A Text Book of MATHEMATICS for Class IX

Kripa Anna Saj B.Ed. Mathematics

Contents

1

CIRCLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1

Introduction

7

1.2

Circles and its construction

8

1.3

Terms related to circle

8

1.4

Angle Subtended by a Chord at a Point

10

1.5

Perpendicular from the Centre to a Chord

13

2

CIRCLE MEASURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.1

Introduction

15

2.2

Perimeter of a circle

16

2.3

Area of circle

17

2.4

Area of a Sector

20

Preface

Mathematics has become a fundamental element of contemporary life and an indispensable tool in most spheres of the thought apart from its dominant role in the field of science and technology. Mathematics has been an inseparable part of school education since the beginning of formal education and it has played a predominant role not only in the advancement of civilizations but also in the development of physical sciences and other disciplines. The book has been based on the SCERT mathematics syllabus of Class 9. This textbook has been designed primarily as a textbook for the students of Class 9. This textbook has been written in a lucid style and simple language. This book contains detailed explanations of concepts. I hope the readers would get as much pleasure in reading the textbook as we have got in writing the same. Kripa Anna Saj

1. CIRCLES

1.1

Introduction

In our environment, you find many things that are round, a wheel, a bangle, a coin, buttons of shirts, etc. In a clock, you might have observed that the second’s hand goes round the dial of the clock rapidly and its tip moves in a round path. This path traced by the tip of the second’s hand is called a circle. A circle is a simple closed curve which is not a polygon. It has some very special properties. In this chapter, you will study about circles, other related terms and some properties of a circle.

Figure 1.1: Examples of Circles in real life

Chapter 1. CIRCLES

8

1.2

Circles and its construction Take a compass and fix a pencil in it. Put its pointed leg on a point on a sheet of a paper. Open the other leg to some distance. Keeping the pointed leg on the same point, rotate the other leg through one revolution. What is the closed figure traced by the pencil on paper? As you know, it is a circle (see Figure 1.2). How did you get a circle? You kept one point fixed and drew all the points that were at a fixed distance from the fixed point.

Figure 1.2: From this construction we can define circle as: The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle.

1.3

Terms related to circle A circle divides the plane on which it lies into three parts. They are: 1. inside the circle, which is also called the interior of the circle; 2. the circle and 3. outside the circle, which is also called theexterior of the circle. The circle and its interior make up the circular region. If you take two points P andQ on a

1.3 Terms related to circle

9

circle, then the line segment PQ is called a chord of the circle. The chord, which passes through the centre of the circle, is called a diameter of the circle. A diameter is the longest chord and all diameters have the same length, which is equal to two times the radius A piece of a circle between two points is called an arc.

The longer one is called the major arc PQ and the shorter one is called the minor arc c and the major arc PQ by PRQ d , where R is PQ. The minor arcPQ is also denoted by PQ some point on the arc betweenP and Q. WhenP and Q are ends of a diameter, then both arcs are equal and each is called a semicircle The length of the complete circle is called itscircumference. The region between a chord and either of its arcs is called a segment of the circle. There are two types of segments also, which are the major segment and the minor segment The region between an arc and the two radii, joining the centre to the end points of the arc is called asector

Chapter 1. CIRCLES

10 Exercise 1.1

1. Fill in the blanks: (a) The centre of a circle lies in

of the circle. (exterior/ interior)

(b) The longest chord of a circle is a (c) An arc is a

of the circle.

when its ends are the ends of a diameter.

(d) Segment of a circle is the region between an arc and

of the

circle. 2. Write True or False (a) Line segment joining the centre to any point on the circle is a radius of the circle. (b) A circle has only finite number of equal chords. (c) A chord of a circle, which is twice as long as its radius, is a diameter of the circle. (d) Sector is the region between the chord and its corresponding arc. 

1.4

Angle Subtended by a Chord at a Point Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR. Then ∠PRQ is called the angle subtended by the line segment PQ at the pointR. In ∠POQ is the angle subtended by the chord PQ at the centre O, ∠PRQ and ∠PSQ are respectively the angles subtended by PQ at points R and S on the major and minor arcs PQ respectively. Let us look at the relationship between the size of the chord and the angle subtended by it

at the centre. For this you may draw different chords of a circle and angles subtended by them at the centre. You will see that the longer is the chord, the bigger will be the angle

