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# NCERT Textbook- Areas Related to Circles Class 10 Notes | EduRev

## Class 10 : NCERT Textbook- Areas Related to Circles Class 10 Notes | EduRev

``` Page 1

AREAS RELA TED TO CIRCLES 223
12
12.1 Introduction
You are already familiar with some methods of finding perimeters and areas of simple
plane figures such as rectangles, squares, parallelograms, triangles and circles from
your earlier classes. 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 (thela),
dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular
paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).
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 a review of
the concepts of perimeter (circumference) and area of a circle and apply this knowledge
in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle)
known as sector and segment. We shall also see how to find the areas of some
combinations of plane figures involving circles or their parts.
Fig. 12.1
AREAS RELATED TO CIRCLES
Page 2

AREAS RELA TED TO CIRCLES 223
12
12.1 Introduction
You are already familiar with some methods of finding perimeters and areas of simple
plane figures such as rectangles, squares, parallelograms, triangles and circles from
your earlier classes. 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 (thela),
dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular
paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).
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 a review of
the concepts of perimeter (circumference) and area of a circle and apply this knowledge
in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle)
known as sector and segment. We shall also see how to find the areas of some
combinations of plane figures involving circles or their parts.
Fig. 12.1
AREAS RELATED TO CIRCLES
224 MATHEMA TICS
12.2 Perimeter and Area of a Circle — A Review
Recall that the distance covered by travelling once around a circle is its perimeter,
usually called its circumference. You also know from your earlier classes, that
circumference of a circle bears a constant ratio with its diameter. This constant ratio
is denoted by the Greek letter p (read as ‘pi’). In other words,
circumference
diameter
= p
or, circumference = p × diameter
= p × 2r (where r is the radius of the circle)
=2pr
The great Indian mathematician Aryabhatta (A.D. 476 – 550) gave an approximate
value of p. He stated that p =
62832
,
20000
which is nearly equal to 3.1416. It is also
interesting to note that using an identity of the great mathematical genius Srinivas
Ramanujan (1887–1920) of India, mathematicians have been able to calculate the
value of p correct to million places of decimals. As you know from Chapter 1 of
Class IX, p is an irrational number and its decimal expansion is non-terminating and
non-recurring (non-repeating). However, for practical purposes, we generally take
the value of p as
22
7
or 3.14, approximately.
You may also recall that area of a circle is pr
2
, where r is the radius of the circle.
Recall that you have verified it in Class VII, by cutting a circle into a number of
sectors and rearranging them as shown in Fig. 12.2.
Fig 12.2
Page 3

AREAS RELA TED TO CIRCLES 223
12
12.1 Introduction
You are already familiar with some methods of finding perimeters and areas of simple
plane figures such as rectangles, squares, parallelograms, triangles and circles from
your earlier classes. 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 (thela),
dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular
paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).
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 a review of
the concepts of perimeter (circumference) and area of a circle and apply this knowledge
in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle)
known as sector and segment. We shall also see how to find the areas of some
combinations of plane figures involving circles or their parts.
Fig. 12.1
AREAS RELATED TO CIRCLES
224 MATHEMA TICS
12.2 Perimeter and Area of a Circle — A Review
Recall that the distance covered by travelling once around a circle is its perimeter,
usually called its circumference. You also know from your earlier classes, that
circumference of a circle bears a constant ratio with its diameter. This constant ratio
is denoted by the Greek letter p (read as ‘pi’). In other words,
circumference
diameter
= p
or, circumference = p × diameter
= p × 2r (where r is the radius of the circle)
=2pr
The great Indian mathematician Aryabhatta (A.D. 476 – 550) gave an approximate
value of p. He stated that p =
62832
,
20000
which is nearly equal to 3.1416. It is also
interesting to note that using an identity of the great mathematical genius Srinivas
Ramanujan (1887–1920) of India, mathematicians have been able to calculate the
value of p correct to million places of decimals. As you know from Chapter 1 of
Class IX, p is an irrational number and its decimal expansion is non-terminating and
non-recurring (non-repeating). However, for practical purposes, we generally take
the value of p as
22
7
or 3.14, approximately.
