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iii
Government of Karnataka
MATHEMATICS English Medium
7SEVENTH STANDARD
Part - II
KARNATAKA TEXT BOOK SOCIETY (R)100 Feet Ring Road, Banashankari 3rd stage,
Bengaluru - 85
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CONTENTSPart - II
Sl. No Chapters Pages
1 Indices 1 - 18
2 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 Data Handling 138 - 165
9 Probability 166 - 176
10Representation of 3 dimensional Objects in 2 dimensional Figures
177 - 194
Answers 195 - 198
ii
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CHAPTER - 1
INDICESAfter studying this chapter you : write repeated multiplication in exponential form 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 prime factors and express it in exponential form,
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 scientific notation and write some numbers in that form.
Introduction of numbers in Exponential form To write numbers in exponential form You know the place value of numbers (numerals) in decimal system as well as meaning of square and cube.
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
10 c
m
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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 the square and side of the cube is 10cm.
We write, 10 square as 10×10 = 102
cube, as 10×10×10=103
Then the length of line segment = 10cm.
Area of square 10×10 = 102 = (10cm)2 = 100 Square cm
In the above example the number written on the right top of '10' indicates how many times 10 is multiplied by 10.In the same way : 5 × 5 × 5 = 53
2 × 2 × 2 × 2 = 24
7 × 7 × 7 × 7 × 7 = 75
( 23 )×( 2
3 )×( 23 )×( 2
3 )×( 23 )×( 2
3 )=( 23 )6
x × x × x × x × x × x × x = x 7
In the above examples repeated multiplication is written in exponential form.
Exponential form as repeated multiplicationWe 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
( 34 )5
= ( 34 )×( 3
4 )×( 34 )×( 3
4 )×( 34 )
(-5)4 = (-5) × (-5) × (-5) × (-5)
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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 10.106 = 10 ×10 ×10 ×10× 10 × 10 means 10 is multiplied six times.
106 Base number
Index/Power/Exponent
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. 3) In 28, 2 is base and 8 is index.
4) In ( 32 )5
, 32 is base and 5 is index.
Observe this Table
Exponential number Base Index Read as
64 6 4 6 to the power of 447 4 7 4 to the power of 7125 12 5 12 to the power of 5
( )33
4 43- 3 ( )34 to the power of 3
x8 x 8 x to the power of 8
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Activity 1 : Take a white sheet. Fold and crease it exactly at the middle so that it is divided into two equal parts. Again fold it into two parts as shown in the figure. Tabulate the rectangular parts obtained at each time as shown below.
First fold Second fold
Number of folds 1 2 3 4 5
Number of rectangles 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 rectangles formed in exponential form.
Method of writing the numbers in Exponential form.
Take any number and write it as product of repeated factors, then write each factor as base and mark index to each of them.
Example 1:
1) Write 125 in exponential form to base 5.
Solution :
1
125 555
255
125 = 5 × 5 × 5
= 53
∴125 = 53
53 is the exponential form of 125.
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Example 2 :
Write 256 in exponential form to base 2, 4, 16, 256.
Solution :
1
25621282642322162824222
256= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ∴256 = 28
2564644164441
256=4 × 4× 4 × 4
∴256 = 44
2561616 16
1
∴256 = 161 × 16
∴256 = 162
2562561
∴256= 2561
Example 3 : Write exponential form of 1331 to base 11.Solution : 133111
1111
121111
1331 = 11 × 11 × 11
∴1331 = 113
Example 4 : Write 1125 in exponential form.Solution : 1125 is completely divisible by 5 and 3.
9
112555533
22545
31
1125 = 5 × 5 × 5 × 3 × 3
1125 = 53 × 32
Example 5: Write 324 in exponential form.Solution : 3242
162281327393331
324 = 2 × 2 × 3 × 3 × 3 × 3 ∴ 324 = 22 × 34
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Exercise : 1.1
I. Read these exponential numbers.
1) 83 2) 136 3)( 47 )10
4) 104 5) (-6)5
II. Write the base and index of these exponential numbers.
1) 35 2) 108 3) (- 23 )6
4) x20
III. Write appropriate answer in the space provided.
1) 3 × 3 × 3 × 3 × 3 = 5
2) 58 × 5
8 × 58 × 5
8 = ( 58 )
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.
1) 38 2) 113 3) ( 52 )6
4) (1.5)6 5) ( pq )4
V. Express the following number as directed.
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).
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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.
21 22 23 24
25 26 27
28 29 210
31 32 33
34 35 36
37 38 39
310
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
By extending this activity to other examples, we notice 104 ×106 = 10 4 + 6 = 1010
x10 × x25 = x 10 + 25 = x35
( 23 )4
×( 23 )12
= ( 23 )4+12
= ( 23 )16
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
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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.
Example :
a) 25 × 24 = 25+4 = 29
b) 106 ×102 = 106+2 = 108
c) ( 23 )5
×( 23 )10
= ( 23 )5+10
= ( 23 )15
d) x15 × x20 = x15+20 = x35
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
b) m3 × m4 × m5 = m × m × m × m × m × m × m × m × m × m × m × m
= m12 or
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
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Exercise : 1.2
I. Simplify using am × an = am+n.
1) 72 × 75 2) (-3)5 × (-3)3 3) ( 52 )3
×( 23 )6
4) 103 × 107 × 105 5) a6 × a4 × a10 6) (2.5)4 × (2.5)8
II. Convert the following to exponential form and apply first law of exponents.
1) 49 × 7 2) 27 × 81 3) 243 × 81 4) 1024 × 16
III. Fill up the space provided with suitable answer.
1) 10 10 108 3# = 2) ( 23 )15
×( 23 )6
= ( 23 )□
3) a a13 3 10#= 4) (25)6 = (25)□ × (25)5
Division of numbers with exponents :
Activity : 3Take any two cards from the box of cards used in the second activity. Divide the numbers with higher exponent by the other number in the cards.
Example :
a) 2 2 2
2 2 22 222 2 24 33
4
'# #
# # #= = = or
22 2 2 23
44 3 1= = =-
b) 3 33
3 3 3 3 3 3 933
3 3 33 3 36 4
4
62'
# # #
# # # # ##= = = = =
33 3 3 94
66 4 2= = =-
c) 38 ÷ 34 = 38-4 = 34
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d) 29 ÷ 25 = 29-5 = 24 = 16 similarly
aa a a
aa a if m n
5 555 5 5 25
10 101010 10 10 10000
>n
mm n
6 44
66 4 2
8 44
88 4 4
4
1515 4 11
'
'
`
= = = =
= = = =
= =
=
-
-
-
-
Observe the following examples a)
22
2 2 22 2 2 13
3
# #
# #= =
As per 2nd law, 22 2 2
2 13
33 3 0
0`
= =
=
-
b) 1xx
x x x x xx x x x x
5
5
# # # #
# # # #= =
As per 2nd law, 1
xx x x5
55 5 0= = =-
By above examples we can generalise that
aan
m
=1 when m=n and a 0]
100 = 1, 1000=1, 50 = 1, ( )zxy 10 =
Note: For any non-zero base, if the power is zero then its value is 1
Let us examine what happens if m<n in aanm
, a 0]
Examples : a) 22
2 2 2 2 22 2 2
21
215 325
3# # # #
# #= = = -
b) 3 3 3
3
33
3 3 3 33 3 3
31
74
3# # # # # #
# # #= =
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From above examples, we can generalise that
,aa
aa m nand1 0 <n
mn m != -
Then, 22 2 2
21
53 3 5 2
2= = =- -
So, , ,m
m41 4 8
81 1
77 2
2 44= = =- -
-
Also 33 3 3
31
74 4 7 3
3= = =- -
,aa
a1 0mm !=-
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
an = am-n , a ≠ 0. This is 2nd law of exponents.
Exercise: 1.3
I. Simplify, using 2nd law, am ÷ an = am-n .
1) 75 ÷ 72 2) (-3)5 ÷ (-3)2 3)( 52 )8
÷( 52 )3
4) (8.5)6 ÷ (8.5)5 5) x11 ÷ x3 6)
xx10
10
7) 4410
5
II. Convert the following numbers to exponential form using am ÷ an = am-n and simplify.
1) 125 ÷ 25 2) 81 ÷ 9 3) 256 ÷ 8 4)27 ÷ 243
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. 1) 54 2)
314 3) 27 4)
x15
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Exponent of numbers with exponents
Look at these examples.
1) (32)4 means 32 is multiplied 4 times.
(32)4 = 32 × 32 × 32 × 32 = 32+2+2+2 = 38 (Ist law of indices)
or (32)4 = 32×4 = 38
Similarly, (53)4 = 53×4 = 512
(x8)5 = x8×5 = x40
(a4)6 = a4×6 = a24
2) (73)2 means 73 is multiplied twice.
(73)2 = 73 × 73 = 73+3=76 (using 1st law of indices)
or (73)2 = 73×2 = 76
We can generalize that
∴ (am)n = am x n
(am)n = amn , a ≠ o.
The exponent of a number in exponential form is equal to the product of exponents.
(am)n = am×n, a ≠ 0.
This is called the 3rd law of exponents.
Example : a) (35)
2 = 35×2 = 310
b) (4x)6 = 4x.6 = 46x
c) (2x)y = 2x.y
Know this: We can extend this law to
(((am)n)p)r
= am×n×p×r = amnpr
Example :((23)2)4
= 23×2×4 = 224
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Exercise : 1. 4
I. Simplify the following, applying (am)n = am×n. 1) (24)
6 2) (83)
2 3) (116)
7 4) (pq)
r 5) ((23)2)4 6) [( 3
2 )4]
5
Think : Which is greater? How?
i) 232, (23)2 ii) 1023, (102)
3 iii) 424, (42)
4
Multiplication of exponents with different bases. Observe this number pattern. 1) (5 × 7)4 = (5 × 7) × (5 × 7) × (5 × 7) × (5 × 7)
= (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
For all non - zero integers, a and b, (a × b)m = am × bm.
This is called the 4th law of exponents.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
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Exercise : 1.5
I. Express the following in (a × b)m = am × bm.
1) (4 × 5)2 2) (8 × 6)6 3) (11 × 5)7
II. Write/express the following in (a × b)m form.
1) 33 × 23 2) 48 × 58 3) 103 × 23
Division of exponents with different bases. Observe the following exponents.
( 43 )5
= 43 × 4
3 × 43 × 4
3 × 43 = 4
5
35
( 27 )6
= 27 × 2
7 × 27 × 2
7 × 27 × 2
7 = 26
76
( 310)
4= 3
10× 310× 3
10× 310 = 34
104
Similarly ( 57 )6
= 56
76
( xy )10
= x10
y10
∴ ( ab )m
= am
bm
( ab )m
= am
bm , a, b ≠ 0.
This is called the 5th law of exponents.
Example : a) (107 )6
= 106
76 b) ( m5 )8
= m8
58 c) (107 )6
= 106
76
Exercise : 1.6
I. Express using ( ab )m
= am
bm .
1) (1213)
6 2) (14
5 )3 3) ( 8
7 )7 4) ( x
z )3
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Application of laws of exponents.
Suhasini and Mary were playing with numbers. Suhasini
wrote 332 and 552 and Mary wrote (33)2 and (55)2. Both of them
were arguing that the value of each number written by each
is greater than the other. To clarify they met Suhasini's uncle,
who is a mathematics teacher. He clarified like this.
Suhasini Mary Conclusion
332 = 33 × 3 = 39
552 = 55 × 5 = 525
(33)2 = 33 × 2 = 36
(55)2 = 55 × 2 = 510
39 > 36 ∴ 332> (33)
2
525 > 510 ∴ 552> (55)
2
Both Suhasini and Mary were satisfied by the uncle's
clarification
Uncle gave a few numbers and helped them to solve.
a) (22)(33) = (22)27 = 22×27 = 254
b) (23)(23) = (23)
8 = 28×3 = 224
c) (33)(22) = (33)
4 = 33×4 = 312
d) (32)23 = (32)
8 = 316
a) Express 27 × 27 × 27 in exponential form of base 3.
Solution : 27 × 27 × 27 = 33 × 33 × 33 = (33)3 = 33 × 3 = 39
b) Simplify ( 26
22 )× 24
Solution : (26-2) × 24 = 24 × 24+4 = 28
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Exercise: 1.7
I. Apply laws of exponents and simplify. 1) (0.7)2 × (0.7)3 2) (103)
2 × (102)
3
3) 3 ×
3 ×3 ×62
2
3 5 3
4) (26 × 34
63 )2 × ( 33 × 25
3 × 8 )3
5) (30 × 25) + 50 6) 2 5 33000 82 2# #
#
II. Simplify by converting into exponential form.
1) 72 × 55 × 45
Clue: 72 = 23 × 32
2) 12×436×9 3) 49 × 121 4)
( )( )
1881
11
32
3
5
# #--
III. Examine the following and state whether it is right or wrong.
1) 100 × 1010 = 105 × 106 2) 43 × 52 = 205
3) (-5)0 = 50 4) 23 = 6
5) (-1)5 = (-1)3 (1)2 6) (-1) × (-1) ...... 13 times = -1
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 the tiny (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.
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Example 1 : Big (Macro numbers)
a) 112 = 1.12 × 102 [ Note: 1 < 1.12 < 10]
b) 236000 = 236 × 103
= 23.6 × 104
= 2.36 × 105 [ Note: 1 < 2.36 < 10]
c) 14567800000 = 145678 × 105
= 14567.8 × 106
= 1456.78 × 107
= 145.678 × 108
= 14.5678 × 109
= 1.45678 × 1010 [ Note: 1 < 1.45678 < 10]
Example 2 : Small (Micro numbers)
a) 0.000342 = 1000000342
10342
6=
= 342 × 10-6
= 34.2 × 10-5
= 3.42 × 10-4 [Note : 1<3.42<10]
b) 0.00045213 = 45213 × 10-8
= 4521.3 × 10-7
= 452.13 × 10-6
= 45.213 × 10-5
= 4.5213 × 10-4 [Note : 1<4.5213<10]
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Example 3:1) The temperature at the interior of the Sun is
approximately 2,00, 00, 000 0K. Scientific notation of this is 2,00,00,0000 K=2 × 10000000= (2 × 107)0K
[Note : 2 > 1 and 2 < 10]2) Weight of a small grain is 0.0005 gram. The scientific
notation of the above is. 0.0005 g = 5/ 10000 = 5 × 10-4 g [Note : 5>1 and 5<10]
Exercise - 1.8
1) Write the scientific notation of velocity of light, velocity of sound, distance between sun and earth, and other astronomical distances which you come across in your science text book.
2) Collect numerical information of population, budget from newspapers and write these numbers in standard scientific notation.
Points to remember :1) a a am n m n# = + (I Law)
2) i),
aa a m n>nm m n= - (II Law)
ii) aaa a a 10m n n nnm= = = =- - ,
m=n
iii)
aa
a1 ,m n<n
mn m= -
3) ( )a am n mn= (III Law)
4) ( )a b a bm m m# #= (IV Law)
5) ( ab )m
= am
bm(V Law)
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CHAPTER - 2
RATIO AND PROPORTION
After studying this chapter 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,
solve the problems based on proportion,
solve the problems on direct proportion,
solve the problems based on inverse proportion.
Unitary method :In our daily life we observe things such as pens,notebook,
etc., that are sold in bundles or packets and 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 to the shopkeeper?
Example 1 : The cost of one dozen bananas is ̀ 36. Somanna buys 20 bananas. Find the amount Somanna has to pay.
Cost of 1 dozen (12) bananas = ` 36
If the cost of 1 dozen (12) bananas is known then, how to calculate the cost of 20 bananas?
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cost of one banana = ( 112)
th the cost of 1 dozen bananas
= ` 36 121#
= ` 3∴cost of 20 bananas = ̀ 3 × 20
= ̀ 60
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 is the same food enough for 40 scouts ?
30 scouts have enough food for 4 days. ∴Thefoodsufficientforonescoutis30×4 = 120 days. ∴Thefoodsufficientfor40 scouts = 120
40 days.
= 3 days.
Exercise 2.1
I. Solve the following problems by unitary method.1) The cost of 3 balls is ̀ 36. Find the cost of 5 such balls.2) The cost of 5 pens is ` 30. Find the cost of 12 pens.3) The cost of 15 oranges is ` 60. Find the cost of
50 such 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 liters of petrol.
5) The cost of 25 m cloth is ` 750. Find the cost of 12 m cloth of same type.
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6) In a school there is enough food for 100 students for 4 days. How long does that food last for 40 students ?
7) 12 persons can reap the crop from a field in 5 days. Calculate the number of days required for 20 persons to do the same job ?
8) 24 workers can build a wall in 15 days. How many days will 9 workers take to build a similar wall?
RatioWe incorporate many concepts of mathematics in our day
to day transactions. We always prefer to buy more things for less amount. So we compare the cost of thing with its quantity.
` 45 500g
` 25250g
You must have seen such tea packets in the shop. Which one of the two packets is more beneficial for the buyer ?
Let us compare the weight and the cost of these two packets.
Let us compare the weight of the packets firstThe 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.
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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
The cost of bigger packet is 1.8 times the cost of smaller packet.
The weight of the bigger packet is 2 times the smaller packet and 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
The quotient of weight of tea powder in two packet. 500250
This comparision by division can be written in the form 500 : 250. This definite relationship is called ratio.The quotient of cost of tea powder in two packets = 25
45 . This comparision by division can be written in the form 45 : 25. This definite relationship is called ratio.
The comparison of two quantities of the same kind by division 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 b
a .
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Representing the ratio in its simplest form.
Ratio is a fraction. The simplification of fraction canbe done by dividing or multiplying the numerator and the denominator by the same number. In the same way the ratio canbesimpifiedbydividingormultiplyingthetwotermsbythe same number. By doing this there will be no change in the value of the ratio.
Example 1 : Write 10:15 in its simplest form.
Which number divides 10 and 15 completely?
Yes, it is divisible by 5.
Divide both 10 and 15 by 5
510 : 5
15 = 12 :13 or 2:3
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 21 : 31 in its simplest form.
The LCM of the denominator ( 2 and 3 ) is 6.
Multiply both 21 and 3
1 by 6
21 × 6 : 3
1 ×6 = 3:2
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.