1.4 Angle Subtended by a Chord at a Point

11

subtended by it at the centre. What will happen if you take two equal chords of a circle? Will the angles subtended at the centre be the same or not? Draw two or more equal chords of a circle and measure the angles subtended by them at the centre. You will find that the angles subtended by them at the centre are equal. We can also be verify this by considering two equal chords AB and CD of a circle with centre O We want to prove that ∠AOB = ∠COD. For that consider the triangles, 4AOB

and4COD. In these triangles, OA, OC, OB, OD are Radii of a circle. Therefore, this two triangle, 4AOB and 4COD are equal. This gives , ∠AOB =∠COD. Thus our general conclusion can be put like the following: Equal chords of a circle subtend equal angles at the centre. Now if two chords of a circle subtend equal angles at the centre, what can you say about the chords? Are they equal or not? Let us examine it: Take a tracing paper and trace a circle on it. Cut it along the circle to get a disc. At its centre O, draw an angle AOB where A, B are points on the circle. Make another angle POQ at the centre equal to ∠AOB. Cut the disc along AB and PQ (see Figure 1.3). You will get two segments ACB and PRQ of the circle. If you put one on the other, what do you observe? They cover each other, so we can say that they are equal. So AB = PQ. Repeat this for other equal angles too. You will observe that the chords turn out to be equal for any measure of angles. You can also be verify this by considering ∠AOB and ∠COD are two equal angles subtended by the chords AB and CD of a circle at its centre O.(see Figure 1.3 ).

Chapter 1. CIRCLES

12

Figure 1.3:

We want to prove that AB = CD. In triangles 4AOB and 4COD,

OA = OC (Radii of a circle) OB = OD (Radii of a circle)

Also we have

∠AOB = ∠COD

Therefore, this two triangle,4AOB and 4COD are equal. This gives AB = CD. Thus our general conclusion it as follows: If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal. Recall that two circles are equal if they have same radii. Exercise 1.2

1. Prove that equal chords of equal circles subtend equal angles at their centres. 2. Prove that if chords of equal circles subtend equal angles at their centres, then the chords are equal. 

1.5 Perpendicular from the Centre to a Chord

1.5

13

Perpendicular from the Centre to a Chord Draw a circle on a tracing paper. Let O be its centre. Draw a chord AB. Fold the paper along a line through O so that a portion of the chord falls on the other. Let the crease cut AB at the point M. Then, ∠OMA = ∠OMB = 90° or OM is perpendicular to AB. Does the point B coincide with A (see Figure 1.4)? Yes it will. So MA = MB. Give a proof

Figure 1.4: yourself by joiningOAandOB and proving the right triangles OMA and OMB to be equal. This example is a particular instance of the following result: The perpendicular from the centre of a circle to a chord bisects the chord. What will be the converse of this? Let us examine it through following :

Figure 1.5:

Chapter 1. CIRCLES

14

Consider AB be a chord of a circle with centre O and O is joined to the mid-point M of AB. You have to verify that OM is perpendicular to AB. For this, join OA and OB (see Figure 1.5). In triangles 4OAMand4OBM, OA = OB (Radii 0f a circle) OM = OM(Common) Also we have AM = BM Therefore, this two triangle, 4OAM and 4 are equal. This gives,∠AMO = ∠BMO. But, ∠AMO + ∠BMO = 180◦. This gives,∠AMO = ∠BMO = 90◦ Thus our general conclusion can be put like this The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Exercise 1.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? 2. Suppose you are given a circle. Give a construction to find its centre. 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. 

2. CIRCLE MEASURES

2.1

Introduction Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow, pizza, dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Figure). So, the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall begin our discussion with the concepts of perimeter or circumference and area of a circle and apply this knowledge in finding the areas of two special parts of a circular region known as sector and segment.

Chapter 2. CIRCLE MEASURES

16

Figure 2.1: Examples of objects having circular shape

2.2

Perimeter of a circle The distance covered by travelling once around a circle is its perimeter, usually called its circumference. The distance around a circular region is known as its circumference. The circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter π (read as ‘pi’). In other words, circumference =π diameter circumference = π × diameter = π × 2r

where r is the radius of the circle

= 2πr

Example 1 What is the circumference of a circle of diameter 10 cm (Take π = 3.14)?

Solution

Diameter of the circle, d = 10 cm Circumference of circle = πd = 3.14 × 10 = 31.4 cm

So, the circumference of the circle of diameter 10 cm is 31.4 cm. Example 2 The radius of a circular pipe is 10 cm. What length of a tape is required to

wrap once around the pipe (π = 3.14)? Solution Radius of the pipe, r = 10 cm

2.3 Area of circle

17

Length of tape required is equal to the circumference of the pipe. Circumference of the pipe = 2πr = 23.14 × 10 = 62.8 cm Therefore, length of the tape needed to wrap once around the pipe is 62.8 cm.