You may also recall that area of a circle is pr
2
, where r is the radius of the circle.
Recall that you have verified it in Class VII, by cutting a circle into a number of
sectors and rearranging them as shown in Fig. 12.2.
Fig 12.2
AREAS RELA TED TO CIRCLES 225
Y ou can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length
1
2
2
r ×p
and breadth r. This suggests that the area of the circle =
1
2
× 2pr × r = pr
2
. Let us
recall the concepts learnt in earlier classes, through an example.
Example 1 : The cost of fencing a circular field at the rate of Rs 24 per metre is
Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m
2
. Find the cost of
ploughing the field (Take p =
22
7
).
Solution : Length of the fence (in metres) =
Total cost
Rate
=
5280
= 220
24
So, circumference of the field = 220 m
Therefore, if r metres is the radius of the field, then
2pr = 220
or, 2 ×
22
7
× r = 220
or, r =
220 × 7
2× 22
= 35
i.e., radius of the field is 35 m.
Therefore, area of the field = pr
2
=
2
7
2

× 35 × 35 m
2
= 22 × 5 × 35 m
2
Now, cost of ploughing 1 m
2
of the field = Rs 0.50
So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925
EXERCISE 12.1
Unless stated otherwise, use p =
22
7
.
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.
3. Fig. 12.3 depicts an archery target marked with its five
scoring areas from the centre outwards as Gold, Red, Blue,
Black and White. The diameter of the region representing
Gold score is 21 cm and each of the other bands is 10.5 cm
wide. Find the area of each of the five scoring regions.
Fig. 12.3
Page 4

AREAS RELA TED TO CIRCLES 223
12
12.1 Introduction
You are already familiar with some methods of finding perimeters and areas of simple
plane figures such as rectangles, squares, parallelograms, triangles and circles from
your earlier classes. 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 (thela),
dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular
paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).
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 a review of
the concepts of perimeter (circumference) and area of a circle and apply this knowledge
in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle)
known as sector and segment. We shall also see how to find the areas of some
combinations of plane figures involving circles or their parts.
Fig. 12.1
AREAS RELATED TO CIRCLES
224 MATHEMA TICS
12.2 Perimeter and Area of a Circle — A Review
Recall that the distance covered by travelling once around a circle is its perimeter,
usually called its circumference. You also know from your earlier classes, that
circumference of a circle bears a constant ratio with its diameter. This constant ratio
is denoted by the Greek letter p (read as ‘pi’). In other words,
circumference
diameter
= p
or, circumference = p × diameter
= p × 2r (where r is the radius of the circle)
=2pr
The great Indian mathematician Aryabhatta (A.D. 476 – 550) gave an approximate
value of p. He stated that p =
62832
,
20000
which is nearly equal to 3.1416. It is also
interesting to note that using an identity of the great mathematical genius Srinivas
Ramanujan (1887–1920) of India, mathematicians have been able to calculate the
value of p correct to million places of decimals. As you know from Chapter 1 of
Class IX, p is an irrational number and its decimal expansion is non-terminating and
non-recurring (non-repeating). However, for practical purposes, we generally take
the value of p as
22
7
or 3.14, approximately.
You may also recall that area of a circle is pr
2
, where r is the radius of the circle.
Recall that you have verified it in Class VII, by cutting a circle into a number of
sectors and rearranging them as shown in Fig. 12.2.
Fig 12.2
AREAS RELA TED TO CIRCLES 225
Y ou can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length
1
2
2
r ×p
and breadth r. This suggests that the area of the circle =
1
2
× 2pr × r = pr
2
. Let us
recall the concepts learnt in earlier classes, through an example.
Example 1 : The cost of fencing a circular field at the rate of Rs 24 per metre is
Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m
2
. Find the cost of
ploughing the field (Take p =
22
7
).
Solution : Length of the fence (in metres) =
Total cost
Rate
=
5280
= 220
24
So, circumference of the field = 220 m
Therefore, if r metres is the radius of the field, then
2pr = 220
or, 2 ×
22
7
× r = 220
or, r =
220 × 7
2× 22
= 35
i.e., radius of the field is 35 m.