The weight of one packet = 2kg = 2000g The weight of the other packet = 500g
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Now both weights are in the same unit.
The ratio of these weights = 2000 : 500
= 5002000 : 500
500
= 4:1
Note :• Ratio is a comparison of two similar quantities. Hence,
they are in number form. Therefore, unit should not be mentioned . .
• In the ratio, the units of similar quantities to be compared must be the same.
• If the units are different then convert into same units.
Remember :
The length and breadth of our Make a list of sit-uations where the concept of ratio is used.
nationalflagshouldbeintheratio3:2
The mixture of cement and sand used for constructing building is usually in the ratio of 1:6
To prepare tasty Idlies the ratio of black gram dal to rice is = 1:2
Reciprocal of ratio
If the antecedent and the consequent of ratio are interchanged, we get the reciprocal ratio of the original.
Example : The reciprocal ratio of 2:3 is 3:2
The reciprocal ratio of a:b is b:a Note : ba
ab 1# =
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Examples:
1) Write 25 and 35 in the form of a ratio
The ratio of 25 to 35 = 25:35
The simplest form = 525 : 5
35 ( divide by 5 )
= 5:7
2) There are 25 boys and 20 girls in a class. Find
i) The ratio of the number of boys to the number of girls.
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 { = 25 : 20
Divide both the terms by 5 = 5 : 4
The ratio of the number of boys to the total number of students { = 25 : 45
Divide both the terms by 5 = 5 : 9
3) Write the reciprocal ratio of 4:9.
The reciprocal ratio of 4:9 is 9:4.
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) 32 :53 6) 5
4 :83 7) 1 2
1 :4 21 8) 3
1 : 2
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II. Express the following in simplest ratio form.
1) 100g and 500g 2) 3 hours and 6 hours
3) 500g and 1kg 4) 30 minutes and 2 hours 5) 25 cm and2 m 6) 200ml and 1l
III. Express the following ratios into their reciprocal form.
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 900areilliterate.Thenfindout 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 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.
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∴ ` 750 is to be distributed between Vinay and Victor in 3 : 2
Vinay gets 53 of the amount, Victor gets 5
2 of the amount
Vinay's share of earning = of53 750 = 5
3 750#
∴ Vinay's share = ` 450
Victor's share of earning = of52 750 = 5
2 750#
∴ Victor's share = ` 300
Theprofitearnedbymorethantwopersonsinbusinesscan also be calculated applying the concept of ratio.
Exercise 2.3
I. Solve the following problems.
1) Damuworks in thefieldofRaju.Theyhaveagreedto share the income obtained in the ratio of 4 : 3. Find the share of income received by each of them, if the income for one year is ` 21,000
2) 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.
3) Roshan and Hameed invested ` 30,000 and ` 40,000 respectively in a business. After one month they earnedaprofitof̀ 2800. According to the ratio of their investment,findtheshareoftheprofitofeach.
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4) 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
Proportion
Example 1: If the cost of one pen is ` 5 find the cost of two pens.
As the number of pens increases, their total cost will also increase. Is it not? 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.
The ratio of number of pens = 1:2The 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
In this way if we write two ratios as equal then that statement is in proportion.
∴ Proportion is an equation in which two ratios are equal to each other.
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a:b=c:d is a proportion. In this proportion there are four terms.
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 ba = d
c (fraction form) We read a:b::c:d as a is to b is as c is to d.
Activity : There are a few cards on which ratios are written. Select the pair of cards having equal ratios.
2 : 3 2 : 8 5 : 3 6 : 3 12 : 6 4 : 6 4 : 16 10 : 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).
Do you find any relationship between the product ofmeans and product of extremes? Observe. By completing the table, draw the conclusion.
Ex proportiona:b = c:d
product of extremes a×d
product of means b×c
1 2:3 = 4:6 2×6 = 12 3×4 = 12
2 6:3=10:6
3
4
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Observation : In the above table, the product of extremes is equal to the product of means.
Rule of Proportion,In any proportion, the product of extremes is equal
to the product of means.
product of means b × c
product of extremes a × d
a : b = c : d
∴ If a:b = c:d then a×d = b×cIf a:b = c:d or b
a = dc
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. The product of extremes = 3 × 8 = 24
The product of means = 4 × 6 = 24
∴ The product of extremes = The product of means ∴ The given numbers are in proportion.
2) If 5 : 2 = 10 : x. Then find the value x 5 × x = 2 × 10
x = 52 10# = 4
The value of x is 4
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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 = ` 470
Let the cost of 8 kg of rice = ` x The ratio of weight of rice = 10:8 The ratio of their cost = 470:x These two ratios are equal. 10 : 8 = 470 : x 10 × x = 8 × 470
x = 108 470# = 376
∴The cost of 8 kg of rice ` (x) = ` 376
Exercise 2.4 I. Fill up the blanks with appropriate numbers.
1) 6 : 8 = : 16 2) 21 : 24 = 7 : 3) : 7 = 4 : 2
4) :32
9 5) :5
4 8 6) 25 : = 5 : 4
II. Verify whether the following numbers are in proportion.1) 5, 6, 10, 12 2) 8, 15, 3, 6
3) 7, 42, 13, 78 4) 1.5, 4.5, 2, 6
III. The cost of 5 kg wheat is ` 127.50
1) Find the cost of 8 kg wheat 2) How much quantity of wheat can be bought for ` 765.
IV. A motor bike can travel 325 km using 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.
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Types of proportion 1) As the number of people in a house increases; the
expenditure for the house also increases. Among the two interdependent quantities, the increase in one has increased the othe. 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 )
2) Here among the two interdependent quantities the increase in one has reduced the other.
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 ` 2Then 6 chocolates for ` 38 chocolates for ` 4Observe the total number of chocolates got and money paid each cases. What is your conclusion?
More Money more chocolatesless money less chocolates
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.
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4 chocolates for ` 2 8 chocolates for ` 4The ratio of the money is 2 : 4, The ratio of number of chocolates is 4 : 8 ∴ 2 : 4 = 4 : 8
Example 1: The cost of 2 m cloth is ` 80. Find the cost of 5m cloth of the same type.
The cost of 2m cloth = ` 80
The cost of 5m cloth = ` xThe ratio of the quantity of the cloth is 2:5The ratio of the cost of the cloth is 80:xAs the quantity of the cloth increases the cost also
increases.It is an example of direct proportion.Hence 2:5 = 80:x
2×x = 5×80 [ The product extreme must be equal to x =
25 80 200# = product of means]
The cost of 5m cloth = ` 200
Exercise 2.5
I. Solve the following problems on direct proportion :1) If the cost of 3 kg of sugar is ` 84,thenfindthecostof
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,thenfindthelengthof
cloth for ` 450.4) One kg of rice is enough for 8 people. Find the quantity
of rice required for 200 people.
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Inverse proportionAs the speed of the bus increases thetimetakentotraveladefinitedistance decreases.
Speed and the distance travelled are related factors.
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 intheformofdirectproportion,eitherthefirstratioorthe 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]solving problems involving inverse proportionA 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.
The time taken by the bus if the speed is 45 km/hr {= 8 hour
Let the time taken by the bus if the speed is 60 km/hr
{= x hour
The ratio of their speed = 45 : 60The ratio of their time taken = 8 : xAs the speed increases the time taken decreases, this is
an example for inverse proportion.
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45:60 = x:8 ( inverse of 8:x )
Hence 45:60 = x:8 ( solving by direct proportion method )
60×x = 45×8
∴ x 6045 8 6#= =
If the speed of the bus is 60 km/hr then the time taken to travel from Mangalore to Bangalore (x) {= 6 hours
Exercise 2.6
I. Solve the following problems.
1) A car travelling with a speed of 50 km/hr reaches
Bengaluru from Hubli in 9 hours. Calculate the time
taken, if the car travels with a speed of 60 km/hr to
reach Bengaluru from Hubli.
2) There is enough food for 15 days for 20 persons in a
residential school. How long does it last for 30 persons?
3) 12cowscangrazeafieldfor10 days. In how many days
will 20cowsgrazethesamefield?
4) 30 people can complete a work in 12 days. How many
days will 20 people take for completing the same work?
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CHAPTER- 3
PERCENTAGE
After studying this chapter you : learn the meaning of percentage, convert the fraction into percentage, convert the percentage into fraction, convert the decimal number into percentage, 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 on account of Deepavali
festival.
85% pass in SSLC examination.
75% voting in Uttar Pradesh Grameena Rural Bank offers 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.
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Who is getting the pen ?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.
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.
Divya's score19 out of 2538 out of 5076 out of 100
Kavya's score16 out of 2032 out of 4048 out of 6064 out of 8080 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.
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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.Percentage is a convenient way of comparing quantities.
Conversion of fraction into percentage.
Example 1 : Convert 5016 into percentage.
5016 means 16 out of 50. Hence, how much is out of 100 ?
5016 × 100 = 32
= 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.
30% means, 30 out of 100.
Hence, we write 30% as 10030
When 10030 is simplified, it becomes 10
3
∴ 30% = 10030
103=
Example 2 : Convert 62.5% into a fraction.
. % . .62 5 10062 5
100 1062 5 10
1000625
4025
85
##= = = = =
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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%
Hence, the number of illiterate in that village 7500 10%
7500 00110
750
#
#
=
=
=Exercise 3.1
I. 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) 21 2) 4
1 3) 43 4) 8
1
4) 52 5) 8
3 6) 258 7) 20
7
III. Answer the following:
1) Kavitha scored 15 out of 25 in a test. Express the marks scored by her in percentage.
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.
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Percentage of profit and percentage of loss in transaction
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. Calculation of loss or profit is done for a standard amount of ̀ 100. This is necessary for the business men.
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 cost price is ` 100 and profit or loss is calucated for the cost price then it is called percentage of profit or loss.
Example 1 : A shopkeeper buys a coconut for ` 10 and sells it for `15. Find the percentage of profit.
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 105 100#=
The percentage of profit can be written in the form of formula.
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Profitcos
Percentage of profitt price
100#= ( )Cost price
Selling price Cost price 100#=-
Like wise,
100Percentage of Loss Cost priceLoss #=
C(C p S )
osos
t pricet rice elling price 100#=
-
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 in biscuit transaction =15 - 10 = ` 5
Business : 2 Profit earned by the shopkeeper in sweet box transaction =60 - 50= ̀ 10
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.In Business : 1, The shopkeeper earned ` 5 as profit by investing ` 10In Business : 2, The shopkeeper earned ` 10 as profit by investing ` 50
∴ Percentage of profit in business 1 = %105 100 50# =
∴ Percentage of profit in business 2 = %5010 100 20# =
It means, if ` 100 is the investment in both the business,` 50 is the profit in business 1` 20 is the profit in business 2Calculation of profit in terms of percentage is very much
essential for the shopkeepers.
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Example 1 : A shopkeeper purchased an old bike for ` 20,000 and then sold it for ̀ 22000. Find the profit percentage earned by the shopkeeper.
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
= ` 2000. 100cosPercentage of profit t price
profit` #=
%200002000 100
1010
1#=
=
Example 2 : 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 by him.
The cost price of 20 kg tomatoes = ` 200
The quantity of rotten tomatoes = 5 kgRemaining tomatoes = 20 - 5 = 15 kgThe amount obtained by selling 15 kg at the rate of ` 12
per kilogram = 15 ×12 = ` 180
If the cost price is more than the selling price, the vegetable vendor incurs loss.
Cost price > Selling price = 200 > 180 ∴ Loss
The loss = cost price- selling price = 200- 180
= ` 20
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%
cosPercentage of loss tpriceloss 100
20020 100
10
` #
#
=
=
=
Exercise 3.2
I. Answer the following:1) 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 ` 18,000. Find out the loss percentage incurred 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.
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 InterestSomanna 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?
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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.
know this: Approximately about 5000 years ago people started collective agricultural activities. In those days they used barter system of trading. They borrowed seeds. They used to get more grains from sowing one grain. While returning the loan they paid more than what they received as loan.
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 institutions 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?
Suppose they keep the money in a bank?
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
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loans to the needy people. In turn we are also involved in the activity of nation building.
• 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 = Principle + Interest. [A = P + I]• 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 their account 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.
Activity : Collect more information about banking from elder people (knowledgeable) or by visiting a bank which is nearby.
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.
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Calculation of simple interestRahim 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).
Rate of interest 12% means R = 12%For every `100 loan, the interest for 1 year = ` 12There are 20 hundreds in ` 2000∴ Interest for 20 hundreds for one year = ` 100
2000 12 1# #
∴ Interest for 20 hundreds for two years = ` 121002000 2# #
= ` 480
Simple interest= x Rate of interest x Time (in Year)Principle
100
( )I P R T years100
# #=
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 yearsThe simple interest got by Sujeeth
5000 8 2
I P R T100
100
# #
# #
=
=
= ` 800
The amount received by Sujeeth after 2 years = Principal + Interest
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= ` 5000 + 800 = ` 5800Using I PTR
100= we can calculate any one of I,P,R,T if the other three are known.
Exercise 3.3
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.
Sl. No
Principle Rate of Interest
Time Simple Interest
Amount
1 ` 2500 12% 2 ` `
2 ` 8450 10% 3 ` `
3 ` 7500 15% 4 ` `
4 ` 12,500 8% 2 ` `
5 ` 2400 9% 3 ` `
4) Rekha obtained a loan of ` 3000 from a cooperative bank
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.
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CHAPTER - 4
SIMPLE LINEAR EQUATIONS
After studying this chapter you : understand the meaning of equality of two mathematical
statements, understand the meaning of equation, inequality and
simple linear equation, convert a verbal statement to equation, solve simple linear equations.
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.
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 to play it?
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.
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Let us find out the number chosen
5x +10 65
by Meera :Let the number chosen by Meera be xOn multiplication of x by 5, we get 5x10 is added to 5x it gives 5x +10
As a result we get 5x + 10 = 65
Here 5x +10 is an algebraic expression and is equal to 65
So, we get the equation 5x + 10 = 65.
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 ?
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.
3x+5 10Both sides of this
mathematical equation are equal
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Observe the above mathematical statements,
'Robert Recorde' used the symbol '=' in his algebra book
during 1557.
i) 5x + 10 = 65
ii) x + 6 = 20
iii) 2y - 10 = 15
In an equation there will be two algebraic expressions or numbers on both sides with '=' sign between.
3x+5 = 5x-1
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.
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 inequalities.
Thus, 4x + 5 > 65 is not an equation, it is an inequality. 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 inequality
3) 7x-8 < 12 is an inequality
4) -5x+2 ≠ 15 is an inequality
Note : An equation remains the same when the expressions on left side and right side are inter-changed.
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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
Thus, x + 10 = 25 is an equationExample 2 : Two times of a number is 40
Solution : Let the number be 'y' Two times y means '2y' 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. Z - 5 is equal to 30 Thus, the equation is z - 5 = 30
Exercise 4.1
I. Convert the following verbal statements into equations.
Statement Equation1) 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
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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.
Degree : The highest power or exponent of the variables in an equation is called its degree.Example :
Equation 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
If the variables of terms of an equation is in first degree, then such equations are called linear equations.Example : 5x - 4 = 0, 3y + 6 = 12, m + 2 = 0, x + y = 15, 3m + 5n = 50
An equation containing only one variable with highest power one is called a simple linear equation.
In the above examples x + y = 15, 3m + 5n = 50 are not simple linear equations.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 its 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.
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Example 1 : Solve the equation x + 3 = 6 by Inspection method.
The given equation is x + 3 = 6. Let us give different values for x. Find the LHS and RHS of the equation. For one particular value of 'x' the RHS is equal to LHS. That value is the required solution.
Value of x LHS (x+3) RHS (6) Is LHS = RHS?
0 3 6 No1 4 6 No2 5 6 No3 6 6 Yes
4 7 6 No
Observe that for x = 3, LHS = RHS in the equationTherefore, 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.
Value of y LHS(2y-3) RHS(5) Is LHS = RHS?
1 -1 5 No2 1 5 No3 3 5 No4 5 5 Yes5 8 5 No
LHS = RHS only when y = 4
Therefore, the solution of the equation 2y - 3 = 5 is y = 4
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Example 3: Solve the equation x3 8 11+ = by trial and error
method using the table.
Value of x LHS RHS Is LHS = RHS?
12345678910
31 8+ 11 No
Exercise 4.2
I. Choose the correct answer
1) The variable in the equation 2z + 6 = 18 is
a) 2 b) z c) 6 d) 18
2) The equation among the following is
a) 2x + 3 = 8 b) 2x + 3 < 8
c) 2x + 3 > 8 d) 2x + 3 ≠ 8
3) The equation with y=4 as solution is
a) 2y + 3 = 0 b) y - 7 = 2
c) y + 3 = 7 d) y + 4 = 0
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II. Match the Following Equation Solutiona) 2t = 16 i) t = 2b) 4 = t + 2 ii) t = 10c) t - 5 = 5 iii) t = 8d) 2-t = 3 iv) t = 0
v) t = -1
III Write the L.H.S. and R.H.S. of the following equations. 1) x - 5 = 8 2) 3y + 6 = -9 3) 14-k = 2k + 4
IV. Solve the following equations by the trial and error method1) x-4=9 2) x8 6= 3) 5y-3=12
4) +6=9 5) 6m-1=29 V. Express the following statements in the form of
equations and solve1) The sum of x and 9 is equal to 15 2) Twice a number decreased by 8 times the same number
is equal to 18 3) When 15 is added to Megha's age the sum is equal to 35
Elimination method of solving the equations 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 get 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.
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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 15 5+10 15
Rules for elimination Rule - 1 : If equals are added to equals, the sums are equal.
x 5
x = 5
x+3 5+3
x + 3 = ?
Example : Solve x - 5 = 15
Solution: x -5 = 15
Adding 5 on both sides
x - 5 + 5 = 15 + 5
x + 0 = 20
Therefore, x = 20
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Rule - 2 : If equals are subtracted from equals, the result is also equal.
x 8
x = 8
x-3 8-3
x - 3 = 8 - 3
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.