2.3

Area of circle Draw a circle and shade one half of the circle. Now fold the circle into eighths and cut along the folds. As in Figure 2.2(a). Arrange the separate pieces as shown, in Figure

Figure 2.2: 2.2(b) which is roughly a parallelogram. The more sectors we have, the nearer we reach an appropriate parallelogram. As done above if we divide the circle in 64 sectors, and arrange these sectors. It gives nearly a rectangle Figure 2.3. What is the breadth of this rectangle? The breadth of this

Figure 2.3:

Chapter 2. CIRCLE MEASURES

18

rectangle is the radius of the circle, r. As the whole circle is divided into 64 sectors and on each side we have 32 sectors, the length of the rectangle is the length of the 32 sectors, which is half of the circumference. Area of the circle = Area of rectangle formed = product of the adjacent sides of the rectangle = Half of circumference × radius 1 = ( × 2π × r) 2 = πr2

The area of the circle with radius r is = πr2 Example 3 Find the area of a circle of radius 30 cm. Solution We have,

Radius, r = 30 cm Area of the circle = πr2 = 3.14 × 302 = 2 cm2 Example 4 Diameter of a circular garden is 9.8 m. Find its area. Solution Diameter, d = 9.8 m. Therefore, radius r = 9.8 2 = 4.9 m

Area of the circle = πr2 22 × (4.9)2 m2 7 22 = × (4.9) × (4.9)m2 7

=

= 75.46m2

2.3 Area of circle

19

Example 5 The cost of fencing a circular field at the rate of 24 rupees per metre is 5280

rupees. The field is to be ploughed at the rate of ‘0.50 rupees per square meter. Find the cost of ploughing the field (Take π =

22 7 ).

5280 cost Solution Length of the fence in metres = Total Rate = 24 = 220

So, circumference of the field = 220 m Therefore, if r metres is the radius of the field, then

2πr = 220 2×

22 × r = 220 7 2207 = 35 r= 222

Hence the radius of the field is 35 m.

Therefore the area of the field = πr2 =

22 × 35 × 35m2 7

= 22 × 5 × 35m2 cost of ploughing 1 m2 of the field = 0.50 rupees So, total cost of ploughing the field = 22 × 5 × 35 × 0.50 rupees = 1925 rupees

Exercise 2.1

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. 2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles. 

Chapter 2. CIRCLE MEASURES

20

2.4

Area of a Sector The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle. The angle formed by the two radii is called a central angle. A sector with a central angle less than 180◦ is called a minor sector. A sector with a central angle greater than 180◦ is called a major sector.

Let us try to find some relations (or formulae) to calculate areas of sector. Let OAPB

be a sector of a circle with centre O and radius r (see figure 2.4). Let the degree measure of ∠AOB = θ . You know that area of a circle is πr2 . In a way, we can consider this circular region to be a sector forming an angle of 360◦ at the centre O. Now by applying the Unitary Method, we can arrive at the area of the sector OAPB as follows: When degree measure of the angle at the centre is 360, area of the sector = πr2 So, when the degree measure of the angle at the centre is 1, area of the sector =

πr2 360 .

2.4 Area of a Sector

21

Therefore, when the degree measure of the angle at the centre is q, area of the sector =

πr2 360

×θ =

θ 360

× πr2

Thus, we obtain the following relation for area of a sector of a circle: Area of the sector of central angle θ =

θ × πr2 , where r is the radius of the circle 360

Can we find the length of the arc APB corresponding to this sector? Yes. Again, by applying the Unitary Method and taking the whole length of the circle as 2πr, we can obtain the required length of the arc APB as = Length of an arc with central angle θ =

θ 360

× 2πr.

θ × 2πr. where r is the radius of the circle 360

Let do, Example 6 Find the area of the sector of a circle with radius 4 cm and of angle 30◦.

Also, find the area of the corresponding major sector Solution Given sector is OAPB

θ × πr2 360 30 = × 3.14 × 4 × 4cm2 360 12.56 2 = cm 3

Area of the sector =

= 4.19cm2

Chapter 2. CIRCLE MEASURES

22 Area of the corresponding major sector = πr2 ˘area of sector OAPB = (3.14 × 16˘4.19) cm2 = 46.05cm2 = 46.1 cm2

Exercise 2.2

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60◦. 2. Find the area of a quadrant of a circle whose circumference is 22 cm. 3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. 4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector.