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he
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Government of Karnataka
©
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MATHEMATICS
English Medium
7
Second Semester
2015
No
t
to
SEVENTH STANDARD
KARNATAKA TEXT BOOK SOCIETY (R)
100 Feet Ring Road, Banashankari 3rd stage,
Bengaluru - 85
i
CONTENTS
Second Semester
Indices
1 - 18
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1
Pages
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Chapter Name
he
Sl. No
Ratio and Proportion
19 - 35
3
Percentage
36 - 47
4
Simple linear equations
48 - 59
5
Congruence
60 - 66
6
Geometrical Construction
67 - 91
7
Mensuration
92 - 137
8
to
2
138 - 165
Probability
166 - 176
Representation of 3 dimensional
Objects in 2 dimensional Figures
177 - 194
Answers
195 - 198
No
t
9
Data Handling
10
ii
UNIT - 1
INDICES
After studying this unit you :
write repeated multiplication in exponential form and
and exponential form as repeated multiplication,

recognise, read as well as write the base, index (power)
of given numbers in exponential form,

convert numbers to its factors and express it in
exponential form,

he
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
d

by writing the pattern of examples generalize laws of
exponents and use them,
understand the standard form adopted to write large
and small numbers in science and write some numbers
in that form.
Introduction of numbers in Exponential form
To write numbers in exponential form
to
You know the place value of numbers (numerals) in decimal
system as well as meaning of square and cube.
No
t
Let us recall line, square and cube studied in geometry
and draw a line segment measuring 10cm, a square and a
cube of side 10 cm.
10 cm
10 cm
10 cm
10 cm
1
10 cm
Length of the line segment is 10 cm.The area covered by
the square is 10 cm ×10 cm = 100 sq cm and the space occupied
by the cube is 10 cm × 10 cm × 10 cm = 1000 cm3.
Here the length of the line segment, side of a square and
side of the cube is 10cm.
he
volume of a cube, as 10×10×10=103
d
We write, 10 to a square as 10×10 = 102
Then the length of line segment = 10cm.
= 10cm.
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Area of square
In the above example the number written on the right top
of '10' indicates how many times 10 is multiplied by 10.
5 × 5 × 5 = 53
In the same way :
2 × 2 × 2 × 2 = 24
7 × 7 × 7 × 7 × 7 = 75
2
2
2
2
2
2
2 6
a3 k # a3 k # a3 k # a3 k # a3 k # a3 k = a3 k
x × x × x × x × x × x × x = x
7
to
In the above examples repeated multiplication is written
in exponential form.
Exponential form as repeated multiplication
No
t
We can write a number in exponential form as repeated
multiplication of that number
Example : 104 = 10 × 10 × 10 × 10
56 = 5 × 5 × 5 × 5 × 5 × 5
a8 = a × a × a × a × a × a × a × a
3 5 3 3 3 3 3
a4 k = 4 # 4 # 4 # 4 # 4
4
_- 5i = (- 5) # (- 5) # (- 5) # (- 5)
2
Base and Index ( exponent /power ) of numbers in
exponential form.
106 is read as ten to the power of six.
It is also read as ten raised to the power of six as well as
sixth power of 10.
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6
Index/Power/Exponent
Base number
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10
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106 = 10 ×10 ×10 ×10× 10 × 10 means 10 is multiplied six times.
When repeated multiplication is written in exponential
form the number which repeats is called the base and the
number of times the base repeats is called index.
Example :
1) 8 × 8 × 8 × 8 × 8 × 8 × 8 = 87
In 87 , 8 is base and 7 is index.
2) In 54, 5 is base and 4 is index.
to
3) In 28, 2 is base and 8 is index.
5
4) In ` 2 j , 2 is base and 5 is index.
3
3
Observe this Table
Exponential number
Index
64
6
4
6
to the power of
4
47
4
7
4
to the power of
7
125
12
5
12 to
the power of
3
`- 4 j
-3
4
3
3
`- 4 j
to the power of
x8
x
8
x
No
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Base
3
3
Read as
to the power of
5
8
3
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Activity 1 : Take a sheet of white paper. Fold and crease
it, in such a way as to get equal parts in each fold, length
wise . Prepare a tabular form of number of folds and number
of rectangles formed as shown below.
Second fold
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First fold
Number of folds
1
2
3
4
5
Number of rectangle formed
2
4
8
16
32
Repeated multiplication form
2×1
2×2
2×4
2×8
2×16
Exponential form
21
22
23
24
25
Think : If the paper is folded '0' times how can you write the
number of rectangle formed in exponential form.
Method to write numbers in Exponential form.
to
Take any number and write it as product of repeated factors,
then write each factor as base and mark index to each of them.
No
t
Example 1:
1) Write 125 in exponential form to base 5.
Solution : 5 125
5 25
5 5
1
125
=5×5×5
= 53
∴125 = 53
53 is the exponential form of 125.
4
Example 2 :
Write 256 in exponential form to base 2, 4, 16, 256.
4
4
4
4
256
64
16
4
1
16 256
16 16
1
256 256
1
∴256 = 16 × 16
1
2
256=4 × 4 × 4 × 4 ∴256 = 16
∴256= 2561
d
128
64
32
16
8
4
2
1
256= 2 × 2 × 2 × 2 ×
2×2×2×2
∴256 = 28
∴256 = 44
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2
2
2
2
2
2
2
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Solution : 2 256
Example 3 : Write exponential form of 1331 to base 11.
Solution : 11 1331
11 121
11 11
1
1331 = 11 × 11 × 11
∴1331 = 113
Example 4 : Write exponential form of 1125.
Solution : 1125 is divisible by 5 and 3.
1125
225
45
9
3
1
No
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to
5
5
5
3
3
1125 = 5 × 5 × 5 × 3 × 3
1125 = 53 × 32
∴1125 = 32 × 53
Example 5: Write exponential form of 324.
Solution : 2 324
2
3
3
3
3
162
81
27
9
3
1
324 = 2 × 2 × 3 × 3 × 3 × 3
∴ 324 = 22 × 34
5
Exercise : 1.1
Read these exponential numbers.
1) 83
4) 104
5) (-6)5
Write the base and index of these numbers.
1) 35
6
3) `- 2 j
3
2) 108
4) x20
d
II.
10
3) ` 4 j
7
2) 136
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I.
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III. Write appropriate answer in the space provided.
1) 3 × 3 × 3 × 3 × 3 =
5
2) 5 # 5 # 5 # 5 = ` 5 j
8 8 8 8
8
3) 4 × 5 × 4 × 5 × 4 × 5 × 4 × 5 × 4 = 4
×5
4) a × a × a × a × ------ n times('a' is multiplied 'n' times) = ---IV. Write the expansion form of these.
2) 113
to
1) 38
6
3) c 5 m
2
4) (1.5)6
p 4
5) c m
q
V. Express the following number as directed.
No
t
1) 81 in exponential form as
a) base 9
b) base 3
2) 15625 in exponential form as
a) base 5
ii) base 25
3) (-243) in exponential form as base (-3).
6
Laws related to operations in exponents
Activity 2 : Let us play a game. Prepare 20 cards having
numbers as shown in the example. Put them in different
boxes.
25
26
27
2
2
2
9
2
4
31
32
33
34
35
36
37
10
38
39
310
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8
2
3
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2
2
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2
1
Shuffle the cards in each box separately. Then ask the
students to pick up two cards from any one box, and write
the product of the numbers in the cards
Example : a) 22 × 23 = 2 × 2 × 2 × 2 × 2 = 25 or
22 × 23 = 22 + 3 = 25
b) 34 × 33 = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 37
34 × 33 = 34+3 = 37
Similarly 28 × 22 = 28+2 = 210
35 × 36 = 35+6 = 311
to
By extending this activity to other examples, we notice
104 ×106 = 10
4+6
= 1010
No
t
x10 × x25 = x 10 + 25 = x35
2 4
2 12
2 4 + 12
2 16
`3 j #`3 j = `3 j = `3 j
By the above examples we can write a × a = a1+1 = a2
a × a × a = a1+1+1 = a3
a5 × a3 = a5+3 = a8
a6 × an = a6+n
Similarly am × an = am+n
7
When two numbers in exponential form with same base
are multiplied, the exponent of the product is the sum of
their exponents.
am × an = am+n, ( a ≠ o) This is called first law of exponents.
b)
c)
106 ×102 = 106+2 = 108
2 5
2 10
2 5 + 10
2 15
`3 j #`3 j = `3 j = `3 j
d)
x15 × x20 = x15+20 = x35
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25 × 24 = 25+4 = 29
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a)
d
Example :
Let us apply this law for multiplying more than two
numbers with exponents.
Example :
a)
23 × 22 × 24 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
or 23 × 22 × 24 = 23+2 × 24 = 25 × 24 = 25+4 = 29
23 × 22 × 24 = 23+2+4 = 29
m3 × m4 × m5 = m × m × m × m × m × m × m × m × m × m × m × m
to
b)
= m12 or
No
t
m3 × m4 × m5 = m3 + 4 + 5 = m12
Similarly: a) 102 × 105 × 107 = 10 2+5+7 = 1014
b) 1008 × 1006 × 10020 = 100 8+6+20 = 10034
c) x8 × x10 × x20 × x12 = x 8+10+20+12= x50
This can be generalized as am ×an × ap × aq= am+n+p+q
8
Exercise: 1. 2
Simplify using am × an = am+n.
a) 72 × 75
b) (-3)5 × (-3)3
3
6
c) c 5 m # c 5 m
2
2
d) 103 × 107 × 105
e) a6 × a4 × a10
f) (2.5)4 × (2.5)8
d
I.
he
II. Convert the following to exponential form and apply
first law of exponents.
b) 27 × 81
c) 243 × 81
d) 1024 × 16
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a) 49 × 7
III. Fill up the space provided with suitable answer.
13
6
b) ` 2 j # ` 2 j = ` 2 j
3
3
3
a) 108 # 103 = 10
c) a13 =
3
# a10
d) ^25h6 = ^25h
# ^25h5
Division of numbers with exponents :
Activity :
to
Take any two cards from the box of cards used in your
previous activity. Divide the numbers with higher exponent
by the other number in the cards.
Example :
No
t
4
2#2#2#2
4-3
1
24
a) 24 ' 23 = 23 =
= 2 or 3 = 2 = 2 = 2
2#2#2
2
2
6
3#3#3#3#3#3
2
b) 36 ' 34 = 34 =
= 3#3 = 3 = 9
3#3#3#3
3
36 36 - 4 3 2 9
=
= =
34
c) 38 ÷ 34 = 38-4 = 34
9
d)
2
9
÷
2
5
=
2
=
9 - 5
2
4
s i m i l a r l y
a)
23 2 # 2 # 2 1
=
=
23 2 # 2 # 2
As per 2nd law,
b)
he
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Observe the following examples
d
6
56 ' 54 = 54 = 56 - 4 = 52 = 25
5
8
108 ' 104 = 104 = 108 - 4 = 104 = 10000
10
15
15 - 4
11
a
=a =a
a4
m
` a n = a m - n if m > n
a
23 23 - 3 20
=
=
23
` 20 = 1
x5 x # x # x # x # x 1
=
=
x5 x # x # x # x # x
5
As per 2nd law, x5 = x5 - 5 = x0 = 1
x
By above examples we can generalise that
to
a m =1 when m=n and a ] 0
an
100 = 1, 1000=1, 50 = 1, (
xy 0
) =1
z
No
t
Note: For any non-zero base, if the power is zero then its
value is 1
m
Let us examine what happens if m<n in a n , a ] 0
a
Examples :
a)
b)
2#2#2
1
1
23
=
= 2 = 5-3
5
2
2
2
2
2
#
#
#
#
2
2
2
3#3#3#3
34
1
=
= 3
7
3
3
3
3
3
3
3
#
#
#
#
#
#
3
3
10
From above examples, we can generalise that
1
am
= n - m , a ! 0 and m < n
n
a
a
3
Then, 25 = 23 - 5 = 2- 2 = 12
So, 1- 7 = 47, 8- 2 = 12 , 1- 4 = m4
8
4
m
he
4
Also 37 = 34 - 7 = 3- 3 = 13
3
3
d
2
2
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a- m = 1m , a ! 0
a
When two numbers in exponential form with same base
are divided by each other, the exponent of quotient is the
difference
between the exponents of dividend and divisor.
am am - n, a ! 0 m > n This is 2nd law of exponents.
n =
a
Exercise: 1.3
I.
Simplify, using 2nd law, am ÷ an = am-n .
to
1) 75 ÷ 72
4) (8.5)6 ÷ (8.5)5
2) (-3)5 ÷ (-3)2
8
3
3) c 5 m ' c 5 m
2
2
5) x11 ÷ x3
10
6) x10
x
5
7) 410
4
3) 256 ÷ 8
4)27 ÷ 243
II. Convert the following numbers to exponential form
No
t
using am ÷ an = am-nand simplify.
1) 125 ÷ 25
2) 81 ÷ 9
III. Express the following in positive index.
1) 3-5
2) 10-7
3) a-10
4) x-12
IV. Express the following in negative index.
a) 54
b) 14
3
c) 27
11
d) 15
x
Exponent of numbers with exponents
Look at these examples.
4
1) (32) means 32 is multiplied 4 times.
4
(32) = 32 × 32 × 32 × 32 = 32+2+2+2 = 38 (Ist law of indices)
4
d
or (32) = 32×4 = 38
he
4
Similarly,
(53) = 53×4 = 512
5
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(x8) = x8×5 = x40
6
(a4) = a4×6 = a24
2
2) (73) means 73 is multiplied twice.
2
(73) = 73 × 73 = 73+3=76
2
or (73) = 73×2 = 76
We can generalized that
n
∴ (am) = am×n
n
(am) = am×n , a ≠ o.
to
The exponent of a number in exponential form is equal
to the product of exponents.
n
No
t
(am) = am×n a ≠ 0 This is called the 3rd law of exponents.
Example :
2
a) (3 ) = 3
5
5×2
=3
10
6
b) (4x) = 4x.6 = 46x
y
Know this We can extend
this law to
r
((
n
(am)
c) (2x) = 2x.y
Example :((23)2) = 23×2×4 = 224
4
12
))
p
= am×n×p×r = a
mnpr
Exercise
: 1. 4
n
Simplify the following, applying (am) = am×n.
I.
6
a) (24)
2
b) (83)
7
c) (116)
2 4
r
d) (pq)
4 5
f) [( 2 )
3
e) ((23) )
Think : Which is greater? How?
2
3
3
4
ii) 102 , (102)
4
iii) 42 , (42)
d
2
i) 23 , (23)
]
(5 × 7)4
= (5 × 7) × (5 × 7) × (5 × 7) × (5 × 7)
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1)
he
Multiplication of exponents with the different bases.
Observe this number pattern.
= (5 × 5 × 5 × 5) × (7 × 7 × 7 × 7)
(5 × 7)4
= 54 × 74
∴ (5 × 7)4
= 54 × 74
Similarly (3 × 11)5 = (3 × 11) × (3 × 11) × (3 × 11) × (3 × 11) × (3 × 11)
= (3 × 3 × 3 × 3 × 3) × (11 × 11 × 11 × 11 × 11)
= 35 × 115
∴ (3 × 11)5 = 35 × 115
Similarly (a × b)m = (a × b) × (a × b) × (a × b) × (a × b)...... m times
= (a × a × a...... m times) × (b × b × b...... m times)
(a × b)m = am × bm
to
For all non - zero integers, a and b (a × b)m = am× bm.
This is called the 4th law of exponents.
No
t
Example :
1) (6×5)3 = 63 × 53
2) (4×6)5 = 45 × 65
This can be extended to the number having more than two
bases. (a × b × c × d)n = a n × b n × c n × d n a, b, c, d ≠ 0
Example :
1) (4×2×3)5 = 45× 25× 35
2) (2abc)8 = 28 a8 b8 c8
13
Exercise
I.
: 1.5
Express the following in (a × b)m = am × bm.
1) (4 × 5)2
2) (8 × 6)6
3) (11 × 5)7
2) 48 × 58
3) 103 × 23
he
1) 33 × 23
d
II. Write/express the following in (a × b)m form.
Division of exponents with different bases.
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Observe the following exponents.
4 5 4 4 4 4 4 45
`3j = 3 # 3 # 3 # 3 # 3 = 5
3
2 6 2 2 2 2 2 2 26
`7j = 7 # 7 # 7 # 7 # 7 # 7 = 6
7
3 4 3
3
3
3
34
`10 j = 10 # 10 # 10 # 10 = 4 Similarly
10
5 6 56
`7j = 6
7
x 10 x10
` y j = 10
y
m
m
` ` a j = am
b
b
to
a m am
` b j = m this is called the 5th law of exponents. a, b ≠ 0.
b
Example : a) `10 j = 106
7
7
No
t
6
I.
8
8
b) ` m j = m8
5
5
6
Exercise
20
20
c) ` 9 j = 920
4
4
: 1.6
m
Express using ` a jm = a m .
b
b
6
a) `12 j
13
5
b) `14 j
5
7
c) ` 8 j
7
14
3
b) ` x j
z
Application of laws of exponents.
Read this.
Suhasini and Mary were playing with numbers. Suhasini
2
2
2
2
wrote 33 and 55 and Mary wrote (33) and (55) . Both of them
d
were arguing that the value of each number written by each
he
is greater than the other. To clarify they met Suhasini's uncle,
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who is a mathematics teacher. He clarified like this.
Suhasini
2
33 = 33 × 3 = 39
2
55 = 55 × 5 = 525
Mary
Conclusion
(33) = 33 × 2 = 36
2
(55) = 55 × 2 = 510
39 > 36
525 > 510
2
Both Suhasini and Mary satisfied by the uncle's clarification
Uncle gave a few numbers and helped them to solve the
value of each of these numbers
3
a) (22)(3 ) = (22)27 = 22×27 = 254
3
8
2
4
to
b) (23)(2 ) = (23) = 28×3 = 224
No
t
c) (33)(2 ) = (33) = 33×4 = 312
3
8
d) (32)2 = (32) = 316
a)
Express 27 × 27 × 27 in exponential form of base 3.
3
Solution :
27 × 27 × 27 = 33 × 33 × 33 = (33) = 33 × 3 = 39
b) Simplify c 22 m # 24
2
6
Solution : ^26 - 2h # 24 = 24 # 24 = 24 + 4 = 28
15
Exercise: 1.7
I. Apply laws of exponents and simplify.
3
3
5
×23
3) 3 ×3
32 ×6
6
4 2
3
5 3
4) c 2 ×33 m × c 3 ×2 m
3×8
6
5) ^30 # 25h + 50
#8
6) 3000
22 # 52 # 3
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2) (103) × (102)
d
2
1) (0.7)2 × (0.7)3
II. Simplify by converting into exponential form.
1) 72 × 55 × 45
Clue: 72 = 23 × 32
2) 36×9
12×4
(- 1)
4) 81 #
#2
18 (- 1) 3 3
3) 49 × 121
5
III. Examine the following and state whether it is right or
wrong.
2) 43 × 52 = 205
3) (-5)0 = 50
4) 23 = 6
5) (-1)5 = (-1)3 (1)2
6) (-1) × (-1) ...... 13 times = -1
to
1) 100 × 1010 = 105 × 106
No
t
7) (-1) × (-1)...... 22 times = -1
Standard form of expressing scientific notation.
It is very useful in science to write and read the huge(Macro)
and tiny small (micro) numbers in exponential form. In
scientific notation, the given number is expressed as a product
of number greater than '1' but less than '10' and an integer
power of 10.
16
Example 1 : (Macro numbers)
a) 112
= 1.12 × 102 [ Note 1.12 > 1, 1.12 < 10]
b) 236000
= 236 × 103
d
= 23.6 × 104
c) 14567800000
= 145678 × 105
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= 14567.8 × 106
he
[ Note. 2.36 > 1, 2.36 < 10]
= 2.36 × 105
= 1456.78 × 107
= 145.678 × 108
= 14.5678 × 109
= 1.45678 × 1010
Example 2 : (Micro numbers)
a) 0.000342
=
342
342
=
1000000 106
to
= 342 × 10-6
No
t
= 34.2 × 10-5
b) 0.00045213
= 3.42 × 10-4
[Note : 3.42>1,3.42<10]
= 45213 × 10-8
= 4521.3 × 10-7
= 452.13 × 10-6
= 45.213 × 10-5
= 4.5213 × 10-4
17
[Note : 3.42>1,3.42<10]
Example 3:
The temperature at the interior of the earth is
approximately 2,00, 00, 000 0C. Scientific notation of this
is 2,00,00,0000 C=2 × 10000000= (2 × 107)0C
[Note : 2 > 1 and 2 < 10]
2) Weight of a small grain is 0.005 gram. The scientific
notation of the above is. 0.0005 g = 5/ 10000 = 5 × 10-4 g
[Note : 5>1 and 5<10]
1)
2)
- 1.8
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Exercise
he
d
1)
Write the scientific notation of velocity of light, velocity
of sound, distance between sun and earth, and other
astronomical distance which you come across in your
science text book.
Collect numerical information of population, budget
from newspapers and write these numbers in standard
scientific notation.
Points to remember :
am # an = am+n
a m a m - n, m > n
=
an
No
t
to
am a m - n a n - n a0 1 , m=n
=
=
= =
an
am
1
= n-m , m < n
an
a
(a m) n = a mn
(a # b) m = a m # b m
a m am
`bj = m
b