Rule - 3 : If equals are multiplied by equals, the products are equal.
x 4
x = 4
x×3 4×3
x ×3 = 4 ×3
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Example : Solve x4 12=
Solution: x4 12= Multiplying both sides by 4,
x
4 × 4 = 12 × 4
∴ x = 48Observe the balance and frame a rule.
x 7 x/3 7/3
Example : Solve 7x = 35Solution: 7x = 35
7x7 7
35= (Dividing both sides by 7 we get)
∴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
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Exercise 4.3
I. Solve the following equations by elimination method:
1) x+8=15 2) x-12=9 3) 15+y=18
4) 2k+6=0 5) m5 =3 6) 2p=p+12
7) k- 21 = 4
3 8) 27 x= 2
105 9) 6x-3=15
10) 10-4x=26 11) 3(x-7)=24 12) 16-5x=6
II. Solve the following mathematical Puzzles
1) I am an integer. If you add 4 times the number of intersecting points of this figure to me you get 46, what is 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 ?
3) I am a number. If you double me only one decade is enough to reach century! If you divide me into 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 the number 8 form me. Divide further by a dozen to come up with a full team for
a game of cricket.
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CHAPTER- 5
CONGRUENCE
After studying this chapter you : identify the congruency of a figure through
superimposition identifythecongruentfigure identifycorrespondingpartsofcongruentfigures definecongruency
Let us take two identical ten rupee coins. Place one on the
other. Doesoneexactlyfitsontheother?
Similarly place a one rupee coin on a ten rupee coin. Observe
whetheroneexactlyfitstheother.Youseethat,itisimpossible
to superimpose one on the other.
Congruency is one of the fundamental concepts in geometry
whichisusedtoclassifythegeometricalfiguresonthebasis
of their shapes.Observe the following patterns
Copythefigureonaplainsheetofpaper.Cuttheshadedpartandun-shadedpartofeachfigure.Superimposethemtoverifywhetheroneexactlyfitsontheother.
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Twogeometricalfiguresaresaidtobecongruent,iftheyhave same shape and size.
Remember: The symbol for congruence is ≅
The symbol for non-congruence is
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.
Fromtheaboveactivity,weobservethatthecurrenciesare of the same shape and the same size.
Check whether the following objects are congruent or not
a)Postalstampsofthesamedenominationandsamefigure
b) Biscuits in the same pack
c) Photos of same measurement
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 circlesarecongruent?
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Activity 3: Identify the angle which is congrment to ∠ABC in 3XYZ.
A
600
600 600B C
X
600
600
600
Y
Z
Congruent Triangles
A
B4
3
6
C
D
E4
3
6
F
Observe the ∆ABC and ∆DEF.Aretheyarecongruent?Theyhavethe same size and same shape. If yousuperimpose,onetriangleonanother,theycoincidewitheachother.
VertexAcoincideswithvertexD,VertexBcoicideswithVertex E and Vertex C coincides with Vertex F.
Similarly,side AB coincides with
sideDE, side BC coincideswith side EF and side AC coincides with side DF.
Know this:Coinciding parts are called corresponding
parts
Thus,theabovetwo triangles have six pairs of corresponding elements namely 3-angles and 3-sides.
∴ ∆ABC ≅ ∆DEF
(Read as triangle ABC congruent to triangle DEF)
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When two congruent figures are to be P
Q R
L
N M
named, it is a convention to name them in such a way that the corresponding components appear in same order. If the two adjoining triangles are congruent then
∠P = ∠L∠Q = ∠M∠R = ∠NPQ = LMQR = NMPR = LN ∴ ∆ PQR ≅ ∆LMN
Congruent Figures and Similar Figures Notice the shape and size of the figures given below.
2cm
2cm
2cm
2cm
2cm
2cm
2cm
2cm
These 2 Squares are congruent.
They have exactly the same size and shape.
3cm 3cm
These 2 circles are congruent.They have exactly the same size and shape.
3cm
4cm
5cm
3cm
4cm
5cm
These 2 triangles are congruent.
They have exactly same size and shape.
The figures which have exactly same size and shape are
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termed as Congruentfigures.Similar Figures have the same shape, but not same size.
Observetherectanglesgiveninthefigure.Doyoufindanydifferenceamongthem?Yestheyhavethesameshapebutdifferent in size. Similarly observe these circles. What is your observation?Againyoufindthatshapeissamebutsizeisdifferent.
These two rectangles are similar. They have the sameshape,butnotthesamesize.
These two circles are similar. They have same shape,butnotsamesize.
These 2 triangles are similar. They are of the sameshape,butnotthesamesize.
Remember : 1.Congruentfiguresareofthesamesizeandshape.2.Similarfiguresareofthesameshapebutdiffer in size. 3.Allcongruentfiguresaresimilarbutitsconversemay
not be true.
Exercise 5.1
I. Choose the correct answer:
1) Which of the following figures are congruent ?
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a) K L
N M
P Q
S R
b)
B C
A P
Q Rc) P Q
5 CM
3 C
M
RS
A B
5 CM
3 C
M
CDd)
2cm 3cm
2) Which of the following figures are similar?a) K L
M N
P Q
RSb) A
600
600 600B C
G
600
400 800H I
c)A B
DC
P
Q
R
S
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d)
3cm 4cm
II. 1) Fill up the blanks with suitable answers.
1) The figures having same shape and same size are called ______________
2) The symbol used to represent congruent is ___________.
3) All congruent figures are always ___________ .
4) The figures which have same shape but may differ in their sizes are called ___________ .
2)∆ABC≅∆DEF.ThecorrespondingpartofACis
a) DE b) DF c) ∠B d) ∠F
III. Constructthefiguresaccordingtothegivendata.VerifywhetherthesetwotrianglesarecongruentandWritetheircorrespondingvertices,sidesandangles.
A
B C
P
R Q3 cm3 cm
4 cm
4 cm
Activity : Identify the congruent and similar figures in your surroundings. Discuss in group.
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CHAPTER- 6
GEOMETRICAL CONSTRUCTIONS
After studying this chapter 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,
using ruler and compasses construct an angle equal to a given angle using
compasses and scale, 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
We know that two lines or rays
One right angle = 900
or segments are said to be perpendicular to each other if the angles formed at the point of intersection of them are right angles.
Where do you find perpendicular lines in your day-to-day life ?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 ?
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Perpendicular bisector
Activity : Take a sheet of 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.
A B AB A B
Make a crease and unfold the paper. Now observe that the line obtained by creasing the 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.
1. Construction of the perpendicular bisector to a given line segment
Example:1 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.
A B8CM
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2) With A as the centre and radius more than half of AB, draw two arcs (using compases) above & below AB as shown in the figure.
A B
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.
Name the point of intersection of the arcs as P and Q as shown.
A B
Q
P
4) Join P and Q. Let PQ cut AB at O
A Bo
Q
P
PQ is the perpendicular bisector of AB.
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?
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2. How do you construct an angle equal to a given angle without using protractor. Example : 2 Construct ∠ABC which is equal to ∠PQR. Given: ∠PQR
To construct: ∠ABC = ∠PQR Steps of construction:
1) Construct ∠PQR of any measure using pencil and scale.
Q R
P
2) Using compases, with Q as 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.
Q R
P
X
Y
3) Draw another ray BC. B C4) 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 point M. B M C
5) With M as the centre and radius equal to (XY), draw an arc to cut the arc drawn in step- 4, using compasses. Name the arcs inter section point as N. B M
N
C
6) Join BN using scale and pencil.extend it to get BA.
B M
AN
C
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Note : Using the protractor, measure ∠ABC and ∠PQR Are the measure of ∠ABC and ∠PQR same ?
3. Construction 600 angle using scale and compasses.
Activity : Draw a circle of fixed radius with 'o' as centre Mark a point 'A' on the circle keeping 'A' as centre and with the same radius 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 can we draw? , ,OA OB OC What is measure of AOB+ ? Are all the angles equal ?
Steps of construction (600 angle) :
1) Draw a ray OB using a scale and a pencil.
O B
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. O P B
3) With P as the centre and with the same radius as in step-2, draw an arc cutting the previous arc. Name the point of intersection as Q. O P
Q
B
4) Join OQ using a scale and a pencil. Extend it to get OA.
O
A
Q
B600
Using the protractor, measure ∠AOB. ∠AOB measures 600.
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Do you know!In the above figure, why is the measure of ∠AOB is 600? Because OP, OQ and PQ are the sides of ∆ POQ which are equal. ∴ ∆ POQ is an equilateral triangle.
4. Construction of an angle of 1200 using scale and compass.
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 compases.Steps of construction:
1) Draw a ray OB. O B
2) With O as the centre, draw an arc with suitable radius that cuts OB . Name the point of intersection as P. O P B
3) 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.
O
Q
P B
4) With Q as the centre and with same radius as in step- 3, draw an arc so as to cut the arc drawn in step - 2. Name the point of intersection as R. O
QR
P B
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5) Join points O and R using a scale and a pencil. Extend it to get ray OA
O
QR
P B1200
Thus formed ∠AOB measures 1200.
Verification: Measure ∠AOB using a protractor.
Exercise 6.1
1) Draw line segments of given length and then draw a perpendicular bisector to each of them using a scale and compasses.
a) 6 cm b) 8 cm c) 7.4 cm d) 66 mm
2) 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.
3) Measure the following angles and construct these angles without using the protractor.
A
B C
Q
RP
L
M N
4) Construct each of the following angles using a scale (ruler) and compasses.
a) ∠XYZ = 600 b) ∠DEF = 1200
A
A
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Bisector of an angleActivity : 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
CBACB
A
CB
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.
5. To construct bisector of a given angleExample 1: Construct an angle bisector to the given ∠AOB = 500
Given: ∠AOB = 500
To construct:
The bisector of ∠AOB.
Steps of construction:
1) Construct ∠AOB = 500 using a pencil, a scale and a protractor
B
AO500
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2) Taking ‘O’ as the centre and with a suitable radius draw an arc which cuts arms OA and OB.
Name the points of intersection on side OA and OB as ‘P’ and ‘Q’ respectively.
B
P AO
Q
3) With P as the centre and radius more than half of PQ, draw an arc in APQB.
4) 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.
5) Join points O and R using a scale and a pencil. Extend it to get the ray OC.
C
OC is the bisector of ∠AOB
Verification: Measure ∠AOC and ∠BOC. What do you infer?
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6 (a). Constructing an angle of 900 using a scale and compasses.
Discussion of Steps of construction:
1) Draw a ray BC. B C
2) With B as the centre, draw an arc with 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 to cut 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.
5) 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)
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6) 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 S.
S
7) Join points B and S using a scale and a pencil. Extend it to get the ray BA.
A
S
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 ?
6 (b) Alternate method to construct 900.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 ABat point P and Q.
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3) With P as the centre and radius more than half of PQ, draw an arc above the line segment ABas shown in figure. A P Q BO
4) 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.
5) 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 ∠COA
Think! Compare construction of 900 with that of perpendicular bisector. What do you infer?
7. Construction of an angle of 300 using a scale and compasesNote 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
A O B
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2) With O as the centre, draw an arc with suitable radius that cuts OB. Name the point of intersection as P. A O P B
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. Imagine QOP .
Q
A O P B
Now bisect ∠QOP using compasses and a scale.
4) With P as the centre and radius more than half of PQ, draw an arc in the interior of ∠QPB as shown in figure.
Q
A O P B
5) With Q as the centre and with same radius as in step- 4, draw another arc, which cuts the previous arc.
Name the point of intersection as C
Q
A O P B
C
6) Join points O and C using a scale and a pencil and extend it.
Q
A O P
C
Thus formed ∠COB measures 300.
Think! What is the measure of ∠COA ? It must be 1500 Why ?
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8. Construction of an angle of 450 using a scale and compasses
Note: bisecting an angle of 900, we get an angle of 450.
Steps of construction:
1) Draw a line segment AB. Mark a point 'O' on it. A O B
2) With 'O' as the centre, draw a semicircle, with any suitable radius, that cuts AB. Name the point of intersection as P and Q.
A Q O P B
3) With P as the centre, draw an arc with same radius as taken in step- 2 cutting the previous arc. Name the point of intersection as L.
A Q
L
O P B
4) With L as the centre, draw an arc with same radius as taken in step- 2 to cut are LQ Name the point of intersection as M. A Q
M L
O P B
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 Q
M L
O P B
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6) 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 intersection as C. A Q
M L
O
C
P B
7) Join points O and C by using a scale and a pencil. Let OC cut the semicircle at point N.
A Q
M NC
L
O P B
Thus formed ∠COB measures 900. Now bisect ∠COB using scale and compasses
8) With P as the centre and radius more than half of PN, draw an arc in the interior of ∠BPC as shown in figure A Q
M NC
L
O P B
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 as D. A Q
M NC
L
O P B
D
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10) Join points O and D using scale and pencil and extend to get OD
A Q
M N
C
L
O P450
B
D
Thus formed ∠BOD measures 450.
Think! What is the measure of ∠DOA ? It should be 1350. Why ?
9. How do you construct an angle of 1500 using scale and compases?
Steps of construction:
1) Draw a line segment AB . Mark a point 'O' on
A O B
2) With O as the centre, draw semicircle with suitable radius that cuts AB at point S and P. A S O P B
3) With P as the centre and with same radius chosen in step- 2, draw another arc, which cuts the previous arc. Name the point of intersection as Q.
Q
A S O P B
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4) With Q as the centre and with same radius chosen in step-2, draw another arc which cuts the semicircle. Name the point of intersection as R.
R Q
A S O P B
5) With R as the centre and radius
more than half of RS , draw an arc in the interior of ∠RSA shown in figure. (imagine ∠RSA)
R Q
A S O P B
6) With S as the centre and with same radius chosen in step- 5, draw another arc, which cuts the previous arc. Name the point of intersection as C.
RC Q
A S O P B
7) Join points O and C by using a scale and a pencil and extend ray OC
RC Q
A S O P B
Thus formed ∠BOC measures 1500.
Think! Can angle 1500 be constructed by constructing 300 ?
Fig 7.11
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10. Construction of an angle of 1350 using a scale and a compases
Steps of construction:
1) Draw a line segment AB . Mark a point 'O' on
A O B
2) With O as the centre, draw semicircle with a suitable radius that cuts line segment AB at point P and Q.
A O BQP
3) With Q as the centre and radius more than half of PQ, draw an arc as shown in the figure. A O BQP
4) With P as the centre and with same radius chosen in step-3, draw another arc which cuts the previous arc. Name the point of intersection as R. A O BQ
R
P
5) Join points O and R using a scale and a pencil. Let it cut the semicircle at S. Thus formed ∠AOR measures 900.
A O
S
R
BQP
Now bisect ∠AOR using scale and compasses.
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6) With P as the centre and radius more than half of
, draw an arc in the interior of APSR as shown in the figure. A O
S
R
BQP
7) 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. A O
C
S
R
BP Q
8) Join points O and C with the help of a scale and a pencil and extend to get ray OC
A O
C
S
R
BP Q
Thus formed BOC measures 1350.
How to construct 22.50 using a compases 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 ?
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Exercise 6.2
1) Construct each of the following angles using the protractor
and draw an angle bisector in each case using a scale
(ruler) and compasses.
a) 800 b) 1100 c) 360 d) 1320
2) Construct the angles of following measurement using
a scale (ruler) and compasses. Verify the angle using
protractor.
a) 900 b) 300 c) 450 d) 1500 e) 1350
11. How to draw a perpendicular at 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
at point P exactly along with the line AB.
A P B A
BP
A P B
Now make a crease and unfold the paper. Now we get the
crease which is perpendicular to line AB at point P.
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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:
1) Draw a straight line AB and mark point P on it
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.
5) Us ing a sca l e and a pencil, join PQ. Now PQ is perpendicular to .
Verification: Measure ∠BPQ.
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12. Drawing a perpendicular to a given line from a point which is outside the line. 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.
A
P
B AB
P
A
P
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.
Steps of construction:
1) Draw a straight line AB and mark a point P lying outside AB. A B
P
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 A X Y B
P
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3) With radius more than half of XY draw an arc below AB as shown in figure and X as centre
BA X Y
4) With Y as centre draw another arc, with same radius chosen in step-3, that cuts the previous arc. Name the point of intersection as Q.
P
BA X Y
Q
5) Using a scale, join PQ. Mark the intersection point of PQ and AB as ‘O’. Now PQ is perpendicular to AB.
P
BA X Y
Q
Verification : Measure ∠BOP.
13. To Construct a line parallel to a given line through a point outside it
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.
P Q
B
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2) Mark any point A on PQ and join AB.
P A Q
B
3) With suitable radius and A as the centre, draw an arc which cuts PQ and AB. Let the arc cut PQ at X and AB at Y. P A QX
Y
B
4) Now with and B as the centre (the same radius as in step-3,) draw an arc which cuts AB as shown in figure Mark intersection point of arc with AB as M.
P A QX
Y
B
M
5) Place the steel point of the compasses at point X and take XY as radius of compases.
6) With M as the centre, draw an arc with XY as radius cutting the arc drawn in step-4.
Let these two arcs cut each other at point N.
P A QX
Y
B
M
N
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7) Join BN and extend on either side to point R and S.
P
S R
A QX
Y
B
M
N
Now, RS is the required parallel line to PQ and passing through the given point B.
Think! In the above construction, ∠BAQ and ∠ABS are constructed so that both the angles are equal and they are on either side of the transversal AB. Will PQ || RS if ∠BAQ and ∠ABR are equal and they are constructed on same side of the transversal AB. Justify your answer.
Exercise 6.3
1) Draw a line segment AB of length 10 cm. Mark a point P outside the AB. Draw perpendicular line passing through point P.
2) Draw PQ of length 8 cm. Mark a point A on the PQ. Draw perpendicular line passing through point A.
3) Construct ∠PQR = 600. Mark point A on line segment PQ and draw a line parallel to QR through that point.
4) Draw a line 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 PQ passing through A and parallel to BC.
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CHAPTER - 7
MENSURATIONAfter studying this chapter you : explaintheconceptofperimeterofclosedfigureslike
asquare,arectangle,atriangle,andaparallelogramandfindouttheirperimeters,
explain the concept of area of closedfigures like asquare,arectangle,a triangle,andaparallelogramandfindouttheirarea,
explaintheconceptofcircumferenceofacircleandareaofacircle,
calculatecircumferenceandareaofcircle, solve the problems related to the perimeter and
circumference, solve the problemsrelatedtotheareainlifesituation.
