Grade 9 Book 1 Unit 1 and 2

SARASAS AFFILIATED SCHOOLS
Mathematics Grade 9

All rights reserved.
No part of this publication may be reproduced, stored in a
retrieval system, or transmitted in any form or by any
means, electronic, mechanical, photocopying, recording or
otherwise, without the prior permission of the publishers.



Foreword:
By Dr. Chamrat Nongmak

When talking about bilingual education in private elementary and secondary schools in Thailand, discussion
invariably tends to focus on Sarasas Affiliated Schools, a large group of private schools that was formerly owned and
administered by the late Mr. Peboon Yongkamol. Mr. Peboon, a well-known and respected educator, was a leading
pioneer of the bilingual education concept here in Thailand. These schools provide a benchmark and a showcase for
bilingual education in Thailand. Many school proprietors, directors, and managers make it a point to visit Sarasas
Affiliated Schools seeking advice and guidance from a successful leader in the field of bilingual education.

Sarasas programs for teaching students both in English and in Thai have been carefully planned and developed
step by step. One of the toughest challenges Sarasas has had to face has been finding the necessary texts and materials
for teaching Thai subjects in English. This is because a bilingual school, like any other Thai school, needs government
approval to operate and must teach the curriculum mandated by the Ministry of Education. Unfortunately, the
materials needed for teaching Thai subjects in English are not readily available. Publishers do little to support this need
because the concept is still fairly new and the demand is small. This has forced Sarasas to create its own texts and
materials.

Sarasas has been more fortunate though, than most other schools struggling to develop their own bilingual
program. It already has bilingual education experience, a qualified staff and the resources necessary to produce its own
high-quality textbooks and materials. The staff, of course, is the most important ingredient. Sarasas’ large, well-
seasoned staff of highly competent, well qualified and dedicated teaching professionals, both Thai and native speakers
of English, gives it a unique advantage over others attempting to produce these much-needed materials.

Even so, it has not been an easy task. The materials have been produced through the careful process of
research and development. Revision or total change is done when those materials are proven to be inadequate. The
people involved in developing these textbooks or workbooks are classroom teachers. They are the ones who will have
to teach with the same materials they have helped to create. Knowing this gives them a greater incentive to do it right.

The teams of classroom teachers responsible for classroom material development operate under the close
supervision of the Sarasas Board of Executives.

One of the roles of the board is to assure proper quality control by collecting and carefully evaluating the actual
work done by students using the new materials. They make certain that the students are able to understand the
material and use them effectively. One of the things they check for, is whether the level of English used in the books is
appropriate for the students’ level of English. The board also makes certain that the content of the material is in
accordance with the approved syllabus. The Mathematics book has been developed through this process.

This book is not only useful for teachers but also for parents and guardians. Its purpose is to encourage and
support students in their quest for knowledge and understanding of Mathematics, as well as stimulate their interest
and improve their skills. Each unit provides an introduction to the material, a list of new words the student will
encounter, problem solving examples to help the student better understand how to do the work, oral exercises, and a
review of what was learned.

This new Maths material will provide students with a greater international perspective and understanding of
Mathematics and Mathematics terminology which will benefit them greatly in later life. This said, I would like to take
this opportunity to thank those responsible for developing this material and offer my sincere congratulations for a job
well done.

Chamrat Nongmak, PhD.

Former Deputy Permanent Secretary
The Ministry of Education

PREFACE

This Mathematics textbook has been developed for bilingual learners based on
the B.E. 2560 (A.D. 2017) revised version of Thailand’s Basic Education Core
Curriculum B.E. 2551 (A.D. 2008). The content of this book follows the Basic
Mathematics for Grade 9 Book written by The Institute of Academic Development.

This textbook aims to enhance students’ abilities needed in the 21st century.
These include analytical skills, problem solving skills, creativity and collaboration
skills.

At the beginning of each unit of this textbook, a useful vocabulary list is
provided to learners. Examples demonstrate to students how to apply the
information they have learnt to solve related problems. Exercises and activities follow
that allow for immediate practice. At the end of each unit, a summary of the key
learning concepts along with a revision exercise is provided to help students
consolidate what they have learnt.

Author,
Sarasas Affiliated Schools

March, 2022

TABLE OF CONTENTS Week 1 to 4

Unit 1 : Surface area and Volume…………..………..………….…………. 11 Week 5 to 7

1. Three-Dimensional Geometric Figures………..….. 22 Week 8 to 10
2. Surface Area and Volume of a Pyramid……….….. 44
3. Surface Area and Volume of a Cone………….…….. 1111
4. Surface Area and Volume of a Sphere…………….. 15
Summary…………………………………………………………….. 21
Revision Exercise ………………………………………………… 22

Unit 2 : System of Linear Equation……………………..…………………… 26

1. Review about System of Linear Equations ……… 27
Slope-intercept Form …………………………………… 28
The Coefficient in Slope-intercept Form………. 29
The y-intercept Form………………………….………… 29

2. System of Linear Equation .…………………………….. 34
By Graphing ………………………………………………… 35
By Substitution Method ……………………………… 42
By Elimination Method ……………………………….. 45

3. Problems Leading to System of Linear Equation 49
Summary…………………………………………………………….. 53
Revision Exercise …………………………………………........ 54

Unit 3 : Factoring Polynomials …………….………………………………….. 57

1. Review about Polynomial ……………………….……… 58
2. Review about Factoring Polynomial……….……….. 60

Factoring by GCF ……………………………………….. 61
Factoring by Grouping ……………………………….. 62
Factoring a General Trinomial ……………………. 63
3. Factoring Higher Degree Polynomials…….……….. 66
Factoring a Perfect Square Trinomial …………. 67
Factoring a Difference of Two Squares……….. 68
Factoring a Sum or Difference of Two cubes 69
Factoring Using Synthetic Division ……………… 70