18


UNIT - 2
RATIO AND PROPORTION
After studying this unit you :
solve the problems using unitary method,

read, write and simplify the ratio,

solve the problems on proportional division,

identify the different situations where the concept of
ratio is used,


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
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
solve the problems based on proportion,
solve the problems on direct proportion,
solve the problems based on inverse proportion.
Unitary method :
to
In our daily life we observe the things such as pens,notebook,
etc are sold in bundles or packets the price will be indicated
on each bundle or packet. If we want to purchase a few things
out of them, then how to calculate the money to be paid for
the shopkeeper?
No
t
Example 1 : The cost of one dozen bananas is ` 36. Somanna
buys 20 bananas. Find the amount Somanna has to pay.
How to calculate the cost of 20
bananas? We know the cost of 1
dozen ( 12 ) bananas.
19
cost of one banana
= ` 1 j the cost of 1 dozen bananas
12
th
1
= ` 36 # 12
=`3
= ` 3 × 20
= ` 60
d
∴cost of 20 bananas
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The method of calculating the cost of one thing and then
calculating the cost of given number of things is known as
unitary method.
Example 2: A school has arranged a scout camp with enough
food for 30 scouts for 4 days. But 40 scouts participated in
the camp. For how many days the same food is enough for
40 scouts ?
30 scouts have enough food for 4 days.
∴ The food sufficient for one scout is 30×4 = 120 days.
∴ The food sufficient for 40 scouts
= 8120 B days.
40
= 3 days.
to
Exercise 2.1
No
t
I. Solve the following problems by unitary method.
1) The cost of 3 balls is ` 36. Find the cost of 5 balls.
2) The cost of 5 pens is ` 30. Find the cost of 12 pens.
3) The cost of 15 oranges is ` 30. Find the cost of 50
oranges.
4) A car uses 12 litres of petrol to travel a distance of
180 km. Find the distance travelled by the car for 20
litre of petrol.
5) The cost of 25 m cloth is ` 750. Find the cost of
12 m cloth of same type.
20
6)
In a school there is enough food for 100 students for 4
days. How long that the food lasts for 40 students ?
7)
12 persons can reap the crop from a field in 5 days.
Calculate the days required for 20 persons to do the
same job ?
Proportion
d
24 workers can build a wall in 15 days. How many days
will 9 workers take to build the similar wall?
he
8)
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We incorporate many concepts of mathematics in our
day to day transactions. We always prefer to buy more things
for less amount, People compare the cost of thing with the
quantity.
` 25
250g
` 45
500g
You must have seen such packets in the shop. Which one
of the two packets is more beneficial for the buyer ?
to
Let us compare the weight and the cost of these two
packets.
No
t
Let us compare the weight of the packets first
The weight of tea powder in the bigger packet
= 500g
The weight of the tea powder in the smaller packet = 250g
500g ÷ 250g = 2
Let us compare the weight of the packets first
∴The weight of the tea powder in the bigger packet is twice
the weight of the tea powder in the smaller packet.
21
Now let us compare the cost of tea powder in two packets.
The cost of tea powder in the bigger packet
= ` 45
The cost of tea powder in the smaller packet
= ` 25
` 45 ÷ ` 25
= 1.8
d
The cost of bigger packet is 1.8 times the cost of smaller
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packet.
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The weight of the bigger packet is 2 times the smaller
packet while the cost of the bigger packet is 1.8 times the
smaller packet.
By comparing this, what conclusion can be drawn ?
It is beneficial to purchase bigger packet
to
The quotient of weight of tea powder in two packet - 500 .
250
This can be written in the form 500 : 250 This
definite relationship is called ratio.
The quotient of cost of tea powder in two packets = 45 .
25
This can be written in the form 45 : 25. This definite
relationship is called ratio.
No
t
The comparison of two quantities of the same kind is
called ratio.
If a and b are of two quantities of the same kind then
the ratio between a and b is written a:b and read a is to b.
In the ratio a:b, a is antecedent and b is consequent.
This ratio a:b can also be written in fraction as a .
b
22
Representing the ratio in its simplest form.
he
Example 1 : Write 10:15 in to simplest form.
d
Ratio is a fraction. The simplification of fraction can
be done by dividing or multiplying the numerator and the
denominator by the same number. Hence by dividing or
multiplying the two terms of a ratio, there will be no change
in the value of the ratio.
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Which number divides 10 and 15 completely?
Yes, it is divisible by 5.
Divide both 10 and 15 by 5
10 : 15 = 2 : 3 or 2:3
5
5
1 1
In other words, if you want to reduce the ratios to their
simplest form, divide the antecedent and consequent by their
HCF.
Example 2 : Write 1 : 1 in its simplest form.
2 3
The LCM of the denominator ( 2 and 3 ) is 6.
Multiply both 1 and 1 by 6
to
2
1 × 6 : 1 ×6 = 3:2
2
3
3
No
t
Example 3 : Express the weights of two tea packets weighing
2kg and 500g in ratio.
The units of weights of two tea packets are different.
Convert them into same unit.
= 2kg
The weight of one packet
= 2000g
The weight of the other packet
23
= 500g
Now both weights are of the same unit.
= 2000 : 500
The ratio of these weights
= 2000 : 500
500
500
= 4:1
d
Note :
Ratio is a comparison of two quantities. Hence, they
are in number form. Therefore, unit should not be
mentioned . .
•
If the units are different then convert it into same units
•
In the ratio, the units of quantities to be compared.
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•
Remember :
Make a list of situations where the
national flag is in the ratio 3:2
concept of ratio is
The mixture of cement and sand used used.
The length and breadth of our
for constructing building is usually in
to
the ratio of 1:6
To prepare tasty Idlies the ratio of black gram dal : rice = 1:2
No
t
Reciprocal of ratio
If the antecedent and the consequent of ratio are
interchanged, we get the reciprocal ratio of the original.
Example : The reciprocal of 2:3 is 3:2
The reciprocal of a:b is b:a
24
Note a # b = 1
b a
Examples:
1) Write 25 and 35 in the form of a ratio
The ratio of 25 and 35 = 25:35
= 25 : 35
The simplest form
5
5
( divide by 5 )
d
= 5:7
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2) There are 25 boys and 20 girls in a class. Find
The ratio of the number of boys to the number of girls.
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i)
ii)
The ratio of the number of boys to the number of total
students in the class.
The ratio of the number of boys
to the number of girls
Divide both the terms by 5
to the total number of students
Divide both the terms by 5
to
= 25 : 20
=5: 4
The ratio of the number of boys
No
t
{
{
= 25 : 45
=5:9
Exercise 2.2
I. Express the following ratios in the simplest form.
1) 6 : 8
2) 21 :24
3) 33 : 77
4) 25 : 125
5) 2 : 3
6) 4 : 3
7) 1 1 : 4 1
8) 1 : 2
3 5
5 8
2
25
2
3
II. Express the following in to ratio (simplest form).
1) 100g and 500g
2) 3 hours and 6 hours
3) 500g and 1kg
4) 30 minutes and 2 hours
d
5) 25 cm and2 m
6) 200ml and 1l
III. Express the following ratios into their reciprocal
form.
to
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1) 5 : 8
2) 21 :23
3) 30 :77
4) 25 :12
IV. Solve the following problems and express the answer
in their simplest form.
1) The total population of Bandipur village is 5,400.
out of them 900 are illiterate. Then find out
i) the ratio of the total population to the number of
illiterates.
ii) the ratio of the number of literates to number of
illiterates.
2) The length and breadth of a play ground is 50m and
90m respectively. Find the ratio of length and breadth
of the play ground .
3) The monthly income of a family is ` 9000 and the
monthly expenditure is ` 7000. Find the ratio of monthly
income to expenditure.
Proportional division :
No
t
Example 1: Vinay and Victor worked together and earned
` 750. Vinay worked for 3 hours and Victor worked for 2 hours,
Calculate the share of amount earned by Vinay and Victor.
Let us work out this problem.
Victor and Vinay worked together and earned ` 750. But
the amount earned cannot be distributed equally because
the number of hours they worked are different. Hence the
earnings should be distributed according to the number of
hours they worked.
26
∴ ` 750 is to be distributed among Vinay and Victor in 3 : 2
Vinay gets 3 of the amount, Victor gets 2 of the amount
5
5
∴ Vinay's share
= ` 450
d
= 3 of 750 = 3 # 750
5
5
Vinay's share of earning
∴ Victor's share
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= ` 300
he
= 2 of 750 = 2 # 750
5
5
Victor's share of earning
The profit earned by more than two persons in business
can also be calculated applying the concept of ratio.
Exercise
2.3
I. Solve the following problems.
The population of Padnur village is 5880. The ratio
of number of females to number of males is 10 : 11.
Calculate the number of females and males in that
village.
No
t
2)
Damu works in the field of Raju. They have agreed to
share the income obtained in the ratio of 4 : 3. Find
out the share of income received by both of them,
if the income for one year is ` 21,000
to
1)
3)
Roshan and Hameed invested ` 30,000 and ` 40,000
respectively in a business. After one month they
earned a profit of ` 2800. According to the ratio of their
investment, find the share of the profit of each.
27
In an alloy the ratio of copper and zinc is 5:3 by weight.
Calculate the weight of copper and zinc in 240 g of the
alloy.
5)
Joky and Jani are weavers of bamboo baskets. One
day Joky and Jani weaved 5 baskets and 4 baskets
respectively. They sold them in the market for ` 540.
Find the share of their earnings.
6)
Distribute ` 642 among A,B and C in the ratio 1:2:3
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4)
Proportion :
Example 1: If the cost of one pen is ` 5 find the cost of two
pens.
The cost of two pens is more because as the number of pens
increases the total cost increases. We can see many situations
in our daily life where quantities are interdependent .
As the number of pens changes from 1 to 2 the cost changes
from ` 5 to ` 10.
to
The ratio of number of pens = 1:2
No
t
The ratio of their cost
= 5:10
= 1:2 ( simplified )
It means the ratio of the number of pens = The ratio of
their cost.
∴ 1:2 = 5:10
∴ Proportion is an equation in which two ratios are equal
to each other.
28
he
d
If four terms a, b, c and d are in proportion then the
relationship can be symbolically written as
a:b = c:d
a and d are called extremes(end terms) and c and b are
called means (middle terms)
a:b = c:d can also be written as a = c (a:b::c:d)
b
d
We read a:b::c:d as a is to b is as c is to d.
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Activity : There are a few cards on which ratios are written.
Select the pair of cards having equal ratios.
2:3
6:3
4 : 16
5:3
2:8
10 : 6
12 : 6
4:6
Write the selected pairs of ratios the form of proportion
in the table given below. Find the product of extremes and
the product of means(middle terms).
proportion
a:b = c:d
product of
extremes a×d
product of
means b×c
2:3 = 4:6
2×6 = 12
3×4 = 12
No
t
Ex
to
Do you find any relationship between the product of
means and product of extremes? Observe by completing the
table.
1
2
3
4
29
Observation : In the above table, the product of extremes
is equal to the product of means.
he
product of extremes a × d
d
In proportion, the product of extremes is equal to the
product of means.
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a:b=c:d
product of means b × c
∴ If a:b = c:d then a×d = b×c
If a:b = c:d or a = c
b
d
then a,b,c and d are in proportion.
Example If 2:3 = 4:6 then 2,3,4,6 are in proportion.
Examples :
1)
Are 3, 4, 6 and 8 in proportion ?
In 3, 4, 6, 8.
= 3 × 8 = 24
The product of means
= 4 × 6 = 24
∴ The product of extremes
= The product of means
No
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to
The product of extremes
∴ The given numbers are in proportion.
2)
If 5 : 2 = 10 : x. Then find the value x
5 × x = 2 × 10
x = 2 # 10 = 4
5
The value of x is 4
30
3) The cost of 10 kg of rice is ` 470. Find the cost of 8 kg of
rice.
The cost of 10 kg of rice
Let the cost of 8 kg of rice
The ratio of weight of rice
The ratio of their cost
These two ratios are equal.
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d
= ` 470
=`x
= 10:8
= 470:x
10 : 8 = 470 : x
10 × x = 8 × 470
470 = 376
x = 8 #10
∴The cost of 8 kg of rice ` (x) = ` 376
Exercise
I.
2.4
Fill up the blanks with appropriate numbers.
1) 6 : 8 =
: 16
4) 2 :
3 9
2) 21 : 24 = 7 :
3)
5) 4 : 8
5
6) 25 :
:7=4:2
=5:4
II. Check whether the following numbers are in
proportion.
2) 8, 15, 3, 6
3) 7, 42, 13, 78
4) 1.5, 4.5, 2, 6
to
1) 5, 6, 10, 12
No
t
III. The cost of 5 kg wheat is ` 127.50
1) Find the cost of 8 kg wheat
2) How much wheat can be bought for ` 765.
IV. A motor bike can travel 325 km for 5 litre of petrol.
Find the number of litre required to travel 130 km.
V. The cost of one litre of oil is ` 75. Find how many
litres of oil can be bought for ` 300.
31
Types of proportion :
d
1) As the number of people in a house increases; the
expenditure for the house also increases. The portion of
savories each would get is less, if the number of people
in the house are more ( Note : the quantity of savories
in stock remain the same )
he
2) Whenever two quantities are related such that when
one increases the other also increases or decreases
according to the situation.
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Hence, there are two types in proportion (a) Direct
proportion. (b) Inverse proportion.
Let us learn how to identify these.
Direct Proportion
If in a shop we get 4 chocolates for
`2
Then 6 chocolates for ` 3
to
8 chocolates for ` 4
Observe the total number of
chocolates got and money paid each
cases. What is your conclusion?
More Money more chocolates
less money less chocolates
No
t
If two quantities are
so related to each other
such that an increase or decrease in the magnitude of one,
results in the increase or decrease in the magnitude of the
other in the same ratio, then the two quantities are in direct
proportion.
The ratio of the money and the ratio of the number of
chocolates we get are equal.
32
4 chocolates for ` 2
8 chocolates for ` 4
The ratio of the money is 2 : 4,
The ratio of number of chocolates is 4 : 8
∴2:4=4:8
Problems on direct proportion
The cost of 2m cloth = ` 80
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The cost of 5m cloth = ` x
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d
Example 1: The cost of 2 m cloth is ` 80. Find the cost of 5m
cloth of the same type.
The ratio of the quantity of the cloth is 2:5
The ratio of the cost of the cloth is 80:x
As the quantity of the cloth increases the cost also
increases.
It is an example of direct proportion.
Hence 2:5 = 80:x
to
2×x = 5×80
[ The product extreme must be equal to
5
80
#
x=
product of means]
= 200
2
The cost of 5m cloth = ` 200
Exercise 2.5
No
t
I. Solve the following problems on direct proportion :
1) If the cost of 3 kg of sugar is ` 84, then find the cost of
5 kg of sugar.
2) The weight of 2 m long iron rod is 6 kg. Find the length
of the iron rod if it weighs 15 kg.
3) If the cost of 5m cloth is ` 150, then find the length of
cloth for ` 450.
4) One kg of rice is enough for 8 people. Find the quantity
of rice required for 200 people.
33
Inverse proportion
As the speed of the bus increases
the time taken to travel a definite
distance decreases.
he
d
Speed and the distance
travelled are related factors.
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When two quantities are so related such that as one
increases the other decreases, then we say that the two
quantities are in inverse proportion.
Note : When we write the two ratios of inverse proportion
in the form of direct proportion, either the first ratio or
the second ratio should be written in inverse form.
If a:b and c:d are in inverse proportion, then a:b = d:c.
[ In the above equation the inverse of c:d written as d:c ]
Problems on inverse proportion
to
A bus with average speed of 45km/hour takes 8 hours
to travel from Mangalore to Bangalore. Suppose the average
speed of the bus is 60km/hr, calculate the time taken by the
bus to reach Bangalore.
No
t
The time taken by the bus if
the speed is 45 km/hr
Let the time taken by the bus
if the speed is 60 km/hr
{ = hour
{ = x hour
8
The ratio of their speed
= 45 : 60
The ratio of their time taken
=8:x
As the speed increases the time taken decreases, this is
an example for inverse proportion.
34
45:60 = x:8 ( inverse of 8:x )
Hence 45:60 = x:8 ( solving by direct proportion method )
60×x = 45×8
{
d
∴ x = 45 # 8 = 6
60
2.6
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Exercise
= 6 hours
he
If the speed of the bus is 60 km/hr then the time
taken to travel from Mangalore to Bangalore (x)
I. Solve the following problems.
1)
A car travelling with a speed of 50 km/hr reaches
Bangalore from Hubli in 9 hours. Calculate the time
taken, if the car travels with a speed of 60 km/hr to
reach Bangalore from Hubli.
2)
There is enough food for 15 days for 20 persons in a
3)
to
residential school. How long does it last for 30 persons?
12 cows can graze a field for 10 days. In how many days
No
t
will 20 cows graze the same field.
4)
30 people can complete a job in 12 days. How many
days will 20 people take for completing the same job?



35


UNIT - 3
PERCENTAGE
After studying this unit you :
learn the meaning of percentage,

convert the percentage into fraction,

convert the fraction into percentage,

convert the decimal number into percentage,



he
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
d

solve problems on percentage,
understand the meaning of cost price, selling price,
percentage of profit or loss,
calculate profit percent or loss percent in financial
transactions,
calculate the simple interest.
Flex board
25% discount sale is on
85% pass in SSLC
in the view of Deepavali
festival.
to
examination.
No
t
Grameena Rural Bank offers
75% voting in Uttar Pradesh 11% rate of interest (on the
deposits)
You might have seen such posters or advertisement
statements, in the news papers or on the road corners. What
is the meaning of 25%, 85%, 75% and 11%?
Dear children the symbol % is read as percent. Percent
means for every hundred.
Let us learn more about percentage in this unit.
36
Who is getting the pen?
he
d
Divya and Kavya are the two children of Ramesh and
Shantha. Ramesh is a Bank employee who has gone to attend
a conference, while returning he brought a beautiful pen.
Divya and Kavya wanted to have this pen and started to argue
between themselves. Mother Shantha pacified the children
saying that whoever gets more marks in the school test will
get the pen.
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The next day Kavya got 16 out of 20 marks and elder sister
Divya got 19 out of 25 marks in the test. Divya who got 19 out
of 25 took the pen.
Ramesh returned from the Bank took back the pen from
Divya and gave it to Kavya who scored 16 out of 20. Divya
pleaded with father that the pen should be given to her, as
she scored more marks than Kavya.
Ramesh suggested, let us first convert the marks scored by
both of you for hundreds and then let us arrive at a conclusion.
Both Kavya and Divya agreed for this suggestion.
Kavya's score
= 16 out of 20
= 32 out of 40
= 48 out of 60
= 64 out of 80
= 80 out of 100
No
t
to
Divya's score
= 19 out of 25
= 38 out of 50
= 76 out of 100
Divya scored 76 out of 100 means (76 percent) 76%
Kavya scored 80 out of 100 means (80 percent) 80%
When father declared the scores in percentage both the
girls were satisfied.
Percentage is a form of fraction. This is represented by the
symbol % When the denominator of a fraction is 100, then the
numerator represents the percentage.
37
In the above problem of "Who is getting the pen", the
marks obtained by Divya and Kavya were in fractional form.
When these fractions are converted into percentage, then the
comparison becomes easy.
Example 1 : Convert 16 into percentage.
he
Conversion of fraction into percentage.
d
Percentage is a convenient way of comparing quantities.
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50
16 means 16 out of 50. Hence, how much out of 100 ?
50
16 × 100 = 32
50
= 32%
If a fraction is multiplied by 100 then, it gets converted into
percentage.
Conversion of percentage into fraction.
Example 1 :
Convert 30% into fraction.
No
t
to
30% means, 30 out of 100.
30
Hence, we write 30% as 100
When 30 is simplified, it becomes 3
100
∴ 30% = 30 = 3
100 10
Example 2 : Convert 62.5% into a fraction.
62.5% = 62.5 = 62.5 # 10 = 625 = 25 = 5
100 # 10 1000 40 8
100
38
10
Example 3 : The population of a village is 7500, 10% of them
are illiterate. Find how many of them are illiterate?
Population of a village
= 7500
Percentage of illiterate
= 10%
d
Hence, the number of illiterate in that village = 7500 # 10%
I.
= 750
10
100
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Exercise 3.1
he
= 7500 #
Convert the following percentage into fractions:
1) 50%
2) 25%
3) 20%
4) 10%
5) 75%
6) 12.5%
7) 87.5%
8) 37.5%
II. Convert the following fractions into percentage:
1) 1
2
2) 1
4
3) 3
4
4) 1
8
4) 2
5
5) 3
8
6) 8
25
7) 7
20
Kavitha scored 15 out of 25 in a test. Express the marks
scored by her in percentage.
No
t
1)
to
III. Answer the following:
2)
50 students from Navodaya school appeared for
S.S.L.C. examination. 45 of them are declared passed
in the examination. Find the percentage of students
passed.
3)
There are 560 students in a school. 320 are boys. Find
the percentage of girls in that school.
39
Percentage of profit and percentage of loss in
transaction
he
d
Traders / Shopkeepers calculate profit and loss in their
transaction. A business may incur a profit or loss, which does
not indicate the quality of a business. Comparison of loss or
profit is to be calculated for a standard amount of ` 100. This
is necessary for the business men.
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Generally, a businessman buys an article for a definite
price (cost price) and sells it for another price (selling price). If
the selling price is more than the cost price, then it is profit.
If the selling price is less then the cost price, then it is loss.
This can be written in the form of a formula as shown below;
Profit = Selling price – Cost price
Loss = Cost price – Selling price
The profit/ loss incurred by a businessman depends on
the cost price. If the average of s percentage of profit or loss.
to
Example 1 : A shopkeeper buys a coconut for ` 10 and sells
it for `15. Find the percentage of profit.
No
t
A shopkeeper invests `10 and gains a profit of `5. If he
invests `100, profit, what he earns is the percentage of profit.
For `10 the profit is ` 5. Then what is the profit for `100. This
calculation is written mathematically as
Percentage of profit = 5 # 100
10
The percentage of profit can be written in the form of formula.
Percentage of profit =
Profit # 100
(Selling price - Cost price)
# 100
=
cos t price
Cost price
40
Like wise,
Percentage of Loss =
Loss
# 100 = (C os t price - Selling price) # 100
Cost price
C os t price
to
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Let us compare the two business carried out by a
shopkeeper.
Example 2 : A shopkeeper carries the following two business
activites.
Business : 1
Buying a packet of biscuits for ` 10 and selling
it for ` 15.
Business : 2
Buying a sweet box for ` 50 and selling it
for ` 60.
In the above business, which is more beneficial for shopkeeper?
Business :1 Profit earned by the shopkeeper
=15 - 10 = ` 5
in biscuit transaction
Business : 2 Profit earned by the shopkeeper
=60 - 50= ` 10
in sweet box transaction
When we calculate the profit earned by the shopkeeper in
both the businesses, it appears to be more in business 2.
But we have to consider the investment and the earned
in both the cases. Let us compare the profit earned in both
the business in terms of percentage.
Business : 1 The shopkeeper earned ` 5 by investing of ` 10
The shopkeeper earned ` 10 by investing ` 50
∴ Percentage of profit in business 1 = 5 # 100 = 50%
10
∴ Percentage of profit in business 2 = 10 # 100 = 20%
No
t
50
It means, if ` 100 is the investment in both the business,
` 50 is the profit in business 1
` 20 is the profit in business 2
Calculation of profit in terms of percentage is very much
essential for the shopkeepers.
Example 3 : A shopkeeper purchased an old bike for ` 20,000
and then sold it for ` 22000. Find the profit percentage earned
by the shopkeeper.
41
The cost price of the bike
= ` 20,000
The selling price of the bike = ` 22,000
The profit
= selling price - cost price
= 22,000 - 20,000
profit
# 100
cos t price
he
` Percentage of profit =
d
= ` 2000.
20001
# 100
2000010
= 10%
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Example 4 : A vegetable vendor bought 20 kg tomatoes for
` 200. Out of which 5 kg tomatoes were rotten. He sold the
remaining tomatoes at ` 12 per kg . Find out the percentage
of profit or loss incurred to him.
The cost price of 20 kg tomatoes = ` 200
The quantity of rotten tomatoes = 5 kg
Remaining tomatoes
= 20 - 5
= 15 kg
The amount obtained by selling 15 kg at the rate of ` 12
per kilogram
= 15 ×12 = ` 180
Cost price > Selling price
= 200 > 180 ∴ Loss
The loss
= cost price- selling price
= 200- 180
= ` 20
If the cost price is more than the selling price, the vegetable
vendor incurs loss.
loss
# 100
cos tprice
20 # 100
=
200
= 10%
` Percentage of loss =
42
Exercise
3.2
I. Answer the following:
Rahul bought a mobile for ` 500 and sold it for ` 625.
Find the percentage of profit earned by Rahul.
2)
Mary bought a scooter for ` 20 , 000 and sold it
for ` 21,000. Find out the profit percentage earned by
Mary in this business.
3)
A shopkeeper bought 1000 coconuts for ` 8500, out of
which 50 coconuts were spoiled. He sold the remaining
coconuts for `12 each. Find the profit and profit
percentage earned by the shopkeeper.
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1)
4)
A computer shop owner bought a computer from the
manufacturer for ` 12,000. He sold it for ` 15,000. Find
the profit percentage earned by him.
Simple Interest
No
t
to
Somanna had grown paddy in
his 4 acres land. The pump which
is attached to the borewell is not
working at present. If he does not
replace the old pump with a new
- one, the paddy crops dry up and
there by incurring a huge loss to
him. He is not having enough
money to buy a new pump.
How to solve this problem?
He obtained a loan from the
village co-operative bank and
bought a new pump. Now, it is
possible for him to supply water to the grown up crops. At
the end of the season, he got good yield as well as profit. He
repaid the bank loan.
43
e
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e.
s
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of
It
Just like the above situation, people need money for their
various activities like building houses, investing money in
a business, buying land, the children's education and for
marriage. He borrows money from friends, banks or financial
institution for a period of time. This is known as loan. After
a fixed period of time he has to repay the amount he had
borrowed along with some extra money. For the usage of
money for some time, the extra money should be given.
We also come across people who have money. They do not
spend unnecessarily, but save it for the future, so that it can
be used for children's education, or building new houses or
for the marriage of their daughter/son. The problem is that
they cannot keep the money with them due to the fear of theft.
What is the solution for this?
to
of
5000 years ago people
started collective agricultural activities. In those days they
used barter system of trading. They borrowed seeds. They use
to get more grains from sowing one grain. While returning
the loan they paid more than what they received as loan.
Suppose they keep the money in a bank?
No
t
s
know this: Approximately about
Yes, money is safe in a bank. We can withdraw the money
whenever we want. We get interest for the money which we
deposit in the bank. The banks use this money for giving
loans to the needy people. In turn we are also involved in the
activity of nation building.
44
•
•
•
•
•
•
d
•
he
•
The money deposited/borrowed is called Principal. It is
denoted by the letter 'P'
The extra money paid on the principal after a period of
time is called interest. It is denoted by the letter I.
The total money paid is called amount.
Thus, Amount = Principal + Interest.
Interest for every `100 for one year is known as rate of
interest per annum. This is denoted by the letter 'R'. It
is denoted by %.
The interest calculated uniformly on the principal alone
throughout the loan period is called simple interest. In
other words, it is the interest paid on the principal alone.
T = time for which the money is kept in the bank. T is
always expressed in years.
The money which is kept in banks is called deposit.
The interest on the deposit is calculated in the same way
as the interest is calculated for loan. (Deposits could be
considered as loan given to the banks by the people)
Usually, the rate of interest for the deposit is less than
the rate of interest for loans.
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Activity :
Collect more information about banking from elder people
(knowledgeable) or by visiting a bank which is nearby.
No
t
We have already understood the terminologies like loan,
deposit, interest etc. Anybody can come across situations like
depositing the money or taking loans. So, one must have the
knowledge about calculating interest for the loans as well as
deposits.
Calculation of simple interest
Rahim has taken a loan of `2000 from a bank for 2 years.
Calculate how much more money he has to pay to the bank
after two years (Rate of interest is 12% per annum).
45
Rate of interest 12% means
R = 12%
For every `100 loan, the interest for 1 year = ` 12
There are 20 hundreds in ` 2000
∴ Interest for 20 hundreds for one year
= ` 2000 # 12 # 1
100
= ` 2000 # 12 # 2
100
d
∴ Interest for 20 hundreds for two years
he
= ` 480
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Simple interest= Pr incipal # Rate of int erest # Time (in Years)
100
I=
P # R # T (years)
100
to
It is easier to calculate the simple interest by using the
above formula.
Example :
Sujeeth kept ` 5000 for 2 years in a bank. The rate of
interest is 8%. Find the simple interest and the amount
received by him after 2 years.
The amount deposited by Sujeeth = P = ` 5000
Rate of Interest
= R = 8%
Period (Time)
= T = 2 years
The simple interest got by Sujeeth
P#R#T
I =
No
t
100
5000 # 8 # 2
=
100
= ` 800
The amount received by Sujeeth after 2 years
= Principal + Interest
= ` 5000 + 800
= ` 5800
PTR
Using I = 100 we can calculate any one of IPRT if the other
three are known.
46
Exercise
3.3
Principal
1
` 2500
12%
2
`
`
` 8450
10%
3
`
`
` 7500
15%
4
`
`
` 12,500
8%
2
`
`
` 2400
9%
3
`
`
3
4
5
Time
Simple
Interest
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No
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Rate of
Interest
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I. Answer the following:
1) Find the simple interest for ` 3000 for 3 years at the rate
of 12% per annum.
2) Find the simple interest for ` 4500 for 2 years at the rate
of 11% per annum.
3) Fill up the table given below.
Amount
to
4) Rekha obtained a loan of ` 3000 from a cooperative bank
No
t
for 2 years at the rate of 7% per annum. Find the simple
interest she has to pay to the bank after 2 years.
5) Vasanth starts a Dairy (milk) business by obtaining a
loan of ` 25,000 from a bank at the rate of 15% per annum.
Calculate the amount to be paid to the bank after 4 years.



47


UNIT - 4
SIMPLE LINEAR EQUATIONS
After studying this unit you :


d
he

understand the meaning of equality of two mathematical
statements,
understand the meaning of equation, inequation and
simple linear equation,
convert a verbal statement to equation,
solve simple linear equations.
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
One day Megha, in mathematics class told her teacher
that she has learnt a game based on the algebraic expressions
which she has learnt in the previous semester.
to
The teacher appreciated her and invited her to present the
game to the whole class. Megha began the game by asking
Meera to choose a number of her choice. She asked to multiply
it by 5 and add 10 to the product. Now she asked Meera to
tell the result. She said it is 65. Immediately Megha told the
number chosen by Meera as 11. When Meera nodded the whole
class was surprised.
Everybody was interested to know the game. Do you know
how it works?
No
t
Megha began to explain. Let the number chosen by
Meera be 'x'. When Meera multiplied the number by 5, she
got 5x, then she adds 10 to the product which gives 5x+10.
The value of (5x+10) depends on the value of x. Thus, if x = 1,
5x +10 = 5 × 1 +10 =15.
This means that if Meera had chosen '1' her result would
be 15. If she had thought of 5, result would be 35. Similarly
by substituting 11, we get 65.
48
he
d
Let us find out the number chosen
by Meera :
Let the number chosen by Meera be x
On multiplication of x by 5, we get 5x
10 is added to 5x it gives 5x +10
5x +10
65
As a result we get 5x + 10 = 65
Here 5x +10 is an algebraic expression and is equal to 65
So, we get the 5x + 10 = 65.
to
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What is an Equation ?
Study the following verbal statements and their
corresponding mathematical equation.
Example 1 : If you add 6 to a number, the sum will be equal to 20
Let us convert this into a mathematical statement
Let the unknown number be 'x'
Add 6 to the unknown number
We get x + 6
Their sum is equal to 20
So , x + 6 = 20 .......... this is a mathematical equation.
Example 2 : When 10 is subtracted from 2 times a number the
result is 15.
How do you convert this into a mathematical equation ?
No
t
Let the unknown number be y.
Two times the number will be
'2y' .
By subtracting 10 from 2y we get
2y-10 .
Their difference is equal to 15.
So 2y-10=15
This is a mathematical equation.
49
3x+5
10
Both sides of this
mathematical equation are
equal
Observe the above mathematical statements,
i) 5x + 10 = 65
ii)
'Robert Recorde' used the
symbol '=' in his algebra book
during 1557.
x + 6 = 20
d
iii) 2y - 10 = 15
=
5x-1
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3x+5
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In an equation there will be two algebraic expressions or
numbers on both sides with '=' sign between.
Algebraic
expression on Left
hand side (L.H.S.)
Equality sign Algebraic expression
on Right hand side
(R.H.S.)
Equation : Two algebraic expressions connected by the
equal sign (=) is called an equation.
No
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Inequation :
If in an algebraic expression instead of = sign, the sign like
< (less than), > (greater than) or ! (not equal) is used, then
such types of algebraic expressions are called inequations.
Thus, 4x + 5 > 65 is not an equation, it is an inequation. It
shows that the value of 4x + 5 is greater than 65.
Example :
1) 3x-7 = 10 is an equation
2) 4x+5 > 10 is an inequation
3) 7x-8 < 12 is an inequation
4) -5x+2 ≠ 15 is an inequation
Note : An equation remains the same when the expressions
on left side and right side are inter-changed.
50
Convertion of verbal statements into equation.
Example 1 : When 10 is added to a number, we get 25
Solution :
Let the number be 'x'
When 10 is added to 'x' we get x + 10.
So, x + 10 is equal to 25
Solution:
Let the number be 'y'
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Two times y means '2y'
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Example 2 : Two times of a number is 40
d
Thus, x + 10 = 25 is an equation
So '2y' is equal to 40
∴ The equation is 2y = 40
Example 3 : 5 is subtracted from a number we get 30
Solution:
Let the number be 'z'
Five subtracted from z gives z - 5.
Thus, the equation is z - 5 = 30
Exercise
4.1
to
I. Convert the following verbal statements into equations.
Statement
No
t
1) When 6 is added to a number we get 18
2) When Twice a number is multiplied by 5 we
get 40
3) When 6 is added three times to a number we
get 30
4) When a number divided by 10 we get
quotient 4 leaving no reminder
51
Equation
Linear Equations
Variable : The literal numbers (unknown) in an equation
are called variables of the equation. These are usually denoted
by small letters of english alphabets such as x, y, z, u, v, w etc.
Example :
he
d
Degree : The highest power or exponent of the variables
in an equation is called its degree.
Highest Power of
Variable
Degree
3x+4=12
One
First degree equation
2y2+6=8
Two
Second degree equation
m3-1=0
Three
Third degree equation
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Equation
If the variables of terms of an equation is in first degree,
then such equations are called linear equations.
Example :
to
5x - 4 = 0, 3y + 6 = 12, m + 2 = 0, x + y = 15, 3m + 5n = 50
No
t
An equation containing only one variable with highest
power one is called a simple linear equation.
Solving simple linear equations.
Finding the value of the unknown variable in an equation is
called solving the equation and the found out value is called
solution.
Inspection method: It is a method of solving the solution by
trying out various values for the variable. This is also called
trial and error method.
52
d
Example 1 :
Solve the equation x + 3 = 6 by Inspection method.
The given equation is x + 3 = 6, Let x have different values.
Write the LHS and RHS of the equation. At one particular
value of 'x' the RHS is equal to LHS.
Left hand side is equal to right hand side.
Value of x
LHS
RHS
0
3
6
1
4
6
2
5
6
No
3
6
6
Yes
4
7
6
No
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Is LHS = RHS?
No
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No
Observe that for x = 3, LHS = RHS in the equation
Therefore, x = 3 is the solution of this equation.
Example 2 : Solve the equation 2y - 3 = 5 by trial and error
method.
Give different values to y and let us prepare the table.
LHS
RHS
Is LHS = RHS?
1
-1
5
No
2
1
5
-
3
3
-
No
4
5
5
-
5
8
5
No
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Value of y
LHS = RHS only when y = 4
Therefore, the solution of the equation 2y - 3 = 5 is y = 4
53
x 8 11 by trial and error
Example 3: Solve the equation 3
+ =
method using the table.
LHS
RHS
Is LHS = RHS?
1
2
3
4
5
6
7
8
9
10
1 8
+
3
11
No
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Value of x
Exercise
4.2
I. Choose the correct answer
1) The variable in the equation 2z + 6 = 18 is
b) z
c) 6
to
a) 2
No
t
2) The equation among the following is?
3)
a) 2 x + 3 = 8
b ) 2x + 3 < 8
c) 2x + 3 > 8
d ) 2x + 3 ≠ 8
The equation with y=4 as solution is
a) 2y + 3 = 0
b) y - 7 = 2
c) y + 3 = 7
d) y + 4 = 0
54
d) 18
Solution
1) t = 2
2) t = 10
3) t = 8
4) t = 0
5) t = -1
d
II. Match the Following
Equation
a) 2t = 16
b) 4 = t + 2
c) t - 5 = 5
d) 2-t = 3
2) 3y + 6 = -9
3) 14-k = 2k + 4
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1) x - 5 = 8
he
III Write the L.H.S. and R.H.S. of the following
equations.
IV. Solve the following equations by the trial and error
method
1) x-4=9
4) p +6=9
2
2) x = 6
8
3) 5y-3=12
5) 6m-1=29
V. Express the following statements in the form of
equations and solve
1) The sum of x and 9 is equal to 15
2) Twice a number decreased by 8 is equal to 18
3) When 22 is added to Megha's age the sum is equal to 35
to
Elimination method of solving the equations
No
t
When there are more terms on both sides of the equation,
the trial and error method to get the solution takes much
time .So method of elimination helps to solve equation in a
short time.
Consider the following example.
Solve x +10 =15
What number is to be added to 10 to give 15 ?
Clearly it shows that the number must be 5 to make both
sides equal.
Hence, x=5 is the only value of the variable, which satisfies
the given equation x+10=15.
Therefore 5 is the solution of the equation.
55
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We see from the figure given below that the two weights in
the pan represents the two sides of the equation which are
being balanced. the equation can be compared to weighing
balance as shown in the figure.
x+10
5+10
15
15
Rules for elimination
Rule - 1 : If equals are added to equals, the sums are equal.
x=5
to
x+3
5
No
t
x
x+3=?
Example : Solve x - 5 = 15
Solution: x -5 = 15
Adding 5 to both sides
x - 5 + 5 = 15 + 5
x + 0 = 20
Therefore, x = 20
56
5+3
Rule - 2 : If equals are subtracted from equals, the result is
also equal.
x=8
x
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x-3=8-3
x-3
8-3
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8
Example : Solve x+8=20
Solution : x +8=20
By subtracting 8 from both sides we get,
x +8-8 =20-8
x + 0 = 12
∴
x =12
Note : The term which is eliminated from one side of the
equation appears on the other side but with opposite sign.
to
Rule - 3 : If equals are multiplied by equals, the products are
equal.
x=4
No
t
x×3=4×3
x
x×3
4
57
4×3
x 12
Example : Solve 4
=
x 12 Multiplying both sides by 4,
Solution:
4=
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Observe the balance and frame a rule.
d
x × 4 = 12 × 4
4
∴ x = 48
x
x/3
7
7/3
Example : Solve 7x = 35
Solution: 7x = 35
7x 35 (Dividing both sides by 7 we get)
=
7
7
No
t
to
∴x = 5
Rule - 4 : If equals are divided by equals, the quotients are
equal.
Remember : The value of the equation does not change if.
a) The same number is added to both the sides of an
equation
b) The same number is subtracted from both the sides of
an equation
c) Both the sides of the equation are multiplied by the
same number
d) Both sides of the equation are divided by the same
non-zero number
58
Exercise
4.3
I. Solve the following equations by elimination method:
2) x-12=9
3) 15+y=18
4) 2k+6=0
5) m =3
5
6) 2p=p+12
7) k- 1 = 3
2 4
8) 7 x= 105
2
2
10) 10-4x=26
11) 3(x-7)=24
d
1) x+8=15
he
9) 6x-3=15
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12) 16-5x=6
II. Mathematical Puzzle
1) I am an integers. If you add 4 times the number
of intersecting points to of this figure me you get 46,
find my value ?
2) I am a two digit number. I am a multiple of 11. When I am
divided by 7, I leave no remainder. When 4 is added to the
quotient 15 is obtained, What is my value ?
to
3) I am a number. If you double me only one decade is enough
No
t
to reach century! If you divide me to three equal parts then
also only one decade is enough to become silver jubilee
What is my value?
4) Tell me who am I ? Take away from me the number eight.
Divide further by a dozen to come up with a full team for
a game of cricket.



59


UNIT- 5
CONGRUENCE
After studying this unit you :
identify the congruency of a figure through
superimposition

identify the congruent figure

identify corresponding parts of congruent figures

define congruency
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
Let us take two identical one rupee coins. Place one on the
other. Does one exactly fits the other?
Similarly place a 50 paise coin on a ten rupee coin. Observe
whether one exactly fits the other. You see that, it is impossible
to superimpose one on the other.
Congruency is one of the fundamental concepts in geometry
to
which is used to classify the geometrical figures on the basis
of their shapes.
No
t
Study the following patterns
Copy the figure on a plain sheet of paper. Cut the shaded
part and un-shaded part of each figure. Superimpose them
to see if one exactly fits with the other.
60
Two geometrical figures are said to be congruent, if they
have same shape and size.
Remember: The symbol for congruence is ,
The symbol for non-congruence is @
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Activity 1:Take two ten rupee notes. Place them one on the
other. What do you observe?
One note covers the other completely and exactly.
From the above activity, we observe that the currencies
are of the same shape and the same size.
Check whether the following objects are congruent or not
a) Postal stamps the of same denomination.
b) Biscuits in the same pack.
to
c) Photos of same measurement
No
t
Activity 2:
Draw two circles on a sheet of transparent paper using a
bangle or by any circular object. Cut the sheets of paper
along the circular path. Superimpose the paper one above
the other write your observation.
Can you say that the two
circles are congruent ?
61
Activity 3: Examine the congruency of 3 ABC with 3 XYZ
triangle given below
A
0
Y
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C
600
Z
d
600
600
0
B 60
600
he
60
X
Congruent Triangles
Observe the ∆ABC and ∆DEF.
6
6
Are they are congruent? They have
4
4
the same size and same shape. If
B 3 C
E 3 F
you superimpose, one triangle on
another, they coincide with each
other.
Vertex A coincides with vertex D, Vertex B coicides with
Vertex E and Vertex C coincides with Vertex F.
A
D
Similarly,
No
t
to
Side AB coincides with
side DE, side BC coincides
with side EF and side AC
coincides with side DF.
Know this:
Coinciding parts are
called corresponding
parts
Thus, the above two triangles have six pairs of corresponding
elements namely 3-angles and 3-sides.
&∆ABC ≅ ∆DEF
(Read as triangle ABC congruent to triangle DEF)
62
L
R N
M
d
When two congruent figures are to be P
named, it is a convention to name them
in such a way that the corresponding Q
components appear in same order. If the
two adjoining triangles are congruent then
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P = L, Q
= M, R = N
PQ = LM, QR
= NM, PR = LN
3 PQR / 3 LMN
Congruent Figures and Similar Figures
Notice the shape and size of the figures given below.
These 2 Squares are congruent.
2cm
2cm
2cm
2cm
2cm
2cm
2cm
2cm
3cm
4cm
m
4cm
These 2 circles are congruent.
They are of the exact same size and
shape.
These 2 triangles are congruent.
5c
m
5c
No
t
to
3cm
3cm
They have exactly the same size and
shape.
They are of exact size and shape.
3cm
The figures which are having exactly same size and shape
are termed as Congruent figures.
63
Similar Figures have the same shape, but not same size.
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These two rectangles are similar. They have the
same shape, but not the same size.
d
Observe the rectangle given in the figure. Do you find any
difference among them? Yes they have the same shape but
different in size. Similarly observe these circles. What is your
observation? Again you find that shape is same but size is
different.
These two circles are similar. They have same
shape, but not same size.
These 2 triangles are similar. They are of the
same shape, but not the same size.
Remember :
1. Congruent figures are of the same size and shape.
2. Similar figures are of the same shape but may differ
in size.
to
3. All congruent figures are similar but its reverse may
not true.
No
t
Exercise
5.1
I. Choose the correct answer:
1) Which of the following figures are congruent ?
a)
K
L
M
N
P
Q
S
R
64
A
P
Q
S
R
Q
A
B
3 CM
P
3 CM
c)
C
D
R
5 CM
C
5 CM
d)
2cm
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3cm
d
B
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b)
2) Which of the following figures are similar.
a)
K
L
P
M
N
S
b)
A
Q
G
600
600
600
to
0
B 60
A
C
C
B
No
t
c)
R
H
Q
P
D
d)
3cm
400
R
S
4cm
65
800
I
3) If ∆ ABC ≅ ∆DEF, the corresponding element of AC is.
a) DE
b)
DF
c) ∠B
d)
∠F
II. Fill up the blanks with suitable answer
d
1) The figures having same shape and same size are called
______________
he
2) The symbol to represent congruent is ___________ .
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3) All congruent figures are always ___________ .
4) The figures which are having same shape but may differ
in their sizes are called ___________ .
III.
Construct the figures according to the given data.
Verify whether these two triangles are congruent and
Write their corresponding vertices, sides and angles.
P
4 CM
3 CM
A
R
C
4 CM
No
t
IV.
3 CM
to
B
Q
Activity : Identify the congruent and similar figures in your
surrounding discuss in group.



66


UNIT- 6
GEOMETRICAL CONSTRUCTIONS
After studying this unit you :
draw a perpendicular bisector to a given line segment,