IntroductionSandesh and Sangeeta purchased pictures of the
Mysore palace and the Jog falls respectively as a tokenofmemory of school tour. Sangeethawanted to laminateand Sandesh wanted to frame the photoes they hadpurchased. The photo of the Mysore palace measures 30 cm × 60 cm. The photo of the Jog falls measures 40 cm × 50cm.Ifthecostoflaminationis̀ 300 persq.m.Findcostoflamination.Also,findthecostofframing,ifthecostofframingis` 200permeter,whohaspaidmoreamount?
Letusrecallwhatwehavelearntabouttheperimeterandareaofrectangleandsquare.Perimeter
Thelengthalongtheboundaryofaclosedfigureinaplaneiscalled theperimeter.Theperimeterofaclosedfigure isdenotedby'P'.
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Perimeter of a rectangleIn the figure 7.1, lengths of the
Fig : 7.1
A
D
Bl
l
b b
C
rectangleABCD are ABand CD.(LetthelengthoftherectangleABCD be ‘l’units.)ThenAB = CD = lunits.Breadthoftherectangle ABCD are BC and DA. Let breadth of the rectangle ABCDbe ‘b’ units.ThenBC = AD=bunits. Perimeteroftherectangle=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) Perimeter of the square
Inthefigure7.2,lengthofthesidesoftheA
a a
a
a B
CD
Fig : 7.2
square ABCD are AB, BC, CD and AD. Letlength of the sides of square ABCDbe ‘a’ units.ThenAB = BC = CD = AD=aunits.
Perimeterofthesquare=P P = AB+BC+CD+AD
= 4×lengthofitsside = 4 × side = 4aunits∴Perimeter of the square (P) = 4a
Do it yourself : Takeasheetofpaper,cutthesheetintorectanglesandsquaresofdifferentsizes.Findtheperim-eterofeachsquareandrectanglesheet.Example 1 : Find the perimeter of a A l =10 m
b =
6 m
B
CD Fig:7.3
rectangular field of length 10 m and breadth 6m.
Solution: Thelengthoftherectangularfield=l = 10 m Thebreadthoftherectangularfield=b=6 m
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Weknowthat,
Perimeteroftherectangularfield=P=2[length+breadth]
= 2 [10+6]=2 × 16
= 32 m
∴Perimeteroftherectangularfield=P=32 m Example 2: Find theperimeter of a squarefield of length 60m.
Solution: Thesideofthesquarefield=a=60 m
PerimeterofthesquareP=4a A
a
a= 6
0 m
a = 60 m
B
CD
= 4 × 60 m
= 240 m
∴Perimeterofthesquarefield=P=240m
Example 3: Findthecostoffencingwiretobewoundfourtimesaroundasquareshapedflowergardenofside80m,iftherateoffencingwireis` 10permeter.Solution:Thesideofasquareshapedflowergarden =a=80 mThecostoffencingpermeter =` 10Perimeterofthesquareflowergarden =4 × a = 4 × 80 = 320 m∴Lengthoffencingwirerequiredforoneroundaroundthegarden=320 mThelengthoffencingwirerequiredforfourroundsaroundthegarden =4 × 320 = 1,280 mThecostoffencingwire=Thelengthwire× Cost per meter = 1,280 × 10 = 12,800∴ The cost of fencing wire four time around flowergarden =` 12,800
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AreaThesurfaceenclosedbyaclosedfigureinaplaneiscalledarea. Theareaofclosedfigureisdenotedby'A'.
Area of a rectangle
In the figure 7.5, lengths of therectangleABCD are AB and CD.LetlengthoftherectangleABCDbe ‘l’units.Then AB = CD = l units.BreadthoftherectangleABCD are BC and DA. Let breadth ofthe rectangleABCDbe‘b’units.Then, BC = AD=bunits.
A
b b
l
l B
CD
Fig 7.5
Areaoftherectangle=A=length×breadth
Areaoftherectangle=A = l × b
= l bsq.units
∴Areaoftherectangle=A= l b
Area of the Square
Inthefigure7.6,lengthofthesidesofsquareABCD are
AB,BC,CD and AD.LetlengthofthesidesofsquareABCD be
‘a’units.Then,AB=BC=CD=AD=aunits.A
a a
a
a B
CDFig 7.6
Areaofthesquare=A=side×side
Areaofthesquare=A=a×a
= a2sq.units
Area of the square = A= a2
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Try this Find the area and perimeter of floor of classroom door,windows,blackboardandtopofthetable.
Units used to measure area. Tomeasure anything we require a definite unit of
measurement.Letusanalysetheunitswhichweknowverywell.
Metreisthefundamentalunitusedtomeasurelength.
Examine the scaleusedby shopkeepers formeasuringthecloth.Itisametrescale.Themetreistheunitoflength.
What is the unit used to measure area of a closed region?
Observethefigure,givenontherightside.It 1 m
1 m
isasquarewithlengthofeachsideas1metre.Themeasure of the region enclosed by asquarewithlengthofeachsideas1 metre is called a squaremetre and is written as 1 squaremetreor1 m2.What do you call the area of a square with length of each side as 1 centimetre?
Themeasure of the region enclosedbya squarewithlengthofeachsideas1centimetreiscalledsquarecentimetreandiswrittenas1squarecentimetreor1 cm2.
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What is the relation between 1 sq. metre and 1 sq. centimetre?
Consider a square2 1 2 3 4 100
1
2
.....
4
5
eachsidemeasuring1metre.The horizontal lines aredrawn to divide it into 100 equalpartsandtheverticallinesaredrawntodivide itinto 100equalparts.
Afterallthehorizontalandverticallinesaredrawn,youwillget100smallsquaresinarowand100smallsquaresinacolumn.Lengthofeachsideofsmallsquarewillbe1cm.The total number of small square will be 100 × 100 = 10,000innumber.
∴ 1 m2 = 10,000 cm2.
Dependingonthelengthoftheunitsquare,theunitofmeasuringareawouldbedifferent.Iflengthofeachsideoftheunitsquaremeasures1decimetre(symbolically1dm)or1millimetre(1mm),thentheunitofmeasurementofareawillbe1squaredecimetre(symbolically1 dm2)and1squaremillimetre(symbolically1 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
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Iflengthofeachsideofthesquaremeasures1 decametre (1dam)or1hectometre(1hm),thentheunitofmeasurementofareawillbe1squaredecametre(symbolically1 dam2)and1squarehectometre(1 hm2)respectively.
The relationship between these metric units can be shown as below.
• 1 dm2 = 100 m2
• 1 hm2 = 10,000 m2
• 1km2 = 10,00,000 m2
Example 1
Findtheareaofarectangularfieldof length10 m and
breadth 6m.
Solution : Thelengthoftherectangularfieldl = 10 m
Thebreadthoftherectangularfield=b=6 m
WeknowthatA l =10 m
b =
6 m
B
CD fig:7.7
AreaofrectangleA=l × b
= 10 m × 6 m
= 60 m2
∴TheAreaoftherectangleA=60 m2
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Example 2 Findareaofasquarefieldoflength60m.
Solution : Thesideofthesquarefield=a=60 m A B
C
a =
60 m
a
a = 60 m
aDFig:7.8
Areaofthesquare =A=a2
= 602
= 60 m × 60 m
= 3600 sq m
∴Areaofthesquare=A=3600 sq m Example 3
The area of a rectangular garden of 150 m long is 9000 sq.m.Findthewidth(breadth)ofthegarden.
Solution: Lengthoftherectangulargarden =l = 150 m
Areaoftherectangulargarden =A=9000sq.m
Width(breadth)oftherectangulargarden=b=?
Areaoftherectangulargarden =A =l × b
9000 = 150 b
150 b = 9000
Solvingtheaboveequationforb
Breadth(width)oftherectanglegarden = b = 601509000 =
∴ Widthofthegarden=b= 60 m
Think! Area of one metre of ribbon is not equal to area of one-metretowel.Why?
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Example 4 : Adooroflength2 m and breadth 1misfixedina squarewalloflength4m.Findthecostofpainting,iftherateofpaintingthewallis̀ 25persquaremetreandthedooris ` 50persquaremetre.Solution: Thelengthofthedoor =l = 2 m
Thebreadthofthedoor =b=1 mThelengthofthesquarewall =l = 4 mThecostofthepaintingwall =` 25 per m2
Theareaofthedoor =l × b = 2 × 1 = 2 m2
Theareaofthesquarewall =a2
= 4 × 4 = 16 m2
Theareaofthewalltobepainted =areaofthewall-areaofthedoor = 16 – 2 = 14 m2
Thecostofpaintingthewall=Area× Cost = 14 × 25
= 350 ∴Thecostofpaintingthewall =` 350
Thecostofpaintingthedoor =Area× Cost = 2 × 50
∴Costofpaintingthedoor =` 100Thetotalcostofpaintingthewallandthedoor =` 350+` 100
∴Thetotalcostofpainting =` 450
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Example 5: Roopahascutasheetofpaperintotheshapeofasquareandarectangle,insuchawaythattheareaofthesquareandtherectangleareequal.Ifthesideofthesquaresheet is 30cmandthebreadthoftherectangularsheet is20 cm,findthelengthoftherectangularsheet.Also,findtheperimeteroftherectangularsheet.
Solution: Thesideofthesquaresheet=a=30 cm
Thebreadthoftherectangularsheet=b=20 cm
Theareaofthesquaresheet =a2
= 302
= 30 × 30
∴TheAreaofthissquaresheet =900 cm2
Asperthegivendata,
Areaoftherectangularsheet=Theareaofthesquare=900 cm2
Areaoftherectangularsheet =900 cm2
A= l × b = 900 cm2
l × 20 = 900 cm2
l = 20900 = 45 cm
∴Thelengthoftherectangularsheet=l = 45cm.
Perimeteroftherectangularsheet =P=2 [l+b]
= 2 [45+20]
= 2 × 65
= 130 cm
∴Perimeteroftherectangularsheet=P=130 cm
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Example 6 :Anilhasasquareshapedchessboardofarea144 cm2.Findthelengthofthesideofthechessboard.
Solution:
Theareaofthesquareshapedchessboard=a2 = 144 cm2.
a2 = 144
a ×a = 12 × 12
a2 = 122
a = 12
∴ Thelengthofthechessboard=a=12cm
Example 7:
Theareaofasquareshapedfieldis64hectare.Findtheperimeterofthefieldinmetre.
Solution : Theareaofthesquareshapedfield=a2 = 64 hectare
a × a = 64× 10,000(∴1 hectare = 10,000 m2)
= 6,40,000
a × a = 800 × 800
∴ a = 800 m
∴Thelengthofthesideofthesquareshapedfield=a=800 m
Theperimeterofthesquareshapedfield =4a
= 4 × 800
= 3,200
∴Theperimeterofthesquareshapedfield=3,200 m
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Exercise 7.1
I. Find the perimeter of the rectangles, with the following measurements.
1) length=8cm,breadth=6 cm
2) length=3m,breadth=2 m
3) breadth=4.5m,length=9.5 m
II. Find the area of the rectangles with the following measurements.
1) length=6cm,breadth=4 cm
2) length=12.5m,breadth=7 m
3) breadth=3.5m,length=6.5 m
III. Find the perimeter of the squares formed with the following measurements.
1)side=6cm 2)side=15m 3)side=5.6 cm
IV. Find the area of the squares formed with the following measurements.
1)side=6m 2)side=12cm 3)side=9.8 cm
V. Solve the following :
1) Theperimeterofarectangularplankis120cm.Ifthelengthis 40cm,finditsbreadthandsurfaceareaoftheplank.
2) Thimmarajuhasaplotoflength12 m and breadth 10m.Ithastobefencedwithfourroundsofwire.Ifthecostofthewireis` 30permeter,findthecostoffencingtheplot.
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3) Awireisbentintheshapeofarectangle.Itslengthis36
cm and breadth is 25cm.Ifthesamewireisbentinthe
shapeofasquare,whatwillbethemeasureofitsside?
Amongthesetwoshapewhichshapeoccupiesmorearea?
4) Awireoflength100cmisbenttotheformofasquare.
Findtheareaoccupiedbythesquareshape.
5) Theareaofthesquareshapedfieldis16hectare.Findthe
perimeterofthefieldinmetre.
(Hint:1hectare=10,000m2)
6) Afarmerhasafieldintherectangularshape.Thelength
of rectangularshapedfield is150 m and its breadth is
100m.Findthecostofploughingthefieldattherateof`
0.2 per m2.
7) Whathappenstotheareaofarectanglewhen
1) itslengthisdoubled,breadthremainingthesame?
2) itslengthisdoubled,breadthishalved?
3) itslengthandbreadtharedoubled?
8) Whathappenstotheareaofsquare
1) Whenitssidesaredoubled?
2) Whenitssidesarehalved?
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Perimeter of a triangleWhoamI?Clue 1: I am a closed
B C
A
a
c b
fig:7.9
geometricalfigureofminimumnumberofsides Clue 2:Ihavethreesides.
Clue 3:Ihavethreeangles.Itisatriangle.Whatisatriangle? A triangle is a closed geometrical figure in a plane
having three sides. ObservetriangleABCinfigure7.9.LetlengthofsidesBC,
ACand ABbea,bandcrespectively.Weknowthattheperimeterofageometricalfigureisthe
sumofallthesidesofgeometricalfigure.ItisdenotedbyP.HerethetrianglePQRhasthreesides.Theperimeterofatriangle=P =BC+AC+AB =a+b+cunits
∴ The perimeter of a triangle (P) = a + b + cIfallthesidesofatriangleareequal,thenthetriangleis
calledequilateraltriangle.Whatistheperimeteroftheequilateral
A
B Cfig:7.10
a
a
atriangle?Infigure7.10,triangleABCisanequilateraltriangle.LetAB=BC=AC=aunits.∴Theperimeteroftheequilateraltriangle= AB+BC+CA =a+a+a = 3aunits
∴ The perimeter of an equilateral triangle (P) = 3a
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Example 1: Findtheperimeterofatrianglewhosesidesare10cm,7cm,and5cm.Solution: Lengthofsidesofthetriangle,a=10cm,b=7cm,and c = 5cm.
Lengthofsidesofthetrianglearedifferent.Theperimeterofthetriangle=P =a+b+c = 10+7+5 ∴ P = 22 cm∴The perimeter of the given triangle = 22 cm
Example 2 : Findtheperimeterofanequilateraltrianglewhose side is 15cm.Solution: Length of each side of the equilateral triangle = a = 15cmTheperimeteroftheequilateraltriangle=P=3a
= 3 × 15
∴ P = 45 cm∴The perimeter of the equilateral triangle = 45 cm
Example 3: Theperimeterofanequilateraltriangularshapedcard board is 201cm.Findthelengthofasideofthecardboard.Solution: Theperimeteroftheequilateraltriangularshapedcardboard = 3a = 201 cm
The length of side of the equilateral triangular shaped
cardboard a = 2013
= 67 cm∴Thelengthofsideoftheequilateraltriangularshaped
card board = 67 cm
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Area of a triangleReena takesarectangularpapersheetABCDoflength
AB= 14 cm and breadth BC = 8cm.Shecutstherectangularsheet ABCDalongthediagonalAC.HowmanypiecesofpaperdoesReenaget?
Al =14 cm
b =
8 cm
B
CD Fig 7.11
Shegetstwopiecesofpaper.Whatistheshapeofeachpaper?Yes,itisintheshapeofatriangle.
Placethesetwotrianglesoneovertheotherinsuchawaythat vertex B coincideswithD and vertex AcoincideswithC.
Arethesetwotrianglessameinshape,sizeandarea?Aretheycongruenttriangles?
Yes,thesetrianglesareofthesameshape,size,andarea.Theyarecongruenttriangles∴Thediagonaloftherectangledividestherectangleinto
twoequaltriangles.LengthoftherectangleABCD is the base of
A
h
b
B
CD
Fig 7.12
triangleADCandbreadthoftherectangleABCDistheheightofthetriangleADC.Let thebaseof the triangleADC be ‘b’ unitsandheightofthetriangleADC be ‘h’units.AreaoftherectangleABCD = 2 ×AreaofthetriangleADC∴ AreaofthetriangleADC = 2
1 (AreaoftherectangleABCD)
= 21 (length×breadth)
= 2
1 b h
∴TheareaofthetriangleADC=A= 2
1 b hThesymbolusedtorepresenttriangleis∆
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Finding the area of a triangle 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?
Fig 7.13 Fig 7.14 Fig 7.15
A A A Q
AAB B
R R R
R
S S
S
S
P P PQ Q Q
Consider a rectangular piece of paper PQRS. Mark any point A on SR. Join PA and QA. ∆ PQA is inscribed in the rectangle PQRS as shown in the Fig. 7.13
Now draw a line AB perpendicular to PQ.
Fig : 7.16
A
B
h h h
RS
P Qb Units
We observe that PS = AB = QR. Let PS = AB = QR= 'h' units and PQ = 'b' units.
Now cut along the lines PA and QA. Super impose two triangles ∆ PAS and ∆ QAR on ∆ PAB and ∆ ABQ respectively as shown in the Fig. 7.15
From figure 7.15, we understand that
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
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2 ×Areaof∆PAQ=AreaoftherectanglePQRS
Areaofthe∆PAQ = 21 × (areaofrectanglePQRS)
= 21 × (length×breadth)
= 21 × (b× h)
= 21 bhsq.units
Areaofanytriangle=A = 21 bhsquareunits
Wherebisthebaseandhistheheightofthetriangle.
ARYABHATA - I
Aryabhata-IagreatIndianmathematicianandastronomer,isbelievedtohavebeenbornintheregionbetweentheNarmadaandGodavaririvers in central India or in south India.AryabhatawenttoKusumapuraforadvancedstudies. Kusumapura is identified asPātaliputra,modernPatna.TheUniversityofNalandawasinPataliputraatthattimeand
had an astronomical observatory. It is speculated thatAryabhatamighthavebeentheheadoftheNalandaUniversity.Aryabhata isalsoreputed tohavesetupanastronomicalobservatoryattheSuntempleinTaregana,Bihar.Aryabhata - I mentionsinthe'Aryabhatiya'aworkofhimthatitwascomposed 3,630years intotheKaliYuga,whenhewas23 yearsold.Hehasgiventheformulaforareaoftriangle.
Itmeans,Theareaofatriangleistheproductofhalfofitsbaseandheight.
