Vedanta Excel in Mathematics Book 9 Final Re_CTP

Mensuration (II): Prism

Facts to remember

1) The area of cross section or base of rectangular prism (cuboid) = l × b
2) The area of cross section or base of square prism = l2

b) Volume of prism

We have already discussed that a cross section of a prism is congruent to its
base. Therefore, area of cross section (or cross sectional area) is, of course; equal
to the area of the base of a prism. Then, the volume of a prism is the product of
its cross sectional area and height.
i.e., Volume of a prism = Area of cross section × height

c) Lateral Surface Area (L.S.A.) of a prism

The lateral surfaces of a prism are the rectangular faces excluding the bases of
the prism.

In the adjoining rectangular based prism (cuboid),

Lateral surface area (L.S.A.) = Area of opposite walls h
along the length + Area of opposite walls along the
breadth
b
= 2lh + 2bh l

= 2h (l + b)

= 2 (l + b) × h
= Perimeter of base × height (h)

L.S.A. of a prism = Perimeter of base × height, i.e. P × h

d) Total Surface Area (T.S.A.) of a prism

The total surface area of a prism is the sum of the lateral surface area and two
times the area of its cross section (or two times the area of a base).

i.e., Total surface area of a prism = lateral surface area + 2 × area of cross section

Worked-out Examples

Example 1: Calculate (i) the area of cross section 5cm 2cm5cm 2cm
(ii) lateral surface area (iii) total surface area

(iv) volume of the adjoining prism. 10 cm 8 cm
Solution:
D E 5 cm C

(i) Here, the area of cross section = Area of (ABCD – FGCE) 2 cm

= 10 × (2 + 2) cm2 – 5 × 2 cm2 FG
2 cm

= 30 cm2 A 10 cm B

Alternative process

Area of cross section = Area of trapeziums (ABCD + ADEF) F 5cm ED2cm5cm C
2cm
= 1 × 2 (10 + 5) cm2 + 1 × 5 (4 + 2) cm2
2 2

= 15 cm2 + 15 cm2 = 30 cm2 A 10 cm B

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Mensuration (II): Prism

(ii) Lateral surface area = Perimeter of cross section × length

= (10 + 2 + 5 + 2 + 5 + 4) cm × 8 cm

= 224 cm2

(iii) Total surface area = lateral surface area + 2 × cross sectional area

= 224 cm2 + 2 × 30 cm2

= 284 cm2

(iv) Volume of the prism = Area of cross section × height

= 30 cm2 × 10 cm

= 300 cm3
Example 2: Calculate (i) the area of cross section (ii) lateral
surface area (iii) total surface area (iv) volume
of the adjoining prism.

Solution:

(i) Area of cross section = (9 × 4) cm2 + (3 × 4) cm2 3 cm

= 48 cm2 4 cm 4 cm
3 cm 3 cm 4 cm
(ii) Lateral surface area = Perimeter of cross section × length

= (9 + 4 + 3 + 4 + 3 + 4 + 3 + 4) cm × 8 cm 9 cm

= 34 cm × 8 cm = 272 cm2

(iii) Total surface area = lateral surface area + 2 × cross sectional area

= 272 cm2 + 2 × 48 cm2

= 368 cm2

(iv) Volume of the prism = Area of cross section × height

= 48 cm2 × 8 cm = 384 cm3

Example 3: A rectangular metallic block is 30 cm long, 25 cm broad and 8 cm high.

How many pieces of rectangular slices each of 5 mm thick can be cast

lengthwise from the block? Alternative process:
Solution:
Here, length of the block = 30 cm 8cm

Breadth of the block = 25 cm 25cm
Height of the block = 8 cm
∴ Volume of block = 30 cm×25 cm×8 cm 30 cm
Also, volume of each slice = 0.5 cm×25 cm× 8cm
Here,

thickness of each slice = 5 mm

∴ Number of slices = volume of block = 0.5 cm
volume of each slice
30 × 25 × 8 Length of the block = 30 cm
0.5 × 25 × 8
= When it is cut lengthwise,

= 60 number of slices

= 30 cm÷0.5 cm = 60

Hence, the required number of rectangular slices is 60.

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Mensuration (II): Prism

Example 4: A cubical vessel of length 10 cm is completely filled with water. If the water

is poured into a rectangular 10 cm
10 cm
vessel 25 cm × 20 cm × 10 cm,

find the height of the water level

in the rectangular vessel. How 10 cm 10 cm 25 cm 20 cm
much more water is poured into
the vessel to fill it completely?

(1 l = 1000 cm3)

Solution:

Here, volume of water in cubical vessel = 10 cm × 10 cm × 10 cm

Volume of water in rectangular vessel = 25 cm × 20 cm × h cm

Now, 25 × 20 cm × h cm = 10 cm × 10 cm × 10 cm

or, h = 10 cm × 10 cm × 10 cm = 2 cm
25 cm × 20 cm

Again, remaining height of rectangular vessel to be filled = 10 cm - 2 cm = 8 cm
Then, volume of empty space of the vessel = 25 cm × 20 cm × 8 cm = 4000 cm3

Now, 1000 cm3 = 1 l ∴ 4000 cm3 = 1 × 4000 l = 4 l
1000

Hence, the required height of water level in the rectangular vessel is 2 cm and 4 l of water is
needed to pour into it to fill it completely.

EXERCISE 6.1

General section

1. a) What is a prism? Define with examples.
b) Define a cross section of a prism. What is the shape of cross section of a cuboid?
c) Define lateral surface of a prism. What are the lateral surfaces of a triangular prism?
d) If the area of a base of a prism is 24 cm2, what is its cross sectional area?
2. a) The area of cross section of a prism is 36 cm2 and its height is 15 cm. Find its

volume.
b) The perimeter of cross section of a prism of length 10 cm is 48 cm. Find its lateral

surface area.
c) The lateral surface area and the cross sectional area of a prism are 175 cm2 and

40 cm2 respectively. Find total surface area of the prism.
d) The perimeter of cross section and the area of cross section of a prism are 36 cm

and 54 cm2 respectively. If the length of the prism is 15 cm, find its total surface
area.
3. a) The volume of a cube is 125 cm3. Find its total surface area.
b) The total surface area of a cube is 96 cm2. Find its volume.
c) The length and the breadth of a cuboid are 15 cm and 8 cm respectively. If the
volume of the cuboid is 720 cm3, find its height and calculate the total surface
area.
d) The area of the rectangular base of a metallic block is 192 cm2 and its volume is
768 cm3. Find its thickness.

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Mensuration (II): Prism

e) A rectangular metallic block is 40 cm long, 24 cm broad and 10 cm high. How
many pieces of rectangular slices each of 8 mm thick can be cast lengthwise from
the block?

Creative section-A

4. Calculate the cross sectional area, the lateral surface area, the total surface area and

volume of each of the following prisms.

a) 8 cm b) 2cm c)
4cm
2cm 4cm
4cm 9 cm
5 cm 6cm

8cm 6 cm 6 cm 3cm
16 cm 10 cm 4cm
5 cm 10cm
15 m
d) e) f) 10cm
2cm 6m 10cm10cm 10cm
3cm 3m 10cm
3m
3cm
2cm
2 cm 2 cm 5m 30 cm
3cm 5 cm 4m
30 cm
6m
4m

5. a) A cubical water tank is filled in 1296 seconds at the rate of 1 litre per 6 seconds.
(i) Calculate the internal volume and the length of side of the tank.
(ii) Calculate the total internal surface area of the tank.
b) A lidless rectangular water tank made of zinc plates is 2 m long, 1.5 m broad and

1 m high.

(i) How many square metres of zinc plates are used in the tank?

(ii) How many litres of water does it hold when it is full?

(iii) Find the cost of zinc plates at Rs 1200 per sq. m.

c) A rectangular carton is 80 cm × 60 cm × 40 cm.

(i) How many packets of soaps can each of 10 cm × 5 cm × 4 cm be kept
inside the carton?

(ii) By how many centimetres should the height of the carton be increased to keep
1200 packets of soaps?

6. a) A cubical wooden block of length 12 cm is cut into 8 equal cubical pieces. Find the
length of edge of each piece.

b) 8 metallic cubical blocks of equal size are melted and joined together to form a
bigger cubical block. If each smaller block is 10 cm thick, find the thickness of the
bigger block.

c) A rectangular metallic block is 50 cm × 20 cm × 8 cm. If it is melted and re-formed
into a cubical block, find the length of the edge of the cube.

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Mensuration (II): Prism

d) Vegetable ghee is stored in a rectangular vessel of internal dimensions
60 cm × 10 cm × 45 cm. It is transferred into the identical cubical vessels. If the
internal length of each cubical vessel is 10 cm, how many vessels are required to
empty the rectangular vessel?

Creative section-B

7. a) The adjoining figure is a rectangular glass vessel of 20 cm

length 40 cm, breadth 30 cm, and height 20 cm. If it

contains some water upto the height of 12 cm, how
20 cm
many litres of water is to be poured into it to fill the 12 cm 40 cm 30 cm
vessel completely? (1 l = 1000 cm3)

b) In the given figure, a cubical vessel 20 cm 15 cm
of length 20 cm is completely filled

with water. If the water is poured

into a rectangular vessel of length 20 cm 32 cm 25 cm
32 cm, breadth 25 cm, and height
15 cm find the height of the water level

in the rectangular vessel. How much more water is required to fill the rectangular

vessel completely? (1l = 1000 cm3)

6.4 Surface area and volume of triangular prisms Cross section

A triangular prism has two congruent and parallel triangular bases.

It has 3 rectangular lateral faces.

In the adjoining triangular prism, Its base is a triangle.
So, it is a triangular prism.
(i) Cross sectional area = Area of triangular base

(ii) Lateral surface area = Area of 3 rectangular surface ac

= l.a. + l.b + l.c b
l
= l (a + b + c)

= perimeter of triangular base × length

= p.l

(iii) Total surface area = Lateral surface area + 2 × Area of triangular base

= p.l + 2D

(iv) Volume of prism = Area of triangular base × length (or height of prism)

Facts to remember

1) The area of cross section or base of triangle prism = 1 × base (b) ×height (h)
2) The area of cross section or base of 2 1 ×p
right angled triangular prism = 2 × b

3) The area of cross section or base of equilateral triangular prism = 43 a2

4) The area of cross section or base of isosceles triangular prism = b 4a2 – b2
4

5) The area of cross section or base of scalene triangular prism = s (s – a) (s – b) (s – c)
where s is the semi-perimeter of the triangle

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Mensuration (II): Prism

Example 1: Find (i) the lateral surface area (ii) total surface area (iii) volume of the
following triangular prism.

a) A ' b) c)
15 cm14 cm
A 10 cm

C' 18 cm

B 6 cm C 15 cm 2 3 cm 13 cm 16 cm

Solution:

a) The triangular base is a right angled triangle.
Here, BC = 6 cm and AC = 10 cm

By using Pythagoras theorem, AB = AC2 – BC2 = 102 – 62 = 64 = 8 cm

Now, perimeter of the triangular base (P) = AB + BC + CA

= 8 cm + 6 cm + 10 cm=24 cm
LAarteeraaol fsutrrifaancgeualraeraboafseth(eAp) r=ism21 P×erpim=ete21r
(i) ×b × 6 cm × 8 cm = 24 cm2 × length (l)
= of the triangular base (P)

= 24 cm × 15 cm

= 360 cm2

(ii) Total surface area of the prism = Lateral surface area+2 × Area of triangular base (A)

= 360 cm2 + 2 × 24 cm2

= 408 cm2

(iii) Volume of the prism (V) = Area of triangular base (A) × length (l)

= 24 cm2 × 15 cm

= 360 cm3

b) The triangular base is an equilateral triangle.

Here, each side (a) = 2 3 cm

Now, perimeter of the triangular base (P) = 3 × 2 3 cm = 6 3 cm
Area of triangular base (A) = 43 × a2 = 43 × (2 3)2 = 43 × 4 × 3 = 3 3 cm2
(i) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l)

= 6 3 cm × 18 cm = 108 3 cm2

(ii) Total surface area of the prism = Lateral surface area+2 × Area of triangular base (A)

= 108 3 cm2+ 2 × 3 3 cm2 = 114 3 cm2

(iii) Volume of the prism (V) = Area of triangular base (A) × length (l)

= 3 3 cm2 × 18 cm = 54 3 cm3

c) The triangular base is a scalene triangle.

Here, a = 13 cm, b = 14 cm and c = 15 cm

Now, perimeter of the triangular base (P) = 13 cm + 14 cm + 15 cm = 42 cm

Semi-perimeter (s) = 13 cm + 14 cm + 15 cm = 21 cm
2

Area of triangular base (A) = s (s – a) (s – b) (s – c)

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Mensuration (II): Prism

= 21(21 – 13)(21 – 14)(21 – 15)

= 21 × 8 × 7 × 6 = 84 cm2

(i) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l)

= 42 cm ×16 cm

= 672 cm2

(ii) Total surface area of the prism = Lateral surface area+2 × Area of triangular base (A)

= 672 cm2+ 2 × 84 cm2

= 840 cm2

(iii) Volume of the prism (V) = Area of triangular base (A) × length (l)

= 84 cm2 × 16 cm

= 1,344 cm3

Example 2: Mr. Gurung is making a wooden birdhouse 5 in. 5 in.
having the dimensions as shown in the figure. 6 in.
If the volume of the birdhouse is 900 cubic
inches, find its length. 8 in.

Solution:

Here, the birdhouse has pentagonal cross-section.

∴Cross-sectional area = area of rectangle + area of triangle.

Now, area of rectangular portion (A1) = l × b = 8 in. × 6 in. = 48 sq. inch
Also, for isosceles triangular portion, base (b) = 8 inch and equal sides (a) = 5 inch

∴ Area of the triangle (A2) = b 4a2 – b2
4

= 8 4 × 52 – 82 = 2 100 – 64 = 2 36 = 2 × 6 = 12 sq. inch.
4

∴ The area of cross section of the birdhouse (A) = A1+A2 = 48 + 12 = 60 sq. inch

Again, volume of the birdhouse (V) = Area of cross section (A) × length (l)

or, 900 = 60 × l

or, l = 900 = 15
60

Hence, the length of the birdhouse is 15 inch.