Therefore, area of the field = pr
2
=
2
7
2

× 35 × 35 m
2
= 22 × 5 × 35 m
2
Now, cost of ploughing 1 m
2
of the field = Rs 0.50
So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925
EXERCISE 12.1
Unless stated otherwise, use p =
22
7
.
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.
3. Fig. 12.3 depicts an archery target marked with its five
scoring areas from the centre outwards as Gold, Red, Blue,
Black and White. The diameter of the region representing
Gold score is 21 cm and each of the other bands is 10.5 cm
wide. Find the area of each of the five scoring regions.
Fig. 12.3
226 MATHEMA TICS
Fig. 12.5
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does
each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
5. Tick the correct answer in the following and justify your choice : If the perimeter and the
area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) p units (C) 4 units (D) 7 units
12.3 Areas of Sector and Segment of a Circle
You have already come across the terms sector and
segment of a circle in your earlier classes. Recall
that the portion (or part) of the circular region enclosed
by two radii and the corresponding arc is called a
sector of the circle and the portion (or part) of the
circular region enclosed between a chord and the
corresponding arc is called a segment of the circle.
Thus, in Fig. 12.4, shaded region OAPB is a sector
of the circle with centre O. ? AOB is called the
angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of
the circle. For obvious reasons, OAPB is called the minor sector and
OAQB is called the major sector. You can also see that angle of the major sector is
360° – ? AOB.
Now, look at Fig. 12.5 in which AB is a chord
of the circle with centre O. So, shaded region APB is
a segment of the circle. You can also note that
unshaded region AQB is another segment of the circle
formed by the chord AB. For obvious reasons, APB
is called the minor segment and AQB is called the
major segment.
Remark : When we write ‘segment’ and ‘sector’
we will mean the ‘minor segment’ and the ‘minor
sector’ respectively, unless stated otherwise.
Now with this knowledge, let us try to find some
relations (or formulae) to calculate their areas.
Let OAPB be a sector of a circle with centre
O and radius r (see Fig. 12.6). Let the degree
measure of ? AOB be ?.
You know that area of a circle (in fact of a
circular region or disc) is pr
2
.
Fig. 12.4
Fig. 12.6
Page 5

AREAS RELA TED TO CIRCLES 223
12
12.1 Introduction
You are already familiar with some methods of finding perimeters and areas of simple
plane figures such as rectangles, squares, parallelograms, triangles and circles from
your earlier classes. 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 (thela),
dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular
paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).
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 a review of
the concepts of perimeter (circumference) and area of a circle and apply this knowledge
in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle)
known as sector and segment. We shall also see how to find the areas of some
combinations of plane figures involving circles or their parts.
Fig. 12.1
AREAS RELATED TO CIRCLES
224 MATHEMA TICS
12.2 Perimeter and Area of a Circle — A Review
Recall that the distance covered by travelling once around a circle is its perimeter,
usually called its circumference. You also know from your earlier classes, that
circumference of a circle bears a constant ratio with its diameter. This constant ratio
is denoted by the Greek letter p (read as ‘pi’). In other words,
circumference
diameter
= p
or, circumference = p × diameter
= p × 2r (where r is the radius of the circle)
=2pr
The great Indian mathematician Aryabhatta (A.D. 476 – 550) gave an approximate
value of p. He stated that p =
62832
,
20000
which is nearly equal to 3.1416. It is also
interesting to note that using an identity of the great mathematical genius Srinivas
Ramanujan (1887–1920) of India, mathematicians have been able to calculate the
value of p correct to million places of decimals. As you know from Chapter 1 of
Class IX, p is an irrational number and its decimal expansion is non-terminating and
non-recurring (non-repeating). However, for practical purposes, we generally take
the value of p as
22
7
or 3.14, approximately.
You may also recall that area of a circle is pr
2
, where r is the radius of the circle.
Recall that you have verified it in Class VII, by cutting a circle into a number of
sectors and rearranging them as shown in Fig. 12.2.
Fig 12.2
AREAS RELA TED TO CIRCLES 225
Y ou can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length
1
2
2
r ×p
and breadth r. This suggests that the area of the circle =
1
2
× 2pr × r = pr
2
. Let us
recall the concepts learnt in earlier classes, through an example.