Example 1 : Findtheareaofthetrianglewhosebaseis10 cmandheightis6cm.
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Solution: Baseofthetriangle =b=10cm.
Heightofthetriangle =h=6cm.
Areaofthegiventriangle= 21bh
= 21 × 10 × 6
= 30sq.cm.
∴Theareaofthegiventriangle =30sq.cm.Example 2:
Findtheheightofatriangularshapedfieldwhoseareais400 m2 and its base is 50m.
Solution:
Theareaofthetriangularshapedfield=A=400 m2.
Thebaseofthetriangularshapedfield=b=50m.
WeknowthatAreaofanytriangle =A= 21 bh
i.e 21 bh=A
21 × 50 × h = 400
25 × h = 400
Solvingtheaboveequationforh
h = 251 × 400
= 16 m
∴ Theheightofthetriangularshapedfield=h = 16 mExample 3: Theratioofthebaseandheightofatriangleis 2 : 3.Iftheareaofthetriangleis1200 m2,findthemeasurementofitsbaseandheight.
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Solution: Theareaofthetriangle=A=1200 m2
Theratioofthebaseandtheheightofatriangle=2 : 3.Thelengthofthebaseandheight=2x : 3x.Weknowthat 2
1 bh =A 2
1 × 2x × 3x = 1200
3x2 = 1200
x2 = 31200
x2 = 400
x2 = 202
∴ x = 20 Therefore,thelengthofthebase=2x = 2 × 20
= 40m. Theheightofthetriangle=3x = 3 × 20 = 60 m
TRY THISIn the adjoining figure,
R C E D S
QBAP
PQ|| RSand∆ABDisrightangledatB.ThelengthofAB = 10 cm and BD = 8cm.Findtheareaof∆ABC,∆ABDand∆ABE.Dotheyhavethesamearea?Ifyesorno,why?Givereason
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EXERCISE 7.2
1) Findtheperimeterofthefollowingtriangles.P
Q R8 cm
6 cm
10 cm
L
M N5 cm
6.5cm
6.5cm
X
Y Z8 cm8 cm8
cm
A
B C4 cm
5 cm
7 cm
2) Calculatetheperimeterofthetrianglewhosesidesare6cm,8 cm and 6cm.
3) Findtheperimeteroftheequilateraltriangleswithsides a)8cm b)13 cm c)11 cm4) Inatriangle,theperimeteris60cmandthelengthoftwo
sidesoftriangleare13 cm and 20cm.Findthelengthofthethirdside.
5) Theperimeterofanequilateraltriangleis45cm.Findthelengthofeachsideoftheequilateraltriangle.
6) A lawn in a garden is in the shape of an equilateraltriangle.Ifthelengthofonesideoftheequilateraltriangleis 75m,findthecostof fencingattherateof` 12 per metre.
7) Findtheareaofthefollowingtriangles.A
B C8 cm
6 cm
10 cm
P
QS R4 cm
6.5cm
M N
L
5 cm
6 cm
6.5cm6.5cm
8) Findtheareaofthetrianglewhosebaseis14 cm and theheightis7cm.
9) Agardenisintheequilateraltriangularshape.Itsbaseis 22 m and breadth is 18m.Findthecostoflevellingthegardenat` 5 per m2.
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10) Theratioofthebaseandtheheightofatriangleare4 : 3.Iftheareaofthetriangleis216 m2,findthelengthandheight.
11) Whathappenstotheareaoftrianglewhen
a)Itsbaseisdoubled,heightremainingthesame?
b)Itsbaseisdoubled,heightishalved?
c)Itsbaseandheightaredoubled?Area of parallelograms
Pallavimakesarectangleusingfoursticksandcyclevalvetubes.Shecutsthestickintoapairofequallengthsforlengthandbreadthofarectangle.Sheinsertedtheendsofstickintocyclevalvetubetomakerectangleasshowninfigure7.17.SheshowsittoMeena.MeenaappreciatesandjustpushespointAsoftlytowardspointB.Nowfigureobtainedisshowninfig7.18.
A B
CD fig-7.17
A B
CD fig-7.18
Howmanypairsofparallellinesarethereinthefig7.18 ?
Ithastwopairsofparallellines.
Whatisthegeometricalshapeoffigure7.18?
Yes, it is termed as parallelogram.Ageometricalenclosedfigureonplanewithtwopairsof
parallellinesiscalledparallelogram.
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How do you find area of a parallelogram?A B
CEDfig-7.19
A B
C FEDfig-7.20
DrawaparallelogramABCDonapapersheetasshownbelowinfigure7.19.DrawalineperpendiculartobaseCD from vertexA.LetitmeetthebaseCD at E.CutthetriangleAEDandattachitinsuchawaythatBCcoincideswithAD.
Whatistheshapeofgeometricalfigureobtained?Itisinrectangularshape.Istheareaoftheparallelogramequaltotheareaofthe
rectangleformed?Yes,theAreaoftheparallelogram=Areaoftherectangle
formed.FromFig.7.21,weobservethatthe
A B
C FE
h
bunits
bunits
h
fig-7.21
lengthoftherectangleformedisequalto the base of the parallelogramandbreadthoftherectangleisequaltotheheightoftheparallelogram.Areaoftheparallelogram=AreaoftherectangleA =(length×breadth)squnits =(base×height)squnits =bhsqunits
∴ Areaofparallelogram=A=bh
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Any side of a parallelogram canbe taken as base of theparallelogram.Theperpendiculardrawnonthatsidefromtheoppositevertexisknownasheight(altitude).
A B
CED
In the paral le logram ABCD, AE is perpendiculartoCD.HereCD is the base and AEistheheightoftheparallelogram.
In this parallelogram ABCD, CF is the
A B
CD
F
perpendiculartooppositesideAD.
Here ADis the base and CFistheheight.oftheparallelogram
Try this! Findtheperimeterofaparallelogramusingthepropertythat,oppositesidesofparallelogramareequal.
Example 1 :
Findtheareaoftheparallelogram havingbase8 cm and altitude5cm.(NoteAltitudemeansheight)
Solution:
Thebaseoftheparallelogram =b=8 cm
Theheight(altitude)oftheparallelogram =h=5 cm
Areaoftheparallelogram =bh
= 8 cm × 5 cm
bh= 40 cm2
∴TheAreaoftheparallelogram =40 cm2
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Example: 2 Findtheheightofaparallelogramshapedfieldwhoseareais108 m2 and base is 12 m.Solution: The area of the parallelogram shaped field =A=108 m2
Thebaseoftheparallelogramshapedfield =b=12 mAreaoftheparallelogramshapedfield =A=bh 108 = 12 × h (Solvingforh) h= 12
108 h = 9 m∴ Theheightoftheparallelogramshapedfield=h=9 m
Example 3: Theratioofthebaseandheightofaparallelogramis 4 : 3.Iftheareaoftheparallelogramis48 m2,findthebaseandheight.Solution:Theareaoftheparallelogram =A=48 m2
Theratioofthebaseandheightofaparallelogram=4 : 3Letthelengthofthebaseandheightofaparallelogrambe4x and 3x respectively
Weknowthat bh =A 3x × 4x = 48
12×x2 = 48 ∴ x2 = 12
48
ie.x2 = 4 ∴ x2 = 22
If powersof exponentsare the same;basesof exponentsmustbeequal ∴ x = 2Thelengthofthebaseofaparallelogram=4x
= 4 × 2 = 8 mThelengthoftheheightofaparallelogram =3x = 3 × 2 = 6 m
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Exercise 7.3
1) Completethefollowingtablewithrespecttoa parallelogram.
Sl.No Base Height Area01 8 cm 6 cm02 15 m 11 m03 14 m 280 m2
04 24 cm 1200 cm2
2) Findtheareaoftheparallelogramhavingbase9 cm and altitude7cm.
3) Findtheareaoftheparallelogramshapedfieldhavinglengthofthebase21mandheight15m.
4) Find thealtitudeof theparallelogramshapedbannerhavingbase12 m and area 108sq.m.
5) Findthebaseoftheparallelogramshapedsitehavingheight1.3 m and area 104sq.m.
6) ABCDisarectanglewithlengthAB = 13 cm and breadth CB = 7cm.PointDispushedtowardspointCtoformparallelogram.FindtheareaofparallelogramABCD.
7) ABCDisaparallelogramwithAB = 15 cm and BC = 18cm.TheheightoftheparallelogramfromvertexA to BC is 9cm.FindtheheightoftheparallelogramfromvertexC to AB?
8) Theratioofthebaseandheightofaparallelogramis 3 : 2.Iftheareaoftheparallelogramis150 m2,findthebaseandheight.
9) Theratioofthebaseandheightofaparallelogramis 5 : 2.Ifareaoftheparallelogramis1000 m2,findthebaseandheight.
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CircleInourdailylife,wecomeacrossanumberofobjectslike
wheels,bangles,coins,rings,giantwheel,papad,compactdisc(C.D.)
Whatistheshapeoftheseobjects?‘Round’
Yes,theyareround.Theseshapesaretermedascircularshapedobjects.
Activity : Takeacardboard.Markthecentreoftheboard,
O A
namethispointas ‘O’.Fixa small nail f i rmly at ‘O ’ , Take a thread o f convenientlengthwhichislessthanthebreadthoftheboard.Tieoneendtightlytothenail.Tietheotherendofthethreadtoasharpenedpencilfirmly.Stretchthethreadcompletelytowardspoint‘A’.Startmoving the pencil, holding the thread
stretched tightly, so that the pencilmarks on the sheet untilthepencilreachesthestartingpoint'A'. Nowobservethepathtracedbythepencil.
Whatistheshapeofthepathtracedbythepencil?Theshapeofthepathtracedisacircle.
A CircleYes,acircleisalocus(path)ofpointwhich
o NM
Pmovesinsuchawaythatitsdistancefromafixedpointisalwaysconstant.
Inthefigure,observethepoint‘O’.Itisatequidistant fromanypointon thecircle.Hence, it is always called centreof the circle.Thedistancebetween thecentre of circle and any point on the circle isconstant.Thisconstantdistanceiscalledradius of the circle.OPistheradiusofthecircle.
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OP,OMandONaretheradii(pluralformofradius)ofthecircle.
DoesMNpassthroughthecentreofthecircleO?Yes,MNpassesthroughthecentreofthecircleO.Anylinesegmenthavingitsendpointsonthecircleand
passingthroughthecentreofthecircleiscalledthediameter of the circle(MN).Inagivencircle,thediameterisalwaystwicetheradiusofthecircle.Inotherwordstheradiusofthecircleishalfofthediameter.
Let‘d’denotethelengthofthediameterand‘r’denotetheradius of the circle. Then above statement is symbolicallywrittenasd=2r or r = d2 .
Do you know!• Howmanydiameterscanbedrawninacircle?• Howmanyradiicanbedrawninacircle?
Circumference of a circle.Does a circle have perimeter ?
Yes,ithasaperimeter.Theperimeterofthecircleiscalledcircumferenceofthecircle.Thecircumferenceistheboundaryofthecircle.Wecannotmeasureaccurately,thecircumferenceofthecircleusingscale.Thiscanbemeasuredbyrollingthecircleontoalineasexplainedbelow.
Tieathreadtightlyaroundthecircumferenceofacircle.markthepointwherethetwoedgesofthethreadmeet.Now,removethethreadandmeasurethelengthofthethreadusingascale.Thisisthecircumferenceofthecircle.
Takeacircleofradius3.5cmusingabangleorathickcardboard. Markapoint 'A'ontherimofthecircle.DrawalineonapageofyournotebookandmarkaninitialpointPonit.NowcoincidepointAofthecirclewithpointPontheline.
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P
A
AA
AA
O O O OO
Q22 cm
SlowlyrollthecircleonthelineuntilpointAagaintouchestheline.MarkthispointasQ.MeasurePQ.Thisgivesthelengthofthecircumferenceofthecircle.Approximatelythecircumferenceofthecircleis22cm.
Relationship between circumference and diameter of the circle
Cutcircleshavingdifferent radii ona thickcardboard.Measurethediameterandradiusofeachcircleusingascale.Tabulate the data in the table given below.Run a threadaroundthecircleandmeasurethelengthofthethreadthatgivesthecircumferenceofthecircle.Recordtheobservationinthetable.Repeatthesestepsforeachcircle.
Findtheratioofcircumferencetodiameterofeachcircle.Whatdoyouconclude?
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Serialnumber
Diameter (d)
Radius(r)
Circumference(C)
( )( )
Diameter dCircumference c
1
2
3
4
5
Itisobservedthattheratioofcircumferenceanddiameteris little more than 3. Approximately its value is 3.14. It isdenotedbyGreekletterπandispronouncedas'pi'.
Circumference(C)
Diameter(d)= r
Try this : Collectone,two,five,andtenrupeescoinsofdifferentradii.Measurethecircumferenceanddiameterofeachcoin.Findthe ratio of circumference todiameterineachcase.
Circumference(C)=π ×(d)Diameter(d) =π× 2r
= 2 πr
∴Circumferenceofacircle=C=2πr
Think !Acircleof radius rcm iscutinto two semicircles. Whatis the circumference of asemicircle?
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Knowthis:Indiancontributiontofindthevalueofπ1. Aryabhata-Iperiodgavethevalueofπ as 3.1416up
to 4decimalplaces.Healsostatedthateventhisvalueisapproximatevalue.
MVwa{YH$_² eV_ð>JwU_ Xÿmf{ð>ñVWm ghòmUm_²
A`wVXÿ` {dîH§$^ñ`mgÞmo d¥Îmn[aUmh… Meaning, “Add four to one hundred, multiply it by eight and then add sixty two thousand; The result is approxi-mately the circumference of a circle of diameter of twenty thousand.”
i.e. π = Circumference
Diameter =
8(100+4)+6200020000 = 62832
20000 = 3.1416
2. Madhava(1350-1425A.D)wasborninvillageSangamagramanearCochininKeralastate.Hecalculatedthevalueof πuptoelevendecimalplaces.Itsvalueis3.14159265359 Modernvalueofπuptotwentydecimalplaceis3.1415926535 8979323846. Using the computer, π value is calculated uptomilliondecimalplaces.Hence,π isanon-recurring,non-terminatingdecimalnumber.
3.Srinivas Ramanujan(1887 - 1920)wasoneofthegreatmathematicians of 20thcentury.Hewasaselftaughtgeniusandhighlycreative,guidedbyimaginationandintuition.Hehasgivenapproximatevalueofas(97 1
2 - 111)
14 = 3.1415926526...
Thevalueofπ in 20thcenturyis3.14159265358976323846.Thevalueofπisnonrecurringandneverendingdecimalnumber.
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Do it yourself: Drawacircleofradius5cm.Drawadiametertothecircleand name their end points as A and B. Then divide thelengthofthediameterABintotenequalpartsi.e.onecmeach.MounttwonailsatpointA and B.TieathreadtonailmountedatA.Nowmountthenailsalongthecircumferenceofthecircleascloseaspossible.RunthethreadaroundthecirclealongwiththecircumferenceuntilitreachespointA.Cutthethreadusingscissor.NowremovethethreadandwindthethreadalongwiththediameterAB.
HowmanycompletewindingscanbedonealongwiththediameterAB?
3completewindings.
How much thread is left after threewindings?
Itisjustmorethanonedivision.
Thisgivesusthevalueofπ.
Therefore,thevalueoftheπis≈3.1
(≈symbolmeansapproximately)Example 1 : Find the circumference of a circle whose radiusis14cm.Theradiusofthecircle=r=14 cm
Note : Forcalculationpurposeweconsider the
approximatevalueof π as 7
22 or 3.14
Circumferenceofthecircle=C=2 πr = 2 7
22 14# #
= 88 cm
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Example 2 : Findthediameterofacirclewhosecircumferenceis 21.98m.(π = 3.14).Solution:Circumferenceofacircle=C =21.98 m
2π r = 21.98
Radiusofacircle= r = 21.98
2π =
21.982×3.14
= 21.986.28
=3.5mDiameter of a circle = d = 2r = 2 × 3.5 = 7 m ∴ Diameter of a circle = 7m.
Example 3 : Thediameterofacircularshapedflowergardenis 70m.Whatisthecostoffencingit,at` 15permetre?Theunitcostoffencingthecircularshapedgardenpermetre= ` 15 per metreThediameterofthecircularshapedflowergarden=d=70 mCircumferenceofcircularshapedflowergarden=πd = 7
22 × 70
= 220 mThecostoffencing=circumference × unitcost = 220 × 15
= ` 3300
∴Thecostof fencing thecircularshapedflowergarden = ` 3300
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Example 4 : Therectangularshapeofthewire16cmlongand6cmbroadisbenttotheformofacircle.Findthediameterofthecircle.Lengthofthewire =Perimeteroftherectangle
= 2(l+b) Lengthoftherectangle
= l = 16 cm
Breadthoftherectangle
= b = 6 cm
= 2(16+6) = 2 × 22
= 44 cm
Nowthesamewireisbenttotheformofacircle.Lettheradiusofthecircleber
Thecircumferenceofthecircle =lengthofthewire c= πd = 44 cmThediameterofthecircle,c = d = 44
d = 4422
x 7= 2 x 7 = 14
∴Thediameter of the circle = d = 14 cmExample 5 : ThecostoffencingMahatmaGandhicircleis` 1,100 at the rate of ` 50permetre.WhatistheradiusoftheMahatmaGandhicircle?
ThecostoffencinginMahatmaGandhicircle =̀ 1,100
Unitcostoffencing =` 50 TheCircumferenceof
MahatmaGandhicircle Unit CostCost of the fencing=
= m501100 22=
∴CircumferenceoftheMahatmaGandhicircle =22 m C = 2πr = 22
r = 222π = 22 × 7
7 × 22 = 7
2 =3.5m
∴TheradiusoftheMahatmaGandhicircle=3.5 m
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Area of the circle
Rehana has a circular shaped compact disc (CD). Shewantstocalculatetheareaof thecompactdisc (CD).Howcanshecalculateit?
Takeacardboard.Drawacircleofconvenientradiusrcm.Cutthecirclewithpairofscissors.Dividethecardboardcircleintosixteenequalsectors.Cutthesesectorsusingapairofscissorsandarrangethemasshownbelow.Thearrangementofsectorslookslikeaparallelogram.