Example 3: Mrs. Rana manages the accommodation for 30 A'
guests in her son’s birthday ceremony. For this A
purpose, she plans to make a tent in the shape
of triangular prism of length 20 m in such a way B ? B' C'
that each person has the space of 4 square meters DC 20 m
on the floor and 8 cubic meters of air to breathe.
What is the height of the tent?

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Mensuration (II): Prism

Solution:
Here, the volume of the tent (V) = 30 × 8 m3 = 240 m3

The area of rectangular floor = 30 × 4 m2 = 120 m2

But, area of the floor = l × b

or, 120 m2 = 20 m × BC

∴ BC = 6 m

Also, area of cross section, D ABC (A) = 1 × BC × AD = 1 × 6 × AD = 3 AD
2 2
Again, the volume of the tent (V) = area of cross section (A) × length (l)

or, 240 = 3AD × 20

or, 60 AD = 240

or, AD = 4

Hence, the height of the tent is 4 m.

EXERCISE 6.2

General section

1. a) From the given triangular prism, write the formulae to find its
(i) cross sectional area (ii) lateral surface area
z
(iii) total surface area (iv) volume y l

x

b) If p, q, r are the three sides of a triangle of a triangular base prism of length h,
write the formulae to find its (i) cross sectional area (ii) lateral surface area (iii) total
surface area (iv) volume

c) The base of a prism of length h cm is an equilateral triangle with each side a cm,
write the formulae to find its (i) cross sectional area (ii) lateral surface area (iii) total
surface area (iv) volume

2. a) A prism has a right-angled triangular base with perpendicular 8 cm and base 6 cm.
Find its cross sectional area.

b) A right prism base is a triangle whose sides are 3 cm, 25 cm and 26 cm. Find the area
of its cross section.

c) If the perimeter of the triangular base of a prism is 18 cm and its length is 15 cm, find
its lateral surface area.

d) The perimeter of the triangular base of a prism is 14.5 cm and its lateral surface area
is 290 cm2. Find the length of the prism.

e) The area of triangular base of a prism is 30.5 cm2 and its lateral surface area is 305
cm2. Find the total surface area of the prism.

f) The perimeter of the triangular base of a prism is 24 cm and the area of its cross
section is 24 cm2. If the prism is 10 cm long, find its total surface area.

3. a) If the area of the triangular base of a prism 25 cm long is 16.4 cm2, find the volume
of the prism.

b) A triangular based prism is 30 cm long. If the length of the sides of its triangular base
are 3 cm, 4 cm and 5 cm, find its volume.

c) In a prism, if the volume is 400 cm3 and area of its triangular base is 50 cm2, find
the length of the prism.

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Mensuration (II): Prism

Creative section-A P'
4. a) Find the volume of the following prisms.
(i) A' (ii) L' (iii)

A L M' P
M 6 cm N Q'
12 cm B' C' N' 8 cm R'
10 3 cm 10 cmM R 30 cm
25 cm Q

B 5 cm C

(iv) 18 cm (v) (vi) 3.5 cm 8 cm 40 cm

17 cm

13 cm 20 cm 12 cm
10 cm 16 cm
15 cm

b) Calculate the lateral surface area and total surface area of the following prisms.

(i) C E (ii) C' (iii) P P'
Q' R'
C
Q
6 cm
R 15 cm
5 cm 7 cm 6 cm 8cm 3cm
G A' 8 cm
A4 cm B 8 cm H A3.4 cm B 8 cm B'
7cm
(iv) A A' (v) A D (vi) A P
12cm
B' E BQ
5cm B C 15cm R
10cm F
C 20cm
C 30cm B 6 cm
10cm
A A'
B'
5. a) The given diagram is the solid prism of triangular base. If B 4cm C
the volume of the prism is 48 cm3, find its height. A C'
A'
b) If the volume of the prism given aside is 432 cm3, find its B B'
height.
C'
C

c) The volume of a prism having its base a right angled triangle is 6,300 cm3. If the
lengths of the sides of the right angled triangle containing the right angle are 20 cm
and 21 cm, find the height of the prism.

6. a) The area of rectangular surfaces of a triangular prism having base sides 9 cm, 10 cm
and 17 cm is 864 cm2. Calculate the height of the prism.

b) A prism with an equilateral triangular base is 18 cm long. If the area of its rectangular
surfaces is 567 cm2, find the length of each side of the triangular base.

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Mensuration (II): Prism

Creative section-B

7. a) The perimeter and area of the cross section of a triangular prism are 44 cm and 66
cm2 respectively. If the area of lateral surface area of the prism is 1,100 cm2,

(i) find its volume (ii) find its total surface area

b) The area of cross section of a triangular prism is 126 cm2 and its volume is 5,040
cm3. If the perimeter of its base is 54 cm, (i) find its lateral surface area (ii) find its
total surface area.

c) The perimeter and area of the cross section of a triangular prism are 24 cm and 24
cm2. If its total surface area is 480 cm2, find (i) the lateral surface area (ii) the
volume

8. a) A trap for insects is in the shape of triangular prism which is 10 cm
shown alongside.

(i) Find the total surface area of trap.

(ii) Find the volume of the trap. 6 3 cm

b) Two campers made a tent shown alongside in the shape of 5 ft.
triangular prism. 5 ft.

(i) Find the amount of fabric required for making it 8 ft. 6 ft.
including the floor.

(ii) How much air is occupied in the tent?

9. a) Mrs. Ghale manages the accommodation for 16 guests in ? 12 m
her daughter’s birthday ceremony. For this purpose, she ? 30m
plans to make a tent in the shape of triangular prism of
length 12 m. If each person has the space of 6 square meters
on the floor and 15 cubic meters of air to breathe in average
inside the tent, what is the height of the tent? Find it.

b) In the marriage ceremony of Suresh’s son, there was
accommodation for 150 people. For this purpose, he
made a tent in the shape of triangular prism in such a
way that each person had the space of 4 square meters
on the floor and 20 cubic meters of air to breathe in
average inside the tent. If the tent was 30 m long, what
was the height of the tent?

10. a) Find the volume of dog house given along side. 3 ft. 3 ft.
4 ft.
5 ft.
5 ft. 6 ft.
5 in. 5 in.
b) Mr. Chemjung is making a wooden birdhouse having 7 in.
the dimensions shown in the figure. If the volume of the
birdhouse is 432 cubic inches, find its length. 6 in.

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Mensuration (II): Prism

Project work and activity section
11. a) Make the group of your 5 friends. Collect some objects in the shape of triangular

prism which are available in your surrounding. Calculate their L.S.A., T.S.A. and
volume.
b) Prepare a model of prism with the base (triangle, L-shape, T-shape or U-shape) by
using woods or cardboard paper. Then, find the cross-sectional area, L.S.A., T.S.A.
and volume.

OBJECTIVE QUESTIONS

Let’s tick (√) the correct alternative.

1. If A, h and V denote the area of cross section, length and volume of a prism, what is
the relation among A, h and V?

(A) V = A×h (B) A =V×h (C) h = V×A (D) V = A/h

2. The cross sectional area of a prism is 30 cm2 and height is 15 cm, what is its volume?

(A) 45 cm3 (B) 450 cm3 (C) 900 cm3 (D) 225 cm3

3. What is the volume of a prism whose area of base is 600 cm2 and length is 30 cm?

(A) 20 cm3 (B) 630 cm3 (C) 570 cm3 (D) 18,000 cm3

4. If L, P and h represent the lateral surface area, perimeter of base and height of a prism
respectively. The relation among L, P and h is

(A) L = P × h (B) P =L × h (C) h = L × P (D) L = P + h

5. A prism of length 20 cm has the sides of base 8 cm, 6 cm and 10 cm. The area of its
rectangular surfaces is

(A) 160 cm2 (B) 120 cm2 (C) 200 cm2 (D) 480 cm2

6. If A and B represent the area of base and the area of cross section of a prism respectively.
What is its total surface area?

(A) 2(A+B) (B) A+2B (C) 2A + B (D) 2AB

7. The area of cross section and the lateral surface area of a prism are 30 cm2 and 300
cm2 respectively. What is the total surface area of the prism?

(A) 330 cm2 (B) 360 cm2 (C) 9000 cm2 (D) 630 cm2

8. The capacity of a water tank of dimension 4 m × 3 m × 2 m is

(A) 12,000 liters (B) 8,000 liters (C) 6,000 liters (D) 24,000 liters

9. A rectangular aquarium is 60 cm long, 40 cm wide and 50 cm high. How much water
can it hold?

(A) 24 liters (B) 150 liters (C) 120 liters (D) 100 liters

10. How many brick each of 15cm × 10cm × 5 cm are required to build a wall of 20 m
long, 3 m high, and 20 cm wide?

(A) 16000 (B) 18000 (C) 20000 (D) 24000

Vedanta ICT Corner
Please! Scan this QR code or
browse the link given below:

https://www.geogebra.org/m/g5mxqvn4

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Unit 7 Mensuration (III): Cylinder and Sphere



7.1 Cylinder –Looking back

Classwork-Exercise
Let's fill in the blanks with correct answers as quickly as possible.

a) The circumference of a circle with radius r units is …..
b) The area of a circle with radius x units is …..
c) The circumference of a circle with radius 7 cm is …..
d) The area of a circle with diameter 28 cm is …..

7.2 Cylinder

Let’s observe a few real-life examples of cylindrical objects which are available in our
surrounding.

madal water jar rods battery wooden log

A cylinder is a three dimensional solid that has two parallel circular bases. The two
circular bases are joined by a curved surface. The distance between the two circular
bases is called the height or length of the cylinder. Bottles, battery, LP-glass cylinder,
rod, pipe, bucket, dustbin, water-jar, pipe, roller, candles, pole etc are a few real life
examples of cylinder.

Facts to remember

1. The line segment joining the centers of circular bases of a cylinder is called its
axis.

2. A right circular cylinder is a solid generated by the revolution of a rectangle about
one of its sides as axis.

3. The opposite circular bases are congruent and parallel.

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Mensuration (III): Cylinder and Sphere

7.3 Area of cylinder

The surface area of a solid object is a measure of the total area that the surface of the
object occupies.
Let’s take a rectangular sheet of paper and revolve it about one of the sides to complete
a full rotation (without overlapping) as shown in the figure.

From the above figures, it is clear that,

(i) The perimeter of the circular base(C) = 2πr
(ii) The area of the circular base(A) = πr2
(iii) The curved surface area (C.S.A) of a cylinder is equal to area of rectangle with

length 2πr and width h. Thus, C.S.A.= 2πr × h = 2πrh

(iv) The total surface area of a cylinder = C.S.A. + Area of two opposite circular bases

Veda nta ICT Corner = 2πrh+ πr2 + πr2
Please ! S can this QR code or = 2πrh+2πr2
brows e th e link given below: = 2πr (r + h)
https: //www .geogebra.org/m/aurfzgtg

where, r is the radius of a circular face and h is the height of the cylinder.

7.4 Volume of cylinder

Let’s take a cylindrical radish or cucumber. Suppose the radius of base is r units and
height is h units. Let’s cut it into 16 equal sections through centers of circular bases
along its height. Such material may be made with wood or bought from the market
and keep in the mathematics laboratory. Then, let’s place the segments alternatively as
shown in the figure to form in to cuboid.






Here, in the newly formed cuboid, length (l) = πr,
Breadth (b) = r and height (h) = h

∴Volume of cuboid (V) = l × b × h = πr × r × h = πr2h
Now, the volume of cylinder (V) = Volume of cuboid = πr2h
Moreover,
The area of circular base of a cylinder = πr2
Then, volume of the cylinder = Area of circular base × height

= πr2 × h = πr2h

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Mensuration (III): Cylinder and Sphere

7.5 Curved surface area, total surface area, and volume of a hollow

cylinder Rr

The figure alongside is a hollow cylinder (or a pipe). Let R be the h
external radius, r the internal radius and h be the height of the hollow
cylinder.

Here, the external curved surface area = 2πRh

The internal curved surface area = 2πrh

∴ The total curved surface area = 2πRh + 2πrh = 2πh(R + r)

Again, the area of a ring-shaped circular base = πR2 – πr2 = π(R + r) (R – r)

∴ The area of two ring-shaped circular bases = 2π(R + r) (R – r)
Now, the total surface area of a hollow cylinder = 2πh(R + r) + 2π(R + r) (R – r)
= 2π(R + r) (h + R – r)
Volume of material contained by a hollow cylinder:

From the above diagram,
The external volume of the cylinder = πR2h
The internal volume of the cylinder = πr2h
∴ Volume of the material contained by the cylinder = πR2h – πr2h

= πh(R2 – r2)

= πh (R + r) (R – r) d=2r
h
7.6 Half cylinder or semicylinder

When a cylinder is cut into half along it's axis it is called a h
half cylinder or a semicylinder. It is a horizontal cylindrical
segment.

A half cylinder has a curved surface, a rectangular flat
surface and two semi-circular bases.

Length of the rectangular flat surface = height = h

Breadth of the rectangular flat surface = diameter of circular base = 2r

∴ Area of the rectangular flat surface = length × breadth = h × 2r = 2rh

Also, the area of curved surface of a half cylinder = 1 × 2prh = prh
2
1
And, the area of two semi-circles = 2 × 2 × pr2 = pr2

Now, lateral surface area of a half cylinder = Area of (curved surface + flat surface)

= prh + 2rh

= rh(p + 2)

Total surface area of a half cylinder = lateral surface area + area of 2 semi-circles

= rh(p + 2) + pr2

Also, volume of a half cylinder = 1 pr2h
2

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Mensuration (III): Cylinder and Sphere

Worked-out Examples

Example 1: The diameter of a cylindrical drum is 35 cm and its 35 cm

height is 30 cm. Find (i) its curved surface area (ii) total 30 cm
surface area (iii) volume.

