NCERT Solutions Class 9 Maths Chapter 14 Math of Space Surface Area and Volume

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 9 · M AT H S

NCERT Solutions
+ ×
Chapter 14: Math of
Space: Surface Area
and Volume

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 123 – 148 12 61 English

Every question Step by step Exam ready
Activities, in-text questions The working shown, not Written to match how
and all exercises. just the final answer. marks are awarded.

Solutions, notes, sample papers & more at 54 pages

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 14: Math of Space: Surface
Area and Volume
Chapter 14 moves from flat shapes to solids. It finds how much surface covers a solid and how much space
it fills, for cuboids and cubes, right circular cylinders, cones, pyramids, spheres and hemispheres. Each
formula is explained by a model — a stack of cards, unrolled paper, cones filled with salt, string wound on a
ball. By the end you can use these formulas to solve problems, compare shapes when lengths are scaled,
and make sensible guesstimates of everyday volumes.

TEXTBOOK BOOK PAGES
Ganita Manjari (Class 9) Part II, 123 – 148

SECTIONS QUESTIONS
12 61

MEDIUM
English

In-text Questions — Page 124
Section 14.1 Cuboids and Cubes

Q1 Compare the formula for the volume of a cuboid with the formula for the area of a
rectangle (length × width).

The two formulas have the same pattern. The volume formula just has one more factor, the
height.

Area of a rectangle = l × w (square units)
Volume of a cuboid = l × w × h = (area of the base rectangle) × h (cubic units)

Area counts the unit squares that cover a flat shape. Volume counts the unit cubes that fill a
solid. A cuboid is a pile of h layers, and each layer is a rectangle of l × w unit cubes. For
example, a 4 cm × 3 cm rectangle has area 12 cm², and a 4 cm × 3 cm × 5 cm cuboid has
volume 12 × 5 = 60 cm³.

Page 1 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Why: an area multiplies two lengths, so its unit is cm × cm = cm². A volume
multiplies three lengths, so its unit is cm × cm × cm = cm³. The rectangle is the
"2D version" of the cuboid, and the cuboid is the "3D version" of the rectangle.

Tip: the power in the unit tells you how many lengths were multiplied: cm for a
length, cm² for an area, cm³ for a volume. If your answer for a volume comes
out in cm², a length is missing.

Think and Reflect — Page 124
Section 14.1 Cuboids and Cubes
THINK AND REFLECT

Q1 Try to work out for yourself why this model explains the formula for the volume of a
cuboid, i.e., volume = area of base × height.

Fig. 14.3
A photograph of a pack of playing cards: one neat stack of cards lies on a table,
and a few loose cards are spread out beside it. The stack has the shape of a
cuboid made of many thin, identical rectangular cards lying flat on top of one
another.

Fig. 14.3, page 124 — the model for the volume of a cuboid (a photograph, described here).

Think of the cuboid as a pack of n identical thin cards, each a rectangle of length l and width w,
and each of very small thickness t.

Page 2 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Volume of one card = (area of its face) × thickness = lw × t
Volume of the pack = n × lw × t = lw × (n × t)

But n × t = height of the pack = h

So volume = lw × h = area of base × height

Every card has exactly the same area as the base, so stacking cards only multiplies the base
area by the height. A taller pack has more cards and so more volume, in exact proportion to its
height.

Why: the same idea works with unit cubes. Fill the bottom of the cuboid with
one layer of cubes: that layer has l × w cubes (the number equals the base
area). There are h such layers, so the total is l × w × h.

Did you know? The model never used the fact that the cards are rectangles. A
stack of identical thin circles (like the khakhras on page 128) gives a cylinder,
and the same reasoning gives volume = area of base × height = πr²h. It works
for any solid whose cross-section is the same all the way up.

Check it yourself: push the pack of cards so that it leans sideways. Each card
still has the same area and the height is the same, so the volume has not
changed.

Q2 Write the formula for the surface area and the volume of a cube.

For a cube of side a:

Total surface area = 6 × (area of one square face) = 6a²

Volume = a × a × a = a³

A cube has 6 faces, and each one is a square of area a². For example, a cube of side 3 cm has
surface area 6 × 9 = 54 cm² and volume 27 cm³.

Page 3 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

co m
m.

o m l ase
g The area of
Tip: ifma.cbox has no lid, it has only 5 faces, so its outer area isa5a².
la sefour side walls alone is 4a².
ag
the

. comhas surface area 6 cm² and volume 1
Check it yourself: a unit cube (side 1 cm)
ag
m the volumes of cubes of side 1, 2, 3, 4 ….
cm³. Cube numbers 1, 8, 27, 64 … eare
glas
a

co m
em.
om — Page 124
In-text Questions
gl as
.c a
a emCuboids and Cubes — the cube as a special cuboid
s
Section 14.1

agl
Observe how these two formulas (total surface area 6a², volume a³) are special cases
s
Q1
omwhen w = h = l. a
.c agl
of the formulas 2(wl + hl + hw) and whl,
se m
la
ag
A cube is a cuboid whose length, width and height are all equal. Put l = w = h = a in the cuboid
formulas: om . c
e m
com gl as
. a
se m
TSA = 2(wl + hl + hw) = 2(a·a + a·a + a·a) = 2 × 3a² = 6a²
a V = whl = a · a · a = a³
agl
se m
So the cube formulas are not new formulas.
c o mThey are the cuboid formulas with all three edges g l a
.
m become three equal pairs of squares. a
e
las
equal: the three different pairs of faces

− Explanation ag

. com
s e m bottom), l × h
Why: in a cuboid the faces come in three pairs: l × w (top and
. co(front a gla pairs are a × a, so 2 ×
m and back) and w × h (two sides). In a cube all three
as em (3 equal areas) = 6a².
agl c
m .
m a s e
. co agl
sem
gla 125–126
Exercise Set 14.1 — Pages
a

co m
m .
m as e
.co

ag l Page 4 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Section 14.1 Cuboids and Cubes

Q1 The volume of a cube is 64 cm³. What is its total surface area?

a³ = 64 ⇒ a = ∛64 = 4 cm
TSA = 6a² = 6 × 4² = 6 × 16 = 96 cm²

The total surface area is 96 cm².

Check it yourself: 4 × 4 × 4 = 64, so the side really is 4 cm. Each face is 4 × 4
= 16 cm², and there are 6 faces.

Q2 How many small cubes with side 20 cm can be packed tight in a cubical box with side 2
m?

First use the same unit: 2 m = 200 cm.

Cubes along one edge = 200 ÷ 20 = 10

Number of cubes = 10 × 10 × 10 = 1000

1000 small cubes can be packed.

Check it yourself: by volume, box = 200³ = 80,00,000 cm³ and one small cube
= 20³ = 8000 cm³; 80,00,000 ÷ 8000 = 1000. The two methods agree because
20 divides 200 exactly, so no space is wasted.

Page 5 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Tip: the common mistake is to divide 2 by 20 without converting metres to
centimetres.

Q3 The dimensions of a godown are 40 m × 25 m × 10 m. If it is filled with cuboidal boxes,
each of dimensions 2 m × 1.25 m × 1 m, then find the number of boxes.

Line the boxes up along each edge of the godown:

Along 40 m: 40 ÷ 2 = 20 boxes

Along 25 m: 25 ÷ 1.25 = 20 boxes
Along 10 m: 10 ÷ 1 = 10 boxes
Number of boxes = 20 × 20 × 10 = 4000

The godown holds 4000 boxes.

Check it yourself: volume of godown = 40 × 25 × 10 = 10,000 m³; volume of
one box = 2 × 1.25 × 1 = 2.5 m³; 10,000 ÷ 2.5 = 4000.

Why both methods agree: each edge of the godown is an exact multiple of the
matching edge of a box. If it were not (say the godown were 41 m long), dividing
the volumes would give a number larger than the boxes that can really fit,
because the leftover strips cannot hold a whole box.

Page 6 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q4 Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the
resulting cuboid.

a³ = 125 ⇒ a = 5 cm

Resulting cuboid: l = 5 + 5 = 10 cm, w = 5 cm, h = 5 cm

TSA = 2(lw + wh + hl) = 2(10×5 + 5×5 + 5×10)
= 2(50 + 25 + 50) = 2 × 125 = 250 cm²

The surface area of the cuboid is 250 cm².

Check it yourself: the two cubes separately have 2 × 6 × 25 = 300 cm². When
they are joined, one face of each cube is hidden: 300 − 2 × 25 = 250 cm².

Tip: do not just double 150 cm². The faces that touch are no longer on the
outside.

Q5 A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface
areas of the original cube and all the cut-out cubes? (Note that there is no change in
volume but a big change in the surface area. This property has major consequences in
the biological world.)

Surface area of the original cube = 6 × 4² = 96 cm²

Number of small cubes = 4 × 4 × 4 = 64

Surface area of one small cube = 6 × 1² = 6 cm²
Total surface area of the small cubes = 64 × 6 = 384 cm²

Ratio = 96 : 384 = 1 : 4

Page 7 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

The ratio is 1 : 4. The volume stays 64 cm³, but cutting makes 4 times as much surface.

Why: every cut opens up new faces that were hidden inside. If a cube of side n
is cut into unit cubes, the surface goes from 6n² to n³ × 6 = 6n³, which is n
times as much. Here n = 4.

