boxes.
Finish with projections, three views, shadows and isometric drawing (last sets).

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you draw Step 2 of the Sierpinski carpet and gasket?
Can you count holes using $H_{n + 1} = H_{n} + R_{n}$ ?
Can you find the remaining area at Step n of each Sierpinski fractal?
Can you find the Koch snowflake's perimeter at Step n?
Can you name fractal-like patterns in art?
Can you count faces, edges and vertices of any prism or pyramid?
Can you tell whether a figure is a net of a cube?
Can you find the shortest path on a box by unfolding it?
Can you draw the front, top and side views of a cylinder or cone?
Can you draw a cuboid on isometric grid paper?

Exploring Some Geometric Themes: frequently asked questions

Which book has Exploring Some Geometric Themes for Class 8?

It is Chapter 4 of Ganita Prakash Part II (Chapter 11 in our list), the NCERT Grade 8 Mathematics textbook (Reprint 2026-27). It has two sections: Fractals and Visualising Solids.

Do I need to draw fractals accurately in the exam?

You should be able to sketch the first two or three steps neatly and, more importantly, explain the rule and count pieces, holes, sides, perimeter or area at each step using a table or formula.

How many nets of a cube should I know?

There are 11, and the book asks you to find them all. Exam questions may also show a figure and ask whether it is a net of a cube, so practise imagining the folding.

Why do we need three views of a solid?

Different solids can have the same projection from one direction. Front, top and side views together usually pin down the solid, which is why engineers use them.

How is the Baudhayana-Pythagoras theorem used in this chapter?

For the shortest path on a box: after unfolding, the path is the hypotenuse of a right triangle. It also explains why a tilted line's projection is shorter than the line.

Based on: NCERT Ganita Prakash, Mathematics Textbook for Grade 8, Part II (Reprint 2026-27); NCERT chapter PDF: Exploring Some Geometric Themes