Solution:

Here, diameter of the cylindrical drum (d) = 35 cm.
c=m325
∴The radius of the drum (r) = 17.5 cm
Height of the drum (h) = 30
22
(i) Now, the curved surface area (C.S.A.) = 2πrh = 2 × 7 × 17.5 × 30 cm = 3,300 cm2

(ii) Also, the total surface area (T.S.A.) = 2πr (r + h)

= 2 × 22 × 17.5 cm (17.5 cm + 30 cm)
7
22
= 2 × 7 × 17.5 cm × 47.5 cm = 5,225 cm2

(iii) And, volume (V) = πr2h =272 × 17.5 cm × 17.5 cm × 30 cm = 28,875 cm3

Example 2: The perimeter of the circular base of a cylinder having height 20 cm is
44 cm. Calculate its:

(i) curved surface area (ii) total surface area (iii) volume

Solution:

Here, the perimeter of circular base (C) = 44 cm.

or, 2pr = 44 cm

or, 2 × 22 × r = 44 cm
7

or, r = 7 cm

Height of the cylinder (h) = 20 cm 22
7
(i) Now, the curved surface area (C.S.A.) = 2prh = 2 × × 7 cm × 20 cm= 880 cm2

(ii) Also, the total surface area (T.S.A.) = 2pr (r + h) = 2 × 22 × 7 cm(7 cm + 20 cm)
7
22
= 2 × 7 × 7 cm × 27 cm = 1,188 cm2

(iii) And, volume (V) = pr2h = 22 × 7 cm × 7 cm × 20 cm = 3,080 cm3
7

Example 3: The radius and height of a cylindrical log are in the ratio

7:9. If its curved surface area is 6,336 sq. inch, find its

volume.

Solution:

Let, the radius of circular base (r) = 7x inch and height (h) = 9x inch.

Now, the curved surface area of the log = 6336 sq. inch

or, 2πrh = 6336
22
or, 2 × 7 × 7x × 9x = 6336
or, 396x2 = 6336

or, x2 = 16

or, x = 4

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Mensuration (III): Cylinder and Sphere

∴ The radius of the log (r) = 7 × 4 inch = 28 inch and height (h) =9 × 4 inch = 36 inch

Again, volume (V) = πr2h = 22 × 28 inch × 28 inch × 36 inch = 88,704 cu. inch.
7

Example 4: If a circular water tank having radius of base 3.5 m can hold 3.08 × 105

liters of water. Find the surface area of the tank.

Solution:

Let, the radius of circular base (r) = 3.5 m

Capacity of the tank = 3.08 × 105 liters = 308000 liters

∴Volume of the tank (V) = 308000 m3= 308 m3 [ 1000 l = 1 m3]
1000

Now, volume of the tank (V) = 308 m3

or, πr2h = 308

or, 22 × 3.5 × 3.5 × h = 308
7

or, 38.5 h = 308

or, h = 8 m

Again, the total surface area (T.S.A.) = 2πr (r + h) = 2 × 22 × 3.5 m (3.5 + 8) m = 253 m2
7

Example 5: Suppose that you are going to buy a can of ghee
in a dairy. The shopkeeper showed you the
ghee of same quality in two cans of different 14 cm
size manufactured by a milk factory. The price 10 cm
of the ghee is not labeled in the can, which can
is more expensive and why? Give reason with
calculation. 10 cm 14 cm
Can - B
Solution: Can - A

Here, for the can-A, base radius (r) = 10 cm = 5 cm and height (h) = 14 cm
2

Now, volume of can-A (V1) = πr2h = 22 × 5 cm × 5 cm × 14 cm= 1100 cm3
7

Also, for the can-B, base radius (r) = 14 cm = 7 cm and height (h) = 10 cm
2

Now, volume of can-B (V2) = πr2h = 22 × 7 cm × 7 cm × 10 cm = 1540 cm3
7

Since, the volume of the ghee in the can-B is more than the volume of ghee in the can-A, the
can-B would be more expensive than can-A.

Example 6: The external and the internal radii of a hollow cylindrical pipe 70 cm long
are 5.4 cm and 5.1 cm respectively. Find (i) the volume of the material
contained by the vessel and (ii) the total surface area of the pipe.

Solution:

Here, the external radius of the pipe (R) = 5.4 cm

The internal radius of the pipe (r) = 5.1 cm

The height of the pipe (h) = 70 cm

(i) Now, the volume of the material = π (R2 – r2) h

= 272× (5.42 – 5.12) × 70 cm3 = 693 cm3

Hence, the required volume of the material contained by the vessel is 693 cm3.

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Mensuration (III): Cylinder and Sphere

(ii) Again, the T.S.A. of the hollow pipe = 22π×(R272+(5r).4(h++5.R1)–(7r)0 + 5.4 – 5.1)
=

= 4,639.8 cm2

Hence, the surface area of the pipe is 4,639.8 cm2.

Example 7: A metal pipe is 1.5 m long. Its internal radius is 48 mm and thickness
is 2 mm. If 1 mm3 of the metal weighs 0.02 gm, find the weight of the pipe.

Solution:
Here, the internal radius of the pipe (r) = 48 mm, thickness (t) = 2 mm
∴ The external radius of the pipe (R) = r + t = 48 mm + 2mm = 50 mm
The length of the pipe (h) = 1.5 m = 1.5 × 100 × 10 mm = 1500 mm

Now, the volume of the material used in the pipe = π (R2 – r2)h
22
= 7 × (502 – 482) × 1500 cm3

= 9,24,000 mm3

Also, 1 mm3 of metal weighs 0.02 gm 18480
1000
∴ 9,24,000 mm3 of metal pipe weighs 9,24,000 × 0.02 gm = 18480 gm = kg

= 18.48 kg
Hence, the weight of the pipe is 18.48 kg.

Example 8: The internal diameter and height of a cylindrical bucket are 14 cm and
35 cm respectively and it is filled with water completely. If the water is
poured into a rectangular glass tray with internal length 28 cm and breadth
17.5 cm and it is completely filled with water, find the height of the tray.

Solution:

Here, the internal diameter of the cylindrical bucket = 14 cm. So, radius (r) = 7 cm

Height of the bucket (h) = 35 cm

Now, the internal volume of the bucket = πr2h = 22 × 7 × 7 × 35 = 5,390 cm3
Thus, the volume of water 7
= internal volume of bucket = 5,390 cm3

Again, volume of the rectangular glass tray = volume of water = l × b × h = 5,390

or, 28 × 17.5 × h = 5,390

or, h = 5,390 = 11 cm
28 × 17.5
Hence, the height of the rectangular glass tray is 11 cm.

Example 9: A roller of diameter 100 cm and length 140 cm takes 450
complete revolutions to level a playground. Calculate the
area of the garden.

Solution:

Here, the diameter of roller (d) = 100 cm
100 cm 50
∴ The radius of roller (r) = 2 = 50 cm = 100 m = 0.5 m

Length of roller (h) = 140 cm = 140 m = 1.4 m
100

Now, the area covered by the roller in 1 revolution = curved surface area of the roller
24π5r0hr=ev2o×lu2t7i2on×s 0.5
= × 1.4 m2 = 4.4m2
the in
∴ Area of ground = Area covered by the roller

= 450 ×4.4m2 = 1,980 m2

Hence, the area of the ground is 1,980 m2.

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Mensuration (III): Cylinder and Sphere

EXERCISE 7.1

General section x cm
1. a) If x be the radius and y be the height of the cylinder given alongside,
y cm
write the formulae to find it’s

(i) curved surface area (C.S.A.)

(ii) Total surface area (T.S.A.) (iii) volume

b) A and a are the external and internal radii and b be the height of a hollow cylinder
respectively. Write the formula to find it’s (i) curved surface area (C.S.A.) (ii) total
surface area (T.S.A.) (iii) volume of material contained by the cylinder.

c) From the given half-cylinder, write the formulae to find

(i) curved surface area (ii) lateral surface area r=pcm q cm
(ii) total surface area (iii) volume

2. a) The perimeter of the base of a cylinder is 44 cm and height is 10 cm, what is its
curved surface area?

b) What is the total surface area of a cylinder in which the circumference of the base is
22 inch and the sum of radius and height is 15 inch?

c) What is the volume of a cylinder having area of base 38.5 cm2 and height 20 cm?

3. Find the (i) curved surface area, (ii) total surface area and (iii) volume of the given
cylinders.
a) b) c) 14 cm d)

28 cm

25 cm 35 in.20cm
6 cm
7cm 50 in.

Creative section-A
4. a) There is a cylindrical wooden log in a meat-shop. If the radius of its

circular base is 21 cm and height is 50 cm,

(i) find the curved surface area of the log.

(ii) find the total surface area of the log.

(iii) find the volume of the log. 1.75 in.
b) From the can of vegetables given alongside.

(i) Find the area of the label on the can. 5 in.
(ii) Find the area of the metal sheet used to make the can.

(iii) Find the volume of the can.

5. a) A cylindrical tank has diameter 1.4 m and height 2m. How many litres of water is
required to fill the tank completely?

b) The diameter and height of a cylindrical vessel are equal. If its radius is 35 cm and
it is completely filled with diesel. How many litres of diesel is filled in the vessel?

c) How many cubic meter of earth must be dug out to make a well 20 m deep and 1.4
m diameter?

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Mensuration (III): Cylinder and Sphere

6. a) The circumference of circular base of a closed cylindrical juice can of height 10 cm
is 22 cm. How much metal sheet is required to make it? Find it.

b) The height of a cylindrical drum is 20 cm and the area of its base is 346.5 sq. cm. If
it is to be covered with leather, find the minimum quantity of leather sheet needed
for it in sq. cm.

7. a) The radius and height of a cylindrical container are in the ratio 5:7 and its curved
surface area is 5500 sq.cm. Find its volume.

b) The radius and height of a cylindrical jar are in the ratio 7:17 and its total surface
area is 4224 sq.cm. Find its volume.

c) A cylinder container water tank contains 4,62,000 litres of water and its radius is
3.5 m. Find its curved surface area.

d) The volume of a cylindrical can is 3.08 litre. If the diameter of its base is 14 cm, find
8. a) the total surface area of the can.
2
The area of curved surface of a solid cylinder is equal to 3 of the total surface area.

If the total surface area is 924 cm2, find the volume of the cylinder.

b) The number of volume of a cylinder is half of its number of the total surface area. If
the radius of its base is 7 cm, find the volume.

9. a) The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the
total surface area of the cylinder is 1,628 cm2, calculate its volume.

b) The circumference of the base of a cylindrical drum is 44 cm and the sum of its
radius and height is 27 cm, find the volume of the cylinder.

c) The curved surface area and volume of a cylinder are 880 cm2 and 3,080 cm3
respectively, find its total surface area.

10. a) Calculate the lateral surface area, total surface area and volume of the 20cm
given half-cylinder.

b) The radius of the semi-circular base of a half-cylinder is 21 cm and its 14cm
height is 50 cm. Find its lateral surface area, total surface area and
volume.
c) Mr. Binod made a vegetable tunnel as shown in the figure
given aside. Find the volume of the tunnel.
6 m 21 m

11. a) A hollow cylindrical metallic pipe is 28 cm long. If the external and internal
diameters of the pipe are 12 cm and 8 cm respectively, find the volume of metal
used in making the pipe.

b) The external and the internal radii of a cylindrical pipe of length 50 cm are 3 cm and
2.6 cm respectively. Find the volume of material contained by the pipe.

c) A metal pipe is 1.4 m long. Its internal diameter is 25 mm and thickness 1 mm. If 1
mm3 of the metal weighs 0.02 gm, find the weight of the pipe.

d) A metal pipe has inner diameter of 5 cm. The pipe is 5 mm thick all round and 2 m long.
What is the weight of the pipe if 1 cm3 of the metal weighs 7.7 g?

12. a) The internal radius of a cylindrical bucket of height 50 cm is 21 cm. It is filled
with water completely. If the water is poured into a rectangular vessel with internal
length 63 cm and breadth 44 cm and it is completely filled with water, find the
height of the vessel.

b) Vegetable ghee is stored in a cylindrical vessel of internal radius 1.4 m and height
1.5 m. If it is transferred into the rectangular tin cans 33 cm × 10 cm × 5 cm, how
many cans are required to empty the vessel?

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Mensuration (III): Cylinder and Sphere

Creative section - B

13. a) Mrs. Thapa has two cylindrical buckets. The bucket-A has radius of base 14 cm and
height 30 cm. Likewise, bucket-B has the radius of base 16 cm and height 28 cm.
Which of these buckets can hold more water and by how many litres?

b) Mr. Rai has fixed two overhead cylindrical water tanks on the roof of his house. The
diameter of the base of tank-X is 2.1 m and height is 3 m. Similarly, the diameter of
the base of tank-Y is 3 m and height is 2.1 m. Which of these tanks holds more water
and by how many litres? Find.

14. a) Mr. Chaudhary drives a tanker in a milk supply company. He supplies 5 tankers
of milk every day. If the milk tank is cylindrical in shape having radius 0.8 m and
length 3.5 m. How many liters of milk does he supply in each day?

b) Chari Maya is a juice seller. She serves the juice completely filled with the cylindrical
glasses; each has diameter 7 cm and height 10 cm. If she sells 50 glasses of juice in
a day, how many litres of juice does she sell in a day?

15. a) A roller of diameter 112 cm and length 150 cm takes 550 complete revolutions to
level a playground. Calculate the area of the ground.

b) A temple has 14 cylindrical pillars made of concrete. If the radius of base of each
pillar is 20 cm and height is 5 m, how much concrete is required to make the pillars?
Find it.

Project work and activity section
16. a) Measure the different dimensions (such as diameter, circumference, height) of water

tank, drum, cylindrical bucket, etc. in your house or in school and calculate their
curved surface area, total surface area and capacity.

b) Measure the circumference of the circular base of a mineral water bottle, then find
its radius. Calculate the approximate capacity of the bottle.

c) Measure the length and breadth of a rectangular sheet of chart paper and find its
area. Now, roll the paper to form a cylinder and find its curved surface area. Then
compare the two areas.

7.7 Area and Volume of a Sphere

A sphere is a perfectly round geometrical object in three-
dimensional space. It is generated by revolving a semi-circle
about its diameter. A football, a cricket ball, etc. are said to have the shape of a
sphere.