Summary…………………………………………………………..… 75 Week 11
Revision Exercise …………………………..…………………… 76

Unit 4 : Quadratic Equation…………..………….….................... 79

1. Quadratic Equation.………………........................ 80 Week 12 to 15
2. Solving Quadratic Equation
Unit 5 :
By Factorization………………………………….. 81
By Completing the Square……………........ 85
By Using Quadratic Formula……………….. 90
3. Problems Leading to Quadratic Equations……. 94
Summary…………………………………….…............... 98
Revision Exercise.……….……………………………….. 99

Quadratic Function.…………..…………........................ 101

1. The Quadratic Function.………………................. 102 Week 16 to 19
2. The Graph of Quadratic Function

Axis of Symmetry..……………………………... 112
Maximum and Minimum Values………..… 113
3. Problems Involving Maximum and Minimum..116
Summary…………………………………….………...........118
Revision Exercise.…………………………..………..……119





Surface Area and Volume of Solid Figures

Mathematical Terms

Center Point จุดศนู ยก์ ลาง Pyramid พรี ะมดิ

Curved surface ผวิ โคง้ Radius/ Radii รัศมี

Three – dimensional ปริภมู ิ 3 มิติ Regular polygon รูปหลายเหล่ียม

Cylinder ทรงกระบอก Similar คล้ายคลงึ

Equidistant ระยะห่างทเี่ ท่ากัน Slant height สูงเอยี ง

Hemisphere ครง่ึ วงกลม Sphere ทรงกลม

Hexagon หกเหลยี่ ม Tetrahedron พรี ะมดิ ฐานสามเหลี่ยม

Parallelograms สเี่ หลย่ี มด้านขนาน Total Surface Area พนื้ ทผี่ ิวทงั้ หมด

Perimeter ความยาวรอบรูป Vertex จดุ ยอดมุม

Prism ปริซึม Volume ปริมาตร

Did you know?

From the beginning of the Dynastic Era (2950 B.C.), royal tombs were
carved into rock and covered with flat-roofed rectangular structures
known as “mastabas,” which were precursors to the pyramids.
The oldest known pyramid in Egypt was built
around 2630 B.C. at Saqqara, for the third dynasty's King Djoser.

Mathematics Book 1 1

1. Three-dimensional Geometric Figures

.

1.1 Review about three-dimensional geometric figures.

In geometry, a three-dimensional figure (3D shape) can be defined as a solid
figure or an object or shape that has three dimensions – length, width and height.
Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.

Below are some examples of 3D shapes.

Prism is a solid geometric figure whose two end faces are similar, equal, and
parallel rectilinear figures, and whose sides are parallelograms.

Triangular prism Rectangular prism

Height Height

width length

Pentagonal prism Hexagonal prism
prism prism
prism vertex

lateral edge side side
base edge

base

A cylinder has two flat ends in the shape of circles. These two faces are
connected by a curved face that looks like a tube. If you make a flat net for a cylinder,
it looks like a rectangle with a circle attached at each end.

height height
base
base

2 Mathematics Book 1

A pyramid is a polyhedron for which the base is a polygon and all lateral faces
are triangles. A pyramid is typically described by the shape of its base. For instance, a
triangular pyramid has a base that is a triangle, and a hexagonal pyramid has a base
that is a hexagon.

apex

Square Pentagonal Tetrahedron
pyramid pyramid

height

base

A cone is a three-dimensional geometric shape that tapers smoothly from a
flat base to a point called the apex or vertex.

vertex radius
slant height
radius
height
vertex

A sphere is a round solid figure, with every point on its surface equidistant
from its center.

center point radius

curved surface

Mathematics Book 1 3

2. Surface Area and Volume of a Pyramid

.2.1 Surface area of a pyramid

The lateral surface area of a regular pyramid is the sum of the areas of its
lateral faces. The total surface area of a regular pyramid is the sum of the areas of
its lateral faces and its base.

Study the figures below.

Vertical slant height Lateral
height slant edge surface

base Square pyramid Lateral base Lateral
surface surface

Lateral
surface

Note:

The formula to get the lateral surface area of any regular pyramid is . . =


where represents the perimeter of the base and is the slant height.

Example 1

Find the lateral surface area of a tetrahedron if each edge of the base measures
10 cm and the slant height is 7 cm.

Solution The perimeter of the base is the sum of the sides.

10 cm Perimeter ( ) = 3(10) = 30

L. S. A = 1 , = 30 , = 7
2

1
L. S. A = 2 (30)(8)

L. S. A = 1 (240) = 140
2

Answer: The L.S.A of a tetrahedron is 120 cm2.

4 Mathematics Book 1

Note:

The formula for the total surface area of a regular pyramid is . . = +


where represents the perimeter of the base, is the slant height and is the

area of the base.

Example 2

Find the total surface area of a regular pyramid with a square base, if each
edge of the base measures 20 cm, the slant height of a side is 22 cm and the vertical
height is 18 centimeters.

Solution

18 cm

The perimeter of the base is 4s since it is a square
Perimeter ( ) = 4(20) = 80

The area of the base, B = s2 = 202 = 400cm2

T. S. A = B + 1 pl , p = 80cm, l = 22cm, B = 400cm2
2

1
T. S. A = 400 + 2 (80)(22)

T. S. A = 400 + (880) = 1280cm2

Answer: The T.S.A of the pyramid is 1280 cm2.

Mathematics Book 1 5

Note:
There is no formula for a surface area of a non-regular pyramid since slant height
is not defined. To find the area, find the area of each face and the area of the base
and add them.

Example 3

Find the total surface area of a rectangular pyramid if its base is 6 cm by 8 cm
and its slant heights are 5.7 cm and 5 cm.

Solution

5 cm

5.7 cm

6 cm

The area of the base,

A = 8 × 6 = 48 cm2

The area of the two triangular faces from the front and back views,

A = 1 × 6 × 5.7) = 34.2 cm2
2 (2

The area of the two triangular faces from the side views,

A = 1 × 8 × 5) = 40 cm2
2 (2

Therefore, the total surface area,

T. S. A = 48 + 34.2 + 40

= .

6 Mathematics Book 1

Example 4

Find the length of the slant edge of a tetrahedron if its base is an equilateral
triangle of side 10 cm and its slant height is 12 cm.

Solution

V

12 cm

c
B

x

A

VABC is a tetrahedron, triangle ABC is the base and VX is the slant height, VA,
VB and VC are slant edges.

Triangle VAB is an isosceles triangle. The right-angled triangle VAX equals to
the right-angled triangle VBX.

In this case we can use the Pythagorean formula to solve for the slant edge of
this tetrahedron.

Given: In a triangle VAX: AX = 10 = 5 cm,
2

VX = 12c m

Pythagorean formula: c2 = a2 + b2, = VA, = AX, = VX

∴ VX2 = AX2 + VX2

= 52 + 122

= 25 + 144

= √169 = 13

∴ The slant edge of this pyramid is 13 centimeters.

Mathematics Book 1 7

Exercise 1A

1. For each of the following solids, calculate the total surface area (T.S.A).

a. 15 cm ________________________________________________________

________________________________________________________

________________________________________________________

________________________________________________________

Regular hexagon

b. 16 cm ________________________________________________________

________________________________________________________
________________________________________________________
8 cm ________________________________________________________

2. A right pyramid of height 9 cm stands on a square base on a side of 8 cm.
Find the following and round off your answer to the nearest hundredths.
a. the slant height

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

b. the slant edge

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
8 Mathematics Book 1

2.2 Volume of a pyramid

The volume of a pyramid is one-third the area of the base ( )times the
height ( ) .

Volume of the pyramid: = 1 ℎ, where is the base area and is the height.
3

Example 5

Find the volume of a rectangular pyramid if its base is 12 cm by 5 cm and its
height is 13 cm.

Solution

13 cm Volume of Pyramid = 1 ℎ
3

= 1 × (12 × 5) × 13

3

= 260 3

Example 6

Find the height of a tetrahedron if its volume is 96 cm3 and its base area is 32 cm2.

Solution

Volume of pyramid = 1 Bh Given: V = 96 cm3, A = 32 cm3
3

Unknown: height (ℎ)

96 = 1 (32)h

3

32h = 288

h = 288 = 9, ∴ The height of the tetrahedron is 9 cm.
32

Note:

The volume of a three-dimensional figure is the amount of space it
occupies. Volume is measured in cubic units (in3 ft3 cm3 m3). Be sure that all the
measurements are in the same units before computing the volume.

Mathematics Book 1 9

Exercise 1B

1. Calculate the volume of the following pyramids:

____________________________________________________

a.

____________________________________________________

____________________________________________________

5 cm

____________________________________________________

10 cm ____________________________________________________
Square pyramid ____________________________________________________
____________________________________________________
b. 10 cm

____________________________________________________

____________________________________________________

____________________________________________________

Regular Tetrahedron

2. The base of a right pyramid is an equilateral triangle on a side of 10 cm,
and its vertical height is 5 cm. find the area of the base and the volume
of this pyramid.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

3. Find the volume of a pyramid with 15 cm high, standing on a square base
whose side is 16 cm.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
10 Mathematics Book 1

3. Surface Area and Volume of a Cone

.

3.1 Surface area of a cone

The total surface area of a cone is the sum of the area of its base and the
curved (side) surface. The curved surface area of a cone is the area of the lateral or
side surface only.

Curved surface

Base

Since a cone is closely related to a pyramid , the formulas for their surface
areas are related.

Slant height Vertical height
Curved surface
Base Slant height
Base

We have learned that the formulas for the curved surface area of a pyramid is


and the total surface area is + . Since the base of a cone is a circle, we

substitute for and for where r is the radius of the base of the cylinder.

. . =



Let is equal to and is the slant height.

C. . = =



. . = +

So, the formula for the curved surface area of a right cone is . . = ,
where is the radius of the circle and is the slant height of the cone.

The formula for the total surface area of a right cone is . . = +

Mathematics Book 1 11

Example 7

Find the total surface area of a solid cone of the base radius 7 cm and the slant
height 15 cm. (take π = )

Solution

Given: π = , r = 7, = 15,



Unknown: . .

Formula: T. S. A of a cone = πrl + πr2

T. S. A of a cone = ( ) (7)(15) + ( )(7)(7) 7 cm



= (22)(15) + (22)(7)

= 330 + 154
. . =

Example 8

Find the base area of a cone with the curved surface area of 44 2and the

slant height of 3.5 . ( Take π = and write your answer in two decimal places)


Solution

Given: . . = 44 2, = 3.5 cm
Unknown: Base area (π 2)

We have to solve the radius first to find the base area of a cone.

Formula: L. S. A = πrl

44 = (22)(r)(3.5)

7

r = 424 × 1 × 7

3.5 22

2×7
r = 3.5 = 4

Formula: Base area = π 2

= 22 × 4 × 4 r
7
= 50.29 2

12 Mathematics Book 1

Exercise 1C

1. From the given cones, calculate the curved and the total surface area.

a. Take π = ____________________________________________________
____________________________________________________

____________________________________________________

____________________________________________________

7 cm ____________________________________________________

b. Take π = 3.14, round off your answer to the two decimal places.

