Unit

The Pyramids at Giza in Egypt are among the

best known pieces of architecture in the world.
The Pyramid of Khafre was built as the final
resting place of the Pharaoh Khafre and is about

MEASUREMENT 136 m high.

Unit Outcomes:
After completing this unit, you should be able to:
solve problems involving surface area and volume of solid figures.
know basic facts about frustums of cones and pyramids.

Main Contents
7.1 Revision on Surface Areas and Volumes of Prisms and Cylinders
7.2 Pyramids, Cones and Spheres
7.3 Frustums of Pyramids and Cones
7.4 Surface Areas and Volumes of Composite Solids
Key Terms
Summary
Review Exercises
Mathematics Grade 10

INTRODUCTION
Recall that geometrical figures that have three dimensions (length, width and height) are
called solid figures. For example, cubes, prisms, cylinders, cones and pyramids are
three dimensional solid figures. In your lower grades, you have learnt how to find the
surface areas and volumes of solid figures like cylinders and prisms. In this unit, you
will learn more about surface areas and volumes of other solid figures. You will also
study about surface areas and volumes of composed solids and frustums of pyramids
and cones.

OPENING PROBLEM
Ato Nigatu decided to build a garage and began by calculating the number of bricks
required. The floor of the garage is rectangular with lengths 6 m and 4 m. The height of
the building is 4 m. Each brick used to construct the building measures 22 cm by 10 cm
by 7 cm.
a How many bricks might be needed to construct the garage?
b Find the area of each side of the building.
c What more information do you need to find the exact number of bricks
required?

7.1 REVISION ON SURFACE AREAS AND

VOLUMES OF PRISMS AND CYLINDERS
There are many things around us which are either prismatic or cylindrical in shape. In
this sub-unit, you will closely look at the geometric solids called prisms and cylinders
and their surface areas and volumes.

Let E1 and E2 be two parallel planes, ℓ a line intersecting both planes, and R be a region

in E1. For each point P of R, let P ' be the point in E2 such that PP ' is parallel to ℓ.

The union of all points P ' is a region ℓ ℓ

R ' in E2 corresponding to the region R' P'P
P'′ E
EE222
EEE22E2
2
R in E1. The union of all the
segments PP ' is called a solid
P R E1
region D. This solid region is R E1

known as a cylinder. See Figure 7.2

Figure 7.1
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 Unit 7 Measurement

ℓ
Some important terms
For the cylinder D, the region R is called its lower base R' E2
or simply base and R' is its upper base.
The line ℓ is called its directrix and the perpendicular D

distance between E1 and E2 is the altitude of D. If ℓ is

R E1
perpendicular to E1, then D is called a right cylinder,
otherwise it is an oblique cylinder. If R is a circular
region, then D is called a circular cylinder. Figure 7.2
Directrix ℓ
Upper base

ℓ ℓ

Lower base

a b c
Oblique cylinder Right cylinder Right circular cylinder
Figure 7.3

Let C be the bounding curve of the base region R.

P'
The union of all the elements PP ' for which P Lateral
belongs to C is called the lateral surface of the surface
cylinder. The total surface is the union of the Total surface

lateral surface and the bases of the cylinder.

P C
Figure 7.4
There are other familiar solid figures that are special cylinders. Look again at the solid
figure D described above in Figure 7.2.

Definition 1.1
If R is a polygonal region, then D is called a prism.
If R is a parallelogram region, then D is a parallelepiped.
If R is a triangular region, then D is a triangular prism.
If R is a square region, then D is a square prism.
A cube is a square right prism whose altitude is equal to the length of the edge
of the base.

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Mathematics Grade 10

Note: Edges of upper base

In the prism shown in Figure 7.5, B

A C
1 AB , BC , CD , DE , EA are edges of
the upper base. E D
A ' B ', B ' C ', C ' D ', D ' E ', E ' A ' are
edges of the lower base. Lateral edge Lateral
2 AA ', BB ', CC ', DD ', EE ' are called B' face
lateral edges of the prism. A' C'
3 The parallelogram regions ABB'A', E' D'
BCC ' B ', AEE ' A ', DCC ' D ', EDD ' E
Edge of the
Edges base
lower base
are called lateral faces of the prism.
Figure 7.5
4 The union of the lateral faces of a
prism is called its lateral surface.
5 The union of its lateral faces and its two bases is called its total surface or simply
its surface.

ACTIVITY 7.1
1 How many edges does the base of the prism shown in Figure 7.5
have? Name them.
2 Identify each of the solids in Figure 7.6, as prism, cylinder, triangular prism, right
prism, parallelepiped, rectangular parallelepiped and cube.

a b c d

e f g
Figure 7.6
3 Are the lateral edges of a prism equal and parallel?
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 Unit 7 Measurement

4 Using Figure 7.7, complete the following blank

B
spaces to make true statements: C
A P'
a The figure is called a ______.
D
b The region ABCD is called a __________.
c AE and CG are called ______.
d The region AEHD is called a ______ .
e ______ is the altitude of the prism. F
G
f If ABCD were a parallelogram, the prism E P
would be called a ______. H
Figure 7.7
g If AE were perpendicular to the plane of
the quadrilateral EFGH, then the prism would be called ______.
5 Consider a rectangular prism with dimensions of its bases l and w and height h.
Determine:
a the base area b lateral surface area c total surface area
If we denote the lateral surface area of a prism by AL, the area of the base by AB, altitude
h and the total surface area by AT, then
AL = Ph; where P is the perimeter of the base and h is the height of the prism.
AT = 2AB + AL
Example 1 Find the lateral surface area of each of the following right prisms.

5
6 cm 5
4
5 cm 3 cm
10
3 cm 5 cm
a b
Figure 7.8
Solution:
a AL = Ph = (3 + 5 + 3 + 5) cm × 6 cm = 16 cm × 6 cm = 96 cm2
b AL = Ph = (5 + 5 + 4) × 10 = 14 × 10 = 140 units2
Similarly, the lateral surface area (AL) of a right circular cylinder is equal to the product
of the circumference of the base and altitude (h) of the cylinder. That is,
AL = 2π rh, where r is the radius of the base of the cylinder.

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Mathematics Grade 10

The total surface area AT is equal to the sum of the

areas of the bases and the lateral surface area. That is,
AT = AL + 2AB h

AT = 2π r h + 2π r2 = 2π r (h + r)
r
Figure 7.9
Example 2 The total surface area of a circular cylinder is 12π cm2 and the altitude is
1cm. Find the radius of the base.
Solution: AT = 2π r (h + r) ⇒ 12π = 2π r (1 + r) ⇒ 6 = r + r2
r2 + r – 6 = 0 ⇒ (r + 3) (r – 2) = 0 ⇒ r + 3 = 0 or r – 2 = 0
⇒ r = –3 or r = 2.
Therefore, the radius of the base is 2 cm. (why?)
The measurement of space completely enclosed by the bounding surface of a solid is
called its volume.
The volume (V) of any prism equals the product of its base
area (AB) and altitude (h). That is,
V = ABh h

AB
Figure 7.10
Example 3 Find the total surface area and volume of the following prism.
6

7
1 6
14
Figure 7.11
6 8
Solution: Taking the base of the prism to be, as shown 4
shaded in the following figure, we have: 7 6
10
1
1  14
AB = (7 × 14) –  × 8 × 6  Figure 7.12
2 
= 98 – 24 = 74 units2
AL = Ph = (7 + 6 + 10 + 14 + 1) × 6
= 38 × 6 = 228 units2
AT = AL + 2AB = 228 + 2 × 74 = 376 units2
V = AB h = 74 × 6 = 444 units3

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 Unit 7 Measurement

Volume of a right circular cylinder r

The volume (V) of a circular cylinder is equal to the product of
the base area (AB) and its altitude (h). That is, h
V = ABh
V = πr2 h, where r is the radius of the base.
Figure 7.13
Example 4 Find the volume of the cylinder whose base circumference is 12π cm
and whose lateral area is 288 π cm2.
Solution: C = 2π r ⇒ 12π = 2π r ⇒ r = 6 cm
AL = 2π r h
288π cm2 = 2π × 6 cm × h ⇒ 288π cm2 = 12π cm × h ⇒ h = 24 cm
Therefore, V = π r2 h = π (6 cm)2 × 24 cm = 36π cm2 × 24 cm = 864π cm3

Exercise 7.1
1 The altitude of a rectangular prism is 4 units and the width and length of its base
are 3 units and 2 units respectively. Find:
a the lateral surface area b the total surface area c the volume
2 The altitude of the right pentagonal prism shown in Figure 7.14 is 5 units and the
lengths of the edges of its base are 3, 4, 5, 6 and 4 units. Find the lateral surface
area of the prism.

5
4 6
3 4 5
Figure 7.14
3 A lateral edge of a right prism is 6 cm and the perimeter of its base is 20 cm. Find
the area of its lateral surface.
4 Find the lateral surface area of each of the solid figures given in Figure 7.15.

2 cm

5 cm
10 cm 4 cm
5 cm 10 cm
7 cm
a b
Figure 7.15

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Mathematics Grade 10

5 Find the perimeter of the base of a right prism for which the area of the lateral
surface is 180 units2 and the altitude is 4 units.
6 The base of a right prism is an equilateral triangle of length 3 cm and its lateral
surfaces are rectangular regions. If its altitude is 8 cm, then find:
a the total surface area of the prism b the volume of the prism.
7 If the dimensions of a right rectangular prism are 7 cm, 9 cm and 3 cm, then find:
a its total surface area b its volume
c the length of its diagonal.
8 Find the total surface area and the volume of each of the following solid figures:

3 cm

8 cm
6 cm
7 cm
6 cm 5 cm
6 cm

a 2 cm b c
Figure 7.16

9 If the diagonal of a cube is 12 cm, find the area of its lateral surface.
10 The radius of the base of a right circular cylinder is 2 cm and its altitude is 3 cm.
Find:
a the area of its lateral surface b the total surface area
c the volume.
11 Show that the area of the lateral surface of a right circular cylinder whose altitude
is h and whose base has radius r is 2πr h.
12 Imagine a cylindrical container in which the height and the diameter are equal.
Find expressions, in terms of its height, for its
a total surface area b volume.
13 A circular hole of radius 5 cm is drilled through the centre of a right circular
cylinder whose base has radius 6 cm and whose altitude is 8 cm. Find the total
surface area and volume of the resulting solid figure.

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7.2 PYRAMIDS, CONES AND SPHERES

Do you remember what you learnt about pyramids, cones and spheres in your previous
grades? Can you give some examples of pyramids, cones and spheres from real life?

Definition 7.2
A pyramid is a solid figure formed when each vertex of a polygon is joined
to the same point not in the plane of the polygon (See Figure 7.17).

B
B C B
A C A C

A D E D
Triangular pyramid Quadrilateral pyramid Pentagonal pyramid
a b c
Figure 7.17

ACTIVITY 7.2
1 What is a regular pyramid?
2 What is a tetrahedron?
3 Determine whether each of the following statements is true or false:
a The lateral faces of a pyramid are triangular regions.
b The number of triangular faces of a pyramid having same vertex is equal to
the number of edges of the base.
c The altitude of a cone is the perpendicular distance from the base to the
vertex of the cone.
V
4 Using Figure 7.18, complete the following to make true statements.
a The figure is called a ______.
b The region VED is called a _______.
c The region ABCDEF is called ______.
A B
F P C
d ______ is the altitude of the pyramid.
E D
e VE and VF are called ______. Figure 7.18

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Mathematics Grade 10

f Since ABCDEF is a hexagonal region, the pyramid is called a _______.

5 Draw a cone and indicate:
a its slant height b its base c its lateral surface.
The altitude of a pyramid is the length of the perpendicular from the vertex to the plane
containing the base.
The slant height of a regular pyramid is the altitude of any of its lateral faces.
Definition 7.3
The solid figure formed by joining all points of a circle to a point not on the
plane of the circle is called a cone.

V
Vertex
V

Slant height
Lateral
face Altitude
Lateral face

Base Base

Figure 7.19 Figure 7.20

The figure shown in Figure 7.19 represents a cone. Note that the curved surface is the
lateral surface of the cone.

A right circular cone (see Figure 7.20) is a cone with the foot of its altitude at the
centre of the base. A line segment from the vertex of a cone to any point on the
boundary of the base (circle) is called the slant height.

ACTIVITY 7.3
1 Consider a regular square pyramid with base edge 6 cm and
slant height 5 cm.
a How many lateral faces does it have?
b Find the area of each lateral face.
c Find the lateral surface area.
d Find the total surface area.
2 Try to write the formula for the total surface area of a pyramid or a cone.

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 Unit 7 Measurement

Surface area
The lateral surface area of a regular pyramid is equal to half the product of its slant
height and the perimeter of the base. That is, V
1
AL = Pℓ,
2
where AL denotes the lateral surface area;
P denotes the perimeter of the base; ℓ
ℓ denotes the slant height. D C
AB
The total surface area (AT) of a pyramid is given by
A B
1
AT = AB + AL = AB + Pℓ, Figure 7.21
2
where AB is area of the base.
Example 1 A regular pyramid has a square base whose side is 4 cm long. The lateral
edges are 6 cm each.
a What is its slant height? b What is the lateral surface area?
V
c What is the total surface area?
Solution: Consider Figure 7.22,
6
a ( VE )2 + ( EC )2 = ( VC )2 ℓ
ℓ2 + 22 = 62
D C
ℓ2 = 32
2
O E
ℓ = 4 2 cm 2
A 4 B
Therefore, the slant height is 4 2 cm. Figure 7.22

b There are 4 isosceles triangles.

Therefore,

1 1 
AL = 4 × BC × VE = 4  × 4 × 4 2  = 32 2 cm2
2 2 
1 1
or AL = Pℓ = (4 + 4 + 4 + 4) 4 2 = 8 × 4 2 = 32 2 cm2
2 2
c AT = AL + AB = 32 2 + 4 × 4
= 32 2 + 16 = 16 (2 2 + 1 ) cm2

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Mathematics Grade 10

The lateral surface area of a right circular cone is equal to half the product of its slant
height and the circumference of the base. That is,
1 1
AL = Pℓ = (2πr) ℓ = πrℓ;
2 2 h
ℓ
ℓ = h2 + r 2
r

Figure 7.23
where AL denotes the lateral surface area, ℓ represents the slant height, r stands for the
base radius, and h for the altitude.
The total surface area (AT) is equal to the sum of the area of the base and the lateral
surface area. That is,
AT = AL + AB = πrℓ + πr2 = πr (ℓ + r)

Example 2 The altitude of a right circular cone is 8 cm. If the radius of the base is 6 cm,
then find its:
a slant height b lateral surface area c total surface area.
Solution: Consider Figure 7.24

a ℓ= h2 + r 2 = 82 + 62 = 100

8
ℓ = 10 cm ℓ

b AL = π rℓ = π × 6 × 10 = 60π cm2
6

c AT = π r (ℓ + r) = π × 6 (10 + 6) = 6π × 16
Figure 7.24
= 96π cm2

Volume
The volume of any pyramid is equal to one third the product V

of its base area and its altitude. That is,

1
V= AB h, h
3
where V denotes the volume, AB the area of the base and AB
h the altitude. Figure 7.25

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 Unit 7 Measurement

Example 3 Find the volume of the pyramid given in Example 1 above.

Solution: Here, we need to find the altitude of the pyramid as shown below:

( V O )2 + ( OE )2 = ( VE )2 ⇒ h2 + 22 = (4 2 )2
h2 + 4 = 32
h2 = 28 ⇒ h = 2 7 cm
1 1 32 3
V = AB h = × (4 × 4) × 2 7 = 7 cm
3 3 3
The volume of a circular cone is equal to one-third of the
product of its base area and its altitude. That is,
h ℓ
1 1
V = AB h = π r 2 h
3 3
r
where V denotes the volume, r the radius of the base
and h the altitude. Figure 7.26
Example 4 Find the volume of the right circular cone given in Example 2 above.
1 2 1
Solution: V = π r h = π (6) 2 × 8 = 96π cm 3
3 3
Example 5 Find the lateral surface area, total surface area and the volume of the
following regular pyramid and right circular cone.
V

h
12 cm

8 2 cm
D C
8 cm
E F
10 cm
A 10 cm B

a b
Figure 7.27

Solution:
a To find the lateral surface area, we must find the slant height ℓ.
In ∆VEF, we have,
(VE)2 + (EF)2 = (VF)2 ⇒ 122 + 52 = (VF)2
169 = (VF)2 ⇒ VF = 13 cm
Therefore, the slant height is 13 cm.
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Mathematics Grade 10

1 1
Now, AL = Pℓ = (10 + 10 + 10 + 10)13 = 260 cm2
2 2
AT = AL + AB = 260 cm2 + 100 cm2 = 360 cm2
1 1
V = AB h = ×100 ×12 = 400 cm3 .
3 3
b Altitude : h = ℓ2 − r 2 = (8 2) 2 − 82 = 128 − 64 = 64 = 8 cm

AL = πrℓ = π × 8 × 8 2 = 64 2π cm2

AT = πr (ℓ + r) = 8π (8 2 + 8) = 64π ( 2 + 1) cm2
1 2 1 512π
V= π r h = π (8) 2 × 8 = cm3
3 3 3

Surface area and volume of a sphere

The sphere is another solid figure you studied in lower grades.

Definition 7.4
A sphere is a closed surface, all points of which are r
equidistant from a point called the centre.
o

Figure 7.28

The surface area (A) and the volume (V) of a sphere of

radius r are given by
r
A = 4 π r2
4
V = π r3
3

Figure 7.29

Example 6 Find the surface area and volume of a spherical gas balloon with a diameter
of 10 m.
d 10
Solution: We know that d = 2r or r = ∴r= =5m
2 2
A = 4π r2 = 4π (5)2 = 100π m2
4 4 500
V = π r 3 = π (5)3 = π m3
3 3 3
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 Unit 7 Measurement

Exercise 7.2
1 Calculate the volume of each of the following solid figures:

20
6
15
3 5
9

10
a b c
Figure 7.30
2 One edge of a right square pyramid is 6 cm long. If the length of the lateral edge
is 8 cm, then find:
a its total surface area b its volume.
3 The altitude of a right equilateral triangular pyramid is 6 cm. If one edge of the
base is 6 cm, then find:
a its total surface area b its volume.
4 A regular square pyramid has all its edges 7 cm long. Find:
a its total surface area b its volume
5 The altitude and radius of a right circular cone are 12 cm and 5 cm respectively. Find:
a its total surface area b its volume.
6 The volume of a pyramid is 240 cm3. The pyramid has a rectangular base with
sides 6 cm by 4 cm. Find the altitude and lateral surface area of the pyramid if the
pyramid has equal lateral edges.
7 Show that the volume of a regular square pyramid whose lateral faces are
s3 2
equilateral triangles of side length s, is .
6
8 The lateral edge of a regular tetrahedron is 8 cm. Find its altitude.
9 Find the volume of a cone of height 12 cm and slant height 13 cm.
10 Find the volume and surface area of a spherical football with a radius of 10 cm.
11 A glass is in the form of an inverted cone whose base has a diameter of 20 cm. If
0.1 litres of water fills the glass completely, find the depth of water in the glass
 22 
 take π ≈  .
 7
12 A solid metal cylinder with a length of 24 cm and radius 2 cm is melted down to
form a sphere. What is the radius of the sphere?
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Mathematics Grade 10

7.3 FRUSTUMS OF PYRAMIDS AND CONES

In the preceding section, you have studied about pyramids and cones. You will now
study the solid figure obtained when a pyramid and a cone are cut by a plane parallel to
the base as shown in Figure 7.31.
Let E be the plane that contains the base and E' be the plane parallel to the base that cuts
the pyramid and cone.
Cross-section

E'
E'

Parallel
plane

E E

a b
Figure 7.31

Definition 7.5
If a pyramid or a cone is cut by a plane parallel to the base, the intersection
of the plane and the pyramid (or the cone) is called a horizontal cross-
section of the pyramid (or the cone).

Let us now examine the relationship between the base and the cross-section.
V
Let ∆ABC be the base of the pyramid lying in
the plane E. Let h be the altitude of the pyramid k
and let ∆A ' B ' C ' be the cross-section at a A' C'
h
D' B'
distance k units from the vertex. E'

Let D and D' be the points at which the

perpendicular from V to E meet E and E ', A
C
respectively. E D
B
We have, Figure 7.32

1 ∆VA'D' ∼ ∆VAD.
This follows from the fact that if a plane intersects each of two parallel planes, it
intersects them in two parallel lines, and an application of the AA similarity
theorem. Hence,
VA ' VD ' k
= =
VA VD h

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 Unit 7 Measurement

2 Similarly, ∆ VD'B' ∼ ∆VDB and hence,

VB ' VD ' k
= =
VB VD h
Then, from 1 and 2 and the SAS similarity theorem, we get,
A ' B ' VA ' k
3 ∆ VA'B' ∼ ∆ VAB. Therefore, = =
AB VA h
By an argument similar to that leading to (3), we have
B 'C ' k A 'C ' k
4 = and =
BC h AC h
Hence, by the SSS similarity theorem,
∆ABC ∼ ∆ A'B'C '

ACTIVITY 7.4
In the pyramid shown in Figure 7.33, ∆ABC is equilateral. A plane
parallel to the base intersects the lateral edges in D, E and F such
1 V
that VE = EB.
3
VF D F
a What is ? E
VC
EF A C
b What is ?
BC
c Compare the areas of ∆VEF and ∆VBC and of B
Figure 7.33
∆DEF and ∆ABC.

Theorem 7.1
In any pyramid, the ratio of the area of a cross-section to the area of the base is
k2
where h is the altitude of the pyramid and k is the distance from the vertex
h2
to the plane of the cross-section.

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Mathematics Grade 10

V V

k
B' k
A' Ac C'
B'
D' A' A C'
c
h E' D' h

B B
A Ab C
C
A Ab
E D
D
Ac area( A ' B ' C ' D ') k 2 area( A ' B ' C ' D ' E ') k 2
= = 2 = 2
Ab area( ABCD) h area( ABCD) h
Figure 7.34
Example 1 The area of the base of a pyramid is 90 cm2. The altitude of the pyramid is
12 cm. What is the area of a horizontal cross-section 4 cm from the vertex?
Solution: Let Ac be the area of the cross-section, and Ab the base area.
A k2 A 42
Then, c = 2 ⇒ c = 2
Ab h 90 12
90 × 16
∴ Ac = cm 2 = 10 cm 2
144
Note that similar properties hold true when a cone is cut by a plane parallel to its base.
Can you state them?

ACTIVITY 7.5
1 The altitude of a square pyramid is 5 units long and a side of the base
is 4 units long. Find the area of a horizontal cross-section at a
distance 2 units above the base.
2 The area of the base of a pyramid is 64 cm2. The altitude of the pyramid is 8 cm.
What is the area of a cross-section 2 cm from the vertex?
3 The radius of a cross-section of a cone at a distance 5 cm from the base is 2 cm. If
the radius of the base of the cone is 3 cm, find its altitude.
When a prism is cut by a plane parallel to the base, each part of the prism is again a
prism as shown in Figure 7.35a.

a b
Figure 7.35
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 Unit 7 Measurement

However, when a pyramid is cut by a plane parallel to the base, the part of the pyramid
between the vertex and the horizontal cross-section is again a pyramid whereas the other
part is not a pyramid (as shown in Figure 7.35b).

Frustum of a pyramid
Definition 7.6
A frustum of a pyramid is a part of the pyramid included between the base and
a plane parallel to the base.

The base of the pyramid and the cross-section made by the plane parallel to it are called
the bases of the frustum. The other faces are called lateral faces. The total surface
of a frustum is the sum of the lateral surface and the bases.
The altitude of a frustum of a pyramid is the perpendicular distance between the bases.
Upper base

Altitude Lateral faces

Lower base
Figure 7.36
Note:
i The lateral faces of a frustum of a pyramid are trapeziums.
ii The lateral faces of a frustum of a regular pyramid are congruent isosceles
trapeziums.
iii The slant height of a frustum of a regular pyramid is the altitude of any one
of the lateral faces.
iv The lateral surface area of a frustum of a pyramid is the sum of the areas of
the lateral faces.

Frustum of a cone

Definition 7.7
A frustum of a cone is a part of the cone included between the base and a
horizontal cross-section made by a plane parallel to the base.

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For a frustum of a cone, the bases are the base of Upper base
the cone and the cross-section parallel to the base.
Lateral face
The lateral surface is the curved surface that
makes up the frustum. The altitude is the Altitude
Slant height
perpendicular distance between the bases.
Lower base
Figure 7.37
The slant height of a frustum of a right circular cone is that part of the slant height of
the cone which is included between the bases.
Can you name some objects we use in real life (at home) that are frustums of cones?
Are a bucket and a glass frustum of cones? Discuss.
Example 2 The lower base of the frustum of a regular pyramid is a square 4 cm long,
the upper base is 3 cm long. If the slant height is 6 cm, find its lateral
surface area.
Solution: As shown in Figure 7.38, each lateral face is a trapezium, the area of
each lateral face is 3

1 1
AL = × h (b1 + b2 ) = × 6(3 + 4) = 21 cm 2
2 2 6
Since the four faces are congruent isosceles trapeziums,
the lateral surface area is 4
2 2
AL = 4 × 21 cm = 84 cm Figure 7.38
Example 3 The lower base of the frustum of a regular A' B'
pyramid is a square of side s units long. D' s' C' s'
The upper base is s' units long. If the slant
height of the frustum is ℓ, then find the ℓ
A B
lateral surface area.
s
D s C
Figure 7.39
Solution: Figure 7.39 represents the given problem. ABCD is a square s units long.
Similarly A'B'C 'D' is a square s' units long.
Lateral surface area:
AL = area (D'C 'CD) + area (C 'B'BC) + area (A'B'BA) + area (D'A'AD)
1 1 1 1
= ℓ (s + s ') + ℓ (s + s') + ℓ (s + s') + ℓ (s + s')
2 2 2 2
1
AL = ℓ (4s + 4s') = 2ℓ (s + s').
2
Observe that 4s and 4s' are the perimeters of the lower and upper bases, respectively.
In general, we have the following theorem:
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 Unit 7 Measurement

Theorem 7.2
The lateral surface area (AL) of a frustum of a regular pyramid is equal
to half the product of the slant height (ℓ) and the sum of the perimeter
(P) of the lower base and the perimeter (P ') of the upper base. That is,
1
AL = ℓ (P + P ')
2

Group Work 7.1

Consider the following figure.

1 Find the areas of the bases.
2 Find the circumferences of the bases of the frustum, c1 and c2. k2
3 Find lateral surface area of the bigger cone. r2 k1

4 Find lateral surface area of the smaller cone.

5 Find lateral surface area of the frustum. r1
6 Give the volume of the frustum.
Figure 7.40
Example 4 A frustum of height 4 cm is formed from a
right circular cone of height 8 cm and base k
radius 6 cm as shown in Figure 7.41. Calculate 8 cm
the lateral surface area of the frustum. Ac

Solution: Let Ab, Ac and AL stand for area of the base of 4 cm

the cone, area of the cross-section and lateral 6 cm Ab
surface area of the frustum, respectively.
2 Figure 7.41
Area of cross-section  k 
= 
Area of the base h
2
Ac  4 
=   , since k = 8 cm – 4 cm = 4 cm
Ab  8 
Ac 1 2 2
= (area of the base = π r = π × 6 = 36π)
36π 4
1
Ac = × 36π = 9π cm2
4
Ac = π (r') 2, where r' is radius of the cross-section
∴ 9π = π (r')2 ⇒ r' = 3 cm
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Mathematics Grade 10

Slant height of the bigger cone is:

ℓ= h2 + r 2 = 82 + 62 = 100 = 10 cm
Slant height of the smaller cone is:
2
ℓ' = k 2 + ( r ′) = 42 + 32 = 25 = 5 cm
Now the lateral surface area of:
the smaller cone = π r'ℓ' = π (3 cm) × 5 cm = 15π cm2

the bigger cone = π rℓ = π (6 cm) × 10 cm = 60π cm2.

Hence, the area of the lateral surface of the frustum is
AL = 60π cm2 – 15π cm2 = 45π cm2.
The lateral surface (curved surface) of a frustum of a circular cone is a trapezium whose
parallel sides (bases) have lengths equal to the circumference of the bases of the frustum
and whose height is equal to the height of the frustum.

Theorem 7.3
For a frustum of a right circular cone with altitude h and slant height ℓ,
if the circumferences of the bases are c and c', then the lateral surface
area of the frustum is given by
1 1
AL = ℓ (c + c') = ℓ (2π r + 2π r') = ℓπ (r + r')
2 2

Example 5 A frustum formed from a right circular cone has base radii of 8 cm and
12 cm and slant height of 10 cm. Find:
a the area of the curved surface
b the area of the total surface. (Use π ≈ 3.14).
Solution:
a AL = πℓ ( r + r') = π × 10 cm (8 + 12) cm = 10π cm × 20 cm
= 200π cm2 = 200 × 3.14 cm2 = 628 cm2
b Area of bases:
AB = Ac + Ab = π (r') 2 + π r2 = π (8 cm)2 + π (12 cm)2 = 64π cm2 + 144π cm2
= 208π cm2 ≈ 208 × 3.14 cm2 ≈ 653 cm2
Total surface area of the frustum:
AT = AL + AB ≈ 628 cm2 + 653 cm2 = 1281 cm2

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 Unit 7 Measurement

Example 6 The area of the upper and lower bases of a

frustum of a pyramid are 25 cm2 and 36 cm2
respectively. If its altitude is 2 cm, then find k
the altitude of the pyramid.
Solution:
2 2
Ac  k  25 2
k
=  ⇒ =
Ab  h  36 (2 + k )2
5 k
⇒ = ⇒ 6k = 5k + 10 Figure 7.42
6 2+k
∴ k = 10
Therefore, the altitude of the pyramid is 2 cm + 10 cm = 12 cm.
Note that the upper and lower bases of the frustum of a pyramid are similar polygons
and that of a cone are similar circles.