Whenmorenumberofsmallersectorsarecutfromthecircle,thefiguretakestheshapeofarectangle.
πr
r
Whenthenumberofsectorsincreasesmoreandmore,inthelimitingcase,thefiguretakestheshapeofarectangle.
Thebaseoftheparallelogramishalfofthecircumferenceofthecircle.
i.e.Thebaseoftheparallelogram= 21 × 2πr
= πr Heightoftheparallelogram=radiusofthecircle(approx) = r cm Areaofthecircle =Areaoftheparallelogram = base ×height = πr × r = πr2 sq.units
Areaofthe circle = πr2 sq.units
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Knowthis:Aryabhatahasgiventheformulatofindareaofcircle.
Itmeans,"Areaofacircleisequaltotheproductofhalfofthecircumferenceandhalfofthediameter".
A= 21 × 2πr × 2
1 ×d
= 21 × 2πr × 2
1 ×2r
= πr × r = πr2
Example 1 : Findtheareaofacirclehavingradiusequalto4.9 cm
Theradiusofthecircle=r=4.9 cmTheareaofthecircle =πr2 = 7
22 × 4.92
= 722 × 4.9 × 4.9
= 75.46sq.cm∴Theareaofthecircle=75.46sq.cm=75.46 cm2
Example 2 : Acowistetheredtoapeginabiggrassyardusingtheropeoflength5m.Findthemaximumareaoffieldinwhichthecowcangraze.(π = 3.14).
Thelengthoftherope =Theradiusofthecircler=5 m
Theareaoffieldthecowcangraze=πr2
= 3.14 × 52
= 3.14 × 5 × 5
= 78.5 m2
∴Themaximumareaoffieldthecowcangraze=78.5 m2
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Example 3 : Findthecircumferenceof thecircularregionhavinganarea616squaremetre.(Useπ = 7
22 )Theareaofthecircularregion=πr2 = 616 m2
722 r2 = 616
r2 = 616 × 227
r2 = 28 × 7 r2 = 196,r2 = 142
Ifpowersofexponentsaresame;basesofexponentsmustbeequal,thatistosay. r=14
∴Theradiusofthecircularregion=r=14 mThecircumferenceofthecircularregion=2πr = 2 × 7
22 × 14 = 2 × 22 × 2∴Thecircumferenceofthecircularregion=88 cm
Example 4 : Thecircumferenceofthecircularshapedbaseofwatersupplyingtankis132m.Findtheareaofthebaseplate.(π = 7
22 )The circumference of the base of water supplying tank=132 m 2 πr = 132 m
Theradiusofthebaseofwatersupplyingtank,r=1322π
r = 2 22132 7##
r = 21 m∴Theradiusofthebaseofwatersupplyingtank=r=21 mTheareaoftheplotofwatersupplyingtank=πr2 = 7
22 × 212
= 722 × 21 × 21
= 22 × 3 × 21
= 1,386 m2
∴ Theareaoftheplotofwatersupplyingtank=1,386 m2
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Example 5 : Awireof88cmlongisbenttoformasquareandthesamewireisbenttoformacircle.Findthedifferencebetweentheareaofthecircleandthesquare?(π= 7
22 ).
Theperimeterofasquare=lengthofthewireTheperimeterofasquare =88 cm 4a = 88
a = 488
a = 22 cmAreaofthesquare =a2 = 222
= 484 sq cm∴Areaofthesquare =484 sq cmThecircumferenceofcircle =lengthofthewire 2πr = 88 cm
Radiusofthecircle =r= 882π
r = 2 2288 7##
r = 14 cm∴Radiusofthecircle=r =14 cmTheareaofthecircle =πr2
= 722 × 14 × 14
= 22 × 2 × 14
= 616 cm2
∴Theareaofthecircle =616 cm2
Differencebetweentheareaofcircleandtheareaofsquare = 616 – 484
= 132 cm2
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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
The radius of the circular = r = .21 8 m
dining table = 0.9 m
The area of the surface of dining table = πr2
= 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.
Exercise 7.4
1) Find the circumference of the circles of following radii. a) 7 cm b) 10.5 cm c) 21 m2) 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.
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5) Findthediameterofacirclewhosecircumferenceis1256
cm.useπ = 3.14
6) Thediameterofacirculargardenis42m.Whatisthecostoffencingitat` 25permetre?
7) Thediameterofcirculargardenis49m.Findthedistancecovered by a runner if hemakes 2 rounds round thegarden.
8) Thesquareshapeofthewireofalength44 cm is bent in theformofacircle.Findthediameterofthecircle.
9) Findtheareaofthecirclesoffollowingradii.
a)7cm b)10.5cm c)21 m
10)Findtheareaofthecirclesoffollowingdiameters.
a)70cm b)56m c)49 cm
11)Thecircumferenceofthecircleis396m.Findtheareaofthecircle.
12)Findthecircumferenceofthecircularregionhavingarea5544squaremeter.
13)Awireof1.76m long isbent to formasquareandthesamewireisbenttoformacircle.Findthedifferenceintheareaofcircleandsquare?
14)Findthecostoflaminatingdecolamtoacirculartableofradius1.4m,iftherateofdecolamis` 50 per m2.
15)Awireisbentintheformofasquareoccupiesanareaof196 m2.Ifthesamewireisbenttoformacircle,findtheareaoccupied.
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Area between the rectangles / Area of rectangular pathwayInthefigure,thesmallerrectangle A B
B
CLDS l
l
R
b
QPenclosedbyrectangularpathway.Thefigurehastworectangles.ThesmallerrectanglePQRS haslengthlunitsandbreadthbunitsandthelargerrectangleABCDhaslengthLunitsandbreadthBunits.The regionbetween the tworectanglesiscalledthe rectangular pathway.
Theareaofrectangularpathwayisequaltodifferenceofareaoflargerandsmallerrectangle.
Theareaofrectangularpathway=TheareaofrectangleABCD-TheareaofrectanglePQRS = LB - lb
Theareaofrectangularpathway=LB-lb
Rashmi has a rectangle of 20 cm length and 12 cm breadth. It ismadeofpinkcolourcardboard.Smithahasbluecolouredcardboardoflength15 cm and breadth 7cm.Theyplacedthebluecolouredcardboardexactlyatthecentreofthepinkcolouredcardboard,sothatitleftuniformpinkpassagearoundthebluecardboard.Howdotheycalculatetheareaofpinkcardboardpassagearoundthebluecardboard?
Areaofthepinkcolouredcardboard=A1 = lb = 20× 15
= 300 cm2
SecondlyfindtheareaofbluecolouredcardboardLengthofthebluecolouredcardboard=l =12 cmBreadthofthebluecolouredcardboard=b=7 cm Areaofthebluecolouredcardboard=A2 = lb = 12 × 7 = 84 cm2
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Nowfindthedifferencebetweentheareasoftworectangles.Theareaofthepinkcardboard(passage)aroundtheblue
cardboard=A1–A2
= 300 – 84
= 216 cm2
∴Thearea of thepink cardboard (passage) around thebluecardboard=216 cm2
Example 1 : Arectangularparkmeasuring19 m by 14 m is to becoveredbyexternalpathwhichis3mwide.Findthecostoflayingastoneslabpathattherateof` 30persquaremetre.
Let PQRS be the rectangular park. LetA B
CDS R
QP3 m
3 m
3 m
3 m
ABCDbetheexternalboundaryofthepathwhichisalsointherectangularshape.
Width of the path = 3 mLengthofthePark =PQ=19 mBreadthofthePark =QR=14 mAreaoftheParkPQRS =A1 = 19 × 14
= 266 m2
Lengthoftheexternalboundaryofthepath=AB=19m+3m+3m = 25 mBreadthoftheexternalboundaryofthepath= BC = 14m+3m+3m = 20 m AreaoftheexternalboundaryofthepathABCD=A2 = 25 × 20 = 500 m2
Areaofthepath =A2-A1
= 500 – 266 = 234 m2
Costofthelayingthestoneslabpath=234 × 30 = ` 7020
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Example 2 : Apaintingismountedona A B
B
CDS R
b
QP
card board of 10 cm × 7cmsizeinsuchawaythatthere isamarginof1 cm alongeachside.Findthetotalareaofthemarginonthecardboard.Let PQRS be the cardboard and ABCD bethepainting.
Widthofthemargin =1 cmLengthofthecardboard =PQ = 10 cmBreadth of the cardboard = QR = 7 cmAreaofthecardboardPQRS=A1 = 10 × 7 = 70 cm2
Lengthofthepainting=AB= 10 cm – 1 cm – 1 cm = 8 cmWidthofthepainting=BC = 7 cm – 1 cm – 1 cm = 5 cmTheareaofthepaintingABCD=A2 = 8× 5 = 40 cm2
Theareaofthemargin=A1–A2 = 70 – 40 = 30 cm2
The area between the two concentric circles / Area of circular pathways
In thefigure, thesmallercirclehaving
OR
r
theradiusrisenclosedbycircularpathway.Thefigurehastwoconcentriccircles.Smallercirclehastheradius‘r’unitsandthelargercircle has the radius ‘R’ units. The regionbetweenthetwocirclesiscalledthe circular pathway.
Theareaofcircularpathwayisequaltothedifferenceofareaoflargerandsmallercircles.
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Area of circular pathway
= Area of thelargercircle
–Areaofthesmaller circle
= πR2 - πr2
∴Areaofthecircularpathway
= π(R2 - r2)
∴ Areaofthecircularpathway= π(R2 - r2)
Example 1 : Iftwoconcentriccircleshavingradii10 cm and 6cmformacircularpath,findtheareaofthecircularpathway.(π = 3.14) Theradiusofthelargercircle =R=10 cm
Theradiusofthesmallercircle =r=6 cmAreaofthecircularpathway =π(R2 - r2) = 3.14(102 – 62) = 3.14(100 – 36) = 3.14 × 64
∴Theareaofcircularpathway =200.96 cm2
Example 2 : Theradiusofacircularshapedpondis7m.Findthecostofcementingthecircularplatformofthewidth3.5 m aroundthepondattherateof` 25 per m2. Unitcostofcementingaroundthepond=` 25 per m2.
Theradiusofthepond =r=7 mTheradiusoftheoutercircle =R=7+3.5 = 10.5 mAreaofthecircularplatformaroundthepond=π(R2 - r2) = 7
22 (10.52 – 72) =
722 (110.25 – 49)
= 722× 61.25
= 22 × 8.75 = 192.5 m2
∴Areaofthecircularplatformaround the pond = 192.5 m2
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Thecostofcementingthecircularplatform=Areaofthepathwayxunitcost =192.5 × 25
= 192.5 × 25
= ` 4,812.5
∴Thecostofcementingthecircularplatform=` 4,812.5
Exercise 7.5
1) Aslateis30cmlongand25cmwide.Itissurroundedbyawoodenframeofwidth2.5cmallaround.Johncolouredtheframe,donotconsiderthethicknessoftheframe.Findtheareaofcolouredframe.
2) Apathofuniformwidth1.5mrunsaroundtheinsideofarectangularplotof80mlongand60mwide.Calculatetheareaofthepatharoundtheplot.
3) Apathofuniformwidthof m121 runsaroundtheborderof
arectangulargardenof55mlongand30mwide.Calculatethecostoflayingthestoneslabinthepathwayattherateof ` 30 per m2.
4) Calculatetheareaofthecircularpathwayformedbytwoconcentriccircleshavingthefollowingradii.
a)R=20cm,r=6cm c)R=42m,r=10.5 m
b)R=15.5m,r=5.5m d)R=18.5cm,r=3.5 cm
5) The running track is enclosed between two concentriccircles of radii 42 m and 49m.findthecostofcementingthetrackattherateof` 25 per m2.
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Points to Remember Perimeter:Thedistancecoveredalongtheboundaryof
theclosedfigureiscalledtheperimeter.ItisdenotedbyP.
Perimeteroftherectangle=p=2(l+b)units,wherel denoteslengthoftherectangleandbdenotesbreadthoftherectangle.
Perimeterofthesquare=4aunits,where“a”islengthofsideofthesquare.
Area:Thesurfaceenclosedbyaclosedfigureiscalledarea.AreaofaclosedfigureisdenotedbyA.
Areaoftherectangle=A=lb Areaofthesquare=A=a2sq.units. Theperimeterofascalenetriangle=P=a+b+c,where
‘a’,‘b’,and‘c’arelengthofsidesofthescalenetriangle. Theperimeteroftheequilateraltriangle=3a,where‘a’
islengthofeachsideofanequilateraltriangle. Theareaofatriangle=2
1 bh,where‘b’isthebaseand‘h’istheheight/altitudeoftriangle.
Thecircumferenceofthecircle=2πr. Theareaofthecircle=πr2. Areaofthecircularpathway=π(R2 - r2) Approximatevalueofπ = 22
7 ≈3.14
Unitsofmeasurementsofarea •1 m2 = 100 dm2
•1 m2 = 10,000 cm2
•1 m2 = 10,00,000 mm2
•1 dam2 = 100 m2
•1 hm2 = 10,000 m2
•1km2 = 10,00,000 m2
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CHAPTER- 8
DATA HANDLINGAfter studying this chapter you : explain data collection and organise data in a systematic
manner, understand the 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.
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.
Name of the activity.
Roll Number of student interested
Total Number of student interested
1) Singing2) Dancing3) Carrom Board4) Chess5) Kho - Kho6) Essay Writing7) Cricket8) Foot ball
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1) How many students are interested in singing?2) In which activity are maximum number of students
interested?3) In which activity are minimum number of students
interested?4) How many students are interested in more than one
activity?Activity 2 : Observe the table given below, and then answer the questions Temperature of the district head quarters in Karnataka as on 25/1/2012 is given belowName of the District Max. temperature Min. temperaturekalaburgi 360C 250CTumkuru 320C 240CShivamogga 280C 210CMadikeri 260C 200CChikkamagaluru 280C 220CKolar 290C 260CMysuru 300 C 270C
1) In which district is the highest maximum temperature recorded?
2) In which district is the lowest minimum temperature recorded?
3) Name the districts in which same maximum temperature is recorded
4) Find the difference between the maximum and minimum temperature of each district
Do This : 1 : With the help of your teacher or elders Collect the tem-perature data of all the days of any one month in this year.2 : Collect some data regarding sports from news papers3 : Try to analyse and conclude from the above data.
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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 Q R S T
Count the beads inserted in each rod and fill in the blanks given below.
P has 4 beads.
Q has ________ beads.
R has ________ beads.
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.
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 T
The rod P has 8 beads.
The rod Q has 6 beads.
The rod R has 4 beads.
The rod S has 4 beads.
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?
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P Q R S T P Q R S T
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
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 Q R S T
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Let us tabulate our observation of the above activities.
Activity
Number of beads in the rods.
Rod P Rod Q Rod R Rod S Rod T
4 8 6 4 4 3
4(after
transfer)7 6 5 4 3
5 5 5 5 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.
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 .
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 4, number of beads are 8, 6, 4, 4, 3 respectively.Average Number of rods
Sumof the beads=
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558 6 4 4 3
525= + + + + = =
∴ the average of the beads is 5.................. (1)In Activity 4, (After transfer) the number of beads are 7, 6, 5, 4, 3
respectively.Average Number of rods
Sumof the beads=
57 6 5 4 3
525 5= + + + + = =
∴ 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∴ Average = 55
5 5 5 5 5525= + + + + = =
In activity 4 we notice that median is 5 and in activities 4 (After transfer) 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.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
Station Name
Maximum temperature
in 0C
Minimum temperature
in 0CNew Delhi 34 26
Kolkata 33 26
Mumbai 31 26
Chennai 33 24
Pune 28 21
Hyderabad 25 21
Trivandrum 27 22
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What are the central tendencies of the above data?Arrange the above data in two tables in ascending order
as shown below
Station'sName
Maximumtemperature
in 0C
Station'sName
Minimumtemperature
in 0C
Hyderabad 25 Hyderabad 21
Trivandrum 27 Pune 21
Pune 28 Trivandrum 22
Mumbai 31 Chennai 24
Kolkata 33 Kolkata 26
Chennai 33 Mumbai 26
New Delhi 34 New Delhi 26
From the above tables, find the mean for maximum and minimum temperature separately.
Mean for maximum temperature
=
.
max
C
Number of StationsSumof all the imum temperature
725 27 28 31 33 33 34
7211 30 140= + + + + + + = =
The Mean maximum temperature of the 7 stations is
= 30.140C.Similarly, mean for minimum temperature.
.
min t
C
Number of StationsSumof all the imum emperature
721 21 22 24 26 26 26
7166 23 710
=
= + + + + + + = =
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To find the median of above given temperature arrange data in ascending order.
maximum temperature
minimum temperature
25, 27, 28, 31, 33, 33, 34
21, 21, 22, 24, 26, 26, 26Since 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 = 310CThe Median of the minimum temperature = 240CNext, to find the Mode round off the most occurred
(repeated) number in maximum and minimum temperature. maximum temperature
minimum temperature
25, 27, 28, 31, 33, 33, 34
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).
Measures of Central Tendency 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
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(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.
MeanLet us take a situation, where Manaswi and Harish received
their unit test marks sheet.
The test has 5 subject, each carrying ten marks and their scores were given in the table
Subject Number. 1 2 3 4 5
Manaswi 9 10 8 6 9
Harish 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.
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.
Manaswi's average score : 9+10+8+6+9 = 542 = 8.4
Harish's average score : 6+5+10+10+10 = 541 = 8.2
Since, Manaswi’s average score is more than that of Harish, teacher convinced Harish that her performance in this test is better than him.
ArithmeticMean Number of quantitiesSumof all the given quantities
=
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Average or Arithmetic mean is a Central Tendency, which is obtained by the sum of scores divided by the number of scores.
Arithmetic mean = .....n
x x x xn1 2 3+ + + +
Mean nnx=
/
Where '∑' is a Greek letter used as a symbol to represent the sum of numbers and is read as 'SIGMA'.
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 April May June July August September
Supplied Milk in liters
215 218 314 340 420 410
Total milk supplied to the dairy in 6 months
= 215 + 218+ 314 + 340 + 420+ 410 = 1917
∴ Mean of the milk supplied= supnumber of monthstotal milk plied
319.56215 218 314 340 420 410
61917 0= + + + + + = = l
Example 2: The water consumption of eight different families is as follows. Calculate the mean consumption of water by the families in a month.