draw an angle bisector for a given angle,

construct angles of 300, 450, 600, 900, 1200, 1350 and 1500,

construct an angle equal to a given angle using
compasses and scale,

he
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
d

draw perpendicular line to a given line through a point
on the line,
draw perpendicular line to a given line from a point
not on the line.
Perpendicular lines
to
We know that two lines or rays
or segments are said to be
perpendicular to each other if the
angles formed at the point of
intersection of them are right
angles.
No
t
Where do you find perpendicular
lines in your day-to-day life?
One right angle = 900
Observe the following :
The angle between wall and the floor, angle between
the plank and the legs of the table, angle between two adjacent
edges of black board, and the angle between two adjacent
edges of the door.
Can you see right angles in these examples?
67
Perpendicular bisector
B
A
B
B
A
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d
Activity : Take a sheet paper preferably semi-transparent .
Draw a line segment AB on it. Fold the paper in such a way
that point B must coincide on point A exactly along with the
line.
Make a crease and unfold the paper. Now observe that the
line obtained by the receive is perpendicular bisector of the
line AB .
The straight line, which is perpendicular and divides a given
line segment into two equal parts, is called the perpendicular
bisector of a given line.
to
Construction of the perpendicular bisector to a given
line segment
No
t
Example: Construct a perpendicular bisector to a line
segment, AB = 8 cm.
Given: AB = 8 cm.
To construct: Perpendicular bisector of AB .
Steps of construction:
1) Draw a line segment AB = 8 cm using
a scale.
68
A
8CM
B
2) With A as the centre and radius more
A
B
P
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3) With B as the centre and the same
radius as in step - 2, draw two arcs to
intersect the arcs drawn in step - 2.
he
d
than half of AB , draw two arcs (using
compass) on both sides of AB .
Name the point of intersection of the
arcs as P and Q as shown.
A
4) Draw a straight line through points
P and Q,
Let PQ cut AB at O
B
Q
P
A
o
B
to
Q
PQ is the perpendicular bisector of AB .
No
t
Verification: Measure ∠POB. Is ∠POB = 900 ?
Measure AO and OB . Are they equal?
You will find that ∠POB is right angle and AO = OB .
Hence, PQ is the perpendicular bisector of AB
Think : Does AB bisects PQ ?
Think! : What would happen if the radius taken for
constructing arcs in step - 2 and 3 is less than half of AB ?
69
How do you construct an angle equal to a given angle
without using protractor?
Construct ∠PQR of any measure
using pencil, scale and protractor.
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1)
he
P
Q
3)
4)
Y
Q
X
B
Draw another ray BC .
Now with B as the centre, draw
an arc with the same radius
chosen in step- 2. Name the point
of intersection of arc and BC as
B
point M.
N
With M as the centre and radius
equal to (XY), draw an arc to cut
the arc drawn in step- 4, using
compass. Name the arcs inter
B
section point as N.
No
t
5)
Using compass, with Q as the
centre, draw an arc with any
suitable radius to cut the arms
QR and QP . Name the point of
intersection of arc and arms as X
and Y.
6)
N
Join BN using scale and pencil.
extend it to get BA .
B
70
R
P
to
2)
d
Example : Construct ∠ABC which is equal to ∠PQR.
Given: ∠PQR
To construct: ∠ABC = ∠PQR
Steps of construction:
R
C
M
C
M
C
A
M
C
Note : Using the protractor, measure ∠ABC. Are the measure
of ∠ABC and ∠PQR same?
Construction
600 angle
using scale and compass?
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d
Activity : Draw a circle of fixed radius with 'o' as centre
Mark a point 'A' on the circle keeping 'A' as centre draw an
arc to cut, the circle at point B. Now draw one more arc
keeping the same radius and 'B' as centre. How many arcs
we can draw? OA, OB, OC What is measure of +AOB ? Are all
the angles equal ?
Steps of construction (600 angle) :
Draw a ray OB using a scale and
a pencil.
2) With O as the centre and with
any suitable radius draw an
arc using compasses, which
cuts OB . Name the point of
intersection as P.
3) With P as the centre and with
the same radius as in step2 , draw an arc cutting the
previous arc. Name the point of
intersection as Q.
O
O
No
t
4)
B
P
B
Q
to
1)
P
O
B
A
Join OQ using a scale and a
pencil. Extend it to get OA .
Q
O
600
B
Using the protractor, measure ∠AOB. ∠AOB measures 600.
71
Do you know!
In the above figure, why is the ∠AOB formed measure 600?
Because OP, OQ and PQ sides of ∆ POQ are equal.
∴ ∆ POQ is an equilateral triangle.
1200
using scale and
d
Construction of an angle of
compass.
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How many 600 makes 1200 angle?
Two 600 angles make one 1200 angle.
So construct 600 angle twice to construct an angle of 1200
using a scale and compass.
Steps of construction:
O
1)
Draw a ray OB .
2)
With O as the centre, draw an
arc with suitable radius that
cuts OB . Name the point of
intersection as P.
4)
With Q as the centre and with
same radius as in step- 3, draw R
an arc so as to cut the arc
drawn in step - 2. Name the
point of intersection as R.
72
P
B
P
B
Q
to
With P as the centre, draw
an arc with the same radius
as in step-2 intersecting the
previous arc. Name the point of
intersection as R.
No
t
3)
O
B
O
Q
O
P
B
Join points O and R using a
scale and a pencil. Extend it to
get ray OA
Q
R
O
1200
A
Thus formed ∠AOB measures 1200.
Draw a line segments of length given below and then
draw a perpendicular bisector to each of them using a
scale and compasses.
a) 6 cm
2)
3)
b) 8 cm
c) 7.4 cm
d) 66 mm
Draw a line segment PQ =10 cm. Divide the line segment
into four equal parts using a scale and compasses.
Measure the length of each part.
Measure the following angles and construct these angles
without using the protractor.
A
Q
to
No
t
B
4)
6.1
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1)
B
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Verification: Measure ∠AOB using protractor.
Exercise
P
d
5)
C
L
R
P
M
N
Construct each of the following angles using a scale (ruler)
and compasses.
a) ∠XYZ = 600
b) ∠DEF = 1200
73
Bisector of an angle
d
Activity : Take a semi-transparent sheet of paper. Draw an
angle ABC on it using scale and protractor. Fold the paper at
point B in such a way that line AB must coincide with the line
BC.
A
A
C
B
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he
A
C
B
C
Make a crease and unfold the paper. The line obtained by
crease is the bisector of ∠ABC.
An angle bisector is a ray or a line which divides the angle
into two equal parts.
To construct bisector of a given angle
Example 1: Construct an angle bisector to the given ∠AOB = 500
Given: ∠AOB = 500
to
To construct:
No
t
The bisector of ∠AOB.
Steps of construction:
1)
B
Construct ∠AOB = 50 using a pencil,
a scale and a protractor
0
O
74
500
A
Taking ‘O’ as the centre and with a
suitable radius draw an arc which
cuts arms OA and OB .
B
Q
With Q as the centre and with the
same radius as in step - 3, draw
another arc, which cuts the previous
arc.
Name the point of intersection as R.
to
4)
With P as the centre and radius more
than half of PQ , draw an arc in APQB.
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3)
P
he
Name the point of intersections
on side OA and OB as ‘P’ and ‘Q’
O
respectively.
No
t
5) Join points O and R using a scale and
a pencil. Extend it to get the ray OR .
OR is the bisector of ∠AOB
Verification:
Measure ∠AOR and ∠ROB. What do you infer?
75
A
d
2)
Constructing an angle of 900 using a scale and compasses.
Discussion of Steps of construction:
1)
Draw a ray BC .
2)
With B as the centre, draw an
arc with any suitable radius
that cuts BC . Name the point
of intersection as P.
3)
With P as the centre, draw an
arc with same radius as in
step-2 cutting the previous arc.
Name the point of intersection
as Q.
4)
With Q as the centre, draw an
arc with same radius as in
step- 2 cutting the arc that
drawn in step-2. Name the point
of intersection as R.
B
d
he
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No
t
5)
C
With Q as the centre and radius
more than half of QR , draw an
arc in the exterior of ∠QBR as
shown in figure
(imagine ∠QBR)
76
With R as the centre, draw
another arc with same radius as
in step-5 that cuts the previous
arc.
Name the point of intersection
as A.
S
Join points B and A using a
scale and a pencil. Extend it to
get the ray BA .
S
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7)
he
A
d
6)
Thus formed ∠ABC measures 900 angle.
Verification: Measure ∠ABC using protractor.
Think!
In this method, we get angle 900 Why?
Is there any other method to construct 900?
to
Alternate method to construct
No
t
Steps of construction:
1)
Draw a line segment AB ,
mark a point 'O' on it.
2)
With O as the centre, draw
semicircle with suitable
radius that cuts line segment
AB at point P and Q.
77
900.
5)
O
Q
B
d
P
he
4)
With P as the centre and
radius more than half of
PQ , draw an arc above the
line segment AB as shown in
A
figure.
With Q as the centre and with
same radius as in step -3 ,
draw another arc, which cuts
the previous arc.
Name the point of intersection
as C.
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3)
Join points O and C using a
scale and a pencil and extend
to get ray OC
Thus formed ∠COB measures 900. What is the measure of
to
∠COA
Think!
No
t
Compare construction of 900 with that of perpendicular
bisector. What do you infer?
Construction of an angle of
compass
300
using a scale and
Note that by bisecting the angle of 600, we get an angle of 300.
Steps of construction :
1)
Draw a line segment AB .
Mark a point 'O' on
78
A
O
B
A
With P as the centre, draw
an arc with same radius
as in step- 2 cutting the
previous arc. Name the point
of intersection as Q. Imagine
QOP .
P
O
B
d
Q
A
he
3)
With O as the centre, draw an
arc with suitable radius that
cuts OB . Name the point of
intersection as P.
O
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2)
P
B
P
B
Now bisect ∠QOP using compasses and a scale.
4)
Q
A
Q
A
No
t
Name the point of intersection
as C
6)
O
With Q as the centre and with
same radius as in step- 4,
draw another arc, which cuts
the previous arc.
to
5)
With P as the centre and
radius more than half of PQ ,
draw an arc in the interior of
∠QPB as shown in figure.
C
P
O
B
Q
Join points O and C using a
scale and a pencil and extend
it.
C
A
O
30
0
P
B
Thus formed ∠COB measures 300.
Think! What is the measure of
Why?
79
∠COA? It must be 1500
Construction of an angle of
compass
450
using a scale and
By bisecting an angle of 900, we get an angle of 450.
Steps of construction:
With 'O' as the centre, draw a
semicircle, with any suitable
radius, that cuts AB . Name
the point of intersection as P A
and Q.
3)
With P as the centre, draw
an arc with same radius as
taken in step- 2 cutting the
previous arc. Name the point A
of intersection as L.
With L as the centre, draw an
arc with same radius as taken
in step- 2 to cut are LQ Name
A
the point of intersection as M.
Q
O
5)
With L as the centre, draw
an arc with radius more than
!
half of ML above the arc ML .
as shown in figure
A
80
P
B
P
B
L
Q
O
L
M
Q
O
No
t
to
4)
B
d
2)
O
he
Draw a line segment AB . Mark
A
a point 'O' on
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1)
B
L
M
Q
P
O
P
B
With M as the centre and with
same radius taken in step-5,
draw another arc, which cuts
the previous arc. as shown
in figure Name the point of
A
intersection as C.
C
L
M
Q
B
P
O
Join points O and C by using
a scale and a pencil. Let the
OC cut the semicircle at point
N.
C
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7)
he
d
6)
N
M
Q
A
O
L
B
P
Thus formed ∠COB measures 900. Now bisect ∠COB using
scale and compasses
With P as the centre and
radius more than half of PN ,
draw an arc in the interior of
∠BPC as shown in figure
9)
With N as the centre and with
same radius as in step - 8,
draw another arc, which cuts
the previous arc.
Name the point of intersection
A
as D.
81
N
M
Q
A
No
t
to
8)
C
L
P
O
C
M
Q
N
O
L
P
B
D
B
C
10) Join points O and D using
M
A
Q
N
O
450
P
B
d
scale and pencil and extend
to get OD
D
L
he
Thus formed ∠BOD measures 450.
Think!
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What is the measure of ∠DOA? It should be 1350. Why?
How do you construct an angle of 1500 using scale and
compass?
Steps of construction:
Draw a line segment AB . Mark
A
a point 'O' on
2)
With O as the centre, draw
semicircle with suitable radius
that cuts AB at point S and P.
A
3)
With P as the centre and with
same radius chosen in step- 2,
draw another arc, which cuts
the previous arc. Name the
A
point of intersection as Q.
82
B
O
S
O
No
t
to
1)
P
B
P
B
Q
S
O
With Q as the centre and with
same radius chosen in step-2,
draw another arc which cuts
the semicircle. Name the point
A
of intersection as R.
S
R
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more than half of RS , draw
an arc in the interior of ∠RSA
shown in figure. (imagine ∠RSA)
A
to
With S as the centre and with
C
same radius chosen in step- 5,
draw another arc, which cuts
the previous arc. Name the
A
point of intersection as C.
7) Join points O and C by using a
No
t
scale and a pencil and extend
ray OC
S
A
S
B
Q
B
P
O
Q
R
S
P
O
R
C
Q
O
Thus formed ∠BOC measures 1500.
Think!
Can Angle 1500 be constructed by constructing 300?
83
B
P
O
Fig 7.11
d
5) With R as the centre and radius
6)
Q
R
he
4)
P
B
Constructi on of an angle of
acompass
1350
using a scale and
Steps of construction:
With O as the centre, draw
semicircle with a suitable
radius that cuts line segment
AB at point P and Q.
3)
With Q as the centre and
radius more than half of
PQ , draw an arc as shown
in the figure.
With P as the centre and with
same radius chosen in step3, draw another arc which
cuts the previous arc. Name
the point of intersection as
R.
A
P
A
P
5)
O
Q
B
O
Q
B
R
A
P
No
t
to
4)
B
O
d
2)
A
he
Draw a line segment AB .
Mark a point 'O' on
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1)
Join points O and R using
a scale and a pencil. Let it
cut the semicircle at S. Thus
formed ∠AOR measures 900.
O
B
Q
B
R
S
A
P
O
Now bisect ∠AOR using scale and compasses.
84
Q
A
d
S
P
Q
O
S
A
P
C
O
Q
B
Q
B
R
S
A
No
t
B
R
C
Join points O and C with the
help of a scale and a pencil
and extend to get ray OC
to
8)
With S as the centre and with
the same radius chosen in
step - 6, draw another arc
which cuts the previous
arc. Name the point of
intersection as C.
R
he
7)
With P as the centre and
radius more than half of
, draw an arc in the
interior of APSR as shown
in the figure.
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6)
P
O
Thus formed BOC measures 1350.
How to construct 22.50 using acompass and a scale?
When you bisect 450 using compasses and a scale, you
will get 22. 50. Think! How to construct 150, 67.50 and 750 using
compasses and a scale?
85
Exercise
1)
6.2
Construct each of the following angles using the protractor
and draw an angle bisector in each case using a scale
c) 360
d) 1320
Construct the angles of following measurement using
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2)
b) 1100
he
a) 800
d
(ruler) and compasses.
a scale (ruler) and compasses. Verify the angle using
protractor.
a) 900
b) 300
c) 450
d) 1500
e) 1350
How to draw a perpendicular to a given point on a line?
Activity:
Take a sheet of semi-transparent paper. Draw a line AB on
it. Mark a point P anywhere on the line AB. Fold the paper
No
t
to
at point P exactly along with the line AB.
A
P
B
A
P
A
P
B
B
Now make a crease and unfold the paper. Now we get the
crease which is perpendicular to line AB at point P.
86
2)
With P as the centre, draw
two arcs with a suitable
radius which cut AB at X
and Y such that P lies in
between X and Y.
3)
With radius more than half
of XY and X as centre draw
an are above AB
4)
Draw another arc that cuts
the previous arc with centre
at Y and with same radius
as in step - 3 . Name the
point of intersection as Q.
to
No
t
5)
he
Draw a straight line AB and
mark point P on it
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1)
d
Procedure of construction
Given: A line AB and any point P lying on it.
To construct: A line through P which is perpendicular
.
to the
Step of construction:
Using a scale and a
pencil, join PQ . Now PQ is
.
perpendicular to
Verification: Measure ∠BPQ.
87
Drawing a perpendicular to a given line from a point
which is outside the line.
P
P
P
A
A
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Activity:
Take a sheet of semi-transparent paper. Draw AB on it.
Mark a point P anywhere outside the line on the paper. Fold
the paper at point P exactly along with the AB .
B
B
B
Now make a crease and unfold the paper. Now we get the
crease which is perpendicular to line AB from point P .
Procedure of construction
Given:
AB is a line and a point P is lying outside AB .
To construct: A line through P which is perpendicular to
the line AB.
Draw a straight line AB and
mark a point P lying outside
AB .
No
t
1)
to
Steps of construction:
2)
With a suitable radius and P as
the centre draw an arc which
cuts AB at X and Y as shown
in the figure
88
P
A
B
P
A X
Y
B
With yas centre draw another
arc, with same radius chosen
in step-3, that cuts the
previous arc. Name the point
of intersection as Q.
5)
Y
P
Using a scale, join PQ . Mark
the intersection point of PQ
and AB as ‘ O’. Now PQ is
perpendicular to AB .
B
d
4)
A X
he
With radius more than half
of XY draw an arc below AB
as shown in figure and X as
centre
A X
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3)
Y
B
Y
B
Q
P
A X
Q
Verification : Measure ∠BOP.
to
To Construct a line parallel to a given line through a
point outside it
No
t
Given: PQ a line segment and the point B is out side PQ .
To construct: A line parallel to PQ through point B
Steps of construction:
1)
Draw a straight line PQ and
mark a point B outside PQ .
89
B
P
Q
B
2)
Mark any point A on PQ and
join AB .
With suitable radius and A
he
3)
B
as the centre, draw an arc
Y
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which cuts PQ and AB . Let
the arc cut PQ at X and AB
P
at Y.
4)
Q
A
d
P
Now with and
A
X
Q
B as the
centre (the same radius as in
B
step-3,) draw an arc which
Y
cuts AB as shown in figure
A
Q
X
to
Mark intersection point of P
arc with AB as M.
M
5) Place the steel point of the compass at point X and take
No
t
XY as radius of compasses.
6)
With M as the centre, draw an
N
arc with XY as radius cutting
the arc drawn in step-4.
Y
Let these two arcs cut each
other at point N.
90
P
A
B
M
X
Q
N
S
7)
Join BN and extend on either
side to point R and S.
Y
A
R
M
X
Q
d
P
B
he
Now, RS is the required parallel line to PQ and passing
through the given point B.
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Think! In the construction above, ∠BAQ and ∠ABS are
constructed so that both the angles are equal and they
are either side of the transversal line AB . Will PQ || RS
if ∠BAQ and ∠ABR are equal and they are constructed
on same side of the transversal line AB. Justify your
answer.
Exercise 6.3
2)
Draw a PQ of length 8 cm. Mark a point A on the PQ .
Draw perpendicular line passing through point A.
Construct ∠PQR = 600. Mark point A on line segment PQ
and draw a line parallel to QR through that point.
No
t
3)
Draw a line segment AB of length 10 cm. Mark a point P
outside the AB . Draw perpendicular line passing through
point P.
to
1)
4)
Draw a XY . Draw a line parallel to XY at a distance of 4
cm from it. (clue : Draw a perpendicular to XY through
a point and continue).
5)
Mark any three non - collinear points A, B and C. Join
them to form a triangle. Draw a PQ passing through A
and parallel to BC .



91


UNIT - 7
MENSURATION
After studying this unit you :
explain the concept of perimeter of closed figures like
a square, a rectangle, a triangle, and a parallelogram
and find out their perimeters,

explain the concept of area of closed figures like a
square, a rectangle, a triangle, and a parallelogram
and find out their area,



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
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
explain the concept of circumference of a circle and
area of a circle,
calculate circumference and area of circle,
solve the problems related to the perimeter and
circumference,
solve the problems related to the area in life situation.
No
t
to
Introduction
Sandesh and Sangeeta purchased pictures of the
Mysore palace and the Jog falls respectively as a token
of memory of school tour. Therefore, Sangetha wanted to
laminate and Sandesh wants to frame the photoes they
have taken. The photo of the Mysore palace measures
30 cm × 60 cm. The photo of the Jog falls measures
40 cm × 50 cm. If the cost of lamination is ` 300 per sq. m. Find
cost of lamination. Also, find the cost of framing, if the cost
of framing is ` 200 per meter, who has paid more amount?
Let us recall what we have learnt about the perimeter and
area of rectangle and square.
Perimeter
The length along the boundary of a closed figure in a plane
is called the perimeter. The perimeter of a closed figure is
denoted by 'P'.
92
to
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Perimeter of a rectangle
l
In the figure 7. 1, lengths of the A
B
rectangle ABCD are AB and CD. (Let the
b
length of the rectangle ABCD be ‘l’ units.) b
Then AB = CD = l units. Breadth of the
C
l
rectangle ABCD are BC and DA. Let D
Fig : 7.1
breadth of the rectangle ABCD be ‘b’
units. Then BC = AD = b units.
Perimeter of the rectangle = P = AB + BC + CD + AD
=l+b+l+b
=2×l+2×b
= 2(l+b) units
∴ Perimeter of the rectangle = P = 2 (l + b)
a
B
A
Perimeter of the square
In the figure 7.2, length of the sides of the
square ABCD are AB, BC, CD and AD. Let
length of the sides of square ABCD be ‘a’ a
a
units. Then AB = BC = CD = AD = a units.
Perimeter of the square = P
P = AB + BC + CD + AD
D
C
a
= 4 × length of its side
Fig : 7.2
= 4 × side
= 4a units
∴Perimeter of the square P = 4a
D
Solution: The length of the rectangular
field= l = 10 m
The breadth of the rectangular field = b = 6 m
93
b=6m
No
t
Do it yourself : Take a sheet of paper, cut the sheet into
rectangles and squares of different sizes. Find the perimeter of each square and rectangle sheet.
l =10 m
Example 1 : Find the perimeter of a A
B
rectangular field of length 10 m and
breadth 6 m.
Fig:7.3
C
We know that,
Perimeter of the rectangle = P
= 2 [length + breadth]
= 2 [10 + 6] m = 2 × 16 m
= 32 m
he
d
∴ Perimeter of the rectangle = P = 32 m
Example 2: Find the perimeter of a square field of length
60 m.
A
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Perimeter of the square = P = 4a
= 4 × 60 m
a
= 240 m
∴ Perimeter of the square field= P = 240m D
a = 60 m
B
a= 60 m
Solution: The side of the square field = a = 60 m
C
No
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Example 3: Find the cost of fencing wire to be wound four
times around a square shaped flower garden of side 80 m, if
the rate of fencing wire is ` 10 per meter.
Solution:
The side of a square shaped flower garden
= a = 80 m
The cost of fencing per meter
= ` 10
Perimeter of the square flower garden
=4×a
= 4 × 80
= 320 m
∴Length of fencing wire required for one round around the
garden = 320 m
The length of fencing wire required for four rounds around
= 1,280 m
the garden
= 4 × 320
The cost of fencing wire = The length wire × Cost per meter
= 12,800
= 1,280 × 10
∴ The cost of fencing wire four time around flower
garden
= ` 12,800
94
Area
The surface enclosed by a closed figure in a plane is called
area. The area of closed figure is denoted by 'A'.
Area of a rectangle
l
he
b
l
C
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In the figure 7.5, lengths of the
rectangle ABCD are AB and CD. Let length b
of the rectangle ABCD be ‘l’ units. Then
AB = CD = l units. Breadth of the rectangle D
ABCD are BC and DA. Let breadth of
the rectangle ABCD be ‘b’ units. Then,
BC = AD = b units.
B
d
A
Fig 7.5
Area of the rectangle = A = length × breadth
Area of the rectangle = A = l × b
= l b sq. units
∴ Area of the rectangle = A= l b
Area of the Square
to
In the figure 7.6, length of the sides of square ABCD are
AB, BC, CD and AD. Let length of the sides of square ABCD be
No
t
‘a’ units. Then, AB = BC = CD = AD = a units. A
a
B
Area of the square = A = side × side
Area of the square = A = a × a
= a2 sq. units
Area of the square = A= a2
95
a
D
a
a
Fig 7.6
C
Try this
Find the area and perimeter of floor of classroom door,
windows, blackboard and top of the table.
Units used to measure area.
he
d
To measure anything we require a definite unit of
measurement. Let us analyse the units which we know very
well.
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Metre is the fundamental unit used to measure length.
Examine the scale used by shopkeepers for measuring
the cloth. It is a metre scale measured in metre is the unit of
length.
What is the unit used to measure area of a closed
Observe the figure, given on the right side. It
is a square with length of each side as 1 metre.
The measure of the region enclosed by a
1m
1m
region?
to
square with length of each side as 1 metre is
called a square metre and is written as 1
No
t
square metre or 1 m2.
What do you call the area of a square with length of each
side as 1 centimetre?
The measure of the region enclosed by a square with
length of each side as 1 centimetre is called square centimetre
and is written as 1 square. centimetre or 1 cm2.
96
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.....
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What is the relation between 1 sq. metre and 1 sq.
centimetre?
1
2
......
98
100
Consider a square of
each side measuring 1 metre. 1
The horizontal lines are
drawn to divide it into 100 2
equal parts and the vertical
lines are drawn to divide it
into 100 equal parts.
After all the horizontal and 98
vertical lines are drawn, you
will get 100 small squares in 100
a row and 100 small squares
in a column. Length of each
side of small square will be 1 cm. The total number of small
square will be 100 × 100 = 10,000 in number.
∴ 1 m2 = 10,000 cm2.
No
t
to
Depending on the length of the unit square, the unit of
measuring area would be different. If length of each side of
the unit square measures 1 decimetre (symbolically 1 dm) or
1 millimetre (1 mm), then the unit of measurement of area
will be 1 square decimetre (symbolically 1 dm2) and 1 square
millimetre (symbolically 1 mm2) respectively.
The relationship between these metric units can be
shown as below.
• 1 m2 = 100 dm2
• 1 m2 = 10,000 cm2
• 1 m2 = 10,00,000 mm2
97
If length of each side of the square measures 1 decametre
(1 dam) or 1 hectometre (1 hm), then the unit of measurement
of area will be 1 square decametre (symbolically 1 dam2) and
1 square hectometre (1 hm2) respectively.
he
d
The relationship between these metric units can be
shown as below.
• 1 dam2 = 100 m2
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• 1 hm2 = 10,000 m2
• 1 km2 = 10,00,000 m2
Example 1
Find the area of a rectangular field of length 10 m and
breadth 6 m.
Solution : The length of the rectangular field = l = 10 m
to
The breadth of the rectangular field = b = 6 m
No
t
We know that
l =10 m
Area of the rectangle = A = l × b
= 10 m × 6 m
= 60 m2
D
∴The Area of the rectangle = A = 60 m2
98
B
b=6m
A
fig : 7.7
C
Example 2
Find area of a square field of length 60 m.
Solution :
a = 60 m
d
= 602
a
he
= 60 m × 60 m
B
a = 60 m
The side of the square field = a = 60 m A
Area of the square = A = a2
= 3600 sq m
D
C
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a
Fig:7.8
∴ Area of the square = A = 3600 sq m
Example 3
The area of a rectangular garden of 150 m long is
9000 sq.m. Find the width (breadth) of the garden.
Solution:
Length of the rectangular garden
= l = 150 m
Area of the rectangular garden
= A = 9000 sq. m
Width (breadth) of the rectangular garden = b = ?
=A =l×b
No
t
to
Area of the rectangular garden
9000 = 150 b
150 b = 9000
Solving the above equation for b
Breadth (width) of the rectangle garden
=
b = 9000 = 60
150
∴ Width of the garden = b = 60 m
Think!
Area of one metre of ribbon is not equal to area of
one-metre towel. Why?
99
Example 4 : A door of length 2 m and breadth 1m is fixed in
a square wall of length 4m. Find the cost of painting, if the
rate of painting the wall is ` 25 per square metre and the door
is ` 50 per square metre.
Solution: The length of the door
=l=2m
=b=1m
The length of the square wall
=l=4m
The cost of the painting wall
The area of the door
= ` 25 per m2
=l×b
he
d
The breadth of the door
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=2×1
= 2 m2
The area of the square wall
= a2
=4×4
= 16 m2
The area of the wall to be painted
= area of the wall - area of the door
= 16 – 2
= 14 m2
to
The cost of painting the wall = Area × Cost
= 14 × 25
No
t
= 350
∴The cost of painting the wall
= ` 350
The cost of painting the door
= Area × Cost
= 2 × 50
∴ Cost of painting the door
The total cost of painting the
= ` 100
wall and the door
∴The total cost of painting
100
= ` 350 + ` 100
= ` 450
d
Example 5: Roopa has cut a sheet of paper into the shape of
a square and a rectangle, in such away that the area of the
square and the rectangle are equal. If the side of the square
sheet is 30 cm and the breadth of the rectangular sheet is
20 cm, find the length of the rectangular sheet. Also, find the
perimeter of the rectangular sheet.
he
Solution: The side of the square sheet = a = 30 cm
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The breadth of the rectangular sheet = b = 20 cm
The area of the square sheet
= a2
= 302
= 30 × 30
∴The Area of this square sheet
As per the given data,
= 900 cm2
Area of the rectangular sheet=The area of the square = 900 cm2
Area of the rectangular sheet
= 900 cm2
A= l × b = 900 cm2
to
l × 20 = 900 cm2
l = 900 = 45 cm
20
No
t
∴The length of the rectangular sheet = l = 45 cm.
Perimeter of the rectangular sheet
= P = 2 [l + b]
= 2 [45 + 20]
= 2 × 65
= 130 cm
∴ Perimeter of the rectangular sheet = P = 130 cm
101
Example 6 : Anil has a square shaped chess board of area
144 cm2. Find the length of the side of the chess board.
Solution:
The area of the square shaped chess board = a2 = 144 cm2.
d
a2 = 144
he
a ×a = 12 × 12
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a = 12
∴ The length of the chess board = a = 12 cm
Example 7:
The area of the square shaped field is 64 hectare. Find the
perimeter of the field in metre.
Solution : The area of square shaped field = a2 = 64 hectare
a × a = 64× 10,000 (a1 hectare = 10,000 m2)
= 6, 40,000
to
a × a = 800 × 800
No
t
∴ a = 800
∴The length of the side of the square shaped field = a = 800 m
The perimeter of the square shaped field
= 4a
= 4 × 800
= 3,200
∴ The perimeter of the square shaped field = 3,200 m
102
Exercise
1)
length = 8 cm, breadth = 6 cm
2)
length = 3 m, breadth = 2 m
3)
breadth = 4.5 m, length = 9.5 m
d
Find the perimeter of the rectangles, of the following
measurements.
he
I.
7.1
II. Find the area of the rectangles of the following
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measurements.
1)
length = 6 cm, breadth = 4 cm
2)
length = 12.5 m, breadth = 7 m
3)
breadth = 3.5 m, length = 6.5 m
III. Find the perimeter of the squares formed by the
following measurements.
1) side = 6 cm
2) side = 15 m
3) side = 5.6 cm
IV. Find the area of the squares formed by the following
to
measurements.
2) side = 12 cm
3) side = 9.8 cm
No
t
1) side = 6 m
V. Solve the following :
1) The perimeter of a rectangular plank is 120 cm. If the length
is 40 cm, find its breadth and surface area of the plank.
2) Thimmaraju has a plot of length 12 m and breadth 10 m.
It has to be fenced with four rounds of wire. If the cost of
the wire is ` 30 per meter, find the cost of fencing the plot.
103
3) A wire is bent in the shape of a rectangle. Its length is 36
cm and breadth is 25 cm. If the same wire is bent in the
shape of a square, what will be the measure of its side?
Among these two shape which shape occupies more area?
he
Find the area occupied by the square shape.
d
4) A wire of length 100 cm is bent to the form of a square.
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5) The area of the square shaped field is 16 hectare. Find the
perimeter of the field in metre.
(Hint: 1 hectare = 10,000 m2)
6) A farmer has a field in the rectangular shape. The length
of rectangular shaped field is 150 m and its breadth is
100 m. Find out the cost of ploughing the field at the rate
of ` 0.2 per m2.
What happens to the area of a rectangle when
1)
its length is doubled, breadth remaining the same?
2)
its length is doubled, breadth is halved?
3)
its length and breadth are doubled?
No
t
to
7)
8)
What happens to the area of square when
1)
When its sides is doubled the area increase by
_________times.
2)
When its side is halved the area becomes _________
times.
104
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Perimeter of a triangle
Who am I?
A
Clue 1 : I am a closed
geometrical figure of minimum
c
b
number of sides
Clue 2: I have three sides.
C
Clue 3: I have three angles. B
a
fig : 7.9
It is a triangle.
What is a triangle?
A triangle is a closed geometrical figure in a plane
having three sides.
Observe triangle ABC in figure 7.9. Let length of sides BC,
AC and AB be a, b and c respectively.
We know that the perimeter of a geometrical figure is the
sum of all the sides of geometrical figure. It is denoted by P.
Here the triangle PQR has three sides.
The perimeter of a triangle = P
= BC + AC + AB
= a + b + c units
∴ The perimeter of a triangle = P = a + b + c
to
If all the sides of a triangle are equal, then the triangle is
called equilateral triangle.
A
No
t
What is the perimeter of the equilateral
triangle?
In figure 7.10, triangle ABC is an equilateral
triangle. Let AB = BC = AC = a units.
∴The perimeter of the equilateral triangle B
= AB + BC + CA
=a+a+a
a
a
a
fig : 7.10
= 3a units
∴ The perimeter of an equilateral triangle = P = 3a
105
C
Example 1: Find the perimeter of a triangle whose sides are
10 cm, 7 cm, and 5 cm.
Solution: Length of sides of the triangle, a = 10 cm, b = 7 cm, and
c = 5 cm.
The perimeter of the triangle = P = a + b + c
he
= 10 + 7 + 5
d
Length of sides of the triangle are different.
∴ P = 22 cm
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∴The perimeter of the given triangle = 22 cm
Example 2 : Find the perimeter of an equilateral triangle whose
side is 15 cm.
Solution: Length of each side of the equilateral triangle
= a = 15cm
The perimeter of the equilateral triangle = P = 3a
= 3 × 15
∴ P = 45 cm
to
∴The perimeter of the equilateral triangle = 45 cm
Example 3: The perimeter of an equilateral triangular shaped
card board is 201 cm. Find the length of a side of the card
board.
No
t
Solution:
The perimeter of the equilateral triangular shaped card board
= 3a = 201 cm
The length of side of the equilateral triangular shaped
cardboard
a = 201
3
= 67 cm
∴The length of side of the equilateral triangular shaped
card board = 67 cm
106
Area of a triangle
Reena takes a rectangular paper sheet ABCD of length
AB = 14 cm and breadth BC = 8 cm. She cuts the rectangular
l =14 cm
sheet ABCD along the diagonal AC.
B
A
b = 8 cm
How many pieces of paper does Reena get?
Yes, it is in the shape of a triangle.
he
What is the shape of each paper?
d
She gets two pieces of paper.
D
Fig 7.11
C
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Place these two triangles one over the other in such a way
that vertex B coincides with D and vertex A coincides with C.
Are these two triangles same in shape, size and area? Are
they congruent triangles?
Yes, these triangles are of the same shape, size, and area.
They are congruent triangles
∴The diagonal of the rectangle divides the rectangle into
two equal triangles.
B
A
No
t
to
Length of the rectangle ABCD is base of
the triangle ADC and breadth of the h
rectangle ABCD is height of the triangle
ADC. Let the base of the triangle ADC be
C
D
b
‘b’ units and height of the triangle ADC
Fig 7.12
be ‘h’ units.
Area of the rectangle ABCD = 2 × Area of the triangle ADC
∴ Area of the triangle ADC = 1 (Area of the rectangle ABCD)
2
= 1 (length × breadth)
2
= 1
2 bh
∴The area of the triangle ADC The symbol used to
=A = 1 bh
represent triangle is ∆
2
107
Finding the area of a triangle
Fig 7.13
R
S
A
R
Q
P
B
Fig 7.14
Q
A
S
A Q
P
SR
B
Fig 7.15
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P
A
R
d
S
he
Arun draws a ∆ PQA in the rectangle PQRS as shown in
figure 7.13. He wants to find the area of ∆ PQA. How does he
find the area of ∆ PQA?
Q
A
Consider a rectangular piece of paper PQRS. Mark any
point A on SR. Join PA and QA. The ∆ PQA is inscribed in the
rectangle PQRS as shown in the Fig. 7.13
S
Now draw a line AB perpendicular to PQ. h
We observe that PS = AB = QR.
Let PS = AB = QR= 'h' units and PQ = 'b' units. P
A
h
B
h
Q
to
b Units
Now cut along the lines PA and QA.
Fig : 7.16
Superpose two triangles ∆ PAS and ∆ QAR
on ∆ PAB and ∆ ABQ respectively as shown in the Fig. 7.15
R
From figure 7.15, we understand that
No
t
Area of ∆PAQ = Area of ∆PAB + Area of ∆ABQ
= Area of ∆PAS + Area of ∆QAR ..... (1)
Area of the Rectangle PQRS = (Area of ∆PAQ +
Area of ∆PAS +Area of ∆QAR)
= Area of ∆PAQ + Area of ∆PAQ a [By using equation (1)]
= 2 × Area of ∆PAQ
108
he
d
2 × Area of ∆PAQ = Area of the rectangle PQRS
Area of the ∆PAQ = 1 × (area of rectangle PQRS)
2
= 1 × (length × breadth)
2
= 1 × (b × h)
2
= 1 bh sq.units
2
Area of any triangle = A = 1 bh squareunits
2
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Where b is the base and h is the height of the triangle.
ARYABHATA - I
No
t
to
Aryabhata-I a great Indian mathematician
and astronomer, is believed to have been born
in the region between the Narmada and
Godavari rivers in central India or in south
India. Aryabhata went to Kusumapura for
advanced studies. Kusumapura is identified
as Pātaliputra, modern Patna. The University
of Nalanda was in Pataliputra at that time
and had an astronomical observatory. It is speculated that
Aryabhata might have been the head of the Nalanda
University. Aryabhata is also reputed to have set up an
astronomical observatory at the Sun temple in Taregana,
Bihar. Aryabhata - I mentions in the 'Aryabhatiya' a work of
him that it was composed 3,630 years into the Kali Yuga,
when he was 23 years old. He has given the formula for area
of triangle.
ef$eYegpem³e HeÀue Mejerjce meceouekeÀesefì YegpeeOe&mecJeie& :~
It means, The area of a triangle is the product of half of
its base and height.
109
Example 1 : Find the area of the triangle whose base is 10
cm and height is 6 cm.
Solution: Base of the triangle
= b = 10 cm.
= h = 6 cm.
Height of the triangle
Area of the given triangle = 1 bh
d
2
2
he
= 1 × 10 × 6
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= 30 sq. cm.
∴The area of the given triangle
Example 2:
= 30 sq. cm.
Find the height of a triangular shaped field whose area is
400 m2 and its base is 50 m.
Solution:
The area of the triangular shaped field = A = 400 m2.
The base of the triangular shaped field = b = 50 m.
No
t
to
We know that Area of any triangle
= A = 1 bh
2
i.e 1 bh = A
2
1 × 50 × h = 400
2
25 × h = 400
Solving the above equation for h
h=
1 × 400
25
= 16 m
∴ The height of the triangular shaped field = h = 16 m
110
Example 3: The ratio of the base and height of a triangle is
2 : 3. If the area of the triangle is 1200 m2, find the measurement
of its base and height.
Solution: The area of the triangle = A = 1200 m2
The ratio of the base and the height of a triangle = 2 : 3.
=A
he
1 bh
2
1 × 2x × 3 x
2
3x2
= 1200
= 1200
= 1200
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We know that
d
The length of the base and height = 2x : 3x.
x2
x
2
x2
∴x
3
= 400
= 202
= 20
Therefore, the length of the base = 2x
= 2 × 20
= 40 m.
The height of the triangle = 3x
to
= 3 × 20
= 60 m
TRY THIS
No
t
In the adjoining figure,
PQ|| RS and ∆ ABD is right at B. R
angled The length of AB = 10 cm
and BD = 8 cm. Find the area of
∆ ABC, ∆ ABD and ∆ ABE. Do they
have the same area? If yes or no,
P
why? Give reason
111
C
E
A
D
S
B
Q
EXERCISE
7.2
1) Find the perimeter of the following triangles.
Y
8 cm
Z
B
7
cm
C
4 cm
d
cm
8
6.5
M 5 cm N
m
6 cm
8c
cm
cm
R
6.5
10
Q 8 cm
A
X
5 cm
L
cm
P
Calculate the perimeter of the triangle whose sides are
6 cm, 8 cm and 6 cm.
3)
Find the perimeter of the equilateral triangles with sides
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2)
a) 8 cm
b) 13 cm
c) 11 cm
In a triangle, the perimeter is 60 cm and the length of two
sides of triangle are 13 cm and 20 cm. Find the length of
the third side.
5)
The perimeter of an equilateral triangle is 45 cm. Find
the length of each side of the equilateral triangle.
6)
A lawn in a garden is in the shape of an equilateral
triangle. If the length of one side of the equilateral triangle
is 75m, find the cost of fencing at the rate of ` 12 per
metre.
7)
Find the area of the following triangles.
8 cm
C
S
cm
L
Q 4 cm
R
M
cm
B
P
6.5
cm
No
t
10
6.5
6 cm
A
6.5 cm
to
4)
6 cm
5 cm
N
8)
Find the area of the triangle whose base is 14 cm and
the height is 7 cm.
9)
A garden is in the equilateral triangular shape. Its base
is 22 m and breadth is 18 m. Find the cost of levelling
the garden at ` 5 per m2.
112
10)
The ratio of the base and the height of a triangle are 4
: 3. If the area of the triangle is 216 m2, find the length
and height.
11)
What happens to the area of triangle when
c) Its base and height are doubled?
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Area of parallelograms
he
b) Its base is doubled, height is halved?
d
a) Its base is doubled, height remaining the same?
Pallavi makes a rectangle using four broomsticks and a
cycle valve tube. She cuts the broomstick into a pair of equal
lengths for length and breadth of a rectangle. She inserted
the ends of broomstick into cycle valve tube to make rectangle
as shown in figure 7.17. She shows it to Meena. Meena
appreciates and just pushes point A softly towards point B.
Now figure obtained is shown in fig 7.18.
A
to
fig-7.17
C
D
fig-7.18
B
C
No
t
D
A
B
How many pairs of parallel lines are there in the fig 7.18?
It has two pairs of parallel lines.
What is the geometrical shape of figure 7.18?
Yes, it is termed as parallelogram.
A geometrical enclosed figure on plane with two pairs of
parallel lines is called parallelogram.
113
How do you find area of a parallelogram?
B
E
fig-7.19
C
D
d
D
B
A
E
C
F
he
A
fig-7.20
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Draw a parallelogram ABCD on a paper sheet as shown
below in figure 7.19. Draw a line perpendicular to base CD from
vertex A. Let it meet the base CD at E. Cut the triangle AED
and attach it in such a way that BC coincides with AD .
What is the shape of geometrical figure obtained?
It is in rectangular shape.
Is the area of the parallelogram equal to the area of the
rectangle formed?
Yes, the Area of the parallelogram = Area of the rectangle
b units
B
A
formed.
No
t
to
From Fig. 7.21, we observe that the
length of the rectangle formed is equal
to the base of the parallelogram and
breadth of the rectangle is equal to the
height of the parallelogram.
Area of the parallelogram
= Area of the rectangle
h
E
h
b units C
fig-7.21
A = (length × breadth) sq units
= (base × height) sq units
= bh sq units
∴ Area of parallelogram = A = bh
114
F
Any side of a parallelogram can be taken as base of the
parallelogram. The perpendicular drawn on that side from
the opposite vertex is known as height (altitude).
In the parallelogram ABCD, AE is
perpendicular to CD. Here CD is the base
and AE is the height of the parallelogram.
C
E
he
D
B
d
A
A
F
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In this parallelogram ABCD, CF is the
perpendicular to opposite side AD.
B
D
Here AD is the base and CF is the height.
of the parallelogram
C
Try this!
Find the perimeter of a parallelogram using the property
that, opposite sides of parallelogram are equal.
Example 1 :
Find the area of the parallelogram having base 8 cm and
to
altitude 5 cm. (Note Altitude means height)
Solution:
= b = 8 cm
The height (altitude) of the parallelogram
= h = 5 cm
Area of the parallelogram
= bh
No
t
The base of the parallelogram
= 8 cm × 5 cm
bh= 40 cm2
∴The Area of the parallelogram
115
= 40 cm2
No
t
to
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Example: 2 Find the height of a parallelogram shaped field
whose area is 108 m2 and base is 12 m.
Solution: The area of the parallelogram shaped field
= A = 108 m2
The base of the parallelogram shaped field = b = 12 m
Area of the parallelogram shaped field
= A = bh
108 = 12 × h
(Solving for h)
h = 108
12
h=9m