____________________________________________________
____________________________________________________
____________________________________________________
____________________________________________________
5 cm ____________________________________________________

2. Find the curved surface area of a right cone if the radius is 44 cm and
the slant height is 55 cm.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

3. Find the total surface area of a right cone if the radius
is 66 inches and the slant height is 10 inches.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

Mathematics Book 1 13

3.2 Volume of a cone

A cone is a three-dimensional figure with one circular base. A curved surface
connects the base and the vertex. We have learned that a cone may be considered as
the result of a sequence of pyramids with an increasing number of faces.

We can say that a cone is a special pyramid with a circular base. We can also
apply the formula for the volume of a pyramid to a cone.

The volume of a cone with radius is one-third the area of the base times
the height ℎ .

= 1 ℎ, where =π 2

3

= , where is the radius and is the height of a cone.



The volumes of a cone and a cylinder are related in the same way as
the volumes of a pyramid and a prism are related. If the heights of a cone and a
cylinder are equal, then the volume of the cylinder is three times as much as the
volume of a cone.

We learned that the = π 2ℎ

Then the = 1π 2ℎ

3

= 1 × volume of a cylinder

3

∴ = ×

radius radius radius radius

height = + +

height height height

14 Mathematics Book 1

Example 9

Find the volume of a solid cone of the base radius 7 cm and height 27 cm.
(Take π = ).



Solution

27 cm Given: = 7, ℎ = 27
7 cm Unknown: Volume of a cone
Formula:

= 1π 2

3

= 1× × 72× 27

3

∴ = ,

Example 10

Find the height of a cone of base radius 21 cm and volume of 3,675 π 3.

Solution

Given: = 21, = 3,675 π 3
Unknown: height of a cone
Formula:

= 1π 2

3

3,675π = 1× π × 21× 21 × ℎ
3

3,675π = 441πℎ

3

3,675π × 3 = 441πℎ
ℎ = 11025π

441π

ℎ = 25

∴

Mathematics Book 1 15

Note:
Volume of a cone is one-third of the volume of a cylinder of the same radius

and the same height.

Exercise 1D

1. Taking π = 272, find the volume of a cone to the nearest cubic centimeter of
the cone in which:
a. radius = 7 cm, slant height = 7.4 cm

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

b. radius = 1.4 cm, vertical height = 4.2 cm

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

2. Find the height of each of the following cones:
a. Radius = 7 m and volume = 392 π 3

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

b. Diameter = 8.4 cm and volume = 23.52 π 3

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

3. Water flows at the rate of 50 3per minute. How long would it take to fill
a conical vessel whose diameter at the surface is 42 cm and depth of 25 cm?

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
16 Mathematics Book 1

4. Surface Area and Volume of a sphere

.

4.1 Surface area of a sphere

A sphere is a perfectly round geometrical 3-dimensional object. It can be
characterized as the set of all points located distance (radius) away from a given
point (center). It is perfectly symmetrical, and has no edges or vertices.

center rr
curved surface

The Greek mathematician Archimedes 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 a height as the length of the diameter of the sphere.

In the diagram, a cylinder is circumscribed about a sphere.

r

r
2r

The curved surface area of a cylinder is equal to 2πrh, where ℎ = 2 .

Therefore,

ℎ =

= 2πr(2r) Mathematics Book 1 17

∴ =

4.2 Volume of a sphere r

The surface of the sphere may be divided into
many small portions of area, as indicated in the diagram.
If these portions diminished indefinitely, each one tends
to become ultimately plane, and may be considered the
base of a pyramid, whose vertex is the center, and whose
height is the radius of the sphere.

The volume of any such pyramid,

= 1(the portion of surface) × r

3

Since the sum of all the portions of surface is the whole surface of the sphere,
and the sum of all the corresponding pyramids is the whole volume of the sphere.

Therefore,

Volume of sphere = 1(surface of sphere) × r
3
= 1× 4π 2 × r
3
= 4π 3
3
∴ 4π 3
3

We can also find the volume of a sphere in relating with the volume of a
cylinder.

Volume of sphere = 2 × volume of cylinder

3

r = 2 × π 2h
3

2r = h = 2 × π 2(2r)

3

= 4 × π 3

3

Volume of sphere = 4 π 3 ,where r is the radius.
3

The volume of a sphere is 2 of the volume of a cylinder with the same radius,

3

and height equal to the diameter.

18 Mathematics Book 1

Example 9

Find to the nearest square centimeter the surface areas and their volumes to
the nearest cubic centimeter of the spheres whose radii are:

a. 2.1 cm b. 10.5 cm

Solution

a. The surface area

Given: = 2.1 π = 22,

7

Unknown: surface area

Formula: Surface area = 4π 2
= 4(22)(02..31)(2.1)

7

= 55.44

= 55 2

The volume

Given: = 2.1 π = 272,

Unknown: volume

Volume = 4 π 3
3
= 4 × 22 × 0.3 × 2.1 × 2.1

37 2.1

= 38.808

= 39 3 (correct to the nearest 3)

b. The surface area 19
Given: = 10.5 π = 272,
Unknown: surface area

Formula: Surface area = 4π 2

= 4 × 22 × 10.5 × 10.5

7

= 1,386 2

The volume
Given: = 10.5 π = 272,
Unknown: surface area

Formula: Surface area = 4π 2

= 4 × 22 × 10.5 × 10.5

7

= 4,851 3
Mathematics Book 1

Exercise 1E

1. Calculate the curved surface area and the volume of each of the following
spheres: (Taking π=3.14, correct to 2 decimal places)
a. radius = 1.4 cm

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

b. diameter = 14 cm

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

2. Find the radius and the surface area in terms of π of the following spheres whose

volume is given:
a. Volume = 288 π 3

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

b. Volume = 85 1π 3
3
________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

3. Find the whole surface and weight of a hemispherical copper bowl 12 cm in
external diameter and 1 cm in thickness. (Taking 1 cm3 of copper weight 8.88g)

__________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________
20 Mathematics Book 1

______________________

Summary

➢ Prism is a solid geometric figure whose two end faces are similar, equal,

and parallel rectilinear figures, and whose sides are parallelograms.

➢ A cone is a three-dimensional geometric shape that tapers smoothly from

a flat base to a point called the apex or vertex.
➢ A sphere is a round solid figure, with every point on its surface equidistant

from its center
➢ A cylinder has two flat ends in the shape of circles and these two faces are

connected by a curved face that looks like a tube.
➢ A pyramid is a polyhedron for which the base is a polygon and all lateral

faces are triangles.
➢ . . stands for Lateral Surface Area
➢ . . stands for Curved Surface Area
➢ . . stands for Total Surface Area
➢ The formula to get the lateral surface area of any regular pyramid is

. . = where represents the perimeter of the base and is the slant


height.
➢ The formula for the total surface area of a regular pyramid is . . = +



where represents the perimeter of the base, is the slant height and

is the area of the base.
➢ There is no formula for a surface area of a non-regular pyramid since slant

height is not defined. To find the area, find the area of each face and the

area of the base and add them.

➢ The formula for the total surface area of a right cone is . . = +

➢ The formula for the curved surface area of a right cone is . . = ,

where is the radius of the circle and is the slant height of the cone.
➢ =
➢ The volume of a three-dimensional figure is the amount of space it

occupies. Volume is measured in cubic units (in3 ft3 cm3 m3). Be sure that

all the measurements are in the same units before computing the volume.
➢ The volume of a pyramid is one-third the area of the base ( )times the

height ( ) .

➢ Volume of a cone is one-third of the volume of a cylinder of the same

radius and the same height. The formula of the cone’s volume is ( )

➢ = ×

➢ The volume of a sphere is


➢ Hemisphere is half of a sphere.

➢ Archimedes discovered the formula of the volume of a sphere.

Mathematics Book 1 21

Revision Exercise

Part 1. (Skills: 1-7)
(1-4) Calculate the total surface area and the volume of the following solids.
(Taking (π = 3.14 ), answer to the nearest hundredths of a centimeters.)

1. __________________________________________________
7.4 cm __________________________________________________
6.4 cm
__________________________________________________
__________________________________________________
__________________________________________________
__________________________________________________

8 cm2. __________________________________________________
__________________________________________________
7.4 cm __________________________________________________
__________________________________________________
__________________________________________________
__________________________________________________

3. __________________________________________________

__________________________________________________

__________________________________________________

10.4 cm __________________________________________________

__________________________________________________

__________________________________________________

4. __________________________________________________

9.5 cm __________________________________________________
11.2 cm

__________________________________________________

__________________________________________________

__________________________________________________

__________________________________________________

22 Mathematics Book 1

Calculate the missing value of the following solids.

5. Cones
a. Volume 33331 3, base area 100 2, height ______________ cm.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

b. Volume 6.6 3, height 0.7 cm, radius ____________________ cm.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

6. Square pyramids
a. Volume 108 3, height 9 cm, length of side of the base ________ cm.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

b. Volume 400 3, length of its base side 10 cm, height __________ cm.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

7. Spheres
a. Volume 36 π 3, radius ____________ cm.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

b. Surface area 36 π 3, volume _____________ 3.

________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

Mathematics Book 1 23

Part 2. Concepts. Circle the letter of the correct answer.

1. Which one is a tetrahedron? 6. Which solid figure is one-third the
volume of a cylinder of the same
a. c. radius and height?
a. cone
b. d. b. pyramid
c. cube
2. Which one can guide you to name d. sphere
the pyramid?
a. the slant edges 7. Which value do we need to know
b. the angle at its face in finding the surface or a sphere?
c. the shape of its base
d. the distance from the vertex a. its radius

3. What is the shape of each face of a b. its diameter
square pyramid?
a. scalene triangle c. its perimeter
b. isosceles triangle
c. equilateral triangle d. its circumference
d. right-angled triangle
8. If the circular base of a cone has a
4. It has two flat ends in the shape of
circles and these two faces are radius F, what is the base area of
connected by a curved face that
looks like a tube. this cone?
a. Cylinder a. πF2
b. Cone
c. Pyramid b. πF
d. Sphere c. πrF
d. πrF2
5. What is a hemisphere? 9. It is a solid geometric figure whose
a. triangular face
b. quarter circle two end faces are similar, equal,
c. half circle
d. horizontal corner and parallel rectilinear figures, and

whose sides are parallelograms.

a. tetrahedron

b. prism

c. sphere

d. cuboid

10.Who is the Greek mathematician

that discovered the surface area of

a sphere is the same as the

lateral surface area of a cylinder

with the same radius and height?

a. Einstien c. Cynodon

b. Newton d. Archimedes

24 Mathematics Book 1

Part 3. Word problems. Circle the letter of the correct answer.