Families 1 2 3 4 5 6 7 8
Water Consumption in kiloliters
7.5 8.2 7.5 20.00 9.10 4.55 6.62 8.24
Total number of families = 8
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Total water consumption in kiloliters
= 7.5 8.20 7.5 20.00 9.10 4.55 6.62 8.24 .71 71+ + + + + + + =
Mean Number of families
Total water consumption by the families=
. 8.968
71 71 = kiloliters
Therefore, the Mean of water consumption by the families is 8.96 kiloliters. That is 8960 litres. 1Kl = 1000l
Try This : How would you find the average of your study hours in a week?
Exercise: 8.1I. 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 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 eachproject.
1) Will the average of the marks change after the addition of the 3 marks for each score?
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2) Calculate the mean of the scores in both the situation.V. The runs scored by two batsmen in the six matches
are given below: Who performed better?Batsman -1 60 65 100 45 50 70Batsman -2 30 100 94 60 50 40
VI. Find the missing number in each of following data : 1) 21,25,29,____31,33 and the mean is 28. 2) 41, 43,____,47, 49,51 and the mean is 46.
MedianConsider this example, the salary of 10 workers in a factory
is as follows.
Staff A B C D E F G H I J Total (10)Monthly salary
` 90
00`
5000
` 70
00`
8000
` 75
00`
6000
` 18
000
` 15
000
` 12
000
` 45
000
` 13
2500
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.
B F C E D A I H G J
5000
6000
7000
7500
8000
9000
1200
015
000
1800
0
4500
0
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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.
/
, ( ) kersth th
Thus middle value salary of the wor
25 6
28000 9000 8500= + = + = -
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 = Sumof the twomiddle scores2
orMedian = n
2n2
* For the given data, if the number of terms is an odd
number then the Median n + 12
term.
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 stands 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.
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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.
Example 2: 15, 21, 22, 18, 20, 17, 12, 11, 10, 8, 7, 19. Find the Median.
Arranging in Ascending order: 7,8,10,11,12, 15,17, 18,19,20,21,22
Here n = 12, which is an even number.
Therefore, Median = n2
n2 term
= (6th term) + (7th term)
2
= (15) + (17)2
= 16
∴ Median of the data = 16
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 is 24.
∴ Median =
n2
2 =
5th term + 6th term2
24 = (2x + 10) + (2x - 6)
2 48 = 4x + 4
4x = 48 - 4
4x = 44
n + 12
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x = 444
∴ x = 11
(i) 2x + 10
= 2 (11) + 10 = 32
(ii) 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 12,10,8,15,____,20,24,29.
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
18 2 1536 15
36 1521
x
xx
xx
2`
#
= +
= += += -=
Exercise : 8.2
I. 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, 5,2,3,7,11.5) 36, 32, 28, 22, 26, 20,18,40.
II. In a cricket match, 11 players scored runs as follows:
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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?
Mode:ACTIVITY 1: The class attendance of 10 students in 7th standard is given below in Table 1.
Table 1 Table 2Name No. of attendance
(for 30 days)Name No of attendance
(for 30 days)Suhasini 25 Rema 10Raja 15 Raja 15Rohit 30 Sanjana 17Gulabi 25 Francis 18Hussain 28 Sandeep 18Rema 10 Suhasini 25Sanjana 17 Gulabi 25Fransis 18 Karthik 25Sandeep 18 Hussain 28karthik 25 Rohith 30
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.
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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 week5 10
6 25
7 36
8 22
9 9
From the above table what do you observe?From the above table it is observed that, number of shoes
of size 7 is sold the most.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.
The "Mode" is the value that occurs most frequently or most often in the data.
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 38 cm, 36 cm, 39 cm, 40 cm, 42 cm.
The following is the record for the week
Size 36 cm 38 cm 39 cm 40 cm 42 cmNo. of shirts sold 8 22 32 37 6
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.
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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 occurs most often?
7, 13, 18, 24, 9, 3, 18 Arranging the score from least to greatestWe get; 3, 7, 9, 13, 18, 18, 24. The score which occurs most often is 18.∴ The Mode is 18.
Example 3 : For what value of 'x' is the Mode of the following data is 9.
17, 9, 12, 17, 18, 11, 19, 21, 9, x- 4Here 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 = 9Therefore x - 4 = 9 x = 9 + 4 x = 13
Exercise: 8.3
I. Find the Mode of1) 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; findtheMode.
168, 165, 163, 160, 163, 160, 161, 162, 164, 163,160, 163, 160, 165, 163, 162, 163, 164, 163, 160,156, 165, 162, 168, 168.
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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.
Graphical representation of the data:You have studied how to represent data in pictograph as
well as bar graph.Name the following graphs:
Banana
Orange
Apple
Mango Grapes
Papaya
fig (i)
Activities
Dan
cin
g
Kh
o-kh
o
Kab
badi
Dra
win
g
No.
of S
tude
nt
510
15
20
2530
35
40
fig (ii)
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.
Look at the graph given below: It shows the number of plants planted in successive weeks.
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12001000
800600400200
1st W
eek
2nd W
eek
3rd
Wee
k 4th
Wee
k
5th W
eek
num
ber
of
plan
ts p
lant
ed
number of weeks
By observing the graph find the number of plants, planted in each week.
Week 1st 2nd 3rd 4th 5th
Number of Plants 1000
In which week are the maximum number of plants planted?In which week are minimum number of plants planted.
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 to 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 :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.]
1cm
Y
X
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Step 2: Along the horizontal l ine (X – axis), choose uniform width of the bars with uniform gap between them to represent the subjects. [Each subject = 1cm]
Y
X1cm
x axis
y ax
is
Step 3: On the vertical line (Y – axis), we have to represent the marks obtained. 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 = cm5
1
Step 4: Calculate the heights of various bars as shown below (along Y axis).
The height of the bar for Kannada = cm51 20 44# =
The height of the bar for English = cm51 15 3# =
The height of the bar for Hindi = cm51 10 2# =
The height of the bar for Mathematics = cm51 25 5# =
The height of the bar for Science = cm51 20 4# =
The height of the bar for Social Science = cm51 15 3# =
Bar Graph of Bhavana's Score
Mar
ks s
core
d→
Subjects →
3025
2015
10
50 English Hindi Maths Science S.ScienceKannada
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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:• Every bar graph should clearly show what it represents.• The bars should be neatly drawn and neatly coloured
or shaded.• 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.
StateAndhra Pradesh
Maharashtra Karnataka KeralaTamil nadu
No. of Athlets
45 65 40 75 30
Method: To draw the bar graph, the following steps are to be followed.Stage 1: Draw two lines on a graph paper, which are perpendicular to each other and mark them as X-axis, and Y-axis.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.
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Therefore, 1 athlete = 1/10 cm.The height of the bar representing Andhra Pradesh is = 45 . cm10
1 4 5# =
The height of the bar representing Maharashtra is = 65 6.5 cm10
1 # = The height of the bar representing Karnataka is = 40 4 cm10
1 # = The height of the bar representing Kerala is = 75 7.5 cm10
1 # = The height of the bar representing Tamil Nadu is
= 30 3 cm101 # =
Stage 5 : Draw bars of equal width and of height calculated in the stage 4.
0
10
20
30
40
50
60
70
80
An
dhra
Pra
desh
Mah
aras
htr
a
Kar
nata
ka
Ker
ala
Tam
il N
adu
State
Nu
mbe
r.of
ath
lete
s
Exercise: 8.4I. The strength of the different classes of Government
higher primary school are given below.
class 4 5 6 7 8strength 90 70 50 40 40
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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 studentsPlaying 35
Reading story books 10
Watching Television 20
Listening to Music 5
Painting 30
III. Read the following bar graph and answer the following questions:Number of students who went abroad for studies
2005 2006 2007 2008 2009 20100
1000
2000
3000
4000
5000
Year
No.
of s
tude
nts
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1) In which year maximum number of students went to study abroad?
2) In which year minimum number of students went to study 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)
Items Expenditure House rent 3500
Food 4000
Education 800
Electricity 1200
Transportation 1000
Miscellaneous 2000
Double Bar Graph (Composite Bar Graph):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.
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Subject Marks of Ramesh Marks of Satish
Kannada 90 80
English 85 70
Hindi 70 90
Mathematics 100 80
Science 90 60
Social Science 80 55
The graphs below represent the marks scored by Ramesh and Satish:
Ramesh Satish120
100
80
60
40
20
0
Kan
nad
a
En
glis
h
Hin
di
Mat
hs
Sci
ence
S.S
cien
ce
Subjects
Mar
ks
120
100
80
60
40
20
0
Kan
nad
a
En
glis
h
Hin
di
Mat
hs
Sci
ence
S.S
cien
ce
Subjects
Mar
ks
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:
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Eng Sci S.SciSubjects
Mar
ks
Kan Hin Maths0
2040
6080
100
120
Ramesh Satish
When we want to compare the same quantity between two different data, we use composite bar graph.
Exercise : 8.5
I. The following table shows the education level of men and women in Karnataka. Represent it in suitable composite graph.
Education level 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
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II. Mysuru Mahanagarapalike spent various amounts 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.
Subject 2010-11 Expenditure
(in lakhs)
2011-12 Expenditure (in lakhs)
Education 60 70
Water supply 35 45
Medical aid 25 30
Transportation 45 50
Sanitation 15 22
Road construction 30 35
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?
Know this : The mode is useful when the most common item, characteristic or value of the data is required.
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CHAPTER - 9
PROBABILITY
After studying this chapter you : explain the meaning of probability, definethemeaningofrandomexperiment, predict the outcome of random experiment, represent outcomes of an event in the graph.
IntroductionIn our day to day life, we come across many situations where
uncertainty plays a vital role. We usually use statements like “thereisachanceofraintoday”or“probablyhewillgetfirstclass 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, inthecaseofastudentgettingfirstclassprobablyindicatesuncertainty.Let us see a few more statements :• Headmasterdoubts about the Kaveri's promotion in the
examination.
• Opinion of the audience is “most probably, Kevinwillstandfirstinthecompetition”.
• ThereisPossibility of increase in cost of vegetables due to hike in price of diesel.
• Thereisa50-50 chance of a school team winning the trophy.
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Thewordsuchas‘probably’,‘doubt’,‘possibilities’,‘chance’,etc., used in the above statements indicate an element of uncertainty.
The word probability is related to the occurrence ofuncertainty.
Probability has been extensively used in the fields ofweather forecasting, share market, commerce, medical sciences, biological sciences, physical sciences etc.
Therefore, ‘probably’ or uncertainty is measurednumerically.Thewordforsuchmeasurementsis‘probability’.Theconceptofprobabilitycanbequantified.
Random ExperimentNow let us consider a situation of the cricket match. Before
the commencement of the match captains of both the teams decide,'Whowillbatfirst'?bytossingacoin.Captainofteam'A' tosses the coin. Suppose you are the captain of team 'B' and choose'Head'tocomeup,areyousurethat'Head'willcomeup?Ifyouchoose'Tail'tocomeup,doesithappen?ExactlyIt is not predictable. Result of the toss cannot be controlled. the chances are 50 : 50.
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 thecolourofthechosenball?No,itisnotpossible.
Similarly, can we predict what will be the number of dots ontheupperfaceofarolleddice?Itwillbeanyoneof1, 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 arecalledRandomExperiments.
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A possible result of a random experiment with a dice is shown below.
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.Theresultofnumber 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 andTrepresentstail.
In the above random experiments the possible outcomes arewrittenusingflowerbrackets{ } and is denoted by S.
∴ 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.
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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.
Exercise 9.1
1) If you spin a coloured disc containing the colours Yellow, Blue, Red, and Green, It stops at the pointer after some time. Write the samplespaceforthecolouratwhichitstops?
2) You spin the wheel shown: make a list of all the possible outcomes of the number at whichpointerstops?
3) Tosstwocoins.Writeallpossibleoutcomesassociatedwith this experiment. S=(H1H2),(H1T2), ___, ____}
4) Fiveflashcardshavingoddnumbersfrom11areshuffledwell and one card is drawn from it. Write the sample space.
Event:
Considertherandomexperimentofadicehavingsixfacesthrown once.
Thepossibleoutcomesfornumbersontopofthediceare{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}.ThisoutcomeisanexampleofanEvent.
Similarly, if we restrict the outcome of the experiment to be a prime number then B = {2, 3, 5}. ThisisalsoanEvent.
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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
Tounderstandthenotionofchance,letustaketheactivityof '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.Sl no Number of
times tossed n (S)
Head comingup n(H)
Tail comingup n(T)
n(H)n(S)
n(T)n(S)
1 15 10 5 1510
155
2 30 18 12 308
3012
3 50 35 15 5035
5015
4 70 40 30 7040
7030
5 80 45 35 8045
8035
6 90 47 43 9047
9043
7 100 52 48 10052
10048
For trial 1, the ratio of getting head turned up to the total
number of times a coin is thrown is n(H)n(S)
= 10
15 Similarly, for
trial 2 it is n(H)n(S)
= 18
30 and for trial 3 it is n(H)n(S)
= 35
50 and so on.
Whatdoyouobservefromtheabovetable?
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Youwillfindthatasthenumber of tosses increased, the
value of the n(H)n(S)
or
n(T)n(S)
come closer to 2
1 or 0.5
Thismeansthat,chanceofgettingheadortailwhena
coin is tossed is 50:50. It also means the probability of getting
head up when a coin is tossed, is 21 or 0.5
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 updot 1 dot 2 dot 3 dot 4 dot 5 dot 6
15
30
90
100
120
Using the above table write the observation in the form of
fraction. Number of times dice thrownNumber of times one dot come up for all the trials.
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 6
1 or 0.167.
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Probability Total number of eventsPosibility of an event
=
Soprobabilitymaybedefinedastheratioofnumberofexpected or favorable outcomes to number of all possible outcomes or number of sample space.
Probability Number of all the possible out comesThe number of favourable out comes=
Example 1 : A dice is rolled, calculate the probability of getting an even number on the top.
Thedicehassixfaceseachnumbered(marked with dots) with 1, 2, 3, 4, 5, and 6.
Theoutcomesofthisexperimentare{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 =
Total number of all possible out comesNumber of favourable out comes
63
21= =
Example 2 : A bag contains, two of each blue, red, green and yellow coloured balls. Randomly one ball is picked, what willbetheprobabilityofgettingablueball?
Totalnumberofballsinthebag = 8
Expectedoutcome of blue coloured ball = 2
Theprobabilityofgetting blue coloured ball Total number of balls in the bag
Number of blue balls in the bag82
41= ==
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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 itsedge,whatistheprobabilityinsuchcases? Such cases, will not be considered under probability
Exercise 9.2
I Write the possible outcomes of rolling a dice. Write the probability of getting the following numbers
1) A number greater than 5.
2) A even number
3) A odd number
4) A number less than 5.
II 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) Number 7.
b) Number greater than 7.
c) Number less than 7.
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.
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Representing the result of a random experiment in graph:
1. Rama tossed a coin 20 times, the outcomes of this experiment is listed below;
Whenacoinistossedonce,thepossibleoutcomesareHeadorTail.
Herethesamplespaceforthisexperimentis{H,T}
Sl. No. Outcome Sl. No. Outcome
1 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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The above data is tabulated as followsOutcomesofanexperiment Number of times it occurred
H 12
T 8
Theabovedatacanberepresentedonagraph,takingeventof an experiment on X – axis, and the number of outcomes on Y- axis,
Scale: X- axis 1 cm, = 1 units. Y-axis 1 cm, = 2 units.
Outcomes
Length of each bar
Head 6.0 cm
Tail 4 cm
Numberofevents→
0
42
6
10
14
8
12
16
H
T
Event→
2. When a dice is rolled 60 times the result of each roll can be recorded as shown in the table :
Score on Dice Tally Frequency1 10
2 6
3 13
4 12
5 8
6 11
TotalFrequency= 60
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Now let us draw a bar graph to illustrate this :Scale :
OnXaxis1 cm = 1 unitOnYaxis1 cm = 1 unit
10
5
10 10
13 12 11
86
15
2 3 4 5 6
Frequency
→
Numbers on the dice →
Exercise 7.3
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 100times.Tablegivenbelowshowstheresultsof 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 C100 26 10 14 20 12 18
2) A disk wheel is divided into 5 sectors and a pointer is placed infrontofit.Then,itisrotatedfor75times.Tablegivenbelowshows the result that the position of the pointer against the numbers. Draw a bar graph to represent the results.
Number of times disk wheel is spin
Thepositionofthepointeragainstthe numbers
1 2 3 4 575 6 14 26 19 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.
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CHAPTER -10
REPRESENTATION OF 3 DIMENSIONAL OBJECTS IN 2 DIMENSIONAL FIGURES
After studying this chapter you : explain the meaning of 2 dimensions and 3 dimensions, drawlinediagramofsolidfigurescube,cuboidand
tetrahedron, recognisevisibleandhiddenfacesofsolidfigures, drawthediagramsofcubeandcuboid, explain the net of cube and know themethod of
drawingthenetsofcube,cuboidandtetrahedron, imaginetheimagesofsolidfiguresmentally,knowthe
numberofedges,facesandvertices.
IntroductionWeknowthatthe'point'isthesmallestfigureinGeometry.
Pointhasnodimension,butithasaposition.Welocatethiswiththesharpesttipofapencil.
Alinecontainsinnumerablepoints.Ifpointsarecollinearthenthelineiscalled'straight line'.Apartofastraightlineiscalledlinesegment. Alinesegmenthassingledimension.(Onlylength)
Aplaneisaflatsurface. Linesegmentswhenjoinedinaplaneformgeometrical
figures. Squares and rectangles are the geometrical figures
whichare formedby the line segments on the sameplane.
Arectanglehas two dimensions (lengthandbreadth).
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Representation of 3 dimensional solids in 2 dimensional figures.
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. Ithasonlytwodimensions,isn'tit? Placetheotherfacesofthesamematchboxonasheet
ofwhitepaperanddrawtheborderlinesasbefore.
Matches Box
fig:1
Matches Box
Borderlinesoffaces
Observe that each surface of the match box has two di-mensions.