k
h h
A'

h'

Figure 7.43

Let h = the height (altitude)of the complete cone or pyramid.

k = the height of the smaller cone or pyramid.
A = the base area of the bigger cone or pyramid (lower base of the frustum)
A' = the base area of the completing cone or pyramid (upper base of the frustum)
h' = h – k = the height of the frustum of the cone or pyramid.
V = the volume of the bigger cone or pyramid.
V ' = the volume of the smaller cone or pyramid (upper part).
Vf = the volume of the frustum
1 1
V= Ah and V ' = A ' k , consequently the volume (Vf) of the frustum of the
3 3
pyramid is
1 1 1
Vf = V – V ' = Ah – A'k = (Ah – A'k)
3 3 3

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Mathematics Grade 10

Using this notion, we shall give the formula for finding the volume of a frustum of a
cone or pyramid as follows:
h'
Vf =
3
(
A + A '+ AA ' )
where A is the lower base area, A' the upper base area and h' is the height of a frustum of
a cone or pyramid.
From this, we can give the formula for finding the volume of a frustum of a cone in
terms of r and r' as follows:
π
Vf = h ' ( r 2 + (r ') 2 + rr ')
3
where r is the radius of the bigger (the lower base of the frustum) cone and r' is the
radius of the smaller cone (upper base of the frustum).
Example 7 A frustum of a regular square pyramid has 2 cm
height 5 cm. The upper base is of side 2 cm 2 cm
and the lower base is of side 6 cm. Find the
5 cm
volume of the frustum.
Solution:
6 cm
Since the upper base and lower base are squares, Figure 7.44
6 cm
A = (6 cm)2 = 36 cm2
A' = (2 cm)2 = 4 cm2
h' 5
Vf =
3
(
A + A '+ AA ' =) 3
( )
36 + 4 + 36 × 4 cm 3

5 5 260
= (40 + 12) cm3 = × 52 cm3 = cm3 .
3 3 3

Exercise 7.3
1 The lower base of a frustum of a regular pyramid is a square of side 6 cm, and the
upper base has side length 3 cm. If the slant height is 8 cm, find:
a its lateral surface area b its total surface area.
2
2 A circular cone with altitude h and base radius r is cut at a heightof the way
3
from the base to form a frustum of a cone. Find the volume of the frustum.
3 The areas of bases of a frustum of a pyramid are 25 cm2 and 49 cm2. If its altitude
is 3 cm, find its volume.

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 Unit 7 Measurement

4 The slant height of a frustum of a cone is 10 cm. If the radii of the bases are 6 cm
and 3 cm, find
a the lateral surface area b the total surface area
c the volume of the frustum.
5 A frustum of a regular square pyramid whose lateral faces are equilateral triangles
of side 10 cm has altitude 5 cm. Calculate the volume of the frustum.
6 The altitude of a pyramid is 10 cm. The base is a square whose sides are each
6 cm long. If a plane parallel to the base cuts the pyramid at a distance of 5 cm
from the vertex, then find the volume of the frustum formed.
7 The bucket shown in Figure 7.45 is in the form of a frustum of right circular cone.
The radii of the bases are 12 cm and 20 cm, and the volume is 6000 cm3. Find its
a height b slant height

Figure 7.45
8 A frustum of height 12 cm is formed from a right circular cone of height 16 cm
and base radius 8 cm. Calculate:
a the lateral surface area of the frustum
b the total surface area of the frustum
c the volume of the frustum.
9 A frustum is formed from a regular pyramid. Let the perimeter of the lower base
be P, the perimeter of the upper base be P ' and the slant height be ℓ. Show that
the lateral surface area of the frustum is
1
AL = ℓ(P + P').
2
10 A frustum of height 5 cm is formed from a right circular cone of height 10 cm and
base radius 4 cm. Calculate:
a the lateral surface area b the volume of the frustum.
11 A frustum of a regular square pyramid has height 2 cm. The lateral faces of the
pyramid are equilateral triangles of side 3 2 cm. Find the volume of the frustum.

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Mathematics Grade 10

12 A container is in the shape of an inverted frustum of a right circular cone as

shown in Figure 7.46. It has a circular bottom of radius 20 cm, a circular top of
radius 60 cm and height 40 cm. How many litres of oil could it contain?
60 cm

40 cm

20 cm

Figure 7.46

7.4 SURFACE AREAS AND VOLUMES OF

COMPOSED SOLIDS
In the preceding sections, you have learned how to calculate the volume and surface
area of cylinders, prisms, cones, pyramids, spheres and frustums. In this section, you
will study how to find the areas and volumes of solids formed by combining the above
solid figures.

ACTIVITY 7.6
1 Give the formula used for:
a finding the lateral surface area of a
i cylinder ii prism iii cone iv pyramid
v sphere vi frustum of a pyramid vii frustum of a cone
b finding the volume of a
i cylinder ii prism iii cone iv pyramid
v sphere vi frustum of a pyramid vii frustum of a cone
2 If the diameter of a sphere is halved, what effect does this have on its volume and
its surface area?
3 What is the ratio of the volume of a sphere whose radius is r units to the cone
having equal radius and height 2r units?
Consider the following examples.
Example 1 A candle is made in the form of a circular cylinder of radius 4 cm at the
bottom and a right circular cone of altitude 3 cm, as shown in Figure
7.47. If the overall height is 12 cm, find the total surface area and the
volume of the candle.

Solution: Slant height of the cone is ℓ = 32 + 42 = 5 cm

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 Unit 7 Measurement

The total surface area of the candle is the sum of the

lateral surface areas of the cone, the cylinder and the
area of the base of the cylinder. That is,
3 cm
3
AT = π rℓ + 2π rh + π r2 = π (4) 5 + 2π (4) 9 + π (4)2

= 20π + 72π + 16π = 108π cm2

The volume of the candle is the sum of the volumes of 12 cm
the cone and cylinder.
1 2
VT = Vcone + Vcylinder = π r hco + π r 2 hcy
3 4 cm
1
= π (4)2 × 3 + π (4)2 × 9 = 16π + 144π = 160π cm3 Figure 7.47
3
Example 2 Through a right circular cylinder whose base
radius is 10 cm and whose height is 12 cm is
drilled a triangular prism hole whose base has
edges 3 cm, 4 cm and 5 cm as shown in Figure
7.48. Find the total surface area and volume of
the remaining solid.
Solution: The total surface area is the sum of the lateral
surface areas of the cylinder and prism, and
the base area of the cylinder, minus the base Figure 7.48
area of the prism.
1 
AT = 2π rh + ph + 2π r2 – 2  ab 
2 
1 
= 2π (10) 12 + (3 + 4 + 5) 12 + 2π (10)2 – 2  × 3× 4 
2 
= 240π + 144 + 200π – 12 = (440π + 132) cm2
The volume of the resulting solid is the difference between the volume of the
cylinder and prism.
1 1
VT = Vcy – Vp = πr2 h – abh = π (10)2 × 12 – × 3 × 4 × 12
2 2
= 1200π cm3 – 72 cm3 = 24 (50π – 3) cm3.
Example 3 A cone is contained in a cylinder so that
their base radius and height are the same, as
shown in Figure 7.49. Calculate the volume
of the space inside the cylinder but outside
the cone. Figure 7.49

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Mathematics Grade 10

Solution: The required volume is equal to the difference between the volume of the
cylinder and the cone. That is,
1 2
V = Vcy – Vco = π r2h – π r 2 h = π r 2 h.
3 3
2
As r = h, then V = π r 3 .
3

Group Work 7.2

1 A cylindrical tin 8 cm in diameter contains water to a

depth of 4 cm. If a cylindrical wooden rod 4 cm in
diameter and 6 cm long is placed in the tin it floats exactly
half submerged. What is the new depth of water?
2 An open pencil case comprises a cylinder of length 20 cm and radius 2 cm and a
cone of height 4 cm, as shown in Figure 7.50. Calculate the total surface area and
the volume of the pencil case.