∴ The height of the parallelogram shaped field = h = 9 m
Example 3: The ratio of the base and height of a parallelogram
is 4 : 3. If the area of the parallelogram is 48 m2, find the base
and height.
Solution: The area of the parallelogram
= A = 48 m2
The ratio of the base and height of a parallelogram = 4 : 3
Let the length of the base and height of a parallelogram be 4x
and 3x respectively
We know that
bh = A
3x × 4x = 48
12×x2 = 48
∴ x2 = 48
12
ie. x2 = 4
∴ x2 = 22
If powers of exponents are the same; bases of exponents must
be equal
∴x=2
The length of the base of a parallelogram
= 4x
=4×2
=8m
The length of the height of a parallelogram = 3x
=3×2
=6m
116
Exercise
7.3
1) Complete the following table with respect to a
parallelogram.
02
03
04
4)
5)
6)
1200 cm2
Find the area of the parallelogram having base 9 cm and
altitude 7 cm.
Find the area of the parallelogram shaped field having
length of the base 21 m and height 15 m.
Find the altitude of the parallelogram shaped banner
having base 12 m and area 108 sq. m.
Find the base of the parallelogram shaped site having
height 1.3 m and area 104 sq. m.
ABCD is a rectangle with length AB = 13 cm and breadth
CB = 7 cm. Point D is pushed towards point C to form
parallelogram. Find the area of parallelogram ABCD.
ABCD is a parallelogram with AB = 15 cm and BC = 18 cm.
The height of the parallelogram from vertex A to BC is
9 cm. Find the height of the parallelogram from vertex
C to AB?
No
t
7)
280 m2
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3)
Area
to
2)
24 cm
Height
6 cm
11 m
14 m
d
01
Base
8 cm
15 m
he
Sl. No
8)
The ratio of the base and height of a parallelogram is
3 : 2. If the area of the parallelogram is 150 m2, find the
base and height.
9)
The ratio of the base and height of a parallelogram is
5 : 2. If area of the parallelogram is 1000 m2, find the base
and height.
117
Circle
he
d
In our daily life, we come across a number of objects like
wheels, bangles, coins, rings, giant wheel, papad, compact
disc (C.D.)
What is the shape of these objects?
‘Round’
Yes, they are round. These shapes are termed as circular
shaped objects.
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Activity : Take a cardboard. Mark the middle of the board,
name this point as ‘O’. Fix a small nail
firmly at ‘O’, sheet Take a thread of
convenient length which is less than the
breadth of the board. Tie one end tightly
A
O
to the nail. Tie the other end of the thread
to a sharpened pencil firmly. Stretch the
thread completely towards point ‘A’. Start
moving the pencil, holding the thread
stretched tightly, so that the pencil marks on the sheet
until the pencil reaches the starting point 'A'.
Now observe the path traced by the pencil.
to
What is the shape of the path traced by the pencil?
The shape of the path traced is a circle.
No
t
A Circle
Yes, a circle is a locus of point which
moves in such a way that its distance from
a fixed point is always constant.
In the figure, observe the point ‘O’. It is
o
at equidistant from any point on the circle. M
Hence, it is always called centre of the
circle. The distance between the centre
of circle and any point on the circle is
constant. This constant distance is called
radius of the circle. OP is the radius of the circle.
118
P
N
OP, OM and ON are the radii (plural form of radius) of the
circle.
Does MN pass through the centre of the circle O?
Yes, MN passes through the centre of the circle O.
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Any line segment having its end points on the circle and
passing through the centre of the circle is called the diameter
of the circle (MN). In a given circle, the diameter is always
twice the radius of the circle. In other words the radius of the
circle is half of the diameter.
Let ‘d’ denote the length of the diameter and ‘r’ denote the
radius of the circle. Then above statement is symbolically
written as d = 2r or r = d .
2
Do you know!
•
•
How many diameters can be drawn in a circle?
How many radii can be drawn in a circle?
Circumference of a circle.
Does a circle have perimeter?
No
t
to
Yes, it has a perimeter. The perimeter of the circle is called
circumference of the circle. The circumference is the boundary
of the circle. We cannot measure accurately, the circumference
of the circle using scale. This can be measured by rolling the
circle on to a line as explained below.
Tie a thread tightly around the circumference of a circle.
mark the point where the two edges of the thread meet. Now,
remove the thread and measure the length of the thread using
a scale. This is the circumference of the circle.
Take a circle of radius 3.5 cm using a bangle or a thick cardboard.
Mark a point 'A' on the rim of the circle. Draw a line on a
page of your notebook and mark an initial point P on it. Now
coincide point A of the circle with point P on the line.
119
A
O
A
O
O
A
O
A
A
P
O
22 cm
Q
he
d
Slowly roll the circle on the line until point A again touches
the line. Mark this point as Q. Measure PQ. This gives the
length of the circumference of the circle. Approximately the
circumference of the circle is 22 cm.
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Another way of measuring the circumference of a circle
may be passing a thread along the circle tightly. Make a mark
on the thread where exactly the two edges of the thread join.
Measure length between mark on the thread and other end
of thread using scale. This length is the circumference of the
circle.
Sanmati has a cardboard cut in the shape of circle. The
diameter of the circular cardboard is 14 cm. She wants to
decorate it by sticking colour thread on the edge of circular
cardboard. what is the length of the thread does she need to
buy from a shop without wasting it?
to
Relationship between circumference and
diameter of the circle
No
t
Cut circles having different radii on a thick cardboard.
Measure the diameter and radius of each circle using a scale.
Tabulate the data in the table given below. Run a thread
around the circle and measure the length of the thread that
gives the circumference of the circle. Record the observation
in the table. Repeat these steps for each circle.
Find the ratio of circumference to diameter of each circle.
What do you conclude?
120
Serial Diameter Radius Circumference Circumference (c)
number
(d)
(r)
(C)
Diameter (d)
1
he
d
2
4
5
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3
It is observed that the ratio of circumference and diameter
is little more than 3. Approximately its value is 3.14. It is
denoted by Greek letter π and is pronounced as 'pi'.
No
t
to
Circumference (c)
=r
Diameter (d)
Try this : Collect one, two, five,
and ten rupees coins of different
radii. Measure the circumference
and diameter of each coin. Find the
ratio of circumference to diameter
in each case.
Circumference (C) = π × (d)
Diameter (d)
= π × 2r
= 2 πr
Think !
A circle of radius r cm is cut
into two semicircles. What
is the circumference of a
semicircle?
∴Circumference of a circle = C = 2πr
121
Know this :Indian contribution to find the value of π
1. Aryabhata-I gave the value of π as 3.1416 up to 4 decimal
places. He also stated that even this value is approximate
value.
d
®elegjefOekeÀce MeleceäietCece otJeeefÿmleLee menñeeCecedt ~
De³edgleodtle³eefJeMkeÀcYem³eemeVees efJe´ÊeHeefjCen :~~
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Meaning, “Add four to one hundred, multiply it by eight
and then add sixty two thousand; The result is approximately the circumference of a circle of diameter of
twenty thousand.”
8 (100 + 4) + 62000 62832
i.er = Circumference =
=
= 3.1416
20000
20000
Diameter
2. Madhava (1350-1425 A.D) was born in village Sangama
to
grama near Cochin in Kerala state. He calculated the value
of π upto eleven decimal places. Its value is 3.14159265359
Modern value of π upto twenty decimal place is 3.1415926535
8979323846 . Using the computer, π value is calculated
upto million decimal places. Hence, π is a non-recurring,
non-terminating decimal number.
3. Srinivas Ramanujan (1887 - 1920) was one of the great
No
t
mathematicians of 20th century. He was a self taught genius
and highly creative, guided by imagination and intuition. He
has given approximate value of as 97 1 1 14 3.1415926526....
`
2
=
11 j
=
The value of π in 20th century is 3.1415926535 8976323846.
The value of π is non recurring and never ending decimal
number.
122
Do it yourself:
Draw a circle of radius 5 cm. Draw a diameter to the circle
and name their end points as A and B. Then divide the
length of the diameter AB into ten equal parts i.e. one cm
d
each. Mount two nails at point A and B. Tie a thread to nail
he
mounted at A. Now mount the nails along the circumference
of the circle as close as possible. Run the thread around the
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circle along with the circumference until it reaches point A.
Cut the thread using scissor. Now remove the thread and
wind the thread along with the diameter AB.
How many complete windings can be done
along with the diameter AB?
3 complete windings.
How much thread is left after three
windings?
It is just more than one division.
to
This gives us the value of π.
Therefore, the value of the π is ≈ 3.1
No
t
(≈ symbol means approximately)
Example 1 : Find the circumference of a circle whose
radius is 14 cm.
Note :
The radius of the circle = r = 14 cm
For calculation
Circumference of the circle = C = 2 πr
purpose we
consider the
= 2 # 22 # 14
7
approximate value
= 88 cm
of π as 22 or 3.14
7
123
Example 2 : Find the diameter of a circle whose circumference
is 21.98 m. (π = 3.14).
Solution:
Circumference of a circle = C
= 21.98 m
= 21.98
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2r
21.98
=
2 # 3.14
21.98
=
6.28
= 3.5 m
he
Radius of a circle = r
d
2π r = 21.98
Diameter of a circle = d = 2r
= 2 × 3.5
=7m
∴ Diameter of a circle
= 7 m.
Example 3 : The diameter of a circular shaped flower garden
is 70 m. What is the cost of fencing it, at ` 15 per metre?
The unit cost of fencing the circular shaped garden per metre
= ` 15 per metre
to
The diameter of the circular shaped flower garden = d = 70 m
Circumference of circular shaped flower garden = πd
No
t
= 22 × 70
7
= 220 m
The cost of fencing = circumference × unit cost
= 220 × 15
= ` 3300
∴The cost of fencing the circular shaped flower garden
= ` 3300
124
he
d
Example 4 : The rectangular shape of the wire 16 cm long and
6 cm broad is bent to the form of a circle. Find the diameter
of the circle.
Length of the wire = Perimeter of the rectangle
Length of the rectangle
= 2(l + b)
= 2(16 + 6)
= l = 16 cm
= 2 × 22
Breadth of the rectangle
= 44 cm
= b = 6 cm
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Now the same wire is bent to the form of a circle.Let the
radius of the circle be r
The circumference of the circle = length of the wire
c= πd = 44 cm
The diameter of the circle, c = d = 44
r
r
d = 44 # 7 = 2 # 7 = 14
22
∴The diameter of the circle = d = 14 cm
Example 5 : The cost of fencing Mahatma Gandhi circle is
` 1,100 at the rate of ` 50 per metre. What is the radius of the
Mahatma Gandhi circle?
to
The cost of fencing in Mahatma Gandhi circle
= ` 1,100
No
t
Unit cost of fencing
= ` 50
Cost
of
the
fencing
The Circumference of
=
Unit Cost
Mahatma Gandhi circle
= 1100 = 22m
50
∴ Circumference of the Mahatma Gandhi circle = 22 m
C = 2πr = 22
r = 22 = 22 # 7
2 # 22
2r
7
=
= 3.5 m
2
∴The radius of the Mahatma Gandhi circle = 3.5 m
125
Area of the circle
Rehana has a circular shaped compact disc (CD). She
wants to calculate the area of the compact disc (CD). How
can she calculate it ?
he
d
Take a cardboard. Draw a circle of convenient radius r cm.
Cut the circle with pair of scissors. Divide the cardboard circle
into sixteen equal sectors. Cut these sectors using a pair of
scissors and arrange them as shown below. The arrangement
of sectors looks like a parallelogram.
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When more number of smaller sectors are cut from the
circle, the figure takes the shape of a rectangle.
r
πr
When the number of sectors increases more and more, in
the limiting case, the figure takes the shape of a rectangle.
to
The base of the parallelogram is half of the circumference
of the circle.
No
t
i.e. The base of the parallelogram = 1 × 2πr
2
= πr
Height of the parallelogram = radius of the circle (approx)
= r cm
Area of the circle
= Area of the parallelogram
= base × height
= πr × r
= πr2 sq.units
Area of the circle = πr2 sq.units
126
Know this : Aryabhata has given the formula to find area
of circle.
meceHeefjCeenm³eeOe¥ efJe<kebÀYeeOe&nlecesJe Je=Êe HeÀueced ~
It means, "Area of a circle is equal to the product of half
of the circumference and half of the diameter".
d
A = 1 × 2πr × 1 ×d
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2
2
= 1 × 2πr × 1 ×2r
2
2
= πr × r = πr2
Example 1 : Find the area of a circle having radius equal to
4.9 cm
The radius of the circle = r = 4.9 cm
The area of the circle = πr2
= 22 × 4.92
7
= 22 × 4.9 × 4.9
7
= 75.46sq. cm
to
∴The area of the circle = 75.46 sq. cm = 75.46 cm2
Example 2 : A cow is tethered to a peg in a big grass yard
using the rope of length 5 m. Find the maximum area of field
in which the cow can graze. (π = 3.14).
No
t
The length of the rope = The radius of the circle r = 5 m
The area of field the cow can graze = πr2
= 3.14 × 52
= 3.14 × 5 × 5
= 78.5 m2
14.2
∴The maximum area of field the cow can graze fi=g 78
.5 m2
127
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Example 3 : Find the circumference of the circular region
having an area 616 square metre. (Use π = 22 )
7
The area of the circular region = πr2 = 616 m2
22 r2 = 616
7
r2 = 616 × 7
22
r2 = 28 × 7
r2 = 196 , r2 = 142
If powers of exponents are same; bases of exponents must
be equal, that is to say.
r = 14
`The radius of the circular region = r = 14 m
The circumference of the circular region = 2πr
= 2 × 22 × 14
7
= 2 × 22 × 2
∴The circumference of the circular region = 88 cm
Example 4 : The circumference of the circular shaped base
of water supplying tank is 132 m. Find the area of the base
plate. ( π = 22 )
7
The circumference of the base of water supplying
tank = 132 m
2 πr = 132 m
The radius of the base of water supplying tank, r = 132
to
2r
132
#7
r=
2 # 22
r = 21 m
No
t
∴The radius of the base of water supplying tank = r = 21 m
The area of the plot of water supplying tank = πr2
= 22 × 212
7
22
=
× 21 × 21
7
= 22 × 3 × 21
= 1,386 m2
∴ The area of the plot of water supplying tank = 1,386 m2
128
Example 5 : A wire of 88 cm long is bent to form a square
and the same wire is bent to form a circle. Find the difference
between the area of the circle and the square? ( π= 22 ).
7
The perimeter of a square = length of the wire
= a2 = 222
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Area of the square
he
4a = 88
a = 88
4
a = 22 cm
d
= 88 cm
The perimeter of a square
= 484 sq cm
∴ Area of the square
= 484 sq cm
The circumference of circle
= length of the wire
Radius of the circle
2πr = 88 cm
= r = 88
2r
88
#7
r=
2 # 22
r = 14 cm
= 14 cm
The area of the circle
= πr2
to
∴Radius of the circle = r
= 22 × 14 × 14
7
No
t
= 22 × 2 × 14
= 616 cm2
∴The area of the circle
= 616 cm2
Difference between the area of circle and the area of square
= 616 – 484
= 132 cm2
129
Example 6 : Find the cost of polishing (top) surface of a
circular dining table of diameter 1.8 m, if the rate of polishing
is ` 20 per m2. (Use π = 3.14).
The diameter of the surface of dining table = d = 1.8 m
= 12.8 m
d
The radius of the above = r
he
= 0.9 m
The area of the surface of dining table = πr2
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= 3.14 × (0.9)2
= 3.14 × 0.9 × 0.9
= 2.5434 m2
The cost of polishing dining table
= Area × unit cost
= 2.5434 × 20
= 50.868
= ` 51
∴The cost of polishing the circular dining table = ` 51.
1)
Find the circumference of the circles of following radii.
No
t
a) 7 cm
2)
7.4
to
Exercise
b) 10.5 cm
c) 21 m
Find the circumference of the circles of following
diameters.
a) 70 cm
b) 56 m
c) 49 cm
3) Find the circumference of a circle whose radius is 6.3 cm.
4) The circumference of the circle is 35.2 m, find the diameter
of the circle.
130
5) Find the diameter of a circle whose circumference is 1256
cm. use π = 3.14
6) The diameter of a circular garden is 42 m. What is the cost
of fencing it at ` 25 per metre?
d
7) The diameter of circular garden is 49 m. Find the distance
he
covered by a runner if he makes 2 rounds round the
garden.
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8) The square shape of the wire of a length of side 44 cm is
bent in the form of a circle. Find the diameter of the circle.
9) Find the area of the circles of following radii.
a) 7 cm
b) 10.5 cm
c) 21 m
10) Find the area of the circles of following diameters.
a) 70 cm
b) 56 m
c) 49 cm
11) The circumference of the circle is 396 m. Find the area of
the circle.
12) Find the circumference of the circular region having area
to
5544 square meter.
13) A wire of 1.76 m long is bent to form a square and the
No
t
same wire is bent to form a circle. Find the difference in
the area of circle and square ?
14) Find the cost of laminating decolam to a circular table of
radius 1.4 m, if the rate of decolam is ` 50 per m2.
15) A wire is bent in the form of a square occupies an area of
196 m2. If the same wire is bent to form a circle, find the
area occupied.
131
Area between the rectangles / Area of rectangular pathway
B
Q
b
B
R
C
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In the figure, the smaller rectangle A
P
enclosed by rectangular pathway. The
figure has two rectangles. The smaller
l
rectangle PQRS has length l units and
S
l
breadth b units and the larger rectangle
ABCD has length L units and breadth D
L
B units. The region between the two
rectangles is called the rectangular pathway.
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The area of rectangular pathway is equal to difference of
area of larger and smaller rectangle.
The area of rectangular pathway = The area of rectangle
ABCD - The area of rectangle PQRS = LB - lb
The area of rectangular pathway = LB - lb
No
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Rashmi has a rectangle of 20 cm length and 12 cm
breadth. It is made of pink colour card board. Smitha has
blue coloured cardboard of length 15 cm and breadth 7 cm.
They placed the blue coloured cardboard exactly at the centre
of the pink coloured cardboard, so that it left uniform pink
passage around the blue cardboard. How do they calculate the
area of pink cardboard passage around the blue cardboard?
Area of the pink coloured cardboard = A1
= lb
= 20× 15
= 300 cm2
Secondly find the area of blue coloured card board
Length of the blue coloured cardboard = l = 12 cm
Breadth of the blue coloured cardboard = b = 7 cm
= lb
Area of the blue coloured cardboard = A2
= 12 × 7
= 84 cm2
132
Now find the difference between the two rectangles.
The area of the pink cardboard (passage) around the blue
cardboard = A1 – A2
= 300 – 84
= 216 cm2
A
P
3m
Let PQRS be the rectangular park. Let
ABCD be the external boundary of the path
which is also in the rectangular shape.
S
3m
3m
Q
R
B
3m
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∴The area of the pink cardboard (passage) around the
blue cardboard = 216 cm2
Example 1 : A rectangular park measuring 19 m by 14 m is
to be covered by external path which is 3 m wide. Find the
cost of laying the stone slab of the path at the rate of ` 30 per
square metre.
No
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to
D
C
Width of the path
=3m
Length of the Park
= PQ = 19 m
Breadth of the Park
= QR = 14 m
Area of the Park PQRS
= A1 = 19 × 14
= 266 m2
Length of the external boundary of the path
= 25 m
= AB = 19 m + 3m + 3m
Breadth of the external boundary of the path
= 20 m
= BC = 14 m + 3m + 3m
Area of the external boundary of the path
ABCD = A2
= 25 × 20
= 500 m2
Area of the path
= A2- A1
= 500 – 266
= 234 m2
Cost of the laying the stone slab of the path = 234 × 30
= ` 7020
133
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Q
Example 2 : A painting is mounted on a P A
B
card board of 10 cm × 7 cm size in such
a way that there is a margin of 1 cm
b
B
along each sides. Find the total area of
the margin on the card board.
D
C
Let PQRS be the cardboard and ABCD S
R
be the painting.
Width of the margin
= 1 cm
Length of the cardboard
= PQ = 10 cm
Breadth of the cardboard = QR = 7 cm
Area of the cardboard PQRS = A1 = 10 × 7
= 70 cm2
Length of the painting = AB = 10 cm – 1 cm – 1 cm
= 8 cm
Width of the painting = BC = 7 cm – 1 cm – 1 cm
= 5 cm
The area of the painting ABCD = A2 = 8× 5
= 40 cm2
The area of the margin = A1 – A2
= 70 – 40
= 30 cm2
The area between the two concentric circles / Area of
circular pathways
In the figure, the smaller circle having
the radius r is enclosed by circular pathway.
The figure has two concentric circles. Smaller
r
R
circle has the radius ‘r’ units and the larger
O
circle has the radius ‘R’ units. The region
between the two circles is called the circular
pathway.
The area of circular path way is equal to the difference of
area of larger and smaller circles.
134
Area of circular path = Area of the – Area of the smaller
way
larger circle
circle
= πR2 - πr2
∴Area of the circular = π (R2 - r2)
path way
∴ Area of the circular path way= π (R2 - r2)
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Example 1 : If two concentric circles having radii 10 cm and
6 cm form a circular path, find the area of the circular path
way. (π = 3.14)
The radius of the larger circle
= R = 10 cm
The radius of the smaller circle = r = 6 cm
Area of the circular pathway
= π (R2 - r2)
= 3.14 (102 – 62)
= 3.14 (100 – 36)
= 3.14 × 64
∴The area of circular pathway
= 200.96 cm2
Example 2 : The radius of a circular shaped pond is 7 m. Find
the cost of cementing the circular platform of the width 3.5 m
around the pond at the rate of ` 25 per m2.
Unit cost of cementing around the pond = ` 25 per m2.
=r=7m
The radius of the outer circle
= R = 7 + 3.5
to
The radius of the pond
No
t
= 10.5 m
Area of the circular platform around the pond = π (R2 - r2)
22
2
2
7 (10.5 – 7 )
= 22 (110.25 – 49)
7
= 22 × 61.25
7
= 22 × 8.75 = 192.5 m2
=
∴Area of the circular platform around the pond = 192.5 m2
135
The cost of cementing the circular platform = Area of the
pathway x unit cost
= 192.5 × 25
= 192.5 × 25
= ` 4,812.5
7.5
he
Exercise
d
∴The cost of cementing the circular platform = ` 4,812.5
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1) A slate is 30 cm long and 25 cm wide. It is surrounded by
a wooden frame of width 2.5 cm all around. John coloured
the frame, need not consider the thickness of the frame.
Find the area of coloured frame.
2) A path of uniform width 1.5 m runs around the inside of
a rectangular plot of 80 m long and 60 m wide. Calculate
the area of the path around the plot.
3) A path of uniform width of 1 1 m runs around the border of
2
a rectangular garden of 55 m long and 30 m wide. Calculate
to
the cost of laying the stone slab in the pathway at the rate
of ` 30 per m2.
4) Calculate the area of the circular pathway formed by two
No
t
concentric circles having the following radii.
a) R = 20 cm, r = 6 cm
c) R = 42 m, r = 10.5 m
b) R = 15.5 m, r = 5.5 m
d) R = 18.5 cm, r = 3.5 cm
5) The running track is enclosed between two concentric
circles of radii 42 m and 49 m. find the cost of cementing
the track at the rate of ` 25 per m2.
136
No
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Points to Remember
 Perimeter: The distance covered along the boundary of
the closed figure is called the perimeter. It is denoted
by P.
 Perimeter of the rectangle = p = 2 (l + b) units, where l
denotes length of the rectangle and b denotes breadth
of the rectangle.
 Perimeter of the square = 4a units, where “a” is length
of side of the square.
 Area: The surface enclosed by a closed figure is called
area. Area of a closed figure is denoted by A.
 Area of the rectangle = A = lb
 Area of the square = A = a2 sq. units.
 The perimeter of a scalene triangle = P = a + b + c, where
‘a’, ‘b’, and ‘c’ are length of sides of the scalene triangle.
 The perimeter of the equilateral triangle = 3a, where ‘a’
is length of each side of an equilateral triangle.
 The area of a triangle = 1 b h, where ‘b’ is the base and
2
‘h’ is the height/ altitude of triangle.
 The circumference of the circle = 2πr.
 The area of the circle = πr2.
 Area of the circular path way = π (R2 - r2)
 Approximate value of π = 22 c 3.14
7
 Units of measurements of area
• 1 m2 = 100 dm2
• 1 m2 = 10,000 cm2
• 1 m2 = 10,00,000 mm2
• 1 dam2 = 100 m2
• 1 hm2 = 10,000 m2
• 1 km2 = 10,00,000 m2



137


UNIT- 8
DATA HANDLING
After studying this unit you :


d
he

explain data collection and organise data in a systematic
manner,
understand meaning and importance of Central
tendency,
find out the Mean, Median and Mode for the given data,
construct a bar graph to represent a data.
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
You know various types of data, collection of data, and
how to tabulate data in a pictograph as well as in a bar graph.
The collection, tabulation and presentation of data
in organized form helps us to draw inference about that
particular data.
Activity 1 : Collect information from your classmates who are
interested in the following activities and complete the table.
to
Name of the
activity.
Roll no. of student
interested
1) Singing
No
t
2) Dancing
3) Carrom Board
4) Chess
5) Kho - Kho
6) Essay Writing
7) Cricket
8) Foot ball
138
No. of Student
interested
How many students are interested in singing?
In which activity maximum number of students are
interested?
3) In which activity minimum number of students are
interested?
4) How many students are interested in more than one
activity?
Activity 2 : Observe the table given below, and then answer
the question
he
d
1)
2)
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Temperature of the district head quarters in Karnataka as
on 25/1/2012 is given below
Name of the District Max. temperature Min. temperature
360C
250C
kalaburgi
320C
240C
Tumkuru
280C
210C
Shivamogga
260C
200C
Madikeri
280C
220C
Chikkamagaluru
290C
260C
Kolar
300 C
270C
Mysuru
1) In which district the highest maximum temperature is
recorded?
2) In which district the lowest minimum temperature is
recorded?
3) Name the districts in which same maximum temperature
is recorded
4) Note the difference between the maximum and minimum
temperature of each district
Do This :
Activity 1 : With the help of your teacher or elders Collect the
temperature data of all the days of any one month in this year.
2 : Collect some data regarding sports from news papers
3 : Try to analyses and conclude from the above data.
139
Activity 3 : There are 25 beads in a box. The teacher instructs
five children to insert these beads in the rods P, Q, R, S, and
T.The children inserted the beads as shown in the figure.
P has 4 beads.
he
Q has ________ beads.
d
Count the beads inserted in each
rod and fill in the blanks given below.
P
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R has ________ beads.
Q
R
S
T
S has ________ beads.
T has ________ beads.
Name the rods which contain equal number of beads.
Name the rod which contain maximum number of beads.
Which rod contains least number of beads.
No
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We can observe that the rod P and S have same number
of beads.
Activity 4 : Let us arrange these beads in the descending
order and count the beads in each rods.
P
Q
R
S
The rod P has 8 beads.
The rod Q has 6 beads.
The rod R has 4 beads.
The rod S has 4 beads.
T
The rod T has 3 beads.
Now transfer one bead from the rod P to the rod R. Now
notice how many beads are there in each rod?
140
d
Q
R
S
T
P
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P
Q
R
S
T
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We can see that the beads are arranged in the descending
order 7, 6, 5, 4, 3
Count how many beads are there in rod R?
The rod R has 5 beads. Mark the number which is in the
middle of 7, 6, 5, 4, 3.
7, 6, 5 , 4, 3 (middle number is marked)
The middle number in the above list is 5
No
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Activity 5 : Arrange the beads so that all the rods have equal
number of beads as shown in the figure.
P
Q
R
S
T
P
141
Q
R
S
T
Let us tabulate our observation of the above activities.
Activity
Number of beads in the rods.
Rod Q
Rod R
Rod S
Rod T
3
8
6
4
4
3
4
7
6
5
4
5
5
5
5
d
Rod P
he
3
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5
5
By observing the table we can have a conclusion as
follows.
In the Activity 4, beads are arranged in the form 8, 6, 4,
4 and 3. These numbers are called data. In this, the most
repeated value is 4. This representative value of data is called
Mode.
Similarly, in the Activity 4, after transfer, beads are
arranged in the form 7, 6, 5, 4, 3. In this the middle number
is 5. This representative value of the above data is called
Median.
to
In the activity 5, the beads are arranged in the form 5, 5,
5, 5, 5. In this case the representative value of the data is 5
which is called Mean or Arithmetic mean .
No
t
Mean is also called average of the given numbers .
To show this let us find the average of beads(quantities)
in all the three activities.
In the activity 3, number of beads are 8, 6, 4, 4, 3 respectively.
Average = Sum of the beads
Number of rods
=
8 + 6 + 4 + 4 + 3 25 5
=
=
5
5
142
∴ the average of the beads is 5.................. (1)
In Activity 4, the number of
respectively.
Average = Sum of the beads
Number of rods
7 + 6 + 5 + 4 + 3 25 5
=
=
5
5
d
=
beads are 7, 6, 5, 4, 3
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∴ the average of the beads is 5...................
( 2)
In Activity 5 we have arranged the beads so that all the
rods have equal number of beads namely five each
+ 5 + 5 = 25 = 5
∴ Average = = 5 + 5 + 5
5
5
In activity 3 we notice that median is 5 and in activities 4
and 5 the Mean and Median are same, that is 5, but it may
not be equal in all the cases. They may differ for different
types of data.
Mean, Median and Mode which are representative values
of the data is called Central Tendency.
Note : The arithmetic Mean is also called as Average.
No
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To clarify this point let us take another example of weather
forecast table as given below in a News paper dated 07-08-2013.
Temperature Table
Stations
Name
New Delhi
Kolkata
Mumbai
Chennai
Pune
Hyderabad
Trivandrum
143
Maximum
Minimum
temperature temperature
in 0C
in 0C
34
26
33
26
31
26
33
24
28
21
25
21
27
22
What are the central tendencies of the above data?
Arrange the above data in two tables in ascending order
as shown below
Station's
Name
Maximum
temperature
in 0C
Station's
Name
Hyderabad
25
Hyderabad
Trivandrum
27
Pune
Pune
28
Trivandrum
22
Mumbai
31
Chennai
24
Kolkata
33
Kolkata
26
Chennai
33
Mumbai
26
New Delhi
34
New Delhi
26
d
Minimum
temperature
in 0C
he
21
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21
From the above tables, find the mean for maximum and
minimum temperature separately.
to
Mean for maximum temperature
Sum of all the max imum temperature
=
Number of Stations
25 + 27 + 28 + 31 + 33 + 33 + 34 211 30.140 C
=
=
7
7
No
t
=
The Mean maximum temperature of the 7 stations is
= 30.140C.
Similarly, mean for minimum temperature.
=
Sum of all the min imum temperature
Number of Stations
=
21 + 21 + 22 + 24 + 26 + 26 + 26 166 23.710 C
=
=
7
7
144
To find the median of above given temperature arrange
data in ascending order.
maximum temperature 25, 27, 28, 31, 33, 33, 34
he
d
minimum temperature 21, 21, 22, 24, 26, 26, 26
Since it is an arrangement of having 7 numbers can you
mark the middle number? yes, it is odd number, we can easily
mark the middle number which gives the Median.
The Median of the maximum temperature = 310C
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The Median of the minimum temperature = 240C
Next, to find the Mode round off the most occurred
(repeated) number in maximum and minimum temperature.
maximum temperature 25, 27, 28, 31, 33, 33, 34
minimum temperature 21, 21, 22, 24, 26, 26, 26
The Mode for the maximum temperature is 330C (2 times
repeated).
The Mode of the minimum temperature is 260C (3 times
repeated).
to
Measures of Central Tendency :
No
t
For a given sample of data we notice difference in Mean,
Median, and Mode.
These three measures show the central tendency of the
given data. Therefore the measures are called the measure
of central tendency.
The average lies between the highest and the lowest value
of the given data, and it is a measure of the Central Tendency
of a group of data.
The central tendency of the data is represented by Average
145
(Mean), Median of Mode depending upon the nature and need
of the data.
Measures of Central Tendency are:
I. Arithmetic Mean.
II. Median and
III. Mode.
d
Mean
he
Let us take a situation, where Manaswi and Harish received
their unit test marks sheet.
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The test has 5 subject, each carrying ten marks and their
score were given in the table
Subject Number.
Manswi
Harish
1
2
3
4
5
9
10
8
6
9
6
5
10
10
10
Harish argued that his progress better because he had
scored 10 marks in more subjects than Manaswi in this
test.
to
But Manaswi did not agree with his argument, so they
met their teacher. The teacher asked them to find the average
marks scored by them. Both of them calculated as follows.
: 9+10+8+6+9 = 42 = 8.4
Manaswi's average score
No
t
5
: 6+5+10+10+10 = 41 = 8.2
Harish's average score
5
Since, Manaswi’s average score is more than that of
Harish, teacher convinced Harish that her performance in
this test is better than him.
Sum of all the given quantities
Arithmetic Mean =
Number of quantities
146
Average or Arithmetic mean is a Central Tendency, which
is obtained by the sum of scores divided by the number of
scores.
Arithmetic mean = x1 + x2 + x3 + ..... + x n
Mean =
/x
n
n
he
d
n
Where '∑' is a Greek letter used as a symbol to represent the
sum of numbers and is read as 'SIGMA'.
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Example 1: The quantity of milk supplied by a farmer to a
dairy as given below. Find the average milk supplied by him
in a month.
Month
Supplied
Milk in litre
April May June July August September
215
218
314
340
420
410
Total milk supplied to a dairy in 6 months
= 215 + 218+ 314 + 340 + 420+ 410 = 1917
∴ Mean of the milk supplied= total milk sup plied
to
number of months
215 + 218 + 314 + 340 + 420 + 410 1917 319.50 l
=
=
=
6
6
Example 2: The water consumption of eight different families
No
t
is as follows Calculate the mean consumption of water by the
families in a month.
Families
1
Water
Consumption
in kiloliters
7.5
2
3
8.2 7.5
4
5
6
7
8
20.00
9.10
4.55
6.62
8.24
Total number of families = 8
147
Total water consumption in kiloliters
= 7.5 + 8.20 + 7.5 + 20.00 + 9.10 + 4.55 + 6.62 + 8.24 = 71.71
Mean =
Total water consumption by the families
Number of families
d
71.71 8.96
=
8
he
Therefore, the Mean of water consumption by the families
is 8.96 kiloliters. That is 8960 litres.
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Try This : How would you find the average of your study
hours in a week?
Exercise: 8.1
In 'one day' cricket matches, the runs scored by a
batsman Mohan are given below.
15, 17, 23,108, 35, 8, 38, 30.
Calculate the average runs scored by Mohan.
II. Find the mean of;
1) First nine prime numbers.
2) First eight natural numbers.
III. The the weights of 10 new born babies in a hospital on
a particular day: (in Kg.)
3.4, 3.5, 4.5, 3.9, 4.2, 3.8, 4.4, 4.5, 3.6, and 4.1 respectively.
Find the mean weight these babies.
IV. An organisation evaluated ten projects and awarded
the following marks.
12, 11, 15, 16, 10, 14, 9, 13, 8, 17.
Later the organisation awarded 3 additional marks to each
projects.
1) Will the average of the marks change after the addition
of the 3 marks for each score?
2) Calculate the mean of the score in both the situation.
No
t
to
I.
148
V. The runs scored by two batsmen in the six matches
are given below: Who performed better?
Batsman -1
Batsman -2
60
65
100
45
50
70
30
100
94
60
50
40
VI. Find the missing number in each of following data :
41, 43,____,47, 49,51 and the mean is 46.
he
2)
d
1) 21,25,29,____31,33 and the mean is 28.
Median
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Consider this example, the salary of 10 workers in a factory
is as follows.
I
` 15000
` 12000
` 18000
` 6000
` 7500
` 8000
` 7000
` 5000
` 9000
A B C D E F G H
J
Total (10)
` 45000
` 132500
Staff
Monthly salary
149
J
45000
15000
18000
I H G
12000
9000
8000
6000
C E D A
7500
F
7000
B
5000
No
t
to
Here salary of most of the workers lies between 5000 to
15000. The Mean salary of these ten workers is ` 13250/- and
it is nearer to the larger salaries. This Arithmetic Mean does
not properly represent the central tendency of the data, that
is the salary of workers.
Therefore, we require a better way to measure the Central
Tendency for the above statistical data.
To deal with such a situation we have one more measure
called the Median.
Arrange the salary of 10 workers in ascending order and
observe the numbers in the middle of the table.
=
5th + 6th 8000 + 9000 8500/ =
=
2
2
d
From the above example, the numbers that lies at the
middle. We have two salaries (values) marked bold, located
in the middle are 8000 and 9000. In the above case, we take
the average of two numbers as the middle value.
Thus, middle value (salary) of the workers
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Here the Mean loses its ability to provide the best central
location for the data because it is nearer towards the higher
values. However the middle value retains central position and
which is not very nearer to any particular value and also this
value divides the data into equal parts. Thus, the value which
lies in the middle of the scores represents MEDIAN of the data.
Note : For the given scores, if the number of terms is an even
number then,
Median = Sum of the two middle scores
2
or
n
n
8 2 B term + 8 2 + 1 B term
th
th
to
Median =
2
* For the given data, if the number
of terms is an odd
th
n
1
+
number then the Median = 8
B term.
2
No
t
The "Median" is the "middle" value of the given data. To
find the Median, the scores have to be listed in ascending
or descending order.
Activity : Let the students of your class stand in row.
Arrange them according to their height. Then choose the
student who stand in the middle, his height gives the concept
of the Median.
Example 1: The scores obtained in a monthly mathematics
test by 7 students is as follows 24, 36, 46, 17, 18, 25, 35. Find
the Median of this data.
150
Let us arrange the data in ascending order, and mark the
middle term.
17, 18, 24, 25, 35, 36, 46.
n=7
From the above data, it is observed that, the 4th term is 25.
Therefore Median of the scores is 25.
he
d
Example 2 : 15, 21, 22, 18, 20, 17, 12, 11, 10, 8, 7, 19.
Find the Median.
Here n = 12, which is an even number.
n
n
8 2 B term + 8 2 + 1 B term
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Therefore, Median
th
=
=
=
th
2
^6 th termh + ^7 th termh
^15h + ^17h
= 16
2
2
∴ Median of the data = 16
to
Example 3: The numbers 50, 42, 41,35, 2x +10, 2x – 6, 12, 11,
8, and 6 are written in descending order. If their median is 24,
find the value of x and calculate 2x + 10 and 2x -6.
Here the number of terms is 10. And Median for these scores
th
is 24.
n th
n
a k term + a + 1 k
No
t
` Median = 2
2
2
(2x + 10) + (2x - 6)
24 =
2
48 = 4x + 4
4x = 48 - 4
4x = 44
x = 44
4
` x = 11
151
=
5 th term + 6 th term
2
(i)
2x + 10
(ii)
= 2 (11) + 10 = 32
2x - 6
= 2 (11) - 6 = 16
Example 4: The median of the given data is 18 in which one
number is missing. Find the missing number for the scores
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12,10,8,15,____,20,24,29.
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Solution: Here, on arranging the scores in ascending order
it would be 8,10, 12, 15, x, 20, 24, 29 In the scores the median
lies between 15 and x.
` 18 = 15 + x
2
18 # 2 = 15 + x
36 = 15 + x
x = 36 - 15
x = 21
Exercise : 8.2
Find the Median of:
1) 7, 4, 25, 1, 4, 0, 10, 3, 8, 5, 9, 2.
2) 20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22.
3) 15, 6, 16, 8, 22, 21,9, 18, 25.
4) 2, 10,9, 9, 5,2,3,7,11.
to
I.
No
t
5) 36, 32, 28, 22, 26, 20,18,40.
II. In a cricket match, 11 players scored runs as follows:
6,15, 120, 50, 100, 80, 10, 15, 8,10,15. Find the Median
III. Find the median of the following data:
19, 25, 59,48,35,31,30,32,51.
If 25 is replaced by 52 then what will be the new Median?
IV. Out of 10 observations arranged in descending order,
the 5th and 6th terms are 13 and 11 respectively. What
is the Median value of all the ten observations?
152
Mode:
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ACTIVITY 1: The class attendance of 10 students in
7th standard is given below in Table 1.
Table 1
Table 2
Name
No. of attendance
Name No of attendance
(for 30 days)
(for 30 days)
10
Suhasini 25
Reema
15
15
Raja
Raja
30
Rohit
Sanjana 17
25
Gulabi
Francis 18
28
Hussain
Sandeep 18
10
Rema
Suhasini 25
25
Sanjana 17
Gulabi
Fransis 18
Karthik 25
Sandeep 18
Hussain 28
30
karthik 25
Rohith
The attendance of students given in table 1 are arranged
in ascending order in table -2
Which number is repeated many times?
Here we observe that the number repeated maximum
times is 25.
No
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Look at the following example:
A man opened a shoe shop. To keep the stock of the store
as per the requirement, he noted down the number of shoes
of different sizes sold during a week.
Size
Number of pair of shoes sold in the week
5
10
6
25
7
36
8
22
9
9
From the above table what do you observe?
153
From the above table it is observed that, number of shoes
of size 7 is sold the most.
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In the above situation, we observe that the another
representative value of the data, which is most repeated. This
representative value of Central Tendency is called "Mode" of
the data.
he
The "Mode" is the value that occurs most frequently or most
often in the data.
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Note : Mode is a mathematical term that refers to
‛‛MAJORITY”.
Example 1: To find out the weekly demand for different sizes
of shirt, a shopkeeper kept records of sales of shirts of sizes
90 cm, 36 cm, 39 cm, 40 cm, 42 cm.
The following is the record for the week
Size
No. of shirts sold
36 cm 38 cm
8
22
39 cm
32
40 cm
37
42 cm
6
No
t
to
From the above record, the shopkeeper decides to postpone
the purchase of shirts of size 36 cm and 42 cm. Why?
Because less number of shirts were sold of size 36 cm and
42 cm.
Which is the most popular size of shirt that is sold the
most?
Clearly, it is 40 cm.
Therefore, the Mode is 40 cm.
Example 2 : The number of points scored in a series of football
games is listed below. Which score occur most often?
7, 13, 18, 24, 9, 3, 18
Arranging the score from least to greatest
We get; 3, 7, 9, 13, 18, 18, 24.
The score which occurs most often is 18.
∴ The Mode is 18.
154
= 13
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Example 3 : For what value of 'x' the Mode of the following
data is 9.
17, 9, 12, 17, 18, 11, 19, 21, 9, x- 4
Here both 17 and 9 are appearing twice and other values
are occurring only once. Since, the Mode is given as 9, it must
appear at least 3 times.
Mode = 9
Therefore x - 4 = 9
x
=9+4
Exercise: 8.3
I.
Find the Mode of
1)
2, 6, 5, 3, 0, 3, 4, 3, 3, 2, 4, 5, 2, 4.
2)
2, 14, 16, 12, 14, 14, 14, 16, 14, 10, 14, 18, 14.
3)
61, 65, 66, 66, 68, 69, 69, 50.
II. The heights (in cm) of the 25 children are given below;
find the Mode.
to
168, 165, 163, 160, 163, 160, 161, 162, 164, 163,
160, 163, 160, 165, 163, 162, 163, 164, 163, 160,
No
t
156, 165, 162, 168, 168.
III. Find the Mode of following data.
2, 16, 14, 14, 13, 16, 19, 14, 12.
IV. For what value of x, the Mode of the following data is 18?
31, 35, 17, 18, 17, 18, 40, x +12.
V. For what value of x the Mode of the following data is 26?
26, 51, 24, 26, 24, 26, 35, x – 1.
155
Graphical representation of the data:
You have studied how to represent data in pictograph as
well as bar graph.
Name the following graphs:
d
Banana
40
10
5
Kho-kho
Papaya
15
Drawing
Grapes
20
Dancing
Mango
Kabbadi
30
25
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Apple
35
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No. of Student
Orange
Activities
fig (i)
fig (ii)
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Yes, figure (1) is a pictograph and figure (2) is a bar graph.
Activity: Collect similar types of pictographs and bar graphs,
that appear in news papers and magazine having different
data.
number of weeks
156
5th Week
4th Week
3rd Week
2nd Week
1200
1000
800
600
400
200
1st Week
plants planted
number of
No
t
Look at the graph given below: It shows the number of
plants planted in successive weeks.
By observing the graph find the number of plants, planted
in each week.
Week
1st
Number of Plants
1000
2nd
3rd
4th
5th
d
In which week the maximum number of plants are planted?
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In which week where minimum number of plants are
planted.
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Bar Graph :
Bar graph consists of rectangular bars which are drawn
on one of the axis at uniform distance.
Height of each bar represents the quantity of the collected
data.
It is easier and interpret the data from the bar graph.
Let us draw the bar graph with an example.
Example 1 : Bhavana's marks in monthly test is given below :
Y
No
t
to
Kannda = 20, English = 15, Hindi = 10,
Mathematics = 25, Science = 20,
Social Science = 15 prepare bar
graph for the data.
Step 1: Take a graph sheet and draw
two perpendicular lines on it.
[The horizontal line is called
X – axis and the vertical line is
called Y – axis.]
157
1cm
X
Step 2: A l o n g t h e h o r i z o n t a l l i n e
to represent the subjects.
X
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[Each subject = 1cm]
x axis
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uniform gap between them
y axis
width of the bars with
1cm
(X – axis), choose uniform
Y
Step 3: On the vertical line (Y – axis), we have to represent the
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number of students. As we have to represent
a maximum of 25 marks, so we have to choose an
appropriate scale: 1 cm = 5 marks. For 1 mark =
1 cm
5
Step 4: Calculate the heights of various bars as shown
below (along Y axis ).
= 1 # 20 4 = 4cm
5
The height of the bar for English
= 1 # 15 = 3cm
5
The height of the bar for Hindi
= 1 # 10 = 2cm
5
The height of the bar for Mathematics
= 1 # 25 = 5cm
5
The height of the bar for Science
= 1 # 20 = 4cm
5
No
t
to
The height of the bar for Kannada
The height of the bar for Social Science = 15 # 15 = 3cm
158
Bar Graph of Bhavana's Score
Marks scored→
30
25
he
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20
15
10
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5
0
Kannada
English
Hindi
Maths Science
Subjects →
S.Science
Observe the above graph which shows the marks of
Bhavana.
Try to answer the following questions.
a) In which subject her score is maximum?
b) In which subject her score is minimum?
c)
Name the subjects in which her score is the same?
Note:
The bars should be neatly drawn and neatly coloured
or shaded.
No
t
•
Every bar graph should clearly show what it represents.
to
•
•
The unit of scale chosen should be clear and marked
properly on X-axis and Y-axis.
Example 2 : The table below shows the number of athletes
selected from 5 states to represent a National competition.
Represent the data in a bar graph.
159
State
Andhra
Pradesh
No. of
Athlets
45
Maharashtra Karnataka Kerala
65
40
75
Tamil
nadu
30
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Draw two lines on a graph paper, which are
perpendicular to each other and mark it as X-axis,
and Y-axis.
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Stage 1:
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Method: To draw the bar graph, the following steps are to be
followed.
Stage 2: Along the X- axis mark the states and along the
Y- axis mark the number of athletes.
Stage 3: On X- axis mark the states with uniform width with
equal gaps.
Stage 4: On Y-axis choose a convenient scale to determine
the height of the bars.
Let us choose a scale 1 cm = 10 athletes.
Therefore, 1 athlete
= 1/10 cm.
to
The height of the bar representing Andhra Pradesh is
= 1 # 45 = 4.5 cm
10
The height of the bar representing Maharashtra is
No
t
= 1 # 65 = 6.5 cm
10
The height of the bar representing Karnataka is
= 1 # 40 = 4 cm
10
The height of the bar representing Kerala is
= 1 # 75 = 7.5 cm
10
The height of the bar representing Tamil Nadu is
= 1 # 30 = 3 cm
10
160
Stage 5 : Draw bars of equal width and of height calculated
in the stage 4.
70
60
d
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Tamil Nadu
Karnataka
30
Maharashtra
40
Andhra Pradesh
50
20
10
0
Kerala
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Number.of athletes
80
State
Exercise: 8.4
I.
The strength of the different classes of Government
higher primary school are given below.
class
strength
4
90
5
6
70 50
7
8
40 40
No
t
to
Draw a bar graph to represent the above data.
II. A survey of 100 school students regarding their
favourite leisure time habits are given below. Draw a
bar graph to represent the above data.
Habit
No. of students
Playing
35
Reading story books
10
Watching Television
20
Listening to the Music
5
Painting
30
161
III. Read the following bar graph and answer the following
questions:
Number of students who went abroad for studies
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4000
3000
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No. of students
5000
2000
1000
0
2005
2006
2007
2008
Year
2009
2010
In which year maximum number of students went to
study in abroad?
2) In which year minimum number of students went to
study in abroad?
IV. The following table shows the monthly expenditure of
Satish’s family on various items. Draw a bar diagram
to represent this data. ( 1cm = ` 500 units)
No
t
to
1)
Items
House rent
Food
Education
Electricity
Transportation
Miscellaneous
Expenditure
3500
4000
800
1200
1000
2000
162
Double Bar Graph (Composite Bar Graph):
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As we know that a bar graph can be formed by drawing
vertical bars (as shown in previous example), or by drawing
horizontal bars.
A graph with vertical bars is called a vertical bar graph.
A graph with horizontal bars is called horizontal bar graph.
We can represent a double bar graph either horizontally
or vertically.
The Marks scored by Ramesh and Satish of 7th standard,
are as follows.
Subject
Marks of Ramesh
90
Kannada
85
English
70
Hindi
100
Mathematics
90
Science
80
Social Science
Marks of Satish
80
70
90
80
60
55
The graphs below represent the marks scored by Ramesh
and Satish:
Satish
to
Ramesh
120
120
100
80
163
S.Science
0
Science
Subjects
Maths
20
Hindi
40
English
60
Kannada
Marks
S.Science
Science
0
Maths
20
Hindi
40
English
60
Kannada
Marks
No
t
100
80
Subjects
Can we represent the same in a single graph?
Yes, we can represent this as follows:
The same graph can be represented compositely and it
is as follows:
d
100
80
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Marks
120
60
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40
20
0
Kan
Eng
Hin
Maths
Subjects
Sci
S.Sci
Ramesh
Satish
When we want to compare the same quantity between two
different data, we use composite bar graph.
Exercise : 8.5
The following table shows the education level of men
and women in Karnataka. Represent it in suitable
composite graph.
to
I.
No. of male
(in ten thousand)
No. of female
(in ten thousand)
Middle
95
92
Secondary
20
10
Higher secondary
45
30
Graduate
30
25
No
t
Education level
164
II. Mysore Mahanagarapalike spent various amounts
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on areas of social need in two consecutive years
2010-11 and 2011-12. Represent the data on a double
bar graph using the table given below.
2010-11
Subject
2011-12
Expenditure
Expenditure
(in lakhs)
(in lakhs)
60
70
Education
35
45
Water supply
25
30
Medical aid
45
50
Transportation
15
22
Sanitation
30
35
Road construction
to
1) On which items did the corporation spend more in
2011-12
2) Which items incurred the same expenditure in both the
year?
3) What is the total expenditure on water supply in both
the years?
4) How much more did the corporation spend in 2011-12
than in the previous year on road construction?
No
t
Know this : The mode is useful when the most common
item, characteristic or value of the data is required.



165


UNIT - 9
PROBABILITY
After studying this unit you :

represent outcomes of an event in the graph.
he
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
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
explain the meaning of probability,
define the meaning of random experiment,
predict the outcome of random experiment,

Introduction
to
In our day to day life, we come across many situations where
uncertainty plays a vital role. We usually use statements like
“there is a chance of rain today” or “probably he will get first
class in the examination” etc. In all these contexts the term
chance or probably indicates uncertainty. In the statement
“there is a chance of rain today”, means it may rain or may
not rain today. We are predicting rain today based on our past
experience when it rained under similar conditions. Similarly,
in the case of a student getting first class probably indicates
uncertainty.
No
t
Let us see a few more statements :
•
Headmaster doubts about the Kaveri's promotion in the
examination.
•
Opinion of the audience is “m o s t p r o b a b l y ,
Kevin will stand first in the competition”.
•
There is Possibilities of increase in cost of vegetables as
there is, due to hike in price of diesel.
•
There is a 50-50 chance of a school team winning the
trophy.
166
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The word such as ‘probably’, ‘doubt’, ‘possibilities’, ‘chance’,
etc., used in the above statements indicate an element of
uncertainty.
The word probability is related to the occurrence of
uncertainty.
Probability has been extensively used in the fields of
weather forecasting, share market, commerce, medical
sciences, biological sciences, physical sciences etc.
Therefore, ‘probably’ or uncertainty is measured
numerically. The word for such measurements is ‘probability’.
The concept of probability can be quantified.
Random Experiment
Now let us consider a situation of the cricket match. Before
the commencement of the match captains of both the teams
decide, 'Who will bat first'? by tossing a coin. Captain of team
'A' tosses the coin. Suppose you are the captain of team 'B' and
choose 'Head' to come up, are you sure that 'Head' will come
up? If you choose 'Tail' to come up, does it happen? Exactly
It is not predictable. Result of the toss cannot be controlled.
the chances are 50 : 50.
No
t
to
Now let us consider another situation of choosing a ball
from the box containing balls of different colours. Without
looking at the colour of the ball, is it possible to say, in advance
the colour of the chosen ball? No, it is not possible.
Similarly, can we predict what will be the number of dots
on the upper face of a rolled dice? It will be anyone of 1, 2,
3, 4, 5, 6. It may or may not be a number of the dots of your
prediction.
In the above experiments the result is not exactly
predictable. It will be more than one, it will be different. Such
experiments whose result is not controllable or predictable
are called Random Experiments.
167
d
A possible result of a random experiment with a dice is
shown below.
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An experiment is said to be random experiment, if it
satisfies the following conditions.
(i) it has more than one possible result, and
(ii) it is not possible to predict the result in advance.
Sample Space
In a random experiment of rolling the dice, the number
visible on the top of the dice is 1, 2, 3, 4, 5 or 6. The result of
number of visible dots on the top of the dice is written as {1,
2, 3, 4, 5, 6}
Similarly when a coin is tossed once, either head or tail
come up. It is expressed as {H, T}. where H represents head
and T represents tail.
to
In the above random experiments the possible outcomes
are written using flower brackets { } and is denoted by S.
No
t
∴ For a random experiment of rolling dice the sample
space is S = {1, 2, 3, 4, 5, 6}
Similarly, for the random experiment of tossing a coin once
the sample space is S = {H, T}.
Activity 1 : Prepare six flash cards having numbers.
3, 5, 7, 11, 13 and 17, shuffle and randomly, pick a card from
the above flash cards. What are the probable outcomes of the
experiment? Write it down.
168
Activity 2 : Put blue, red, yellow and green balls in a box,
randomly take a ball from the box. What are the possible
outcomes? List them.
9.1
1) If you spin a coloured disc containing the
d
Exercise
Green Red
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colours Yellow, Blue, Red, and Green, It stops Yellow
Blue
at the pointer after some time. Write the
sample space for the colour at which it stops?
2) You spin the wheel shown: make a list of
all the possible outcomes of the number at
which pointer stops ?
3) Toss two coins. Write all possible outcomes associated
with this experiment. S = (H1 H2), (H1 T2), ___, ____}
4) Five flash cards having odd numbers from 11 are shuffled
well and one card is drawn from it. Write the sample
space.
Event:
to
Consider the random experiment of a dice having six faces
thrown once.
No
t
The possible outcomes for numbers on top of the dice are
{1, 2, 3, 4, 5, 6}
Suppose, we restrict the outcome to be an even number,
then .The possible outcomes satisfying this condition is
A= {2, 4, 6}. This outcome is an example of an Event.
Similarly, if we restrict the outcome of the experiment to
be a prime number then B = {2, 3, 5}. This is also an Event.
169
When a coin is tossed getting only head or tail is also an
example of an event.
An EVENT of a random experiment is a particular
outcome of the sample space and denoted by E.
Notion of chance or probability
he
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To understand the notion of chance, let us take the activity
of 'tossing a coin'. We know that sample space will be {H, T}.
Let us toss the coin again and again and write the result as
shown in the table below.
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Sl no N u m b e r o f Head coming Tail coming n^Hh n^ Th
n^Sh n^Sh
times tossed up n (H)
up n (T)
n (S)
2
3
4
5
10
5
10
15
5
15
30
18
12
8
30
12
30
50
35
15
35
50
15
50
70
40
30
40
70
30
70
80
45
35
90
47
43
45
80
47
90
35
80
43
90
52
48
52
100
48
100
No
t
6
15
to
1
7
100
For trial 1, the ratio of getting head turned up to the total
n^Hh 10
number of times a coin is thrown is n^Sh 15 Similarly, for
n^Hh 18
and for trial 3 it is n^Hh 35 and so on.
trial 2 it is
n^Sh 30
n^Sh 50
What do you observe from the above table?
170
You will find that as the number of tosses increased, the
value of the n^Hh or n^ Th come closer to 1 or 0.5
n^Sh
2
n^Sh
This means that, chance of getting head or tail when a
head up when a coin is tossed, is 1 or 0.5
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coin is tossed is 50:50. It also means the probability of getting
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Activity 1: Throw a dice 15 times and note down the number
of dots 1, 2, 3, 4, 5 or 6 come up. Record your observations in
the form of a table given below. Repeat the activity for 30, 90,
100, 120 and more times and complete table.
Table
Number of times
a dice is thrown
Number of times the dots comes up
dot 1
dot 2
dot 3
dot 4
dot 5 dot 6
15
30
90
to
100
120
No
t
Using the above table write the observation in the form of
fraction.
Number of times one dot come up
for all the trials.
Number of times dice thrown
Similarly, Find the above ratio for 2 dots, 3 dots, 4 dots,
5 dots, 6 dots. It is possible to notice that, by increasing the
number of trials the chance or probability of one dot / any
dot to be on top, of the dice thrown, is 1 or 0.167.
6
171
Pr obability =
Posibility of an event
Total number of events
he
Pr obability = The number of favourable out comes
Number of all the possible out comes
d
So probability may be defined as the ratio of number of
expected or favorable out comes to number of all possible
outcomes or number of sample space.
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Example 1 : A dice is rolled, calculate the probability of getting
an even number on the top.
The dice has six faces each numbered (marked with dots)
with 1, 2, 3, 4, 5, and 6.
The outcomes of this experiment are {1, 2, 3, 4, 5, 6}.
So number of possible outcomes = 6
A dice has three even numbers, they are 2, 4 and 6.
So number of favourable outcomes = 3
∴ the possibility of getting the even number =
to
Number of favourable out comes
3 1
= =
Total number of all possible out comes 6 2
No
t
Example 2 : A bag contains, two of each blue, red, green
and yellow coloured balls. Randomly one ball is picked, what
will be the probability of getting a blue ball?
Total number of balls in the bag
=8
Expected outcome of blue coloured ball
=2
The probability of getting Number of blue balls in the bag
2 1
=
= Total number of balls in the bag = 8
blue coloured ball
4
172
Think! When a coin is tossed, imagine that if
the coin stands vertically on the ground and
when a dice is rolled, assume that it stands on
its edge, what is the probability in such cases?
Write the possible outcomes of rolling a dice. Write
the probability of getting the following numbers
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9.2
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Exercise
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Such cases, will not be considered under probability
1) A number greater than 5.
2) A even number
3) A odd number
4) A number less than 5.
Numbers 2 to 8 are written on separate slips, which are
kept in a box and shuffled well. One slip is chosen from
the box without looking at it. What is the probability
of getting the following?
a)
b)
c)
Number 7.
Number greater than 7.
to
II
Number less than 7.
No
t
III Numbers 1 to 25 are printed on balls and were kept in
a box. If one ball is drawn randomly, then what is the
probability of getting the following?
a) an even number.
b) a multiple of 5.
c)
a factor of 24.
173
Representing the result of a random experiment in graph:
1. Rama tossed a coin 20 times, the outcomes of this
experiment is listed below;
Outcome
Sl. No.
Outcome
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Sl. No.
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When a coin is tossed once, the possible outcomes are Head
or Tail.
Here the sample space for this experiment is {H, T}
H
11
H
2
H
12
T
3
T
13
H
4
H
14
H
5
T
15
H
6
H
16
T
7
H
17
H
8
H
18
T
9
T
19
T
10
T
20
H
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1
174
The above data is tabulated as follows
Outcomes of an experiment Number of times it occurred
H
12
T
8
Length of
each bar
Head
6.0 cm
Tail
4 cm
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Out
comes
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The above data can be represented on a graph, taking event
of an experiment on X – axis, and the number of outcomes
on Y- axis,
Y-axis 1 cm, = 2 units.
Scale: X- axis 1 cm, = 1 units.
Number of events →
16
14
12
10
8
6
4
2
0
H
T
Event →
2. When a dice is rolled 60 times the result of each roll
can be recorded as shown in the table :
to
Score on Dice
Tally
Frequency
;;;; ;;;;
10
2
;;;; ;
6
3
;;;;;;;
;;;;;;;
;;;; ;;;;;
13
4
;;;;;;; ;;
;;;; ;;;;;
12
5
;;;;
;;;;
;;;;;;;;;;; ;
8
6
;;;; ;;;;;
;;; ;
11
No
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1
Total Frequency =
175
60
13
10
10
12
On X axis
11
1 cm = 1 unit
8
On Y axis
6
5
d
0
1 cm = 1 unit
1 2 3 4 5 6
Numbers on the dice →
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Exercise
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Frequency →
Now let us draw a bar graph to illustrate this :
Scale :
15
9.3
1) A dice marked with the letters A, B, and C on three of its
faces and numbers 1, 2 and 3 on remaining three faces. It
is rolled for 100 times. Table given below shows the results
of the face that come up. Draw a bar graph to illustrate
the results.
Number of times a dice is thrown Number of times these
scores turn up
1
2
3
A
B
26 10 14 20 12
100
C
18
2) A disk wheel is divided into 5 sectors and a pointer is placed
in front of it. Then, it is spin for 75 times. Table given below
to
shows the result that the position of the pointer against
the numbers. Draw a bar graph to represent the results.
No
t
Number of times disk The position of the pointer against
wheel is spin
the numbers
75
1
6
2
14
3
26
4
19
5
10
Activity : Take a glass bowl. Put 3 red, 5 blue and 2 green
marbles into it. Without looking into the bowl randomly take
a marble and record it. Repeat the experiment fifty times.
Tabulate the data and draw a bar graph to illustrate the
results.
176
CHAPTER 10
REPRESENTATION OF 3 DIMENSIONAL
OBJECTS IN 2 DIMENSIONAL FIGURES
After studying this unit you :




d
he

explain the meaning of 2 dimensions and 3 dimensions,
draw line diagram of solid figures cube, cuboid and
tetrahedron,
recognise visible and hidden faces of solid figures,
draw the diagrams of cube and cuboid,
explain the net of cube and know the method of
drawing the nets of cube, cuboid and tetrahedron,
imagine the images of solid figures mentally, know the
number of edges, faces and vertices.
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
Introduction
No
t
to
We know that the 'point' is the smallest figure in Geometry.
Point has no dimension, but it has a position. We locate this
with the sharpest tip of a pencil.
A line contains innumerable points. If points are collinear
then the line is called 'straight line'. A part of a straight line
is called line segment.
 A line segment has single dimension. (Only length)
 A plane is a flat surface.
 Line segments when joined in a plane form geometrical
figures.
 Squares and rectangles are the geometrical figures
which are formed by the line segments on the same
plane.

A rectangle has two dimensions (length and breadth).
177
Representation of 3 dimensional solids in
dimensional figures.
2
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Activity 1 :
Take an empty match box and count the number of surfaces.
Place it on a sheet of white paper and draw its borderline
with a pencil. Remove the match box and observe the border
lines drawn.
 It has only two dimensions, isn't it?
 Place the other faces of the same match box on a sheet
of white paper and draw the borderlines as before.
Matches Box
Matches Box
fig : 1
Borderlines of faces
Observe that each surface of the match box has two dimensions.
to
The figure drawn from border lines of the match box has no
thickness. But match box has thickness. Therefore, thickness is
another dimension. Hence match box has three dimensions. That
is length, breadth and height.
No
t
Objects having three dimensions are called solid objects. Solid
objects occupy one part of the space.
A shadow is formed when the light falls on the opaque objects.
The shadows formed by using our hands, may appear as, shadow
of flying bird, dog, rabbit etc. Our hand is three dimensional but
the shadows are of two dimensions. Forming shadows by using
different solids figures is a type of entertainment. Here, three
dimensional objects are shown in two dimension in different
types.
178
he
fig : 2
d
Observe these shadows formed by hands and try this.
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These are the figures of solid
models made by wood.
These figures are called solid
figures. They are named based
on their shapes. Solid objects are
called three dimensional objects.
We can see and also touch the
exterior parts of the solids. The
exterior parts are called surfaces
fig : 3
of solids. Solids may have flat
surface as well as curved surface.
Faces, edges and vertices of solids :
to
Cuboid :
E
No
t
D
A
C
H
G A brick is an example for a cuboid.
F Compare this with match box. A
brick has six plane surfaces, each
plane surface is called plane face
fig : 4
fig : 5
or, simply, face. Look at the fig (4). It is a brick. We can see
the top and two side faces of it from one direction [DHGE at
the top, ABHD and BFGH, at the sides]. The other three sides
[ABFC at the bottom, ACED and CFGE, at the sides] are
hidden.
B
179
Activity 1 :
Keep a match box on a table as
shown in the figure and
observe it.
How many faces can be seen?
Look at the match box from
different sides of a table.
List the number of faces that can
be seen (visible) in each situation.
The maximum number of
visible faces of the match box at a time is only three. Then,
what can you say about remaining faces? They are not
visible. Faces which are not visible in a solid object are
called hidden faces.
Cube
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Matches Box
Number of faces of this solid is
also six. But, the shape of the
H
D
each face is a square, i.e, length,
C
breadth, height are equal to each
F
other. This is called cube. ABHD
A
B
is one of its faces. Name the
remaining faces of the cube.
Let us see how many adjacent faces are joined together.
In this solid figure, for each face, there are four adjacent
faces.
How are the adjacent faces joined to each other?
These faces are joined by line segments.
ABHD and BFGH are joined by line segment BH .
Similarly, line segment BF joins the faces BFGH and ABFC.
The line segment which joins the adjoining faces is called
'Edge'. Cube has such 12 line segments or edges.
If BH and BF are the other two edges, name the remaining
edges.
G
No
t
to
E
180
Now observe how these edges meet each other. For example
edges HB and FB meet each other at the point B. Similarly edge
AB also meet at the point B.
In a solid figure, the point at which three or more edges
he
point C is the vertex. B and C are the vertices.
d
meet is called vertex. Edges AC, EC, CF meet at C. Therefore,
In the same way name the other vertices.
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By this you can understand that the above solid figure
has 6 faces, 12 edges and 8 vertices.
Activity :
Draw the figures of cuboid and cube. Name the faces, edges
and vertices. Write the visible and hidden faces.
Triangular based Pyramid
Construct equilateral triangle of convenient
measurement on a paper. Let it be ∆PQR.
P
Inside the ∆PQR mark a point 'S' on a
paper. Join PS, QS and RS.
to
Now you will get a diagram as shown
in the figure.
S
No
t
Q
R
This figure is 2D figure of the solid
object called Triangular based Pyramid.
You know this :
A solid figure having many faces is called Polyhedron. Poly
means many and hedron means face. Polyhedra is plural of
Polyhedron.
Note : VERTEX - singular, VERTICES - plural
181
P
Look at the triangular based pyramid.
How many faces has it?
How many faces are hidden in this?
Q
R
d
Count the number of edges and number of vertices.
he
This solid has 4 faces, In this QRS is base, and the other
three are side faces. (Adjacent faces)
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In a triangular based solid if all the faces are equilateral
triangles, then it is called Tetrahedron.
In the above figure we can observe PQ, QR, RP, PS, QS and RS
are edges and P, Q, R and S are vertices.
Therefore, tetrahedron has 6 edges and 4 vertices.
Square based Pyramid
Look at this figure.
to
Here is a figure of a pyramid
No
t
The base of this is a square and other sides are in the
shape of isosceles triangle.
If the base of the pyramid is a square, then it is called a
in square based pyramid.
From the above figure, we observe that a square based
pyramid has 5 faces, 8 edges and 5 vertices.
Similarly, if the base of the pyramid is a pentagon then it
is called a pentagonal based pyramid.
182
Write the number of faces, the number of vertices and total
number of edges in the base of the solid given below in
the table.
Number
of edges
in the
base
Number of Number of
Total
faces
vertices
number
of edges
d
Name of the
solid
he
Tetrahedron
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Square based
Pyramid
Pentagonal
based
pyramid
Hexagonal
based
pyramid
In the same way generalise to n-based pyramid.
Cylinder
to
Look at this figure, it is a figure of a cylinder.
No
t
It has 3 faces, one is curved surface and the
other two are circular plane surfaces. Circular plane
surfaces are joined with curved surface, its edges are
circular. Do the edges meet each other?
No. Therefore, cylinder has no vertex.
In a cylinder one circular face and a part of curved surface
are visible.
183
Cone
Observe the diagram of the cone.
A cone has two surfaces, in which one is curved,
that ends at a point, and another one is plane. In a
cone base and a part of curved surface is hidden.
Number
of hidden
faces
he
Number Number of Number
of faces vertices of edges
No
t
to
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Figure of the
solid
d
Complete the following table.
184
Drawing of Polyhedral nets for making solid models:
No
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to
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5 cm
Sushrutha wants to have a
5 cm
small cardboard box and like
to keep his pen, sharpener,
Fold
and rubber in it. So he asked
his grand father to help him
to do one. His grandfather
took (30cm × 25cm dimension) 5 cm
5 cm
5 cm
cardboard sheet, pen, pencil, Fold
Fold
Fold
scissors and gum. He told
Sushrutha to observe how he
would make a box. His Grandfather said now I draw a sketch
for the box required. Since, the length of the box is 3cm. Let
the edge of the box be 5cm. Then he makes a box by following
procedure.
Step 1 : He drew a rectangle with dimension (20cm × 5cm)
on the card board and marks the parts as shown in the figure.
Step 2 : Grand father folds the dotted lines, and paste the
shaded parts with gum, except partial shaded part. Now a
box is ready.
Tejaswi said he wanted a box having its length 10 cm. by
seeing this box. Then Grand father drew the skecth as shown
below.
10cm
10cm
Fold
5 cm
Fold
10cm
10cm
Fold
Fold
185
Then he prepared a box by folding along the doted line on
the sheet and pasted on shaded part. Both Sushrutha and
Tejaswi were happy with their boxes.
This type of sketch to built 3 dimensional objects is called
polyhedral nets.
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So, a net is a 2 dimensional figure that can be folded to
form a 3 dimensional object. Here each face of 3 dimensional
figure is represented as 2 dimensional figure.
Nets for some solid are given below.
Nets
Name of the solid / figure
a
a
a
Cube
to
4
1
2
No
t
3
Triangular based pyramid
5
4 1 2
3
Square based pyramid
186
1
4
2
3
5
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Triangular based prism
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B
A
C
Cuboid
A
C
B
Easy way of drawing solid figures on a flat Surface
In previous activities figures of solids are drawn on flat
surfaces. They appear as 3-D figures.
No
t
to
Among the faces of a solid drawn on a sheet of paper, some
faces are exactly having the same shape as that of the solid
face, but not all the faces. Even though the three visible faces
are not similar, we can recognise that figures as a solid and
find out what type of solid the are such sketches are termed
as oblique sketches.
To practice oblique sketches, there are some easy ways;
such as
i) Using square lined are dotted paper.
ii) using Isometric sheets of lines or dots.
187
Know this : In a square lined, sheet each small box is a
square. In a square lined dotted sheet we can get a square
by joining dots. By joining the dots on a isometric dot sheet
we get an equilateral triangle.
d
Isometric graph (grid)
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Observe the following figures :
Square Paper
Isometric Paper
Note: Isometric dot sheets are useful to draw sketches, in
which measurements also agree with those of solid.
Activity: How to mark 4 × 4 × 4 unit solid (cube) on square
lined sheet? Take a square lined sheet.
Draw the opposite face which is also
square of 4 units. To do this, mark the
mid point of the front face, mark 4 units
towards right side of this and complete
the square as shown in fig.
Join the corresponding corners.
to
step 1:
Draw the front face that is 4 units square
with pencil.
No
t
step 2:
step 3:
step 4: Retrace the sketch with pen.
188
Note : It is conventional to mark hidden edges by dotted lines.
ly draw a sketch of cuboid of 6 × 4 × 4 unit.
he
Draw a rectangle measuring
6×4
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step 2 :
d
step 1: Take Squared lined sheet of paper
Draw one more rectangle by marking
step 3 : the center of first rectangle of
6× 4 as in pervious activity.
step 4 : Join the corners and retrace it.
No
t
to
Know this: The term isometric refers to equal measure. An
isometric sheet consists of equilateral triangles. The lines
on the paper are in the three directions, each of which
represents a dimension.
The lines up and down represent the
vertical dimension (height). The other
two dimension represent the horizontal
dimension ( length and breadth ) .
Therefore to draw a 3-D figures we use
a isometric graph sheet
The isometric dot sheet , on which
dots are marked at equal distances is
also used to draw 3-D figure such sheet
is called isometric dot sheet.
189
Activity : Draw a sketch of a solid figure on an isometric
graph (grid) and on an isometric dot sheet.
Consider a solid of measurement 2 × 2 × 4 units
he
d
Step 1:
Take an isometric graph (grid) and an isometric dot
sheet. Draw the edges AB and BC on the bottom of the solid,
such that AB = BC = 2 units
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C
A
A
B
C
B
Step 2:
Draw vertical lines at the vertices of the base such
that AF = BE = CD = 4 units.
F
D
F
E
C
A
to
A
B
C
B
Join EF and ED.
No
t
Step 3:
E
D
F
F
D
E
A
C
E
A
B
C
B
190
D
Step 4: Count two units from D, E, F and G join FG and DG.
Now go through the lines - bold along the visible edges and
draw dotted lines for hidden edges.
G
F
D
F
E
C
A
B
C
B
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G
Exercise : 10.1
I. Fill in the blanks.
Solid
a) Cuboid
Faces Edges Vertices
6
12
to
b) Cube
c) Triangular based
No
t
pyramid
d) Square based
pyramid
e) Triangular based
prism
191
8
Shape of
faces
Rectangle
II. Answer the following questions.
Mention the number of curved surface, curved edge and
vertices in a cone.
2)
Mention the number of curved surface, curved edge and
vertices in a cylinder.
3)
Mention the number of curved surface, curved edge and
vertices in a sphere.
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1)
III. Match the following.
Name of the
solid
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Sl.
No.
Nets of solids
Answers
5
1
Cuboid
4 1 2
a)
3
B
Cube
b)
A
C
Triangular
c)
based Pyramid
4
Square based
d)
Pyramid
4
No
t
3
A
B
to
2
1
3
192
2
C
IV. Construct solids having the following dimensions on
a squared sheet (check sheet).
a) Cuboid having 3 × 2 × 2 unit dimension.
b) Cube having 4 × 4 × 4 unit dimension.
Sample sheet
No
t
to
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Isometric Grid
193
No
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Isometric Dot paper
194
Unit - 1 Indices
=3×3×3×3×3×3×3×3×3
he
IV a) 38
d
Exercise1.1 : I 1) 8 to the power of 3 2)13 to the power of 6
c) 4 to the power of 10 d) 10 to the power of 4 e) (-6) to the
7
power of 5
II a) 5 is index -3 is base number
b) 8 is index - 10 is base number c) 6 index -` - 2 j is base num3
ber d) 20 index - 3 is base number
III. a) 3 b) 4 c) 5,4 d) an
b) 113 = 11 × 11 × 11
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6
c) c 5 m = c 5 m # c 5 m # c 5 m # c 5 m # c 5 m # c 5 m
2
2
2
2
2
2
2
d) (1.5)6= (1.5) × (1.5) × (1.5) × (1.5) × (1.5)
4
e) c p m = c p m # c p m # c p m # c p m
q
q
q
q
q
V a) (i) 92
(ii) 34
b) (i) 56
(ii) 254
c) (-3)5
3+6
Exercise 1.2 : I 1) 72+5 2) -35+3 3) c 5 m 4) 103+7+5 5) a6+4+10
2
7) 2.54+8
II. 1) 73 2) 37 3) 39 3) 214 III 1) 11 2) 19 3) a 4) 1
5
Exercise 1.3 : I 1) 73 2) (-3)3 3) c 5 m 4) (8.5) 5) x8 6)1 7) 15 = 4- 5
II 1) 5
IV 1) 1
5
to
5- 4
2
1
2) 3 3) 2 III 1) 5
2) 1 7 3) 110
3
10
a
2) 3-4 3) 1
4) x -5
2- 7
2
4) 112
x
4
Exercise 1.4 : II 1) 324
2) 86 3) 1142
4) P qre) 224 5) ` 2 j
3
Exercise 1.5 : I 1) 4252
2) 86 66
3) 1111 511
II
1) (3 × 2)3
2) (4 × 5)8 3) (10 × 2)3
6
5
7
3
Exercise 1.6 : I 1) 126
2) 145
3) 87
4) x3
13
5
7
z
Exercise 1.7 : I 1) (0.7)5 2) (10)12
3) 3525
4) 212 38
II 1) 23345211 2) 34 2-2 3) 72 112 4) 3 III 1) 1
2) 32 3) 80
Exercise 1.8 : 1) Velocity of light = 3 × 108 m/s
2) Velocity of sound = 343.2 m/s 3) Distance between sun
and earth = 149600000, other astronomical distances = 1.496
×109 m/s
No
t
20
195
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Unit - 2 Ratio and Proportion
2) ` 72
3) ` 200
4) 300 km
Exercise 2.1 : I 1) ` 60
5) ` 360
6) 10 days
7) 3 days
3) 3 : 7
4) 1:5
Exercise 2.2 : I. 1) 3 : 4 2) 7 : 8
5) 10 : 9 6) 32 : 5 7) 1 : 3
8) 1 : 6
II. 1) 1 : 5 2) 1 : 2
3) 1 : 2 4) 1 : 4 5) 1 : 8 6) 1 : 5 III. 1) 8 : 5
2) 23 : 21
3) 77 : 30 4) 12 : 25 IV. a)6 : 1 b) 5 : 1 2) 5 : 9 3) 9 : 2
Exercise 2.3 : 1 Raju - ` 12,000, Dhamu - `9,000
2. Females - 2,800 Male - 3080
3. Roshan - ` 1,200 Hameed - ` 1,600
4. Copper - 150 g Zinc- 90 g 5. ` 300; ` 240 6.107 : 214 : 321
Exercise 2.4 : I 1) 12 2) 8 3) 14 4) 6 5) 10 6) 20
II 1) Yes 2) No
3) Yes
4) No
III 1) ` 204 2) 30 kg IV
2l
V. 4 l
3) 15 m
4) 25 kg
Exercise 2.5 : I 1) ` 140 2) 5 m
1
hour 2) 10 days 3) 6 days 4) 18 days
Exercise 2.6 : I 1) 7
2
Unit - 3 Percentage
No
t
to
Exercise 3.1 : I1) 1 2) 1 3) 1 4) 1 5) 3 6) 1 7) 7 8) 3
2
4
5
10
4
8
8
8
II 1) 50% 2) 25%
3) 75%
4) 12.5% 5) 40% 6) 37.5%
7) 32%
8) 35%
III 1) 60% 2) 90%
3) 42.85%
2) 5%
3) ` 34.11%
4) 25%
Exercise 3.2 : 1) 25%
2) ` 990
4) ` 420
5) ` 40,000
Exercise 3.3 : 1) ` 108
Unit - 4 Simple Linear Equation
Exercise 4.2 : I.1, b) Z, 2) a) 2x + 3 =8, 3) c) y + 3 = 7
II a) - (iii) b) - (i) c) -(ii)
III
LHS
RHS
1) 1) x - 5
8
2) 3y + 6
-9
3)14 - k
2k + 4
IV) 1) x = 13
2) x = 48 3) y = 35 4) p = 6
5) m = 5
V) x + 9 = 15
ii) 8x -2x = 18 = 3x
iii) x + 15 = 35 x=20
Exercise 4.3 : I 1) x = 7 2) x = 21 3) y = +3 4) k = -3 5)
6) p = 12 7) k = 5
8) x = 105 9) x = 310) x = -4
m = 15
4
7
11) x = 15 12) x = 2 II 1) Equation 4x + 6 = 46 or x = 10
2) Equation x + 4 = 15 or x = 77
7
196
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Unit - 5 Unit Congruence
Exercise 5.1 : I
i) a
ii) c
iii) a
II 1) sides : KL , PQ, ML , RQ, KM , PR
Unit - 7 Geometrical Constructions
2) 10 m
3) 28 m
Exercise 7.1 : I 1) 28 cm
2
2
2) 87.5 m
3) 22.75 m
II 1) 24 m
III 1) 24 cm
2) 60 m
3) 22.4 cm
2
2
2) 144 cm
3) 96.04 cm2
IV 1) 36 m
V 1) 20 cm 2) 44, ` 5280 3) Side of the Square = 30.5 cm
4) 625 cm2 5) 1600 m 6) 15000 sqm2, 3000 Rs
7) 4.20
8) a) Area will be doubled b) same as it is c) double 4 times
9) a) area is decreased by 2 times
Exercise 7.2 : I 1) 24 cm, 18 cm, 16 cm 2) 20 cm
3) a) 24 cm
b) 39 cm
c) 33 cm 4) a) 27 cm
5) 15 cm 6) 225 m and ` 2600 7) 24 cm2, 13cm2, 15cm2
8) 49 cm2 9) 198 m & ` 1990
10) Base = 24 Height = 18 m
11) a) area is doubled
b) Remains same
c) 4 times
Sl. no
Exercise 7.3 :
base height
area
1
2
3
4
8 cm
15 cm
20 cm
24 cm
6 cm
11 m
14 m
50 m
48 cm2
165 m2
280 m2
1200 cm2
No
t
2) 63 cm2 3) 315 m2 4) 9 m 5) 80 m 6) 91 cm 7) 10.8 m
8) base = 10 m, length = 20 m 9) base = 50 m, length = 20 m
Exercise 7.4 : 1)a) 44 m b) 66 cm c) 132 m 2)a) 220 cm b) 176 m
c) 154m 3)39.6 cm 4) 11.2 m 5) 400 cm, 6)` 3300, 7) 14 cm
9)a)154 cm2 b) 3146 cm2 10)a) 3850 cm2 b) 2464 m2 c) 1886 cm2
11)12474 m2 12) 264 cm1 3)0.1936 m2, 015 211 m2, 0.04149
14)77 15) 154 m2
Exercise 7.5 : 1) 300 m2 b) 411 m2 c) 246 m2 `7380 4)
b) 660 m2 c) 5197.5 m2 d) 1037.1428 cm2
a) 1584 cm2
5)2002 m2
`50050
197
7
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be T
B
re S
pu
bl
is
he
d
Unit - 8 Data Handling
II 1) 11.00 2) 4.5
III
3.99 kg
Exercise 8.1 : I 34.25
IV 1) 12.5, 2) 15.5 V bat man -1 65 run bat man -2 30 run
bat men -1 Good VI a)
29
b) 45
Exercise 8.2 : I 1) 4.5 2) 16 3) 16 4) 7 II. 27 III 15 IV
24
V Median 32, New Median 35 V 12
Exercise 8.3 : I 1) 3 2) 14 3) 66 & 69 II 163 III 14 IV 6 V27
Unit - 9 Probability
Exercise 9.1 : 1)S = {Yellow, Blue, Red And Green}
2) S = {1, 2, 3, 4, 5, 6, 7, 8} 3) S = { Head, Tail} 4) S = { }
5) S = {11, 12, 13, 14, 15}
Exercise 9.2 : I. a) {1, 3, 5, 7, 11, 13 ...} b) {2, 4, 6, 8, 10...}
II
a) 1 b) 5
c) { 1, 3, 5, 7, 9, 11...} d)6 e) 1, 2, 3, 4
7
7
c) 1 III a) 12 b) 5 c) 6
25
25
25
1
Exercise 9.3 :I. a) 2 b) 12 c) 12 d) 16 e) 2
3
1
5
1
12
5
6
II. a) 7
b) 7 c) 7 III. a) 25 b) 25
c) 25
Unit - 10 Representation of 3 diomensional objects in 2
Solid Faces Edges
a)
6
12
8
b) 6
12
8
c) 4
6
6
d) 5
8
5
e) 5
6
9
No
t
to
Exercise 10.1 : I
dimensional figure
Vertices shape of focus
rectangle
square
triangle
square
triangle and rectangle
II Name of the solid
Curved Faces
Curved Edge Vertices
1)
1
1
1
2)
1
2
3)
1
III 1 → d
,
2 → a,
3→b
,
198
4→C