Did you know? Small things have a lot of surface for their volume. This is why
living cells stay tiny (food and oxygen enter through the surface), why crushed
sugar dissolves faster than a lump, and why small animals lose body heat faster
than large ones.

Q6 The surface areas of the three faces of a cuboid that meet at one of the corners of the
cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?

The three faces at a corner are l × w, w × h and h × l. Let lw = 6, wh = 15, hl = 10.

(lw)(wh)(hl) = 6 × 15 × 10
l²w²h² = 900

(lwh)² = 900 ⇒ lwh = 30

The volume of the cuboid is 30 cm³.

Check it yourself: the edges are 2 cm, 3 cm and 5 cm: 2 × 3 = 6, 3 × 5 = 15, 5 ×
2 = 10, and 2 × 3 × 5 = 30.

Why it works: each edge appears in exactly two of the three faces, so
multiplying the three areas gives every edge twice, i.e. (lwh)². Taking the square
root gives the volume directly, without finding each edge.

Page 8 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

m
cohow
.
A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes,
em
*Q7
m
many of these 1 cm³ cubes have
as
.co three faces painted? a gl
emexactly two faces painted?
(i) exactly
a s
agl (iii) exactly one face painted?
(ii)

(iv) no face painted?
co m
m . ag
la se
ag

co m
em.
m l as
m .co ag
la se
ag
1 cm
m a s
m .co agl
1 cm la se
ag
5 cm 1 cm
co m
m .
as e
com l
1 cm
. ag
a se m
agl 1 cm

se m
1 cmom1 cm 1 cm 1 cm 1 cm
g l a
m .c a
se 5 cm
a gla
Figure on page 126 — the painted 5 cm cube, marked into 1 cm slices.

co m
m .
m la se
.co
ag
a s emThere are 5 × 5 × 5 = 125 small cubes in all.
agl c
m .
m a s e
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 9 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

(i) Three faces painted: only the corner cubes → 8

(ii) Two faces painted: cubes on an edge but not at a corner
each edge has 5 − 2 = 3 such cubes, 12 edges → 12 × 3 = 36

(iii) One face painted: inner cubes of each face

each face has 3 × 3 = 9, 6 faces → 6 × 9 = 54
(iv) No face painted: the hidden inner cube 3 × 3 × 3 → 27

Check: 8 + 36 + 54 + 27 = 125.

Why: a small cube gets one painted face for each outer face of the big cube it
touches. A corner touches 3 faces, an edge cube (not at a corner) touches 2, a
cube in the middle of a face touches 1, and the cubes inside touch none.
Removing the outer layer from every side leaves a cube of side 5 − 2 = 3.

Tip: for a cube of side n cut into unit cubes the counts are 8, 12(n − 2), 6(n − 2)²
and (n − 2)³. With n = 5 these give 8, 36, 54 and 27.

*Q8 Find a cuboid with edges whose lengths are integers (in cm), given that it has a total
surface area of exactly 100 cm².
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.

2(lw + wh + hl) = 100, so lw + wh + hl = 50. Take l ≥ w ≥ h ≥ 1.
(i) Yes. (ii) Exactly two: 16 × 2 × 1 and 7 × 4 × 2 (in cm).
(iii) Each product is at least h², so 3h² ≤ 50 and h ≤ 4. Adding h² to both sides of lw + h(l + w) =
50 gives

Page 10 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

(l + h)(w + h) = 50 + h², where both factors are at least 2h

h = 1: 51 = 3 × 17 → w = 2, l = 16

h = 2: 54 = 6 × 9 → w = 4, l = 7 (2 × 27 and 3 × 18 have a factor below 4)
h = 3: 59 is prime → none

h = 4: 66 = 6 × 11, 3 × 22, 2 × 33, 1 × 66 → no factor reaches 8 → none

All cases are checked, so there are no others.

Check it yourself: 16×2 + 2×1 + 1×16 = 32 + 2 + 16 = 50, and 7×4 + 4×2 + 2×7
= 28 + 8 + 14 = 50. Doubling gives 100 cm² in both cases.

Why the factorising trick works: (l + h)(w + h) = lw + lh + hw + h² = (lw + wh +
hl) + h² = 50 + h². This turns a sum we cannot easily search into a product, and
a product of whole numbers can only be split in a few ways.

Why both factors are at least 2h: we chose w ≥ h and l ≥ h, so w + h ≥ 2h and l
+ h ≥ 2h.

Tip: a cube is not possible: 6a² = 100 gives a² = 16.67, which is not a perfect
square.

Think and Reflect — Page 126

Page 11 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Section 14.1 Cuboids and Cubes
THINK AND REFLECT

Q1 Suppose a swimming pool is being built in your school. Two choices are available for
the dimensions of the pool: (A) 8 ft (depth) × 50 ft × 20 ft and (B) 6 ft (depth) × 100 ft ×
15 ft. Which option would be suitable/appropriate for your school? Why? What
parameters/aspects did you consider to make the choice? [1 cubic foot ≈ 28.3 litres]

POOL A POOL B

Depth 8 ft ≈ 2.4 m 6 ft ≈ 1.8 m

Water (volume) 8 × 50 × 20 = 8000 ft³ ≈ 2,26,400 6 × 100 × 15 = 9000 ft³ ≈
L 2,54,700 L

Water surface (land 50 × 20 = 1000 ft² 100 × 15 = 1500 ft²
needed)

Area to tile (floor + 4 walls) 1000 + 2(50 + 20) × 8 = 2120 ft² 1500 + 2(100 + 15) × 6 = 2880 ft²

Option B suits a school better. 8 ft is far above the head of a Class 9 student (about 5 ft tall),
which is unsafe for learners; B is 2 ft shallower. B is also 100 ft (about 30 m) long, good for
swimming laps and races, and its larger surface lets more students swim at once. The price is
about 28,300 L more water, 50% more land and more tiling.
Parameters considered: safety (depth), purpose (lessons, laps), water needed, land available,
and cost of digging and tiling.

Why surface area matters too: volume decides how much water you must fill
and treat. The floor area decides the land needed, and the floor + wall area
decides how many tiles you must buy. A pool is a cuboid open at the top, so it
has 5 faces, not 6.

Check it yourself: 8000 × 28.3 = 2,26,400 litres and 9000 × 28.3 = 2,54,700
litres. The difference, 1000 ft³, is about 28,300 litres.

Page 12 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Your own answer: you may choose differently. If water is scarce in your area or
the school has little open ground, A uses less of both. If the pool is meant for
diving, A's extra depth is needed. Real school pools often have a sloping floor,
shallow at one end and deep at the other.

Exercise Set 14.2 — Pages 129–130
Section 14.2 Right Circular Cylinder

Q1 Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A,
and the height of cylinder B is half that of cylinder A. Find the ratio of the curved
surface area of A to the curved surface area of B. Also find the ratio of the volume of A
to the volume of B.

Let cylinder A have radius r and height h. Then B has radius 2r and height h/2.

CSA of A = 2πrh

CSA of B = 2π(2r)(h/2) = 2πrh
CSA of A : CSA of B = 1 : 1

Volume of A = πr²h
Volume of B = π(2r)²(h/2) = π × 4r² × h/2 = 2πr²h
Volume of A : Volume of B = 1 : 2

Why the answers differ: the curved surface depends on r × h, so doubling r
and halving h cancel out. The volume depends on r² × h: doubling r multiplies it
by 4, halving h divides it by 2, leaving a factor of 2.

Page 13 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Check it yourself: take r = 1, h = 4 for A and r = 2, h = 2 for B. CSA: 2π ×c4om
. = 8π
sem a
com
and 2π × 4 = 8π. Volume: 4π and 8π.
m . agl
la se
ag
m and their heights are in the ratio 3:2. Find
The radii of two cylinders are in the ratioo2:3,
. c ag
Q2
(a) the ratio of their volumes, and (b)mthe ratio of their curved surface areas.
la se
ag
Let the radii be 2k and 3k, and the heights 3m and 2m.
co m
em.
om= π(2k)²(3m) : π(3k)²(2m)
(a) V₁ .: cV₂ gl as
em a
la s
ag
= 12k²m : 18k²m = 2 : 3

m a s
(b) CSA₁ : CSA₂ = 2π(2k)(3m) : 2π(3k)(2m)
m .co agl
la se
ag
= 12km : 12km = 1 : 1

com
m .
o m l ase
g ratio)² × (height
.c you can work with ratios directly. Volume ratio = (radius
a
se mTip:
a
agl
ratio) = (4/9) × (3/2) = 2/3. CSA ratio = (radius ratio) × (height ratio) = (2/3) ×
(3/2) = 1.
se m
com g l a
m . a
e
ag las
Q3 The edge of a cube measures r cm. The largest possible right circular cylinder is cut
out of the cube. What do you think is the volume of the cylinder (inom
cm³)?
.c
s e m
. com a gla
emThe largest cylinder stands on one face of the cube. Its circular base just touches the four

a s
agl sides of that square face, so its diameter is r and its radius is r/2. Its height is the edge, r. c
m .
m a s e
. co agl
sem
V = π × (r/2)² × r = π × r²/4 × r = πr³/4 cm³

a g la
That is about 0.785 r³, roughly 78.5% of the cube.

co m
m .
m as e
.co

ag l Page 14 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Tip: here r is the edge of the cube, not the radius of the cylinder. The common
mistake is to write πr²·r = πr³.

Why π/4: the base circle fills π/4 of the square face (area π(r/2)² against r²),
and since both solids have the same height r, the cylinder fills the same fraction
π/4 of the cube.

*Q4 The radius of the base of a cylinder is increased by 10%. At the same time, the height
of the cylinder is decreased by x%. Given that the volume of the cylinder remains
unchanged, find the value of x.

Let the original radius be r and height h. New radius = 1.1r, new height = h(1 − x/100).

π(1.1r)² × h(1 − x/100) = πr²h
1.21 × (1 − x/100) = 1

1 − x/100 = 1/1.21 = 100/121
x/100 = 21/121

x = 2100/121 = 1743⁄121 ≈ 17.36

The height must decrease by about 17.36% (exactly 2100/121 %).

Why not 10%: the radius is squared in the volume, so a 10% increase in radius
makes the base area 21% larger (1.1² = 1.21). The height must shrink enough to
cancel a factor of 1.21, which is a cut of 21/121 ≈ 17.4%, not 21% and not 10%.

Check it yourself: 1.21 × (1 − 0.17355) = 1.21 × 0.82645 ≈ 1.0000.

Page 15 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q5 A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each
having radius 2 cm and height 12 cm. Find:
(i) the volume of the cube,
(ii) the volume of one cylindrical rod,
(iii) the approximate number of complete cylindrical rods that can be formed. (π ≈
22/7)

(i) Volume of cube = 12³ = 1728 cm³

(ii) Volume of one rod = πr²h = 22/7 × 2² × 12 = 1056/7 ≈ 150.86 cm³
(iii) Number of rods = 1728 ÷ (1056/7) = 1728 × 7/1056 ≈ 11.45

So 11 complete rods can be formed. The metal left over (about 68.6 cm³) is not enough for a
twelfth rod.

Why volume: melting and recasting changes the shape, not the amount of
metal, so the total volume stays the same.

Check it yourself: 11 rods use 11 × 150.86 ≈ 1659.4 cm³; 1728 − 1659.4 ≈ 68.6
cm³ remains, less than one rod (150.86 cm³). Do not round 11.45 up to 12: the
question asks for complete rods.

Exercise Set 14.3 — Pages 133–134

Page 16 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Section 14.3 Cones (use π ≈ 22/7 or π ≈ 3.14)

Q1 Find the total surface area of a cone, if its slant height is 21 m and diameter of its base
is 24 m.

r = 24 ÷ 2 = 12 m, l = 21 m
TSA = πr(l + r) = 22/7 × 12 × (21 + 12)
= 22/7 × 12 × 33 = 8712/7 ≈ 1244.57 m²

The total surface area is about 1244.57 m².

Tip: the formula needs the radius, not the diameter. Using 24 instead of 12 is
the most common slip.

Check it yourself: with π ≈ 3.14 the answer is 3.14 × 12 × 33 = 1243.44 m². The
small difference comes only from the value of π used.

Q2 Find the curved surface area of a right circular cone whose slant height is 10 cm and
base radius is 7 cm.

CSA = πrl = 22/7 × 7 × 10 = 220 cm²

The curved surface area is 220 cm².

Tip: when the radius is a multiple of 7, π ≈ 22/7 makes the arithmetic exact.

Page 17 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q3 The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface
area and the total surface area of the cone.

l = √(h² + r²) = √(16² + 12²) = √(256 + 144) = √400 = 20 cm
CSA = πrl = 3.14 × 12 × 20 = 753.6 cm²

TSA = πr(l + r) = 3.14 × 12 × (20 + 12) = 3.14 × 384 = 1205.76 cm²

CSA = 753.6 cm² and TSA = 1205.76 cm².

Why find l first: the surface formulas use the slant height l, but the question
gives the vertical height h. The height, the radius and the slant height form a
right triangle (Baudhāyana–Pythagoras theorem), so l² = h² + r².

Check it yourself: with π ≈ 22/7, CSA = 240π ≈ 754.29 cm² and TSA = 384π ≈
1206.86 cm².

Q4 A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.

V = (1/3)πr²h
1570 = (1/3) × 3.14 × r² × 15 = 15.7 r²

r² = 1570 ÷ 15.7 = 100
r = 10 cm

The radius of the base is 10 cm.

Page 18 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

com
m.

m as e
c o
. = 3.14 × 500 is a hint that π ≈ 3.14 is intended here. ag l
e m
Tip: 1570
g las
a
Check it yourself: (1/3) × 3.14 × 100 × 15 = 3.14 × 500 = 1570 cm³.
co m
m . ag
la se
ag
m
The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the
.co
Q5

em
radius of the base, (ii) the total surface area of the cone.
m as
em
.c o agl
s
a
agl
(i) πrl = 308
m a s
22/7 × r × 14 = 308 ⇒ 44r = 308 ⇒ r = 7 cm
m .co agl
la se
ag
(ii) TSA = CSA + area of base = 308 + 22/7 × 7² = 308 + 154 = 462 cm²

co m
.

se m
om l a
m.c it yourself: πr(l + r) = 22/7 × 7 × (14 + 7) = 22 ×a21g = 462 cm².
se
Check
a
agl
se m
com g l a
. a
ema right circular cone of base radius 7 cm and height 24
A joker’s cap is in the formsof
a
aglsheet required to make 10 such caps.
Q6
cm. Find the area of the

co m
.

se m
m a
A cap has no base, so only its curved surface is made of sheet.
.co ag l
a s em l = √(24² + 7²) = √(576 + 49) = √625 = 25 cm
agl c
m .
e
CSA of one cap = πrl = 22/7 × 7 × 25 = 550 cm²
m a s
Sheet for 10 caps = 10 × 550 = 5500 cm²
. co agl
la sem
ag cm² (0.55 m²).
The sheet required is 5500

co m
m .
m as e
.co

ag l Page 19 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Tip: the height 24 cm is not the slant height. The sheet wraps along the slant,
so you must first find l = 25 cm.

*Q7 What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and
base radius 6 m? Assume that the extra length of material that is required for stitching
margins and wastage in cutting is 20 cm.

The tent has no floor, so the tarpaulin covers only the curved surface.

l = √(8² + 6²) = √100 = 10 m

CSA = πrl = 3.14 × 6 × 10 = 188.4 m²
Length of tarpaulin = area ÷ width = 188.4 ÷ 3 = 62.8 m
Add 20 cm = 0.2 m for margins and wastage: 62.8 + 0.2 = 63 m

The length of tarpaulin required is 63 m.

Why divide by the width: the tarpaulin is a long rectangle 3 m wide. Its area is
length × 3, and this must equal the curved surface area of the tent.

Check it yourself: with π ≈ 22/7, CSA = 60π ≈ 188.57 m², length ≈ 62.86 m, and
with the margin about 63.06 m. Either way, about 63 m.

Tip: change 20 cm into metres (0.2 m) before adding.

Page 20 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q8 A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side
of 8 cm. Find the volume and the curved surface area of the solid so formed.

6² + 8² = 36 + 64 = 100 = 10², so the right angle is between the 6 cm and 8 cm sides. Turning
about the 8 cm side gives a cone with h = 8 cm, r = 6 cm, l = 10 cm.

V = (1/3)πr²h = (1/3) × 3.14 × 36 × 8 = 96π ≈ 301.44 cm³
CSA = πrl = 3.14 × 6 × 10 = 60π ≈ 188.4 cm²

Volume ≈ 301.44 cm³, curved surface area ≈ 188.4 cm².

Why: the side we turn about stays fixed and becomes the height (axis). The
other short side sweeps out the circular base, so it is the radius. The
hypotenuse sweeps out the slanting surface, so it is the slant height.

Check it yourself: with π ≈ 22/7: V = 96 × 22/7 ≈ 301.71 cm³ and CSA = 60 ×
22/7 ≈ 188.57 cm².

Page 21 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

*Q9 Suppose you have a cup in the shape of a right circular cone. Fill it with water to half
the depth of the cone. What fraction of the volume of the cup is occupied by the
water?

depth of cup

water: half
the depth

Picture on page 134 — a conical cup, point down, filled with water to half its depth.

Let the cup have radius r and depth h. The water forms a smaller cone of the same shape with
depth h/2. By similar triangles its radius is r/2.

Volume of water = (1/3)π(r/2)²(h/2) = (1/3)πr²h × (1/4) × (1/2)

= (1/8) × volume of the cup

The water fills only 1/8 of the cup.

Why so little: halving every length of a cone (both the radius and the depth)
multiplies its volume by (1/2)³ = 1/8. The water is a half-size copy of the cup.

Page 22 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Did you know? The top half of the cup holds the other 7/8. To fill the cup
halfway by volume, the water must reach about 0.79 of the depth, because
0.79³ ≈ 1/2.

Tip: this assumes the cup is held point down, as conical cups are. If a cone
stood on its flat base and were filled to half its height, the water would fill 7/8 of
it.

In-text Question — Page 137
Section 14.5 Spheres and Hemispheres

Q1 Archimedes found the formula for the volume of a sphere, V = (4/3)πr³. You may be
able to find your own derivation of the formula by connecting it with the formula for
volume of a cone; namely, by writing it as V = (1/3) × (4πr²) × r, i.e., volume of sphere =
(1/3) × surface area of sphere × radius of sphere. Try to work out the connecting links
on your own.

Compare with the cone formula, V = (1/3) × area of base × height. The two match if the sphere
is built out of many thin cones:
Divide the surface of the sphere into very tiny pieces. Join every piece to the centre O. Each
piece becomes the base of a thin cone whose apex is O and whose height is the radius r.

Volume of sphere = sum of volumes of all the thin cones
= (1/3) × b₁ × r + (1/3) × b₂ × r + … + (1/3) × bₙ × r
= (1/3) × (b₁ + b₂ + … + bₙ) × r

= (1/3) × (surface area of sphere) × r
= (1/3) × 4πr² × r = (4/3)πr³

Page 23 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

co m
m.

o m l ase
Whym c
it .works: a gout as a
se factor, just like the slant height l came out in the cone's surface area
all the cones have the same height r, so r comes
la
ag
common
on page 131. The base areas then add up to the whole surface, 4πr².

. com ag
Did you know? The same trick gives
a s emthe area of a circle. Cut a circle into thin
agl has height r, so area = (1/2) × circumference × r
triangles from the centre: each
= (1/2) × 2πr × r = πr². For the circle the factor is 1/2 (triangles); for the sphere it
m
.co
is 1/3 (cones).
se m
om l a
g pieces, the
Tip:.cthe tiny bases are slightly curved, not flat. The smallerathe
e m
ag lascloser they are to flat, and the result becomes exact. The book shows this in the
box "Visualising Volume Formula" on page 138.
m a s
m .co agl
la se
ag
Exercise Set 14.4 — Pages 140–141
co m
.
Section 14.5 Spheres and Hemispheres

se m
o m g l a
Q1m.cA ball bearing has a radius of 0.7 cm. Find its surface area.
a
e
las
ag ANSWER
se m
c om g l a
SA = 4πr² = 4 × 22/7 × 0.7 × 0.7 e=m . a
a s
4 × 22 × 0.1 × 0.7 = 6.16 cm²

agl
The surface area is 6.16 cm².
co m
m .
se
m la
.co ag
a s em
agl
Tip: cancel 7 against 0.7 first: 22/7 × 0.7 = 2.2. Then 4 × 2.2 × 0.7 = 6.16.
.c
s e m
m a
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 24 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

*Q2 Two solid spheres made of the same metal have weights 5920 g and 740 g.
Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.

Both spheres are of the same metal, so weight is proportional to volume.

V(large) : V(small) = 5920 : 740 = 8 : 1

(4/3)πR³ : (4/3)πr³ = 8 : 1 ⇒ R³ / r³ = 8 ⇒ R/r = 2
r = 5 ÷ 2 = 2.5 cm ⇒ R = 2 × 2.5 = 5 cm

The radius of the larger sphere is 5 cm.

Why cube root: volume grows as the cube of the radius. A sphere 8 times as
heavy has 8 times the volume, so its radius is ∛8 = 2 times as large, not 8 times.

Tip: the smaller sphere's diameter is given; halve it first.

Q3 The diameter of the moon is approximately one fourth the diameter of the earth. Given
that the moon and earth are both roughly spherical, find the ratio of their surface
areas.

If the moon's diameter is 1/4 of the earth's, its radius is also 1/4 of the earth's: r(moon) = R/4.

SA(moon) : SA(earth) = 4π(R/4)² : 4πR²
= 1/16 : 1 = 1 : 16

Page 25 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Why squared: area depends on the square of the radius. One-fourth the size
means (1/4)² = 1/16 of the surface. (Volume would be (1/4)³ = 1/64.)

Q4 Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius
21 cm.

(i) CSA = 2πr² = 2 × 22/7 × 21 × 21 = 2 × 22 × 3 × 21 = 2772 cm²
(ii) TSA = 3πr² = 3 × 22/7 × 21 × 21 = 3 × 22 × 3 × 21 = 4158 cm²

Why 3πr²: the total surface is the curved half of a sphere (2πr²) plus the flat
circular base (πr²).

Check it yourself: TSA − CSA = 4158 − 2772 = 1386 = 22/7 × 441, the area of
the base circle.

Page 26 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q5 Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal
dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with
preservative liquid. Find the volume of this liquid. Round your answer to the nearest
integer.

Volume of box = 16 × 8 × 8 = 1024 cm³
Volume of one sphere = (4/3)π × 2³ = 32π/3 cm³
Volume of 16 spheres = 16 × 32π/3 = 512π/3 ≈ 512 × 3.14 ÷ 3 ≈ 535.9 cm³ (π ≈ 3.14)
Volume of liquid = 1024 − 535.9 ≈ 488.1 cm³

The volume of the liquid is about 488 cm³.

Why the liquid fills the rest: the liquid takes up all the space in the box that the
spheres do not.

Check it yourself: the 16 spheres fit exactly. Each has diameter 4 cm, so 16 ÷ 4
= 4 fit along the length, 8 ÷ 4 = 2 along the width and 2 in height: 4 × 2 × 2 = 16.
With π ≈ 22/7 the spheres take 536.4 cm³ and the liquid 487.6 cm³, which also
rounds to 488 cm³.

Did you know? The spheres fill only 535.9 ÷ 1024 ≈ 52% of the box. This is the
"first configuration" packing of page 143, where each ball sits in its own cube.

Q6 The radius of a sphere is increased by 10%. Show that the volume increases by
approximately 33.1%.

Let the radius be r. The new radius is 1.1r.

Page 27 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

New volume = (4/3)π(1.1r)³ = 1.331 × (4/3)πr³
Increase = 1.331V − V = 0.331V
Percentage increase = 0.331 × 100 = 33.1%

So the volume increases by 33.1%, as required.

Why not 10% or 30%: all three dimensions grow by 10%, and the growth
compounds: 1.1 × 1.1 × 1.1 = 1.331.

Check it yourself: 1.1² = 1.21 and 1.21 × 1.1 = 1.331.

*Q7 The radius of a sphere is increased by x%. The volume of the sphere increases by
72.8%. Find the value of x.

New volume = 1.728 × old volume
(1 + x/100)³ = 1.728
1 + x/100 = ∛1.728 = 1.2 (since 12³ = 1728)
x/100 = 0.2 ⇒ x = 20

The radius was increased by 20%.

Check it yourself: 1.2³ = 1.2 × 1.2 × 1.2 = 1.44 × 1.2 = 1.728, an increase of
72.8%.

Page 28 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

. com
Tip: 1.728 = 1728/1000, and 1728 = 12³, 1000 = 10³. Spotting cube numbers
sem a
com
makes the cube root easy.
m . agl
la se
ag
Q8 The hemispherical dome of a building needs
. c om to be painted (see Fig. 14.16). If the ag
s
circumference of the base of the dome
e m is 35.2 m, find the cost of painting it, given the
a
cost of painting is ₹ 10 per 100lcm².
ag

co m
em.
m l as
m .co ag
la se
ag
m a s
m .co agl
la se hemispherical
ag dome

co m
m .
as e
. com agl
a se m
agl
se m
com g l a
m . a
e
g las
a page 141 — a building topped by a hemispherical dome.
Fig. 14.16,

co m
m .
m la se
.co
ag
a s em
agl 2πr = 35.2 ⇒ r = 35.2 × 7/44 = 5.6 m
c
m .
a s e
om agl
Area to paint = CSA = 2πr² = 2 × 22/7 × 5.6 × 5.6 = 197.12 m²
. c
s e m cm²
= 197.12 × 10,000 cm² = 19,71,200
a
ag×l ₹10 = ₹1,97,120
Cost = 19,71,200 ÷ 100

co m
m .
m as e
.co

ag l Page 29 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

The cost of painting the dome is ₹1,97,120.

Why only the curved surface: the paint goes on the outside of the dome. The
flat base of the hemisphere sits on the building and is not painted.

Tip: 1 m² = 100 cm × 100 cm = 10,000 cm², not 100 cm². The rate is ₹10 per 100
cm², which is ₹1000 per m², so cost = 197.12 × 1000 = ₹1,97,120.

In-text Questions — Page 141
Section 14.6 Areas and Volumes Around Us
ACTIVITY

Q1 We learned about different 3D solids. Identify and list objects or structures around you
that are cubical, cuboidal, cylindrical, conical, spherical, or pyramidal in shape, or
closely resemble these shapes.

SHAPE OBJECTS AND STRUCTURES AROUND US

Cube dice, Rubik’s cube, ice cubes, sugar cubes, some cardboard boxes

Cuboid bricks, books, matchbox, almirah, classroom, shoe box, packet of milk or juice,
water tank on a roof (some)

Cylinder water pipes, gas cylinder, drums, candles, pillars, tumblers, tube light, coins (very
short cylinders)

Cone ice-cream cone, birthday cap, funnel, traffic cone, the roof of a round hut

Sphere / balls, marbles, globe, laddoo, oranges; bowls, domes and igloos are hemispheres
hemisphere

Pyramid the pyramids of Egypt, a tetrahedral puzzle, some roofs and temple towers, tents
with a square base, peaks such as Shivling

Page 30 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Your own answer: look around your home, school and market and add objects
you actually see. Many objects are combinations: a pencil is a cylinder (or
hexagonal prism) with a cone at the end; a capsule tablet is a cylinder with two
hemispheres; a temple may have a cuboid hall and a pyramidal or conical top.

Tip: "closely resemble" is enough. A bucket is wider at the top than at the
bottom, so it is not a true cylinder, but treating it as one gives a useful estimate.

Q2 Activity: Find the approximate volumes of a gas cylinder, a bucket, any cylindrical
vessel, a cupboard, a water tank that you can find near you (use 1000 cm³ = 1 litre).

Method: measure each object with a measuring tape, model it as a cylinder or a cuboid, use V
= πr²h (π ≈ 3.14) or V = lwh, and divide cm³ by 1000 to get litres.

OBJECT (SAMPLE MEASUREMENTS) MODEL AND WORKING VOLUME

Gas cylinder: diameter 30 cm, height of the 3.14 × 15² × 50 = 35,325 cm³ ≈ 35 L
straight part 50 cm

Bucket: top diameter 30 cm, bottom 24 cm, cylinder of average diameter 27 cm: 3.14 ≈ 17 L
height 30 cm × 13.5² × 30 ≈ 17,168 cm³

Steel vessel: diameter 20 cm, height 15 cm 3.14 × 10² × 15 = 4710 cm³ ≈ 4.7 L

Cupboard: 90 cm × 50 cm × 180 cm 90 × 50 × 180 = 8,10,000 cm³ 810 L

Cylindrical water tank: diameter 110 cm, 3.14 × 55² × 120 = 11,39,820 cm³ ≈ 1140 L
height 120 cm

These are sample values: a tape and a formula are enough to estimate any container.

Why these are approximate: the gas cylinder has rounded ends, the bucket
narrows towards the bottom, and walls have thickness. The model ignores
these details, so the answer is an estimate, which is all the activity asks for.

Page 31 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Tip: for the bucket, using the average diameter is a good trick. The exact
formula for a bucket shape (a frustum of a cone) gives about 17,240 cm³ here,
very close to our 17,168 cm³.

Your own answer: replace the sample numbers with your own measurements.
A water tank sold as a "1000-litre tank" makes a good check of your working.

In-text Question — Page 143
Section 14.6.1 Guesstimates: Did You Ever Wonder? (Example 6, tennis balls in a classroom)

Q1 The classroom is 30 ft × 30 ft × 15 ft (about 900 cm × 900 cm × 450 cm) and the tennis
balls have radius 3 cm. Dividing volumes gives an upper bound of about 33,45,132
balls. Using the first configuration of Fig. 14.17, 150 × 150 × 75 = 16,87,500 balls fit.
Using the other configurations, we might be able to fit more balls because of higher
packing efficiency. Make an estimate based on the second and third configurations.

Fig. 14.17: These are some of the ordered ways of packing spheres in a
cuboidal structure
Three pictures, labelled A, B and C. Each shows identical blue balls in two layers
on a flat square base; in each layer the balls touch in straight rows, 3 balls by 2.
In A every ball of the upper layer rests directly on top of a ball below. In B the
whole upper layer is shifted sideways by half a ball, so each upper ball rests in
the groove between two balls below. In C the balls look the most tightly packed:
the upper layer nestles into the hollows of the layer below.

Fig. 14.17, page 143 — a picture, described here.

Count as the book does for the first configuration (room 900 × 900 × 450 cm, ball diameter 6
cm), but the layers now sit lower.

Page 32 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Second (B): balls rest in the grooves; layer gap = √(6² − 3²) ≈ 5.2 cm
Layers = (450 − 6) ÷ 5.2 + 1 ≈ 86, alternately 150 × 150 and 149 × 150 balls
Total ≈ 43 × 22,500 + 43 × 22,350 ≈ 19,28,550

Third (C): balls rest in the hollows of four; layer gap = √(6² − (3√2)²) ≈ 4.24 cm
Layers = (450 − 6) ÷ 4.24 + 1 ≈ 105, alternately 150 × 150 and 149 × 149 balls

Total ≈ 53 × 22,500 + 52 × 22,201 ≈ 23,46,950

About 19 lakh balls (second) and 23 lakh (third): more than 16.9 lakh, less than the 33.5 lakh
bound.

Why the layers are closer: in A a ball sits on the top of the ball below, so the
centres are 6 cm (one diameter) apart vertically. In B and C the ball drops into a
gap. The slanting line joining its centre to the centre of a ball it touches is still 6
cm, so by the Baudhāyana–Pythagoras theorem the vertical gap is shorter: √(6²
− 3²) when it is 3 cm sideways (B, a groove), √(6² − 18) when it is 3√2 ≈ 4.24
cm sideways (C, the middle of a 6 cm square, half its diagonal).

Why 149 in the shifted layers: a shifted layer starts half a ball in from the wall,
so along that direction one ball fewer fits (centres from 6 cm to 894 cm).

Check it yourself (packing efficiency): the fraction of space the balls fill is π/6
≈ 52.4% for A, π/(3√3) ≈ 60.5% for B and π/(3√2) ≈ 74.0% for C. The room is
900 × 900 × 450 = 36,45,00,000 cm³, room for 36,45,00,000 ÷ 113.1 ≈ 32.2
lakh balls if there were no gaps. Then 52.4% gives 16,87,500 (exactly the
book's count for A), 60.5% gives about 19.5 lakh and 74.0% about 23.9 lakh.
Counting gives slightly less, because of the leftover space near the walls.

Tip: the book's bound 33,45,132 uses 1 ft³ ≈ 28,000 cm³, while its count uses 1
ft ≈ 30 cm (900 cm). Use one conversion throughout. The book does not name
the three arrangements; the descriptions above are read from the picture.

Page 33 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

. com
Did you know? Arrangement C is how fruit sellers pile oranges. In 1611 Kepler
m of
guessed that no arrangement of equal balls can fill more than aboutse74%
m
co was finally proved by Thomas Hales in 1998. agl
space; .this
a
se m
la
ag

. com ag
End-of-Chapter Exercises — Pages
a s em144–147
l a guess first before solving)
agmake
Whole chapter (for guesstimate problems,

m
cocontainer
.
Estimate how many scoops of ice cream can be obtained from a cuboidal of
sem
Q1

com a
dimensions 10 cm × 15 cm × 20 cm.

m . agl
la se
ag
Guess first: perhaps about 50.

m a s
Volume of ice cream = 10 × 15 × 20 =m .co cm³ agl
se
3000
la
ag 5 cm (r = 2.5 cm)
Assume a scoop is a ball of diameter

m
Volume of a scoop = (4/3) × 3.14 × 2.5³ ≈ 65.4 cm³

. co
Number of scoops ≈ 3000 ÷ 65.4 ≈ 46
se m
om l a
m.c45 scoops can be obtained (a little is lost in the corners).ag
la se
About
ag − Explanation
se m
com g l a
m . a
Why no gaps here: unlike sthe
a e tennis balls, the ice cream in the box is one solid
g
block. Each scoop isapressedl into a ball, so dividing the volumes works directly.
In practice a little is lost in the corners and on the scoop.
co m
m .
m the answer depends on the scoop size you assume.asAelarger scoop of
oTip:
l your assumption
.c radius 3 cm (≈ 113 cm³) gives only about 26 scoops.agState
a s em
agl
clearly.
.c
s e m
m a
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 34 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q2 A cube of integer side length a is made of unit cubes. Write an expression giving the
number of unit cubes to be added to make a cube of side length a + 1.

Cubes needed = (a + 1)³ − a³
= (a³ + 3a² + 3a + 1) − a³
= 3a² + 3a + 1

Picture it: add three square slabs a × a × 1 on three faces (3a²), three rods a × 1 × 1 along the
edges where the slabs meet (3a), and one corner cube (1).

Check it yourself: a = 1: 8 − 1 = 7 = 3 + 3 + 1. a = 2: 27 − 8 = 19 = 12 + 6 + 1. a =
3: 64 − 27 = 37 = 27 + 9 + 1.

Why the geometry matches the algebra: growing a cube by 1 in all three
directions needs a layer on three faces, then the gaps along the three edges
where those layers meet, then the single corner where all three meet.

*Q3 Solve the following:
(i) Could a person drink enough water in a lifetime to fill an entire room the size of
your classroom?
(ii) Estimate the number of bricks used to build the walls of your classroom.

Sample classroom: 8 m × 6 m × 3 m.

(i) Room = 8 × 6 × 3 = 144 m³ = 1,44,000 L (1 m³ = 1000 L)
Lifetime drinking ≈ 2 L a day × 365 × 70 years ≈ 51,100 L

No. A lifetime of drinking fills only about one-third of the room. It would take about 200 years.

Page 35 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

(ii) Wall area = perimeter × height = 2(8 + 6) × 3 = 84 m²
Minus 1 door (1 × 2.1) and 3 windows (1.5 × 1.2 each): 84 − 7.5 = 76.5 m²
Wall volume (thickness 20 cm) = 76.5 × 0.2 = 15.3 m³ = 1,53,00,000 cm³
One brick with mortar ≈ 20 × 10 × 10 = 2000 cm³

Bricks ≈ 1,53,00,000 ÷ 2000 ≈ 7650

About 7,500 to 8,000 bricks.

Why subtract doors and windows: those openings have no bricks. Ignoring
them would overestimate by about 10%.

Tip: a standard modular brick in India measures about 19 cm × 9 cm × 9 cm;
with a 1 cm layer of mortar it takes up about 20 cm × 10 cm × 10 cm.

Your own answer: measure your own classroom, count its doors and windows,
and repeat the working. If a wall is shared with the next room, count it once for
the whole building.

Q4 You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred
chocolate cubes of side 10 mm. Which option gives you more chocolate?

(A) 50³ = 1,25,000 mm³ = 125 cm³
(B) 100 × 10³ = 100 × 1000 = 1,00,000 mm³ = 100 cm³

Option A gives more chocolate, 25 cm³ (25%) more, even though B "sounds" bigger.

Page 36 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Why: the big cube is 5 times as long, so it holds 5³ = 125 small cubes, not 5 or
25. B gives only 100.

Did you know? B has far more surface: 100 × 6 × 10² = 60,000 mm² against 6 ×
50² = 15,000 mm² for A. B needs four times as much wrapper for less
chocolate.

Q5 If all the people in the world are crowded together in one location, how much area
would it cover?

Guess first, then model it.

World population ≈ 800 crore = 8 × 10⁹ people
Space for one standing person in a tight crowd ≈ 50 cm × 50 cm = 0.25 m²
Area ≈ 8 × 10⁹ × 0.25 = 2 × 10⁹ m²
1 km² = 10⁶ m² ⇒ area ≈ 2000 km²

All of humanity would fit, standing, in about 2000 km²: a square about 45 km on each side
(√2000 ≈ 45), about one and a third times the area of the National Capital Territory of Delhi
(about 1,480 km²).

Why 0.25 m²: a crowd with about 4 people in each square metre is packed
shoulder to shoulder. Children take less room, so this is a fair average.

Tip: the trap is units. 2 × 10⁹ m² sounds enormous, but a square kilometre holds
a million square metres.

Page 37 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Your own answer: if you allow 1 m² per person (a loose crowd), the area
becomes 8000 km², still smaller than many Indian districts.

Q6 A school provides milk to its students in cylindrical glasses, each with a diameter of 7
cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are
needed to serve 1600 students? Use 1000 cm³ = 1 litre.

r = 7 ÷ 2 = 3.5 cm, h = 12 cm

Milk in one glass = πr²h = 22/7 × 3.5 × 3.5 × 12 = 462 cm³
For 1600 students = 1600 × 462 = 7,39,200 cm³
= 7,39,200 ÷ 1000 = 739.2 L

739.2 litres of milk are needed.

Tip: 22/7 × 3.5 = 11, so 11 × 3.5 × 12 = 462. Halve the diameter first.

Q7 The surface area of a sphere of radius 5 cm is five times the area of the curved surface
of a cone of radius 4 cm. Find the height and the volume of the cone.

SA of sphere = 4π × 5² = 100π cm²
CSA of cone = 100π ÷ 5 = 20π cm²
π × 4 × l = 20π ⇒ l = 5 cm
h = √(l² − r²) = √(25 − 16) = 3 cm

V = (1/3)πr²h = (1/3) × π × 16 × 3 = 16π ≈ 50.29 cm³

Page 38 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

The height of the cone is 3 cm and its volume is 16π ≈ 50.29 cm³ (π ≈ 22/7).
co m
e m.
− Explanation m
l as
.co ag
a s em
aglTip: keep π as a letter until the end. It cancels when finding l, and you only
replace it in the last line.
co m
m . ag
la setriangle, so the cone's height, radius and
Check it yourself: 3, 4, 5 is a right
slant height fit l² = h² + r². ag

co m
em.
m l as
.co ag
a sem
l
Take Earth to be a perfect sphere with radius 6370 km. Take Jupiter to be a perfect
ag
Q8
sphere with radius 69,900 km. Take the Sun to be a perfect sphere with radius
6,95,700 km. Compute approximately:
(i) the ratio of the volume of Jupiter tom a s
c o agl
the volume of the Earth;
(ii) the ratio of the volume of them . to the volume of the Earth.
(Calculator can be used.) las
e Sun

ag
om
The ratio of volumes of two spheres is the cube of the ratio of their radii,m .c (4/3)π cancels.
since
as e
. com agl
m69,900 ÷ 6370 ≈ 10.97 ⇒ ratio ≈ 10.97³ ≈ 1321 : 1
se(i)
a
agl (ii) 6,95,700 ÷ 6370 ≈ 109.2 ⇒ ratio ≈ 109.2³ ≈ 13,03,000 : 1
se m
o m l a
m .c and about 13 lakh Earths inside the Sun (by volume). ag
se
About 1300 Earths would fit inside Jupiter,
l a
− Explanation ag
o m
Why cube: V(J) : V(E) = (4/3)πRJ³ : (4/3)πRE³ = (RJ / RE)³. m.c
om la se
em
.c ag
a s
agl
Did you know? Jupiter's radius is only about 11 times Earth's, yet its volume is
over 1300 times as large. Small ratios of length turn into huge ratios of volume.
.c
s e m
m a
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 39 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q9 Show that the volume of a sphere is equal to 2/3 of the volume of the smallest cylinder
which encloses it.

The smallest cylinder that encloses a sphere of radius r fits it tightly: its radius is r and its height
is the diameter, 2r.

Volume of cylinder = πr² × 2r = 2πr³
(2/3) × 2πr³ = (4/3)πr³ = volume of sphere

Hence volume of sphere = (2/3) × volume of the enclosing cylinder.

Did you know? This is the same cylinder Archimedes used for the surface area
(Fig. 14.12): the sphere has the same surface area as the cylinder's curved
surface, and 2/3 of its volume. Archimedes is said to have asked for a sphere
inside a cylinder to be carved on his tomb.

Check it yourself: the surface areas also give 2 : 3. Cylinder total surface = 2πr
× 2r + 2πr² = 6πr²; sphere = 4πr², and 4πr² ÷ 6πr² = 2/3.

*Q10 Suppose the Earth is perfectly spherical. A string is wrapped tightly around the
equator of the Earth. Another string, 1 metre longer, is placed around the Earth so
that it forms a larger circle, staying the same distance above the ground everywhere.
How high above the ground is the second string?
Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the
three answers? Why or why not?

Let the radius be R (in metres) and the gap be x.

Page 40 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

First string = 2πR
Second string = 2π(R + x) = 2πR + 1
2πx = 1 ⇒ x = 1/(2π) ≈ 0.159 m ≈ 16 cm

The second string is about 16 cm above the ground. R cancelled out, so the answer is the same
16 cm for (i) the Moon, (ii) Jupiter and (iii) a volleyball.
Most people are surprised. 1 m is a tiny addition to the Earth's 40,000 km equator, so we expect
a microscopic gap; for a volleyball it more than doubles the string, so we expect a big gap. But
circumference grows by exactly 2π for every unit of radius, whatever the size of the circle.

Why: circumference is directly proportional to radius, C = 2πR. Adding x to the
radius always adds 2πx to the circumference, no matter how large R is. So
adding 1 m of string always lifts it by 1/(2π) m.

Check it yourself: a volleyball has radius about 0.105 m, so the first string is 2π
× 0.105 ≈ 0.66 m and the second is 1.66 m. Its radius is 1.66 ÷ 2π ≈ 0.264 m,
which is 0.264 − 0.105 ≈ 0.159 m above the ball.

*Q11 We have a cylinder with a base radius of r cm and height h cm. A square pyramid is
fitted inside it. The square base of the pyramid lies on the base of the cylinder, its
corners on the boundary of the cylinder. The apex of the pyramid lies on the top of
the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.

The square has its corners on the circle, so its diagonal is a diameter, 2r. If its side is a, then a²
+ a² = (2r)².

2a² = 4r² ⇒ area of base = a² = 2r²
Volume of pyramid = (1/3) × 2r² × h = (2/3)r²h
Volume of cylinder = πr²h
Ratio = (2/3)r²h : πr²h = 2 : 3π (≈ 0.21)

Page 41 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

The pyramid fills about 21% of the cylinder.

a

2r

Top view: the square base of the pyramid inside the circular base of the cylinder; the diagonal
of the square is the diameter 2r.

Why the height is h: the apex lies on the top face of the cylinder, so the
pyramid is exactly as tall as the cylinder.

Page 42 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Check it yourself: a cone in the same cylinder would have ratio 1 : 3. The
pyramid's ratio 2 : 3π is smaller by the factor 2/π ≈ 0.64, which is how much of
the circle the square covers.

Q12 What is the change in volume when:
(i) the length of a cuboid with dimensions l, w, h is increased by 1 unit?
(a) 1 cubic unit (b) (lwh + 1) cubic unit (c) wh cubic units (d) lw cubic units (e) lh
cubic units
(ii) the radius of a cylinder with dimensions r, h is increased by 1 unit?
(a) 1 cubic unit (b) πr²h cubic units (c) πr² cubic units (d) 2πrh + 2πh cubic units
(e) 2πrh + πh cubic units
*(iii) the radius of a sphere is decreased by 1 unit?

(i) (l + 1)wh − lwh = lwh + wh − lwh = wh → (c) wh cubic units

(ii) π(r + 1)²h − πr²h = πh(r² + 2r + 1 − r²) = πh(2r + 1)
= 2πrh + πh → (e)

(iii) (4/3)πr³ − (4/3)π(r − 1)³
= (4/3)π[r³ − (r³ − 3r² + 3r − 1)]
= (4/3)π(3r² − 3r + 1)
= 4πr² − 4πr + (4/3)π cubic units (a decrease)

Why (i) is wh: lengthening a cuboid by 1 unit adds one slab of size 1 × w × h at
the end.

Tip: in (ii) option (d) is a trap: it comes from wrongly writing (r + 1)² = r² + 2r +
2. Remember (r + 1)² = r² + 2r + 1.

Page 43 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

. com=
Check it yourself (iii): r = 2: (4/3)π(8 − 1) = 28π/3; formula: 16π − 8π + 4π/3
sem a
com
8π + 4π/3 = 28π/3.
m . agl
la se
ag
*Q13 omtotal surface area S in terms of V.
Given a cube with volume V, express its
. c ag
s e m
a
ANSWER agl

co m
m.
a³ = V ⇒ a = ∛V

m as e
.co
S = 6a² = 6(∛V)²
agl
a sem
agSl= 6(∛V)² = 6V2/3
m a s
.co agl

a s em
gl a = 2, S = 6 × 4 = 24; and 6 × (∛8)² = 6 × 4 = 24.
Check it yourself: V = 8agives

co m
m .
om l ase
g
.c Given a cube with total surface area S, express its volume
a
e m
s*Q14
a
V in terms of S.
agl
se m
com g l a
m . a
e
6a² = S ⇒ a = √(S/6)
ag las
V = a³ = (√(S/6))³ = (S/6)√(S/6)

co m
m .
V = (S/6)√(S/6) = (S/6)3/2
m la se
.co ag
a s em − Explanation
agl c
m .
Check it yourself: S = 24 gives S/6 = 4, a = 2, V = 8; and (24/6)√(24/6) = 4 × 2
m a s e
= 8. .co m agl
l a se
a g

co m
m .
m as e
.co

ag l Page 44 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Q15 A cylindrical glass of height 25 cm and radius 4 cm has water up to a height of 16 cm.
A crow wants to drink water from this glass. The water must be at a height of 20 cm
for the crow to reach it. There are some marbles lying around. How many marbles of
radius 1 cm should the crow drop into the glass to make the water reach the required
height?

level needed: 20 cm

water now: 16 cm

25 cm

marbles, r = 1 cm

r = 4 cm
Drawn to scale from the question (the picture on page 146 shows a crow beside the glass
and some marbles).

The marbles sink and push the water up. Their total volume must equal the volume of water
needed to raise the level by 20 − 16 = 4 cm.

Page 45 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Volume of the rise = πr²h = π × 4² × 4 = 64π cm³
Volume of one marble = (4/3)π × 1³ = (4/3)π cm³
Number of marbles = 64π ÷ (4/3)π = 64 × 3/4 = 48

The crow should drop 48 marbles.

Why it works: a sunk marble takes up space that water used to fill, so the water
rises by the marble's volume spread over the glass's cross-section. π cancels,
so no value of π is needed.

Check it yourself: the marbles stay fully under water. About 11 marbles fit side
by side on the floor of an 8 cm wide glass, so 48 marbles make about 4 to 5
layers, a pile at most about 10 cm high, well below the 20 cm water level.

Did you know? This is the old fable of "The Thirsty Crow", which dropped
pebbles into a pitcher. Mathematics tells us exactly how many pebbles it
needed.

Q16 Find the volume of ink in a new ball point pen. Take the necessary measurements and
make approximations, as needed.

Method: take out the refill. The ink fills a long thin cylinder inside it. Measure the length of the
ink column with a ruler and estimate the inner diameter of the tube (look at the open end, or
subtract the wall from the outer diameter).

Sample measurements: ink column length h ≈ 10 cm, inner diameter ≈ 2 mm (r = 0.1 cm)
V = πr²h = 3.14 × 0.1 × 0.1 × 10 = 0.314 cm³

A new pen holds about 0.3 cm³ of ink, i.e. about 0.3 millilitre (1 cm³ = 1 mL): only a few drops.

Page 46 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Why the inner diameter matters most: the radius is squared. If the inner
diameter is really 1.6 mm instead of 2 mm, the volume falls to 3.14 × 0.08² × 10 ≈
0.2 cm³. A small error in the radius makes a large error in the volume.

Tip: to measure such a thin refill, lay 10 refills side by side, measure the total
width and divide by 10. That gives the outer diameter; the inner diameter is a
little smaller.

Your own answer: pens differ. Use the numbers from your own pen and keep
the same method.

Q17 Solve:
(i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis
can be made from it?
*(ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible
coconut flesh/fruit by taking the necessary measurements.

(i) Dough = (4/3)π × 6³ = 288π cm³ ≈ 905 cm³
Assume one chapati is a disc of diameter 18 cm (r = 9 cm) and thickness 2 mm (0.2 cm)
One chapati = π × 9² × 0.2 = 16.2π cm³ ≈ 51 cm³
Number ≈ 288π ÷ 16.2π ≈ 17.8

(i) About 17 or 18 chapatis.
(ii) Half a coconut's flesh is a hemispherical shell: a big hemisphere minus the hollow inside.
Sample: outer radius of the flesh R = 5.5 cm, thickness 1 cm, so inner radius r = 4.5 cm.

Flesh = (2/3)π(R³ − r³) = (2/3) × 3.14 × (166.375 − 91.125)
= (2/3) × 3.14 × 75.25 ≈ 157.5 cm³

About 160 cm³ of flesh in half a coconut (sample values).

Page 47 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Why π cancels in (i): both the dough ball and the chapati are measured with π
in their volume, so the ratio needs no value of π. The dough's volume does not
change when it is rolled out.

Muskmelon instead: sample outer radius of the fruit flesh R = 7 cm and seed
hollow radius r = 3 cm. Edible part in half a melon ≈ (2/3) × 3.14 × (343 − 27) ≈
661 cm³.

Your own answer: measure with a ruler across the cut face: the full width gives
2R, the width of the hollow gives 2r. Use your own numbers.

Q18 Looking at the Ganita Manjari, Grade 9, Part 2 textbook, Sheela wonders:
(i) If all the pages of this textbook were laid out side-by-side on the floor would they
cover the entire classroom floor?
(ii) What is the maximum number of textbooks that can fit in an empty storeroom of
dimensions 15 ft × 20 ft × 30 ft?

Sample measurements: a page is about 20 cm × 27 cm; the book has about 150 printed pages
(this chapter ends on page 148), i.e. about 75 sheets; the book is about 1 cm thick.

(i) One page = 20 × 27 = 540 cm² = 0.054 m²
Even counting all 150 pages separately: 150 × 0.054 ≈ 8.1 m²
Classroom floor (8 m × 6 m) = 48 m²

(i) No. The pages would cover only about one-sixth of the floor.

(ii) Storeroom = 15 × 20 × 30 = 9000 ft³ ≈ 9000 × 28,317 ≈ 25,48,50,000 cm³
One book = 20 × 27 × 1 = 540 cm³
Upper bound = 25,48,50,000 ÷ 540 ≈ 4,72,000
Packing books upright (457 cm ÷ 1, 610 cm ÷ 20, 914 cm ÷ 27): 457 × 30 × 33 = 4,52,430

Page 48 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

(ii) About 4.5 lakh textbooks.
com
e m.
− Explanation m
l as
.co ag
a s em
aglWhy two answers in (ii): dividing volumes gives the most that could fit if no
space were wasted. Books are cuboids, so they pack well, but the walls leave
. com wall gives the real figure, a little
thin leftover gaps; counting rows along each
ag
lower. em as
agl
Tip: 1 ft = 30.48 cm, so 15 ft ≈ 457 cm, 20 ft ≈ 610 cm, 30 ft ≈ 914 cm. Convert
m
before dividing. .co
m a s em
.co agl
a sem own answer: measure your own book and classroom; the method stays
Your
ag l the same.

m a s
em
.co agl
a s
*Q19 agl decided to climb into one giant cube, how long would
If the entire human population
the side have to be?
co m
m .
o m l ase
g m × 0.5 m × 2 m = 0.5 m³
.cfirst. Then model each standing person as needing a boxa0.5
m
se stand on many floors, as in the picture).
Guess
la
ag
(people

se m
Population ≈ 8 × 10⁹
com g l a
.
Volume needed ≈ 8 × 10⁹ × 0.5 =e4m× 10⁹ m³
a
g l as
Side = ∛(4 × 10⁹) ≈ 1587am

co m
The cube would need a side of about 1.6 km.
m .
m la se
.−coExplanation ag
a s em
agl c
m .
e
Why so small: a cube 1.6 km wide has room for about 800 floors, each 2 m
om1 crore people. Volume grows as the cube of a s
c agl
high, and each floor holds about
.
emgives a huge volume.
the side, so a modest side
a s
agl

co m
m .
m as e
.co

ag l Page 49 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Tip: if instead you count only the volume of human bodies (about 0.07 m³ for an
adult, since the body has roughly the density of water), the cube shrinks to
∛(5.6 × 10⁸) ≈ 820 m. So a sensible answer is between about 0.8 km and 1.6 km,
depending on the assumption.

*Q20 The Earth’s surface has an estimated volume of 1.38 billion km³ of water. Suppose
the Earth is a perfect sphere and all this water forms a uniform layer completely
covering the Earth’s surface, like a thin water bubble. Estimate the thickness of this
water layer. The Earth’s radius is ~6371 km.
(i) Write an expression that gives the thickness of this water layer.
(ii) Simplify the expression in (i) using a calculator.

(i) The layer is very thin compared with the Earth, so its volume ≈ surface area × thickness.

V ≈ 4πR² × t ⇒ t ≈ V / (4πR²), with V = 1.38 × 10⁹ km³, R = 6371 km

(ii)

4πR² = 4 × 3.1416 × 6371² ≈ 5.10 × 10⁸ km²
t ≈ 1.38 × 10⁹ ÷ 5.10 × 10⁸ ≈ 2.7 km

The water layer would be about 2.7 km thick.

Exact expression: the water is the shell between a sphere of radius R + t and
one of radius R: (4/3)π[(R + t)³ − R³] = V, so t = ∛(R³ + 3V/(4π)) − R. With a
calculator this also gives about 2.70 km. The two answers agree because t is
tiny compared with R.

Page 50 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Did you know? 2.7 km is less than one-two-thousandth of the Earth's radius.
On a school globe of radius 15 cm, all the world's water would be a film only
about 0.06 mm thick.

*Q21 Give the dimension of a cuboid whose volume is halved when its surface area is
doubled.

We need a cuboid that can be reshaped so that its surface area doubles while its volume
halves. One answer: a cuboid 14 cm × 14 cm × 8 cm, reshaped to 28 cm × 28 cm × 1 cm.

Original: V = 14 × 14 × 8 = 1568 cm³
S = 2(14×14 + 14×8 + 8×14) = 2(196 + 112 + 112) = 840 cm²
New: V = 28 × 28 × 1 = 784 cm³ = 1568 ÷ 2 (halved)

S = 2(28×28 + 28×1 + 1×28) = 2(784 + 28 + 28) = 1680 cm² = 2 × 840 (doubled)

In general, an a × a × (4a/7) cuboid works: doubling the length and width and making the
height one-eighth halves the volume and doubles the surface area.

About the book: the wording is unclear. A cuboid's surface area cannot simply
be "doubled" without saying how its shape changes, and the question does not
say. We read it as "find a cuboid that can be reshaped into another cuboid with
double the surface area and half the volume". Many pairs work; the one above
has small whole-number edges. Another possible reading compares the
numbers only: half the volume equals double the surface area (V ÷ 2 = 2S).
Then lwh = 8(lw + wh + hl), i.e. 1/l + 1/w + 1/h = 1/8, and a cube of side 24 cm
works (V ÷ 2 = 13,824 ÷ 2 = 6912 and 2S = 2 × 3456 = 6912), as does a 16 × 32
× 32 cuboid.

How it was found: start with a square-based cuboid a × a × h and change it to
2a × 2a × h/8. Volume: 4a² × h/8 = a²h/2, halved for any h. Surface: new =
2(4a² + 2 × 2a × h/8) = 8a² + ah; double the old = 2 × 2(a² + 2ah) = 4a² + 8ah.
Setting them equal: 4a² = 7ah, so h = 4a/7. With a = 14, h = 8.

Page 51 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Check it yourself: making a solid flatter and wider spreads it out, giving more
surface for less volume, just as cutting the cube into small cubes did in Exercise
Set 14.1, Q5.

Q22 Project: Find the volume of your house making necessary approximations. Present
how you solved it.

Plan: treat each room as a cuboid, measure its length and width (pace it out or use a tape), take
the ceiling height, find each volume, and add.

ROOM (SAMPLE FLAT, CEILING 3 M) FLOOR AREA VOLUME (M³)

Living room 5 × 4 = 20 m² 20 × 3 = 60

Bedroom 1 4 × 3.5 = 14 m² 42

Bedroom 2 3.5 × 3 = 10.5 m² 31.5

Kitchen 3 × 2.5 = 7.5 m² 22.5

Two bathrooms 2 × (2 × 1.5) = 6 m² 18

Total 58 m² 174 m³

Sample result: the inside of the house is about 174 m³ (1,74,000 litres).
Approximations: rooms taken as perfect cuboids; wall thickness, cupboards and the balcony
ignored; one ceiling height for all rooms.

Why add floor areas first: when all rooms have the same height, total volume =
(total floor area) × height = 58 × 3 = 174 m³. This saves work and is a good
check.

Tip: if you measure by pacing, first find the length of your pace: walk 10 steps
along a 10 m tape and divide.

Page 52 of 54

Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

Your own answer: replace the sample with your own rooms. If your house has a
sloping roof, model that part as a triangular prism (area of the triangle × length).

Chapter at a glance
Cuboid (l, w, h): total surface area = 2(lw + wh + hl), volume = lwh = area of base × height
(a pack of cards). Cube of side a: 6a² and a³. Example 1: a 6 cm cube and a 9 × 6 × 4 cm
cuboid both hold 216 cm³, but the cube needs less covering (216 cm² against 228 cm²).
Right circular cylinder (r, h): curved surface area = 2πrh (it unrolls into a rectangle), closed
at both ends 2πr(h + r), open at one end πr(2h + r); volume = πr²h. Example 2: the
kaleidoscope needs 550 cm² of chart paper.
Right circular cone: slant height l² = h² + r²; curved surface area = πrl (it unrolls into a
sector); total surface area = πr(l + r); volume = (1/3)πr²h. Three cones of salt fill a cylinder
of the same base and height. Example 3: the 5-12-13 triangle turned about 12 cm gives
100π cm³.
Pyramid: volume = (1/3) × area of base × height, whatever the shape of the base; its
surface area is the sum of the areas of its faces. A cone is a pyramid with a circular base.
Sphere: surface area = 4πr² (Archimedes: equal to the curved surface of the cylinder that
fits tightly round it; string wound on a ball fills 4 circles of radius r); volume = (4/3)πr³ =
(1/3) × surface area × r (many thin cones from the centre).
Hemisphere: curved surface area 2πr², total surface area 3πr², volume (2/3)πr³. Example
4: a bowl of radius 3.5 cm holds about 89.8 cm³.
Scaling: multiply every length by k and surface areas multiply by k², volumes by k³. Cutting
a solid into small pieces keeps the volume but increases the surface.
Guesstimates: guess first, model the object by a simple solid, state assumptions,
approximate, then compare with your guess. Balls leave gaps, so dividing volumes only
gives an upper bound.

Page 53 of 54

as e
ag l
Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume AglaSem · NCERT Solutions

co m
m.
Quick revision
m l ase
.co IN SYMBOLS / ag STUDENTS SLIP
em
IDEA WHAT IT SAYS WHERE
a s
agl
STATEMENT

Cuboid surface TSA = 2(lw + wh + hl) Three pairs of equal Adding only three faces, or
area
co m
rectangular faces counting a lid on an open box
m .
e of unit cubes that ag
Cuboid volume
la sNumber Mixing units (m with cm)
ag fill it
V = lwh = base area ×
height before multiplying

Cube The cuboid formulas with l =
m
Writing 6a for the area or a²
co
TSA = 6a², V = a³

m.
w=h=a for the volume

m as e
.co CSA = 2πrh
Cylinder: curved Unrolls into a rectangle 2πr
agl
Using the diameter for r
se m
a
surface long and h high

ag l
Cylinder: total 2πr(h + r) closed; πr(2h + Curved part plus one or two Adding two circles to a
surface circles
m
tumbler or a pipe
a s
r) open at one end

. c o agl
sem
Cylinder volume V = πr²h Stack of identical circles Squaring h instead of r
a
agl
(khakhras)

Cone slant l² = h² + r² h, r and l form a right Using the height h in place of l
height triangle in πrl
co m
m .
a e the base for an open
sAdding
com l
ag object like a cap or a tent
Cone surface Unrolls into a sector of
.
CSA = πrl, TSA = πr(l +

se m
r) radius l
a and
agl Cone V = (1/3) × base area × One third of the prism or Forgetting the 1/3

m
pyramid volume cylinder with the same base
se
height

com
and height
g l a
.
m Equal to the curved surface a
e
las
Sphere surface S = 4πr² Writing 4πr³ or πr²
ag of the tight cylinder

m
Sphere volume V = (4/3)πr³ (1/3) × surface × r; 2/3 of Using the diameter, or
the tight cylinder
. co
forgetting the cube
m
se Forgetting the base in the
m la
.co ag
Hemisphere CSA 2πr², TSA 3πr², V = Half a sphere plus a flat

a s em (2/3)πr³ circular base total surface

agl Scaling c
.
lengths × k ⇒ areas × k², Volume grows fastest when Thinking a 10% bigger radius
volumes × k³ a solid is enlarged gives 10% more volume (it is
s e m
com a
. agl
33.1%)

a s em ÷ Dividing volumes gives an
Estimating
ofg
l
number ≤ total volume
volume a one upper bound when there are
Treating the upper bound as
the real count
gaps

co m
m .
m as e
.co

ag l Page 54 of 54