Thefiguredrawnfromborderlinesofthematchboxhasnothickness.Butmatchboxhasthickness.Therefore,thicknessisanotherdimension.Hencematchboxhasthreedimensions.Thatislength,breadthandheight.
Objectshavingthreedimensionsarecalledsolidobjects.Solidobjectsoccupyonepartofthespace.
Ashadowisformedwhenthelightfallsontheopaqueobjects.Theshadowsformedbyusingourhands,mayappearas,shadowofflyingbird,dog,rabbitetc.Ourhandisthreedimensionalbuttheshadowsareoftwodimensions.Formingshadowsbyusingdifferent solids figures is a type of entertainment.Here, threedimensionalobjectsareshown in two dimension in different types.
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Observetheseshadowsformedbyhandsandtrythis.
fig:2
fig:3
These are the figures of solid models made by wood.
These figures are called solidfigures. They arenamedbasedontheirshapes.Solidobjectsarecalledthreedimensionalobjects.Wecanseeandalsotouchtheexteriorpartsofthesolids.Theexteriorpartsarecalledsurfacesof solids. Solidsmay have flat
surfaceaswellascurvedsurface.
Faces, edges and vertices of solids :Cuboid :
A BF
GED H
C
fig:4 fig:5
Abrickisanexampleforacuboid.Compare thiswithmatchbox.Abrickhassixplanesurfaces,eachplanesurfaceiscalledplaneface
or,simply,face.Lookatthefig(4).Itisabrick.Wecanseethetopandtwosidefacesofitfromonedirection[DHGEatthetop,ABHDandBFGH,atthesides].Theotherthreesides[ABFC at the bottom, ACED andCFGE, at the sides] arehidden.
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Activity 1 : Matches Box 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
A B
F
GE
D H
C
Number of faces of this solid isalso six. But, the shape of theeachfaceisasquare,i.e,length,breadth,heightareequaltoeachother.Thisiscalledcube.ABHDis one of its faces. Name theremainingfacesofthecube.
Letusseehowmanyadjacentfacesarejoinedtogether.Inthissolidfigure,foreachface,therearefouradjacent
faces.Howaretheadjacentfacesjoinedtoeachother?Thesefacesarejoinedbylinesegments.ABHD and BFGH are joined by line segment BH.
Similarly,linesegmentBFjoinsthefacesBFGH and ABFC.Thelinesegmentwhichjoinstheadjoiningfacesiscalled
'Edge'.Cubehassuch12linesegmentsoredges.IfBH and BF are the other two edges, name the remaining
edges.
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Nowobservehowtheseedgesmeeteachother.Forexampleedges HB and FBmeeteachotheratthepointB.SimilarlyedgeAB also meet at the point B.
Inasolidfigure,thepointatwhichthreeormoreedgesmeetiscalledvertex.EdgesAC,EC,CF meet at C.Therefore,point Cisthevertex.B and Carethevertices.
Inthesamewaynametheothervertices.
Bythisyoucanunderstandthattheabovesolidfigurehas 6faces,12 edges and 8vertices.
Activity :Draw the figures of cuboid and cube. Name the faces, edges and vertices. Write the visible and hidden faces.
Triangular based Pyramid
Q R
P
S
Constructequilateraltriangleofconvenientmeasurementonapaper.Letitbe∆PQR.
Insidethe∆PQRmarkapoint'S' on a paper.JoinPS, QS and RS.
Nowyouwillgetadiagramasshowninthefigure.
This figure is 2D figure of the solidobjectcalledTriangularbasedPyramid.
Youknowthis:
AsolidfigurehavingmanyfacesiscalledPolyhedron.Polymeansmanyandhedronmeansface.PolyhedraispluralofPolyhedron.
Note: VERTEX -singular,VERTICES-plural
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Lookatthetriangularbasedpyramid.
Q
P
R
SHowmanyfaceshasit?
Howmanyfacesarehiddeninthis?
Countthenumberofedgesandnumberofvertices.
This solid has 4faces,InthisQRSisbase,andtheotherthreearesidefaces.(Adjacentfaces)
Inatriangularbasedsolidifallthefacesareequilateraltriangles,thenitiscalled Tetrahedron.IntheabovefigurewecanobservePQ, QR, RP, PS, QS and RS
are edges and P, Q, R and Sarevertices.
Therefore, tetrahedron has 6 edges and 4vertices.
Square based Pyramid
Lookatthisfigure.
Hereisafigureofapyramid
Thebaseof this isasquareandothersidesare in theshapeofisoscelestriangle.
Ifthebaseofthepyramidisasquare,thenitiscalledainsquarebasedpyramid.
Fromtheabovefigure,weobservethatasquarebasedpyramidhas5faces,8 edges and 5vertices.
Similarly,ifthebaseofthepyramidisapentagonthenitiscalledapentagonalbasedpyramid.
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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.
Name of the solid
Numberof edges in the base
Numberoffaces
Numberofvertices
Total numberof edges
Tetrahedron
SquarebasedPyramid
Pentagonal basedpyramid
Hexagonalbasedpyramid
Inthesamewaygeneraliseton-basedpyramid.
Cylinder
Lookatthisfigure,itisacylinder.
It has 3 faces, one is curved surface and theothertwoarecircularplanesurfaces.Circularplanesurfacesarejoinedwithcurvedsurface,itsedgesare
circular.Dotheedgesmeeteachother?
No.Therefore,cylinderhasnovertex.
Inacylinderonecircularfaceandapartofcurvedsurfacearevisible.
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Cone Observethediagramofthecone.
Aconehastwosurfaces,inwhichoneiscurved,thatendsatapoint,andanotheroneisplane.Inaconebaseandapartofcurvedsurfaceishidden.
Complete the following table.
Figureofthesolid
Numberoffaces
Numberofvertices
Numberof edges
Numberof hidden
faces
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Drawing of Polyhedral nets for making solid models:Sushruthawants to have a
5 cm 5 cm 5 cm
5 cm
5 cm
Fold
FoldFold Fold
smallcardboardboxandliketo keephis pen, sharpener,andrubberinit.Soheaskedhis grand father to help him to do one. His grandfathertook(30cm × 25cmdimension) cardboardsheet,pen,pencil,scissors and gum. He toldSushruthatoobservehowhewouldmakeabox.HisGrandfathersaidnowIdrawasketchfortheboxrequired.Since,thelengthoftherubberis3cm.Let the edge of theboxbe5cm.Thenhemakes aboxbyfollowingprocedure.
Step 1:Hedrewarectanglewithdimension(20cm × 5cm) onthecardboardandmarksthepartsasshowninthefigure.
Step 2:Grandfatherfoldsthedottedlines,andpastetheshadedpartswithgum,exceptpartialshadedpart.Nowaboxisready.
Tejaswisaidhewantedaboxhavingitslength10cm.byseeingthisbox.ThenGrandfatherdrewtheskecthasshownbelow.
5cm
10cm
10cm10cm10cm
Fold
FoldFoldFold
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Thenhepreparedaboxbyfolding along the doted line on thesheetandpastedonshadedpart.BothSushruthaandTejaswiwerehappywiththeirboxes.
Thistypeofsketchtobuilt3dimensionalobjectsiscalledpolyhedralnets.
So, a net is a 2dimensionalfigurethatcanbefoldedtoform a 3dimensionalobject.Hereeachfaceof3 dimensional figureisrepresentedas2dimensionalfigure.Nets for some solid are given below.
Nets Nameofthesolid/figure
a
a
a
Cube
Triangularbasedpyramid
1
3
4 2
Squarebasedpyramid
1
324
5
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Triangularbasedprism
214
5
3
Cuboid
A C CA
B
B
Easy way of drawing solid figures on a flat Surface
Inpreviousactivitiesfiguresofsolidsaredrawnonflatsurfaces.Theyappearas3-Dfigures.
Amongthefacesofasoliddrawnonasheetofpaper,somefacesareexactlyhavingthesameshapeasthatofthesolidface,butnotallthefaces.Eventhoughthethreevisiblefacesarenotsimilar,wecanrecognisethosefiguresasasolidandfindoutwhattypeofsolidtheyare.Suchsketchesaretermedasobliquesketches.
Topracticeobliquesketches,therearesomeeasyways;suchas
i)Usingsquarelinedordottedpaper.
ii)usingIsometricsheetsoflinesordots.
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Know this : Inasquarelined,sheeteachsmallboxisasquare.Inasquarelineddottedsheetwecangetasquarebyjoiningdots.Byjoiningthedotsonaisometricdotsheetwegetanequilateraltriangle.
Observe the following figures :SquarePaper IsometricPaper Isometricgraph(grid)
Note: Isometricdotsheetsareusefultodrawsketches,inwhichmeasurementsalsoagreewiththoseofsolid.
Activity: How to mark 4 × 4 × 4 unit solid (cube) on square lined sheet? Take a square lined sheet.
step 1: Drawthefrontfacethatis4unitssquarewithpencil.
step 2:
Draw the opposite facewhich is alsosquareof4units.Todothis,markthemidpointofthefrontface,mark4unitstowardsrightsideofthisandcompletethesquareasshowninfig.
step 3:
Jointhecorrespondingcorners.step 4: Retracethesketchwithpen.
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Note : It is conventional to mark hidden edges by dotted lines.
Similarlydrawasketchofcuboidof6 × 4 × 4unit.
step 1: TakeSquarelinedsheetofpaper
step 2 : Draw a rectangle measur ing 6 × 4
step 3 :Drawonemorerectanglebymarkingthe center of first rectangle of 6× 4asinperviousactivity.
step 4 : Jointhecornersandretraceit.
Know this: Thetermisometricreferstoequalmeasure.Anisometricsheetconsistsofequilateraltriangles.Thelineson the paper are in the three directions, each of whichrepresentsadimension.Thelinesupanddownrepresentthe vertical dimension (height). The othertwo dimension represent the horizontal dimension (length and breadth ).Therefore to draw a 3-Dfiguresweuseaisometricgraphsheet
Theisometricdotsheet,onwhichdotsaremarkedatequaldistances isalsousedtodraw3-Dfiguresuchsheetiscalledisometricdotsheet.
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Activity : Draw a sketch of a solid figure on an isometric graph (grid) and on an isometric dot sheet.
Considerasolidofmeasurement2 × 2 × 4units
Step 1: Takeanisometricgraph(grid)andanisometricdotsheet.DrawtheedgesAB and BConthebottomofthesolid,suchthatAB = BC = 2units
A
B
CA
B
C
Step 2: Drawverticallinesattheverticesofthebasesuchthat AF=BE=CD=4units.
A
F DE
BC
B
EDF
CA
Step 3: JoinEF and ED.
A
F DE
BC
B
EDF
CA
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Step 4:CounttwounitsfromD,E,FandGjoinFG and DG.Nowgothroughthelines-boldalongthevisibleedgesanddrawdottedlinesforhiddenedges.
A
F DG
E
BC
B
ED
GF
CA
Exercise : 10.1
I. Fill in the blanks.
Solid Faces Edges Vertices Shape of faces
a)Cuboid 6 12 8 Rectangle
b)Cube
c) Triangular based pyramid
d) Square based pyramid
e) Triangular based prism
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II. Answer the following questions.1) Mentionthenumberofcurvedsurface,curvededgeand
verticesinacone.2) Mentionthenumberofcurvedsurface,curvededgeand
verticesinacylinder.3) Mentionthenumberofcurvedsurface,curvededgeand
verticesinasphere.
III. Match the following.
Sl.No.
Name of the solid Nets of solids Answers
1 Cuboid a) 1
324
5
2 Cube b) A C CA
B
B
3 T r i a n g u l a rbasedPyramid c)
1
3
4 2
4 SquarebasedPyramid d)
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IV. Construct solids having the following dimensions on a squared sheet (check sheet).a)Cuboidhaving3 × 2 × 2unitdimension.
b)Cubehaving4 × 4 × 4 unitdimension.Sample sheetIsometric Grid
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Isometric Dot paper
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Chapter - 1 IndicesExercise1.1 : I 1) 8 to the power of 3 2)13 to the power of 6 c) 7
4 to the power of 10 d) 10 to the power of 4 e) (-6) to the power of 5 II a) 5 is index -3 is base number b) 8 is index - 10 is base number c) 6 index - 3
2-` jis base num-ber d) 20 index - 3 is base number III. a) 3 b) 4 c) 5,4 d) an
IV a) 38 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 b) 113 = 11 × 11 × 11
c) ( 52 )6
= ( 52 )×( 5
2 )×( 52 )×( 5
2 )×( 52 )×( 5
2 ) d) (1.5)6= (1.5) × (1.5) × (1.5) × (1.5) × (1.5) e) ( p
q )4 = ( p
q )×( pq )×( p
q )×( pq )
V a) (i) 92 (ii) 34 b) (i) 56 (ii) 254 c) (-3)5
Exercise 1.2 : I 1) 72+5 2) -35+3 3) ( 52 )3+6
4) 103+7+5 5) a6+4+10 7) 2.54+8 II. 1) 73 2) 37 3) 39 3) 214 III 1) 11 2) 19 3) a 4) 1Exercise 1.3 : I 1) 73 2) (-3)3 3) ( 5
2 )54) (8.5) 5) x8 6)1 7)
41 45
5= -
II 1) 5 2) 32 3) 25 III 1) 315 2)
101
7 3) a110 4)
x112
IV 1) 514- 2) 3-4 3)
217- 4) x -5
Exercise 1.4 : II 1) 324 2) 86 3) 1142 4) P qre) 224 5)( 23 )20
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
Exercise 1.6 : I 1) 1312
6
6
2) 514
5
5
3) 787
7
4) zx3
3
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) W 2) W 3) W 4) W 5) R 6) R 7) W
Exercise 1.8 : 1) Velocity of light = 3 × 108 m/s2) Velocity of sound = 343.2 m/s 3) Distance between sun and earth = 149600000, other astronomical distances = 1.496 ×109 m/s
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Chapter - 2 Ratio and ProportionExercise 2.1 : I 1) ` 60 2) ` 72 3) ` 200 4) 300 km 5) ` 360 6) 10 days 7) 3 days Exercise 2.2 : I. 1) 3 : 4 2) 7 : 8 3) 3 : 7 4) 1:5 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 : 2Exercise 2.3 : 1 Raju - ` 12,000, Dhamu - `9,0002. Females - 2,800 Male - 30803. Roshan - ` 1,200 Hameed - ` 1,6004. Copper - 150 g Zinc- 90 g 5. ` 300; ` 240 6.107 : 214 : 321Exercise 2.4 : I 1) 12 2) 8 3) 14 4) 6 5) 10 6) 20II 1) Yes 2) No 3) Yes 4) NoIII 1) ` 204 2) 30 kg IV 2 l V. 4 lExercise 2.5 : I 1) ` 140 2) 5 m 3) 15 m 4) 25 kgExercise 2.6 : I 1) 7 2
1 hour 2) 10 days 3) 6 days 4) 18 days Chapter - 3 PercentageExercise 3.1 : I1) 2
1 2) 41 3) 5
1 4) 101 5) 4
3 6) 81 7) 8
7 8) 83
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%Exercise 3.2 : 1) 25% 2) 10% 3) ` 34.11% 4) 25%Exercise 3.3 : 1) ` 108 2) ` 990 4) ` 420 5) ` 40,000
Chapter - 4 Simple Linear EquationExercise 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 RHS1) 1) x - 5 8 2) 3y + 6 -9 3)14 - k 2k + 4IV) 1) x = 13 2) x = 48 3) y = 35 4) p = 6 5) m = 5V) x + 9 = 15 ii) 8x -2x = 18 = x = 3 iii) x + 15 = 35 x=20Exercise 4.3 : I 1) x = 7 2) x = 21 3) y = +3 4) k = -3 5) m = 15 6) p = 12 7) k = 4
5 8) x = 7105 9) x = 310) x = -4
11) x = 15 12) x = 2 II 1) Equation x+4x6 = 46 or x = 22
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2) Equation 4 15x7 + = or x = 77 3) x = 45 year 4.140
Chapter - 5 Unit CongruenceExercise 5.1 : I i) c ii) c iii) d III DF
Chapter - 7 Geometrical ConstructionsExercise 7.1 : I 1) 28 cm 2) 10 m 3) 28 mII 1) 24 m2 2) 87.5 m 3) 22.75 m2
III 1) 24 cm 2) 60 m 3) 22.4 cmIV 1) 36 m2 2) 144 cm2 3) 96.04 cm2
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) 4.208) a) Area will be doubled b) same as it is c) double 4 times9) a) area is decreased by 2 timesExercise 7.2 : I 1) 24 cm, 18 cm, 16 cm 2) 20 cm3) a) 24 cm b) 39 cm c) 33 cm 4) a) 27 cm5) 15 cm 6) 225 m and ` 2600 7) 24 cm2, 13cm2, 15cm2
8) 49 cm2 9) 198 m & ` 1990 10) Base = 24 Height = 18 m11) a) area is doubled b) Remains same c) 4 times
Exercise 7.3 : Sl. no base height area1 8 cm 6 cm 48 cm2
2 15 cm 11 m 165 m2
3 20 cm 14 m 280 m2
4 24 cm 50 m 1200 cm2
2) 63 cm2 3) 315 m2 4) 9 m 5) 80 m 6) 91 cm 7) 10.8 m8) base = 10 m, length = 20 m 9) base = 50 m, length = 20 mExercise 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) 308 cm & 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) a) 1584 cm2 b) 660 m2 c) 5197.5 m2 d) 1037.1428 cm2 5)2002 m2 `50050
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Chapter - 8 Data Handling Exercise 8.1 : I 34.25 II 1) 11.00 2) 4.5 III 3.99 kgIV 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 Chapter - 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...} c) { 1, 3, 5, 7, 9, 11...} d)6 e) 1, 2, 3, 4 II a) 7
1 b) 75
c) 71 III a) 25
12 b) 255 c) 25
6
Exercise 9.3 :I. a) 21 b) 2
1 c) 21 d) 6
1 e) 32
II. a) 71 b) 7
5 c) 71 III. a) 25
12 b) 255 c) 25
6
Unit - 10 Representation of 3 diomensional objects in 2 dimensional figure
Exercise 10.1 : I Solid Faces Edges Vertices shape of focusa) 6 12 8 rectangleb) 6 12 8 squarec) 4 6 6 triangled) 5 8 5 squaree) 5 6 9 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, 4 → C
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