4 cm
20 cm
4 cm
Figure 7.50
3 A ball is placed inside a box into which it will fit tightly.
If the radius of the ball is 8 cm, calculate:
i the volume of the ball
ii the volume of the box
Figure 7.51
4 An ice-cream consists of a hemisphere and a cone.
Calculate its volume and total surface area. 6 cm

10 cm

20 cm
30 cm
15 cm

4 cm
Figure 7.52
Figure 7.53
5 A torch 20 cm long is in the form of a right circular cylinder of height 15 cm and
radius 4 cm. Joined to it is a frustum of a cone of radius 6 cm. Find the volume of
the torch.

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 Unit 7 Measurement

Exercise 7.4
1 Find the volume of each of the following.
2 cm

5 cm
3 cm 2 cm

4 cm
8 cm
10 cm
4 cm

a b
Figure 7.54
2 A storage tank is in the form of cylinder with one hemispherical end, the other
being flat. The diameter of the cylinder is 4 m and the overall height of the tank is
9 m. What is the capacity of the tank?
3 An iron ball 5 cm in diameter is placed in a cylindrical tin of diameter 10 cm and
water is poured into the tin until its depth is 6 cm. If the ball is now removed,
how far does the water level drop?
4 From a hemispherical solid of radius 8 cm, a conical part is removed as shown in
Figure 7.55. Find the volume and the total surface area of the resulting solid.

6
4

Figure 7.55 Figure 7.56 Figure 7.57

5 The altitude of a frustum of a right circular cone is 20 cm and the radius of its
base is 6 cm. A cylindrical hole of diameter 4 cm is drilled through the cone with
the centre of the drill following the axis of the cone, leaving a solid as shown in
Figure 7.56. Find the volume and the total surface area of the resulting solid.
6 Figure 7.57 shows a hemispherical shell. Find the volume and total surface area of
the solid. 8 cm
7 A cylindrical piece of wood of radius 8 cm and height 18
cm has a cone of the same radius scooped out of it to a
18 cm
depth of 9 cm. Find the ratio of the volume of the wood
scooped out to the volume of wood which is left. (See
Figure 7.58)
Figure 7.58

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Mathematics Grade 10

Key Terms
cone lateral edge regular pyramid

cross–section lateral surface slant height

cylinder prism sphere

frustum pyramid volume

Summary
Prism
AL = Ph
AT = 2Ab + AL
V = Ab h Figure 7.59

Right circular cylinder

AL = 2π rh h
2
AT = 2πr + 2π rh = 2π r (r + h)
V = π r2 h r

Figure 7.60

Regular pyramid V
1
AL = Pℓ
2
h ℓ
1
AT = Ab + Pℓ
2
O
1
V = Ab h Figure 7.61
3 V

Right circular cone

AL = π rℓ
h ℓ
AT = π r2 + π rℓ = π r (r + ℓ)
r
1 O
V = π r 2h
3 Figure 7.62

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 Unit 7 Measurement

Sphere
A = 4 π r2
4 r
V = π r3
3

Frustum of a pyramid
Figure 7.63
1 A′b
AL = ℓ (P + P')
2
1
AT = ℓ (P + P') + Ab + A'b h'
2
V=
1
3
(
h' Ab + A 'b + Ab A 'b ) Ab

Figure 7.64 r'

Frustum of a cone
1
AL = ℓ(2π r + 2π r') = ℓ π (r + r') h’
ℓ
2
1
AT = ℓ(2π r + 2π r') + π r2 + π (r')2 = ℓπ (r + r') +π (r2 + r ' 2 ) r
2
1
V = h' π (r2 + ( r ') 2 + rr') Figure 7.65
3

Review Exercises on Unit 7

1 Find the lateral surface area and volume of each of the following figures.

12
5 6
12

5 2
6
12

a b c d
Figure 7.66
2 A lateral edge of a right prism is 6 cm and the perimeter of its base is 36 cm. Find
the area of its lateral surface.
3 The height of a circular cylinder is equal to the radius of its base. Find its total
surface area and its volume, giving your answer in terms of its radius r.

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Mathematics Grade 10

4 What is the volume of a stone in an Egyptian pyramid with a square base of side
100 m and a slant height of 50 2 m for each of the triangular faces.
5 Find the total surface area of a regular hexagonal pyramid, given that an edge of
the base is 8 cm and the altitude is 12 cm.
6 Find the area of the lateral surface of a right circular cone whose altitude is 8 cm
and base radius 6 cm.
7 Find the total surface area of a right circular cone whose altitude is h and base radius
is r. (Give the answer in terms of r and h)
8 When a lump of stone is submerged in a rectangular water tank whose base is
25 cm by 50 cm, the level of the water rises by 1 cm. What is the volume of the
stone?
9 A frustum whose upper and lower bases are circular regions of radii 8 cm and
6 cm respectively, is 25 cm deep. (See Figure 7.67). Find its volume.
8

25

Figure 7.67 Figure 7.68

10 A cylindrical metal pipe of outer diameter 10 cm is 2 cm thick. What is the
diameter of the hole? Find the volume of the metal if the pipe is 30 cm long.
11 A drinking cup in the shape of frustum of a cone with bottom diameter 4 cm and
top diameter 6 cm, can contain a maximum of 80 cm3 of coffee. Find the height of
the cup.
12 The slant height of a cone is 16 cm and the radius of its base is 4 cm. Find the
area of the lateral surface of the cone and its volume.
13 The radius of the base of a cone is 12 cm and its volume is 720π cm3. Find its
height, slant height, and lateral surface area.
14 If the radius of a sphere is doubled, what effect does this have on its volume and
its surface area?
15 In Figure 7.68, a cone of base radius r and altitude 2r and a hemisphere of radius r
whose base coincides with that of the cone are shown. A is the part of the
hemisphere which lies outside the cone and B is the part of the cone lying outside
the hemisphere. Prove that the volume of A is equal to the volume of B.

300
 Table of Trigonometric Functions

Table of Trigonometric Functions

sin cos tan cot sec csc
0°° 0.0000 1.0000 0.0000 ..... 1.000 ..... 90°°
1°° 0.0175 0.9998 0.0175 57.29 1.000 57.30 89°°
2°° 0.0349 0.9994 0.0349 28.64 1.001 28.65 88°°
3°° 0.0523 0.9986 0.0524 19.08 1.001 19.11 87°°
4°° 0.0698 0.9976 0.0699 14.30 1.002 14.34 86°°
5°° 0.0872 0.9962 0.0875 11.43 1.004 11.47 85°°
6°° 0.1045 0.9945 0.1051 9.514 1.006 9.567 84°°
7°° 0.1219 0.9925 0.1228 8.144 1.008 8.206 83°°
8°° 0.1392 0.9903 0.1405 7.115 1.010 7.185 82°°
9°° 0.1564 0.9877 0.1584 6.314 1.012 6.392 81°°
10°° 0.1736 0.9848 0.1763 5.671 1.015 5.759 80°°
11°° 0.1908 0.9816 0.1944 5.145 1.019 5.241 79°°
12°° 0.2079 0.9781 0.2126 4.705 1.022 4.810 78°°
13°° 0.2250 0.9744 0.2309 4.331 1.026 4.445 77°°
14°° 0.2419 0.9703 0.2493 4.011 1.031 4.134 76°°
15°° 0.2588 0.9659 0.2679 3.732 1.035 3.864 75°°
16°° 0.2756 0.9613 0.2867 3.487 1.040 3.628 74°°
17°° 0.2924 0.9563 0.3057 3.271 1.046 3.420 73°°
18°° 0.3090 0.9511 0.3249 3.078 1.051 3.236 72°°
19°° 0.3256 0.9455 0.3443 2.904 1.058 3.072 71°°
20°° 0.3420 0.9397 0.3640 2.747 1.064 2.924 70°°
21°° 0.3584 0.9336 0.3839 2.605 1.071 2.790 69°°
22°° 0.3746 0.9272 0.4040 2.475 1.079 2.669 68°°
23°° 0.3907 0.9205 0.4245 2.356 1.086 2.559 67°°
24°° 0.4067 0.9135 0.4452 2.246 1.095 2.459 66°°
25°° 0.4226 0.9063 0.4663 2.145 1.103 2.366 65°°
26°° 0.4384 0.8988 0.4877 2.050 1.113 2.281 64°°
27°° 0.4540 0.8910 0.5095 1.963 1.122 2.203 63°°
28°° 0.4695 0.8829 0.5317 1.881 1.133 2.130 62°°
29°° 0.4848 0.8746 0.5543 1.804 1.143 2.063 61°°
30°° 0.5000 0.8660 0.5774 1.732 1.155 2.000 60°°
31°° 0.5150 0.8572 0.6009 1.664 1.167 1.942 59°°
32°° 0.5299 0.8480 0.6249 1.600 1.179 1.887 58°°
33°° 0.5446 0.8387 0.6494 1.540 1.192 1.836 57°°
34°° 0.5592 0.8290 0.6745 1.483 1.206 1.788 56°°
35°° 0.5736 0.8192 0.7002 1.428 1.221 1.743 55°°
36°° 0.5878 0.8090 0.7265 1.376 1.236 1.701 54°°
37°° 0.6018 0.7986 0.7536 1.327 1.252 1.662 53°°
38°° 0.6157 0.7880 0.7813 1.280 1.269 1.624 52°°
39°° 0.6293 0.7771 0.8098 1.235 1.287 1.589 51°°
40°° 0.6428 0.7660 0.8391 1.192 1.305 1.556 50°°
41°° 0.6561 0.7547 0.8693 1.150 1.325 1.524 49°°
42°° 0.6691 0.7431 0.9004 1.111 1.346 1.494 48°°
43°° 0.6820 0.7314 0.9325 1.072 1.367 1.466 47°°
44°° 0.6947 0.7193 0.9667 1.036 1.390 1.440 46°°
45°° 0.7071 0.7071 1 .0000 1.000 1.414 1.414 45°°
cos sin cot tan csc sec

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Mathematics Grade 10

Table of Common Logarithms

n 0 1 2 3 4 5 6 7 8 9
1.0 0.0000 0.0043 0.0086 0.0128 0.0170 0.0212 0.0253 0.0294 0.0334 0.0374
1.1 0.0414 0.0453 0.0492 0.0531 0.0569 0.0607 0.0645 0.0682 0.0719 0.0755
1.2 0.0792 0.0828 0.0864 0.0899 0.0934 0.0969 0.1004 0.1038 0.1072 0.1106
1.3 0.1139 0.1173 0.1206 0.1239 0.1271 0.1303 0.1335 0.1367 0.1399 0.1430
1.4 0.1461 0.1492 0.1523 0.1553 0.1584 0.1614 0.1644 0.1673 0.1703 0.1732

1.5 0.1761 0.1790 0.1818 0.1847 0.1875 0.1903 0.1931 0.1959 0.1987 0.2014
1.6 0.2041 0.2068 0.2095 0.2122 0.2148 0.2175 0.2201 0.2227 0.2253 0.2279
1.7 0.2304 0.2330 0.2355 0.2380 0.2405 0.2430 0.2455 0.2480 0.2504 0.2529
1.8 0.2553 0.2577 0.2601 0.2625 0.2648 0.2672 0.2695 0.2718 0.2742 0.2765
1.9 0.2788 0.2810 0.2833 0.2856 0.2878 0.2900 0.2923 0.2945 0.2967 0.2989

2.0 0.3010 0.3032 0.3054 0.3075 0.3096 0.3118 0.3139 0.3160 0.3181 0.3201
2.1 0.3222 0.3243 0.3263 0.3284 0.3304 0.3324 0.3345 0.3365 0.3385 0.3404
2.2 0.3424 0.3444 0.3464 0.3483 0.3502 0.3522 0.3541 0.3560 0.3579 0.3598
2.3 0.3617 0.3636 0.3655 0.3674 0.3692 0.3711 0.3729 0.3747 0.3766 0.3784
2.4 0.3802 0.3820 0.3838 0.3856 0.3874 0.3892 0.3909 0.3927 0.3945 0.3962

2.5 0.3979 0.3997 0.4014 0.4031 0.4048 0.4065 0.4082 0.4099 0.4116 0.4133
2.6 0.4150 0.4166 0.4183 0.4200 0.4216 0.4232 0.4249 0.4265 0.4281 0.4298
2.7 0.4314 0.4330 0.4346 0.4362 0.4378 0.4393 0.4409 0.4425 0.4440 0.4456
2.8 0.4472 0.4487 0.4502 0.4518 0.4533 0.4548 0.4564 0.4579 0.4594 0.4609
2.9 0.4624 0.4639 0.4654 0.4669 0.4683 0.4698 0.4713 0.4728 0.4742 0.4757

3.0 0.4771 0.4786 0.4800 0.4814 0.4829 0.4843 0.4857 0.4871 0.4886 0.4900
3.1 0.4914 0.4928 0.4942 0.4955 0.4969 0.4983 0.4997 0.5011 0.5024 0.5038
3.2 0.5051 0.5065 0.5079 0.5092 0.5105 0.5119 0.5132 0.5145 0.5159 0.5172
3.3 0.5185 0.5198 0.5211 0.5224 0.5237 0.5250 0.5263 0.5276 0.5289 0.5302
3.4 0.5315 0.5328 0.5340 0.5353 0.5366 0.5378 0.5391 0.5403 0.5416 0.5428

3.5 0.5441 0.5453 0.5465 0.5478 0.5490 0.5502 0.5514 0.5527 0.5539 0.5551
3.6 0.5563 0.5575 0.5587 0.5599 0.5611 0.5623 0.5635 0.5647 0.5658 0.5670
3.7 0.5682 0.5694 0.5705 0.5717 0.5729 0.5740 0.5752 0.5763 0.5775 0.5786
3.8 0.5798 0.5809 0.5821 0.5832 0.5843 0.5855 0.5866 0.5877 0.5888 0.5899
3.9 0.5911 0.5922 0.5933 0.5944 0.5955 0.5966 0.5977 0.5988 0.5999 0.6010

4.0 0.6021 0.6031 0.6042 0.6053 0.6064 0.6075 0.6085 0.6096 0.6107 0.6117
4.1 0.6128 0.6138 0.6149 0.6160 0.6170 0.6180 0.6191 0.6201 0.6212 0.6222
4.2 0.6232 0.6243 0.6253 0.6263 0.6274 0.6284 0.6294 0.6304 0.6314 0.6325
4.3 0.6335 0.6345 0.6355 0.6365 0.6375 0.6385 0.6395 0.6405 0.6415 0.6425
4.4 0.6435 0.6444 0.6454 0.6464 0.6474 0.6484 0.6493 0.6503 0.6513 0.6522

4.5 0.6532 0.6542 0.6551 0.6561 0.6571 0.6580 0.6590 0.6599 0.6609 0.6618
4.6 0.6628 0.6637 0.6646 0.6656 0.6665 0.6675 0.6684 0.6693 0.6702 0.6712
4.7 0.6721 0.6730 0.6739 0.6749 0.6758 0.6767 0.6776 0.6785 0.6794 0.6803
4.8 0.6812 0.6821 0.6830 0.6839 0.6848 0.6857 0.6866 0.6875 0.6884 0.6893
4.9 0.6902 0.6911 0.6920 0.6928 0.6937 0.6946 0.6955 0.6964 0.6972 0.6981

5.0 0.6990 0.6998 0.7007 0.7016 0.7024 0.7033 0.7042 0.7050 0.7059 0.7067
5.1 0.7076 0.7084 0.7093 0.7101 0.7110 0.7118 0.7126 0.7135 0.7143 0.7152
5.2 0.7160 0.7168 0.7177 0.7185 0.7193 0.7202 0.7210 0.7218 0.7226 0.7235
5.3 0.7243 0.7251 0.7259 0.7267 0.7275 0.7284 0.7292 0.7300 0.7308 0.7316
5.4 0.7324 0.7332 0.7340 0.7348 0.7356 0.7364 0.7372 0.7380 0.7388 0.7396

302
 Table of Common Logarithms

5.5 0.7404 0.7412 0.7419 0.7427 0.7435 0.7443 0.7451 0.7459 0.7466 0.7474
5.6 0.7482 0.7490 0.7497 0.7505 0.7513 0.7520 0.7528 0.7536 0.7543 0.7551
5.7 0.7559 0.7566 0.7574 0.7582 0.7589 0.7597 0.7604 0.7612 0.7619 0.7627
5.8 0.7634 0.7642 0.7649 0.7657 0.7664 0.7672 0.7679 0.7686 0.7694 0.7701
5.9 0.7709 0.7716 0.7723 0.7731 0.7738 0.7745 0.7752 0.7760 0.7767 0.7774

6.0 0.7782 0.7789 0.7796 0.7803 0.7810 0.7818 0.7825 0.7832 0.7839 0.7846
6.1 0.7853 0.7860 0.7868 0.7875 0.7882 0.7889 0.7896 0.7903 0.7910 0.7917
6.2 0.7924 0.7931 0.7938 0.7945 0.7952 0.7959 0.7966 0.7973 0.7980 0.7987
6.3 0.7993 0.8000 0.8007 0.8014 0.8021 0.8028 0.8035 0.8041 0.8048 0.8055
6.4 0.8062 0.8069 0.8075 0.8082 0.8089 0.8096 0.8102 0.8109 0.8116 0.8122

6.5 0.8129 0.8136 0.8142 0.8149 0.8156 0.8162 0.8169 0.8176 0.8182 0.8189
6.6 0.8195 0.8202 0.8209 0.8215 0.8222 0.8228 0.8235 0.8241 0.8248 0.8254
6.7 0.8261 0.8267 0.8274 0.8280 0.8287 0.8293 0.8299 0.8306 0.8312 0.8319
6.8 0.8325 0.8331 0.8338 0.8344 0.8351 0.8357 0.8363 0.8370 0.8376 0.8382
6.9 0.8388 0.8395 0.8401 0.8407 0.8414 0.8420 0.8426 0.8432 0.8439 0.8445

7.0 0.8451 0.8457 0.8463 0.8470 0.8476 0.8482 0.8488 0.8494 0.8500 0.8506
7.1 0.8513 0.8519 0.8525 0.8531 0.8537 0.8543 0.8549 0.8555 0.8561 0.8567
7.2 0.8573 0.8579 0.8585 0.8591 0.8597 0.8603 0.8609 0.8615 0.8621 0.8627
7.3 0.8633 0.8639 0.8645 0.8651 0.8657 0.8663 0.8669 0.8675 0.8681 0.8686
7.4 0.8692 0.8698 0.8704 0.8710 0.8716 0.8722 0.8727 0.8733 0.8739 0.8745

7.5 0.8751 0.8756 0.8762 0.8768 0.8774 0.8779 0.8785 0.8791 0.8797 0.8802
7.6 0.8808 0.8814 0.8820 0.8825 0.8831 0.8837 0.8842 0.8848 0.8854 0.8859
7.7 0.8865 0.8871 0.8876 0.8882 0.8887 0.8893 0.8899 0.8904 0.8910 0.8915
7.8 0.8921 0.8927 0.8932 0.8938 0.8943 0.8949 0.8954 0.8960 0.8965 0.8971
7.9 0.8976 0.8982 0.8987 0.8993 0.8998 0.9004 0.9009 0.9015 0.9020 0.9025

8.0 0.9031 0.9036 0.9042 0.9047 0.9053 0.9058 0.9063 0.9069 0.9074 0.9079
8.1 0.9085 0.9090 0.9096 0.9101 0.9106 0.9112 0.9117 0.9122 0.9128 0.9133
8.2 0.9138 0.9143 0.9149 0.9154 0.9159 0.9165 0.9170 0.9175 0.9180 0.9186
8.3 0.9191 0.9196 0.9201 0.9206 0.9212 0.9217 0.9222 0.9227 0.9232 0.9238
8.4 0.9243 0.9248 0.9253 0.9258 0.9263 0.9269 0.9274 0.9279 0.9284 0.9289

8.5 0.9294 0.9299 0.9304 0.9309 0.9315 0.9320 0.9325 0.9330 0.9335 0.9340
8.6 0.9345 0.9350 0.9355 0.9360 0.9365 0.9370 0.9375 0.9380 0.9385 0.9390
8.7 0.9395 0.9400 0.9405 0.9410 0.9415 0.9420 0.9425 0.9430 0.9435 0.9440
8.8 0.9445 0.9450 0.9455 0.9460 0.9465 0.9469 0.9474 0.9479 0.9484 0.9489
8.9 0.9494 0.9499 0.9504 0.9509 0.9513 0.9518 0.9523 0.9528 0.9533 0.9538

9.0 0.9542 0.9547 0.9552 0.9557 0.9562 0.9566 0.9571 0.9576 0.9581 0.9586
9.1 0.9590 0.9595 0.9600 0.9605 0.9609 0.9614 0.9619 0.9624 0.9628 0.9633
9.2 0.9638 0.9643 0.9647 0.9652 0.9657 0.9661 0.9666 0.9671 0.9675 0.9680
9.3 0.9685 0.9689 0.9694 0.9699 0.9703 0.9708 0.9713 0.9717 0.9722 0.9727
9.4 0.9731 0.9736 0.9741 0.9745 0.9750 0.9754 0.9759 0.9763 0.9768 0.9773

9.5 0.9777 0.9782 0.9786 0.9791 0.9795 0.9800 0.9805 0.9809 0.9814 0.9818
9.6 0.9823 0.9827 0.9832 0.9836 0.9841 0.9845 0.9850 0.9854 0.9859 0.9863
9.7 0.9868 0.9872 0.9877 0.9881 0.9886 0.9890 0.9894 0.9899 0.9903 0.9908
9.8 0.9912 0.9917 0.9921 0.9926 0.9930 0.9934 0.9939 0.9943 0.9948 0.9952
9.9 0.9956 0.9961 0.9965 0.9969 0.9974 0.9978 0.9983 0.9987 0.9991 0.9996

303