Example 1 : The cost of fencing a circular field at the rate of Rs 24 per metre is
Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m
2
. Find the cost of
ploughing the field (Take p =
22
7
).
Solution : Length of the fence (in metres) =
Total cost
Rate
=
5280
= 220
24
So, circumference of the field = 220 m
Therefore, if r metres is the radius of the field, then
2pr = 220
or, 2 ×
22
7
× r = 220
or, r =
220 × 7
2× 22
= 35
i.e., radius of the field is 35 m.
Therefore, area of the field = pr
2
=
2
7
2

× 35 × 35 m
2
= 22 × 5 × 35 m
2
Now, cost of ploughing 1 m
2
of the field = Rs 0.50
So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925
EXERCISE 12.1
Unless stated otherwise, use p =
22
7
.
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.
3. Fig. 12.3 depicts an archery target marked with its five
scoring areas from the centre outwards as Gold, Red, Blue,
Black and White. The diameter of the region representing
Gold score is 21 cm and each of the other bands is 10.5 cm
wide. Find the area of each of the five scoring regions.
Fig. 12.3
226 MATHEMA TICS
Fig. 12.5
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does
each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
5. Tick the correct answer in the following and justify your choice : If the perimeter and the
area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) p units (C) 4 units (D) 7 units
12.3 Areas of Sector and Segment of a Circle
You have already come across the terms sector and
segment of a circle in your earlier classes. Recall
that the portion (or part) of the circular region enclosed
by two radii and the corresponding arc is called a
sector of the circle and the portion (or part) of the
circular region enclosed between a chord and the
corresponding arc is called a segment of the circle.
Thus, in Fig. 12.4, shaded region OAPB is a sector
of the circle with centre O. ? AOB is called the
angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of
the circle. For obvious reasons, OAPB is called the minor sector and
OAQB is called the major sector. You can also see that angle of the major sector is
360° – ? AOB.
Now, look at Fig. 12.5 in which AB is a chord
of the circle with centre O. So, shaded region APB is
a segment of the circle. You can also note that
unshaded region AQB is another segment of the circle
formed by the chord AB. For obvious reasons, APB
is called the minor segment and AQB is called the
major segment.
Remark : When we write ‘segment’ and ‘sector’
we will mean the ‘minor segment’ and the ‘minor
sector’ respectively, unless stated otherwise.
Now with this knowledge, let us try to find some
relations (or formulae) to calculate their areas.
Let OAPB be a sector of a circle with centre
O and radius r (see Fig. 12.6). Let the degree
measure of ? AOB be ?.
You know that area of a circle (in fact of a
circular region or disc) is pr
2
.
Fig. 12.4
Fig. 12.6
AREAS RELA TED TO CIRCLES 227
In a way, we can consider this circular region to be a sector forming an angle of
360° (i.e., of degree measure 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 = pr
2
So, when the degree measure of the angle at the centre is 1, area of the
sector =
2
360
r p
·
Therefore, when the degree measure of the angle at the centre is ?, area of the
sector =
2
360
r p
×? =
2
360
r
?
×p
.
Thus, we obtain the following relation (or formula) for area of a sector of a
circle:
Area of the sector of angle ? ? ? ? ? =
×
2
360
?
pr
,
where r is the radius of the circle and ? the angle of the sector in degrees.
Now, a natural question arises : 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
(of angle 360°) as 2pr, we can obtain the required
length of the arc APB as
2
360
r
?
×p
.
So, length of an arc of a sector of angle ? ? ? ? ? =
×
?
p 2
360
r
.
Now let us take the case of the area of the
segment APB of a circle with centre O and radius r
(see Fig. 12.7). You can see that :
Area of the segment APB = Area of the sector OAPB – Area of ? OAB
=
2
– area of OAB
360
r
?
×p ?
Note : From Fig. 12.6 and Fig. 12.7 respectively, you can observe that :
Area of the major sector OAQB = pr
2
– Area of the minor sector OAPB
and            Area of major segment AQB = pr
2
– Area of the minor segment APB
Fig. 12.7
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