Total surface area of a sphere

Archimedes (c. 287 – 212 BCE) was an ancient Greek scientist and
engineer, and one of the greatest mathematicians of all time. He
discovered many concepts of calculus and worked in geometry,
analysis and mechanics. While taking a bath, Archimedes
discovered a way to determine the volume of irregular objects
using the amount of water they displaced when submerged.
He was so excited by this discovery that he ran out on the street, still undressed,
yelling “Eureka!” (Greek for “I have found it!”). He discovered that the surface area of
a sphere is the same as the lateral surface area of a cylinder having the same radius
as the sphere and the height equal to the diameter of the sphere.

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Mensuration (III): Cylinder and Sphere

Let, the radius of the given sphere is equal to the radius of the cylinder
and the diameter of the sphere is equal to the height of the cylinder.
Then, the sphere could easily be fitted in the cylinder.

Here,

Surface area of the sphere = curved surface area of the cylinder = 2πrh
Now, let’s replace h by 2r, we get
Surface area of the sphere = 2πr × 2r = 4πr2.
Volume of a sphere
Step 1: Let’s take a hollow sphere and two identical hollow cylinders in which the

diameter of base and height are equal to the diameter of the sphere.
Step 2: Fill the hollow sphere with sand once and empty it into one of the cylinders.
Also, fill the hollow sphere second time with sand and empty it into the second

cylinder.
Step 3: Again, fill the hollow sphere third time and empty it into the remaining spaces

of the cylinders.

rr
h = 2r
h = 2r

Vedanta ICT Corner

From the above experiment,
the volume of 3 spheres = volume of 2 cylinders

= 2 × πr2h Please! Scan this QR code or
= 2π × r2 × 2r [ h = 2r] browse the link given below:

= 4πr3 4 https://www.geogebra.org/m/nvznmnek
3
Thus, the volume of 1 sphere = πr3

7.8 Hemisphere and great circle r

When a sphere is cut into two halves, one of the half
portions is called hemisphere. A hemisphere has a
plane circular face and a curved surface. The radius
of the circular face is equal to the radius of the sphere.
The circular face of a hemisphere is called a great
circle.

The area of a great circle of a sphere = πr2 or πd2
4

The circumference of the great circle = 2πr or πd

Also, the surface area of a sphere = 4πr2 1 × 4πr2
∴ The curved surface area of a hemisphere = 2

= 2πr2 or π d2
2

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Mensuration (III): Cylinder and Sphere

And, the total surface area of a hemisphere = Area of curved surface + Area of great circle

= 2πr2 + πr2

= 3πr2 or 3πd2
4
1
Similarly, the volume of a hemisphere = 2 of the volume of sphere

= 1 × 4 πr3
2 3
2 πd3
= 3 πr3 or 12

Worked-out Examples

Example 1: The diameter of volleyball given alongside is 21 cm. 21 cm
Find (i) its surface area (ii) volume.
Solution:
Here, the volleyball is in the shape of a sphere.
The diameter of the volleyball (d) = 21 cm.

∴ The radius of the volleyball (r) = 21 cm = 10.5 cm
2

(i) Now, the surface area = 4πr2= 4 × 22 × 10.5 cm × 10.5 cm = 1,386 cm2
7

(ii) And, volume (V) = 4 πr3 = 4 × 22 × 10.5 cm × 10.5 cm × 10.5 cm = 4,851 cm3
3 3 7
Hence, the required surface area of the volleyball is 1,386 cm2 and its volume is 4,851 cm3.

Example 2: The surface area of a concrete sphere is 5544 cm2, calculate its diameter.

Solution:

Here, the surface area of the sphere = 5544 cm2, diameter (d) =?

Now, the surface area of the sphere = 5544

or, 4πr2 = 5544

or, 4 × 22 × r2 = 5544 Vedanta ICT Corner
7 Please! Scan this QR code or
or, r2 = 441 browse the link given below:

or, r = 21 https://www.geogebra.org/m/zacgudyp

∴ The diameter (d) = 2r = 2 × 21 cm = 42 cm

Hence, the required diameter is 42 cm.

Example 3: Given that 1 cm3 of iron weighs 7.8 gm. What is the weight of shot-put if the
circumference of its great circle is 33 cm?

Solution:
Here, circumference of the great circle = 33 cm

or, 2πr = 33

or, 2 × 22 × r = 33
or, 7 r = 5.25
cm

Again, volume (V) = 4 πr3 = 4 × 22 (5.25 cm)3 = 606.375cm3
3 3 7

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Mensuration (III): Cylinder and Sphere

We have, 1 cm3 of iron weighs 7.8 gm

∴ 606.375 cm3 of iron weighs 7.8 × 606.375 gm = 4,729.725 gm

= 4,729.725 kg
1000

= 4.729 kg
Hence, the weight of the shot-put is 4.729 kg.

Example 4: Find the curved surface area, total surface area and

volume of the hemisphere given alongside. 8.4 cm
Solution:

Here, the radius of the hemisphere (r) = 8.4 cm
22
Now, the curved surface area = 2πr2 = 2 × 7 × 8.4 × 8.4 cm = 443.52 cm2

Also, the total surface area = 3πr2 = 3 × 22 × 8.4 × 8.4 cm = 665.28 cm2
7
2 2 22
And, volume (V) = 3 πr3 = 3 × 7 × 8.4 cm × 8.4 cm × 8.4 cm = 1241.856 cm3

Hence, the curved surface area of the hemisphere is 443.52 cm2, its total surface area is
665.28 cm2 and volume is 1241.856 cm3.

Example 5: Three metallic spheres of radii 3 cm, 4 cm, and 5 cm are melted and reformed
into a single sphere. Find the radius of the new sphere.

Solution:

Solution: 3 cm

Here, the radius of the first sphere (r1) = 3 cm

The radius of the second sphere (r2) = 4 cm 4 cm ?

The radius of the third sphere (r3) = 5 cm 4 πr13 = 4 π × 33
Now, the volume of the first sphere (V1) = 3 3

The volume of the second sphere (V2) = 4 πr23 = 4 π × 43 5 cm
3 3
4 4
The volume of the third sphere (V3) = 3 πr33 = 3 π × 53

Again, the volume of new single sphere (V) = V1 + V2+V3

or, 4 πr3 = 4 π × 33 + 4 π × 43 + 4 π × 53
3 3 3 3
4 4
or, 3 πr3 = 3 π (33 + 43 + 53)

or, r3 = 216

∴ r = 6 cm
Hence, the radius of new sphere is 6 cm.

Example 6: A solid metallic sphere of radius 6 cm is melted and drawn into a cylindrical
wire of diameter 4 mm. Find the length of the wire.
Solution:
4
Here, the radius of the cylindrical wire (r) = 2 mm = 2 mm = 0.2 cm

Now, the volume of the cylindrical wire = πr2h = π(0.2 cm)2 × h = 0.008πh cm2

Also, the volume of the metallic sphere = 4 πR3 = 4 π(6 cm)3 = 288π cm3
3 3

Again, the volume of the cylindrical wire = the volume of the sphere

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Mensuration (III): Cylinder and Sphere

or, 0.008πh cm2 = 288 π cm3

∴ h = 36000 cm = 36000 m= 360 m
100

Hence, the required length of the wire is 360 m.

Example 7: A shopkeeper has one spherical laddu of radius 5 cm. How many laddus of

radius 2.5 cm can be made with the same amount of material?

Solution:

Here, the radius of bigger laddu (R) = 5 cm
4 4 500π
∴Volume of bigger laddu (V) = 3 πR3 = 3 π(5 cm)3 = 3 cm3

Also, the radius of each smaller laddu (r) = 2.5 cm

∴ Volume of each smaller laddu (v) = 4 πr3 = 4 π(2.5 cm)3 = 62.5π cm3
3 3 3
500π
V 3
Again, the number of smaller laddus = v = 62.5π = 8

Hence, required number of smaller laddus is 8. 3

EXERCISE 7.2

General section

1. a) Write the formula to find (i) surface area and (ii) volume of x cm
the given sphere.

b) The radius of a hemisphere is p cm. Write the formulae to find its

(i) curved surface area (ii) total surface area (iii) volume.

c) If the radius of a sphere is 1 cm, what are its surface area and the volume?

2. a) Find the surface area and volume of the spheres given below.

(i) (ii) (iii) (iv)

8.4 cm

7 cm 6.3cm 42 cm

b) Calculate the curved surface area, total surface area and volume of the following
hemi-spheres.

(i) 21 cm (ii) (iii) 16.8 c m (iv)

14 cm 5.6 cm

3. a) If the radius of a spherical orange is 3.5 cm, find its surface area and volume.
b) The diameter of a basketball is 21 cm, find its surface area and the volume.
c) The diameter of the moon is 3476 km. Find the volume of the moon.

Creative section - A
4. a) The diameter of spherical iron ball is 14 cm. If 1 cm3 of iron weighs 7.8 gm, what is

the weight of iron ball?

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Mensuration (III): Cylinder and Sphere

b) A spherical watermelon has diameter 21 cm. If it contains 90% of water, find the
volume of water in it.

c) The diameter of a cricket ball is 7 cm. How much sticker is required 7 cm
to paste on the surface of the ball?

d) How much leather sheet is required to make a football with diameter 21 cm?

5. a) If the radius of a hemispherical cake is 10.5 cm, find its total surface area and volume.

b) The diameter of the hemispherical top of a mushroom is 4.2 cm, find its curved
surface area and volume.

6. a) A hemispherical bowl has radius 7 cm. If the bowl is completely filled with vegetable
soup, find the volume of the soup.

b) A hemispherical ‘Damaha’ has its diameter 35 cm. If it is tightly covered with a
plastic, how much plastic is used to cover it?

7. a) If the perimeter of the great circle is p cm, find the volume of the hemisphere.

b) If the circumference of the great circle is 44 cm, find the total surface area of the
hemisphere.

c) The total surface area of a hemisphere is 243p cm2, find its volume.

8. a) A solid metallic sphere of radius 7 cm is cut into two halves. Find the total surface
area of the two hemispheres so formed.

b) A solid metallic sphere of diameter 42 cm is cut into two halves. Find the total
surface area of the two hemispheres so formed.

9. a) If the volume of a spherical object is 9π cm3, find its diameter.
2
b) If the total surface area of a solid sphere is 616 cm2, what is its radius?

c) If the volume of a hemisphere is 19,404 cm3, find its radius.

10. a) If the surface area of a globe is 5,544 cm2, find its volume.

b) If the volume of a spherical ball is 38,808 cm3, find its surface area.

11. a) If the circumference of the great circle of a solid hemisphere is 132 cm, find

(i) its curved surface area (ii) total surface area (iii) volume

b) If the total surface area of a hemisphere is 1,848 cm2, find:

(i) its curved surface area (ii) volume

c) If the volume of a hemisphere is 2425.5 cm3, find its curved surface area and total
surface area.

12. a) If the radius of a sphere is doubled, by how much does its surface area increase?

b) The radius of the earth is four times the radius of the moon, find the ratio of
(i) their surface areas (ii) their volumes.

c) If the volumes of two spherical objects are in the ratio of 8:27, find the ratio of
(i) their radii (iii) their surface areas.

13. a) Three metallic spheres of radii 1 cm, 6 cm and 8 cm respectively are melted and
re-formed to a single sphere. Find the radius of the new sphere.

b) Three silver hemispheres of radii 6 cm, 8 cm and 10 cm respectively are melted and
re-formed to a single hemisphere. Find the radius of the new hemisphere.

c) 8 metallic spheres each of radius 2 cm, are melted and cast into a single sphere.
Calculate the radius of the new sphere.

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Mensuration (III): Cylinder and Sphere

Creative section - B
14. a) A hemi-spherical bowl of internal radius 9 cm is completely filled with water. If the

water is poured into a cylindrical vessel of radius 6 cm and height 20 cm, what is
the height of empty space in the vessel?
b) A solid iron sphere of diameter 42 cm is dropped into a cylindrical drum partly
filled with water. If the radius of the drum is 1.4 m, by how much is the surface of
the water raised?
c) A cylindrical jar of radius 6 cm contains water. How many iron solid spheres each
of radius 1.5 cm are required to immerse into the jar to raise the level of water by
2 cm
15. a) A solid metallic sphere of radius 3 cm is melted and drawn into a cylindrical wire of
radius 5 mm. Find the length of the wire.
b) The diameter of a solid spherical metallic ball is 12 cm. If it is melted and drawn
into a cylindrical wire of length 288 cm, find the thickness of the wire.
c) How many solid spheres each 6 cm diameter can be made from a solid metallic
cylinder of diameter 4 cm and height 45 cm?
16. a) Mrs. Thapa wishes to grow flowers on the hanging bowls at her home. She brings 16
hemispherical bowls from the market having internal diameter 21 cm and uniform
thickness 0.5 cm each and fills the bowls with compost completely.
(i) If 1 cm3 of compost weighs 2.31 gm, find the total weight of required compost.
(ii) If she covers outer curved surface of each bowl with polythene sheet, find the

total amount of polythene.
b) Assume that the earth as a sphere of radius 6400 km.
(i) If the average density of the earth is 5500 kg per cubic meter, estimate the mass

of the earth.
(ii) If 70% of the earth is covered by water, find the earth’s surface area covered by

the land.
Project work and activity section
17. a) Measure the radius or diameter of few spherical objects which are available in your

house or school. Then, calculate their surface area and volume.
b) Collect a few hemispherical objects then, calculate their curved surface area, total

surface area and volume.

7.11 Cost estimation

Let A = Area of floor, carpet, four walls, ceiling etc.

R = Rate of cost of carpeting, plastering, coloring, papering etc.

C = Total cost of carpeting, plastering, painting etc.

N = Required number of pieces of carpets, paper etc.

a = Area of each piece of carpet, paper, bricks.