1. A right pyramid of a height 8 cm 5. A cake decorator rolls a piece of stiff
stands on a regular hexagonal base paper to form a cone. She cuts off the
on a side of 6 cm. What is each slant tip of the cone and uses it as a funnel
surface to the nearest tenth of 1 to pour decorative sprinkles into
square centimeter? small containers. The cone has a
a. 28 cm2 radius of 6 cm and a height of 18 cm.
b. 48 cm2 What is the volume of the cone
c. 53.4 cm2 before the end is cut off?
d. 78.5 cm2 a. 2,034.72 cm3
b. 898.2 cm3
2. How many solid spheres 6 cm in c. 1,102.6 cm3
diameter could be molded from a d. 678.24 cm3
solid metal cylinder whose length is e. None of these
45 cm and diameter of 4 cm?
a. 5 6. What is the whole surface area of a
b. 6 cone with the same height and base
c. 36 of a cylinder whose height is 2.4 cm
d. 90 and diameter of 1.4 cm ?
a. 3.96 cm2
3. The gift shop at a science museum b. 5.77 cm2
sells lollipops that are made to look c. 6.23 cm2
like the planets. Each lollipop is d. 7.04 cm2
shaped like a sphere and has a e. None of these
radius of 12 mm. What is the
volume of each lollipop? 7. A sphere of radius 3 cm is dropped
a. 192π 3 into a cylindrical vessel partly filled
b. 2,304π 3 with water. The diameter of the
c. 508π 3 vessel is 10 cm. If the sphere is
d. 1,307π 3 completely submerged, by how
much will the surface of the water
4. Air is leaking from a spherical- be raised?
shaped advertising balloon at the a. 1.33 cm
rate of 26 cubic feet per minute. If b. 1.44 cm
the radius of the ball is 7 feet, how c. 2.54 cm
long would it take for the balloon to d. 3.49 cm
empty fully? e. None of these
a. 55 minutes
b. 60 minutes Mathematics Book 1 25
c. 1 hour and 5 minutes
d. 1 hour and 20 minutes

System of Linear Equations

Mathematical Terms

Coordinate พกิ ัด Protractor โปรแทรกเตอร์

Eliminate กาจัด Quantity ปรมิ าณ

Graphing กราฟ Remainder เศษ

Infinite solution Simplest form รูปอยา่ งง่าย

สมการทม่ี ีผลเฉลยไม่สนิ้ สุด Simultaneous equation

Intersection เส้นตัด สมการเชิงเสน้ สองตัวแปร

Linear equation สมการเชงิ เสน้ Slope intercept form

Numerator ตัวเศษ สมการเสน้ ตรงในรูปความชนั y = ax + b

Ordered pairs ค่อู ันดบั Solution วิธที า/การแกป้ ัญหา

Plot สร้าง Straight line เส้นตรง

Substitute แทนค่า, แทนท่ี

Parallel เสน้ ขนาน Suitable เหมาะสม

Did you know?

Linear equations were invented in 1843 by Irish mathematician
Sir William Rowan Hamilton. He was born in 1805 and died
in 1865. Through his algebraic theory, Sir Hamilton made
important contributions to mathematics, and his work found
applications in quantum mechanic.

26 Mathematics Book 1

1. Review of Linear Equations in Two Variables

.Linear equations in two variables are equations which can be expressed in

general form as Ax + By + C = 0, where A, B and C are real numbers and both A, and
B are not zero. The solution of such equations is a pair of values for x and y which
makes both sides of the equation equal.

Also, the equation is called linear if the variables are only to the first power,
are only in the numerator and there are no products of variables in any of the
equations.

Take a look at some examples below.

1. − 7 − 10 = 0, where A = 1, B = -7 and C = -10

2. + = 8, where A = 1, B = 1 and C = -8

3. = 3 − 4, where A = 3, B = -1 and C = -4

4. 2 + 4 = 6, where A = 2 , B = 4 and C = -6
3
3

5. 8 + 12 = 0, where A = 8, B = 0 and C = 12

Before we discuss further on how to solve the system of linear equations, we
should first talk about the solution to the system of equation. A solution to a system
of equations is a value of x and y written in an ordered pair (x, y) such that when we
substitute the values of x and y into the equation it will satisfy both the equations at
the same time.

Example 1

Find the solutions of the equation 2 + − 4 = 0

Solution

There are many solutions to this equation but in this case we will consider the
values of ordered pair (x,y) as written below.

Table of values:
x -2 -1 0 1 2 3
y 8 6 4 2 0 -2

Mathematics Book 1 27

From the table of values, the ordered pairs are (-2,8), (-1,6), (0,4), (1,2), (2,0)
and (3,-2). These ordered pairs are also the solutions of the equation 2 + − 4 = 0.
We can also show the solutions of the equation by using graphs. Take a look at the
graph below.

8
7

5 12345678
4
3 -1
2 -2
1 -3
-4
-8 -7 -6 -5 -4 -3 -2 -1 -5
-6
-7
-8
-9

From the graph, you see that it is a straight line with infinite solutions to the
equation 2 + − 4 = 0. The ordered pair (3, -2) is part of a straight line therefore
it is also the solution of the equation 2 + − 4 = 0. We can easily graph the
solutions of linear equation if we are familiar about the slope intercept form.

The slope-intercept form

Slope intercept is a specific form of linear equation. It is in general form as

= + , where m and b are any two real numbers.

These are linear equations in slope intercept form:

1. = 2 + 1 2. = 3 − 4 3. = 1 − 8
2

On the other hand, these linear equations are not in slope intercept form:

1. 2 + = 1 2. − 2 = 4( − 1) 3. = 4 − 2

Slope-intercept is the most prominent form of linear equation.

28 Mathematics Book 1

The coefficients in slope-intercept form

Besides being neat and simplified, slope-intercept form's advantage is that it gives
two main features of the line it represents:

1. The slope is (m).
2. The y-coordinate of the y-intercept is (b).

slope

= + y-intercept

For example, the line = 3 − 2 has a slope of 3 and the y-intercept of (0,-2).

4

Slope = 3 2 2 4
-4 -2
RiseRun

0

-2

-4 y-intercept
(0,-2)

Let us discuss deeper about the slope and y-intercept.

The y-intercept (b) on = +

At the y-intercept, the x-value is always zero. So, if we want to find the y-intercept
of = 3 − 2 , we should let x = 0, and solve for y.

Let x = 0, = 3 − 2

= 3(0) − 2

= −2

We see at y-intercept 3 becomes zero, therefore we are left with y = -2.

The slope (m) on = +

Slope is the ratio of the change in y over the change in x between any two points
on the line. It can be denoted as

= ℎ ( 2− 1)

ℎ ( 2− 1)

Mathematics Book 1 29

Now, let’s try to observe the values of y in the equation = 3 − 2 as the
values of x constantly increases by 1.
Let = 0, = 3(0) − 2 = −2

= 1, = 3(1) − 2 = 1
= 2, = 3(2) − 2 = 4
= 3, = 3(3) − 2 = 7
= 4, = 3(4) − 2 = 10

We see that each time x increases by 1 unit, y increases by 3 units. This is
because x determines the multiple of 3 in the calculation of y.

As stated above, the change in y that corresponds to x increasing by 3 units is
equal to the slope of the line. For this reason, the slope is 3.

Example 2

Find the slope and y-intercept of the following straight lines:

a. = 3 + 5 b. 3 − 2 + 2 = 0

Solution

a. Compare the equation = 3 + 5

with = +

We have the slope, m = 3

and y-intercept, b=5

b. Write the equation 3 − 2 + 2 = 0 in the form = +

We get 2 = 3 + 2

Compare with 2 = 3 +2
We have the slope
22
30 Mathematics Book 1
= 3 + 1

2

= +

= 11 and the y-intercept, = 1

2

Example 3

Write down the equation of the straight line whose

a. Slope is 2 and y-intercept is -2,

b. Slope is − 1 and y-intercept is zero.
2

Solution

a. Given: slope, = 2
y-intercept = −2

Substitute in the form of straight line is = + ,
We will get = 2 − 2,
b. Given: slope, = − 1,

2

y-intercept, = 0
Substitute in the form of straight line is = + ,
We will get = − 1

2

Example 4

Given the slope = −1 and the y-intercept = 5, find the following
a. The equation of the line
b. The next 5 points in the graph

Solution

a. Substitute the values of and in the slope form = + . 4
We get = −1 + 5 or + = 5 1

b. Since the slope form is = −1 + 5
We will substitute the values of x to get the values of y.
x0 1 2 3
y5 4 3 2

Therefore, the next 5 points in the graph given the slope is -1 and the
y-intercept 5 are (0,5), (1,4), (2,3), (3,2), (4,1).

Mathematics Book 1 31

Example 5

For each given straight line, find its equation in the form of = + .
a. b.

4 5
-10 -5 0
2 2
-2 0 5 10
-5
-1
-10

-15

Solution

a. Find its slope m and y-intercept c.

Choose any two points on the line, say (4,0) and (2,0).

Slope = ℎ ( 2− 1)

ℎ ( 2− 1)

= 0−4 = −4 = −2

2−0 2

From the graph, y-intercept, = 4

Therefore, the equation of the straight line is = −2 + 4

b. Find its slope m and y-intercept c.
Choose any two points on the line, say (10,0) and (0,-15).

Slope = ℎ ( 2− 1)

ℎ ( 2− 1)

= −15−0 = −15 = 3

0−10 −10 2

From the graph, y-intercept, = −15

Therefore, the equation of the straight line is = 3 − 15
2

32 Mathematics Book 1

Exercise 2A

1. Find the slope and y-intercept of the following straight lines:

a. 3 = + 2

b. + = 5

c. 4 + = 12

d. 5 − 2 = 6

e. + 3 − 5 = 0

2. Write down the equation of the straight line when slope and y-intercept

are shown in the table.

Slope y-intercept Answer

a. 2 -5

b. 1 0
2 7

c. -4

d. 3 1
−4
2
e. 2 1
−4
3

3. Draw on the same axes the graphs of
3 + 2 = 5, 2 + = 4, 4 = 2 − 5 .

Mathematics Book 1 33

2. System of Linear Equation

.A system of linear equations is a system made up of two linear equations. To

solve the system of equations, you need to find the exact values of x and y that will
satisfy both equations.

Consider the linear equation + = 5, which contains two unknowns.

There are many pairs of values x and y which will satisfy this equation,

In + = 5; = 1, = 4 gives 1 + 4 = 5

= 2, = 3 gives 2 + 3 = 5

= 5, = 0 gives 5 + 0 = 5

= 6, = −1 gives 6 + −1 = 5

= −2, = 7 gives −2 + 7 = 5

etc…

Now then linear equation − = 5 which also contains two unknowns.

There are many pairs of values x and y which will satisfy this equation,

In − = 5; = 6, = 1 gives 6 − 1 = 5

= 5, = 0 gives 5 − 0 = 5

= 1, = −4 gives 1 − (−4) = 5

= −2, = −7 gives (−2) − (−7) = 5

= 0, = −5 gives 0 − (−5) = 5

etc…

Sometimes we need to find the pair of values of x and y which will satisfy both
equations. We call this solving system of linear equation.

In the above examples, only x = 5 and y = 0 will satisfy the equations. When
two or more equations are satisfied by the same value of the unknowns, they are
called simultaneous equations. There are many methods of solving system of linear
equations. Some of algebraic method that we will study in this section are:

1. graphing
2. equating expressions method
3. substitution method
4. elimination method

34 Mathematics Book 1

2.1 Solving System of Linear Equation by Graphing

To solve a system of linear equations graphically we graph both equations in
the same coordinate system. The solution to the system will be in the point where
the two lines intersect.

Before you can graph a linear equation, you need to make sure that it is written
in slope-intercept form. The slope-intercept form of a linear equation is y = mx + b.
Here are the steps in graphing linear equations:

1. It must be in a slope intercept form.
2. Plot the y-intercept. The y-intercept is the point where the line crosses the

y axis.
3. Look at the slope. The slope is a ratio of how far the line goes up in

the y direction divided by how far it goes over in the x direction.
4. Use the slope to plot a second point and then use a ruler to connect the

points and make a straight line.

Example 6

Find the solution of the following equations by graphing:
+ = 4 and − = 4

Solution

First, we must write the equation in a slope-intercept form.
For + = 4
= − + 4, where, = −1 and = 4.
From the equation = − + 4 ,the y-intercept is at (0,12) and a slope of -1

means that you should go down 1 unit in the y direction for every 1 unit you go over
in the x direction.

For − = 4
= − 4, where, = 1 and = −4.

From the equation = − 4, the y-intercept is at (0,-4) and a slope of 1 means
that you should go up 1 unit in the y direction for every 1 unit you go over in
the x direction. Please see the graph on the next page to determine the solution of
systems of linear equation.

Mathematics Book 1 35

The graph of + = 4 and − = 4

y-intercept
(0,4)

Slope = -1

Slope = 1 Intersection (4,0)

y-intercept
(0,-4)

As you can see from the graph, the point where the two lines intersect is at
(4,0) .

Therefore, the solution to the system of equation is (4,0).

Example 7

Solve the following pair of simultaneous equations.
5 − = 6 and 2 + = 8

Solution

Convert into slope-intercept form:
For 5 − = 6, = 5 − 6 = 5, = −6
For 2 + = 8, = −2 + 8 = −2, = 8

8

-2 ---200000040000040600000000000000062000000000 (2,4) is the solution to the
system of linear equation.

-6 -4 246
000
36 Mathematics Book 1 000
000
000
000
000
000
000

Example 8

Find the solution to the system of linear equations = + 1, and = 2 − 2.

Solution

Since the two equations are already in slope-intercept form, we need to find
the intersections on the x and y axes.

Equations: = + 1 1

= 2 − 2. 2

Consider the intersection on the X-axis and the Y-axis to graph the two equations.

For equation = + 1

Find the intersection on the X-axis, substituting y = 0
= + 1
0 = + 1

−1 =
The intersection point at X-axis is (-1,0)
Find the intersection on the Y-axis substituting x = 0

= + 1
= 0 + 1
= 1
The intersection point at X-axis is (0,1)

For equation = 2 − 2 37

Find the intersection on the X-axis substituting y = 0
= 2 − 2
0 = 2 − 2
1 =

The intersection point at X-axis is (1,0)
Find the intersection on the Y-axis substituting x = 0

= 2 − 2
= 0 − 2
= −2
The intersection point at X-axis is (0,-2)
See the graph of these two equations on the next page.

Mathematics Book 1

Graph the = 2 − 2 and = + 1 on the same coordinate system as follows:

6

4 (3,4)

2

-8 -6 -4 -2 24 6
0

-2

= + 1 -4

-6

= 2 − 2

From the graph above, we see that the two lines intersect at point (3,4).
Therefore, the solution to the system of linear equation is (3,4).

A system of linear equations has infinite solutions when the graphs are the
exact same line.

Example 9

Solve the system of the equations + = 3 and 2 + 2 = 6.

Solution

For + = 3

Find the intersection on the X-axis, Find the intersection on the Y-axis,

substituting y = 0 substituting x = 0

+ = 3 = − + 3 + = 3 = − + 3

0 = − + 3 = −(0) + 3

3 = = 3

The intersection point at X-axis is (3,0) The intersection point at X-axis is (0,3)

For 2 + 2 = 6

Find the intersection on the X-axis, Find the intersection on the X-axis,

substituting y = 0 substituting y = 0

2 + 2 = 6 2 = −2 + 6 2 + 2 = 6 2 = −2 + 6

0 = −2 + 6 2 = −2(0) + 6

3 = = 3

The intersection point at X-axis is (3,0) The intersection point at X-axis is (0,3)

38 Mathematics Book 1

Graph the = 2 − 2 and = + 1 on the same coordinate system as follows:

6

4

2 2 + 2 = 6

-8 -6 -4 -2 24 6
0

-2 + = 3

-4

-6

The graph above shows that the two equations are on the same line. Therefore,
there are infinite solutions to the system of equations + = 3 and 2 + 2 = 6.

A system of linear equations has no solution when the graphs are parallel.

Example 9

Solve the system of the equations − 2 = 8 and 5 − 10 = 20.

Solution

For − 2 = 8

Find the intersection on the X-axis, Find the intersection on the Y-axis,

substituting y = 0 substituting x = 0

− 2 = 8 2 = − 8 − 2 = 8 2 = − 8

0 = − 8 2 = 0 − 8

8 = = −4

The intersection point at X-axis is (8,0) The intersection point at X-axis is (0,-4)

For 5 + 10 = 20

Find the intersection on the X-axis, Find the intersection on the X-axis,

substituting y = 0 substituting y = 0

5 − 10 = 20 10 = 5 − 20 5 − 10 = 20 10 = 5 − 20

0 = 5 − 20 10 = 0 − 20

4 = = −2

The intersection point at X-axis is (4,0) The intersection point at X-axis is (0,-2)

Mathematics Book 1 39

Graph the − 2 = 8 and 5 − 10 = 20 on the same coordinate system as follows:

6

4

2

-8 -6 -4 -2 24 6
0

5 − 10 = 20 -2 − 2 = 8
-4

-6

The graph above shows the two lines are parallel. Therefore, there is no
solution to the system of equations − 2 = 8 and 5 − 10 = 20.

Exercise 2B

1. Consider the graphs of a system of linear equations in two variables in each of
the following problems. Do they each have no solution, one solution, or
infinitely many solutions? Write your answer on the blank provided.

a. y b. y

xx

c. y d. y
x
x

40 Mathematics Book 1