NCERT Solutions Class 8 Maths Chapter 11 Exploring Some Geometric Themes

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 8 · M AT H S

NCERT Solutions

Chapter 11: Exploring Some
Geometric Themes

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 70 – 102 19 83 English

Solutions, notes, sample papers & more at 78 pages

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 11: Exploring Some
Geometric Themes
Chapter 4 of Ganita Prakash Part II follows two geometric themes. The first is the fractal — a shape built by
applying one rule again and again, so that the same pattern reappears at smaller and smaller scales. The
second is visualising solids — nets, shortest paths on a box, projections, and isometric drawing. What links
them is a single habit of mind: watch one step of a construction carefully, write down how a quantity changes
from step n to step n + 1, and the whole sequence follows.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) Part II, 70 – 102

SECTIONS QUESTIONS

19 83

MEDIUM

English

In-text Questions — Page 71
4.1 Fractals — Sierpinski Carpet

Q1 Draw the initial few steps (at least till Step 2) of the shape sequence that leads to
the Sierpinski Carpet.

Start with one square. At every step, cut each square that is still present into 9 equal sub-
squares and rub out the middle one.

...

Step 0 Step 1 Step 2

Step 0 is a full square. Step 1 has one hole. Step 2 has that hole plus one new hole in each of the 8
remaining squares.

Page 1 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Tip: Draw on squared paper with a 27 × 27 square. Then a Step-1 sub-square is 9 × 9
and a Step-2 sub-square is 3 × 3, so every line falls on a grid line.

Q2 Do you see any pattern in the number of holes and squares that remain at each
step?

Yes. The number of squares that remain is multiplied by 8 at every step, and the number of
holes goes up by exactly the number of squares that were present before the step.

Step n 0 1 2 3

Squares remaining Rn 1 8 64 512

Holes Hn 0 1 9 73

Why it happens: One square is cut into 9 sub-squares; 1 is removed and 8 survive.
So every single square of Step n turns into 8 squares of Step (n + 1) — hence Rn+1 =
8Rn. At the same time that one square produces exactly one new hole, so the
number of new holes added equals the number of squares present, Rn. Old holes
are never filled in, so Hn+1 = Hn + Rn.

Q3 Can this be used to get a formula for Rₙ?

Yes — Rn = 8n.

R0 = 1

R1 = 8 × R0 = 8

R2 = 8 × R1 = 8 × 8 = 8²

R3 = 8 × R2 = 8³

In general, Rn = 8n

Page 2 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Why the rule Rn+1 = 8Rn gives a power: the rule says “multiply by 8 once for every
step”. Going from Step 0 to Step n takes n steps, so 8 has been used as a factor n
times — and n factors of 8 is what 8n means. A rule of the form “multiply by a fixed
number each time” always produces a power like this.

Q4 Similarly, how do we find the number of holes at a given step?

Add up all the squares that ever existed before that step:

Hn = Hn−1 + Rn−1

so Hn = R0 + R1 + … + Rn−1

= 1 + 8 + 8² + … + 8n−1

This sum can be closed up. Multiply it by 7 and watch the middle terms cancel:

7Hn = (8 − 1)(1 + 8 + … + 8n−1)

= (8 + 8² + … + 8n) − (1 + 8 + … + 8n−1)

= 8n − 1

So Hn = (8n − 1)⁄7

Check: H1 = (8 − 1)⁄7 = 1

H2 = (64 − 1)⁄7 = 9 = 1 + 8 ✓

H3 = (512 − 1)⁄7 = 73 = 1 + 8 + 64 ✓

Did you know? Hn = (8n − 1)⁄7 is exactly the number written as n eights in base 9 —
the same trick that makes 1 + 10 + 100 = 111.

In-text Question — Page 72

Page 3 of 78

ase
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

4.1 Fractals — Sierpinski Gasket
co m
e m.
m l as
Q1
.co a g
Show that by joining the midpoints of an equilateral triangle, we divide it into 4
m
l a se
identical equilateral triangles. [Hint: Note that the corner triangles are isosceles.]

a g

co m
. ag
Let the triangle be ABC with every side of length a, and let P, Q, R be the midpoints of BC, CA, AB.
em
g l as
a
A
co m
em.
m l as
m .co a g
l a se
a g
R m Q a s
m .co agl
l a se
a g

co m
m .
m as e
.co a g l
se m B C
g l a P
a
se m
com
Joining the three midpoints makes the shaded middle triangle PQR and three corner triangles.
g l a
m . a
e
astriangle ARQ. Since R and Q are midpoints,
g l
The three corner triangles. Take
a
AR = AB⁄2 = a⁄2 and AQ = AC⁄2 = a⁄2
co m
m .
as e
comthe angle between them is ∠A = 60°. l
So triangle ARQ is isosceles with AR = AQ,
.and a g
se m
g l a
a Base angles = (180° − 60°)⁄2 = 60° each.
c
m .
s e
. com a gla
All three angles are 60°, so ARQ is equilateral with side a⁄2. The same argument at B and at C
m side a⁄2.
makes BRP and CQP equilateralewith
s
a
l the midpoints of AB and AC, so by the midpoint theorem RQ =
agjoins
The middle triangle. RQ
BC⁄2 = a⁄2, and likewise QP = a⁄2 and PR = a⁄2. So PQR is equilateral with side a⁄2 too.

co m
m .
m ase
.co


a g l Page 4 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

All four triangles are equilateral with side a⁄2, hence identical (congruent).

Why this matters for the fractal: because the four pieces are identical and each is
an exact half-scale copy of the original, removing the middle one and repeating the
construction gives a shape that looks the same at every scale. Self-similarity is built
in from this one fact.

Figure it Out — Page 72
Sierpinski Gasket

Q1 Draw the initial few steps (at least till Step 2) of the shape sequence that leads to
the Sierpinski Triangle.

Begin with a filled equilateral triangle. At every step, join the midpoints of each remaining
triangle and remove the middle one.

...

Step 0 Step 1 Step 2

Step 1 removes one middle triangle; Step 2 removes a middle triangle from each of the 3 that
survived.

Check it yourself: Step 2 should show 9 small filled triangles and 4 white holes —
one large hole and three small ones.

Page 5 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q2 Find the number of holes, and the triangles that remain at each step of the shape
sequence that leads to the Sierpinski Triangle.

...

Step 0 Step 1 Step 2

Page 72 — Step 0, Step 1 and Step 2 of the shape sequence that leads to the Sierpinski
Triangle.

Each remaining triangle gives 3 remaining triangles and 1 new hole at the next step.

Tn+1 = 3 Tn, with T0 = 1 ⇒ Tn = 3n

Hn+1 = Hn + Tn, with H0 = 0

⇒ Hn = 1 + 3 + 3² + … + 3n−1

Closing up the sum the same way as for the carpet:

2Hn = (3 − 1)(1 + 3 + … + 3n−1) = 3n − 1

Hn = (3n − 1)⁄2

Step n 0 1 2 3 4 n

Triangles remaining 1 3 9 27 81 3n

Holes 0 1 4 13 40 (3n − 1)⁄2

Page 6 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Why the two rules are different in shape: triangles are replaced at every step, so
their count is multiplied — that gives a power. Holes are never removed, so their
count is a running total — that gives a sum. Recognising which of the two is
happening is the whole skill here.

Q3 Find the area of the region remaining at the nth step in each of the shape
sequences that lead to the Sierpinski fractals. Take the area of the starting
square/triangle to be 1 sq. unit.

...

Step 0 Step 1 Step 2

Page 70 — Step 0, Step 1 and Step 2 of the sequence that leads to the Sierpinski Carpet.

...

Step 0 Step 1 Step 2

Page 72 — Step 0, Step 1 and Step 2 of the sequence that leads to the Sierpinski Triangle
(Gasket).

Carpet: (8⁄9)n sq. units. Gasket: (3⁄4)n sq. units.

Sierpinski Carpet. Each square is cut into 9 equal parts, so a square of Step n has area (1⁄9)n, and
there are 8n of them.

Page 7 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Arean = 8n × (1⁄9)n = (8⁄9)n

Area1 = 8⁄9, Area2 = 64⁄81, Area3 = 512⁄729 ≈ 0.70

Sierpinski Gasket. Each triangle is cut into 4 identical parts, so a triangle of Step n has area
(1⁄4)n, and there are 3n of them.

Arean = 3n × (1⁄4)n = (3⁄4)n

Area1 = 3⁄4, Area2 = 9⁄16, Area3 = 27⁄64 ≈ 0.42

A second route to the same answer: at each step a fixed fraction of whatever is left is thrown
away — 1⁄9 of it for the carpet, 1⁄4 for the gasket. So the area is multiplied by 8⁄9 (or 3⁄4) every step.

Why it happens: both 8⁄9 and 3⁄4 are less than 1, so the area shrinks at every step and
keeps shrinking. (3⁄4)10 ≈ 0.056 and (3⁄4)50 is under a millionth. The true fractal — the
shape you approach after infinitely many steps — has zero area, even though
something is still there at every point of it. Notice also that the carpet shrinks more
slowly than the gasket, because 8⁄9 is bigger than 3⁄4.

Figure it Out — Page 73
Koch Snowflake

Q1 Draw the initial few steps (at least till Step 2) of the shape sequence that leads to
the Koch Snowflake.

Start with an equilateral triangle. Replace every straight side by the four-segment ‘bump’: divide
the side into three equal parts, build an equilateral triangle on the middle part, and rub out that
middle part.

Page 8 of 78

as e
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co a g
l a se
a g
→
co m
e m . ag
as
one side becomes 4 sides
a g l
The generating rule. The dashed middle third is removed, so 1 side of length s turns into 4 sides of
length s⁄3.
co m
se m.
o m
Step 0 .c An equilateral triangle — 3 sides. g l a
m a
l a se1 The six-pointed star (Star of David outline) — each of the 3 sides has become 4, giving 12 sides.
ag Step

m a s
.co agl
Step 2 A small bump is raised on each of those 12 sides, giving 48 sides — the outline already looks lacy.

a s em
l 9 cm. Then Step 1 has sides of 3 cm and Step 2 has
Tip: Draw Step 0 with a sidegof
a
sides of 1 cm, so you can measure every bump with an ordinary ruler.

co m
m .
o m l a se
g
Q2 .cFind the number of sides in the nth step of the shape sequence
a
m
e Koch Snowflake.
that leads to the

aglas
se m
com g l a
m . a
ase
agl ...

co m
m .
m l a seStep 2
m .co Step 0 Step 1
ag
l a se
ag
Page 73 — Step 0, Step 1 and Step 2 of the shape sequence that leads to the Koch
Snowflake.
.c
s e m
m a
e m . co agl
g l as
Sn = 3 × 4 n .
a

co m
m .
m as e
.co


a g l Page 9 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Every side is replaced by 4 sides, so Sn+1 = 4 Sn

S0 = 3

S1 = 4 × 3 = 12

S2 = 4 × 12 = 48

S3 = 4 × 48 = 192

In general, Sn = 3 × 4n

Why 4 and not 3: the middle third is removed but two new sides of the raised
triangle take its place. So of the three thirds, two survive and the removed one is
replaced by two — 2 + 2 = 4 sides where there was 1.

Q3 Find the perimeter of the shape at the nth step of the sequence. Take the starting
equilateral triangle to have a sidelength of 1 unit.

...

Step 0 Step 1 Step 2

Page 73 — Step 0, Step 1 and Step 2 of the shape sequence that leads to the Koch
Snowflake.

Pn = 3 × (4⁄3)n units.

Length of one side at Step n = (1⁄3)n

Number of sides at Step n = 3 × 4n

Pn = 3 × 4n × (1⁄3)n = 3 × (4⁄3)n

Page 10 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Step n 0 1 2 3 4

Sides 3 12 48 192 768

Side length 1 1⁄3 1⁄9 1⁄27 1⁄81

Perimeter 3 4 16⁄3 ≈ 5.33 64⁄9 ≈ 7.11 256⁄27 ≈ 9.48

Why it happens: four times as many sides, each one-third as long, means the total
length is multiplied by 4 × 1⁄3 = 4⁄3 at every step. Since 4⁄3 > 1 the perimeter grows
without limit — P20 is already more than 900 units. Yet the whole snowflake always
stays inside a small circle drawn round the first triangle. An unlimited boundary
enclosing a limited region is one of the surprises fractals hold.

Compare: for the two Sierpinski fractals the multiplier (8⁄9 and 3⁄4) was less than 1 and
the area died away. Here the multiplier 4⁄3 is greater than 1 and the perimeter blows
up. Same kind of rule, opposite behaviour — the multiplier decides everything.

Build it in Your Imagination — Pages 75–77
4.2 Visualising Solids

Q1 Picture your name, then read off the letters backwards. Make sure to do this by
sight, not by sound — really see your name! Now try with your friend's name.

Hold the written name in your mind as a picture, then read the picture from right to left.

MEENA → A N E E M
RAGHAV → V A H G A R

SUBRAMANIAN → N A I N A M A R B U S

Notice the difference between the two ways of doing it. If you go by sound you have to say the
name to yourself and then work backwards, which is slow and easy to lose track of. If you go by
sight the letters are all there at once and you simply scan them the other way.

Page 11 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Why this exercise opens the chapter: everything that follows — nets, hidden faces,
three views of a solid — asks you to hold a picture steady in your head and then turn
it round or look at it from a new direction. This is that skill on a flat, easy object first.

Q2 Cut off the four corners of an imaginary square, with each cut going between
midpoints of adjacent edges. What shape is left over? How can you reassemble the
four corners to make another square?

What is left is a square — standing on its corner, with half the area of the original.

→

cut off 4 corners a square, half the area
The four corner triangles are congruent right isosceles triangles; the inner square is the one that
survives.

Why it is a square. Each cut joins midpoints of two adjacent sides, so all four cuts have the
same length, and each corner cut removes a right isosceles triangle with legs a⁄2. At each vertex
of the inner shape two 45° angles are removed from a 180° straight angle, leaving 90°.
Reassembling the four corners. The four triangles each have legs a⁄2 and hypotenuse a⁄√2. Put
the four right angles together at one point; the legs match up in pairs and the four hypotenuses
form the boundary. You get a square of side a⁄2.

Original area = a²

Inner square = a²⁄2 (half)

Four corners = 4 × ½ × (a⁄2) × (a⁄2) = a²⁄2 (the other half) ✓

Page 12 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q3 Mark the sides of an equilateral triangle into thirds. Cut off each corner of the
triangle, as far as the marks. What shape do you get?

A regular hexagon.

Each side is divided into thirds; cutting the three corners leaves a hexagon with all six sides equal to
one-third of a side.

Let each side be 3 units, so the marks are 1 unit apart. Each corner triangle has two sides of 1
unit with a 60° angle between them, so it is equilateral with side 1. Cutting all three corners
leaves six sides, and each is 1 unit long: three of them are the middle thirds of the original sides,
three are the cut edges.

Each interior angle of the hexagon = 180° − 60° = 120°

All six sides = 1 unit ⇒ regular hexagon

Area left = 1 − 3 × (1⁄9) = 2⁄3 of the triangle

Page 13 of 78

ase
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
m.
Mark the sides of a square into thirds and cut off each of its corners as far as the
e
Q4

m l as
.co
marks. What shape is left?

a g
se m
g l a
a

An octagon — eight sides, but not a regular one.

co m
ag
Take the square with side 3, so the marks are 1 unit apart. Each corner cut removes a right

m .
e
isosceles triangle with legs 1, so its hypotenuse is √2.

g l as
The 8 sides go: 1, √2, 1, √2, 1, √2, 1, √2
a

co m
m.
1 ≠ √2 ≈ 1.414, so the sides are not all equal

m as e
.co
All 8 angles are equal (135° each)
a g l
se m
a
Area left = 9 − 4 × ½ × 1 × 1 = 9 − 2 = 7 sq. units, i.e. 7⁄9 of the square

a g l

om cut off is equilateral and its third side is a s
agl
Why the triangle gives a regular figure and the square does not: in the triangle
the corner angle is 60°, so the corner.cpiece
a s em
aglthan the sides it replaced. To make the octagon
also 1. In the square the corner angle is 90°, so the piece cut off is right-angled and
its hypotenuse is √2 — longer

m
regular you would have to cut at a different distance, not at the thirds.
. co
e m
m l as
m .co
Try This: Where should you mark a square of side 3 so that the octagon comes out
a g
a s eregular? You need the cut length x to satisfy x√2 = 3 − 2x, which gives x = 3⁄(2 + √2) ≈

agl 0.879 — not one-third.
se m
com g l a
m . a
ase
agl
Q5 A solid whose profile has a square outline

co m
m .
se
A cube, seen straight on so that one face is parallel to the wall.

com l a
m .Solid aggives a square
se
Viewpoint that

g l a
a c
.
Cube Any direction perpendicular to a face

s e m
m a
.co agl
Cuboid with a square cross-section Looking along its length

se m
g l a
Cylinder whose height equals its diameter Looking at it from the side

Square pyramid
a Looking straight down from above

co m
m .
m as e
.co


a g l Page 14 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Why so many answers: the profile only records the outline of the shadow.
Everything about the depth of the solid is thrown away, so completely different
solids can leave the same hole in the wall.

Q6 A solid whose profile has a circular outline

A sphere — and from every viewpoint, which no other solid manages.

Cylinder — looked at along its axis (end on).
Cone — looked at from directly below (or above) along its axis.
Hemisphere — looked at along its axis of symmetry.

Did you know? The sphere is the only solid whose profile is a circle from every
direction. That is why a ball rolls equally well whichever way you push it.

Q7 A solid whose profile has a triangular outline

A cone, seen from the side.

Cone — from the side the outline is an isosceles triangle: the base circle appears as a
segment and the two slant edges as the other two sides.
Square pyramid or triangular pyramid — from the side.
Triangular prism — looked at along its length, the outline is the triangular face itself.

Q8 A solid with a rectangular profile from one viewpoint and a circular profile from
another viewpoint

A cylinder — a tin of ghee, a candle, a piece of chalk.

Page 15 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

cylinder from the side from the top
One solid, two very different profiles — the whole point of taking more than one view.

Side view: rectangle, height h by diameter 2r

Top view: circle of radius r

Q9 A solid with a circular profile from one viewpoint and a triangular one from another
viewpoint

A cone — a party hat, a heap of grain, a road-work cone.

From directly above (along the axis): a circle of radius r

From the side (perpendicular to the axis): an isosceles triangle of base 2r and height h

A hemisphere will not do — from the side it gives a semicircle, not a triangle. The straight slant
edges of the cone are what make the side profile triangular.

Q10 A solid with a rectangular profile from one viewpoint and a triangular one from
another viewpoint

A triangular prism — the shape of a glass prism, or of a tent.

Page 16 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Along its length: the triangular end face

From the side: a rectangle, length by height

A square pyramid also works: from directly above it is a square (a special rectangle), and from
the side it is a triangle.

Q11 A solid with a trapezium shaped profile from one viewpoint and a circular one from
another viewpoint

A frustum of a cone — a cone with its top sliced off parallel to the base. A bucket, a matka, a
lampshade or a tumbler has this shape.

From above: a circle (the wider rim)

From the side: a trapezium — the two circular rims give the parallel sides, the slanting wall

gives the other two

Why a plain cone will not do: the cone comes to a point, so its side profile is a
triangle, not a trapezium. You need the top to be cut off flat so that the profile has
two parallel sides.

Q12 A solid with a pentagonal profile from one viewpoint and a rectangular one from
another viewpoint

A pentagonal prism — the shape of many pencils and of a five-sided pillar.

Along its length: the regular pentagon end face

From the side: a rectangle, length by the width of the pentagon

Another answer: a house-shaped solid — a cuboid with a triangular prism roof on top. From the
front it gives a pentagon (a square with a triangular cap); from the side, a rectangle.

Page 17 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q13 Are there unique solids for each of the conditions, or can you come up with
multiple possibilities?

Not unique — every one of these conditions is met by many different solids.

Condition One answer Another answer

Square profile Cube Square pyramid seen from above

Circular profile Sphere Cylinder seen end on

Rectangle + circle Cylinder Sphere squashed into an oval (a lemon)

Circle + triangle Cone A spinning top

Trapezium + circle Frustum of a cone A bucket, a flower pot

Why uniqueness fails: a profile is a shadow. It records the outline only — nothing
about what is behind it, nothing about hollows or dents, nothing about the depth.
Two solids whose outlines happen to agree from one direction are indistinguishable
in that view. This is exactly the reason engineers never draw one view of a machine
part: they draw three. Even then, as Fig. 4.6 shows, three views can still be shared by
different objects.

In-text Questions — Page 79
Making Solids — prisms and pyramids

Q1 If the congruent polygons of a prism have 10 sides, how many faces, edges and
vertices does the prism have? What if the polygons have n sides?

A decagonal prism has 12 faces, 30 edges and 20 vertices.

Faces = 2 decagons + 10 side faces = 12

Vertices = 10 on the top + 10 on the bottom = 20

Edges = 10 (top) + 10 (bottom) + 10 (joining) = 30

Page 18 of 78

as e
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

For an n-sided polygon, count the same way:
co m
e m.
m l as
m .co
Faces = n + 2
a g
l a se
g
Edges = 3n
aVertices = 2n

com
e m . ag
g l as
Why the edges come in three groups: a prism is two copies of the same polygon,
a
one above the other, joined vertex to vertex. The n sides of the top polygon, the n

m
sides of the bottom polygon, and the n vertical joins are the only edges there are.
co
em.
m l as
.co a g
Check it yourself: F + V − E = (n + 2) + 2n − 3n = 2 for every n. For a cube (n = 4): 6 + 8
m
l a se
− 12 = 2 ✓. This is Euler’s relation, and it holds for every one of these solids.

ag
m a s
m.co
If the base of a pyramid has 10 sides, how many faces, edges and vertices does the
agl
se
Q2

l a
pyramid have? What if the base is an n-sided polygon?
g
a
m

. co
m
A decagonal pyramid has 11 faces, 20 edges and 11 vertices.

m as e
.co a g l
a s em Faces = 1 base + 10 triangles = 11

agl Vertices = 10 on the base + 1 apex = 11

se m
com l a
Edges = 10 (base) + 10 (up to the apex) = 20
. a g
m
ase
agl
For an n-sided base:

Faces = n + 1
co m
m .
as e
com = n + 1 l
Edges = 2n
.Vertices a g
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 19 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Solid Faces Edges Vertices F+V−E

Triangular pyramid (tetrahedron), n = 3 4 6 4 2

Square pyramid, n = 4 5 8 5 2

Decagonal pyramid, n = 10 11 20 11 2

Why faces and vertices are equal for a pyramid: every side face sits on one base
edge, and every base edge ends at one base vertex — so side faces, base edges and
base vertices all come in the same number n. Add the base face to one column and
the apex to the other and the two stay level at n + 1.

In-text Questions — Page 80
Making Solids — nets

Q1 What is a net of a cube?

A net of a cube is the flat shape you get by cutting the cube along some of its edges and
unfolding it onto a plane — six squares joined edge to edge, which can be folded back up to
give the cube.

Page 20 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Fig. 4.1 — one net of a cube
Six squares: a vertical strip of four, with one square on each side of the third square from the top.

Two things are needed. There must be exactly six squares, because a cube has six faces; and
they must be joined so that folding brings every free edge against exactly one other free edge,
with no two faces landing on top of each other.

Tip: When you actually build one, add small flaps on alternate free edges so you can
glue them down — but the flaps are not part of the net. The net is only the unfolded
surface itself.

Page 21 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q2 Visualise how it can be folded to form a cube.

Fold along every internal line, all in the same direction, keeping one square flat on the table.

1. Keep the third square of the long strip (the one with a neighbour on each side) flat — this
becomes the base.
2. Fold the left and right squares up. They become two opposite walls.
3. Fold the square below the base up. It becomes a third wall.
4. Fold the strip going upwards: the square just above the base becomes the fourth wall, and
the one above that folds over to become the lid.

How to check without cutting: in a finished cube, opposite faces never touch. In
this net, take the vertical strip of four squares — folding a strip of four wraps it right
round the cube, so squares 1 and 3 are opposite, and so are 2 and 4. The two side
squares are the remaining pair. Three pairs, six faces, nothing repeated: it folds.

Figure it Out — Pages 80–81
Nets of a cube

TRY THIS

Q1 Which of the following are the nets of a cube? First, try to answer by visualisation.
Then, you may use cutouts and try.

Writing each figure on squared paper, with a dot for an empty cell:

Figure Arrangement of the six squares Net of a cube?

(i) Two squares on top, four in a row below, the row starting under the second square No

(ii) A row of three, and a second row of three below it overlapping in one column Yes

(iii) A staircase — three rows of two, each row stepped one place to the right Yes

(iv) A row of four with one square above and one square below the third column Yes

(v) A row of four with two squares hanging one below the other from the second column No

(vi) A row of four with one square above the third column and one below the second Yes

Page 22 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

So (ii), (iii), (iv) and (vi) fold into a cube; (i) and (v) do not.
Why (iv) and (vi) work. They are of the ‘1 – 4 – 1’ kind. Roll the row of four into a band — that
band is the four side walls. One extra square is above the row and one below, so one closes the
top and the other the bottom. It does not matter which column each extra square hangs from.
Why (v) fails. Its two extra squares are both below the row, stacked in the same column. Roll
the row of four into a band; the first extra square becomes the bottom face. The second is
attached to the far edge of that bottom face, and that edge already belongs to the wall opposite
the one it came from. So the sixth square lands on a wall that is already there, and the top of the
cube is left open.
Why (i) fails. Same trouble. Both extra squares are on one side of the row of four. Fold the row
into a band and let the square directly above the row become the lid. The last square is joined to
that lid along an edge which the lid already shares with one of the four walls — so it folds down
on top of a wall, and the base is never covered.

The quick test: find a row of four. It must become the band of side walls, so the
remaining two squares must lie one on each side of that row — one to close the top
and one to close the bottom. If both are on the same side, you get an overlap and an
open face. That single check settles (i), (iv), (v) and (vi) at a glance.

Q2 A cube has 11 possible net structures in total. In this count, two nets are considered
the same if one can be obtained from the other by a rotation or a flip. For example,
the following nets are all considered the same — Find all the 11 nets of a cube.

The tidiest way to hunt them down is by the length of the longest straight row. Below, X
marks a square and a dot marks an empty cell.
Type 1 – 4 – 1 (six nets). A row of four, with one square above and one below. The one below
can sit under any of the four columns, and the one above under any of the four — but rotations
and flips cut the sixteen possibilities down to six.

1 X... 2 X... 3 X...
XXXX XXXX XXXX
...X ..X. .X..

4 .X.. 5 .X.. 6 .X..
XXXX XXXX XXXX
..X. ...X .X..

Type 2 – 3 – 1 (three nets). The longest row has three squares; two squares sit above it and one
below (or the mirror image).

Page 23 of 78

ase
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

com
X.
7 XX.. 8 XX.. 9 XX..
. X Xm
m a. s. .eX
.XXX .XXX

. co agl
.X.. ..X.

e m
g l as
a
Type 3 – 3 (one net). Two rows of three, overlapping in a single column — this is figure (ii) of
Question 1.

co m
em . ag
as
XXX..

..XXX a g l

co m
m.
Type 2 – 2 – 2 (one net). The staircase — figure (iii) of Question 1.

m ase
. c o a g l
X Xm
s e ..
la
ag . X X .
..XX
m a s
m .co agl
l a se
6 + 3 + 1 + 1 = 11 nets a g

. c om
Why there is no 5 – 1 or 6 net: a straight row of five would have to
s e mwrap round a
. c monly four walls, so two of its squares would land on
cubeoof
a la same face. And a
gthe
m
ase
straight row of six is worse still. That is why the longest row in any cube net is four.

agl
se m
com
Draw a net of a cuboid having.sidelengths: g l a
m a
se
Q3 (i) 5 cm, 3 cm, and 1 cm (ii) 6 cm, 3 cm,
and 2 cm
l a
ag
m

. co
m
Use the same ‘1 – 4 – 1’ layout as the cube net, but now the six rectangles come in three

o m
matching pairs.
l a se
c cm × 3 cm × 1 cm. Lay a horizontal strip of four rectangles
(i).5 a gthat wraps round the 5 cm
se m
a
agl
length, then attach the two 3 cm × 1 cm ends.

.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 24 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

5×1

5×3 5×3

3×1

3×1
5×1

Net of a 5 cm × 3 cm × 1 cm cuboid: two 5 × 3 faces, two 5 × 1 faces and two 3 × 1 faces.

(ii) 6 cm × 3 cm × 2 cm. Exactly the same layout with the numbers changed: two rectangles 6 ×
3, two 6 × 2 and two 3 × 2.

Surface area (i) = 2(5×3 + 5×1 + 3×1) = 2(15 + 5 + 3) = 46 cm²

Surface area (ii) = 2(6×3 + 6×2 + 3×2) = 2(18 + 12 + 6) = 72 cm²

Check it yourself: in any correct net of a cuboid, edges that will be glued together
must be equal in length. Run your finger round the boundary of your drawing and
pair up the free edges — every pair should match.

In-text Questions — Page 81
Nets of other solids

MATH TALK

Q1 What is a net of a regular tetrahedron? Which of the following are nets of a regular
tetrahedron?

A net of a regular tetrahedron is four equilateral triangles joined edge to edge so that they fold
up into the solid. The first figure (the big triangle) and the third figure (the parallelogram)
are nets. The second and the fourth are not.

Page 25 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Figure What it is Net?

1st A large equilateral triangle cut by its midlines into 4 small triangles Yes

2nd A strip of 3 triangles with the 4th hanging below the end triangle No

3rd A strip of 4 triangles forming a parallelogram Yes

4th Five triangles — a strip of 4 with one more hanging below No

Why the fourth is out at once: a tetrahedron has only 4 faces, and that figure has 5 triangles.
One triangle would have nowhere to go.
Why the second fails. Look at the strip of three, up–down–up. All three meet at one point, and
each contributes 60°, so the angles round that point add to 180°. Folding closes that point into a
vertex of the tetrahedron, and doing so brings the two outer edges of the strip together —
those are the two halves of the long base line. The fourth triangle in that figure is stuck onto
one of exactly those two edges, so folding would make it collide with the other. It has been
attached to the wrong edge.

3 triangles round a point: 60° + 60° + 60° = 180°

180° < 360°, so the flat fan folds into a cone point — a vertex
The 4th triangle must fill the triangular hole left over, so it must be joined to one of the three

free edges of the fan, not to the pair that get glued

net 1 — big triangle net 2 — parallelogram

The only two nets of a regular tetrahedron. In the left one the fourth triangle sits on the middle
triangle of the strip; in the right one the strip simply carries on.

Page 26 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q2 Are there any other possible nets?

No — a regular tetrahedron has exactly 2 nets, the big triangle and the parallelogram.
Here is the argument. Whatever the net, it is a chain or fan of 4 triangles. Start from the strip of
three (up–down–up), which is forced: any three triangles joined in a line look like that. Its free
edges are the two slanting outer edges and the top edge of the middle triangle — but folding
glues the two halves of the base together, so the three edges available for the fourth triangle
are:

the top edge of the middle triangle — this gives the big triangle net;
the outer edge of the left triangle, or
the outer edge of the right triangle — either of these gives the parallelogram, and the two
are mirror images, so they count as the same net.

Why so few: a cube has 11 nets and a tetrahedron only 2 because the tetrahedron
has just 4 faces and every face touches every other face. There is almost no freedom
left in how you may unfold it.

Q3 Draw a net with appropriate measurements that can be folded into a regular
tetrahedron. Verify if it works by making an actual cutout.

Take the edge to be 6 cm. Then a net is a single equilateral triangle of side 12 cm, with its three
midpoints joined.

Big triangle: each side 12 cm

Mark the midpoint of each side (at 6 cm)

Join the midpoints — this makes 4 identical equilateral triangles of side 6 cm

Fold up along the three midlines; the three corners meet at the apex

How to draw it accurately. Draw a 12 cm segment. With the compass set at 12 cm, cut arcs
from both ends; their meeting point is the third vertex. Join up, mark the midpoints with a ruler
and join them.

Page 27 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Check it yourself: before folding, measure — all six little edges of the inner triangle
and the corner triangles must read 6 cm, and every angle 60°. After folding, the
three corner vertices should meet exactly at a point with no gap and no overlap. If
there is a gap, your midpoints were not accurate.

Tip: Add 1 cm flaps on alternate outer edges so you can glue the model. The flaps
are for building; they are not part of the net.

Q4 Draw a net with appropriate measurements that can be folded into a square
pyramid. Verify if it works by making an actual cutout.

Take a square base of side 6 cm and slant edges of 8 cm. The net is that square with an
isosceles triangle built outwards on each of its four sides.

Square base: 6 cm × 6 cm

Each triangle: base 6 cm, the other two sides 8 cm each

Slant height (height of a triangular face) = √(8² − 3²) = √55 ≈ 7.4 cm

Page 28 of 78

as e
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co a g
l a se
a g

om
8.ccm ag
a sem
agl

co m
em.
m l as
m .co 6 cm square a g
l a se
a g
m a s
m.co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
.
Net of a square pyramid: the base square with four congruent isosceles triangles folded up around it.
m a
ase
agl
Why any slant edge longer than 3√2 cm works. The apex must sit above the centre of the
square, and the centre is at a distance of half the diagonal, 6√2⁄2 = 3√2 ≈ 4.24 cm, from each

co m
.
corner. So the slant edge has to be more than 4.24 cm. With 8 cm the pyramid stands up tall:

se m
o m l a
c of the pyramid = √(8² − (3√2)²) = √(64 − 18) = √46 a≈ g6.8 cm
m .Height
l a se
ag
.c
Check it yourself: fold the four triangles up. Their apexes should meet at a single
s e m
om a
. c agl
point, and the four slant edges should close with no gap. If they do not meet, the

a s em
four triangles were not congruent.

agl

co m
m .
m ase
.co


a g l Page 29 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

In-text Questions — Page 82
Nets of a cylinder and a cone

MATH TALK

Q1 What is the net of a cylinder?

Two circles and one rectangle. Unroll the curved surface after cutting it once along the height,
and lay the two circular ends flat.

length = circumference = 2πr h

The curved surface flattens into a rectangle; the two ends are circles of radius r.

Why the curved surface flattens perfectly: a cylinder is bent in only one direction.
Cut it along a line parallel to the axis and it opens out flat with nothing stretched or
torn — like unrolling a mat.

Q2 What are the sidelengths of the rectangle obtained?

One side is the height h of the cylinder; the other is the circumference of the base, 2πr.

Page 30 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Rectangle = h × 2πr

Curved surface area = 2πrh

Total surface area = 2πrh + 2 × πr² = 2πr(h + r)

Why the circumference and not the diameter: the cut edge of the rectangle was
originally wrapped once right round the circular rim. Unrolling does not change its
length, so the rectangle is exactly as long as the rim is round — that is 2πr.

Check it yourself: wrap a strip of paper once round a tin, mark where it overlaps
and cut. Now measure the strip and the diameter of the tin. The strip is always about
3.14 times the diameter.

Q3 How will the net of a cone look?

A circle for the base and a sector of a larger circle for the curved surface.
The sector has radius equal to the slant height l of the cone, and its arc is exactly as long as the
base circle, 2πr.

Sector radius = l

Arc length = 2πr

Angle of the sector = (2πr ⁄ 2πl) × 360° = (r⁄l) × 360°

For example, a cone with r = 3 cm and l = 9 cm opens out into a sector of angle (3⁄9) × 360° =
120°, that is one-third of a full circle.

Q4 If the cone is slit open along the line l and then unrolled, what will we get?

A sector of a circle with centre O, where O is the apex of the cone.
Every point on the rim of the base is joined to O by a slant line, and all these slant lines have the
same length l. Slitting along one of them and unrolling does not change any of those lengths, so
in the flat figure every rim point is still at distance l from O. Points at a fixed distance from O lie

Page 31 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

on a circle centred at O — so the curved boundary of the net is an arc of a circle with centre O.

Sector radius = l for every rim point

Arc length = 2πr, the circumference of the base

Curved surface area = ½ × arc × radius = ½ × 2πr × l = πrl

Why unrolling preserves lengths: like the cylinder, a cone bends in only one
direction. Rolling and unrolling stretches nothing, so every distance measured along
the surface is the same before and after. This is exactly the property the chapter will
use again for the ant and the laddu.

Page 32 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q5 What surface do you construct by using the above net, in which O is not the centre
of the boundary circle? Make a physical model to help you answer this question!

O

Page 82 — the net referred to: the cone slit open along l and unrolled, so its boundary is
a part of a circle with centre O.

You get an oblique cone — a cone that leans over, with its apex not above the centre of its base.
In the ordinary net, every point of the boundary is the same distance l from O, so when it is
rolled up all the slant lines are equal and the apex sits directly above the centre of the base. That
is a right circular cone.
Now take a net whose boundary is a circle whose centre is somewhere other than O. Different
boundary points are now at different distances from O, so when the surface is rolled up the
slant lines have different lengths — long on one side, short on the other. The apex is pulled
towards the short side, and the cone tilts.

Page 33 of 78

ase
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
m.
All boundary points at distance l from O Distances from O unequal

m All equal as e
Slant lines
.co a g l
Different lengths

se m
g l a
Solid formed Right circular cone Oblique (slanting) cone
a
Apex sits Above the centre of the base Off to one side

co m
m . ag
l a se
Tip: Cut a paper disc, mark a point O well away from its centre, cut along a line from
agthe two cut edges. The paper will curl into a tilted
O to the boundary and overlap
funnel — a leaning ice-cream cone.

co m
em.
m l as
.co a g
a
Q6
s em Draw a net with appropriate measurements that can be folded into a triangular

ag l prism. Verify that it works by making an actual cutout.

om a s
c agl

Take a prism whose ends are equilateralm .
a s e triangles of side 5 cm and whose length is 9 cm.

agl
Net = 3 rectangles of 9 cm × 5 cm, joined in a row

co m
+ 2 equilateral triangles of side 5 cm, one on each end of the middle rectangle
m .
m as e
.co
Total length of the strip = 3 × 5 = 15 cm
a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 34 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

9 cm × 5 cm

9 cm × 5 cm

9 cm × 5 cm

triangles: equilateral, side 5 cm

Three rectangles wrap round to make the sides; the two triangles fold in to close the ends.

Surface area = 3 × (9 × 5) + 2 × (√3⁄4) × 5²
= 135 + 2 × 10.83 ≈ 156.7 cm²

Check it yourself: the three rectangles must be equal in width (5 cm each) or the
prism will not close, and the triangle side must equal that width exactly. Roll the strip
into a triangular tube first, then fold the ends in.

In-text Questions — Page 83

Page 35 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

The octahedron and the dodecahedron

MATH TALK

Q1 Can you visualise its net? This is one of its nets.

An octahedron has 8 equilateral triangular faces, so its net is made of 8 triangles. The one
shown in the book is a strip of 8 triangles arranged as two rows of four, the rows offset by half a
triangle.

Octahedron: F = 8, E = 12, V = 6

Check: F + V − E = 8 + 6 − 12 = 2 ✓

How to see it. The octahedron is two square pyramids glued base to base. Unfold the top
pyramid outwards to give a ring of 4 triangles, do the same for the bottom pyramid, and slide
the two rings together into a zig-zag strip. Every free edge in the strip has exactly one partner to
be glued to.

Why the faces meet 4 at a vertex: four equilateral triangles round a point give 4 ×
60° = 240°, which is less than 360°, so the flat fan can close into a cone point. That is
the vertex where the two pyramids meet. At the apex of each pyramid, the same
thing happens.

Page 36 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q2 Taking all the triangles in the net to be equilateral, make a cutout of the net and
fold it to form an octahedron.

Page 83 — the net of the octahedron printed in the book: eight triangles, six in a strip
with one more above and one below.

Use equilateral triangles of side 5 cm.

1. Draw a strip 8 triangles long, alternately pointing up and down, all with side 5 cm.
2. Rearrange it into the book’s net: two rows of four, offset so that each triangle in the top row
shares a full edge with one below.
3. Cut it out, adding 1 cm flaps on alternate outer edges.
4. Fold along every internal line in the same direction and tape the free edges in pairs.

Check it yourself: when it is finished, exactly 4 triangles should meet at each of the
6 vertices, and the solid should have 12 edges. Hold it by two opposite vertices and
spin it — a correctly built octahedron spins smoothly, like a top.

Page 37 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Did you know? Like the cube, the octahedron has exactly 11 nets. This is not a
coincidence — the cube and the octahedron are ‘dual’ solids: the cube has 6 faces
and 8 vertices, the octahedron 8 faces and 6 vertices, and both have 12 edges. The
dodecahedron, with 12 pentagonal faces, has 43,380 nets.

In-text Questions — Pages 84–85
Shortest Paths on a Cube

MATH TALK

Q1 Net of a sphere? Experiment and see if you can make a paper cutout that can
perfectly wrap around a ball without leaving any wrinkles, gaps or overlaps.

It cannot be done. A sphere has no net.
A cylinder and a cone flatten out because they are curved in only one direction — every point of
them lies on a straight line drawn on the surface. A sphere carries no straight lines at all; it
curves in every direction at once. So paper, which will bend but not stretch, can never lie flat
against it.

Why paper refuses: take the two ends of a strip of paper and try to press them onto
a ball. Near the middle the paper touches, but at the edges it must either be
stretched (it tears) or gathered up (it wrinkles). Distances on a sphere simply do not
match distances on a plane: on a plane, a circle of radius r has circumference 2πr; on
a sphere, a circle drawn at distance r from a point has circumference less than 2πr.
Something has to give.

Did you know? This is why every flat map of the Earth is wrong somewhere.
Greenland looks as big as Africa on many wall maps, though Africa is about 14 times
larger. It is also why a football is not made from one piece of leather but from many
small panels, and why an orange peel will not lie flat however carefully you press it.

Page 38 of 78

ase
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
m.
What is the shortest path for the ant to reach the laddu?
e
Q2

m l as
.co a g
a s em
l should walk in a straight line on the net — that is, straight up the side face and then
Thegant
a
straight across the top face, the two straight bits meeting the top edge at the same point.

. c om
The idea. Unfold the box so that the side face carrying the ant and the top face carrying the
ag
laddu lie flat next to each other. On this flat m
s e figure, join the ant to the laddu by a straight line.

a g
Now fold the box back up: the line bendsla over the edge and becomes a path on the surface —
and its length has not changed.

co m
On the surface: a bent path, hard to compare
se m.
o m l a
g path between two
c a straight segment — and a straight segment is the shortest
On the.net:
m a
se on a plane
apoints
agl
m a s
m.co
Why this proves it is shortest: every path on the box turns into a path of exactly
agl
l a se
the same length on the net, and every path on the net turns back into a path of the
a g
same length on the box. So the two problems have exactly the same set of lengths.
On the flat net we already know the winner — the straight line. That is why the

co m
answer must be the path that straightens out.
m .
m as e
.co a g l
a s em
a gl Q3 What about in the following case?
se m
com g l a
. a

m
ase
agl
The same method works when the laddu sits at the middle of an edge — but now the laddu is
on the boundary between two faces, so two different unfoldings are open to the ant.

Go across the side face and then over the top face to the edge.

co m
Go across the side face and then over the front face to the same edge.
m .
o m l a se
.c at the midpoint of the edge and the ant at the centre aofgthe face, the two come out equal
Unfold each way, draw the straight line, measure both, and take the shorter. When the laddu sits

se m
exactly
a
agl
— but in general they will not, and the ant must compare.

.c
s e m
m a
agl
Tip: Never settle for the first unfolding that looks reasonable. A point on an edge
. co
e m
belongs to two faces, and a point at a corner belongs to three.

g l as
a

co m
m .
m ase
.co


a g l Page 39 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q4 If we think that a certain path is the shortest, how can we be sure that it truly is,
among all the infinite possibilities?

By turning the question into one we can already answer. Unfold the box; the path becomes a
path on a flat net, with its length unchanged. On a plane we know that the straight line is
shortest — so if our path unfolds into a straight line, nothing can beat it.

path on cuboid ⟷ path of the same length on the net
shortest path on the net = straight line

so shortest path on the cuboid = the one that unfolds straight

Two things make this a proof rather than a guess:

The correspondence works both ways. Any rival path on the box also becomes a path on the
net, and it cannot be shorter than the straight line there.
Lengths are exactly preserved. Unfolding bends the surface but never stretches it, so no
length is lost or gained.

Why we cannot simply try lots of paths: there are infinitely many, so testing is
hopeless. Mathematics gets round this by changing the setting to one where the
answer is already known — that is the real lesson of this section.

Q5 For example, are either of these the shortest path?

Only the first one is. Draw the net and see what each path looks like on it.

Path On the net it becomes Shortest?

The first (red) path A straight segment from ant to laddu Yes

The second path A bent line — it changes direction at the edge No

The test is simple: unfold, and look at whether the path comes out straight. A path that still has
a kink in it on the flat net can always be shortened by pulling it taut.

Page 40 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q6 What does this show?

It shows that the problem of the shortest path on a cuboid is the same problem as the
shortest path on its net.

Every path on the surface → a path of equal length on the net

Every path on the net → a path of equal length on the surface

Because the two collections of paths match up length for length, the shortest in one is the
shortest in the other. That is why the first path in the picture is shortest — it becomes the
straight line joining the ant and the laddu — while the second is not, because it becomes a bent
line.

Why lengths survive the unfolding: unfolding only turns faces about the edges
they share. Nothing is stretched, squeezed or torn, so a length measured along the
surface is the same length after it is laid flat. This is the same fact that let us find the
curved surface area of a cone by flattening it into a sector.

Q7 Have we now completely analysed the problem of finding the shortest path
between two points on a cuboid?

No, not yet — and the next two pages show exactly what is missing.
We have proved that the shortest path must be a straight line on some net. But a cuboid can be
unfolded in many different ways, and:

on some nets the straight segment between the two points runs outside the net, so it does
not correspond to any path on the box at all;
different nets give straight segments of different lengths.

Same box, same two points, three unfoldings:

one gives 42 cm, another gives 40 cm, a third gives something else

So the method is only finished when we list all the sensible unfoldings, work out each straight-
line length, and take the smallest. That is what the worked examples on pages 86 and 87 do.

Page 41 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

In-text Questions — Pages 86–87
Shortest Paths on a Cube — worked cases

TRY THIS

Q1 Find the shortest path between the ant and the laddu in the following case:

The box is 8 cm × 4 cm × 4 cm. The ant is at the centre of the 4 cm × 4 cm end face, and the
laddu is on the bottom front edge, 2 cm from that end.
Unfold the end face and the front face into one plane. Measure from the vertical edge where
they meet, taking that edge as the zero line.

On the end face: the ant is 2 cm from the edge (half of 4) and 2 cm up

On the front face: the laddu is 2 cm from the same edge, on the floor line

Horizontal separation = 2 + 2 = 4 cm

Vertical separation = 2 − 0 = 2 cm

d² = 4² + 2² = 16 + 4 = 20

d = √20 = 2√5 ≈ 4.47 cm

front face end face

ant

2√5 cm

laddu

Page 42 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

The two faces unfolded flat. The straight segment is the ant’s shortest route, and the Baudhayana
Theorem gives its length.

Why the book’s first attempt failed. In the cross-shaped net drawn first, the straight segment
joining the ant to the laddu passes outside the net, over empty paper. A line that leaves the net
does not correspond to any walk on the box, so that unfolding is useless here. Re-unfolding the
box so that the ant’s face is hinged onto the face the laddu is on puts the whole segment inside
the net, and then it is a genuine path.

Check it yourself: the laddu sits on the edge shared by the front face and the
bottom face, so the ant could equally well go across the bottom. Unfold that way
and you get 2 cm and 4 cm again — the same 2√5 cm. Both routes tie.

Why the way you unfold matters: unfolding is a choice, and each choice tests one
family of routes. A route is only found if some unfolding lays out exactly the faces it
crosses, side by side, with the whole segment inside them.

Q2 So what do we do now?

Unfold the cuboid a different way — one that lays out the faces the ant would actually walk
over, so that the straight segment stays inside the net.

1. Decide which faces the path could cross.
2. Unfold so that exactly those faces lie flat, side by side, in the order the ant would meet them.
3. Join the two points with a straight line and check that it stays inside the net.
4. Measure it with the Baudhayana Theorem.
5. Repeat for every other reasonable set of faces, and keep the smallest length.

Why a segment leaving the net is meaningless: the net is the box’s surface,
opened up. Paper outside the net is not part of the box. A line crossing it does not
correspond to any walk the ant could take, so its length tells us nothing.

Page 43 of 78

as e
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
m.
What is the length of the shortest path between the ant and the laddu?
e
Q3

m l as
.co a g
a
s em
l is 30 cm long with 12 cm × 12 cm ends. The laddu is stuck on the back end face, on its
Thegbox
a
centre line, 1 cm above the bottom; the ant is on the front end face, on its centre line, 1 cm

com
ag
below the top.

m .
The shortest path is 40 cm.
as e
a g l
Unfold so that the ant goes down over the bottom, round one side, over the top, and on to the
far face. Lay out the four long faces as one strip: bottom, side, top — that is 12 + 12 + 12 = 36 cm
across the strip. Hinge the laddu’s face onto the bottom and the ant’s face onto the top.
co m
em.
m l as
.co
Along the length of the box:
a g
em(down the back face to the bottom edge) + 30 cm (the box) + 1 cm (up the front face to
1scm
a
agl
the top edge)
m a s
.co agl
= 32 cm

se m
g l a
Across the strip: a

com
.
the laddu is 6 cm along the bottom edge, the ant is 6 cm along the top edge
e m
m l as
.co
6 (across the bottom) + 12 (across the side) + 6 (across the top) = 24 cm
a g
se m
g l a
a d² = 32² + 24² = 1024 + 576 = 1600
se m
d = √1600 = 40 cm
com g l a
m . a
ase
agl
Compare the unfoldings. The book shows two of them; there are others, and they are all worth
testing.

co m Length
m .
as e
Route Legs of the right triangle

comface → bottom → top → front face (round one side)
.Back 24lcm
32 cm and g
a
sem
40 cm — shortest

a
agl Back face → bottom → a side → front face 37 cm and 17 cm √1658 ≈ 40.7 cm
c
m .
m a s e
co gl
Back face → bottom → front face (straight along) 42 cm and 0 42 cm

m . √1864 ≈ 43.2 cm a
e
las
Back face → one side → front face 42 cm and 10 cm

ag

co m
m .
m as e
.co


a g l Page 44 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Why the winning route looks so roundabout: going straight along the bottom
seems obvious, but it makes the ant climb 1 cm down at one end and 11 cm up at
the other — the two ends fight each other. Curling round the box lets the 1 cm at
each end be used in the same direction, and the extra 24 cm travelled sideways buys
a much shorter straight line. That is the surprise of this problem: on a box, the
natural-looking route is often not the shortest, and only a careful list of unfoldings
finds the winner.

Check it yourself: 24, 32, 40 is the 3, 4, 5 triangle multiplied by 8 — a Baudhayana
triple, so the answer comes out a whole number.

In-text Questions — Pages 89–91
Representation of Solids on a Plane Surface — Projections

MATH TALK

Q1 What happens to the length of a line in its projection?

It never gets longer. The projection is at most as long as the segment itself, and usually
shorter.
In Fig. 4.3 the segment AB has length l and its projection DC has length p. Drawing AE
perpendicular to BC makes AECD a rectangle (AD and EC are both perpendicular to the plane, so
AD ∥ EC and AD = EC; that forces AE ∥ DC and AE = DC). So AE = p, and ∠AEB = 90°.

In right triangle AEB, AB is the hypotenuse

AB = l, AE = p

l² = p² + BE² ⇒ l² ≥ p²

p≤l

Why the hypotenuse is the longest side: BE² is a square, so it is never negative. It
is zero only when B and E coincide — that is, only when the segment already lies
parallel to the plane. Every tilt out of the plane adds something to BE² and therefore
makes l strictly bigger than p.

Page 45 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q2 Can you now compare the lengths p and l?

p ≤ l, always.

l² = p² + BE²

so l² − p² = BE² ≥ 0

hence p ≤ l

Concretely: if a 10 cm pencil is tilted at 60° to the wall it is being projected on, its shadow
measures 10 × cos 60° = 5 cm. Tilt it more and the shadow gets shorter still; hold it flat against
the wall and the shadow is the full 10 cm.

Tip: In general p = l cos θ, where θ is the angle the segment makes with the plane.
Since cos θ is never more than 1, p can never beat l.

Q3 When is the length of the projected line equal to its actual length?

Exactly when the segment is parallel to the plane (or already lying in it).

p = l ⇔ BE = 0 ⇔ B coincides with E

⇔ AB ⊥ AD, i.e. AB is perpendicular to the projecting direction

⇔ AB is parallel to the plane

In that case ABCD is itself a rectangle, so DC = AB exactly.

Why this is the one and only case: tilting the segment out of the plane means one
end is further from the plane than the other. Their projections then move closer
together than the ends themselves are, and length is lost. Only when both ends are
at the same distance from the plane is nothing lost.

Page 46 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q4 What do you think are the different possible projections of a square that we get
based on its orientation?

The projection of a square can be a square, a rectangle, a parallelogram, or a line segment
— but never anything else.

Orientation of the square Projection

Parallel to the plane A square of the same size

Tilted about one of its sides A rectangle — one pair of sides keeps its length, the other pair shortens

Tilted about a diagonal, or generally A parallelogram (a rhombus in the diagonal case)

Perpendicular to the plane A line segment — the square collapses

Why it is always a parallelogram: the square has two pairs of parallel, equal sides.
Projection carries parallel segments to parallel segments and equal parallel
segments to equal parallel segments. So the two pairs stay parallel and stay equal in
the picture — and a quadrilateral with both pairs of opposite sides parallel is a
parallelogram.

Check it yourself: cut a square from card and hold it in sunlight over a sheet of
paper. Turn it slowly. You will see the square stretch into rectangles and lean into
parallelograms, and just before it disappears the shadow is a thin line — but the
opposite sides stay parallel throughout.

Q5 What do you think is the projection of a parallelogram under different orientations?
Can this ever be a quadrilateral that is not a parallelogram?

The projection of a parallelogram is always another parallelogram (or, in the flattest case, a
line segment). It can never be a quadrilateral that is not a parallelogram.
Start where the hint suggests, with a pair of parallel lines. Two parallel lines and the direction of
projection lie in two parallel planes. Those planes meet the projection plane in two parallel lines.
So parallel lines project to parallel lines.

Page 47 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

AB ∥ DC ⇒ their projections A′B′ ∥ D′C′

AD ∥ BC ⇒ their projections A′D′ ∥ B′C′

Both pairs of opposite sides parallel ⇒ A′B′C′D′ is a parallelogram

So a trapezium, a kite or an ordinary quadrilateral can never be the shadow of a parallelogram.

Try This: Cut out a parallelogram and look at its shadow in sunlight, which is as good
as a projection because the Sun’s rays are effectively parallel. However you turn it,
the shadow stays a parallelogram — it may become a rectangle, a square, a
rhombus or a very thin sliver, but the opposite sides never stop being parallel.

Q6 What can you say about the projection of an n-sided regular polygon? [Hint:
Projection of a polygon is composed of the projections of its sides.]

It is again a convex polygon with n sides — but it is normally not regular. It is regular only
when the polygon is parallel to the plane.
Following the hint: the projection of the polygon is built from the projections of its n sides. Each
side projects to a segment, corners project to corners, and the pieces stay joined in the same
order. So the picture is an n-sided polygon (unless the polygon is perpendicular to the plane,
when it flattens to a segment).
What survives and what does not:

Property Survives projection?

Number of sides Yes — still n

Sides being parallel Yes

Convexity Yes

Midpoints staying midpoints Yes

All sides equal No — sides across the tilt shrink more

All angles equal No

For an even n the regular polygon has n⁄2 pairs of opposite sides that are parallel and equal,
and both facts survive. So the projection of a regular hexagon is a hexagon whose opposite
sides are still parallel and equal, even though it looks squashed.

Page 48 of 78

as e
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
m.
Why regularity is lost: tilting shortens lengths by different factors in different

m l a se
directions — a side pointing along the tilt shrinks most, a side across the tilt not at
o
c that point in different directions therefore end upagunequal.
all. Equal .sides
se m
l a
ag

com
ag
How would the projections of a cube and a cone look?
.
Q7

e m
g l as
a
Cube (Fig. 4.4). Held with one face parallel to the plane, its projection is a square — the near

co m
m.
face and the far face land exactly on top of each other, and the four side faces project onto the

se
edges.
o m l a
Cone (Fig.c4.5). Held with its axis parallel to the plane, its projection isga triangle — the base
m . a slant lines give the
e
as two sides.
l
circle flattens to a segment (the base of the triangle) and the two extreme

a g
other

m a s
.co agl
Solid Orientation Projection

se m
Cube
g l a
A face parallel to the plane Square

Cube
a
Tilted about one edge direction Rectangle

co m
m .
e
Cube Balanced on a corner Regular hexagon

m l as
.co a g
em
Cone Axis parallel to the plane Isosceles triangle

a s
agl Cone Axis perpendicular to the plane Circle

se m
com g l a
. a
Why the outline is all we get: the projection records where the solid blocks the
m
ase
projecting rays. Two points of the solid on the same ray land on the same spot, so

agl
everything behind the outline is lost. That is why a solid and a hollow shell of the
same shape have identical projections.

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 49 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q8 See Figures 4.2 – 4.5. In each case, see if you can visualise another object that gives
the same projection.

line in space
projection projection

line in space

Fig. 4.2

Fig. 4.2, page 88 — the projection of a line on a plane, drawn for two different lines.

C
p

D E
p

B
A l

Fig. 4.3

Fig. 4.3, page 89 — a line AB of actual length l and its projection DC of length p on the
plane; AE ⊥ BC, so AE = DC = p.

Page 50 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Fig. 4.4 Fig. 4.5

Fig. 4.4 and Fig. 4.5, page 89 — the projection of a cube and the projection of a cone on a
vertical plane.

Yes — in every case there are many others. Fig. 4.6 shows two families of examples.

Figure Object shown Another object with the same projection

4.2 A segment tilted to the Any longer segment tilted more steeply — a 5 cm segment at 53° and a 4 cm
plane segment at 41° both project to 3 cm

4.3 A segment with The same segment slid anywhere parallel to the plane, or turned about the
projection p projecting direction

4.4 A cube, projecting to a A cuboid of any depth with the same square face; a square pyramid seen
square from above; a hollow box

4.5 A cone, projecting to a A triangular prism seen end on; a square pyramid seen from the side; a flat
triangle triangular cutout

Cuboids 1, 2 and 3 units deep — all give the same square front view

Lines of different lengths and tilts — all give the same segment

Why this is worth noticing: it is the reason engineers never trust one drawing. A
single projection throws away everything along the direction of projection. Take
three projections on mutually perpendicular planes and far less is lost — though, as
Question 9 shows, even three views do not always pin the object down.

Page 51 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q9 Find another object that makes the same projection as that of a given cone.

Held with its axis parallel to the plane, a cone projects to an isosceles triangle. So does each of
these:

A triangular prism whose end face is that triangle, seen along its length.
A square pyramid of the same height and base width, seen from the side.
A flat triangular sheet of card of exactly that shape, held parallel to the plane.
A hollow cone, or a cone of any material at all — the projection cannot tell.

Why so many: the projection is only the outline of the shadow. Any solid that fits
neatly inside the ‘tube’ of rays which the cone fills, and blocks all of it, casts the same
shadow. Depth, thickness and what is inside are all invisible. That is precisely the loss
of information the three standard views are designed to reduce.

Figure it Out — Pages 92–93

Page 52 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Front view, top view and side view

MATH TALK

Q1 Observe the front view, top view and side view of the different lines in Fig. 4.6. Is
there any relation between their lengths?

Front View Top View Side View

part of Fig. 4.6

Left: part of Fig. 4.6 — three lines of different lengths that cast the same shadow on the
vertical plane. Right: the front view, top view and side view of each of those three lines,
one line per row.

Yes. All three lines were chosen so that they have the same front view, and their front views are
horizontal segments. For each line separately, the three view-lengths obey the Baudhayana
Theorem.

Line Front view Top view Side view

1 A horizontal segment A horizontal segment of the same length A point

2 The same horizontal segment A slanting segment, a little longer A short segment

3 The same horizontal segment A more steeply slanting, longer segment A longer segment

Take a segment whose ends differ by Δx across, Δy in depth and Δz in height. Then

Page 53 of 78

ase
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

co m
e m.
front view f = √(Δx² + Δz²)
m l as
.co
top view t = √(Δx² + Δy²)
m a g
l a se
g
side view s = √(Δy² + Δz²)
aactual length l = √(Δx² + Δy² + Δz²)

. c om ag
s e
Adding the first three squares counts each ofmΔx², Δy², Δz² exactly twice:
a
agl
f² + t² + s² = 2 l²

co m
em.
m l as
All three lines in Fig. 4.6 lie in a horizontal plane, so Δz = 0 for each. Then f = |Δx| — the same
. c o a g
m
for all three, which is why the front views agree — and s = |Δy|, so:

as e
g l
a t² = Δx² + Δy² = f² + s²
a s
. com of a right triangle whose legs are the front agl
em
ss= 0, so t = f; as the line is swung round in the horizontal
So for these lines the top view is the hypotenuse
view and the side view. Line 1 hasa
gl
plane, s grows and t grows withait, while f stays fixed.

co m
Why no single view is enough: all three lines have identical front views but
m .
m as e
c o
. only all three together give l, through f² + t² + s² = 2l².a g l
different actual lengths. Only when the side view is added can you tell them apart —

e m
and
as
agl
se m
com g l a
. a
Q2 Find the front view, top view and side view of each of the following solids, fixing its
m
ase
orientation with respect to the vertical, horizontal and side planes: cube, cuboid,

agl
parallelepiped, cylinder, cone, prism, and pyramid. If needed, see the next problem
for clues.

co m
m .
e

m l as
.co a g
Fix each solid in its natural position — flat faces parallel to the planes, axes vertical — and read

s e moff the three outlines.
agla
.c
s e m
m a
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 54 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Solid (orientation) Front view Top view Side view

Cube, faces parallel to the Square Square Square
planes

Cuboid l × b × h, edges Rectangle l × h Rectangle l × b Rectangle b × h
along the axes

Parallelepiped, one face on Parallelogram Parallelogram Parallelogram
the floor

Cylinder standing upright Rectangle 2r × h Circle of radius r Rectangle 2r × h

Cone standing on its base Isosceles triangle, base Circle of radius r The same isosceles
2r, height h triangle

Triangular prism lying Triangle Rectangle Rectangle
along the side direction

Square pyramid standing on Isosceles triangle Square (with the diagonals Isosceles triangle
its base showing the four slant edges)

Why the front and side views of a cylinder and a cone are identical: both solids
are symmetric about a vertical axis, so they look the same from every horizontal
direction. Their front and side views must therefore agree. A cuboid has no such
symmetry, so its three views are three different rectangles.

Tip: The three views are not independent. The height shown in the front view must
match the height in the side view; the width in the front view must match the width
in the top view; and the depth in the top view must match the depth in the side view.
This is the standard check draughtsmen use.

Q3 Match each of the following objects with its projections.

Take each object, look along the F, T and S arrows drawn beside it, and find the row of three
outlines that fits. The arrows are not the same for every object, so the direction called ‘front’
changes from picture to picture.

Page 55 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Object Front view Top view Side view

Measuring jug Body with a V-shaped spout Circle with the spout and Body with the handle in
/ mug notch at the rim handle sticking out profile

Funnel Downward triangle with a Circle with a small circle The same triangle with stem
thin stem (the stem) at the centre

Hammer Narrow head above a straight Short bar — the head seen The familiar hammer profile
handle end on

Car Bonnet, grille and headlamps Roof and bonnet outline Full side of the car with both
wheels

Slide with a Tall narrow rectangle with the Long rectangle (the ramp) Ladder and sloping ramp,
box on it box as a small square with the box inside it with the box on the slope

Chair Back and two front legs Seat as a square with the Back, seat and legs in profile
back edge

Ceiling fan Rod, motor and blades edge The three blades spread The same as the front view
on out

Cooking pot / Body with the lid on top Circle with the handle bar Body with the handle
cooker across it sticking out

Why the fan is the easiest and the hammer the hardest: the fan has a clear axis,
so its top view (three blades) is unmistakable and quite unlike its side view. The
hammer looks like a plain bar from two directions and only shows its character from
the third. Match the most distinctive view first, then confirm with the other two.

Tip: Two of the three views of a symmetric object are often identical — the fan and
the funnel are examples. When two views agree, the object almost always has an
axis of symmetry.

In-text Questions — Page 94

Page 56 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Shadows

TRY THIS

Q1 What do you see?

A shadow on the wall whose shape is very close to the projection of the object on that wall —
but usually larger, and sometimes stretched or slightly distorted.

Projection Torch shadow

Rays Parallel and perpendicular to the wall Spreading out from one point

Size Same size as the object’s outline Bigger than the object

Shape Exact Similar, but stretched if the object is tilted

Q2 Observe what happens to the size of the shadow as you vary the distance between
your torch and your object.

Bring the torch closer and the shadow grows; move it back and the shadow shrinks,
settling down towards the size of the object’s outline and never getting smaller than that.

Torch at 20 cm from an object 40 cm from the wall → a big, blurred shadow

Torch at 2 m → a shadow only a little larger than the object

Torch very far away → the shadow becomes the projection

Check it yourself: if the torch is at distance d from the object and the wall is D
behind the torch, the shadow is enlarged by the factor D⁄d. Doubling d roughly
halves the enlargement.

Page 57 of 78

Class 8 Maths Chapter 11 Exploring Some Geometric Themes AglaSem · NCERT Solutions

Q3 Why does this happen?

Because the light leaves the torch as a cone of spreading rays, not as a parallel beam.
Every ray that grazes the edge of the object carries on and lands further out on the wall. The
closer the torch is to the object, the more steeply those edge rays are spreading, and the further
out they land — so the shadow is bigger. Pull the torch back and the rays reaching the object are
much more nearly parallel, so the outline they carve out on the wall is much closer to the
object’s true outline.

Why the Sun makes a perfect projection: the Sun is about 15 crore km away. By
the time its rays reach us they are, for all practical purposes, parallel. So when
sunlight falls perpendicular to a wall, the shadow it casts is the projection — same
size, same shape. That is why the book says you can settle the question about the
parallelogram by taking a cardboard cutout outside and looking at its shadow.

Did you know? This is also why your shadow at noon is short and sharp but long
and fuzzy near sunset, and why the shadow of a hand held near a wall is crisp while
one held near the torch is huge and blurred.

Figure it Out — Pages 95–97
Views of solids built from cubes

MATH TALK

Q1 Draw the top view, front view and the side view of each of the following
combinations of identical cubes.

The method. Write the solid down as a plan with heights — a small grid, one cell for each
square of floor the solid stands on, with the number of cubes stacked there written in. Then
each view is read off mechanically:

Top view — the plan itself: shade every cell where the height is at least 1.
Front view — for each column of the plan (left to right), draw a bar as tall as the largest
height in that column.
Side view — for each row of the plan (front to back), draw a bar as tall as the largest height
in that row.

Page 58 of 78

as e
Class 8 Maths Chapter 11 Exploring Some Geometric Themes
a g l AglaSem · NCERT Solutions

The word largest is the whole trick: a view is a silhouette, so a tall stack hides everything shorter
co m
behind it.
e m.
m l as
.co a g
Solid (i) — 3 cubes, all one layer high: two side by side at the front and one behind the right-

se
hand cube. m
g l a
a
Plan (front row at the bottom): back → 0 1 ; front → 1 1

com
Front view: a 2 × 1 rectangle
e m . ag
g l as
a
Top view: an L of 3 squares

Side view: a 2 × 1 rectangle (2 deep, 1 high)

. com
m a s em column
gl
Solid (ii) — 4 cubes: a one-cube-wide arm running two cubes back, with a two-cube
co the back cube of the arm.
standing .beside
a
a s em
agl Plan with heights: back → 1 2 ; front → 1 0
m a s
.co agl
Front view: L-shaped — left bar 1 high, right bar 2 high

se m
l a
Top view: an L of 3 squares
a g
Side view: L-shaped — front 1 high, back 2 high

co m
m .
e
Solid (iii) — 4 cubes, all one layer high: a row of three at the back with one more sticking out in

m l as
.co g
front of the left-hand end.

em a
a s
agl Plan: back → 1 1 1 ; front → 1 0 0
se m
com a
Front view: a 3 × 1 rectangle

. a g l
m
ase
Top view: the L-tetromino above

Side view: a 2 × 1 rectangle
agl

co m
.
Solid (iv) — 6 cubes: a 2 × 2 block of floor, the back row stacked 2 high and the front row 1 high

em
as
(a two-step staircase).
m l
.co
m Plan with heights: back → 2 2 ; front → 1 1 a g
l a se
ag
.c
m
Front view: a 2 × 2 square (the back wall hides the front step)

m a s e
. co agl
Top view: a 2 × 2 square
e m
l as
Side view: a staircase — 1 high at the front, 2 high at the back
g
a

co m
m .
m ase
.co


a g l Page 59 of 78