L = Total length of carpet/paper

Now,

(i) Total cost (C) = Area × Rate C = A × R . Then, A = C and R = C
R A

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Mensuration (III): Cylinder and Sphere

(ii) Total cost (C) = Length × Rate C = l × R
(iii) Total cost (C) = Number of pieces of carpets/ paper × Rate C = N × R

Worked-out Examples

Example 1: A rectangular room is 9 m long and 6 m broad. The floor of the room is
covered with the carpet each of 4 m long and 1.5 m wide.
(i) Find the cost of carpeting the floor at the rate of Rs 125 per sq. meter.
(ii) Find the cost of carpeting the floor at the rate of Rs 750 per piece.
(iii) Find the cost of carpeting the floor at the rate of Rs 185 per meter.

Solution:

Here, Length of the room (l) = 9 m

Breadth of the room (b) = 6 m

Width of the carpet (b1) = 1.5 m
Length of the carpet (l1) = 4 m
(i) Now, the area of the floor of the room (A) = l × b = 9 m × 6 m = 54 m2
Hence, the cost of carpeting the floor (C) = Area × Rate = 54 ×Rs 125=Rs 6,750

(ii) Also, the area of each piece of carpet (a) = l1 × b1 = 4 m × 1.5 m = 6 m2

Now, the number of pieces of carpet = A = 54 m2 = 9
a 6 m2

Hence, the cost of carpeting the floor (C) = Number of pieces of carpet × Rate

= 9 × Rs 750=Rs 6,750

(iii) Again, let x m be the total length of carpet required to cover the floor.
Here, area of carpet = Area of the floor

or, x × 1.5 m = 54 m2

or, x = 54 m2 = 36 m
1.5 m

Thus, the total cost of carpeting the floor = length × rate = 36 × Rs 185 = Rs 6,660.

Example 2: The length, breadth, and height of a rectangular room are 12 m, 10 m and
6 m respectively.

a) Find the cost of painting its wall and ceiling at Rs 75 per sq. m.
b) Find the cost of carpeting its floor at Rs 120 per sq. m.

Solution:
Here, the length of the room (l) = 12 m

The breadth of the room (b) = 10 m
The height of the room (h) = 6 m
The rate of cost of painting (R) = Rs 75 per sq. m.
The rate of cost of carpeting the floor (R') = Rs 120 per sq. m.
a) Now, the area of 4 walls and ceiling (A) = 2h (l + b) + lb
= 2 × 6 (12 + 10) + 12 × 10 = 384 m2

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Mensuration (III): Cylinder and Sphere

Again, the cost of painting its wall and ceiling = A × R = 384 × Rs 75 = Rs 28,800
Also, the cost of carpeting the floor = Area floor × Rate = (12 × 10) × Rs 120 = Rs 14,400
Hence, the required cost of painting the walls and the ceiling of the room is Rs 28,800 and
the cost of carpeting the floor is Rs 14,400.

Example 3: A rectangular room is 15 m long, 10 m broad and 5 m high. If it contains
Solution: two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m, find
the cost of plastering its walls at Rs 50 per sq. metre.

Here, the length of the room (l) = 15 m

The breadth of the room (b) = 10 m

The height of the room (h) = 5 m

The rate of plastering walls (R) = Rs 50 per sq. m.

Now, the area of its 4 walls = 2h (l + b) = 2 × 5 m (15 m + 10 m) = 250 m2

Also, the area of 2 windows = 2 (2 m × 1.5 m) = 6 m2

The area of a door = 1 m × 4 m = 4 m2

\ The area of 4 walls excluding windows and door = (250 – 6 – 4) m2 = 240 m2
Again, the cost of plastering its walls = A × R = 240 × Rs 50 = Rs 12,000

Hence, the required cost of plastering its walls is Rs 12,000.

Example 4: A room is 12 m long and 5.5 m high. If the cost of carpeting its floor at
Rs 80 per sq. metre is Rs 7,680, find the cost of colouring its walls at
Rs 35.50 per sq, m.
Solution:
Here, the length of the room (l) = 12 m
the height of the room (h) = 5.5 m
cost of carpeting the floor
Now, area of the floor = Rate of cost

or, l × b = 7680 m2
80

or, 12 × b = 96 m2
96
or, b = 12 m =8 m

Again, area of 4 walls of the room (A) = 2h (l + b) = 2 × 5.5 m (12 m + 8 m) = 220 m2

\ The cost of colouring the walls = Area × Rate = 220 × Rs 35.50 = Rs 7810
Hence, the required cost of colouring the walls is Rs 7810.

Example 5: The cost of carpeting a square room at the rate of Rs 75 per sq. metre is

Rs 10,800. If the cost of plastering its walls at Rs 25 per sq. metre is

Solution: Rs 6000, find the height of the room.

Here, the area of the floor of the square room = cost of carpeting
Rate of cost

or, l2 = 112078m500anmd,2 = 144 m2 m
or, l = also b = 12
cost of plastering the walls
Again, area of 4 walls = Rate of cost

or, 2h (l + b) = 6000 m2
25

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Mensuration (III): Cylinder and Sphere

or, 2h (12 + 12) = 240 m2

or, h = 5 m

So, the required height of the room is 5 m.

Example 6: The cost of carpeting the floor of a room, whose breadth is twice the height
and the length is twice its breadth, at the rate of Rs 80 per sq. metre is
Rs 10,240. What is the cost of plastering its walls at Rs 30 per sq. metre?

Solution:
Let, the height of the room (h) = x m
Then, the breadth of the room (b) = 2x m
And, the length of the room (l) = 2b = 2 × 2x m = 4x m

Now, the area of the floor = cost of carpeting the floor
Rate of cost
10240
or, l × b = 80 m2

or, 4x × 2x = 128 m2

or, x2 = 16 m2

or, x = 4 m

\ The height of the room (h) = x = 4 m, the breadth of the room (b) = 2x = 2 × 4 m = 8 m

The length of the room (l) = 4x = 4 × 4 m = 16 m

Again, area of 4 walls = 2h (l + b) = 2 × 4 m (16 m + 8 m) = 192 m2

\ The cost of plastering 4 walls = Area × Rate = 192 × Rs 30 = Rs 5760

So, the required cost of plastering its 4 walls is Rs 5760.

Example 7: A square room contains 180 m3 of air. The cost of plastering its four walls
at Rs 60 per sq. metre is Rs 7,200. Find the height of the room.
Solution:
Let, the length (l) = breadth (b) = x m

Here, volume of the room = Volume of the air

or, l × b × h = 180 m3

or, x × x × h = 180 m3
180
or, h = x2 m ... equation (i)

Again, area of 4 walls = Cost of plastering of 4 walls
Rate of cost
7200
or, 2h (l + b) = 60 m2

or, 2h (x + x) = 120 m2

or, h = 30 m ... equation (ii)
x

From equation (i) and (ii) we get,

180 = 30
x2 x

or, x = 6 m

Now, putting the value of x in equation (ii), we get,
30
h = 6 = 5 m

Hence, the required height of the room is 5 m.

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Mensuration (III): Cylinder and Sphere

EXERCISE 7.3
General section

1. a) If the area of floor of a room is x m2, what is the cost of carpeting the room at Rs y
per sq. m?

b) A room requires P pieces of carpet. What is the cost of carpeting the room at Rs Q
per piece of carpet?

c) The total length of carpet required for a room is A ft. What is the cost of carpeting
the room at Rs B per feet?

d) What is the cost of papering the four walls of a room having area 180 m2 at Rs 120
per square meter?

e) The area of four walls and ceiling of a rectangular room is 260 m2. What is the cost
of white painting the walls and the ceiling at Rs 150 per square meter?

2. a) The length of a hall is 30 ft. and the width is 18 ft. Find the cost of carpeting the
room at Rs 65 per square feet.

b) A rectangular hall is 12 m long and 10 m broad. Find the length of carpet 2 m wide
required for covering its floor. If the rate of cost of carpet is Rs 110 per metre, find
the cost of carpeting the floor.

c) 6 pieces of carpet are required for a room. What is the cost of carpeting the floor at
Rs 750 per piece?

Creative section - A

3. a) A rectangular room is 10 m long and 9 m broad. The floor of the room is covered
with the carpet each of 4 m long and 1.5 m wide.
(i) Find the cost of carpeting the floor at the rate of Rs 150 per sq. meter.
(ii) Find the cost of carpeting the floor at the rate of Rs 850 per piece.
(iii) Find the cost of carpeting the floor at the rate of Rs 200 per meter.

b) The guest room in Sunayana’s house is 24 ft. long and 15 ft. wide. The floor of the
room is completely covered with a few numbers of pieces of carpet each of 9 ft. long
and 5 ft. wide

(i) Find the cost of carpeting the floor if the rate of cost of carpet is Rs 30 per sq. ft.
(ii) Calculate the cost of carpeting the floor at Rs 1250 per piece.
(iii) If the rate of cost of carpet is Rs 160 per ft, find the cost of carpeting the floor.
4. a) A rectangular room is 10 m long, 8 m wide, and 5 m high. Find the cost of colouring
its walls and ceiling at Rs 65 per sq. metre.

b) A square room is 15 feet long and 10 feet high. Find the cost of plastering its walls,
ceiling and floor at Rs 18 per sq. ft.

c) The cost of plastering the walls and the ceiling of a room at Rs 20 per sq. feet is
Rs 14,400. Find the cost of colouring the walls and ceiling at Rs 16 per sq. feet.

5. a) A rectangular room is 8 m long, 6 m broad and 4 m high. It contains two windows
of size 2 m × 1.5 m each and a door of size 1 m × 4 m. Find the cost of painting its
walls and ceiling at Rs 54 per sq. metre.

b) A square hall is 15 m long and 5 m high. It contains three square windows each
of 2 m long and two doors of size 1.5 m × 4 m. Find the total cost of plastering
and colouring its walls and ceiling at Rs 50 per sq. metre and Rs 45 per sq. metre
respectively.

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Mensuration (III): Cylinder and Sphere

Creative section - B
6. a) A rectangular room is 10 m long and 8 m wide. It has two windows each of 2 m × 2 m

and a door of size 1.5 m × 4 m. If the cost of plastering its walls at Rs 55 per sq. metre
is Rs 9,130, find the height of the room.
b) A square room is 4 m high. The cost of carpeting its floor at Rs 95 per sq. metre is
Rs 6,080. Find the cost of painting its walls and ceiling at Rs 60 per sq. metre.
c) A rectangular room is 10 m long and 5 m high. If the cost of paving marbles at Rs 180
per sq. metre is Rs 10800, find the cost of papering its walls at Rs 45 per sq. metre.
d) The cost of carpeting a square room at Rs 110 per sq. metre is Rs 5,390. If the cost of
plastering its walls at Rs 56 per sq. metre is Rs 7,840, find the height of the room.
e) The cost of plastering the floor of a room, which is 10 m long, at Rs 54 pr sq. metre
is Rs 4,320. If the cost of painting its walls at Rs 48 per sq. metre is Rs 10,368, find
the height of the room.
7. a) A rectangular room is twice as long as it is broad and its height is 4.5 m. If the cost of
papering its walls at Rs 40 per sq. metre is Rs 6,480, find the cost of paving marbles
on its floor at Rs 150 per sq. metre.
b) A rectangular room is three times as long as it is broad and its height is 4.6 m. If the
cost of carpeting its floor at Rs 85 per sq. metre is Rs 6,375, find the cost of colouring
the walls at Rs 50 per sq. metre.
8. a) A room is 12 m long and 8 m broad and it contains 480 cu. metre of air. Find the cost
of colouring its walls at Rs 54 per sq metre.
b) A room contains 600 cu. m. of air. If the cost of carpeting its floor at Rs 90 per sq.
metre is Rs 13,500, find the height of the room.
c) A square room contains 288 m3 of air. The cost of carpeting the room at Rs 105 per
sq. metre is Rs 6,720. Find the cost of painting its walls at Rs 45 per sq. metre.
d) A square room contains 256 cu. m. of air. If the cost of painting its 4 walls at Rs 50
per sq. metre is Rs 6,400, find the cost of carpeting its floor at Rs 99 per sq. metre.
e) A square room contains 220.5 cu. m. of air. If the cost of colouring its 4 walls at
Rs 55 per sq. metre is Rs 6,930, find the height of the room.
9. a) A room is two times longer than its breadth and it contains 360 m3 of air. If the cost
of plastering its floor at Rs 40 per sq. metre is Rs 2,880, find the cost of plastering its
walls at Rs 45 per sq. metre.
b) The length of a room is two times its breadth and it contains 210 m3 of air. If the cost
of painting its walls at Rs 54 per sq. m. is Rs 6,804, find the height of the room.

Project work and activity section
10. a) Measure the length, breadth, height, and thickness of walls of your classroom. Also

measure the length and height of windows and doors of the classroom.
(i) Find the volume of wall excluding windows and doors.
(ii) Find the average volume of a brick available in your surroundings.
(iii) Find the number of bricks required to construct the wall excluding windows

and doors.
(iv) Estimate and cost of bricks required to construct the walls as per the local rate

of cost.

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Mensuration (III): Cylinder and Sphere
b) Measure all the dimensions of a part of the compound wall of your school or your

house. Calculate the volume of the part of the wall. Find the volume of a brick and
calculate the number of such bricks required to construct the wall. Also, calculate
the cost of bricks as per the local rate of cost.

OBJECTIVE QUESTIONS

Let’s tick (√) the correct alternative.
1. What is the curved surface area of a cylindrical object whose radius of base is x cm

and height is y cm?

(A) πxy cm2 (B) 2πxy cm2 (C) πx (x + y) cm2 (D) πx2 cm2

2. If the area of base of a cylindrical object having curved surface area p cm2 is q cm2,
what is its total surface area?

(A) (p + q) cm2 (B) (2p + q) cm2 (C) (p + 2q) cm2 (D) 2(p + q) cm2

3. What is the volume of a cylinder whose radius of base is 7 cm and height is 13 cm?

(A) 2002 cm3 (B) 572 cm3 (C) 880 cm3 (D) 154 cm3

4. The internal and external radii of a cylindrical pipe of length ‘h’ cm are ‘r’ cm and ‘R’
cm respectively, what is the volume of material used to make the pipe?

(A) π (R2 + r2) h cm3 (B) π (R2 – r2) h cm3

(C) 2π (R2 – r2) h cm3 (D) 2π(R+r)(R-r+h) cm3

5. The lateral surface area of a half-cylinder having the base radius r cm and height h
cm is

(A) πr2 cm2 (B) πrh cm2 (C) rh (π + 2) cm2 (D) πr (r + h) cm2

6. The surface area of a sphere having radius x cm is

(A) πx2 cm2 (B) 2πx2 cm2 (C) 3πx2 cm2 (D) 4πx2 cm2

7. The diameter and height of a cylinder are equal to the diameter of a sphere. If the
volume of cylinder is V1 and the volume of sphere is V2 then what is the relation
between V1 and V2 is

(A) V1 = V2 (B) V1 = 2V2 (C) 2V1 = 3V2 (D) 3V1 = 2V2

8. If the area of great circle of a sphere is A cm2, what is the surface area of the sphere?

(A) A cm2 (B) 2A cm2 (C) 3A cm2 (D) 4A cm2

9. The surface area of a hemisphere having radius x cm is

(A) πx2 cm2 (B) 2πx2 cm2 (C) 3πx2 cm2 (D) 4πx2 cm2

10. The radius of a single sphere reformed by melting three metallic spheres of radii 1 cm,
6 cm and 8 cm is

(A) 9 cm (B) 15 cm (C) 18 cm (D) 30 cm

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Unit 8 Sequence and Series

8.1 Sequence - Introduction
Let’s observe the patterns in the following figures.

(i) (ii) (iii) 10cm
10cm
,, , , ,, 10cm
10cm

In figure (i), the numbers of squares in each design are 1, 5, 9, …
In figure (ii), the numbers of dots in each design are 3, 6, 9, 12, …
In figure (iii), the total heights of the steps from the base are 10 cm, 20 cm, 30 cm, ….
Let’s study the following lists of numbers and say the next two numbers (if possible) of each
list.

(i) 2, 5, 8, 11, … (ii) 3, 6, 12, 24, … (iv) 1, 4, 9, 16, .., (iv) 5, 6, 10, 19, …

In (i), each number is formed by adding 3 to the just preceding number.

In (ii), each number is formed by multiplying the just preceding number by 2.

In (iii), each number is square of consecutive natural numbers.

In (iv), the set of numbers does not follow any rule. So, we cannot guess the number next
to 19.

Here, the patterns of numbers in (i), (ii) and (iii) are in a certain rule. So, these lists of numbers
are sequences of numbers. But, the list of number in (iv) does not form any sequence.

Facts to remember
1. An ordered list of numbers or objects following a definite rule is known as a sequence.
2. Each element of the sequence is called a term. The terms of the sequence are denoted

by t1, t2, t3, …
3. A sequence having a finite number of terms is called a finite sequence. For example,

100, 90, 80, … 10 is a finite sequence.
4. If the number of terms of a sequence is infinite then it is called an infinite sequence.

For example, 2, 4, 6, 8, … is an infinite sequence.

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Sequence and Series

Worked-out Examples

Example 1: Find the next two terms of the following sequences.
Solution: (i) 3, 8, 13, 18, 23, … (ii) 5, 2, –1, –4, –7, … (iii) 2, 4, 8, 16, …
(i)
8, 13, 18, 23
3,

+5 +5 +5 +5
Here, each term is greater than its preceding term by 5.
The required next two terms are, t6 = 23 + 5 = 28 and t7 = 28 + 5 = 33

(ii) 2, –1, –4, –7
5,

–3 –3 –3 –3
Here, each term is decreased from preceding term by 3.
The required next two terms are, t6 = –7 – 3 = –10 and t7 = –10 – 3 = –13

(iii) 4, 8, 16, 32

×2 ×2 ×2
Here, each term is 2 times its preceding term.
The required next two terms are, t5 = 32 × 2 = 64 and t6 = 64 × 2 = 128

8.2 General term of sequence

In a sequence 7, 11, 15, 19, … ; the numbers 7, 11, 15 are the first, second and third terms
respectively. The nth term of the sequence is called its general term. It is denoted by tn or an.
Let’s observe the following examples to investigate the ideas of finding the general term of
the sequences.

Finding the nth term of the sequences

In 2, 4, 6, 8, … In 5, 9, 13, 17, … In 1, 4, 9, 16, … In 2, 6, 12, 20, … In 6, 12, 24, 48, ...

t1 = 2 = 2×1 t1 = 5 = 4×1 + 1 t1 = 1 = 12 t1 = 2 = 12 + 1 t1 = 6 = 3×21
t2 = 4= 2×2 t2 = 9 = 4×2 + 1 t2 = 4 = 22 t2 = 6 = 22 + 2 t2 = 12 = 3×22
t3 = 6= 2×3 t3 = 13 = 4×3 + 1 t3 = 9 = 32 t3 = 12 = 32+ 3 t3 = 24 = 3×23
t4 = 8= 2×4 t4 = 17 = 4×4 + 1 t4 = 16 = 42 t4 = 20 = 42 + 4 t4 = 48 = 3×24
…………… …………………… …………… …………… ……………

tn = 2n ∴tn = 4n + 1 ∴tn = n2 ∴tn = n2 + n ∴tn = 3×2n

Example 2: Find the first four terms of a sequences whose nth terms are given by
(i) tn = 2n + 3 (ii) tn = (–1) n+1(n + 7)
Solution:
(i) The given general term (tn) = 2n + 3.

Putting n = 1, 2, 3, and 4 to get t1, t2, t3 and t4.

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Sequence and Series

The first term (t1) = 2 × 1 + 3 = 5
The second term (t2) = 2 × 2 + 3 = 7
The third term (t3) = 2 × 3 + 3 = 9
The fourth term (t4) = 2 × 4 + 3 = 11
Thus, the required the first four terms 5, 7, 9, 11

(ii) The given general term (tn) = (–1) n + 1 (n + 7)
The first term (t1) = (–1) 1 + 1 (1 + 7) = (–1)2 (8) =1 × 8 = 8
The second term (t2) = (–1) 2 + 1 (2 + 7) = (–1)3 (9) = –1 × 9 = –9
The third term (t3) = (–1) 3 + 1 (3 + 7) = (–1)4 (10) =1 × 10 = 10
The fourth term (t4) = (–1) 4 + 1 (4 + 7) = (–1)5 (11) = –1 × 11 = –11
Thus, the required the first four terms are 8, –9, 10 and –11.

The general term of linear sequence

The sequence in which each pair of successive terms have a constant difference is called

a linear sequence. The general term of linear sequence is given by tn = an + b; where
a = common difference and b = first term – common difference

Example 3: Find the nth term of the sequence 7, 10, 13, 16, …

Solution: 7 10 13 16
Here, the sequence 7, 10, 13, 16, … has the common difference 3. +3 +3 +3
So, it is a linear sequence.

Let, the nth term (tn) be tn = an + b Alternatively,
Since, the common difference = 3. So, a = 3 In 7, 10, 13, 16, …
t1 = 7 = 3×1 + 4
Now, the first term (t1) = 7 t2 = 10 = 3×2 + 4
t3 = 13 = 3×3 + 4
or, 3 × 1 + b = 7 [∵tn = an + b] ……………………
∴tn = 3n + 4
or, b = 4

Hence, tn = 3n + 4

The general term of quadratic sequence

In quadratic sequence the first difference between each pair of consecutive term is not
constant. However, the second differences are constant. The general term of quadratic
sequence is given by

tn = an2 + bn + c.
Example 4: Find the general term of the sequence 0, 5, 12, 21, 32, …

Solution: 0 5 12 21 32
1st diff→ 5 7 9 11
Let, tsheeconnthdtecromns(ttan)not fdtihffeerseenqcueen=ce2.be tn = an2 + bn + c
The 2nd diff→ 2 2 2
So, it is quadratic sequence.
Now, 2a = 2 ∴a = 1
Also, the first term (t1) = 0

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Sequence and Series

or, 1 × 12 + b × 1 + c = 0 [∵an2 + bn + c] Alternatively,
or, b = – c – 1 … (i) 0, 5, 12, 21, 32, ...
t1 = 0 = 12 - 1 = 12 + (2×1 – 3)
Again, the second term (t2) = 5 t2 = 5 = 22 + 1=22 + (2×2 – 3)
or, 1 × 22 + b × 2 + c = 5 [∵an2 + bn + c] t3= 12 = 32 + 3 = 32 + (2×3 – 3)
or, 2b + c = 1 … (ii) t4 = 21 = 42 + 5 = 42 + (2×4 – 3)
Putting the value of b from equation (i) in (ii), we get …………………………………
2 (–c – 1) + c = 1
or, –2c –2 + c = 1 ∴tn = n2 + 2n – 3
or, c = –3
Putting the value of c in equation (i), we get
b = 3 – 1 = 2
Hence, tn = n2 + 2n – 3

Facts to remember

1. The general term can be generated by using formula or making the patterns of
numbers.

2. (i) If the second common difference, 2a = 1 then tn = an2 +bn + c contains 1 n2.
2
(ii) If the second common difference, 2a = 2 then tn = an2 +bn + c contains n2.

(iii) If the second common difference, 2a = 3 then tn = an2 +bn + c contains 23n2
and so on.

Example 5: The nth term of a tseerqmuesnocfethisegsievqeunebnycea.n = an-1 + an-2, n>2. If a1 = a2 = 1,
find the first six

Solution:

Here, an = an-1 + an-2 and a1 = a2 = 1

On putting n = 3, a3 = a3-1 + a3-2 = a2 + a1 = 1 + 1 = 2
On putting n = 4, a4 = a4-1 + a4-2 = a3 + a2 = 2 + 1 = 3
On putting n = 5, a5 = a5-1 + a5-2 = a4 + a3 = 3 + 2 = 5
On putting n = 6, a6 = a6-1 + a6-2 = a5 + a4 = 5 + 3 = 8
Hence, the first six terms are 1, 1, 2, 3, 5 and 8.

Example 5: Study the pattern in the given figures and answer the questions.

(i) Draw one more figures in the
same pattern.

(ii) Find the nth term of the sequence
of number of dots.

(iii) Find the 7th term of the sequence. , , , ...

Solution:

(i) One more figures in the same patterns is given alongside

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Sequence and Series

(ii) The sequence of number of dots is 1, 5, 9, …

Constant difference (d) = 5 – 1 = 9 – 5 = 4 1, 5, 9, ...

Now, t1= 1 = 4 × 1 – 3 +4 +4
t2 = 5 = 4 × 2 – 3
t3= 9 = 4 × 3 – 3
……………………

∴tn = 4n – 3

(iii) The 7th term (t7) = 4 × 7 – 3 = 25

EXERCISE 8.1

General section

1. Define the following terms with an example.

a) Sequence b) Finite sequence c) Infinite sequence

2. What are the next two terms of each of the following sequences?

a) 4, 9, 14, 19, … b) 20, 17, 14, 11, … c) 1, 2, 4, 8, …

d) 96, 48, 24, 12, … e) 1, 4, 9, 16, … f) 14, 29, 136, 4 , ...
25
3. Find the first four terms of the sequences whose general terms are given below.

a) tn = 3n + 2 b) tn = 5n – 7 c) an = n2 + n d) an = (-1)n. (n+1)

e) tn = n(n + 1) f) tn = n – 1 g) tn = 2n – 1 h) tn = (–1)n.n 1
2 n + 1 (n + 1)2 n2 + n +

Creative section

4. Find the nth terms of the following sequences. Also, find the 6th and 9th terms of each
sequence.

(a) 4, 8, 12, 16, … (b) 6, 11, 16, 21, … (c) 9, 5, 1, –3, … (d) 1, 4, 9, 16, …

(e) 2, –4, 8, –16, …. (f) 1 , 23, 43, 54, ... (g) 1 , 49 , 196, 2156 (h) 3, 6, 11, 18, …
2 4
5. a) a1, a2, a3,… are the terms of a sequence. If a1 = 3, an = 2an–1, find the first five terms
of the sequence.

b) If t1 = 4, t2 = –3 and tn = tn–1 + tn–2, n≥ 3, n∈N, find the first six terms of a sequence.
{
c) The general term of a sequence is defined as an = 2n + 1 when n ∈ N is odd .
Find its 7th and 10th terms. n2 + 1 when n ∈ N is even

6. Study the following patterns in each of the figures and answer the following questions
for each case.

(i) Draw one more figures in the same pattern.

(ii) Find the general term of the sequence of number of dots.

(iii) Find the 10th term of the sequence.

a) , , , ... b)
,
, , ...

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Sequence and Series

c) , d) , , , ...
, , ...

8.3 Series

The sum of the terms of a sequence is called the series corresponding to the sequence.
For examples
a) The series corresponding to the sequence 2, 5, 8, 11, 14 is 2 + 5 + 8 + 11 + 14.
b) The series corresponding to the sequence 5, 3, 1, -1, -3 is 5 + 3 + 1 – 1 – 3

Facts to remember

(i) If a series has finite number of terms, then it is called a finite series.
For example, 2 + 4 + 6+ …+ 20 is a finite series.
(ii) If a series has infinite number of terms, then it is called an infinite series.
For example, 1 + 3 + 5+ … is an infinite series.

8.4 Sigma notation

If a1, a2, a3, …, an is a finite sequence, then the series corresponding to this sequence is

Sn = a1 + a2 + a3 + … + an

This sum can represented by using sigma notation as n 1an , where the Greek letter ‘Σ’ is
n
n =

the summation notation and n = 1an read as ‘summation of an , when n runs from 1 to n’.

8.5 Partial sum
n

The sum of the finite series a1, a2, a3, …, an is Sn = a1 + a2 + a3 + … + an = n = 1an
The sum of limited terms of the series is called the partial sum.

Here, S2 = a1 + a2 = 2
S1 = a1, n = 1an,
3
S3 = a1 + a2 + a3 = n = 1an,

4n
S4 = a1 + a2 + a3+a4 = n = 1an and so on. are the partial sum of n = 1an.

Let’s take 3rd and 4th partial sums of Sn.

Then, S4 = a1 + a2 + a3+a4 and S3 = a1 + a2 + a3

Now, S4 – S3 = (a1 + a2 + a3+a4) – (a1 + a2 + a3) = a4.

It shows that the nth term of the series is calculated by using the following relation.

an = Sn – Sn – 1, for n>1

Facts to remember

q

1. In partial sum n = pan , the number of terms = (q – p) + 1
2. The general term (an) = Sn – Sn-1, for n>1.

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Worked-out Examples

7

Example 1: Find the value of (3n + 2) by expanding.
Solution: n=4

7
Here, (3n + 2) = (3 × 4+2) + (3 × 5+2) + (3 × 6+2) + (3 × 7+2)
n=4
= 14 + 17 + 20 + 23 = 74

7

Hence, the value of (3n + 2) is 74.
n=4

Example 2: Find the nth term and then express the following series in sigma notation.

(i) 7 + 11 + 15 + 19 +23 (ii) –1+2–4+16 –32+…to 9 terms
1 3 5 7
(iii) 2+5+10+17+…to 10 terms (iv) 4 + 9 + 16 + 25 + ... to 15 terms

Solution:

(i) Here, the constant difference = 11– 7 = 4

The 1st term (t1) = 7 = 4 × 1 + 3
The 2nd term (t2) = 11 = 4 × 2 + 3
The 3rd term (t3) = 15 = 4 × 3 + 3
The 4th term (t4) = 19 = 4 × 4 + 3
…………………………………………..

∴The nth term (tn) = 4n + 3
5

Hence, 7 + 11 + 15 + 19 + 23 = (4n +3)
n=1

(ii) Here, the terms are powers of 2 having – and + sign in alternate places.

The 1st term (t1) = –1 = (–1)1.20 = (–1)1.21–1
The 2nd term (t2) = 2 = (–1)2.21 = (–1)1.22–1
The 3rd term (t3) = –4 = (–1)3.22 = (–1)1.23–1
The 4th term (t4) = 8 = (–1)4.23 = (–1)1.24–1
………………………………………………………

∴The nth term (tn) = (–1)n.2n–1
9

Hence, –1 + 2 – 4 + 8 – 16 +… to 9 terms = (–1)n.2n – 1
n=1

(iii) Here, the first differences are, 5 – 2 = 3, 10 – 5 = 5, 17 – 10 = 7, ….

The second differences are, 5 – 3 = 2, 7 – 5 = 2. Since, the second differences are
constant. So, 2 + 5 + 10 + 17 + … is a quadratic series.

The 1st term (t1) = 2 = 1+1 = 12 + 1 2 5 10 17
The 2nd term (t2) = 5 = 4 +1= 22 + 1 357
The 3rd term (t3) = 10 = 9 +1= 32 + 1 22
The 4th term (t4) = 17 = 16 + 1=42 + 1
………………………………………………………

∴The nth term (tn) = n2 + 1
10

Hence, 2 + 5 + 10 + 17 + … to 10 terms = (n2 + 1)
n=1

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(iv) Here, the sequence of numerators is 1, 3, 5, 7, … which has the constant difference
of 2 and the sequence of denominators is 4, 9, 16, 25, … which are perfect square
numbers.
1 2 ×1 – 1 2×1–1
The 1st term (t1) = 4 = 22 = (1 + 1)2

The 2nd term (t2) = 3 = 2 ×2 – 1 = 2×2–1
9 32 (2 + 1)2
5 2 ×3 – 1 2×3–1
The 3rd term (t3) = 16 = 42 = (3 + 1)2

The 4th term (t4) = 7 = 2 ×4 – 1 = 2×4–1
25 52 (4 + 1)2

………………………………………………………

∴The nth term (tn) = 2n – 1
(n + 1)2

Hence, 1 + 3 + 5 + 7 to 15th terms = 15 2n – 1
4 9 16 25 n =1 (n + 1)2

EXERCISE 8.2
General section

1. Define series with an example. d) 8 + 4+0 – 4 – 8 …
2. Identify the following as sequence or series.
a) 2, 4, 8, 10, … b) 1+ 3+ 5+ 7+ … c) 3, –1, –5, –9, ..

3. Write down the series associated with the following sequences.

a) 5, 10, 15, 20, 25 b) 1, –2, 4, –8, 16 c) –7, –3, 1, 5, … d) 1, 1 , 14, 1 , ...
4. Find the number of terms of the following partial sums. 2 8

5 68 10
a) (2n + 1) b) 5n c) (–1)k.k
n=1 n=3 n=4 d) (i2 – i)
5. Evaluate the following sums by expanding. i=3

5 7 4 5

a) (3n + 2) b) (5n – 1) c) (n2 + 3) d) 2k
n=2 n=3 n=1 k=1
5 8 5

e) (3n – n3) f) (–1)i.i g) [1 + (–1)n]
n=1 i=4 n=1

Creative section

5

6. a) If (2n + k) = 45, find the value of k.
n=1
5

b) If (2n + m) = 48, find the value of m.
n=2

7. tn and Sn are the nth term and sum of the first n terms of a series respectively.
a) If Sn n(n + 1)
= 2 , find S3, S4 and t4.

b) If Sn = n(n + 1) (n + 2) , find S4, S5 and t5.
6

8. Find the general term of each of the following series and express them in sigma notation.

a) 2 + 5 + 8 + 11 + 14 + 17 b) –10 – 6 – 2 + 2 + 6 + 10

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c) 3 + 9 +27 + 81 + 243 d) 1 + 4 + 16 + 64 + …. to 9 terms

e) 1 – 2 + 3 – 4 + 5 - ….. to 10 terms f) –2 + 4 – 8 + 16 – 32 + … to 12 terms

g) 12 + 2 + 3 + 4 + ... to 7 terms h) 1 + 4 + 9 + 16 + ... to 9 terms
3 4 5 4 9 16 25
i) 23 + 3 4 5 5 10 15 20
5 + 7 + 9 + ... to 8 terms j) 4 – 9 + 16 – 25 + ... to 8 terms

k) 2 + 6 + 12 + 20 + 30 + 42 + 56 l) 6 + 11 + 18 + 27 + … to 10 terms

8.6 Arithmetic sequence

Let’s make the following figures using match sticks

,, ,

(i) What is the sequence of numbers of match sticks required for each figure? 3, 5, 7, 9, …

(ii) Can we find the difference between the successive terms? 5 – 3 = 2, 7 – 5 = 2,
9 – 7 = 2 Here, the difference between successive terms is always 2.

This illustration leads us to get idea of an arithmetic sequence.

Facts to remember

1. A sequence in which the difference between any two consecutive terms remains
constant is known as arithmetic sequence (A.S.) or arithmetic progression (A.P.)

2. The constant difference between the consecutive terms of an A.P. is called common
difference.

3. The common difference of an A.P. can be positive, negative or zero.
4. Common difference (d) = t2 – t1 or t3 – t2 or t4 – t3

8.7 Terms and Common Difference of an A.P.

Let, ‘a’ be the first term of an A.P. and ‘d’ be the common difference.
Then, the terms of the A.P. can be written as follows.
t1­= a = a + (1 – 1) d
t2 = a + d = a + (2 – 1) d +d +d +d

t3 = a + 2d = a + (3 – 1) d t1 = a t2 = a + d t3 = a + 2d t4 = a + 3d
…t4­=……a …+…3d…=…a…+…(4…–…1…) d

tn = a + (n – 1) d
Thus, the general term (nth) term, (tn) = a + (n – 1) d

Worked-out Examples

Example 1: Check whether the following sequences are in A.P. or not.
(i) 4, 7, 10, 13, 16, … (ii) 1, 5, 25, 125, …
Solution:
Here, to check whether the given sequence is an A.P. or not, it is enough to check the
differences between the consecutive terms.

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(i) to2n–. t1 = 7 – 4 = 3, t3 – t2 = 10 – 7 = 3, t4 – t3 = 13 –10 = 3, t5 – t4 = 16 – 13 = 3 and so
Since, the difference between any two consecutive terms is equal, the given sequence

is an A.P.
(ii) t2 – t1 = 5 – 1 = 4, t3 – t2 = 25 – 5 = 20, t4 – t3 = 125 –25 = 100 and so on.

Since, the difference between any two consecutive terms is not equal, the sequence is
not an A.P.

Example 2: Find the general term (tn) of the arithmetic sequence -2, 1, 4, 7, …
Solution:
Here, the first term (a) = -2, common difference (d) =t2 – t1 = 1 – (-2) = 3, nth term (tn) =?
Now, tn = a + (n – 1) d

= -2 + (n – 1) × 3
= -2 + 3n – 3
= 3n – 5
Hence, the general term of the sequence (tn) = 3n – 5.

Example 3: Find the 10th term of an arithmetic sequence 7, 11, 15, …
Solution:
Here, the first term (a) = 7, common difference (d) = t2 – t1 = 11 – 7 = 4, 10th term (t10) =?
Now, tn = a + (n – 1) d

∴t10 = 7 + (10 – 1) × 4
= 7 + 36
= 43

Hence, the 10th term of the sequence is 43.

Example 4: How many terms are there in the arithmetic series 3+15+27+…+123?

Solution:

Here, the first term (a) = 3, common difference (d) = t2 – t1 = 15 – 3 = 12, nth term (tn) =123,

Number of terms (n) =? Checking

Now, tn = a + (n – 1) d Last term = 11th term

or, 123 = 3 + (n – 1) × 12 = a + 10d

or, 123 = 3 + 12n – 12 = 3 + 10 × 12

or, 132 = 12n = 123

or, n = 11 Which is given in the question.

Hence, the required number of terms is 11.

Example 5: Is 88 a term of the series 11+ 18+ 25 + ……?

Solution:

Here, the first term (a) = 11, common difference (d) = t2 – t1 = 18 – 11 = 7
Let, nth term (tn) =88

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Now, tn = a + (n – 1) d Sequence and Series
or, 88 = 11 + (n – 1) × 7
or, 88 = 11 + 7n – 7 Checking
Last term = 12th term
or, 84 = 7n = a + 11d
= 11 + 11 × 7
or, n = 12 = 88
Which is given in the question.
Yes, 88 is the 12th term of the series.

Example 6: The third and the thirteenth terms of an arithmetic series are -40 and 0
respectively. Find the 28th term.

Solution:

Let ‘a’ and ‘d’ be the first term and the common difference of an AP respectively.

Then, 3rd term (t3) = –40 ∴a + 2d = –40 … (i)

Also, 13th term (t13) = 0 ∴a +12d = 0 … (ii)

Now, subtracting (i) from (ii), we get Checking

a + 12d – (a + 2d) = 0 – (–40) Third term = a + 2d = –48 + 2 × 4 = –40

or, a + 12d – a – 2d = 40 Thirteenth term = a + 12d = –48 + 12 × 4 = 0

or, 10d = 40 Which are given in the question.

∴ d=4

Also, putting the value of ‘d’ in (i), we get

a + 2×4 = –40 ∴ a = –40 – 8 = –48

Again, 28th term (t28) = a + 27d = –48 + 27× 4 = –48 +108 = 60

Example 7: If 6 times the 6th term of an A.P. is equal to 9 times its 9th term, find the 15th
term.

Solution:
Let ‘a’ and ‘d’ be the first term and the common difference of an AP respectively.

Then, 6th term (t6) = a + 5d and 9th term (t9) = a + 8d
According to question, 6t6 = 9t9
or, 6 (a + 5d) = 9 (a + 8d)

or, 6a + 30d = 9a + 72d
or, –3a = 42d

or, a = –14d … (i)
Again, 15th term (t15) = a + 14d = –14d + 14d = 0 [Using (i)]

Example 8: A taxi-meter shows Rs 14 at the time of starting and then runs up by Rs 36 for
each additional kilometer travelled during 6:00 am to 9:00 pm. If Mr. Gurung
travelled a journey of 7 km, how much fare did he pay for the journey?

Solution:

Here, the sequence of taxi fare for the journey of 1 km, 2 km, 3 km, … is Rs (14 + 36),
Rs (14 + 2 × 36), Rs (14 + 3 × 36), … i.e., Rs 50, Rs 86, Rs 122, …

The first term (a) = Rs 50, common difference (d) = Rs 36, number of terms (n) = 7

Now, tn = a + (n – 1) d
∴t8 = 50 + (7 – 1) × 36 = 266

Hence, he paid Rs 266 for travelling a journey of 7 km.

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Example 9: Every year, a tree grows 10 cm less than it does in the preceding year. If it
grows by 2 m in the first year, in how many years will it have stop growing?

Solution:

Here, the sequence of heights of tree in 1st year, 2nd year, 3rd year, … is 200 cm, (200 – 10)
cm , (200 – 2 × 10) cm, … i.e., 200 cm, 190 cm, 180 cm, …..

The first term (a) = 200 cm, common difference (d) = 190 – 200 = –10,

Last term (=tn)a=+0(,nn–u1m)bder of terms (n) =?
Noorro,, w , t00n = 200 + (n – 1) × (–10)
= 200 – 10n + 10

or, 10nn0 = 210 – 10n
or, = 210
or, = 21

Hence, the tree would stop growing after 21 years.

EXERCISE 8.3

General section

1. Define: (a) arithmetic sequence. (b) common difference

2. (a) Write down the formula for finding (tnh)eonftahntearrmith(mtn)etwichseenqufiersntcteeramre (a), common
difference (d) and number of terms given.

b) What is the 6th term of an A.P. having the first term ‘a’ and the common difference
‘d’?

3. Examine whether the following sequences form an A.P. or not.

a) 2, 5, 8, 11, … b) 7, 3, –1, –5, … c) 2, 4, 8, 16, … d) 0.2, 0.22, 0.222, ….

4. The first term (a) and common difference (d) of the arithmetic sequences are given
below. Find their general terms.

a) a = 3, d = 2 b) a = –5, d = 4 c) a = 10, d = –7 d) a = –2, d = –3

5. a) Find the 7th term of the sequence 3, 6, 9, 12, …

b) What is the 10th term of the sequence 7, 13, 19, 25, …?

c) Find the 15th term of an arithmetic sequence 125, 120, 115, 110, ...

6. a) If the fifth term of an arithmetic sequence with first term 2 is 14, find the common
difference.

b) The first and 8th terms of an A.P. are -5 and 23, find the common difference.

c) The common difference of an arithmetic series is 5 and its tenth term is 40, find its
first term.

d) The first term of an AP is 14, common difference is 7 and last term is 98, find the
number of terms.

7. a) How many terms are there in the sequence 7, 12, 17, …., 82?

b) How many terms are there in the sequence 3, 6, 9, ..., 111?

c) Find the number of terms of the series 25 + 50 + 75 + 1000.

8. a) Which term of the series 2 + 5 + 8 + .. is 56?

b) Which term of the sequence 6, 9, 12, 15… is 66?

9. a) Is 49 a term of the sequence 1, 9, 17, 25, …?

b) Is 111 a term of the A.P. 3, 6, 9, 12, ….?

c) Does 36 belongs to the sequence 8, 11, 14, 17, …?

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Creative section - A

10. a) If the fifth and the tenth terms of an arithmetic sequence are 14 and 29 respectively.
Find the first term and the common difference. Also, find the 17th term.

b) If the third and the ninth terms of arithmetic sequence are 15 and 39 respectively.
Find the 15th term.

c) If the 5th and the 12th terms of an arithmetic series are -16 and 5 respectively, find
its 20th term.

11. a) If the fifth and the tenth terms of an arithmetic sequence are 23 and 48 respectively,
find the series.

b) The fourth and the sixth terms of an arithmetic sequence are 41 and 35 respectively,
find the series.

12. a) If the sixth and the ninth terms of an arithmetic sequence are 52 and 76 respectively,
which term is 100? Find it.

b) If the 5th and the 10th terms of an arithmetic sequence are 30 and 55 respectively,
which term is 80? Find it.

13. a) If 6 times the 6th term of an arithmetic sequence is equal to 9 times its 9th term,
show that its 15th term is zero.

b) If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, find the 18th term.

Creative section - B

14. a) The starting salary of a lady working as receptionist in an office is Rs 15,000 per
month. If she gets an increment of Rs. 2,000 every year, what will be her salary in
the 5th year?

b) A taxi-meter shows Rs 14 at the time of stating and then runs up by Rs 36 for each
additional kilometer travelled during 6:00 am to 9:00 pm. If Mrs. Rai travelled a
journey of 5 km by the taxi, how much fare did she pay for journey?

c) A cinema hall has managed 20 rows of seats and there are 12 seats in the front row.
If each successive row contains two additional seats than its front row, how many
seats are there in the last row?

15. a) Suppose that you are attending an interview for a job and the company gives you
the following two offers for 5 years.

Offer A: Rs 20,000 salary to start with followed by an annual increment of
Rs 2,500.

Offer B: Rs 25,000 salary to start with followed by an annual increment of Rs 1,500.

Which offer would you choose and why?

b) A dealer sold 50 computers in B.S. 2077, 70 computers in B.S. 2078, 90 computers
in B.S. 2079 and so on forming an arithmetic sequence.

(i) How many computers will it sell in B.S. 2082?

(ii) In which year, will the dealer sell 250 computers?

16. a) Four people can sit in each table in a restaurant. When two tables are placed
together, 6 people can be seated. When three tables are placed together to form
a long table, 8 people can be seated. How many such tables should be placed
together to form a long table such that 16 people could have the dinner together?

b) On the first day strike of physicians in a hospital, the attendance of the OPD was
450 patients. As the strike continued, the attendance declined by 50 patients every
day, find from which day of strike, the OPD would have no patient?

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c) Every year, a tree grows 5 cm less than it did in the preceding year. If it grows by
1 m in the first year, in how many years would it stop growing?

d) A construction company will be penalized each day of delay in construction for
bridge. The penalty will be Rs 25,000 for the first day and increases by Rs 5,000
for each following day. Based on this budget, the company can afford to pay a
maximum of Rs 1,00,000 toward penalty. Find the maximum number of days by
which the completion of work can be delayed.

Project work and activity section
17. a) Think the values of the first term (a) and common difference (d) of your choice for

the arithmetic sequence. Then, find the general term and the first five terms of the
sequence.
b) Measure the heights of the first four steps of the stair/ladder from its base which is
available in your house/school. Find the general term formula for the heights of the
steps from the base and the height of the last step.

8.8 Geometric sequence

Let’s consider that a viral disease is spreading in a way
such that at any stage two new persons get affected from
an infected person. At first stage, one person is infected, at
second stage two persons are infected and is spreading to
four persons and so on. Then, number of persons infected
at each stage are 1, 2, 4, 8, ... where except the first term,
each term is twice the previous term. This idea helps us to
get the concept of geometric sequence.

Facts to remember
1. A sequence in which each term, except first term, is obtained on multiplying the

preceding term by a fixed non-zero number is called a geometric sequence. The fixed
number is called common ratio and denoted by r.
2. A sequence in which the ratios between any two consecutive terms are constant, is
known as geometric sequence (G.S.) or geometric progression (G.P.)
3. Common ratio (r) = tt12 or tt32 or tt43 and so on.

8.9 Terms and Common Ratio of a G.P.

Let, ‘a’ be the first term of a G.P. and ‘r’ be the common ratio.

Then, the terms of the G.P. can be written as follows.

t1­= a = ar0 = ar1-1 ×r ×r ×r
t2 = ar= ar2-1 t3 = ar2
t1 = a t2 = ar t4 = ar3
t3 = ar2= ar3-1
t4­= ar3= ar4-1
………………

tn = arn-1

Thus, the general term or (nth) term, (tn) = arn-1, r ≠ 0

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Sequence and Series

Worked-out Examples

Example 1: Examine whether the following sequences are in G.P. or not.
(i) 2, 6, 18, 54, … (ii) 3, 33, 333, 3333,… (iii) 8, 12, 18, 27, …
Solution:
Here, to check that the given sequence is in G.P., it is sufficient to check if the ratios
between the consecutive terms are equal or not.

(i) tt21 = 6 = 3, tt23 = 18 = 3, tt43 = 54 = 3, …
2 6 18

Since, the ratios between the two consecutive terms are equal, the given sequence is a
G.P.

(ii) tt12 = 33 = 11, tt32 = 333 = 11111, tt34 = 3333 = 1111 , …
3 33 333 111

Since, the ratios between the two consecutive terms are not equal, the given sequence
is not a G.P.

(iii) tt21 = 12 = 3 , tt32 = 18 = 3 , tt34 = 27 = 3 , …
8 2 12 2 18 2

Since, the ratios between the two consecutive terms are equal, the given sequence is a
G.P.

Example 2: Find the 7th term (t7) of the geometric sequence 120, 60, 30, 15,…
Solution:
tt12 60 1
Here, the first term (a) = 120, common ratio (r) = = 120 = 2 , number of terms (n) = 7,
7th term (t7) =?

Now, tn = arn-1

∴ t7 = 120 × ( 1 )7 – 1
2

= 120 × ( 1 )6
2

= 120 × 1 = 15
64 8

Hence, the 7th term is 185.

Example 3: The fourth term of a GP whose first term is 2 is 54, find the common ratio.

Solution:

Here, the first term (a) = 2, 4th term (t4) = 54, common ratio (r) =?,
Now, tn = arn-1
or, t4 = 2 × r4-1
or, 54 = 2 × r3

or, 27 = r3

or, r3 = 33

or, r = 3

Hence, the common ratio is 3.

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Sequence and Series

Example 4: How many terms are there in the geometric series 16 + 24 + 36 + …+ 81?

Solution: 24 23,
16
Here, the first term (a) = 16, common ratio (r) = = nth term (tn) =81,
number of terms (n) =?

Now, tn = arn-1 ( 3 )n – 1 Checking
or, 81 = 16 × 2
Last term = 5th term
or, 81 = ( 3 )n – 1
16 2 = ar4
3 3 = 16 × (23)4
or ( 2 )4 = ( 2 )n – 1 = 81

or, n – 1 = 4 Which is given in the question.

or, n = 5

Hence, there are 5 terms in the given series.

Example 5: Which term of GP 1 + 1 + 1 +….. is 1128?
Solution: 2 4

1 1
2
Here, the first term (a) = 1, common ratio (r) = 2 =
1
1
Let, nth term (tn) = 128 Checking

Now, tn = arn-1 Last term = 8th term

or, 1 = 1× ( 1 )n – 1 = ar7
128 2
1 1 = 1× ( 1 )7
or, ( 2 )7 = ( 2 )n – 1 = 2
1
or, n – 1 = 7 128
Which is given in the question.
or, n = 8
1
Hence, 128 is the 8th term of the series.

Example 6: The fifth and the eighth terms of a G.P. are 8 and 64 respectively. Find the
10th term.

Solution:
Let ‘a’ and ‘r’ be the first term and the common ratio of a G.P. respectively.

Then, 5th term (t5) = 8 ∴ar4 = 8 … (i)

Also, 8th term (t8) = 64 ∴ ar7 = 64 … (ii)

Now, dividing equation (ii) by equation (i), we get
ar7 64
ar4 = 8 Checking

or, r3 = 8 Fifth term = ar4 = 1 × (2)4 = 8
2
or, r = 2 1
Also, putting the value of ‘r’ in (i), we get Eighth term = ar7 = 2 × (2)7 = 64

a × 24 = 8 Which are given in the question.

or, 16a = 8
8
∴ a = 16 = 1
2

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Sequence and Series

Again, 10th term (t10) = ar9 = 1 × 29 = 1 × 512 = 256
2 2

Hence, the 10th term of the G.P. is 256.

Example 7: A man joined a company as the post of a manager. The company gave him a
starting salary of Rs 80,000 and agreed to increase his salary 10% annually.
What will be his salary after 5 years?

Solution:
Here, the sequence of monthly salary in the 1st, 2nd, 3rd years is Rs 80,000, Rs (110% of
Rs 80,000), Rs (110% of 110% of Rs 80,000), …. i.e., Rs 80,000, Rs 88,000, Rs 96,800,…

The first term (a) = Rs 80,000, common ratio (r) = 88000 = 1110, number of terms (n) = 5
80000

Now, tn = arn-1 × (1110)5 – 1
∴t7 = 80000

= 80000 × (1101)4

= 80000 × 14641 = 1,17,128
10000

Hence, his salary will be Rs 1,17,128 after 5 years.

EXERCISE 8.4 (b) common ratio
General section

1. Define: (a) geometric sequence.

2. a) Write down tnhuemfboremr ouflatefromr sfi(nnd) ionfgatgheeonmtehtrtiecrmseq(tune)nwceheanrefigrisvtetne.rm (a), common
ratio (r) and
b) A G.P. has the first term x and common ratio y, what will be its 4th term.

3. Examine whether the following sequences form a G.P. or not.

a) 2, 4, 8, 16, … b) 1 , 3, 9, 27, … c) 5, –20, 80, –320, … d) 6, 12, 18, 24, …
3
4. Find the first three terms of the G.P. whose first term (a) and common ratio (r) are given
below.
a) a = 2, r = 3 b) a = –9, r =3 c) a = 1 , r = –2 d) a =27 , r = 1
8 3
5. a) Find the 8th term of the sequence 3, 6, 12, 24, …

b) What is the value t7 in the sequence 729, 243, 81,…?
c) The first and the second terms of a geometric sequence are 9 and 18 respectively.
What is the fifth term?

6. a) In a G.P., the common ratio is 2 and the 8th term is 256, find its first term.
b) If the sixth term of a geometric sequence with common ratio 3 is 972, find the first

term.
c) If the fourth term of a GP with first term 2 is 54, find the common ratio.
d) A geometric sequence has the first term 8. If its fourth term is 1, what is its common

ratio?
7. a) How many terms are there in the geometric sequence 3, 6, 12, … , 192?
b) How many terms are there in the geometric series 16 + 24 + 36 + …+ 81?

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Sequence and Series

8. a) Which term of GP 2, 4, 8, 18, .. is 256? 1
81
b) Which term of the series 27 + 9 + 3 + ….is ?

9. a) Is 486 a term of a geometric sequence 2, 6, 18, 54, …?

b) Is 1 a term of a geometric progression 1 + 1 + 1 + ...?
128 2 4

Creative section - A

10. a) The 3rd and the 8th terms of a geometric sequence are 12 and 384 respectively, find
the 10th term.

b) In a geometric sequence, the 2nd term is 6 and the 5th term is 162, what is the 7th
term?

c) If the 6th and the 9th terms of a G.P. are 1 and 8 respectively, find its 10th term.
11. a) If the third and the sixth terms of a geometric series are 45 and 1215 respectively,

find the series.
b) If the fifth term and the ninth term of a G.P. are 144 and 2304 respectively, find the

geometric series.

Creative section - B

12. a) A person has two parents (father and mother), 4 grandparents and so on. Find the
number of his ancestors of 6th generations preceding his own.

b) A bookshelf has 5 shelves. The top shelf has 4 books, and each shelf below
has 3 times as many books as the shelf directly above it. If all of the books are
different in the bookshelf and you want to read a book from the bottom shelf, how
many choices would you have?

c) The number of bacteria in a certain culture doubles every hour. There were 20
bacteria present in the culture originally.

(i) How many bacteria will be present at the end of 2nd hour?

(ii) How many bacteria will be present at the end of of 4th hour?

13. a) The man joined a company as an assistant manager. The company gave him a
starting salary of Rs 60,000 and agreed to increase his salary 10% annually. What
will be his salary after 5 years?

b) The present value of a scanner machine is Rs 50,000 and its value depreciates each
year by 20%. What will be the value of the machine after 5 year?

c) The population of a town grows by 15% per year. If the population of the town was
2,00,000 at the end of 2021 A.D., find the projected population of the town at the
end of 2031 A.D.

d) Shaswat purchased a motorcycle for Rs 4,00,000 this year. The value of a motor
cycle depreciates at the rate of 15% per year. What will be the value of the motor
cycle 4 year hence?

14. If the numbers a, b and c are in an A.P., show that 5a, 5b and 5c form a G.P.

Project work and activity section

15. a) Think the values of the first term (a) and common ratio (r) of your choice for the
geometric sequence. Then, find the general term and the first five terms of the
sequence.

Vedanta Excel in Mathematics - Book 9 146 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur