Mathematics
Classes 9-10

Chapter One

Real Number
Mathematics is originated from the process of expressing quantities in symbols or numbers. The history of numbers is as ancient as the history of human civilization.
Greek Philosopher Aristotle According to the formal inauguration of mathematics occurs in the practice of mathematics by the sect of priest in ancient Egypt. So, the number based mathematics is the creation of about two thousand years before the birth of Christ.
After that, moving from many nations and civilization, numbers and principles of numbers have gained an universal form at present.
The mathematicians in India first introduce zero (0) and 10 based place value system for counting natural numbers, which is considered a milestone in describing numbers.
Chinese and Indian mathematicians extended the idea zero, real numbers, negative number, integer and fractional numbers which the Arabian mathematicians accepted in the middle age. But the credit of expressing number through decimal fraction is awarded to the Muslim Mathematicians. Again they introduce first the irrational numbers in square root form as a solution of the quadratic equation in algebra in the 11th century.
According to the historians, very near to 50 BC the Greek Philosophers also felt the necessity of irrational number for drawing geometric figures, especially for the square root of 2. In the 19th century European Mathematicians gave the real numbers a complete shape by systematization. For daily necessity, a student must have a vivid knowledge about ‘Real Numbers’. In this chapter real numbers are discussed in detail.
At the end of this chapter, the students will be able to –
Classify real numbers
Express real numbers into decimal and determine approximate value
Explain the classification of decimal fractions
Explain recurring decimal numbers and express fractions into recurring decimal numbers Transform recurring decimal fraction into simple fractions
Explain non-terminating non-recurring decimal fraction
Explain non-similar and similar decimal fraction
Add, subtract multiply and divide the recuring decimal fraction and solve various problems related to them.

2

Math

Natural Number
1, 2, 3, 4......... etc. numbers are called natural number or positive whole numbers.
2, 3, 5, 7......... etc. are prime numbers and 4, 6, 8, 9,......... etc. are composite numbers.
Integers
All numbers (both positive and negative) with zero (0) are called integers i.e. .......
3, 2, 1, 0, 1, 2, 3......... etc. are integers.
Fractional Number p If p, q are co-prime numbers ; q 0 and q 1 , numbers expressed in form are q called fractional number.
1 3
5
Example : , , etc. are fractional numbers.
2 2 3
If p q , then it is a proper fraction and if p q then it is an improper fraction :
1 1 2 1
3 4 5 5
Example , , , , ......... etc. proper and , , , , .... etc. improper fraction.
2 3 3 4
2 3 3 4
Rational Number p is called rational
If p and q are integers and q 0 , number expressed in the form q number. For example :

3
1

3,

11
2

5.5,

5
1.666... etc. are rational numbers.
3

Rational numbers can be expressed as the ratio of two integers. So, all integers and all fractional numbers are rational numbers.
Irrational Number p form, where p, q are integers and q 0 are
Numbers which cannot be expressed in q called Irrational Numbers. Square root of a number which is not perfect square, is an irrational number. For example:

2

1.414213.....,

3

1.732.....,

5
2

1.58113....

etc. are irrational numbers. Irrational number cannot be expressed as the ratio of two integers. Decimal Fractional Number
If rational and irrational numbers are expressed in decimal, they are known as decimal fractional numbers. As for instance, 3 3 0, 5 2 5, 10 3 3333......., 3 1 732......... etc.
3
2 are decimal fractional numbers. After the decimal, if the number of digits are finite, it is terminating decimals and if it is infinite it is known as non-terminating decimal number.
For
example,
0.52,
3.4152 etc. are terminating decimals and 1 333......., 2 123512367........... etc. are non-terminating decimals. Again, if the digits

3

Math

after the decimal of numbers are repeated among themselves, they are known recurring decimals and if they are not repeated, they are called non-recurring decimals. For example : 1 2323........, 5 654 etc. are the the recurring decimals and
0 523050056........, 2 12340314........ etc. are non-recurring decimals.
Real Number
All rational and irrational numbers are known as real numbers. For example :

0, 1, 2, 3,..........

1
,
2

3
,
2

4
,........
3

2 , 3, 5, 6......

1 23, 0 415, 1 3333......., 0 62, 4 120345061.......... etc. are real numbers.
Positive Number
All real numbers greater than zero are called positive numbers. As for instance
1, 2,

1 3
, ,
2 2

2 , 0 415, 0 62, 4 120345061.............. etc. are positive numbers.

Negative Number
All real numbers less than zero are called negative numbers. For example,

1,

2,

1
,
2

3
,
2

2,

0 415,

0 62,

4 120345061..............

etc.

are

negative numbers.
Non-Negative Number
All positive numbers including zero are called non-negative numbers. For example,
1
0, 3, , 0 612, 1 3, 2 120345.............. etc. are non-negative numbers.
2
Classification of real Number.
Real

Rational
Integer
Positive

Fractional

1

Fraction
0

Composite

Negative

Proper

Improper

Simple

Mixed

Decimal

Terminating

Irrational

Recurring

Nonrecurring

4

Math

Activity : Show the position of the numbers

3
4

, 5,

7, 13 , 0, 1,

9
7

, 12 , 2

4
,
5

. .

1 1234......, .3 2 3 in the classification of real numbers.
Example 1. etermine the two irrational numbers between
D

3

Solution : H ere, 3 and 4 .

1.7320508......

Let,

a

2.030033000333.....

and

b

2.505500555.......

Clearly :both a and b are real numbers and both are greater than

3 and less than 4.
3 2.03003300333......... 4 and 3 2.505500555................ 4
Again, a and b cannot be expressed into fractions. a and b are the two required irrational numbers.
Basic characteristics of addition and multiplication over a real number :
1. If a, b are real numbers, i a b is real and ii ab is a real number
2. If a, b are real numbers, i a b b a and ii ab ba
3. If a, b, c are real numbers, i a b c a b c and ii ab c a bc
4. If a is a real number, in real numbers there exist only two number 0 and 1 where i 0 1 ii a 0 a iii a.1 1.a a
1
5. If a is a real number, i a ( a ) 0 ii If a 0 , a.
1
a
6. If a, b, c are real numbers, a(b c ) ab ac
7. If a, b are real numbers, a b or a b or a b
8. If a, b, c are real numbers and a b , a c b c
.
9 If a, b, c are real numbers and a b , (i ) ac bc where c 0
(ii ) If ac bc , c 0
Proposition : 2 is an irrational number. e know,
W
i.e.,

1 2 4
1
2
4
or, 1
2 2
2
Proof : 1 1,
2
2, 2 2 4
Therefore, the value of 2 is greater than 1 and less than 2.

5

Math

2 is not an integer.
2 is either a rational number or a irrational number. If 2 is a rational number p let, 2
; where p and q are natural numbers and co-prime to each other and q 1 q or, 2

p2
; squaring q2 or, 2q

p2
; multiplying both sides by q . q Clearly 2q is an integer but

p2 is not an integer because p and q are co-prime natural q numbers and q 1

2q and alue of
V

p2 q p2 cannot be equal, i.e., 2q q 2 cannot be equal to any number with the form

p
i.e.,
q

2

p q 2 is an irrational number.
Example 2. P rove that, sum of adding of 1 with the product of four consecutive natural numbers becomes a perfect square number.
Solution : Let four consecutive natural numbers be x, x 1, x
B adding 1 with their product we get, y x x 1 x 2 x 3 1 xx 3 x 1 x x 2 3 x x 2 3x 2 1 aa 2

1;

[ x 2 3x

2

2, x 3 respectively.

1

a]

a ( a 2) 1 ;

a2 2a 1

a 1

2

2

x 2 3x 1 ;

which is a perfect square number.
If we add 1 with the product of four consecutive numbers, we get a perfect square number. Activity : roof that,
P

3 is an irrational number

Classification of Decimal Fractions
Each real number can be expressed in the form of a decimal fraction.

6

Math

or example,
F

2

2 0,

2
5

0.4,

1
3

0.333.... etc. There are three types of decimal

fractions :terminating decimals, recurring decimals and non-terminating decimals.
Terminating decimals : In terminating decimals, the finite numbers of digits are in the right side of a decimal points. or example, 0.12,1.023,7.832,54.67,.......... etc. are
F
terminating decimals.
Recurring decimals : In recurring decimals, the digits or the part of the digits in the right side of the decimal points will occar repeatedly. or example,
F
etc. are recurring decimals.
3.333....., 2.454545......, 5.12765765
Non-terminating decimals : In non-terminating decimals, the digits in the right side of a decimal point never terminate, i.e., the number of digits in the right side of decimal point will not be finite neither will the part occur repeatedly. Fr o example.1.4142135......, 2.8284271....... etc. are non-terminating decimals.
Terminating decimals and recurring decimals are rational numbers and non-terminating decimals are irrational numbers. The value of an irrational number can be determined upto the required number after the decimal point. If the numerator and denominator of a fraction can be expressed in natural numbers, that fraction is a rational number.
Activity : Classify the decimal fractions stating reasons :

1.723, 5.2333........,
0.450123.......

0.0025,

2.1356124.......,

Recurring decimal fraction :
23
into decimal fractions, we get,
Expressing the fraction
6
23 = 6 ) 23 ( 3 833
6
18
50
48
20
18
20
18
20
18
2

0.0105105........

and

7

Math

Observe : It is found that, the process of division is not ended at the time of dividing a numerator of a fraction by its denominator. To convert it into decimal fraction in the quotient ‘ ’occurs repeatedly. H
3
ere,
3.8333..... is a recurring decimal fraction.
If digit or successive digit of a decimal fractions a in the right side of the decimal point appear again and again, these are called recurring decimal fractions.
If a digit or successive digits of decimal fractions.
In recurring decimal fractions, the portion which occurs again and again, is called recurring part. In recurring decimal fraction, if one digit recurs, the recurring point is used upon it and if more than one digits recurs, the recurring point is used only upon the first and the last digits.
As for example : 2.555....... is written as 2.5 and 3.124124124........ is written as 3.124 .
In recurring decimal fractions, if there is no other digit except recurring one, after decimal point it is called pure recurring decimal and if there is one digit or more after decimal point in addition to recurring one, it is called mixed recurring decimal. or
F
example, 1.3 is a pure recurring decimal and 4.23512 is a mixed recurring decimal.
If there exists prime factors other than 2 and 5 in the denominator of the fraction, the numerator will not be fully divisible by denominator. As the last digit of successive divisions cannot be other than1, 2, ......, 9 , at one stage the same number will be repeated in the remainder. The number in the recurring part is always smaller than that of the denominator.
Example 3. Express 3 into decimal fraction.

9
Example 4. Express 5 into decimal fraction.

11

37

Solution :
11) 30 (0.2727
22
80
77
30
22
80
77
3

Solution :
37 )9 (2.56756
5
74
210
185
250
222
280
259
210
185
250
222
28

8

Math

Required decimal fraction = 0.2727 ......
Required decimal fraction =
= 0.27
2.56756...... 2.567
Conversion of Recurring Decimal into Simple Fraction
Determining the value of recurring fraction :
Example 5. Express 0.3 into simple fraction.
Solution : 0.3 0.3333........
0.3 10 0.333....... 10 and 0.3 1
0.333....... 1

3.333......
0.333......

subtracting, 0.3 10

3

0.3 1

3 or 0.3 9 3
1
Therefore, 0.3
3
1
Required fraction is
3
Example 6. Express 0.24 into simple fraction. or, 0.3

10 1

3
9

Solution : 0.24 = 0 24242424........
So,

0.24 100

and

0.242424....... 100

0.24 1

0.242424....... 1

Subtracting 0.24 100 1 or, 0.24 9

24

24.2424......

0.242424......

24

or, 0.24

24
9

8
33

8
33
Example 7. Express 5.1345 into simple fraction.
Required fraction is

Solution : 5.1345 = 5.1345345345.........
So,

5.1345 10000

5.1345345...... 10000

and

5.1345 10

5.1345345....... 10

Subtracting, 5.1345 9
0
So, 5.1345 =

51345 51
9
0

51345.345......
51.345......

51345 51
5129
4
9
0

8549
1665

5

224
1665

Math

Required fraction is 5

9

224
1665

Example 8. Express 42.3478 into simple fraction.
Solution : 42.3478 = 42.347878........
So,
42.3478 10000 42.347878......... 10000 42348.7878 and 42.3478 100 = 42.347878........ 100 = 4234.7878
Subtracting, 42.3478

0
90
423478 4234
Therefore, 42.3478 =
90
0
Required fraction is 42

41944
2
90
0

= 423478 4234
3497
3
287
42
825
825

287
825

Explanation : rom the examples 5, 6, 7 and 8 , it appears that,
F
The recurring decimal has been multiplied by the number formed by putting at the right side of 1 the number of zeros equal to the number of digits in the right side of decimal point in the recurring decimal.
The recurring decimal has been multiplied by the number formed by putting at the right side of 1 the number of zeros equal to the number of digits which are nonrecurring after decimal point of the recurring decimal. the second product has been subtracted from the first product. y subtracting the
B
second product from the first product the whole number has been obtained at the right side. ere it is observed that, the number of non-recurring part has been
H
subtracted from the number obtained by removing the decimal and recurring points of recurring decimal fraction.
The result of subtraction has been divided by the number formed by writing the same number of 9 equal to the number of digits of recurring part at the left and number of zeros equal to the number of digits of non-recurring part at the right.
In the recurring decimals, converting into fractions the denominator is the number of
9equal to the number of digits in the rec urring part and in right side of all 9
’
s number of zeros equal to the number of digits in the non-recurring part. And the numerator in the result that is obtained by subtracting the number of the digits formed by omitting the digits of recurring part from the number formed by removing the decimal and recurring points of recurring decimal.
Remark : Any recurring decimal can also be converted into a fraction. All recurring decimals are rational numbers.
Example . Express 5.23457 into simple fraction.
9

10

Math

Solution : 5.23457 = 5.23457457457.........
So,
5.23457 100000 and 5.23457 100
Subtracting, 5.23457 90
0
Therefore,

= 523457.457457
= 523.457457
= 52294
3
52294
3
261467
5.23457 =
90
0
490
9
5

Required fraction is 261467
490
9
5

Explanation : ere in the decimal part the recurring decimal has been multiplied first
H
by 100000 ( zeros at the right side of 1) as there are two digits at the left side of
5
recurring part in the decimal portion, the recurring decimal has been multiplied by 100
(wo zeros at the right side of 1) The seco nd product has been subtracted from the first t . product. In one side of the result of subtraction is a whole number and at the other side of the result is (100000 100) = 00 times of the value of the given recurring
9
decimal. D ividing both the sides by 00
, the required fraction is obtained.
9
Activity : Express 0.41 and 3.04623 into fractions.
Rules of Transformation of Recurring Decimals into Simple Fractions
N
umerator of the required fraction = the result by subtracting the number obtained from exempting the decimal point of the given decimal point and the non-recurring part.
D
enominator of the required fraction = umbers formed by putting the number of 9
N
equal to the number of digits in the recurring part of the from the number of zeros equal to the number of digits in the non-recurring part. H the above rules are directly ere applied to convert some recurring decimals into simple factions.
Example 10. Express 45.2346 into simple fraction.
452346 452 45189
4
22597
4
Solution : 45.2346 =
9
0
9
0
49
9
5
1172
Required fraction is 45
49
9
5

45

1172
49
9
5

Example 11. Express 32.567 into simple fraction.
Solution : 32.567

32567 32
9

Required fraction is 32

32535
9

21
3615 1205
32
37
111
37

21
.
37

Activity : Express 0.012 and 3.3124 into fraction.

11

Math

Similar recurring decimals and Non-similar Recurring decimals :
If the numbers of digits in non-recurring part of recurring decimals are equal and also numbers of digits in the recurring parts are equal, those are called similar recurring decimals. O recurring decimals are called non-similar recurring decimals. Fr ther o example : 12.45 and 6.32 ; 9.453 and 125.897 are similar recurring decimals. Again,
0.3456 and 7.45789; 6.4357 and 2.89345 are none-similar recurring decimals.
The Rules of Changing Non-Similar Recurring Decimals into Similar Recurring
Decimals
The value of any recurring decimals is not changed, if the digits of its recurring part are written again and again, or Example, 6.4537 6.453737
F
6.45373 6.453737
H each one is a recurring decimal, 6.45373737......... is a non-terminating decimal. ere It will be seen that each recurring decimal if converted into a simple fraction has the same value.

4537 645 6389
2
90
0
90
0
6453737 645 645309
2
6.453737
90
0
90
0
6.4537

6.453737

6453737 64537
9000
0

6389
2
90
0

638900
2
9000
0

6389
2
90
0

In order to make the recurring decimals similar, number of digits in the non-recurring part of each recurring decimal is to be made equal to the number of digits of nonrecurring part of that recurring decimal in which greatest number of digits in the nonrecurring part exists and the number of digits in the recurring part of each recurring decimal is also to be made equal to the lowest common multiple of the numbers of digits of recurring parts of recurring decimals.
Example 12. Convert 5.6, 7.345 and 10 78423 into similar recurring decimals.
Solution : The number of digits of non-recurring part of 5 6, 7 345 and 10.78423 are ere 0, 1 and 2 respectively. H the number of dig its in the non-recurring part occurs in
10.78423 and that number is 2 . Therefore to make the recurring decimals similar the number of digits in the non-recurring part of each recurring decimal is to be made 2 .
Again, the numbers of digits to recurring parts of 5.6, 7.345 and 10.78423 are 1, 2 and 3 respectively. The lowest common multiple of 1, 2 and 3 is 6 . So the number of digits in the recurring part of each recurring decimal would be 6 in order to make them similar. 12

Math

So, 5.6

5.66666666, 7.345

Required similar recurring
10.78423423 respectively.

7.34545454 and 10.78423 10.78423423 decimals are 5.66666666, 7.34545454 ,

Example 13. Convert 1.7643, 3.24 and 2.78346 into similar recurring decimals.
Solution : In 1.7643 the number of digits in the non-recurring part means 4 digits after decimal point and here there is no recurring part.
In 3.24 the number of digits in the recurring and non-recurring parts are respectively 0 and 2.
In 2.78346 the number of digits in the recurring and non-recurring parts are respectively 2 and 3.
The highest number of digits in the nonrecurring parts is 4 and the L.C.M. of the numbers of digits in the recurring parts i.e. 2 and 3 is 6. The numbers of digits is the recurring and nonrecurring parts of each decimal will be respectively 4 and 6.
1.7643= 1.7643000000 ; 3.24

3.2424242424 ; 2.78346

2.7834634634

Required recurring similar decimals are 1.7643000000 , 3.2424242424
2.7834634634

and

Remark : In order to make the terminating fraction similar, the required number of zeros is placed after the digits at the extreme right of decimal point of each decimal fraction. The number of non-recurring decimals and the numbers of digits of nonrecurring part of decimals after the decimal points are made equal using recurring digits.
After non-recurring part the recurring part can be started from any digit.
Activity : Express 3.467, 2.01243 and 7.5256 into similar recurring fractions.
Addition and Subtraction of Recurring Decimals
In the process of addition or subtraction of recurring decimals, the recurring decimals are to be converted into similar recurring decimals. Then the process of addition or subtraction as that of terminating decimals is followed. If addition or subtraction of terminating decimals and recurring decimals together are done, in order to make recurring decimals similar, the number of digits of non-recurring part of each recurring should be equal to the number of digits between the numbers of digits after the decimal points of terminating decimals and that of the non-recurring parts of recurring decimals.
The number of digits of recurring part of each recurring decimal will be equal to L.C.M. as obtained by applying the rules stated earlier and in case of terminating decimals, necessary numbers of zeros are to be used in its recurring parts. Then the same process of addition and subtraction is to be done following the rules of terminating decimals. The sum or the difference obtained in this way will not be the actual one. It should be

13

Math

observed that in the process of addition of similar decimals if any number is to be carried over after adding the digits at the extreme left of the recurring part of the decimals then that number is added to the sum obtained and thus the actual sum is found. In case of subtraction the number to be carried over is to subtract from the difference obtained and thus actual result is found. The sum or difference which is found in this way is the required sum or difference.
Remark (a) : The sum or difference of recurring decimals is also a recurring decimal. In this sum or difference the number of digits in the non-recurring part will be equal to the number of digits in the non-recurring part of that recurring decimal, which have the highest number of digits in its non-recurring part. Similarly, the number of digits in the recurring part of the sum or the result of subtraction will be the equal to L.C.M. of the numbers of digits of recurring parts of recurring decimals. If there is any terminating decimals, the number of digits in the non-recurring part of each recurring decimal will be equal to the highest numbers of digits that occurs after the decimal point.

(b) Converting the recurring decimals into simple fractions, addition and subtraction may be done according to the rules as used in case of simple fractions and the sum or difference is converted into decimal fractions. But this process needs more time.
Example 14. Add : 3 89, 2 178 and 5 89798
Solution : Here the number of digits in the non-recurring part will be 2 and the number of digits in the recurring part will be 6 which is L.C.M. of 2,2 and 3.
At first three recurring decimals are made similar.
3 89

= 3 89898989

2 178

= 2 127878787

5 89798

= 589798798

11 97576574
[ 8 8 7 2 25 , Here 2 is the number to
2 be carried over, 2 of 25 has been added.]
11 97576576
The required sum is 11 97576576 or 11 97576
Remark : In the sum the number in the recurring part is 575675 . But the value is not changed if 576 is taken as the number of recurring part.
Note : To make clear the concept of adding 2 at the extreme right side, this addition is done in another method :

14

3 89
2 178
5 89798

Math

=

3 89898989 | 89
= 2 17878787 | 87
= 5 89798798 | 79
11.97576576 | 55

Here the number is extended upto 2 more digits after the completion of the recurring part. The additional digits are separated by drawing a vertical line. Then 2 has been carried over from the sum of the digits at the right side of the vertical line and this 2 is added to the sum of the digit at the left side of the vertical line. The digit in the right side of the vertical line are the same and the recurring point. Therefore both the sums are the same.
Example 15. Add : 8 9478, 2 346 and 4 71 .
Solution : To make the decimals similar, the number of digits of non-ecurring parts r would be 3 and that of recurring parts would be 6 which is L.C of 3 and 2.
.M.

8 9478
2 346
4 71

= 8 947847847

2 346000000
= 4 717171717
16 011019564
1

[ 8 0 1 1 10 , Here the digit in the second place on the left is 1 which is to be carried over. Therefore 1 of 10 is added.]

16 011019565
The required sum is 16 011019565 .
Activity : Add :1. 2 097 and 5 12768

2. 1 345, 0 31576 and 8 05678

Example 16. Subtract 5 24673 from 8 243 .
Solution : Here the number of digits in the non-recurring part would be 2 and that of recurring part is 6 which is L.C of 2 and 3. Now making two decimal numbers
.M.
similar, subtraction is done.

8 243
5 24673

= 8 24343434
= 5 24673673

2 99669761
1
2 99669760
The required sum is 2 99669760 .

[Subtracting 6 from 3, 1 is to be carried over.]

15

Math

Remark : If the digit at the beginning place of recurring point in the number from which deduction to be made is smaller than that of the digit in the number 1 is to be subtracted from the extreme right hand digit of the result of subtraction.
Note : In order to make the conception clear why 1 is subtracted, subtraction is done in another method as shown below :

8 243

= 8 24343434 | 34

5 24673

= 5 24673673 | 67

2 99669760 | 67
The required difference is 2 99669760 | 67
Here both the differences are the same.
Example 17. Subtract 16 437 from 24 45645 .
Solution :

24 45645
16 437

= 24 45645
= 16 43743

8 01902 [7 is subtracted from 6.1 is to be carried
1 over.]
8 01901
The required difference is 8.01901
Note :

24 45645

16 437

= 24 45645 | 64
= 16 43743 | 74

8 01901 | 90
Activity : Subtract :1. 10 418 from 13 12784 2. 9 12645 from 23 0394
Multiplication and Division of Recurring Decimals : onverting recurring decimals into simple fraction and completing the process of their
C
multiplication or division, the simple fraction thus obtained when expressed into a decimal fraction will be the product or quotient of the recurring decimals. In the process of multiplication or division amongst terminating and recurring decimals the same method is to be applied. But in case of making division easier if both the divident and the divisior are of recurring decimals, we should convert them into similar recurring decimals.
Example 18. Multiply 4 3 by 5 7 .

16

Math

Solution : 4 3

5 7

4 3

43 4
9
57 5
9

5 7 =

39
9
52
9

13 52
3 9

13
3

676
27

25 037

The required product is 25 037
Example 19. Multiply 0 28 by 42 18 .

8 2 26 13
90
90 45
4218 42 4176 464
42 18
99
99
11
13 464 6032
12 185
=
45 11
495
The required product is 12 185

Solution : 0 28

Example 20. 2 5 4 35 1 234

?

25 5
10 2
435 43 392
4 35
90
90
1234 12 1222
1 23 4
990
990
5 392 611 196 611
2 90 495
8910
The required product is 13 44062
Activity : 1. Multiply 1 13 by 2 6 . 2.

Solution : 2 5

611
495
119756
8910

13 44062... .

0 2 1 12 0 081

Example 21. Divide 7 32 by 0 27 .
Solution : 7 32

0 27
7 32

0 27

732 7 725
99
99
27 2 25 5
90
90 18
725 5
99 18

725 18
99 5

290
11

26 36

?

Math

17

The required quotient is 26 36
Example 22. Divide 2 2718 by 1 912

22718 2 22176
9999
9999
1912 19 1893
1 912
990
990
22716 1893 22716 990
2 2718 1 912 =
9999
990
9999 1893
The required quotient is 1 1881
Solution : 2 2718

120
101

1 1881

Example 23. Divide 9 45 by 2 863 .

945
100
189 99
2 2835
The required quotient is : 3 3

Solution : 9 45

2 863 =

2863 28
990
33
3 3
10

945
100

990
2835

Remark : P roduct of recurring decimals and quot ient of recurring decimals may be or may not be a recurring decimal.
Activity : 1. Divide 0 6 by 0 9 .

2. Divide 0 732 by 0 027

Non Terminating Decimals
There are many decimal fractions in which the number of digits after its decimal point is unlimited, again one or more than one digit does not occur repeatedly. Such decimal fractions are called non-erminating decimal fractions, For example, t 5.134248513942307 ............ is a non-erminating decimal number. The square root of 2 t is a non terminating decimal. Now we want to find the square root of 2.
1 2
1 4142135........
1
24

100
96

281 400
281
2824 11900
11296

18

Math

28282 60400
56564
282841 383600
282841
2828423
28284265

10075900
8485269
159063100
141421325
17641775

If the above process is continued for ever, it will never end.

2 1 4142135...... is a non terminating decimal number.
The value upto the definite number of decimal places and the approximate value upto some decimal places.
It is not the same to find the value of non-erminating decimals upto definite number of t decimal place and the approximate value into some decimal places. For example,
5 4325893...... upto four decimal places will be 5 4325, but the approximate value of the decimal, 5 4325893.... upto four decimal places will be 5 4326 . Here, the value upto 2 decimal places and the approximate value upto 2 decimal places are the same. This value is 5 43 . In this way the approximate terminating decimals can also be found. Remark : When it is needed to find the value upto some decimal places, the digits that occur in those places are to be written without any alternatives of those digits. If approximate values are to be identified, we should add 1 with the last digit when there is 5, 6, 7, 8 or 9 after the decimal places. But if it is 1, 2, 3 or 4 digits remain unchanged. In this case, correct value upto decimal place or approximate value upto decimal place are almost equal. We should find the value upto 1 place more to the required wanted value.
Example 24. Find the square root of 13 and write down the approximate value upto 3 decimal places.
Solution : 3 ) 13 ( 3 605551........
9

66

400
396

19

Math

7205
72105

40000
36025
3697500
3605525

7211101

9197500
7211101

1986399
The required square root is 3 605551........ .
The required approximate value upto 3 decimal places is 3 606 .
Example 25. Find the value and approximate value of 4 4623845....... upto 1, 2, 3, 4 and 5 decimal places.
Solution : The value of 4 4623845 upto 1 decimal place is 4 4 and approximate value
1 ,,
,, ,,
4 5
Value upto 2
,,
,, ,,
4 46 and approximate value upto 2 ,,
,, ,, 4 46
Value upto 3
,,
,, ,, 4 462 and approximate value upto 3 ,,
,, ,, 4 462
Value upto 4
,,
4 4623 and approximate value upto 4 ,,
,, ,, 4 4624
Value upto 5
,,
,, ,, 4 46238 and approximate value upto
5 ,,
,, ,, 4 46238
Activity : Find the square root of 29 and find the value upto two decimal places and the approximate value upto two decimal place.

Exercise 1
(b) 7
(c) 10 is an irrational number.
5
2. (a) Find the two irrational numbers between 0.31 and 0.12 .
1
(b) Find a rational and an irrational numbers between and 2 .
2

1. P that, (a) rove 3. (a) P that, square of any odd integer number is an odd number. rove (b) P that, the product of two consecu tive even numbers is divisible by 8. rove 20

Math

4. Express into recurring decimal fractions :(a)

1
6

(b)

7
11

(c) 3

2
9

(d) 3

8
15

5. Express into simple fractions :
(a) 0 2 (b) 0 35 (c) 0 13 (d) 3 78 (e) 6 2309
6. Express into similar recurring fractions :
(a) 2 3, 5 235

(b) 7 26, 4 237

(c) 5 7, 8 34, 6 245

(d) 12 32, 2 19, 4 3256

7. Add :(a) 0 45 0 134 (b) 2 05 8 04 7 018 (c) 0 006 0 92 0 0134
8. Subtract :
(a) 3 4 2 13

(b) 5 12 3 45

(c) 8 49 5 356

(d) 19 345 13 2349

9. Multiply:
(a) 0 3 0 6

(b) 2 4 0 81

(c) 0 62 0 3

(d) 42 18 0 28

(c) 2 37 0 45

(d) 1 185 0 24

10. Divide :
(a) 0 3 0 6

(b) 0 35 1 7

11. Find the root (upto three decimal places) and write down the approximate values of the square roots upto two decimal places :
(a) 12 (b) 0 25 (c) 1 34 (d) 5 1302
12. Find the rational and irrational numbers from the following numbers :
2
27
8
6
(a) 0 4 (b) 9 (c) 11 (d)
(e)
(f)
(g) 3 (h) 5 639
3
7
48
3
7
13. Simplify :
(a) ( 0 3

0 83 )

0 5 0 1 + 0 35 0 08

(b) 6 27 0 5
0 25

14.

0 5
0 1

0 75

0 75

8 36

21 3

0 5

5 and 4 are two real numbers.
(a) Which one is rational and which one is irrational.
(b) Find the two irrational numbers between
(c) P That, rove 5 is an irrational number.

5 and 4.

Chapter Two

Set and Function
The word set’ is familiar to us, such as dinner set, set of natural numbers, set o f
‘
rational numbers etc. As a modern weapon of mathematics, the use of set is extensive. The erman mathematician eor ge C
G
G antor (1844 –1918) first explaine d his opinion about set. He created a sensation in mathematics by expressing the idea of infinite set and his conception of set is known as set theory’ In this chapter, the
‘
. main obj ctives are to solve problems through using mathematics and symbols from e the conception of set and to acquire knowledge about function.
At the end of this chapter, the students will be able to :
Explain the conception of set and subset and express them by symbols
Describe the method of expressing set
Explain the infinite set and differentiate between finite and infinite set
Explain and examine the union and the intersection of set
Explain power set and form power set with two or three elements
Explain ordered pair and cartesian product
P
rove the easy rules of set by example and Venn Diagram and solve various problems using the rules of set operation
Explain and form sets and functions
Explain what are domain and range
Determine the domain and range of a function
Draw the graph of the function.
Set
Well defined assembling or collection of obj cts of real or imaginative world is e called sets, such as, the set of three textbook of Bangla, English and Mathematics, set of first ten natural odd numbers, set of integers, set of real numbers etc.
Set is generally expressed by the capital letters of english alphabets,
A, B, C,..........X , Y , Z . For example, the set of three numbers 2, 4, 6 is A {2, 4, 6}
Each obj ct or member of set is called set element. Such as, if B {a, b} , a and b e are elements of B . The sign of expressing an element is ' ' . a B and read as a belongs to B

b B and read as b belongs to B no element c is in the above set B.

22

Math

c

B and read as c does not belong to B .

Method of describing Sets :
Set is expressed in two methods :(1) o ster Method or Tabular Method and (2) Set
R
Builder Method.
(1) Tabular Method : In this methods, all the set elements are mentioned
}
particularly by enclosing them within second bracket { , and if there is more than one element, the elements are separated by using a comma (,).
For example : A {a, b} , B {2, 4, 6} , C

Niloy, Tisha, Shuvra}
{
etc.

(2) Set Builder Method : In this methods, general properties are given to determine the set element, without mentioning them particularly :
,
Such as, A {x : x is a natural odd number} B {x : x denotes the first five students of class IX
}
etc.
Here, by ’such as’or in short such that’is indicated. Since in this method, set rule o
:‘
‘
‘
condition is given to determining the set elements of this method, is called rule method.
Example 1. Express the set A {7, 14, 21, 28} by set builder method.
Solution : The elements of set A are 7, 14, 21, 28
Here, each element is divisible by 7, that is, multiple of 7 and not greater than 28.
A {x : x multiple of 7 and x 28} .
Example 2. Express the set B
Solution : Here, 28 1 28

by
{x : x, factors of 28} tabular method.

= 2 14
=4 7 factors of 28 are 1, 2, 4, 7, 14, 28 equired set B {1, 2, 4, 7, 14, 28}
R
Example 3. Express C {x : x is a positive integer and x2
Solution : Psitive integers are 1, 2, 3, 4, 5, ........... o Here if x 1 , x 2 12 1 if x 2 , x 2 2 2 4 if x 3 , x 2 32 9 if x 4 , x 2 4 2 16 if x 5 , x 2 5 2 25 ; which is greater than 18.

18} by tabular method.

r

23

Math

The acceptable positive integers by the condition are 1, 2, 3, 4 equired set is
R

C {1, 2, 3, 4}
.

Activity : 1. Express the set C
2. Express the set Q

{ 9, 6, 3, 3, 6, 9} by set builder method.
{ y : y is an integer and y 3

27} by tabular method.

Finite set : The set whose numbers of elements can be determined by counting is called finite set. For example, D = {x, y, z}E = 3, 6, 9,.......,60}
{
,
F {x : x is a prime number and 30 x 70} etc. are finite set. Here D set has 3 elements, E set has 20 elements and F set has 9 elements.
Infinite set : The set whose numbers of elements can not be determined by counting is called infinite set. For example :, A {x : x is natural odd numbers}set of natural
,
number N {1, 2, 3, 4, ........} , set of integers Z {
....... 3, 2, 1, 0, 1, 2, 3.......} , set of rational numbers Q P : p and q is as integer and q 0} , set of real numbers = q R etc. are infinite set.
Example 4. Show that the set of all natural numbers is an infinite set.
Solution : Set of natural number N {1, 2, 3, 4, 5, 6, 7, 8, .......}
Taking odd natural numbers from set N , the formed set A {1, 3, 5, 7, .......}
,, ,, N , the formed set B {2, 4, 6, 8, .......}
,, even ,,
,,
The set of multiple of 3 C {3, 6, 9, 12, .......} etc.
Here, the elements of the set A, B, C formed from set N can not be determined by counting. So A, B, C is an infinite set.
N is an infinite set.

Activity : Write the finite and infinite set from the sets given below :
1. {3, 5, 7} 2. {1, 2, 2 2 , .......210 } 3. {3, 32 , 33 , .......} 4. {x : x is an integer and x

p p 5.
: p and q are co- rime and q 1} q 6. { y : y

4}

N and y 2 100 y 3 } .

Empty set : The set which has no element is called empty set. Empty set is expressed by {}or φ . Such as, set of three male students of Holycross school
{x N : 10 x 11} {x N : x is a prime number and 23 x 29} etc.
,
Venn-Diagram : J hn Venn (1834- 883) introduced set activities by diagram. Here o 1 the geometrical figure on the plane like rectangular area, circular area and triangular area are used to represent the set under consideration. These diagrams are named
Venn diagram after his name.

24

Math

Subset : A {a, b} is a set. By the elements of set A , the sets {a, b} {a } , { b} can
,
be formed. Again, by not taking any element φ set can be formed.
Here, each of {a, b} {a } , { b} , φ is subset of set A .
,
So, the number of sets which can be formed from any set is called subset of that set.
The sign of subset is . If B is the subset of A , it is read as B A . B is a subset of A. From the above subsets, {a, b} set is equal to A .
Each set is the subset of itself.
Again, from any set, φ set can be formed. φ is a subset of any set.

Q {1, 2, 3} and R {1, 3} are two subsets of P

Q

P and R

{1, 2, 3} . Again P Q

P.

Proper Subset :
If the number of elements of any subset formed from a set is less than the given set,
`
it is called the proper subset. For example : A {3, 4, 5, 6} and B {3, 5} are two sets. Here, all the elements of B exist in set A .
B A
Again, the number of elements of B is less then the number of elements of A .

B is a proper subset of A and expressed as B

A.

Example 5. Write the subsets of P {x, y, z} and find the proper subset from the subsets. Solution : iven,
G

P {x, y, z}

Subsets of P are {x, y, z} , {x, y} , {x, z} , { y, z} , {x} , {y} , {z} , φ . roper subsets of
P

P are {x, y} , {x, z} , { y, z} , {x} , {y} , {z}

Equivalent Set :
If the elements of two or more sets are the same, they are called equivalent sets. Such as, A {3, 5, 7} and B {5, 3, 7} are two equal sets and written as A B .
Again, if A {3, 5, 7} , B {5, 3, 3, 7} and C are equivalent. That is, A B C

{7, 7, 3, 5, 5} , the sets A, B and C

Math

25

It is to be noted if the order of elements is changed or if the elements are repeated, there will be no change of the set.
Difference of Set : Suppose, A {1, 2, 3, 4, 5} and B {3, 5} . If the elements of set
B are discarded from set A , the set thus formed is {1, 2, 4} and written as A \ B or
A B {1, 2, 3, 4, 5} {3, 5} = {1, 2, 4}
So, if a set is discarded from any set, the set thus formed is called different set.
Example 6. If P {x : x , factors of 12}and Q {x : x , multiples of 3 and x 12} , find P Q .
Solution : iven, P {x : x , factors of 12}
G
Here, factors of 12 are 1, 2, 3, 4, 6, 12

P {1, 2, 3, 4, 6, 12}
Again, Q {x : x , multiple of 3 and x 12}
Here, multiple of 3 upto 12 are 3, 6, 9, 12
Q {3, 6, 9, 12}
P Q {1, 2, 3, 4, 6, 12) {3, 6, 9, 12} {1, 2, 4} equired set {1, 2, 4}
R
Universal Set :
All sets related to the discussions are subset of a definite set. Such as, A {x, y} is a subset of
B {x, y, z} . Here, set B is called the universal set in with respect to the set A .
So, if all the sets under discussion are subsets of a particular set, that particular set is called the universal set with respect to its subsets. niversal set is generally expressed by U . But universal set can
U
be expressed by other symbols too. Such as, if set of all even natural numbers C {2, 4, 6........} and set of all natural numbers N {1, 2, 3, 4,...........} , the universal set with respect to C will be N .
Complement of a Set :

U
U is an universal set and A is the subset of . The set formed by all the elements excluding the elements of set A is called the complement of set A . The complement of the set A is expressed by Ac or A . Mathematically, Ac U \ A
Let, P and Q are two sets and the elements of Q which are not elements of P are called complement set of Q with respect to P and written as
Q c P \ Q.

26

Math

Example 7. If U {1, 2, 3, 4, 6, 7} , A {2, 4, 6, 7} and B {1, 3, 5} , determine Ac and B c .
Solution : Ac U \ A {1, 2, 3, 4, 6, 7} \ {2, 4, 6, 7} {1, 3, 5} and B c

U \ B {1, 2, 3, 4, 6, 7} \ {1, 3, 5} {2, 4, 6, 7}

equired set Ac {1, 3, 5} and B c {2, 4, 6, 7}
R
Union of Sets :
The set formed by taking all the elements of two or more sets is called union of sets.
Let, A and B are two sets. The union of A and B set is expressed by A B and read as A union B . In the set builder method A B {x : x A or x B} .
Example 8. If C {3, 4, 5} and D {4, 6, 8} , determine C D .
Solution : iven that, C {3, 4, 5} and D {4, 6, 8}
G

C

D {3, 4, 5} {4, 6, 8} {3, 4, 5, 6, 8}

Intersection of Sets:
The set formed by the common elements of two or more sets is called intersection of sets. Let, A and B are two sets. The intersection of A and B is expressed by
A B and read as A intersection B . In set building method,
A B {x : x A and x B} .
Example 9. If P {x N : 2 x numbers and x 8} , find P Q .
Solution : iven that,
G
and Q {x
P

P

{x

6} and Q {x

N :2

x

N : x are even

4
6

6} = {3, 4, 5, 6}

N : x are even numbers and x 8} {2, 4, 6, 8}

Q {3, 4, 5, 6} {2, 4, 6, 8} {4, 6}

equired set is {4, 6}
R
Disjoint Sets:
If there is no common element in between two sets, the sets are disj ined sets. Let , o o
A and B are two sets. If A B φ , A and B will be mutually disj int sets.
Activity : If U

{1, 3, 5, 7, 9, 11} , E {1, 5, 9} and

F {3, 7, 11} , find, E c

F c and E c

Fc .

Power Sets :
A {m, n} is a set. The subsets of A are {m n} {m), {n} φ . Here, the set of subsets
,
,
{ m n} {m), {n} φ} is called power set of set A . The power set of A is expressed
,
, as P(A) . So, the set formed with all the subsets of any set is called the power set of that set.

27

Math

Example 10. A { } B {a} C {a, b} are three sets.
,
,
Here, P( A) {φ}
The number of elements of set A is 0 and the number of elements of its power set = 1 20
Again, P( B) { a} φ}
,
The number of elements of set B is 1 and the number of elements of its power set is = 2 21 and P(C) { a, b} {a} {b} φ}
, , ,
The number of elements of set C is 2 and the number of elements of its power set is = 4 22
So, if the number of elements is n of a set, the number of elements of its power set will be 2n .
Activity : If G {1, 2, 3} , determine P(G) and show that the number of elements of P(G) supports 2n .
Ordered pair :
Amena and Sumona of class VIII stood 1st and 2nd respectively in the merit list in the final examination. According to merit they can be written (Amena, Sumona) as a pair. In this way, the pair fixed is an ordered pair.
Hence, in the pair of elements, the first and second places are fixed for the elements of the pair and such expression of pair is called ordered pair.
If the first element of an ordered pair is x and second element is y , the ordered pair will be ( x, y) . The ordered pair ( x, y) and (a, b) will be equal or ( x, y) = (a, b) if` x a and y b .
Example 11. If (2x

y, 3) = (6, x

Solution : iven that,
G

(2x

y) , find ( x, y) .

y, 3) = (6, x

y)

According to the definition of ordered pair, 2x and x
Adding equation (1) and (2), we get, 3x u Ptting the value of

y 6........(1) y 3.........(2)

9 or, y

x in equation (1), we get, 6

3

y

6 or, y

0

( x, y) = (3, 0) .

Cartesian Product :
Wangsu decided to give layer of white or blue colour in room of his house at the inner side and red or yellow or green colour at the outer side. The set of colour of inner wall A white, blue}
{
and set of colour of outer wall B red, yellow,
{

28

Math

green} Wangsu can apply the colour of his room in the form of ordered pair as
.
(white, red), (white, yellow), (white, green), (blue, red), (blue, yellow), (blue, green).
The given ordered pair is written as.
{
A B (white, red), (white, yellow), (white, green), (blue, red), (blue, yellow), (blue, green)}
This is the cartesian product set.
In set builder method, A B { x, y); x A and y B}
(
A B is read as A cross B .
Example 12. If P {1, 2, 3} , Q {3, 4} and R P Q , determine P R and R Q .
Solution : iven that, P {1, 2, 3} , Q {3, 4}
G
and R P Q = {1, 2, 3} {3, 4} {3}
(
P R = {1,2,3, } {}
3 = {1 , 3).(2, 3), (3, 3)} and R Q = {3} {3, 4} { 3, 3), (3, 4)}
(
Activity : 1. If

x
2

y
,1
3

1,

x
3

y
, find ( x, y) .
2

2. If P {1, 2, 3} Q {3, 4} and R {x, y} , find (P
,

Q) R and (P

Q) Q .

Example 13. Find the set where 23 is reminders in each case when 311 and 419 are divided by the natural numbers.
Solution : The numbers when 311 and 419 are divided by rational numbers and 23 is reminder, will be greater than 23 and will be common factors of 311 – = 288 and
23
419 – = 396.
23
Let, the set of factors of 288 greater than 23 is A and the set of factors of 396 is B .
Here,
288 1 288 2 144 3 96 4 72 6 48 8 36 9 32 12 24 16 18
A {24, 32, 36, 48, 72, 96, 144, 288}
Again,
396 1 396 2 198 3 132 4 99 6 66 9 44 11 36 12 33 18 22
B {33, 36, 44, 66, 99, 132, 198, 396}
A

B {24, 32, 36, 48, 72, 96, 144, 288}

{33, 36, 44, 66, 99, 132, 198, 396} = {36}

equired set is {36}
R
Example 14. If U {1, 2, 3, 4, 5, 6, 7, 8} , show that,
(i)
C {4, 5, 6, 7} ,
( A B) C ( A C) (B C)
Solution : (i)

A {1, 2, 6, 7} ,
( A B) A

B {2, 3, 5, 6} and B

and
(ii)

Math

29

In the figure, U by rectangle and the sets of A and B are denoted by two mutually intersecting circle sets. set Elements
A B
1, 2, 3, 4, 5, 6, 7
( A B)
4, 8
A
3, 4, 5, 8
1, 4, 7, 8
B
A ∩ B
A B
4, 8
( A B) A B
Solution : (ii) In figure, U by rectangle and sets of A, B, C are denoted by three mutually intersecting circles.
Observe ,
Set
Elements
A B
2, 6
2, 4, 5, 6, 7
( A B) C
A C
1, 2, 4, 5, 6, 7
B C
2, 3, 4, 5, 6, 7
( A C) (B C) 2, 4, 5, 6, 7

( A B) C ( A C) (B C)
Example 15. Among 100 students, 92 in Bangla, 80 in Math and 70 have passed in both subj cts in any exam. Express the information by Venn diagram and find how e many students failed in both subj cts. e Solution : In the Venn diagram, the rectangular region denotes set U of 100 students. and the set of passed students in Bangla and Math are denoted by B and o M . So, the Venn diagram is divided into four disj int sets which are denoted by
P, Q, R, F .
Here, the set of passed students in both subj cts Q B M whose numbers of e elements are 70 .
P the set of passed students in Bangla only, whose number of element = 92 70 18
R the set of passed student in Math only, whose number of elements = 80 70 10 e P Q R B M , the set of passed students in one and both subj cts, whose number of elements = 18 10 70 98 e F the set of students who failed in both subj cts, whose number of elements =
100 98 2
2 students failed in both subj cts. e 30

Math

Exercise 2.1
1. Express the following sets in tabular method :
(a) {x N : x2 9 and x3 130}
(b) {x Z : x2 5 and x3 36}
.
(c) {x N : x, factors of 36 and multiple of 6}
4
3
(d) {x N : x 25 and x 264}
2. Express the following sets in set builder method :
(b) {1, 2, 3, 4, 6, 9, 12, 18, 36}
(a) {3, 5, 7, 9, 11}
(d) { 4, 5, 6}
(c) {4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
3. If A {2, 3, 4} , B {1, 2, a} and C {2, a, b} , determine the sets given below:
(a) B \ C (b) A B
(c) A C
(d) A (B C) (e) A (B C) u 4. If U {1, 2, 3, 4, 5, 6, 7} , A {1, 3, 5} , B {2, 4, 6} and C {3, 4, 5, 6, 7} , j stify the followings :
(i) ( A B) A B
(ii) (B C) B C
(iii ) ( A B) C ( A C ) ( B C ) (iv)
( A B) C ( A C ) ( B C )
5. If Q {x, y} and R {m, n, l} , find P(Q) and P(R) .
6. If A {a, b} , B {a, b, c} and C A B , show that the number of elements of
P(C) is 2n , where n is the number of element of C .
7. (a) If ( x 1, y 2) ( y 2, 2 x 1) , find the value of x and y .
(b) If (ax cy, a2 c2) (0, ay cx) , find the value of ( x, y) .
(c) If (6x y, 13) (1, 3x 2 y) , find the value of ( x, y) .
8. (a) If P {a} , Q {b, c} then, find P Q and Q P .
(b) If A {3, 4, 5} , B {4, 5, 6} and C {x, y} , find ( A B) C .
(c) If P {3, 5, 7} , Q {5, 7} and R P \ Q , find (P Q) R .
9. If A and B are the sets of all factors of 35 and 45 respectively, find A B and A B .
10. Find the set of the number where 31 is the remainder in each case when 346 and
556 are divided by natural numbers.
11. Out of 30 students of any class, 20 students like football and 15 students like cricket. The number of students who like any one of the two is 10. Show with the help of Venn diagram, the number of students who do not like two of the sports.
12. Out of 100 students in any exam, 65% in Bangla, 48% in both Bangla and
English have passed and 15% have failed in both subj cts. e 31

Math

(a) Express the above information by Venn diagram along with brief description.
(b) Find the numbers who have passed only in Bangla and English.
(c) Find the union of two sets of the prime factors of the numbers who have passed and failed in both subj cts. e Relation
We know, the capital of Bangladesh is Dhaka and that of India is Delhi and Pkistan a is Islamabad. Here, there is a relation of capital with the country. The relation is country- apital relations. The above relation can be shown in set as follows: c Relation
Country
Bangladesh
India
akistan
P

Capital
Dhaka
Delhi
Islamabad

That is, country- apital relation = {Bangla desh, Dhaka), (India, Delhi), (Pkistan, c ( a Islamabad)}
.
z
C
If A and B are two sets, the non- ero subset of R of the artesian product A B of the sets is called relation of B from A .
Here, R is a subset of A B set, that is, R A B .
Example 15. Suppose, A {3, 5} and B {2, 4}
A B {3, 5} {2, 4} { 3, 2), (3, 4), (5, 2), (5, 4)}
(
R

(
{ 3, 2), (3, 4), (5, 2), (5, 4)}

If the condition is x

y , R { 3, 2), (5, 2), (5, 4)}
(

and if the condition is x y , R {3, 4}
If an element of set A is x and that of the set B is y and ( x, y) R , we write x R y and read as x is related to y . That the element x is R related to element y .
Again, if the relation of a set, from set A that is R A A , R is called A related. z So, if the relation is given between set A and B , non- ero subset of ordered pair
( x, y) with y B related to x A , is a relation.
Example 16. If P {2, 3, 4} , Q {4, 6} and y 2x is relation under consideration between the elements of P and Q , find the relation.

P {2, 3, 4} and Q {4, 6}
According to the question, R { x, y) : x P, y Q and y
(
Solution : iven that,
G

2x}

Here, P Q {2, 3, 4} {4, 6} { 2, 4), (2, 6), (3,4)(3, 6), (4, 4), (4, 6)}
(
R { 2, 4), (3, 6)}
(

32

Math

equired relation is { 2, 4), (3, 6)}
R
(
Example 17. If A {1, 2, 3} , B {0, 2, 4} and the relation x consideration between elements of C and D , find the relation.
Solution : iven that, A {1, 2, 3} , B {0, 2, 4}
G
According to the question, relation R { x, y) : x
(
Here, A B {1, 2, 3}

A, y

B and x

y 1 is under

y 1}

{0, 2, 4}

= { 1, 0), (1, 2), (1, 4), (2, 0), (2, 2), (2, 4), (3, 0), (3, 2), (3, 4)}
(
R { 1, 2), (3, 4)}
(

Activity : If C {2,5,6} D {4,5} and the relation x
,
between elements of C and D , find the relation.

y is under consideration

Functions
Let us observe the relation between sets A and B below :
Here, When y x 2 ,
1
y 3 for x 1
2
y 4 for x 2
3
y 5 for x 3

3
4
5

That is, for each value of x , only one value of y is obtained and the relation between x and y is made by y x 2 . Hence two variable x and y are so related that for any value of x , only one value of y is obtained even y is called the function of x . The function of x is generally expressed by y, f (x), g (x) , F (x) etc.
Let, y x2 2x 3 is a function. Here, for any single value of x , only one value of y is obtained. Here, both x and y are variables but the value of y depends on the value of x . So, x is independent variable and y is dependent variable.
Example 18. If f ( x)

x2 4x 3 , find f ( 1) .

Solution : iven that,
G

f ( x)

x2 4x 3

f ( 1) = ( 1)2 4( 1) 3 1 4 3 8

Example 19. If g ( x)

x3

ax2
3

3x 6 , for what value of a will be g ( 2)

Solution : iven that, g ( x) x ax
G
g ( 2) ( 2)3 a( 2)2 3( 2) 6
= 8 4a 6 6
= 8 4a = 4a 8
But g ( 2) 0

2

3x 6

0?

Math

33

4a 8 0 or, 4a 8 or, a 2 if a 2 , g ( 2) 0 .
Domain and Range
The first set of elements of the ordered pair of any relation is called its domain and the set of second elements is called its range.
Let R from set A to set B be a relation, that is, R A B . The set of first elements included in the ordered pair of R will be domain of R and the set of second elements will be range of R . The domains of R is expressed as Dom R and range is expressed as ange R .
R
Example 20. R elation S { 2, 1), (2, 2), (3, 2), (4, 5)} . Find the domain and range of
(
the relation.
Solution : iven that, S { 2, 1), (2, 2), (3, 2), (4, 5)}
G
(

In the relation S , the first elements of ordered pair are 2, 2, 3, 4 and second elements are 1, 2, 2, 5 .
Dom S {2, 3, 4} and ange S = {, 2, 5}
R
1
Example 21. If A {0, 1, 2, 3} and R { x, y) : x A, y A and y x 1} , express
(
R
R in tabular method and determine Dom R and ange R .
Solution : iven that, A {0, 1, 2, 3} and R { x, y) : x A, y A and y x 1}
G
(
From the stated conditions of R we get, y x y

Since 4

x 1

A we find the value of y

Now, for each x

A , (3, 4)

0
1

1
2

2
3

x 1.

3
4

R

R { 0, 1), (1, 2), (2, 3)}
(
R
Dom R {0, 1, 2} and ange

R {1, 2, 3}

Activity :
1. If S { 3, 8), ( 2, 3), ( 1, 0), (0, 1), (1, 0), (2, 3)} , find domain and range of
(

S.
2.

If S {x, y ) : x A, y A and y
R
Dom. S and ange S .

x 1} , where A

{ 3, 2, 1, 0} , find

Graphs :
The diagrammatic view of function is called graphs. In order to make the idea of function clear, the importance of graph is immense. French philosopher and

34

Math

mathematician ene Descartes (1596 1650) at first played a vital role in establishing
R
a relation between algebra and geometry. He introduced the modern way to coplanar geometry by defining the position of point on a plane by two intersecting perpendicular function. He defined the two intersecting perpendicular lines as axes and called the point of intersection origin. On any plane, two intersecting perpendicular straight lines XOX and YOY are drawn. The position of any point on this plane can be completely known by these lines. Each of these straight lines are called axis. Horizontal line e XOX is called x axis, Prpendicular line YOY is called y axis and the point of intersection of the two axes O is called origin.
The number with proper signs of the perpendicular distances of a point in the plane from the two axis are called the Cordinates of othat point. Let P be any point on the plane between the two axes. PM and PN are drawn perpendicular from the point P to XOX and YOY axes respectively. As a result, PM ON is the perpendicular distance of the point P from YOY and
PN OM is the perpendicular distance of P from the XOX . If PM x and
PN y , the coordinates of the point P is ( x, y ) . Here, x is called abscissa or x co- rdinate and y is called ordinate or y co- rdinate. o o
In artesian co- rdinate, the geometrical figure is shown easily. For this reason,
C
o generally we put the independent value along the x axis and the dependent value along the y axis. For some values of independent variables from the domain, we find the similar values of dependent variables and form ordered pair to draw the j graphs of the function y f (x) . Then place the ordered pair under (x, y) and oint the obtained points in freehands which is the graph of the function y f (x) .
Example 22. Draw the graph of the function y 2 x ; where 3 x 3 .
Solution : In the domain 3 x 3 , for some values of x , we determine some values of y and form a table :

x y 3
–

2
–

1
–

0

1

2

3

6
–

4
–

2
–

0

2

4

6

On the graph paper, taking the length of square as unit, we identify points of the table on the place and j int the points in free hand. o 35

Math

3x 1
, find the value of
3x 1

Example 23. If f ( x)

1

.
1

3x 1
3x 1

f ( x)

Solution : iven
G

1 x 1 f x

f

1
3
1
1
3 x x x
=
[multiplying the numerator and the
1
3
3 x
3
1
1
x x denomination by x ]
1
1 f 3 x 3 x x , [By componendo –
Dividendo]
or,
1
3 x 3 x
1
f x 6 3
=
2x x
3
equire value is
R
x
1
y3 3y2 1
Example 24. If f ( y )
, show that f f (1 y ) y y (1 y )
3

1 f x

f

=

1 y 1 y again, f (1 y )
=

3

3

2

1 y 1

1
1
1 y y

1 3y y3 y3 y3 3y2 1 y (1 y )

f ( y)

Solution : iven,
G

y2 y 1

1 3y y3 y3 y 1 y2 1 3y y3 y ( y 1)

(1 y ) 3 3(1 y ) 2 1
(1 y ){1 (1 y )}

1 3 y 3 y 2 y 3 3(1 2 y
(1 y )(1 1 y )

y2 ) 1

36

Math

=

1 3y 3y2
1 3y y3 y (1 y )

=
=
f

y3 3 6 y 3y2 1 y (1 y )

1 3y y3 y ( y 1)

1 y (1 3 y y 3 ) y ( y 1)

f (1 y ) .
Exercise 2⋅2

1.

Which one is the set of factors of 4 ?
(a) {8,16,24...}
(b) (1,2,3,4,8} (c) 2, 6, 8}
{

2.

If a relation of set B from set C is R , which one of the following is right ?
(a) R C
(b) R B
(c) R C B
(d) C B R
If A {6, 7, 8, 9, 10, 11, 12, 13} , answer the following questions :

3.

(d) 1, 2}
{

(i ) Which one is builder method of set A ?

(a) {x

N: 6

x 13}

(c) {x

N: 6

x 13} (d) {x

(b) {x

N: 6

x 13}

N: 6

x 13}

(ii ) Which one is the set of prime numbers ?

(a) {6, 8, 10, 12} (b) {7, 9, 11, 13}

(c) {7, 11, 13} (d) A {9, 12}

(iii ) Which is the set of multiple of 3 ?

(b) {6, 11}

(a) {6, 9}

(c) {9, 12}

(d) {6, 9, 12}

(iv) Which is the set of factor of greater even number ?
4.
5.
6.
7.
8.

(a) {1, 13}
(b) {1, 2, 3, 6}
(c) {1, 3, 9}
(d) {1, 2, 3, 4, 6, 12}
If A {3, 4} B {2, 4} , find the relation between elements of A and B
,
considering x y .
If C {2, 5} D {4, 6} and find the relation between element of C and D
,
considering relation x 1 y .
1
If f ( x) x 4 5 x 3 , the find value of f ( 1), f (2) and f
.
2
If f ( y )
If f ( x)

y 3 ky 3 4 y 8 , for which value of k will be f ( 2) x 3

6x

2

11x 6 , for which value of

x

0.

will be f ( x) 0 .

37

Math

1

9.

If f ( x)

f 2
2x 1 x , find the value of
2x 1
1
f

x2

1

.
1

1 x2 x4
1
, show that, g 2 g(x2 ) x2 x
11. Find the domain and range of the following relations :
(b) S { 2, 4), ( 1, 1), (0, 0), (1, 1), (2, 4)}
(a) R { 2, 1), (2, 2), (2, 3)}
(
(

10. If g ( x)

(c) F

1
5
, 0 , (1, 1), (1, 1), , 2 ,
2
2

5
,
2

2

12. Express the relations in tabular method and find domain and range of following relations :
(a) R { x, y ) : x A, y A and x y 1} where A { 2, 1, 0, 1, 2}
(
(b) F { x, y ) : x C , y C and x 2 y} where C { 1, 0, 1, 1, 3}
(
1 5 on graph paper.
13. P the points ( 3, 2), (0, 5), lot ,
2 6
14. P the three points (1, 2), ( 1, 1), (11, 7) on graph paper and show the three lot points are on the same straight line.
15. niversal set U {x : x N and x is an odd number}
U
A {x N : 2 x 7 }
B {x N : 3 x 6 }

N : x 2 5 and x 3 130 }
Express A in tabular method.
Find A and C B .
Find B C and P ( A C ) .
C {x

(a)
(b)
(c)

Chapter Three

Algebraic Expressions
Algebraic formulae are used to solve many algebraic problems. Moreover, many algebraic expressions are presented by resolving them into factors. That is why the problem solved by algebraic formulae and the contents of resolving expressions into factors by making suitable for the students have been presented in this chapter.
Moreover, different types of mathematical problems can be solved by resolving into factors with the help of algebraic formulae. In the previous class, algebraic formulae and their related corollaries have been discussed elaborately. In this chapter, those are reiterated and some of their applications are presented through examples.
Besides, extension of the formulae of square and cube, resolution into factors using remainder theorem and formation of algebraic formulae and their applications in solving practical problems have been discussed here in detail.
At the end of the chapter, the students will be able to –
Expand the formulae of square and cube by applying algebraic formulae
Explain the remainder theorem and resolve into factors by applying the theorem Form algebraic formulae for solving real life problems and solve the problems by applying the formulae.
3⋅1 Algebraic Expressions
Meaningful organization of operational signs and numerical letter symbols is called algebraic expression. Such as, 2a 3b 4c is an algebraic expression. In algebraic expression, different types of information are expressed through the letters a, b, c, p, q, r, m, n, x, y, z, ..... etc. These alphabet are used to solve different types of problems related to algebraic expressions. In arithmetic, only positive numbers are used, where as, in algebra, both positive and negative numbers including zero are used. Algebra is the generalization form of arithmetic. The numbers used in algebraic expressions are constants, their values are fixed.
The letter symbols used in algebraic expressions are variables, their values are not fixed, they can be of any value.
3⋅2 Algebraic Formulae
Any general rule or resolution expressed by algebraic symbols is called Algebraic
Formula. In class VII and VIII, algebraic formulae and related corollaries have been discussed. In this chapter, some applications are presented on the basis of that discussion. 39

Math

Formula 1. (a b)2

a2 2ab b2

Formula 2. (a b) 2

a2

2ab b 2

Remark : It is seen from formula 1 and formula 2 that, adding 2ab or 2ab with a2 b2 , we get a perfect square, i.e. we get, (a b)2 or (a b)2 . Substituting b instead of b in formula 1, we get formula 2 :
{a ( b)}2 a2 2a( b) ( b)2
That is, (a b)2 a2 2ab b2 .
Corollary 1. a2 b2 (a b)2 2ab
Corollary 2. a2 b2 (a b)2 2ab
Corollary 3. (a b)2 (a b)2 4ab
Proof : (a b)2 a2 2ab b2
= a2 2ab b2 4ab
= (a b)2 4ab
Corollary 4. (a b)2 (a b)2 4ab
Proof : (a b)2 a2 2ab b2
= a2 2ab b2 4ab
= (a b)2 4ab
( a b) 2 ( a b) 2
Corollary 5. a 2 b 2
2
Proof : From formula 1 and formula 2, a2 2ab b2

(a b)2

a2 2ab b2

(a b)2

Adding, 2a2 or, Hence,

2b2

(a b)2 (a b)2

2(a2 b2 ) (a b)2 (a b)2

(a 2 b 2 )

Corollary 6. ab

( a b) 2

( a b) 2
2

a b

2

a b

2

2
2
Proof : From formula 1 and formula 2, a2 2ab b2

(a b)2

a2 2ab b2

(a b)2

Subtracting,

4ab

(a b)2

(a b)2

40

Math

or,

( a b)
4

ab

2

a b

Hence, ab

( a b)
4
2

2

a b

2

2
2
Remark : roduct of any two quantities can be expressed as the difference of two
P
squares by applying the corollary 6.
Formula 3. a2 b2 (a b)(a b)
That is, the difference of the squares of two expressions = sum of two expressions difference of two expressions.
Formula 4. ( x a)(x b) x2 (a b) x ab
That is, ( x a)( x b) x2 (algebraic sum of a and b ) x (the product of a and b )
Extension of Formula for Square
There are three terms in the expression a b c . It can be considered the sum of two terms (a b) and c .
Therefore, by applying formula 1, the square of the expression a b c is,
(a b c)2

{ a b) c}2
(

= (a b)2

2(a b)c c2

= a2 2ab b2 2ac 2bc c2
= a2 b2 c2 2ab 2bc 2ac .
Formula 5. (a b c)2 a2 b2 c2 2ab 2bc 2ac .
Corollary 7. a2 b2 c2

(a b c)2 2(ab bc ac)

Corollary 8. 2(ab bc ac) (a b c)2 (a2 b2 c2 )
Observe : Applying formula 5, we get,
(i) (a b c)2 {a b ( c)}2
= a2 b2 ( c)2 2ab 2b( c) 2a( c)
= a2 b2 c2 2ab 2bc 2ac
(ii) (a b c)

2

{a ( b) c}2

= a2 ( b)2 c2 2a( b) 2( b)c 2ac
= a2 b2 c2 2ab 2bc 2ac
(iii) (a b c) {a ( b) ( c)}2
2

= a2 ( b)2 ( c)2 2a( b) 2( b)( c) 2a( c)

Math

41

= a2 b2 c2 2ab 2bc 2ac
Example 1. What is the square of (4x 5 y) ?
Solution : (4 x 5 y ) 2

(4 x) 2

2 (4 x) (5 y ) (5 y ) 2

= 16 x 2 40 xy 25 y 2
Example 2. What is the square of (3a 7b) ?
Solution : (3a 7b) 2

(3a ) 2

2 (3a ) (7b) (7b) 2

= 9a 2 42ab 49b 2
Example 3. Find the square of 996 by applying the formula of square.
Solution : (996)2 (1000 4)2
= (1000) 2 2 1000 4 (4) 2
= 1000000 8000 16
= 1000016 8000
= 992016
Example 4. What is the square of a b c d ?
Solution : (a b c d ) 2 { a b) (c d )}2
(
= ( a b) 2
= a2

2(a b)(c d ) (c d ) 2

2ab b 2

2(ac ad bc bd ) c 2

2cd d 2

= a 2 2ab b 2 2ac 2ad 2bc 2bd c 2 2cd d 2
= a 2 b 2 c 2 d 2 2ab 2ac 2ad 2bc 2bd 2cd
Activity : Find the square with the help of the formulae :
1. 3 xy 2ax
2. 4 x 3 y
3. x 5 y 2 z
Example 5. Simplify :
(5 x 7 y 3 z ) 2 2(7 x 7 y 3 z )(5 x 7 y 3 z ) (7 x 7 y 3 z ) 2
Solution : Let, 5 x 7 y 3 z a and 7 x 7 y 3 z b iven expression =
G

a 2 2.b.a b 2
= a 2 2ab b 2
= ( a b) 2

= { 5 x 7 y 3 z ) (7 x 7 y 3 z )}2
(
[substituting the values of a and b ]
= (5 x 7 y 3 z 7 x 7 y 3 z ) 2
= (12 x) 2
= 144x 2

42

Math

Example 6. If x

2 and xy

y

Solution : ( x y ) 2 x y
100

( x y) 2
10

24 , what is the value of x

4 xy

( 2) 2

4 24

y ?

4 96 100

Example 7. If a 4 a 2b 2 b 4 3 and a 2 ab b 2 3 , what is the value of a2 b2 ?
Solution : a 4 a 2 b 2 b 4 (a 2 ) 2 2a 2 b 2 (b 2 ) 2 a 2 b 2
= (a 2 b 2 ) 2 (ab) 2
= (a 2 b 2 ab)(a 2 b 2 ab)
= (a 2 ab b 2 )(a 2 ab b 2 )
2
2
3 3(a ab b ) [substituting the values]
3
or, a 2 ab b 2
1
3
Now adding, a 2 ab b 2 3 and a 2 ab b 2 1 we get, 2(a 2 b 2 ) 4
4
or, a 2 b 2
2
2 a2 b2 2
Example 8. P that, (a b) 4 rove Solution : (a b) 4

( a b) 4

= { a b) 2
(
= 2( a 2

( a b) 4

8ab(a 2

b2 )

(
{ a b) 2 }2 { a b) 2 }2
(

( a b) 2 } a b) 2
{
(

( a b) 2 }

b 2 ) 4ab

[ ( a b) 2
= 8ab(a

( a b) 4

2

( a b) 2

2(a 2

b 2 ) and (a b) 2 (a b) 2

4ab]

2

b )

( a b) 4

8ab(a 2

b2 )

Example 9. If a b c 15 and a 2 b 2 c 2 83 , what is the value of ab bc ac ?
Alternative method,
Soluton :
We know ,
Here, 2(ab bc ac)
(a b c) 2 (a 2 b 2 c 2 ) 2(ab bc ac)
2
2
2
2
(a b c) (a b c ) or, (15) 2 83 2(ab bc ac)
= (15) 2 83
= 225 83
= 142
142
71 ab bc ac
2

or, 225 83 2(ab bc ac) or, 2(ab bc ac) 142
142
ab bc ac
71
2

43

Math

Example 10. If a b c 2 and ab bc ac 1 , what is the value of
(a b) 2 (b c) 2 (c a ) 2 ?
Solution : (a b) 2

(b c) 2

(c a ) 2

= a 2 2ab b 2 b 2 2bc c 2 c 2 2ca a 2
= (a 2 b 2 c 2 2ab 2bc 2ca) (a 2 b 2 c 2 )
= (a b c) 2
2

(
{ a b c) 2

2(ab bc ac)}

2

= (2) (2) 2 1
=4 4 2 8 2 6
Example 11. Express (2 x 3 y )(4 x 5 y ) as the difference of two squares.
Solution : Let, 2 x 3 y a and 4 x 5 y b iven expression =
G
= 2x 3y 4x 5 y

2

2

=

6x 2 y
2

= 2(3x y )
2

2

a b
2

ab

8y

2x

2

a b
2

2x 3y 4x 5 y
2

2

2

[substituting the values of a and b ]

2

2
2

2(4 y x)
2

= (3 x y ) 2 (4 y x) 2
(2 x 3 y )(4 x 5 y )

2

(3 x

y) 2

(4 y

x) 2

Activity : 1. Simplify : (4 x 3 y ) 2 2(4 x 3 y )(4 x 3 y ) (4 x 3 y ) 2
2. If x y z 12 and x 2 y 2 z 2 50 , find the value of ( x y ) 2 ( y z ) 2 ( z x) 2 .
Exercise 3⋅1
1.

Find the square with the help of the formulae :
1
2
(d) a
(e) 4 y 5 x
(a) 2a 3b (b) 2ab 3bc (c) x 2 a y2

(f) ab c

(g) 5 x 2 y (h) x 2 y 4 z (i) 3 p 4q 5r (j 3b 5c 2a (k) ax by cz
)
(l) a b c d (m) 2a 3 x 2 y 5 z (n) 101
(o) 997 (p) 1007
2. Simplify :
(a) (2a 7) 2 2(2a 7)(2a 7) (2a 7) 2

44

Math

(b) (3 x 2 y ) 2

2(3 x 2 y )(3 x 2 y ) (3 x 2 y ) 2

(c) (7 p 3r 5 x) 2

2(7 p 3r 5 x)(8 p 4r 5 x) (8 p 4r 5 x) 2

3n p) 2 (2m 3n p) 2 2(2m 3n p)(2m 3n
6 35 2 6 35 3 65 3 65 3 65
5874 3774 3774 7548 5874
7529 7519 7519
7529 7519
2345 2345 759 759
(h)
2345 759
3. If a b 4 and ab 60 , what is the value of a b ?
4. If a b 7 and ab 12 , what is the value of a b ?
5. If a b 9m and ab 18m 2 , what is the value of a b ?
6. If x y 2 and xy 63 , what is the value of x 2 y 2 ?
1
1
7. If x
4 , prove that, x 4
322 . x4 x
2
1
?
8. If 2 x
3 , what is the value of x 2 x x2
1
1
1
9. If a
.
a4
2 , show that, a 2
2
a a4 a
7 and a b
5 , prove that, 8ab(a 2 b 2 ) 24 .
10. If a b
(d) (2m
(e) 6 35
(f) 5874
7529
(g)

p)

11. If a b c 9 and ab bc ca 31 , find the value of a 2 b 2 c 2 .
12. If a 2 b 2 c 2 9 and ab bc ca 8 , what is the value of (a b c) 2 ?
13. If a b c 6 and a 2 b 2 c 2 14 , find the value of
(a b) 2 (b c) 2 (c a ) 2 .
14. If x y z 10 and xy yz zx 31 , what is the value of
( x y ) 2 ( y z ) 2 ( z x) 2 ?
,
15. If x 3, y 4 and z 5 find the value of
2
2
2
9 x 16 y 4 z 24 xy 16 yz 12 zx .
16. rove that,
P

x

y
2

2

x

y
2

2

2

x2

y2

2

x2

2

y2
2

17. Express (a 2b)(3a 2c) as the difference of two squares.
18. Express ( x 7)( x 9) as the difference of two squares.
19. Express x 2 10 x 24 as the difference of two squares.

2

.

45

Math

20. If a 4 a 2b 2 b 4 8 and a 2 ab b 2
3⋅3 Formulae of Cubes
Formula 6. (a b) 3 a 3 3a 2 b 3ab 2 b 3
= a3
Proof : (a b)

3

b3

3ab(a b)

(a b)(a b)
= (a b)(a 2
= a(a 2

4 , find the value of (i ) a 2

2

2ab b 2 )

2ab b 2 ) b(a 2

2ab b 2 )

= a 3 2a 2 b ab 2 a 2 b 2ab 2 b 3
= a 3 3a 2b 3ab 2 b 3
= a 3 b 3 3ab(a b)
Corollary 9. a 3 b 3

(a b) 3 3ab(a b)

Formula 7. (a b) 3

a3
=a

Proof : (a b)

3

3a 2 b 3ab 2
3

b

3

3ab(a b)

(a b)(a b)
= (a b)(a 2

b3

2

2ab b 2 )

= a (a 2 2ab b 2 ) b(a 2 2ab b 2 )
= a 3 2a 2b ab 2 a 2b 2ab 2 b 3
= a 3 3a 2 b 3ab 2 b 3
= a 3 b 3 3ab(a b)
Observe : Substituting b instead of b in formula 6, we get formula 7 :
{a ( b)}3 a 3 ( b) 3 3a( b){a ( b)}
That is, (a b) 3

a 3 b 3 3ab(a b)

Corollary 10. a 3 b 3

(a b) 3 3ab(a b)

Formula 8. a 3

b3

(a b)(a 2

Proof : a 3 b 3

(a b) 3 3ab(a b)
= (a b){ a b) 2
(

ab b 2 )
3ab}

= (a b)(a

2

2ab b 2 3ab)

= (a b)(a

2

ab b 2 )

Formula 9. a 3 b 3
Proof : a 3

b3

(a b)(a 2

( a b) 3

ab b 2 )

3ab(a b)

= (a b){ a b) 2 3ab}
(

b 2 , (ii ) ab .

46

Math

= (a b)(a 2

2ab b 2

2

3ab)

2

= (a b)(a ab b )
Example 12. Find the cube of 2 x 3 y .
Solution : (2 x 3 y ) 3

(2 x) 3

3(2 x) 2 3 y 3 2 x(3 y ) 2

= 8x3

3 4x2 3y 3 2x 9 y 2

3

2

= 8 x 36 x y 54 xy
Example 13. Find the cube of 2 x y .
Solution : (2 x y ) 3

2

27 y

(2 x) 3 3 (2 x) 2 y 3 2 x y 2
= 8 x 3 3 4 x 2 y 6 xy 2
= 8x

3

2

12 x y 6 xy

2

(3 y ) 3
27 y 3

3

y3

y3 y3 Activity : Find the cube with the help of the formulae.
2. 3 x 4 y
3. 397
1. 3 x 2 y
Example 14. If x 37 , what is the value of 8 x 3 72 x 2
Solution : 8 x 3 72 x 2 216 x 216
= (2 x) 3 3 (2 x) 2 6 3 2 x.(6) 2 (6) 3

216 x 216 ?

= ( 2 x 6) 3
= (2 37 6) 3 [substituting the values]
= (74 6) 3
= (80) 3
= 512000
8 and xy

Example 15. If x

y

Solution : x 3

8( x

y3

= ( x y)

3

=8

3

y3

8( x

y) 2

3 xy ( x y ) 8{ x y ) 2
(

= (8) 3 3 5 8 8(8 2
3

5 , what is the value of x 3

4 5)

15 8 8(64 20)

= 8 15 8 8 84
= 8(8 2 15 84)
= 8(64 15 84)
= 8 163
= 1304

4 xy}
[substituting the values]

y) 2 ?

47

Math

Example 16. If a 2

3a 1 0 , what is the value of a 3

Solution : iven that,
G

a2

or, a 2 1 or, a2 a a2 1 a 1 or, a a 1 a3 a3
3

1 a = a
3

or,

3

iven expression =
G

=

3a 1 0

3a

1 a 1 a 1
[ a a 3
3

1 a a

3a

3

3 3

3]

=3 3 3 3
=0
Example 17. Simplify :
(a b)(a 2 ab b 2 ) (b c)(b 2 bc c 2 ) (c a )(c 2
Solution : (a b)(a 2
= a3
=0

b3

c3

a

3

3

2
3

3

1 a3 18 3.

2

[multiplying numerator and denominator by
2
3

3

2

2

2

3

3

2

=

2

3
2
3 2

2

3

= a a3

2

=

1 a =

ca a 2 )

1

3
2

=

c3

2 , prove that, a 3

3

Solution : iven that,
G
1 a ca a 2 )

ab b 2 ) (b c)(b 2 bc c 2 ) (c a )(c 2

b3

Example 18. If a

1
?
a3

2

3

3

2

2

2 3

3

2 ]

48

Math

Now, a 3

1 a3 a

= 2 3
= 23

3

1 a 3 a
3

3 2 3
3

3

1 a a

1 a [ a

1 a 2 3]

3 2 3

=8 3 3 6 3
= 24 3 6 3
= 18 3 (proved)
Activity : 1. If x
2 , what is the value of 27 x 3 54 x 2 36 x 8 ?
2. If a b 5 and ab 6 , find the value of a 3 b 3 4(a b) 2 .
1
5
3 , find the value of x 3
3. If x
.
x3
Exercise 3⋅2
1.

2.

Find the cube with the help of the formulae :
(b) 2 x 2 3 y 2 (c) 4a 5 x 2 (d) 7 m 2 2n (e) 403 (f) 998
(a) 2 x 5
(g) 2a b 3c (h) 2 x 3 y z
Simplify :
(a) (4a 3b) 3 3(4a 3b) 2 (2a 3b) 3(4a 3b)(2a 3b) 2 (2a 3b) 3
(b) (2 x

y ) 3 3(2 x

y ) 2 (2 x y ) 3(2 x

y )(2 x y ) 2

(2 x y ) 3

(c) (7 x 3b) 3 (5 x 3b) 3 6 x(7 x 3b)(5 x 3b)
(d) ( x 15) 3 (16 x) 3 3( x 15)(16 x)
(e) (a b c) 3 (a b c) 3 6(b c){a 2 (b c) 2 }
(f) (m n) 6 (m n) 6 12mn(m 2
(g) ( x

y )( x 2

xy

(h) (2 x 3 y 4 z )

3

n2 )2

y 2 ) ( y z )( y 2
(2 x 3 y 4 z )

3

yz z 2 ) ( z x)( z 2
12 x{4 x

2

zx x 2 )

2

(3 y 4 z ) }

3.
4.
5.

If a b 5 and ab 36 , what is the value of a 3 b 3 ?
If a 3 b 3 513 and a b 3 , what is the value of ab ?
If x 19 and y
12 , find the value of 8 x 3 36 x 2 y 54 xy 2

6.
7.

If a 15 , what is the value of 8a 3 60a 2 150a 130 ?
If a 7 and b
5 , what is the value of

27 y 3 .

Math

49

(3a 5b) 3
8.
9.

If a b
If` x

(4b 2a) 3 m, a

2

b

2

3(a b)(3a 5b)(4b 2a) ? n and a 3

y 1 , show that, x

3

y

3

b3 xy p 3 , show that, m 3
( x y)

2 p3

3mn .

2

3 and ab 2 , find the value of (a) a 2 ab b 2 and (b) a 3 b 3 .
5 and ab 36 , find the value of (a) a 2 ab b 2 and (b) a 3 b 3 .
1
. a , find the value of m 3 m3 1
.
p , find the value of x 3 x3 1
1 , show that, a 3
4.
a3 c 0 , show that,
(b c) 2 (c a ) 2 (a b) 2
(a) a 3 b 3 c 3 3abc
(b)
1.
3bc
3ca
3ab
If p q r , show that, p 3 q 3 r 3 3 pqr

10. If a b
11. If a b
1
12. If m m 1
13. If x x 1
14. If a a 15. If a b

16.

17. If 2 x

2 x 18. If a

6

5 , find the value of

a6 1
.
a3

19. If x 3

1 x3 18 3 , prove that, x

3

20. If a 4

a 2 1 0 , prove that, a 3

1 a3 3 , show that, 8 x 3

1 x3 63 .

2.
0.

3⋅4 Resolution into Factors
If an expression is equal to the product of two or more expressions, each of the latter expressions is called a factor of the former expression.
After finding the possible factors of any algebraic expression and then expressing the expression as the product of these factors are called factorization or resolution into factors. The algebraic expressions may consist of one or more terms. So, the factors may also contain one or more terms.
Some process of resolving expressions into factors :
(a) If any polynomial expression has common factor in every term, at first they are to be found out. For example,
(i ) 3a 2 b 6ab 2 12a 2 b 2 3ab(a 2b 4ab)

50

Math

(ii ) 2ab( x y ) 2bc( x y ) 3ca( x y ) ( x y )(2ab 2bc 3ca)
(b) Expressing an expression in the form of a perfect square ;
Example 1. esolve into factors : 4 x 2 12 x 9 .
R
Solution : 4 x 2 12 x 9 (2 x) 2 2 2 x 3 (3) 2

= (2 x 3) 2
Example 2. R esolve into factors :

(2 x 3)(2 x 3)

9x2

30 xy

25 y 2 .

Solution : 9 x 2 30 xy 25 y 2
= (3 x) 2

2 3 x 5 y (5 y ) 2

= (3 x 5 y ) 2 (3 x 5 y )(3 x 5 y )
(c) Expressing an expression as the difference of two squares and then applying the formula a 2 b 2 (a b)(a b) :
Example 3. R eslove into factors : a 2 1 2b b 2 .
Solution : a 2 1 2b b 2 a 2 (b 2 2b 1)
= a 2 (b 1) 2 {a (b 1)} a (b 1)}
{
= (a b 1)(a b 1)
Example 4. R esolve into factors : a 4 64b 4 .
Solution : a 4 64b 4 (a 2 ) 2 (8b 2 ) 2
= (a2)2

2 a2 8b2

= (a2 8b2)2
= (a2 8b2
= (a 2

(8b2)2 16a2b2

(4ab)2

4ab)(a2 8b2

4ab)

4ab 8b 2 )(a 2 4ab 8b 2 )

Activity : R esolve into factors :
2
1. abx acx3 adx4
2. xa2 144xb2
(d) Using the formula x

2

(a b) x ab

3. x2

2xy 4 y 4

( x a)(x b) :

2

Example 5. R esolve into factors : x 12x 35 .
Solution : x2 12x 35 x2 (5 7) x 5 7
= ( x 5)(x 7)
In this method, a polynomial of the form x2 px q can be factorized, if two integers a and b can be found so that, it is a b p and ab q . For this, two factors of q with their signs are to be taken whose algebraic sum is p . If q 0 , a and b will be of same signs and if q 0 , a and b will be of opposite signs.

51

Math

Example 6. R esolve into factors : x2 5x 6 .
Solution : x2 5x 6 x2 ( 2 3) x ( 2)( 3)
= ( x 2)(x 3)
Example 7. R esolve into factors :
Solution : x 2 2 x 35

x2

2 x 35 .

x2

x 20 .

= x2 ( 7 5) x ( 7)( 5)
= ( x 7)( x 5)
Example 8. R esolve into factors :
2

Solution : x x 20
= x 2 (5 4) x (5)( 4)
= ( x 5)(x 4)
(e) By middle term break-up method of polynomial of the form of ax2 bx c :
If ax2 bx c (rx p)(sx q) ax 2 bx c rsx 2 (rq sp) x pq
That is, a rs, b rq sp and c pq .
Hence, ac rspq (rq)( sp) and b rq

sp

Therefore, to determine factors of the polynomial ax2 bx c , ac that is, the product of the coefficient of x2 and the term free from x are to be expressed into two such factors whose algebraic sum is equal to b , the coefficient of x .
Example 9. R esolve into factors : 12x2 35x 18 .
Solution : 12x2 35x 18
Here, 12 18 216 27 8 and
12 x 2 35 x 18 12 x 2 27 x
= 3x(4x 9)
= (4x 9)(3x

27 8 35
8 x 18
2(4x 9)
2)

Example 10. R esolve into factors : 3x2 x 14 .
Solution : 3x2 x 14 3x2 7 x 6x 14
= x(3x 7) 2(3x 7)
= (3x 7)(x 2)
Activity : R esolve into factors :
2
2. 16x3 46x2 15x
3. 12x2 17 x 6
1. x x 56
(f) Expressing the expression in the form of perfect cubes :

52

Math

8x3 36x2 y 54xy2

Example 11. esolve into factors :
R
Solution : 8x3 36x2 y 54xy2

27 y3

= (2x)3 3 (2x)2 3 y 3 2x (3 y)2
= (2x 3 y)3

(a b)(a 2

Solution : (i) 8a3 27b3
= (2a 3b){ 2a)2
(
= (2a 3b)(4a2

b3

4

a2

and a4

4a2 16

b 2 ) and

ab

(ii) a6 64

(2a)3 (3b)3
2a 3b (3b)2}

6ab 9b2)

= (a2 4){ a2 )2 a2 4 (4)2}
(
But, a2

b)( a 2

(i) 8a3 27b3

(ii) a6 64 = (a 2 ) 3 (4) 3
4)(a4

(a

ab b 2 ) :

Example 12. esolve into factors :
R

= (a2

(3 y)3

(2x 3 y)(2x 3 y)(2x 3 y)

3
(g) Applying the formulae : a

a3 b3

27 y3 .

= (a 3 8)(a 3 8)

4a2 16)

22

(a
(a2)2

2)(a 2)
(4)2

= (a2

4)2

2(a2)(4) 4a2

= (a2

4)2
4)2

4 2a)(a2

2

2

2 3 )(a 3

= (a 2)(a 2

23 )
2a 4) (a 2)(a 2

= (a 2)(a 2)(a

2

2a 4)(a

2

2a 4)
2 a 4)

(2a)2

= (a2

4a2

= (a 3

4a2

= (a2

Alternative method : a 6 64 = (a 3 ) 2 8 2

= (a

2a 4)(a

4 2a)

2a 4)

6

a 64
(a 2)(a 2)(a 2 2a 4)(a 2 2a 4)
Activity : esolve into factors:
R
4
1. 2x 16x 2. 8 a3 3a2b 3ab2 b3
3. (a b)3

(a b)3

(h) Factors of the expression with fractional coefficients :
Factors of the expressions with fraction may be expressed in different ways.
1
1
1 2 a 1
For example, a3 a3 a a 27
3
3 9
33
1
1
1
Again, a3
(27a3 1)
(
{ 3a)3 (1)3}
27 27
27

53

Math

1
(3a 1)(9a2 3a 1)
27
Here, in the second solution, the factors involving the variables are with integral coefficients. This result can be expressed as the first solution :
1
(3a 1)(9a2 3a 1)
27
1
1
= (3a 1)
(9a 2 3a 1)
3
9
1
a 1
= a a2 3
3 9

=

Example 13. R esolve into factors :

x3 6x2 y 11xy 2 6 y3 .

Solution : x3 6x2 y 11xy 2 6 y3
= {x 3 3 x 2 2 y 3 x( 2 y ) 2
= ( x 2 y)3

2 y3

y 2( x 2 y)

= ( x 2 y ){ x 2 y ) 2
(
= ( x 2 y)(x 2 y

y 2}

y)(x 2 y

= ( x 2 y)(x 3 y)(x
= (x

(2 y ) 3 } xy 2

y)

y)

y)(x 2 y)(x 3 y)

Activity : R esolve into factors:
1 2 7
1
1
2. a3
1. x x 2
6
3
8

3. 16 x 2 25 y 2 8 xz 10 yz
Exercise 3⋅3

R esolve into factors (1 to 43) :
1.
3.

2. a2 ab ac bc ab a b 1
( x y )( x y ) ( x y )( y z ) ( x y )( z x) 4. ab( x y) bc( x y)

5.
7.

9x2 24x 16 x4 6x2 y 2 y 4

6.
8.

9.
11.

4a2 12ab 9b2 4c2 a 2 6a 8 y 2 2 y

10. 9x4 45a2 x2 36a4
12. 16x2 25 y 2 8xz 10 yz

13.

2b 2 c 2 2c 2 a 2 x4 x2 20

15.

a4

27 a 2 1

(a2 b2 )( x2

y 2 ) 4abxy

2a 2 b 2 a 4 b 4 c 4 14. x2 13x 36
16. a2 30a 216

54

Math

17.

x 6 y 6 x3 y 3 6

18. a8 a4 2

19.
21.
23.

a2b2 8ab 105
4x4 25x2 36
9x2 y 2 5xy 2 14 y 2

x2 37a 650
22. 12 x 2 38 x 20
24. 4x4 27 x2 81

25.

ax2 (a2 1) x a

26. 3(a2 2a)2 22(a2 2a) 40

27.

14( x z)2 29( x z)( x 1) 15( x 1)2

28.

(4a 3b)2

29.

(a 1) x2 a2 xy (a 1) y 2

30. 24x4 3x

31.

(a2 b2 )3 8a3b3

32.

33.

a3 6a2 12a 9

34. a3 9b3 (a b)3

35.

8x3 12x2 6x 63

36. 8a3

37.

a3

39.
41.
43.

2(4a 3b)(a

20.

2b) 35(a

1
8

38.

1
1
2 4a
2
a
4a
( x 5)( x 9) 15
( x 1)( x 3)( x 5)( x 7) 64
4a2

2b)2

x3 3x2 3x 2 b3 27

a6 b6 27

40. (3a 1)3 (2a 3)3
42. ( x 2)( x 3)( x 4)( x 5) 48

44. Show that, x3 9x2 26x 24 ( x 2)( x 3)( x 4)
45. Show that, ( x 1)( x 2)(3x 1)(3x 4) (3x2 2x 1)(3x2 2x 8)
3⋅5 Remainder Theorem
We observe the following example :
If 6x2 7 x 5 is divided by x 1 , the what is quotient and remainder?
Dividing 6x2 7 x 5 by x 1 in common way, we get, x 1 ) 6x2 7 x 5 ( 6x 1
6x2 6x x 5 x 1

4
Here, x 1 is divisor, 6x2 7 x 5 is dividend, 6 x 1 is quotient and 4 is remainder.
We know, dividend = divisor quotient + remainder 55

Math

Now, if we indicate the dividend by f (x) , the quotient by h(x) , the remainder by r and the divisor by ( x a) , from the above formula, we get, f ( x) ( x a) h( x) r .......... (i) this formula is true to all values of a .
Substituting x a in both sides of (i), we get, f ( a ) ( a a ) h( a ) r 0 h( a ) r r
Hence, r f (a) .
Therefore, if f (x) is divided by ( x a) , the remainder is f (a) . This formula is known as remainder theorem. That is, the remainder theorem gives the remainder when a polynomial f (x) of positive degree is divided by ( x a) without performing actual division. The degree of the divisor polynomial ( x a) is 1. If the divisor is a factor of the dividend, the remainder will be zero and if it is not zero, the remainder will be a number other than zero.
Proposition : If the degree of f (x) is positive and a 0 , f (x) is divided by b . a Proof : Degree of the divisor ax b , (a

(ax b) , remainder is f

0) is 1.

Hence, we can write, f ( x) (ax b) h( x) r f ( x)

x

b a a x

b a h( x) r

a h( x ) r

Observe that, if f (x) is divided by x b , quotient is a h(x) and remainder is r . a Here, divisor = x

b a Hence, according to remainder theorem, r

f

b a b
.
a
Corollary : ( x a) will be a factor of f (x) , if and only if f (a) 0 .
Proof : Let, f (a) 0
Therefore, according to remainder theorem, if f (x) is divided by ( x a) , the remainder will be zero. That is, ( x a ) will be a factor of f (x) .
Cnversely, let, ( x a) is a factor of f (x) . o Therefore, f ( x) ( x a) h( x) , where h(x) is a polynomial.
Ptting x a in both sides, we get, u Therefore, if f (x) is divided by (ax b) , remainder is f

56

Math

f (a)

(a a) h(a) 0 f (a) 0 .
Hence, any polynomial f (x) will be divisible by ( x a) , if and only if f (a) 0 .
This formula is known as factorisation theorem or factor theorem.
Corollary : If a 0 , the polynomial ax b will be a factor of any polynomial b f (x) , if and only if f
0.
a

Proof : a

0, ax b = a x

b a will be a factor of f (x) , if and only if x

b a =

b b is a factor of f (x) , i.e. if and only if f
0 . This method of a a determining the factors of polynomial with the help of the remainder theorem is also called the Vanishing method.
Example 1. esolve into factors : x3 x 6 .
R
Solution : Here, f ( x) x3 x 6 is a polynomial. The factors of the constant 6 are 1, 2, 3 and 6.
Ptting, x 1, 1 , we see that the value of f (x) is not zero. u But putting x 2 , we see that the value of f (x) is zero. x i.e., f (2) 2 3 2 6 8 2 6 0
Hence, x 2 is a factor of f (x) f (x) = x 3

x 6

= x 3 2 x 2 2 x 2 4 x 3x 6
= x2( x 2) 2x( x 2) 3( x 2)
= ( x 2)(x2

2x 3)

Example 2. esolve into factors : x3 3xy2 2 y3 .
R
Solution: Here, consider x a variable and y a constant.
We consider the given expression a polynomial of x .
Let, f ( x) x3 3xy2 2 y3
Then, f ( y)
(x

y3 3 y y 2 2 y3

3 y3 3 y3

0

y) is a factor of f (x) .

Now, x3 3xy2
= x3

x2 y

2 y3 x2 y

x2

Again let, g ( x) xy2 2xy2

2 y3

g ( y)

y2

y2

xy 2 y 2

2y2

0

Math

57

( x y) is a factor of g (x)

= x2 ( x y) xy( x y) 2 y 2 ( x y)
2

x2

2

= (x

y)(x

= (x
= (x
= (x

y)(x2 2xy xy 2 y 2)
y){x( x 2 y) y( x 2 y)}
y)(x 2 y)(x y)

xy 2 y )

xy 2 y 2

= x 2 xy 2 xy 2 y 2
= x( x y ) 2 y ( x y )
= ( x y )( x 2 y ) x3 = ( x y) 2 ( x 2 y)

3 xy 2

2 y3

(x

y) 2 ( x 2 y)

Example 3. R esolve into factors : 54x4 27x3a 16x 8a .
Solution : Let, f ( x) 54x4 27 x3a 16x 8a
1
a
2

then, f

54

1 a 2

4

27 a

1 a 2

3

1 a 2

16

8a

27 4 27 4 a a 8a 8a 0
8
8 a i.e., 2 x a is a factor of f (x) . x 2

=
1
a
2

x

Now, 54x4 27 x3a 16x 8a

27 x3(2x a) 8(2x a) (2x a)(27 x3 8)

(2 x a ){ 3 x) 3 (2) 3 } = (2x a)(3x 2)(9x2 6x 4)
(
Activity : R esolve into factors :
3
2. 2x3 3x2
1. x 21x 20

3. x3

3x 1

6x2 11x 6

Exercise 3⋅4 esolve into factors :
R
1.
6x2 7 x 1
3.
x3 7 xy2 6 y3
5.
7.
9.
11.
13.
15.
17.
19.

2x x3 a3 a3 a3 x3 2

x 3
2 x 2 5x 6
3a 36 a2 10a 8
7a2b 7ab2 b3
6x2 y 11xy2 6 y3

4x4 12x3 7 x2 3x 2
4x3 5x2 5x 1

2.
4.

3a 3

6.

2

x

2

3x

2a 5

5x 6

8. x3 10. a 4
12. x3
14. x3
16. 2 x 4

7x 6
4x2 x 6
4a 3
3x2 4x 4 x 24
3x 3 3x 2

x6 x5 x4 x3 x2
20. 18x3 15x2 x 2

18.

x

58

Math

3⋅6 Forming and applying algebraic formulae in solving real life problems
In our daily business we face the realistic problems in different time and in different ways. These problems are described linguistically. In this section, we shall discuss the formation of algebraic formulae and their applications in solving different problems of real surroundings which are described linguistically. As a result of this discussion the students on the one hand, will get the conception about the application of mathematics in real surroundings, on the other hand, they will be eager to learn mathematics for their understanding of the involvement of mathematics with their surroundings. Methods of solving the problems :
(a)
(b)

(c)
(d)
(e)

At first the problem will have to be observed carefully and to read attentively and then to identify which are unknown and which are to be determined.
One of the unknown quantities is to be denoted with any variable (say x ).
Then realising the problem well, express other unknown quantities in terms of the same variable (x) .
The problem will have to be splitted into small parts and express them by algebraic expressions. sing the given conditions, the small parts together are to be expressed by an equation.
U
The value of the unknown quantity x is to be found by solving the equation.

Different formulae are used in solving the problems based on real life. The formulae are mentioned below :
(1) Related to Payable or Attainable : ayable or attainable, A Tk. qn
P
where, q = amount of money payable or attainable per person,

n = number of person.
(2) Related to Time and Work :
If some persons perform a work,
Amount of work done, W qnx where, q = portion of a work performed by every one in unit of time.

n = number of performing of work x = total time of doing work
W portion of a work done by n persons in time x .
(3) Related to Time and Distance :
Distance at a definite time, d vt .

59

Math

where,
(4)

v = speed per hour t = total time.
Related to pipe and water tank :
Amount of water in a tank at a definite time, Q(t) Qo qt

where,

Qo = amount of stored water in a tank at the time of opening the pipe

q = amount of water flowing in or flowing out by the pipe in a unit time. t = time taken. sign, at the time o f
Q(t ) amount of water in the tank in time t (+ flowing of water in and –sign, at the time of flowing of water out are to be used).
(5) Related to percentage : p br ,

where,

b = total quantity

s s% 100 p = (rate of) percentage by parts = s% of b

r = (rate of) percentage by fraction

(6) Related to profit and loss :
S C(I r) ; in case of profit, S in case of loss, S

C(I

r)

C(I r)

where,

(7)

S (Tk.) = selling price
C (Tk.) = cost price
I = profit r = rate of profit or loss
Related to investment and profit :
In the case of simple profit,
I = Pnr (taka)
A P I P Pnr P(1 nr) (taka)

In the case of compound profit,
A P(1 r ) n where, I = profit after time n n = specific time
P = principal

60

Math

r = profit of unit principal at unit time
A = principal with profit after time n .
Example 1. For a function of Annual Sports, members of an association made a budget of Tk. 45000 and divided that every member would subscribe equally. But 5 members refused to subscribe. As a result, amount of subscription of each member increased by Tk. 15 per head. How many members were in the association ?
Solution : Let the number of members of the association be x and amount of subscription per head be Tk. q Then total amount of subscription A = Tk. qx .
Actually numbers of members were ( x 5) and amount of subscription per head became Tk. (q 15) .
Then, total amount of subscription = Tk. ( x 5)(q 15)
By the question, qx and qx

( x 5)(q 15)..........(i)

45,000..........(ii)

From equation (i), we get,

qx

or, qx

( x 5)(q 15)

qx 5q 15x 75

or, 5q 15x 75 q 5(3x 15)

3x 15...........(iii)

Ptting the value of q in equation (ii), u (3x 15) x

or, 3x2 15x or, x

2

45000

45000

5x 15000 [dividing both sides by 3]

or, x2 5x 15000

0

2

or, x 125x 120x 15000 0 or, x( x 125) 120( x 125) 0 or, ( x 125)(x 120)

0

x 125 0 or, x 120
If x 125 0 , x 125
Again, if ( x 120) 0 , x

0
120

Since the number of members i.e,, x cannot be negative, x
120 . x 125
Hence, number of members of the association is 125.
Example 2. R can do a work in 10 days and shafiq can do that work in 15 days. afiq In how many days do they together finish the work ?

61

Math

Solution : Let, afiq and Shafiq together can finish the wok in
R
Let us make the following table :
Name

d days.

Number of days for doing the work

Part of the work done in 1 day

Work done in d days

R afiq 10

1
10

d
10

Shafiq

15

1
15

d
15

By the questions,

d
10

d
15

or, d 1

1
10 15

or, d or, 3 2
30

1

1

1

5d
1
30

or, d 30 6
5

They together can finish the work in 6 days.
Example 3. A boatman can go x km in time t1 hour against the current. To cover that distance along the current he takes t2 hour. How much is the speed of the boat and the current.
Solution : Let the speed of the boat in still water be u km per hour and that of the current be v km per hour.
Then, along the current, the effective speed of boat is (u v) km per hour and against the current, the effective speed of boat is (u v) km per hour.
According to the question, distance traversed x ] u v
.......(i) [ speed time t2 x .......(ii) t1 Adding equations (i) and (ii) we get,
1 1 x x
2u
x t1 t 2 t1 t 2

and u v

62

or, u

Math

x 1
2 t1

1 t2 Subtracting equation (ii) from equation (i) we get,
1 1
2v x t 2 t1 or, v

x 1
2 t2

1 t1 Hence, speed of current is x 1
2 t2

and speed of boat is

x 1
2 t1

1 t2 1 t1 km per hour

km per hour.

Example 4. A pipe can fill up an empty tank in 12 minutes. Another pipe flows out
14 litre of water per minute. If the two pipes are opened together and the empty tank is filled up in 96 minutes, how much water does the tank contain ?
Solution : Let x litre of water flows in per minute by the first pipe and the tank can contain y litre of water.
According to the question, the tank is filled up by first pipe in 12 minutes,

y 12x .......(i)
Again, the empty tank is filled up by the two pipes together in 96 minutes. y 96x 96 14 .........(ii)

y
12
putting the value of x in equation (ii) , we get,
From equation (i) , we get, x

y 96

y
96 14
12

or, y 8 y 96 14 or, 7 y 96 14
96 14
192
7
Hence, total 192 litre of water is contained in the tank.

or, y

63

Math

Activity :
1. For a picnic, a bus was hired at Tk. 2400 and it was decided that every passenger would have to give equal fare. But due to the absence of 10 passengers, fare per head was increased by Tk. 8. How many passengers did go by the bus and how much money did each of the passengers give as fare?
2. A and B together can do a work in p days. A alone can do that work in q days. In how many days can B alone do the work ?
3. A person rowing against the current can go 2 km per hour. If the speed of the current is 3 km per hour, how much time will he take to cover 32 km, rowing along the current ?
Example 5. rice of a book is Tk. 24.00. This price is 80% of the actual price. The
P
overnment subsidize the due price. How much money does the G subsidize for
G
ovt. each book ?
Solution : Market price = 80% of actual price
We know, p br
Here, p Tk. 24 and r 80%
24 b

80
100

80
100

6 5
24 100 or, b b 30
80
4
1
Hence, the actual price of the book is Tk. 30. amount of subsidized money = Tk. (30 24)
= Tk. 6. subsidized money for each book is Tk. 6.
Example 6. The loss is r% when n oranges are sold per taka. How many oranges are to be sold per taka to make a profit of s% ?
Solution : If the cost price is Tk. 100, the selling price at the loss of r% is Tk.
(100 r) .
If selling price is Tk. (100 r) , cost price is Tk. 100
,,

,,

,,

,, Tk. 1

,,

,,

,, Tk. 100

100 r

64

Math

Again, if cost price is Tk. 100, selling price at the profit of s% is Tk. (100 s)
,, ,,

,,

,, Tk. 1

,, ,, ,, ,,

100
,, ,, ,,
100 r

,, ,, ,, ,, Tk.

,, ,,
,, ,,

,,
,,

,, Tk.

,, Tk.

100 s
100

100 s
100
100
100 r

100 s
100 r
100 s in Tk.
, number of oranges is to be sold = n
100 r

= Tk.

in Tk. 1,

,,

,,

,,

,,

,,

,, ,, n

100 r
100 s

Hence, n(100 r) oranges are to be sold per taka.
100 s

Example 7. What is the profit of Tk. 650 in 6 years at the rate of profit Tk. 7 percent per annum ?
Solution : We know , I Pnr .
Here, P Tk. 650, n 6, s 7 r I

s
100

7
100
7
650 6
100

273

Hence, profit is Tk. 273.
Example 8. Find the compound principal and compound profit of Tk. 15000 in 3 years at the profit of 6 percent per annum.
Solution : We know, C compound profit. iven, G

P Tk. 15000 , r

C 15000 1

6
100

= 15000 53
50

3

P(1 r)n , where C is the profit principal in the case of

6%

3

15000 1

6
, n = 3 years
100
3
50

3

Math

65

3
15 53 53 53 3 148877
=
25
125
25
446631
=
17865 24
25

compound principal is Tk. 17865 24 compound profit = Tk. (17865 24 15000)
= Tk. 2865 24
Activity :
1. The loss is n% when 10 lemons are sold per taka. How many lemons are to be sold per taka to make the profit of z% ?
2. What will be the profit principal of Tk. 750 in 4 years at the rate of simple
1
profit 6 percent per annum ?
2
3. Find the compound principal of Tk. 2000 in 3 years at the rate of compound profit of Tk. 4 percent per annum.
Exercise 3⋅5
1.

Which one of the following is the factorized form of x2
(a) ( x 2)(x 3)
(b) ( x 1)( x 8)
(c) ( x 1)( x 6)

2.

3.

4.

5.

7x 6 ?

(d) ( x 1)( x 6)

If f ( x) x2 4x 4 , which one of the following is the value of f (2) ?
(a) 4
(b) 2
(c) 1
(d) 0
If x y x y , which one of the following is the value of y ?
(b) 0
(a) 1
(c) 1
(d) 2 x2 3x3
Which one of the following is the lowest form of
?
x 3x2
(b) x
(a) x2
(c) 1
(d) 0
1 x2
?
Which one of the following is the lowest form of
1 x
(a) 1
(b) x

66

6.

Math

Which one of the following is the value of

1
(
{ a b)2
2

(b) a2 b2

(a) 2(a2

7.

8.

b2 )

(c) 2ab
2
If x x (a) 1

(d) 4ab
8
?
3 , what is the value of x3 x3 (b) 8
(c) 9

(d) 16

Which one of the following is the factorized form of p4
(a) ( p2

p 1)( p2

2

2

(c) ( p
9.

(a b)2} ?

p 1)( p

p 1)

(b) ( p 2

p 1)( p 2

p 1)

p 1)

2

2

p 1)

What are the factors of x
(a) ( x 1), ( x 4)

2

(d) ( p
5x

p 1)( p

4?
(b) ( x 1), ( x 4)

(d) ( x 5), ( x 1)

(c) ( x 2), ( x 2)
10. What is the value of ( x 7)( x 5) ?
(a) x2 12x 35

(b) x2 12x 35

(c) x2 12x 35

(d) x2 12x 35
2 9 2 9 11 11
?
11. What is the value of
2 9 11
(b) 1 9
(a) 1 8
(c) 2
(d) 4
12. If x

3 , what is the value of x2 ?

2

(b) 7 4 3
1
(d)
2 3

(a) 1
(c) 2

3

13. If f ( x)

p2 1 ?

x2 5x 6 and f ( x)

0 , x = what ?

(a) 2, 3

(b)

(c)

(d) 1, 5

2, 3

14.

x

6

x

x2

6x

5

5x

30

5, 1

67

Math

Which one of the following is the total area of the table above ?
(a) x 2 5 x 30

(b) x 2

2

x 30

2

(c) x 6 x 30
(d) x x 30
15. A can do a work in x days, B can do that work in 3 x days. In the same time, how many times does A of B work ?
1
(a) 2 times
(b) 2 times
2
(c) 3 times
(d) 4 times
16. If a b c , a 2 2ab b 2 is expressed in terms of c , which one of the following will be ? c2 (a)
17. If x

(b) c 2

y

3, xy

(d) ca

(c) bc

2 , what is the value of x

(a) 9
(c) 19

3

3

y ?

(b) 18
(d) 27

18. Which one is the factorized form of 8 x 3

27 y 3 ?

(a) (2 x 3 y )(4 x 2

6 xy 9 y 2 )

(b) (2 x 3 y )(4 x 2

6 xy 9 y 2 )

(c) (2 x 3 y )(4 x 2

9y2 )

(d) (2 x 3 y )(4 x 2

9y2 )

19. What is to be added to 9 x 2 16 y 2 , so that their sum will be a perfect square ?
(a) 6 xy

(b) 12 xy

(c) 24 xy

(d) 144 xy

20. If x

4 , which one of the following statements is correct ?

y

(a) x 3

y3

4 xy

64

(b) x 3

y 3 12 xy 12

(c) x 3

y3

3 xy

64

(d) x 3

y 3 12 xy

21. If x

4

(1) x 2
(a) 4

x

2

1 0,

1
= what ? x2 (b) 2

(2) What is the value of x
(a) 4
(c) 2

64

(c) 1
1
x

2

?
(b) 3
(d) 1

(d) 0

68

Math

(3) x 3

22.

23.
24.
25.

26.
27.

28.

1
= what ? x3 (a) 3
(b) 2
(c) 1
(d) 0
A can do a work in p days and B can do it in 2 p days. They started to do the work together and after some days A left the work unfinished. B completed the rest of the work in r days. In how many days was the work finished ?
50 persons can do a work in 12 days by working 8 hours a day. Working how many hours per day can 60 persons finish the work in 16 days. ?
Mita can do a work in x days and R can do that work in y days. In how ita many days will they together complete the work ?
A bus was hired at Tk. 57000 to go for a picnic under the condition that every passenger would bare equal fare. But due to the absence of 5 passengers, the fare was increased by Tk. 3 per head. How many passengers availed the bus ?
A boatman can go d km in p hours against the current. He takes q hours to cover that distance along the current. What is the speed of the current and the boat ?
A boatman plying by oar goes 15 km and returns from there in 4 hours. He goes 5 km at a period of time along the current and goes 3 km at the same period of time against the current. Find the speed of the oar and current.
Two pipes are connected with a tank. The empty tank is filled up in t1 minutes by the first pipe and it becomes empty in t 2 minutes by the second pipe. If the two pipes are opened together, in how much time will the tank be filled up ?
(here t1 t 2 )

29. A tank is filled up in 12 minutes by a pipe. Another pipe flows out 15 litre of water in 1 minute. When the tank remains empty, the two pipes are opened togethr and the tank is filled up in 48 minutes. How much water does the tank contain ?
30. If a pen is sold at Tk. 11, there is a profit of 10%. What was the cost price of the pen ?
31. Due to the sale of a note- ook at Tk. 36, there was a loss. If the note- oo k b b would be sold at Tk. 72, there would be profit amounting twice the loss. What was the cost price of the note- ook ? b 32. Divide Tk. 260 among A, B and C in such a way that 2 times the share of A, 3 times the share of B and 4 times the share of C are equal to one another.
33. Due to the selling of a commodity at the loss of x% such price is obtained that due to the selling at the profit of 3x% Tk. 18 x more is obtained. What was the cost price of the commodity ?

Math

69

34. If the simple profit of Tk. 300 in 4 years and the simple profit of Tk. 400 in 5 years together are Tk. 148, what is the percentage of profit ?
35. If the difference of simple profit and compound profit of some principal in 2 years is Tk. 1 at the rate of profit 4%, what is the principal ?
36. Some principal beomes Tk. 460 with simple profit in 3 years and Tk. 600 with simple profit in 5 years. What is the rate of profit ?
37. How much money will become Tk. 985 as profit principal in 13 years at the rate of simple profit 5% per annum ?
38. How much money will become Tk. 1248 as profit principal in 12 years at the rate of profit 5% per annum ?
39. Find the difference of simple profit and compound profit of Tk. 8000 in 3 years at the rate of profit 5%.
40. The Value Added Tax (VAT ) of sweets is x% . If a trader sells sweets at Tk.
P including VAT, how much VAT is he to pay ? If x 15, P 2300 , what is the amount of VAT ?
41. Sum of a number and its multiplicative inverse is 3.
(a) Taking the number as the variable x , express the above information by an equation. 1
.
(b) Find the value of x 3 x3 1
(c) P that, x 5 rove 123 . x5 42. Each of the members of an association decided to subscribe 100 times the number of members. But 7 members did not subscribe. As a result, amount of subscription for each member was increased by Tk. 500 than the previous.
(a) If the number of members is x and total amount of subscription is Tk. A , find the relation between them.
(b) Find the number of members of the association and total amount of subscription. 1 of total amount of subscription at the rate of simple profit 5% and rest of
(c)
4 the money at the rate of simple profit 4% were invested for 2 years. Find the total profit.

Chapter Four

Exponents and Logarithms
Very large or very small numbers or expressions can easily be expressed in writing them by exponents. As a result, calculations and solution of mathematical problems become easier. Scientific or standard form of a number is expressed by exponents.
Therefore, every student should have the knowledge about the idea of exponents and its applications.
Exponents beget logarithms. Multiplication and division of numbers or expressions and exponent related calculations have become easier with the help of logarithms.
U of logarithm in scientific calculatio ns was the only way before the practice of se using the calculator and computer at present. Still the use of logarithm is important as the alternative of calculator and computer. In this chapter, exponents and logarithms have been discussed in detail.
At the end of the chapter, the students will be able to –
Explain the rational exponent
Explain and apply the positive integral exponents, zero and negative integral exponents Solve the problems by describing and applying the rules of exponents
Explain the n th root and rational fractional exponents and express the n th root in terms of exponents
Explain the logarithms
P and apply the formulae of logarithms rove Explain the natural logarithm and common logarithm
Explain the scientific form of numbers
Explain the characteristic and mantissa of common logarithm and
Find common and natural logarithm by calculator.
4.1 Exponents or Indices
In class VI, we have got the idea of exponents and in class VII, we have known the exponential rules for multiplication and division.
Expression associated with exponent and base is called exponential expression.
Activity : Fill in the blanks
Power or
Successive multiplication of the
Exponential
Base exponent same number or expression expression 3
2 2 2
23
2
3 3 3 3
3
a a a a3 b b b b b
5

71

Math

If a is any real number, successive multiplication of n times a ; that is, n a a a ..... a is written in the form a , where n is a positive integer. n a a a ..... a ( n times a ) = a . n index or power
Here a base Again, conversely, an = a a a ..... a ( n times a ) . Exponents may not only be positive integer, it may also be negative integer or positive fraction or negative fraction. That is, for a
R (set of real numbers) and n Q (set of rational numbers), an is defined. Besides, it may also be irrational exponent. But as it is out of curriculum, it has not been discussed in this chapter.
4.2 Formulae for exponents
Let, a
R; m, n
N.
am an

Formula 1.

am
Formula 2.

a

am

n am 1 ,

n

an m

n

when m
, when

n n m, a 0

Fill in the blanks of the following table : am , an a 0

m n
5, n 3

m

am an

a5 a3

a5

3

a3 a5 5

a an n m
3, n 5

a3 a 5 =

(a a a a a) (a a a) a a a a a a a a a8 m

m

a a3 a a a a a a a a
1

1

a2 a and

m

n

a = a

am an m

a5

3

n

n am 1 , when m n an m

,

when n m

n

Formula 3. (ab) an bn
We observe, (5 2) 3 (5 2) (5 2) (5 2) [ a 3
5 2 5 2 5 2
(5 5 5) (2 2 2)
In general, (ab)n

a a a; a 5 2]

53 23 ab ab ab ....... ab [Successive multiplication of n times ab ]

72

Math

(a a a ....... a) (b b b ........ b) anbn n

a b Formula 4.

an a a

0)

5
2

5 5 5
2 2 2

a a a
[Successive multiplication of n times ]
........
b b b

a b a a a ...... a b b b ...... b
1, (a 0)

an

n

53
23

an bn a0

n

n

a

Again,

, (b

5
2

a b Formula 5. a0
We have,

5
2

n

a b In general,

bn
3

5
2

We observe,

an

n

a a a ..... a a a a ..... a

n

a

1 a 0 1.
Formula 6. a

a

1 a

a

n

n

a

n

1

n

a n , (a

[multiplying both num. and denom. by an ] a n

a

a
R
emark : a n
Formula 7.

am

n

n

ao n a0 an n

n

ao an 1

am

0)

n

1

an

1

an n a

n

a mn

a m a m a m ......... a m

am

m m ......... m

[successive multiplication of n times am ]

[in the power, sum of n times of exponent m ]

Math

73

= a

(a m ) n

n m

a mn

a mn
52
53

Example 1. Find the values (a)
Solution : (a)

52
= 52
53

2
(b)
3

5

3

1
51

51
5

2
3

2
=
3

5 5

2
3

54 8 16
2

5

1

3 2n

(b)

1

5
(b)

3 2

n

2n

4 2
2n

n 2

1

125

2

2

2
5 4

3 2

n

2n

5

2 2n
2n 2

Solution : (a )

p (q r )

q r p

(a )

4

3

2

22

5

1

54

5

27

3

5

54

3

n 2

3 2n

2

2n

2n

(3 1) 2 n
1 n
2
2

r

(a q ) r

r

5

2

(a )

q (r p)

(ii) 5

53

p

(a r ) p

1
2

2 2n
1 n
2
2

[ (a m ) n

55

q

2 2 4.

p q

a a a r ( p q) pq pr qr pq a a a pr qr pq pr qr pq pr qr a a 0 1.
Activity : Fill in the blank boxes :
(i) 3 3 3 3 3

3

1

3 2n 2n
1
1
2n
2 p q r

54 23

2n

2

20
2

Example 3. Show that (a p ) q

4 2n

2n

54 23 2 4
5

5

0

25 125
Solution :(a)

2
3

1
5

5 4 8 16

Example 2. Simplify :(a)

5

2
3

(b)

1 a mn ]

(iii) a 2 a

a

3

(iv)

4
4

(v) ( 5)0

1

27

5

74

Math

4.3 n th root
1

1

52

2

1

We notice, 5 2

52

1

1

1 1
2

Again, 5 2 5 2

5

52

5

1
2

2

5

2

1
2

5.

1
Square (power 2) of 5 2

1

5 and square root (second root) of 5 = 5 2

1

5 2 is written as

5 in terms of the sign
1

1

Again, we notice 5 3 5 3 5 3

1

1

of square root.

3

53

1

1

Again, 5 3 5 3

1

53

5

1 1 1
3 3

53

5

3

1
3

5

3

1
3

5.
1

1

ube (power 3) of 5 3 = 5 and cube root (third root) of 5 = 5 3 .
C1
5 3 is written as 3 5 in terms of the sign 3 of cube root.
In the case of n th root,
1

1

1

1

an

an

an

....... a n

a

1 n 1

an

n

n

.
1

1

Again, a n a n

1

1

a n ....... a n

1 1 1
1
.......... n n n n

an a n

1 n 1 n an a

a.
1

1

n th power of a n

a and n th root of a

an

1 n 75

Math

i.e. n th power of

1 n an 1 an 1

a is written as n a .
3
1
Example 4. Simplify :(a) 7 4 7 2
Solution : (a) 7

3
4

7

1
2

7

1

a n = n a . n th root of

a and n th root of a = (a) n

3 1
4 2

7

(b) (16)

3
4

(16)

1
2

(c) 10

3
4

2
3

5
4

3
3
4

(b) (16)

(16)

1
2

(16)
3
4

2

2 3
4

10 2

Example 5. Simplify :(a) (12)
Solution : (a) (12)

1
2

3

(2 2 3)

1
2

1
2

3

1
(2 3)

1
2

=

27

(b) (

(3 2)

1
3

2
21

1
2

3)3

1

1
1
2 2

31
3

1
2

3

1
2

1

1
2

1

1
3

3

2

1
= ( 3)( 3)( 3)
2
1
4

4

1
4

2.

2

1

1

(2 )
1
3

(2)

(54) 3

1
2

1
3

2

(b) ( 3)3

54

1

54

(33 2)

1

(2 4 ) 4

10 .

(12)

1

1

(16) 4

(16) 4

1
2

1

10 3

(c) 10 3

3 1
2

(16) 4

(33 ) 3 2 3
1

1

32

32

2
3

1
3

2

1
2

4

3

3
.
4

1
2

27
4
5

5

3

24 22
2 2
2 2
Activity : Simplify : (i)
(ii)
(iii) 8 4
32
3
3
To be noticed :
U
1. nder the condition a 0, a 1 , if a x a y , x y
U
2. nder the condition a 0, b 0, x 0 , if a x b x , a b
Example 6. Solve : 4 x 1 32 .

1

82

76

Math

Solution :
4x
2 x or (2 )

1

32

1

32 ,

or, 2 2 x

2

2x 2 5, or, 2x 5 2 , or, 2x
3
x
2

ax

25

ay x

y

3

3
2

Solution is x

Exercise 4.1
Simplify (1 – 10) :
1.

33 35
36

5. (2
8.

1

2.

5 1)

x 1y

1

6. (2a

y 1z

3.

1

3b 1 )

z 1 x , ( x 0, y

1

0, z

73 7 3
3 34

7.

53 8
2 4 125

a 2b 1 a 26

0) 9.

2n

4.
2

3m 1
4 2n 1
10. m m 1
2
(2 )

4

2

3 72 3 7
7

n 2

3m

1

m 1 m 1

(3

)

Prove (11 – 18) :
11.

4n 1
2n 1

a
13.
am a 15. xb

x

xp
17. q x n

1 ab 2n 1

am an xb xc p q r

12. m xq xr xc xa 32 p q 5 p q 6 p
6 10 p 2 15 q
1

14.

ap q a 2r

1

16.

xa xb 1 ca q r p

18. If a x b, b y c and c z
Solve (19 – 22) :

xr xp 1
50

q

1

an a 1 bc 2p

aq r a2 p a b

xb xc ar p a 2q b c

1 xc xa

c a

1

r p q

1

a , show that xyz 1 x 1

2x 1

3
20. 22x 1 128
21. 3
22. 2 x 21 x 3
19. 4 x 8
3
4.4 Logarithm
Logarithm is used to find the values of exponential expressions. Logarithm is written
. roduct, quotient, etc. of large numbers or quantities can easily be in brief as ‘ Log ’ P determined by the help of log.

77

Math

We know, 23 8 ; this mathematical statement is written in terms of log as log 2 8 3 . Again, conversely, if log2 8 3 , it can be written in terms of exponents as 23

8 . That is, if 23 8 , then log 2 8 3 and conversely, if log2 8 3 , then
1
1 1
23 8 . Similarly, 2 3 can be written in terms of log as log 2
3.
3
8
2
8
If a x N , ( a 0, a 1 ), x loga N is defined as a based log N .
To be noticed : Whatever may be the values of x , positive or negative, a x is always positive. So, only the log of positive numbers has values which are real ; log of zero or negative numbers have no real value.
Activity-1 : Express in terms of Activity-2 : Fill in the blanks : log : in terms of exponent in terms of log
(i) 10 2 100
(ii) 3
(iii) 2
(iv)

1
9

2

1
2

100

1

1
2

e0

.....

loge 1 ......

a0

1

...... = ......

24

101 10
1

e ...
....... = ......
Formulae of Logarithms :
Let, a 0, a 1; b 0, b 1 and M 0, N
Formula 1. (a) loga 1 0, (a 0, a 1)
(b) loga a 1, (a

0.

0, a 1)

Proof : (a) We know from the formula of exponents, a0 1 from the definition of log, we get, loga 1 0 (proved )
(b) We know, from the formula of exponents, a1 a from the definition of log, we get, loga a 1 (proved).
Formula 2. loga (MN )
Proof : Let, loga M
M

x

loga M x, loga N

a ,N

a

y

loga N y; log10 1 0

log10 10 1
....... = ....... loga a 1

78

Math

Now, MN a x a y a x y log a ( MN ) x y, or log a ( MN ) log a M loga (MN )

loga M

Note 1. loga (MNP.....)
Note 2. loga (M

N)

log a N [putting the values of x, y ]

loga N. (proved) loga M

loga N

loga M

loga P .........

loga N

M log a M log a N
N
Proof : Let loga M x, loga N y ;

Formula 3. log a

M

a x, N

M ax ax N ay
M
loga x y
N

Now,

loga

M
N

ay y loga M

loga N (proved).

Formula 4. loga M r

r loga M .

Proof : :Let loga M

x;

(M ) r

ax

M

(a x ) r ; or M r

log a M r log a M

r

a rx

rx; or log a M r

r log a M

r log a M . (proved ).

r

N.B. : (log a M ) r log a M
Formula 5. loga M = logb M loga b, (change of base)
Proof : Let, log a M x, log b M y ax ax or b

x y or , x

M ,by M 1 x b y , or (a x ) y

1

(b y ) y

ay

loga b, or x

y loga b

y loga b, or log a M

Corollary : log a b

1
, or, logb a log b a

log b M log a b (proved).
1
loga b

79

Math

Proof : We know, loga M

logb M loga b [formula 5]

Ptting M a we get, u loga a logb a loga b or 1 log b a log b b;
1
log a b

log b a

or loga b

1
(proved).
logb a
(b) log3

Example 7. Find the value : (a) log10 100
Solution:
(a) log10 100 log10 10 2

2 log10 10 [ log10 M r r log10 M ]
2 1 [ log a a 1] = 2

1
1
log 3 2 log 3 3
9
3
2 1[ log a a 1] = 2

(b) log 3

(c) log

3

81 log

3

34

1
(c) log 3 81
9

log 3 {
(

8 log 3 3[ log a M r
8 1, [ log a a 1]
8

2

2 log 3 3 [ log a M r

3 ) 2 }4

log

3

3

8

r log a M ]

Example 8. (a) What is the log of 5 5 to the base 5 ?
(b) log 400 4 ; what is the base ?
Solution : (a) 5 5 to the base 5
1

3

log 5 5 5 log 5 (5 5 2 ) log 5 5 2
3
log 5 5,[ log a M r r log a M ]
2
3
1,[ log a a 1]
2
3
2
(b) Let the base be a. by the question, loga 400 4 a 4 400

r log a M ]

80

Math

or a 4

(20)2

or a 4

(2 5)

a

(
{ 2 5)2}2

(2 5)

4

4

ax

2 5

b x a b]

the base is 2 5
Example 9. Find the value of x :
(b) log x 324
(a) log10 x
2
Solution :
(a) log10 x
2

4
(b) log x 324

4

4

2

x 10
1
1 x 10 2 100 x 0 01

x 324 3 3 3 3 2 2
34 2 2 34 ( 2 ) 4

0 01

or x 4

(3 2)

4

x 3 2.
Example 10. rove that, 3 log10 2
P

log10 5 log10 40
Solution :Left hand side = 3 log10 2 log10 5 log10 2 3 log10 5,[ log a M r

r log a M ]

log10 8 log10 5

log10 (8 5), [ log a ( MN ) log a M log10 40 = ight hand side (proved).
R
Example 11. Simplify :
Solution :

log10 27

log10 27

log10 8 log10 1000 log10 1 2

log10 8 log10 1000 log10 1 2

1

1

log 10 (33 ) 2

log 10 2 3 log 10 (10 3 ) 2
12
log 10
10

3

log a N ]

3

log10 3 2 log10 23 log10 10 2 log10 12 log10 10
3
3 log10 3 3 log10 2 log10 10
2
2
2
log10 (3 2 ) log10 10

Math

81

3
(log10 3 2 log10 2 1)
2
(log10 3 2 log10 2 1)

[ log10 10 1]

3
.
2
1.

Exercise 4.2
Find the value :(a) log3 81 (b) log5 3 5 (c) log4 2 (d) log2 5 400
(e) log 5

3

5

5

2.

Find the value of x :(a) log5 x

3.

Show that,
(a) 5 log10 5 log10 25

3

(b) log x 25

2

(c) log x

1
16

2

log10 125

50 log10 2 2 log10 5 log10 3 2 log10 7
147
(c) 3 log10 2 2 log10 3 log10 5 log10 360
Simplify :
10
25
81
(a) 7 log10
2 log10
3 log10
9
24
80
(b) log 7 5 7 7 log 3 3 3 log 4 2

(b) log10

4.

a 3b 3 b 3c 3 c 3d 3 log e 3 log e 3 3 log e b 2 c c3 d a 4.5 Scientific or Standard form of numbers
We can express very large numbers or very small numbers in easy and small form by exponents. s s Such as, velocity of light
= 300000 km/ ec = 300000000 m/ ec s s
= 3 100000000 m/ ec. = 3 108 m/ ec.
Again, radius of a hydrogen atom = 0 00000000037 cm
37
cm = 37 10 10 cm
=
10000000000
= 3 7 10 10 10 cm = 3 7 10 9 cm
For convenience, very large number or very small number is expressed in the form a 10n , where 1 a 10 and n Z . The form a 10n of any number is called the scientific or standard form of the number.
Activity : Express the following numbers in scientific form :
(b) 0 000512
(a) 15000
(c) log e

82

Math

4.6 Logarithmic Systems
Logarithmic systems are of two kinds :
(a) Natural Logarithm :
The mathematician J hn Napier (1550 –1617) of Scotland first published the book o on logarithm in 1614 by taking e as its base. e is an irrational number, e 2 718....... . Such logarithm is called Napierian logarithm or e based logarithm or natural logarithm. loge x is also written in the form ln x .
(b) Common Logarithm :
The mathematician Henry Briggs (1561 – 1630) of England prepared log table in
1624 by taking 10 as the base. Such logarithm is called Briggs logarithm or 10 based logarithm or practical logarithm.
N.B. : If there is no mention of base, e in the case of expression (algebraic) and 10, in the case of number are considered the base. In log table 10 is taken as the base.
4.7 Characteristic and Mantissa of Common Logarithm
(a) Characteristics :
Let a number N be expressed in scientific form as N
N 0,1 a 10 and n Z .
Taking log of both sides with base 10, log10 N log10 (a 10 n ) log10 a log10 10 n log10 a n log10 10 log10 N n log10 a
[ log10 10 1]

log10 N

a 10n , where

n log10 a

Suppressing the base 10, we have,

log N n log a n is called the characteristic of log N .
We observe : Table-1
N
Form a 10 m of N

Exponent

Number of digits on the left of the decimal point Caracteristic h 6237

6 237 103

3

4

4 1 3

623 7

6 237 102

2

3

3 1 2

62 37

1

1

2

2 1 1

0

1

1 1 0

0

0 1

6 237
0 6237

6 237 10

0

6 237 10
6 237 10

1

1

1

83

Math

We observe : Table-2
Form
N

Exponent Number of zeroes between decimal point and its next first significant digit

a 10 m of N

Caracteristic h 0 6237

6 237 10

1

1

0

(0 1)

1

0 06237

6 237 10

2

2

1

(1 1)

2

6 237 10

3

3

2

(2 1)

3

0 006237

We observe from table-1 :
As many digits are there in the integral part of a number, characteristic of log of the number will be 1 less than that number of digits and that will be positive.
We observe from table 2 :
If there is no integral part of a number, as many zeroes are there in between decimal point and its next first significant digit, the characteristic of log of the number will be 1 more than the number of zeroes and that will be negative.
N. B. 1. Caracteristic may be either positive or negative, but the mantissa will h always be positive.
N. B. 2. If any characteristic is negative, not placing ‘ sign on the left of the
–
characteristic, it is written by giving ‘ (b ar sign) over the characteristic. Such as,
–
characteristic –3 will be written as 3 . Otherwise, whole part of the log including mantissa will mean negative.
Example 12. Find the characteristics of log of the following numbers :
(i) 5570
(iv) 0 000435
(ii) 45 70
(iii) 0 4305
3
Solution : (i) 5570 5 570 1000 5 570 10 haracteristic of log of the number is 3 .
C
Otherwise, number of digits in the number 5570 is 4 .
Caracteristic of log of the number is h 4 1 3
Caracteristic of log of the number is 3. h 1
(ii) 45 70 4 570 10
Caracteristic of log of the number is 1 . h Otherwise, there are 2 digits in the integral part (i.e. on left of decimal point) of the number.
Caracteristic of the log of the number is h 2 1 1
Caracteristic of log of the number is 45 70 is 1. h 1
(iii) 0 4305 4 305 10
Caracteristic of log of the number is h 1
Otherwise, there is no significant digit in the integral part (before the decimal point) of the number or there is zero digit.
0 1
1 1
Caracteristic of log of the number h 84

Math

Again, there is no zero in between decimal point and its next first significant digit of the number 0 4305 , i.e. there is 0 zeroes.
Caracteristic of log of the number is h (0 1)
1 1
Caracteristic of log of the number 0 4305 is 1 h (iv) 0 000435 4 35 10 4
Caracteristic of log of the number is h 4 or 4
Otherwise, there are 3 zeroes in between decimal point and its next 1st significant digit.
Caracteristic of log of the number is h (3 1)
4 4
Caracteristic of log of the number is 0 000435 is 4 h (b) Mantissa :
Mantissa of the Cmmon Logarithm of any number is a non- egative number less o n than 1. It is mainly an irrational number. But the value of mantissa can be determined upto a certain places of decimal.
Mantissa of the log of a number can be found from log table. Again, it can also be found by calculator. We shall find the mantissa of the log of any number in 2nd method, that is by calculator.
Determination of common logarithm with the help of calculator :
Example 13. Find the characteristic and mantissa of log 2717 :
Solution :We use the calculator :
AC

log

2717

=

3 43408

Caracteristic of log 2717 is 3 and mantissa is 43408 h Example 14. Find the characteristic and mantissa of log 43 517 .
Solution :We use the calculator :
= 1 63866
AC
log
43 517
Caracteristic of log 43 517 is 1 and mantissa is 63866 h Example 15. What are the characteristic and mantissa of the log of 0 00836 ?
Solution :We use the calculator :

AC

log

0 00836

= 3 92221
–

= 3 92221

Caracteristic of log 0 00836 is – or 3 and mantissa is 92221 h 3
Example 16. Find log e 10
Solution : log e 10
=

1 log10 e

1
[taking the value of e upto five decimal places] log10 2 71828

1
[using calculator]
0 43429

= 2 30259 (approx).

85

Math

Alternative : We use the calculator : ln AC

10

= 2 30259

(approx).

Activity : Find the logarithm of the following numbers (each with the base 10 and
e) by using calculator : (i) 2550 (ii) 52 143 (iii) 0 4145
(iv) 0 0742
Exercise 4.3
1.
2.

0

On what condition a 1 ?
b. a 0
a. a 0
6

5

5.
6.

3

5

c.

5

3

5

3

3

1

5 ?
d.

3

25

On what exact condition log a a
a. a

4.

b.

3

d. a

0

Which one of the following is the value of
a.

3.

c. a

0
If log x 4
a. 2

1?
b. a 1
c. a 0, a 1
2 , what is the value of x ?
b. 2
c. 4

d. a

0, a 1

d. 10

What is the condition for which a number is to be written in the form a 10 n ?
b. 1 a 10
c. 1 a 10
d. 1 a 10
a. 1 a 10
Observe the following information :

i. log a ( m) p

plog a m

ii. 2 16 and log 2 16 4 are synonymous. iii. log a ( m n) log a m log a n
4

Which of the above information are correct ?
b. ii and iii
c. i and iii
a. i and ii

d. i, ii and iii

7.

What is the characteristic of the common log of 0 0035 ?

8.

b. 1
c. 2
d. 3
a. 3 onsidering the number 0 0225 , answer the following questions :
C
(1) Which one of the following is of the form a n of the number ?
a. ( 2 5) 2
b. ( 015) 2
c. (1 5) 2
d. ( 15) 2
(2) Which one of the following is the scientific form of the number ?
b. 22 5 10 3
c. 2 25 10 2
d. 225 10
a. 225 10 4
(3) What is the characteristic of the common log of the number ?
a. 2

b. 1

c. 0

d. 2

1

86

Math

9.

Express into scientific form :
(b) 60 831
(c) 0 000245
(a) 6530
(e) 0 00000014
10. Express in the form of ordinary decimals :
(a) 105
(b) 10 5
(c) 2 53 10 4

11.

12.

13.

14.

15.

(d) 37500000

(d) 9 813 10

3

(e) 3 12 10 5
Find the characteristic of common logarithm of the following numbers (without using calculator) :
(b) 72 245
(c) 1 734
(d) 0 045
(a) 4820
(e) 0 000036
Find the characteristic and mantissa of the common logarithm of the following numbers by using calculator :
(a) 27
(b) 63 147
(c) 1 405
(d) 0 0456
(e) 0 000673
Find the common logarithm of the product/ uotient (approximate value upto q five decimal places) :
(a) 5 34 8 7
(b) 0 79 0 56 (c) 22 2642 3 42
(d) 0 19926 32 4
If log 2 0 30103, log 3 0 47712 and log 7 0 84510 , find the value of the following expressions :
(a) log 9
(b) log 28
(c) log 42 iven, x 1000 and y 0 0625
G
a. Express x in the form a n b n , where a and b are prime numbers.
b. Express the product of x and y in scientific form.
c. Find the characteristic and mantissa of the common logarithm of xy .

Chapter five

Equations with One Variable
We have known in the previous class what equation is and learnt its usage. We have learnt the solution of simple equations with one variable and acquired knowledge thoroughly about the solution of simple equations by forming equations from real life problems. In this chapter, linear and quadratic equations and Identities have been discussed and their usages have been shown to solve the real life problems.
At the end of the chapter, the students will be able to −
Explain the conception of variable
Explain the difference between equation and identity
Solve the linear equations
Solve by forming linear equations based on real life problems
Solve the quadratic equations and find the solution sets
Form the quadratic equations based on real life problems and solve.
5⋅1 Variables
We know, x 3 5 is an equation. To solve it, we find the value of the unknown quantity x . Here the unknown quantity x is a variable. Again, to solve the equation x a 5 , we find the value of x , not the value of a . Here, x is assumed as variable and a as constant. In this case, we shall get the values of x in terms of a . But if we determine the value of a , we shall write a 5 x ; that is, the value of a will be obtained in terms of x . Here a is considered a variable and x a constant. But if no
G
direction is given, conventionally x is considered a variable. enerally, the small letters x, y, z , the ending part of English alphabet are taken as variables and a, b, c , the starting part of the alphabet are taken as constants.
The equation, which contains only one variable, is called a linear equation with one variable. Such as, in the equation x 3 5 , there is only one variable x . So this is the linear equation with one variable.
We know what the set is. If a set S {x : x R, 1 x 10} , x may be any real number from 1 to 10. Here, x is a variable. So, we can say that when a letter symbol means the element of a set, it is called variable.
Degree of an equation : The highest degree of a variable in any equation is called the degree of the equation. Degree of each of the equations x 1 5, 2 x 1 x 5, y 7 2 y 3 is 1 ; these are linear equations with one variable.

88

Again, the degree of each of the equations x2 5x 6 0, y 2

Math

y 12, 4x2 2x 3 6x

is 2 ; these are quadratic equations with one variable. The equation 2x3 x2 4x 4 0 is the equation of degree 3 with one variable.
5⋅2 Equation and Identity
Equation : There are two polynomials on two sides of the equal sign of an equation, or there may be zero on one side (mainly on right hand side). Degree of the variable of the polynomials on two sides may not be equal. Solving an equation, we get the number of values of the variable equal to the highest degree of that variable. This value or these values are called the roots of the equation. The equation will be satisfied by the root or roots. In the case of more than one root, these may be equal or unequal. Such as, roots of x2 5x 6 0 are 2 an 3. Again, though the value of x in the equations ( x 3)2 0 is 3, the roots of the equation are 3, 3.
Identity` : There are two polynomials of same (equal) degree on two sides of equal sign. Identity will be satisfied by more values than the number of highest degree of the variable. There is no difference between the two sides of equal sign ; that is why, it is called identity. Such as, ( x 1)2 ( x 1)2 4x is an identity ; it will be satisfied for all values of x . So this equation is an identity. Each algebraic formula is an identity. Such as, (a b) 2 a 2 2ab b 2 , (a b) 2 a 2 2ab b 2 , a 2 b 2

(a b)(a b) , (a b) 3 a 3 3a 2 b 3ab 2 b 3 etc. are identities.
All equations are not identities, In identity ' ' sign is used instead of equal (=) sign.
But as all identities are equations, in the case of identity also, generally the equal sign is used. Distinctions between equation and identity are given below :
Equation
Identity
1. Two polynomials may exist on both 1. Two polynomials exist on two sides of equal sign, or there may be sides. zero on one side.
2. Degree of the polynomials on both 2. Degree of the polynomials on both sides may be unequal. sides is equal.
3. The equality is true for one or more 3. enerally, the equality is true for all
G
values of the original set of the values of the variable. variable. 4. The number of values of the variable 4. Equality is true for infinite number does not exceed the highest degree of of values of the variable. the equation
5. All equations are not formulae.
5. All algebraic formulae are identities.

Math

89

1. What is the degree of and how many roots has each of the following equations ?
2 y y 1 3y
(ii )
(i ) 3 x 1 5
5
3
2
2. Write down three identities.
5⋅3 Solution of the equations of first degree
In case of solving equations, some rules are to be applied. If the rules are known, solution of equations becomes easier. The rules are as follows :
1. If the same number or quantity is added to both sides of an equation, two sides remain equal.
2. If the same number or quantity is subtracted from both sides of an equation, two sides remain equal.
3. If both sides of an equation are multiplied by the same number or quantity, the two sides remain equal.
4. If both sides of an equation are divided by same non- ero number or quantity, the z two sides remain equal.
The rules stated above may be expressed in terms of algebraic expressions as follows: x a
If x a and c 0 , (i ) x c a c (ii ) x c a c (iii ) xc ac (iv) c c
Besides, if a, b and c are three quantities, if a b c , a b c and if a c b , a b c.
This law is known as transposition law and different equations can be solved by applying this law.
If the terms of an equation are in fractional form and if the degree of the variables in each numerator is 1 and the denominator in each term is constant, such equations are linear equations.
5x 4 x 2
Example 1. Solve :
7 5 5 7
5x x 4 2
5x 4 x 2
Solution : or, [by transposition]
7 5 5 7
7 5 5 7
25x 7 x 28 10
18 x 18 or, or,
35
35
35 35 or, 18x 18 or, x 1
Solution is x 1.
Now, we shall solve such equations which are in quadratic form. These equations are transformed into their equivalent equations by simplifications and lastly the equations is transformed into linear equation of the form ax b . Again, even if there are variables in the denominator, they are also transformed into linear equation by simplification. Activity :

90

Math

Example 2. Solve : ( y 1)( y 2) ( y 4)( y 2)
Solution : ( y 1)( y 2) ( y 4)( y 2) or, y 2 y 2 y 2 y 2 4 y 2 y 8 or, y 2 2 y 8 or, y 2 y
8 2 [by transposition] or, y
6
or, y 6
Solution is y 6
Example 3. Solve and write the solution set :

6x 1 2x 4
15
7x 1

2x 1
5

6x 1 2x 4 2x 1
15
7x 1
5
6x 1 2x 1 2x 4
[by transposition] or, 15
5
7x 1

Solution :

or,

6x 1 6x 3
15

2x 4
7x 1

or, 4

15

2x 4
7x 1

m or, 15(2x 4) 4(7 x 1) [by cross- ultiplication] or, 30x 60 28x 4 or, 30x 28x 60 4 [by transposition] or, 2x 56, or x 28
Solution is x 28 and solution set is S {28} .
1
1
1
1
Example 4. Solve : x 3 x 4 x 2 x 5
1`
1
1
1 x 3 x 4 x 2 x 5 x 4 x 3 x 5 x 2 or, ( x 3)( x 4) ( x 2)( x 5)

Solution :

2x 7
2x 7 x 2 7 x 12 x 2 7 x 10
Values of the fractions of two sides are equal. Again, numerators of two sides are equal, but denominators are unequal. In this case, if only the value of the numerators is zero, two sides will be equal.
2 x 7 0 or, 2 x 7

or, x

7
2

Solution is x

7
2

or,

Math

91

2x 3 5 2
Alternative method :
Solution : 2x 3 5 2
2x 3 5 2 or, 2x 3 2 5
[by transposition]
2
or, 2x 3 2 5 or, 2x 3
3 2 [squaring both sides]
3
or, 2x 3 or, 2x 3 9
Square root of any real quantity or, 2x 12 cannot be negative. or, x 6
The equation has no solution.
Since there is the sign of square root,
Solution set is :S = φ verification of the correctness is necessary.

Example 5. Find the solution set :

Ptting x 6 in the given equation, we get, u 2 6 3 5 2 or 9 5 2 or, 3 5 2 or, 8 2 , which is impossible.
The equation has no solution.
Solution set is :S = φ
Activities : 1. If 5 1 x 4 4 5 , show that, x 6 2 5
2. Solve and write the solution set : 4x 3 5 2
5⋅4 Usage of linear equations
In real life we have to solve different types of problems. In most cases of solving these problems mathematical knowledge, skill and logic are necessary. In real cases, in the application of mathematical knowledge and skill, as on one side the problems are solved smoothly, on the other side in daily life, solutions of the problems are obtained by mathematics. As a result, the students are interested in mathematics.
Here different types of problems based on real life will be expressed by equations and they will be solved.
For determining the unknown quantity in solving the problems based on real life, variable is assumed instead of the unknown quantity and then equation is formed by the given conditions. Then by solving the equation, value of the variable, that is the unknown quantity is found.
Example 6. The digit of the units place of a number consisting of two digits is 2 more than the digit of its tens place. If the places of the digits are interchanged, the number thus formed will be less by 6 than twice the given number. Find the number.
Solution : Let the digit of tens place be x . Then the digit of units place will be x 2. the number is 10 x ( x 2) or, 11x 2.
Now, if the places of the digits are interchanged, the changed number will be
10( x 2) x or 11x 20

92

Math

By the question, 11x 20 2(11x 2) 6 or, 11x 20 22 x 4 6 or, 22 x 11x 20 6 4 [by transposition] or, 11x 22 or, x 2
The number is 11x 2 11 2 2 24 given number is 24 .
Example 7. In a class if 4 students are seated in each bench, 3 benches remain vacant. But if 3 students are seated on each bench, 6 students are to remain standing.
What is the number of students in that class ?
Solution : Let the number of students in the class be x .
Since, if 4 students are seated in a bench, 3 benches remain vacant, the number of x benches of that class
3
4
Again, since, if 3 students are seated in each bench, 6 students are to remain x 6 standing, the number of benches of that class
3
Since the number of benches is fixed, x x 6 x 12 x 6 or, 3=
3
4
3
4 or, 4x 24 3x 36 , or, 4x 3x 36 24 or, x 60 number of students of the class is 60 .
Example 8. Mr. abir, from his Tk. 56000, invested some money at the rate of profit
K
12% per annum and the rest of the money at the rate of profit 10% per annum. After one year he got the total profit of Tk. 6400. How much money did he invest at the rate of profit 12% ?
Solution : Let Mr. abir invest Tk. x at the rate of profit 12%.
K
he invested Tk. (56000 x) at the rate of profit 10%.
12x
12
Now, profit of Tk. x in 1 year is Tk. x
1 , or, Tk.
100
100
Again, profit of Tk. (56000 or, Tk.

10(56000
100

By the question ,

x) in 1 year is Tk. (56000

x)

12x 10(56000 x)
+
= 6400
100
100

or, 12x 560000 10x 640000 or, 2x 640000 560000

x)

10
,
100

93

Math

or, 2x 80000 or, x 40000
Mr. abir invested Tk. 40000 at the rate of profit 12%.
K
Activity : Solve by forming equations :
1. What is the number, if any same number is added to both numerator and
3
4 denominator of the fraction , the fraction will be ?
5
5
2. If the difference of the squares of two consecutive natural numbers is 151, find the two numbers.
3. If 120 coins of Tk. 1 and Tk. 2 together are Tk. 180, what is the number of coins of each kind ?
Exercise 5⋅1
Solve (1-10) :
1. 3(5 x 3) 2( x 2)

2.

ay b by a a2

b2

3.

( z 1)( z 2) ( z 4)( z 2)
7x 3 2x 4
4
9
25
4.
5.
6.
3 5 5 3
2 x 1 3x 2 5 x 4
1
1
1
1 x 1 x 4 x 2 x 3 a b a b x a x b x 3a 3b
7.
8.
0
x a x b x a b b a a b x a x b a2 b2 b2 a2
10. (3
3 ) z 2 5 3 3.
Find the solution set (11 - 19) :

11. 2 x( x 3)
14.

z 2 z 1

2

2 x 2 12
1

12. 2 x
15.

z 1 m n m n m x n x m n x
1
1
1
1
17.
x 2 x 5 x 4 x 3 x 2b 2 c 2 x 2c 2 a 2
19.
a b b c

1 x 2

3x 4 3 2

1

2

x 1

x 1

2t 6 15 2t
9
12 5t x 2a 2 b 2
0
c a

18.

13.

9.

x a x b

16.

4t 15
18

x a x c

94

Math

Solve by forming equations (20 - 27) :
2
times of another number. If the sum of the numbers is 98, find
20. A number is
5
the two numbers.
21. Difference of num. and denom. of a proper fraction is 1. If 2 is subtracted from
1
numerator and 2 is added to denominator of the fraction, it will be equal to .
6
Find the fraction.
22. Sum of the digits of a number consisting of two digits is 9. If the number obtained by interchanging the places of the digits is less by 45 than the given number, what is the number ?
23. The digit of the units place of a number consisting of two digits is twice the digit of the tens place. Show that, the number is seven times the sum of the digits. 24. A petty merchant by investing Tk. 5600 got the profit 5% on some of the money and profit of 4% on the rest of the money. On how much money did he get the profit of 5% ?
25. Number of passengers in a launch is 47 ; the fare per head for the cabin is twice that for the deck. The fare per head for the deck is Tk. 30. If the total fare collected is Tk. 1680, what is the number of passengers in the cabin ?
26. 120 coins of twenty five paisa and fifty paisa together is Tk. 35. What is the number of coins of each kind ?
27. A car passed over some distance at the speed of 60 km per hour and passed over the rest of the distance at the speed of 40 km per hour. The car passed over the total distance of 240 km in 5 hours. How far did the car pass over at the speed of 60 km per hour ?
5⋅5 Quadratic equation with one variable
Equations of the form ax 2 bx c 0 [where a, b, c are constants and a 0 ] is called the quadratic equation with one variable. Left hand side of a quadratic equation is a polynomial of second degree, right hand side is generally taken to be zero.
Length and breadth of a rectangular region of area 12 square cm. are respectively x cm. and ( x 1) cm. area of the rectangular region is = x( x 1) square cm.
By the question, x( x 1) 12, or x 2 x 12 0 . x is the variable in the equation and highest power of x is 2.
Such equation is a quadratic equation. The equation, which has x the highest degree 2 of the variable, is called the quadratic equation.
In class VIII, we have factorized the quadratic expressions with one variable of the forms x2 px q and ax2 bx c . Here, we shall solve the equations of the forms x 2 px q 0 and ax 2 bx c 0 by factorizing the left hand side and by finding the value of the variable.

5
9

Math

An important law of real numbers is applied to the method of factorization. The law is as follows :
If the product of two quantities is equal to zero, either ony one of the quantities or both quantities will be zero. That is, if the product of two quantities a and b i.e., ab 0 , a 0 or, b 0 , or, both a 0 and b 0 .
Example 9. Solve : ( x 2)( x 3) 0
Solution : ( x 2)( x 3) 0 x 2 0 , or, x 3 0
If x 2 0 , x
2
Again, if x 3 0 , x 3 solution is x
2 or, 3 .
Example 10. Find the solution set : y 2
3y
Solution : y 2
3y
B or, y 2
3 y 0 [ y transposition, right hand side has been done zero] or, y ( y
3) 0 y 0 , or y
3 0
If y
3 0, y
3
Solution set is {0, 3} . x 4
Example 11. Solve and write the solution set : x 4
,x 0. x x 4
Solution : x 4 x [ y cross-multiplication] b or, x( x 4) x 4 b or, x( x 4) ( x 4) 0 [ y transposition] or, ( x 4)( x 1) 0 x 4 0, or, x 1 0
If x 4 0 , x 4
Again, if x 1 0 , x 1
Solution is : x 1 or, 4 and the solution set is {1, 4} .

Solution :

x a x a

2

x a x a

Example 12. Solve :
2

5

5

x a x a

x a y x a
Then from (1) , we get, y 2 or, y 2 2 y 3 y 6 0

x a x a

6

6 0..........(1)

et,
L

5y 6

0

0

6
9

Math

or, y ( y 2) 3( y 2) 0 or, ( y 2)( y 3) 0 y 2 0 , or, y 3 0
If y 2 0 , y 2
If y 3 0 , y 3 ow, when y 2 ,
N
x a 2
[ utting the value of y ] p x a 1 x a x a 2 1
[ y componendo and dividendo] b or, x a x a 2 1
2x 3 or, 2a 1 or, x 3a x a 3
Again, when y 3 , x a 1 x a x a 3 1 or, x a x a 3 1
2x 4 or, 2a 2 x 2 or, a 1 or, x 2a
Solution is : x 3a , or, 2a
Activity :
. Comparing the equation x 2 1 0 with the equation ax 2 bx c 0 , write
1
down the values of a, b, c .
2 W is the degree of the equation
. hat
( x 1) 2 0 ? How many roots has the equation and what are they?
5⋅6 Usage of quadratic equations
Many problems of our daily life can be solved easily by forming linear and quadratic equations. Here, the formation of quadratic equations from the given conditions based on real life problems and techniques for solving them are discussed.
Example 13. Denominator of a proper fraction is 4more than the numerator. If the fraction is squared, its denominator will be 0 more than the numerator. Find the
4
fraction. x Solution : et the fraction be
L
. x 4

Math

7
9

Square of the fraction =

x x 4

2

x2
( x 4) 2

2

x2

x2
8x 1
6

2

Here, numerator = x and denominator = x 8 x 1 .
6
2
2
By the question, x 8 x 1
6
x 4
0
or, 8 x 1
6
0
4
or, 8 x 4 1
0 6 or, 8 x 2
4
or, x 3 x 4 3 4 7 x 3
3
x 4 3 4 7
3
the fraction is
7
Example 14. A rectangular garden with length 5 metre and breadth 0 metre has o f
0
4 equal width all around the inside of the garden. If the area of the garden except the path is 00 square metre, how much is the path wide in metre ?
2
1
Solution : et the path be x metre wide.
L
W ithout the path, ength of the garden = (5 2 x) metre and
L
0 its breadth = (4 2 x) metre
0
x
.
W ithout the path, area of the garden = (5 2 x) × (4 2 x) square metre.
0
0
By the question, (5 2 x) (4 2 x) 10
0
0
2
0 or, or, or, or, or, or,

0
200 8
0
2
4x 1
0
8 x2 4 x
5
x 2 5x x( x 5)
( x 5)( x

x 00 x
1
x 80
0
20 0
0
4 x 20
0
0
4 ( x 5)
0
0
4 ) 0

4 x 2 10
0
2
0
[ ividing by 4 d ]
0
0

0 x 5 0, or x 4 0
If x 5 0 , x 5
If x 4 0 , x 4
0
0
But the breadth of the path will be less than 4 metre from the breadth of the garden.
0
x 4 ;
0
x 5 the path is 5 metres wide.

8
9

Math

Example 15. Shahik bought some pens at Tk. 4. If he would get one more pen in
2
0 that money, average cost of each pen would be less by Tk. . How many pens did he
1
buy ?
Solution : et, Shahik bought x pens in total by Tk. 4. Then each pen costs Tk.
L
0
2
0
4
2
. If he would get one more pen, that is, he would get ( x 1) pens, the cost of x 0
4
2 each pen would be Tk.
.
x 1
2
0
4
0
4
2 x 0
4
2
0
4
2
By the question, ,
1, or x 1 x x 1 x b or, 2 x ( x 1)(2
0
4
0
4
x) [ y cross-multiplication] or, 2 x 2 x 2
4
0
0
4
0
4 x2 x b or, x 2 x 2
0
4
0 [ y transposition] or, x 2 1 x 1 x 2
6
5
4
0
0
or, x( x 1 ) 1 ( x 1 ) 0
6
5
6
or, ( x 1 )( x 1 ) 0
6
5
5
x 1
6
0 , or, x 1 0
If x 1
6
0, x
6
1
Again, if x 1
5
0, x 1
5
But the number of pen x , cannot be negative. x 6
1 ; x 1
5
Shahik bought 1
5
pens.
Activity : Solve by forming equations :
. If a natural number is added to its squ are, the sum will be equal to nine times
1
of exactly its next natural number. W is the number ? hat . ength of a perpendicular drawn from the centre of a circle of radius 1 cm.
2 L
0
to a chord is less by 2cm. than the semi-chord. Find the length of the chord by drawing a probable picture.
Example 16. In an examination of class IX of a school, total marks of x students obtained in mathematics is 1. If at th e same examination, marks of a new student
9
5
0
in mathematics is 34and it is added to the former total marks, the average of the marks become less by .
1
a. W down the average of the obtained marks of all students including the new rite student and separately x students in terms of x .
b. By forming equation from the given conditions, show that, x 2 35 x 1
0
5
9
0
c. By finding the value of x , find the average of the marks in the two cases.
0
5
9
1
Solution : a. Average of the marks obtained by x students = x Average of the marks obtained by ( x 1) students including

9

Math

34 1
9
8
4
x 1 x 1
0
5
9
1
4
8
9
1
b. By the question,
1
x x 1
0
5
9
1
4
8
9
1 or, 1 x x 1
0
5
9
1
0
5
9
4
8
9
1
x 1 x or,
1
x( x 1) or, x 2 x 1
[ y cross-multiplication] b 9
5
0 x 1
9
8
4
x 1
9
5
0
or, x 2 x 1
0
5
9
34 x s x 2 35 x 0
5
9
1
0 [ howed]
2
c. x 35 x 1
9
5
0
0 or, x 2 6 x 30 x 1
5
0
5
9
0
or, x( x 6 ) 30( x 6 ) 0
5
5 or, ( x 6 )( x 30) 0
5
x 6
5
0 , or, x 30 0
If x 6 0 , x
5
5
6
Again, if x 30 0 , x 30
Since the number of students, i.e., x cannot be negative, x
5
6 x 30 .
1
9
5
0 in the first case, average =
6
5
30
1
9
8
4
and in the second case, average =
6 .
4
31

the new student =

1
9
5
0

Exercise 5⋅2
1
.

.
2

Assuming x as the variable in the equation a2 x b 0 , which one of the following is the degree of the equation ?
a. 3
b. 2
c. 1
d. 0
W one of the following is an identity ? hich a. ( x 1) 2 ( x 1) 2 4 x
b. ( x 1) 2 ( x 1) 2 2( x 2 1)
c. (a b) 2 (a b) 2

3.
.
4

2ab

d. (a b) 2

a2

2ab b 2

How many roots are there in the equation ( x 4) 2 0 ?
b. 2
c. 3
d. 4
a. 1
W one of the following are the two roots of the equation hich x2

x 1
2

0?

0
10

Math

W is the coefficient of hat x in the equation 3 x 2 x 5 0 ?
a. 3
b. 2
c. 1
d. 1
. bserve the following equations :
6 O x ii.
2
1
i. 2 x 3 9 iii. 2 x 1 5
2
W are of the above equations equivalent ? hich b. ii and iii
c. i and iii
d. i, ii and iii
a. i and ii
7 W
. hich one of the following is the solution set of the equatio n x 2 (a b) x ab 0 ?
b. {a, b}
c. { a, b}
d. { a, b}
a. {a, b}
. The digit of the tens place of a numbe r consisting of two digits is twice the
8
digit of the units place. In respect of the information, answer the following questions :
(1 If the digit of the units place is x , what is the number ?
)
b. 3 x
c. 1 x
d. 2 x
a. 2 x
2
1
(2 If the places of the digits are interchanged, what will be the number ?
)
a. 3 x
b. 4 x
c. 1 x
d. 2 x
2
1
(3) If x 2 , what will be the difference between the original number and the number by interchanging their places?
b. 2
c. 34
d. 36
a. 1
8
0
Solve (9−18) :
.
9 ( x 2)( x
0
.
3 ) 0 1. ( 2 x 3)( 3 x 2) 0 1 y ( y 5) 6
3
4
1 ( y 5)( y 5) 2
2
.
3
1. 2( z 2 9) 9 z 0
.
4
1
4
2
2 z 1 5z 1
4
x 2 6( x 2) x a x b
.
5
1
.
6
1
.
7
0
1 x 4 5 1
1
x 2 x 6 a x b x
0
1 x 4 x a x b a b
.
8
1
x b x a b a
Find the solution set (19−25):
3
4 x 7 2x 6
.
9
1
2.
0
.
1
2
2
5 x 1 2x 1 x x 1
1 1 1
1
x a b x a b ax b cx d
1
.
2
3. x
2
.
4 2 x 2 4ax 0
2
2 a bx c dx x .
5

Math

0
11

( x 1) 3 ( x 1) 3
2
( x 1) 2 ( x 1) 2
Solve by forming equations (26−31) : r .
6
2 Sum of the two digits of a number consisting of two digits is 5and thei
1
product is 6find the number.
5
;
. Area of the floor of a rectangular room is 2
7
2
9
1 square metre. If the length of the floor is decreased by 4metre and the breadth is increased by 4 metre, the are a remains unchanged. Find the length and breadth of the floor.
. ength of the hypotenuse of a right angled triangle is 1
8 L
2
5 cm. and the difference of the lengths of other two sides is 3 cm. Find the lengths of those two sides.
. The base of a triangle is 6cm. more than twice its height. If the area of the
9
2 triangle is 1 square cm., what is its height ?
0
8
30. As many students are there in a class, each of them contributes equal to the t number of class-mates of the class and thus total Tk. 0 was collected. W
2
4 ha is the number of students in the class and how much did each student contribute ?
31 As many students are there in a class, each of them contributed 30 paisa more
.
than the number of paisa equal to the number of students and thus total Tk. 7
0
was collected. W is the number of students in that class ? hat 32 Sum of the digits of a number consisting of two digits is 7 If the places of the
.
. digits are interchanged, the number so formed is 9 more than the given number.
a. W down the given number and the number obtained by interchangin g rite their places in terms of variable x .
b. Find the given number.
c. If the digits of the original number indicate the length and breadth of a rectangular region in centimetre, find the length of its diagonal. Assuming the diagonal as the side of a square, find the length of the diagonal of the square.
33. The base and height of a right angle triangle are respectively ( x 1) cm. and x cm. and the length of the side of a square is equal to the height of the triangle. Again, the length of a rectangular region is ( x 3) cm. and its breadth is x cm.
a. Show the information in only one picture.
b. If the area of the triangular region is 1 square centimetre, what is its height?
0
c. Find the successive ratio of the areas of the triangular, square and rectangular regions. .
5
2

Chapter Six

Lines, Angles and Triangles
Geometry is an old branch of mathematics. The word ‘geometry’ comes from the
Greek words ‘geo’, meaning the ‘earth’, and ‘metrein’, meaning ‘to measure’. So, the word ‘geometry’ means ‘the measurement of land.’ Geometry appears to have originated from the need for measuring land in the age of agricultural based civilization. However, now a days geometry is not only used for measuring lands, rather knowledge of geometry is now indispensible for solving many complicated mathematical problems. The practice of geometry is evident in relics of ancient civilization. According to the historians, concepts and ideas of geometry were applied to the survey of lands about four thousand years ago in ancient Egypt. Signs of application of geometry are visible in different practical works of ancient Egypt,
Babylon, India, China and the Incas civilisation. In the Indian subcontinent there were extensive usages of gometry in the Indus Valley civilisation. The excavations at
Harappa and Mohenjo-Daro show the evidence of that there was a well planned city.
For example, the roads were parallel to each other and there was a developed underground drainage system. Besides the shape of houses shows that the town dwellers were skilled in mensuration. In Vedic period in the construction of altars (or vedis) definite geometrical shapes and areas were maintained. Usually these were constituted with triangles, quadrilaterals and trapeziums.
But geometry as a systematic discipline evolved in the age of Greek civilization. A
Greek mathematician, Thales is credited with the first geometrical proof. He proved logically that a circle is bisected by its diameter. Thales’ pupil Pythagoras developed the theory of geometry to a great extent. About 300 BC Euclid, a Greek scholar, collected all the work and placed them in an orderly manner it in his famous treatise,
‘Elements’. ‘Elements’ completed in thirteen chapters is the foundation of modern geometry for generations to come. In this chapter, we shall discuss logical geometry in accordance with Euclid.
At the end of this chapter, the students will be able to
Describe the basic postulates of plane geometry
Prove the theorems related to triangles
Apply the theorems and corollaries related to triangles to solve problems.
6⋅1 Concepts of space, plane, line and point
The space around us is limitless. It is occupied by different solids, small and large. By solids we mean the grains of sand, pin, pencil, paper, book, chair, table, brick, rock, house, mountain, the earth, planets and stars. The concepts

Math

103

of geometry springs from the study of space occupied by solids and the shape, siz, e location and properties of the space.

A solid occupies space which is spread in three directions. This spread in three directions denotes the three dimensions (length, breadth and height) of the solid. ence every solid is three- imensional.
H
d
For example, a brick or a box has three dimensions (length, breadth and height). A sphere also has three dimensions, although the dimensions are not distinctly visible.
The boundary of a solid denotes a surface, that is, every solid is bounded by one or more surfaces. For example, the six faces of a box represent six surfaces. The upper face of a sphere is also a surface. But the surfaces of a box and of a sphere are different. The first one is plane while the second is one curved.
Two-dimensional surface : A surface is two dimensional; it has only length and breadth and is said to have no thickness. eeping the two dimension of a box
K
unchanged, if the third dimension is gradually reduced to zro, we are left with a face e or boundary of the box. In this way, we can get the idea of surface from a solid. hen two surfaces intersect, a line is form ed. For example, two faces of a box meet
W
at one side in a line. This line is a straight line. Again, if a lemon is cut by a knife, a curved line is formed on the plane of intersection of curved surface of the lemon.
One-dimensional line : A line is one- imensional; it has only length and no breadth d or thickness. If the width of a face of the box is gradually receded to zro, we are lef t e with only line of the boundary. In this way, we can get the idea of line the from the idea of surface.
The intersection of two lines produces a point. That is, the place of intersection of two lines is denoted by a point. If the two edges of a box meet at a point. A point has no length, breadth and thickness. If the length of a line is gradually reduced to ero at z last it ends in a point. Thus, a point is considered an entity of zro dimension. e 6⋅2 Euclid’s Axioms and Postulates
The discussion above about surface, line and point do not lead to any definition – they are merely description. This description refers to height, breadth and length, neither of which has been defined. Wonly can represent them intuitively. The e definitions of point, line and surface which Euclid mentioned in the beginning of the first volume of his Elements’ are incomple te from modern point of view. A few of
‘
Euclid’s axioms are given below:
1. A point is that which has no part.
2. A line has no end point.

104

Math

3. A line has only length, but no breath and height.
4. A straight line is a line which lies evenly with the points on itself.
5. A surface is that which has length and breadth only.
6. The edges of a surface are lines.
7. A plane surface is a surface which lies evenly with the straight lines on itself.
It is observed that in this description, part, length, width, evenly etc have been accepted without any definition. It is assumed that we have primary ideas about them. The ideas of point, straight line and plane surface have been imparted on this assumption. As a matter of fact, in any mathematical discussion one or more elementary ideas have to be taken granted. Euclid called them axioms. Some of the axioms given by Euclid are:
(1) Things which are eqal to the same thing, are eqal to one another. u u
(2) If eqals are added to eqals, the wholes are eqal. u u u (3) If eqals are subtracted from eqals, the remainders are eqal. u u u (4) Things which coincide with one another, are eqal to one another. u (5) The whole is greater than the part.
In modern geometry, we take a point, a line and a plane as undefined terms and some of their properties are also admitted to be true. These admitted properties are called geometric postulates. These postulates are chosen in such a way that they are consistent with real conception. The five postulates of Euclid are:
Postulate 1: A straight line may be drawn from any one point to any other point.
Postulate 2: A terminated line can be produced indefinitely,
Postulate 3: A circle can be drawn with any centre and any radius.
Postulate 4: All right angles are equal to one another.
Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.
After Euclid stated his postulates and axioms, he used them to prove other results.
Then using these results, he proved some more results by applying deductive reasoning. The statements that were proved are called propositions or theorems.
Euclid is his elements’ proved a total of 465 propositions in a logical chain. This is
‘
the foundation of modern geometry.
Note that there are some incompletion in Euclid’s first postulate. The drawing of a uniqe straight line passing through two distinct points has been ignored. u Postulate 5 is far more complex than any other postulate. On the other hand,
Postulates 1 through 4 are so simple and obvious that these are taken as self- vident
‘ e truths’. owever, it is not possible to prove them. So, these statements are accepted
H

Math

105

without any proof. Since the fifth postulate is related to parallel lines, it will be discussed later.
6.3 Plane Geometry
It has been mentioned earlier that point, straight line and plane are three fundamental concepts of geometry. Although it is not possible to define them properly, based on our real life experience we have ideas about them. As a concrete geometrical conception space is regarded as a set of points and straight lines and planes are considered the subsets of this universal set.
Postulate 1. Space is a set of all points and plane and straight lines are the sub- ets s of this set. From this postulate we observe that each of plane and straight line is a set and points are its elements. H owever, in ge ometrical description the notation of sets is usually avoided. For example, a point included in a straight line or plane is expressed by the point lies on the straight line or plane’ or the straight line or plane
‘
‘ passes through the point’. Similarly if a straight line is the subset of a plane, it is expressed by such sentences as ‘he straigh t line lies on the plane, or ‘he plane t t passes through the straight line’.
It is accepted as properties of straight line and plane that,
Postulate 2. For two different points there exists one and only one straight line, on which both the points lie.
Postulate 3. For three points which are not collinear, there exists one and only one plane, on which all the three points lie.
Postulate 4. A straight line passing through two different points on a plane lie completely in the plane.
Postulate 5. (a) Space contains more than one plane
(b) In each plane more than one straight lines lie.
(c) The points on a straight line and the real numbers can be related in such a way that every point on the line corresponds to a uniqe real u number and conversely every real number corresponds to a uniqe u point of the line.
Remark: The postulates from 1 to 5 are called incidence postulates.
The concept of distance is also an elementary concept. It is assumed that,
Postulate 6 : (a) Each pair of points (P, Q) determines a uniqe real number which u is known as the distance between point P and Q and is denoted by PQ.
(b) If P and Q are different points, the number PQ is positive. Otherwise, PQ=
0.
(c) The distance between P and Q and that between Q and P are the same, i.e.
PQ=QP.

106

Math

According to postulate 5(c) one to one correspondence can be established between the set of points in every straight line and the set of real numbers. In this connection, it is admitted that,
Postulate 7 : One-o- ne correspondence can be established between the set of t o points in a straight line and the set of real numbers such that, for any points P and Q,
PQ = – , where, the one-o- ne correspondence associates points a b|
|
t o
P and Q to real numbers a and b respectively.
If the correspondence stated in this postulate is made, the line is said to have been reduced to a number line. If P corresponds to a in the number line, P is called the graph point of P and a the coordinates of P. To convert a straight line into a number line the co- rdinates of two points are taken as 0 and 1 respectively. Thus a unit o distance and the positive direction are fixed in the straight line. For this, it is also admitted that,
Postulate 8: Any straight line AB can be converted into a number line such that the coordinate of A is 0 and that of B is positive.
Remark: Postulate 6 is known as distance postulate and Postulate 7 as ruler postulate and Postulate 8 ruler placement postulate. as Geometrical figures are drawn to make geometrical description clear. The model of a point is drawn by a thin dot by tip of a pencil or pen on a paper. The model of a straight line is constructed by drawing a line along a ruler. The arrows at ends of a line indicate that the line is extended both ways indefinitely. By postulate 2, two different points A and B define a uniqe straight line on which the two points lie. u This line is called AB or BA line. By postulate 5(c) every such straight line contains infinite number of points.
According to postulate 5(a) more than one plane exist. There is infinite number of straight lines in every such plane. The branch of geometry that deals with points, lines and different geometrical entities related to them, is known as plane Geometry. In this textbook, plane geometry is the matter of our discussion. ence, whenever something is not mentioned
H
in particular, we will assume that all discussed points, lines etc lie in a plane.
Proof of Mathematical statements
In any mathematical theory different statements related to the theory are logically established on the basis of some elementary concepts, definitions and postulates.
Such statements are generally known as propositions. In order to prove correctness of statements some methods of logic are applied. The methods are:
(a) Method of induction
(b) Method of deduction

107

Math

Proof by contradiction
Philosopher Aristotle first introduced this method of logical proof. The basis of this method is:
A property can not be accepted and rejected at the same time.
The same object can not possess opposite properties.
One can not think of anything which is contradictory to itself.
If an object attains some property, that object can not unattain that property at the same time.
6⋅4 Geometrical proof
In geometry, special importance is attached to some propositions which are taken, as theorems and used successively in establishing other propositions. In geometrical proof different statements are explained with the help of figures. But the proof must be logical.
In describing geometrical propositions general or particular enunciation is used. The general enunciation is the description independent of the figure and the particular enunciation is the description based on the figure. If the general enunciation of a proposition is given, subject matter of the proposition is specified through particular enunciation. For this, necessary figure is to be drawn.
Generally, in proving the geometrical theorem the following steps should be followed :
(1) General enunciation.
(2) Figure and particular enunciation.
(3) Description of the necessary constructions and
(4) Description of the logical steps of the proof.
If a proposition is proved directly from the conclusion of a theorem, it is called a corollary of that theorem. Besides, proof of various propositions, proposals for construction of different figures are considered. These are known as constructions.
By drawing figures related to problems, it is necessary to narrate the description of construction and its logical truth.

Exercise 6.1
1.
2.
3.
4.
5.
6.

Give a concept of space, surface, line and point.
State Euclid’s five postulates.
State five postulates of incidence.
State the distance postulate.
State the ruler postulate.
Explain the number line.

10
8

Math

. S the postulat e of ruler placement.
7 tate
. Define intersecting straight line and parallel straight line.
8
Line, Ray and Line Segment
By postulates of plane geometry, every point of a straight line lies in a plane. et AB
L
be a line in a plane and C be a point on it. The point C is called internal to A and B if the points A, C and B are different points on a line and AC + CB = AB. The points A,
C and B are also called collinear points. The set of points including A and B and all the internal points is known as the line segment AB. The points between A and B are called internal points.
Angles
W two rays in a plane meet at a point, an angle hen is formed. The rays are known as the sides of the angle and the common point as vertex. In the figure, two rays OP and OQ make an angle POQ at their common point O. O is the vertex of the angle. The set of all points lying in the plane on the Q side of OP and P side of OQ is the known as the interior region of the POQ. The set of all points not lying in the interior region or on any side of the angle is called exterior region of the angle. Straight Angle
The angle made by two opposite rays at their common end point is a straight angle. In the adjacent figure, a ray AC is drawn from the end point A of the ray AB. Thus the rays AB and AC have formed an angle BAC at their common point
A. BAC is a straight angle. The measurement of a right angle is 2 right angles or 18 0.
0
Adjacent Angle
If two angles in a plane have the same vertex, a common side and the angles lie on opposite sides of the common side, each of the two angles is said to be an adjacent angle of the other. In the adjacent figure, the angles BAC and CAD have the same vertex A, a common side AC and are on opposite sides of AC. BAC and CAD are adjacent angles.

Math

Perpendicular and Right Angle
The bisector of a straight angle is called perpendicular and each of th related adjacent angle a right angle. In the adjacent figure two angles BAC and CAD are produced at the point A of BD. The angles BAC and CAD are equal and lie on opposite sides of the common side AC. Ech of the a angles BAC and CAD is a right angle and the line segments BD and AC are mutually perpendicular.
Acute and Obtuse Angles
An angle which is less than a right angle is called an acute angle and an angle greater than one right angle but less than two right angles is an obtuse angle. In the figure, AOC is an acute angle and AOD is an obtuse angle
Reflex Angle
An angle which is greater than two right angles and less than four right angles is called a reflex angle. In the figure, AOC is a reflex angle.
Complementary Angles
If the sum of two angles is one right angle, the two angles are called complementary angles. In the adjacent figure, AOB is a right angle. The ray OC is in the inner side of the angle and makes two angles AOC and COB. S taking together the o measurement of these two angles is one right angle.
The angles AOC and COB are complementary angles. Supplementary Angles
If the sum of two angles is 2 right angles, two angles are called supplementary angles. The point O is an internal point of the line AB. OC is a ray which is different from the ray OA and ray OB. As a result two angles AOC and COB are formed. The measurement of this two angles is equal to the measurement of the straight angle
AOB i.e., two right angles. The angles AOC and
COB are supplementary angles.

10
9

110

Math

Opposite Angles
Two angles are said to be the opposite angles if the sides of one are the opposite rays of the other.
In the adjoining figure OA and OB are mutually opposite rays. S are the rays OC and OD. The o angles AOC and BOD are a pair of opposite angles. S imilarly, BOC and DOA are another pair of opposite angles. Therefore, two intersecting lines produce two pairs of opposite angles.
Theorem 1
The sum of the two adjacent angles which a ray makes with a straight line on its meeting point is equal to two right angles. et, the ray OC meets the straight line AB at O . As a
L
result two adjacent angles AOC and COB are formed. Draw a perpendicular DO on AB .
AOC + COB =
Sm of the adjacent two angles = u AOD + DOC + COB
= AOD + DOB
= 2 right angles. [oved]
P
r
Theorem 2
When two straight lines intersect, the vertically opposite angles are equal. et AB and CD be two straight lines, which intersect
L
at O . As a result the angles AOC , COB , BOD ,
AOD are formed at O. AOC = opposite BOD and COB = opposite AOD .
6⋅4 Parallel lines
Alternate angles, corresponding angles and interior angles of the traversal

In the figure, two straight lines AB and CD are cut by a straight line EF at P and Q.
The straight line EF is a traversal of AB and CD. The traversal has made eight angles 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 with the lines AB and CD. Among the angles Math

111

(a) 1 and 5 , 2 and 6 , 3 and 7 , 4 and 8 are corresponding angles,
(b) 3 and 6 , 4 and 5 are alternate angles,
(c) 4 , 6 are interior angles on the right
(d) 3 , 5 are interior angles on the left.
In a plane two straight lines may intersect or they are parallel. The lines intersect if there exists a point which is common to both lines. therwise, the lines are parallel.
O
ote
N that two different straight lines ma y atmost have only one point in common.
The parallelism of two straight lines in a plane may be defined in three different ways: (a) The two straight lines never intersect each other ( even if extended to infinity)
(b) very point on one line lies at equal smallest distance from the other.
E
(c) The corresponding angles made by a transversal of the pair of lines are equal.
According to definition (a) in a plane two straight lines are parallel, if they do not intersect each other. Two line segments taken as parts of the parallel lines are also parallel. According to definition (b) the perpendicular distance of any point of one of the parallel lines from the other is always equal. erpendicular distance is the length of
P
the perpendicular from any point on one of the lines to the other. Conversely, if the perpendicular distances of two points on any of the lines to the other are equal, the lines are parallel. This perpendicular distance is known as the distance of the parallel lines. The definition (c) is equivalent to the fifth postulate of Eclid. This definition is u more useful in geometrical proof and constructions. bserve that, through a point not on a line, a unique line parallel to it can be drawn.
O
Theorem 3
When a transversal cuts two parallel straight lines,
(a) the pair of alternate angles are equal.
(b) that pair of interior angles on the same side of the transversal are supplementary. In the figure AB ll CD and the transversal PQ intersects them at E and F respectively. Therefore, b (a) PEB = corresponding EFD [ y definition]
(b) AEF= alternate EFD
(c) BEF + EFD == 2 right angles.
Activity:
1. sing alternate definitions of parallel lines prove the theorems related to parallel
U
straight lines.

Math

112

Theorem 4
When a transversal cuts two straight lines, such that
(a) pairs of corresponding angles are equal, or
(b) pairs of alternate interior angles are equal, or
(c) pairs of interior angles on the same side of the transversal are or equal to, the sum of two right angles the lines are parallel.
In the figure the line PQ intersects the straight lines
AB and CD at E and F respectively and
(a) PEB = alternate EFD or, (b) AEF = Corresponding EFD or, (c) BEF
EFD = 2 right angles.
Therefore, the straight lines AB and CD are parallel.
Corollary 1. The lines which are parallel to a given line are parallel to each other. Exercise 6.2
1.
2.
3.
4.

Define interior and exterior of an angle.
If there are three different points in a line, identify the angles in the figure.
Define adjacent angles and locate its sides.
Define with a figure of each: opposite angles, complementary angle, supplementary angle, right angle, acute and obtuse angle.
6.5 Triangles
A triangle is a figure closed by three line segments. The line segments are known as sides of the triangle. The point common to any pair of sides is the vertex. The sides form angles at the vertices. A triangle has three sides and three angles. Triangles are classified by sides into three types: equilateral, isosceles and scalene. By angles triangles are also classified into three types: acute angled, right angled and obtuse angled.
The sum of the lengths of three sides of the triangle is the perimeter. By triangle we also denote the region closed by the sides. The line segment drawn from a vertex to the mid-point of opposite side is known as the median. Again, the perpendicular distance from any vertex to the opposite side is the height of the triangle.
In the adjacent figure ABC is a triangle. A, B, C are three vertices. AB , BC , CA are three sides and
BAC , ABC , BCA are three angles of the triangle.
The sum of the measurement of AB , BC and CA is the perimeter of the triangle.

Math

Equilateral Triangle
An equilateral triangle is a triangle of three equal sides.
In the adjacent figure, triangle ABC is an equilateral triangle; because, AB = BC = CA i.e., the lengths of three sides are equal.
Isosceles Triangle
An isosceles triangle is triangle with two equal sides. In the adjacent figure triangle ABC is an isosceles triangle; because AB = AC BC i.e., the lengths of only two sides are equal.
Scalene Triangle
Sides of scalene triangle are unequal. Triangle ABC is a scalene triangle, since the lengths of its sides AB , BC , CA are unequal.

Acute Angled Triangle
A triangle having all the three angles acute is acute angled triangle. In the triangle ABC each of the angles
BAC ,
ABC and
BCA is acute i.e., the measurement of any angle is less than 90 . So ABC is acute angled.
Right Angled Triangle
A triangle with one of the angles right is a right angled triangle. In the figure, the DFE is a right angle; each of the two other angles DEF and EDF are acute.
The triangle DEF is a right angled triangle.
Obtuse angled triangle
A triangle having an angle obtuse is an obtuse angled triangle. In the figure, the GKH is an obtuse angle; the two other angles GHK and HGK are acute.
GHK is an obtuse angled triangle.

3
1

4
1

Math

9.3 Interior and Exterior Angles
If a side of a triangle is produced, a new angle is formed. This angle is known as exterior angle. Except the angle adjacent to the exterior angle, the two others angles of the triangle are known as opposite interior angles.
In the adjacent figure, the side BC of ABC is produced to D. The angle ACD is an exterior angle of the triangle.
ABC,
BAC and
ACB are three interior angles.
ACB is the adjacent interior angle of the exterior angle
ACD . Each of ABC and BAC is an opposite interior angle with respect to ACD .
Theorem 5
The sum of the three angles of a triangle is equal to two right angles.

2 et ABC be a triangle. In the triangle
L
BAC + ABC + ACB =right angles.
Corollary 1: If a side of a triangle is produced then exterior angle so formed is equal to the sum of the two opposite interior angles.
Corollary 2: If a side of a triangle is produced, the exterior angle so formed is greater than each of the two interior opposite angles.
Corollary 3: The acute angles of a right angled triangle are complementary to each other. Activity :
. Prove that if a side of a triangle is produced, the exterior angle so formed is
1
greater than each of the two interior opposite angles.
Congruence of Sides and Angles
If two line segments have the same length, they are congruent. Conversely, if two line segments are congruent, they have the same length.
If the measurement of two angles is equal, the angles are congruent. Conversely, if two angles are congruent, their measurement is the same.
Congruence of Triangles
If a triangle when placed on another exactly covers the other, the triangles are congruent. The corresponding sides and angles of two congruent triangles are equal.

5
1

Math

In the figure, ABC and DEF are congruent.
If two triangles ABC and DEF are congruent, by superposition of a copy of ABC on DEF we find that each covers the other completely. . Hence, the line segments as well angles are congruent.
We would express this as ABC
ABC .
Theorem 6 (SAS criterion)
If two sides and the angle included between them of a triangle are equal to two corresponding sides and the angle included between them of another triangle, the triangles are congruent. et ABC and DEF be two triangles in
L
which AB = DE, AC = DF and the included BAC =the included
EDF . Then ABC
DEF .
Theorem 7
If two sides of a triangle are equal, the angles opposite the equal sides are also equal. Suppose in the triangle ABC, AB = AC, then ABC = ACB .
Theorem 8
If two angles of a triangle are equal, the sides opposite the equal angles are also equal. In the triangle ABC
ABC
be proved that AB AC .

ACB . It is to

Proof
Steps
(1 If AB
)

J stification u AC is not equal to AB ,

(i) AB

AC or, (ii) AB AC .
Suppose, AB AC . Cut from AB a part AD equal to AC . Now, the triangle ADC is an isosceles triangle. So,

ADC

ACD

DBC Exterior angle ADC > ABC
ACD > ABC Therefore, ACB > ABC
But this is against the given condition.

In

[ he base angles of an
T
isosceles triangles are equal]

[ xterior angle is greater than
E
each of the interior opposite angles] 6
1

Math

(2 Similarly, (ii) If AB AC , it can be proved
)
that ABC > ACB .
But this is also against the condition,
(3) So neither AB AC nor AB
AB AC (Proved)

AC

Theorem 9 (SSS criterion)
If the three sides of one triangle are equal to the three corresponding sides of another triangle, the triangles are congruent.
In ABC and and BC=EF,

ABC

DEF , AB = DE, AC = DF

ABC .

Theorem 10 (ASA criterion)
If two angles and the included side of a triangle are equal to two corresponding angles and the included side of another triangle, the triangles are congruent. et ABC and DEF be two triangles in which
L
the B = E , C = F . and the side BC
= the corresponding side EF, then the triangles are congruent, i.e. ABC
DEF .
Theorem 11 (HSA criterion)
If the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, the triangles are congruent.

et ABC and DEF be two right angled triangles, in which the hypotenuse AC=
L
hypotenuse DF and AB = DE, then ABC
DEF . .
Theorem 12
If one side of a triangle is greater than another, the angle opposite the greater side is greater than the angle opposite the lesser sides. et ABC be a triangle whose AC >AB, then
L
ABC > ACB .

7
1

Math

Theorem 13
If one angle of a triangle is greater than another, the side opposite the greater angle is greater than the side opposite the lesser. et ABC be a triangle in which
L

ABC >

ACB.

It is required to prove that, AC > AB.
Proof :
Steps

J stification u (1If the side AC is not greater than AB,
)
(i) AC = AB or, (ii) AC < AB
(i) if AC=AB ABC = ACB which is against the [ he base angles of
T
supposition, since by supposition ABC > ACB isosceles triangle are equal]
(ii) Again if AC < AB,

ABC <

ACB

But this is also against the supposition.

[ The angle opposite to smaller side is smaller]

(2 Therefore, the side AC is neither equal to nor
)
less than AB. AC > AB (Proved).
There is a relation between the sum or the differece of the lengths of two sides and the length of the third side of a triangle.
Theorem 14
The sum of the lengths of any two sides of a triangle is greater than the third side. et ABC be a triangle. Then any two of its sides
L
are together greater than the third side. et BC to
L
be the greatest side. Then AB+AC > BC.
Corollary 1. The difference of the lengths of any two sides of a triangle is smaller than the third side. et ABC be a triangle. Then the difference of the lengths of any two of its sides is
L
smaller than the length of third side i.e. AB AC < BC.
Theorem 15
The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and in length it is half.

8
1

Math

et ABC be a triangle and D and E are respectively midL points of the AB and AC. It is required to prove
1
that DE | BC and DE
BC
2
Construction: oin D and E and extend to F so that EF = DE.
J
Proof :
Steps

J stification u (1 Between ADE and CEF , AE
)
DE EF
AED
CEF
ADE
CEF

ADE

EFC and

DAE

[given ]
[ y construction] b [ pposite angles] o [ AS theorem]
S

EC

ECF .

DF | BC or, DE | BC .

(2 Again, DF
)

BC

DE

EF

BC

DE

DE

BC

DE

1
BC
2

Theorem 16 (Pythagoras theorem)
In a right-angled triangle the square on the hypotenuse is equal to the sum of the squares of regions on the two other sides.
1
et in the triangle ABC, ABC =right
L
angle and AC is the hypotenuse.
Then AC 2 AB 2 BC 2 ,

Exercise 6.3
. The lengths of three sides of a triang le are given below. In which case it is
1
possible to draw a triangle?
(a) 5 cm, 6 cm and 7 cm (b) 3 cm, 4 cm and 7 cm (c) 5 cm, 7 and 4 cm (a) 2 4 cm and 8 cm 1 cm, cm
. Consider the following information:
2
(i) A right angled triangle is a triangle with each of three angles right angle.
(ii) An acute angled triangle is a triangle with each of three angles acute.
(iii) A triangle with all sides equal is an equilateral triangle.

Math

Which one of the following is correct?
(a) i and ii
(b) i and iii
(c) ii and iii se the figure below to answer questions 3 and 4.
U

9
1

(d) i, ii and iii

3. Which one is a right angle?
(a) BOC (b) BOD (c) COD (d) AOD
4. What is the angle complementary to BOC?
(a) AOC (b) BOD (c) COD (d) AOD
5. Prove that, the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral.
6. Prove that, the three medians of an equilateral triangle are equal.
. Prove that, the sum of any two exterior angles of a triangle is greater than two righ t
7
angles.
8 D is a point inside a triangle ABC. Prove that, AB + AC > BD + DC.
.
9. If D is the middle point of the side BC of the triangle ABC, prove that, AB + AC
> 2AD.
0. Prove that, the sum of the three medians of a triangle is less than its perimeter.
1
. A is the vertex of an isosceles triangle ABC, and the side BA is produced to D
1
BCD =right angle.
1
such that BA = AD; prove that
. The bisectors of the angles
2
1
B and C of a triangle ABC intersect at O. Prove that BOC = +
90° ½
A.
1. If the sides AB and AC of a triangle ABC are produced and the bisector of the
3
90°
½ A. exterior angles formed at B and C meet at O, prove that, BOC =
4. In the adjoining figure,
1
C =right angle and
1
B=
2
A.
Prove that, AB = 2BC.

5. Prove that, the exterior angle so formed by producing any side of a triangle is
1
equal to the sum of the interior opposite angles.

0
2
1

Math

6. Prove that, the difference between any two sides of a triangle is less than the
1
third.
B =right angle and
1
. In the adjoining figure,
7
1
D is the middle point of the hypotenuse AC of the triangle ABC. Prove that. BD = AC.
½

. In the
8
1
ABC, AB > AC and the bisector AD of A intersects BC at D. Prove that ADB is an obtuse angle.
9. Show that, any point on the perpe ndicular bisector of a line segment is
1
equidistant from the terminal points of that line segment.
0. In the right- ngled triangle
2
a
A =right angle and D is the mid point of BC.
1
a. Draw a triangle ABC with given information.
b. Prove that AB + AC > 2 AD
1
c. Prove that, AD
BC
2

Chapter Seven

Practical Geometry
In the previous classes geometrical figures were drawn in proving different propositions and in the exercises. There was no need of precision in drawing these figures. But sometimes precision is necessary in geometrical constructions. or
F
example, when an architect makes a design of a house or an engineer draws different parts of a machine, high precision of drawing is required. In such geometrical constructions, one makes use of ruler and compasses only. We have already learnt how to construct triangles and quadrilaterals with the help of ruler and compasses. In this chapter we will discuss the construction of some special triangles and quadrilaterals. At the end of the chapter, the students will be able to –
Explain triangles and quadrilaterals with the help of figures
Construct triangle by using given data
Construct parallelogram by using given data.
7⋅1 Construction of Triangles
Every triangle has three sides and three angles. But, to specify the shape and size of a triangle, all sides and angles need not to be specified. or example, as sum of the
F
three angles of a triangle is two right angles, one can easily find the measurement of the third angle when the measurement of the two angles of the triangle given. Again, from the theorems on congruence of triangles it is found that the following combination of three sides and angles are enough to be congruent. That is, a combination of these three parts of a triangle is enough to construct a unique triangle.
In class seven we have learnt how to construct triangles from the following data:
(1 Three sides
)

(2 Two sides and their
)
included angle.

2
1

Math

(3) Two angles and their adjacent sides

(4) Two angles and an opposite side

(5) Two sides and an opposite angle

(6) Hypotenuse and a side of a right- ngled a triangle

Observe that in each of the cases above, three parts of a triangle have been specified.
But any three parts do not necessarily specify a unique triangle. As for example, if three angles are specified, infinite numbers of triangles of different sizes can be drawn with the specified angles (which are known as similar triangles).

Sometimes for construction of a triangle three such data are provided by which we can specify the triangle through various drawing. Construction in a few such cases is stated below.

3
2
1

Math

Construction 1
The base of the base adjacent angle and the sum of other two sider of a triangle are given. Construct the triangle. et the base a, a base adjacent angle x and the
L
sum s of the other two sides of a triangle ABC be given. It is required to construct it.
Steps of construction :
(1 F any ray BE cut the line segment BC
) rom equal to a. At B of the line segment BC, draw an angle CBF = x.
(2 Cut a line segment BD equal to s from
)
the ray BF.
(3) J in C,D and at C make an angle DCG o equal to BDC on the side of DC in which B lies. (4) et the ray CG intersect BD at A.
L
Then, ABC is the required triangle.
Proof : In ACD , ADC b ACD [ y construction]
AC
AD.
Now, In ABC , ABC b x, BC a, [ y construction] and BA AC BA AD BD s . Therefore, ABC is the required triangle.
Alternate Method et the base a, a base adjacent angle x and the
L
sum s of the other two sides of a triangle ABC be given. It is required to construct the triangle.
Steps of construction:
(1 F any ray BE cut the line segment BC
) rom equal to a. At B of the line segment BC draw an angle CBF = x.
(2 Cut a line segment BD equal to s from
)
the ray BF.
(3) J in C,D and construct the perpendicular o bisector PQ of CD.
(4) et the ray PQ intersect BD at A. J in A, C.
L
o
Then, ABC is the required triangle.
Proof: In

ACR and

ADR , CR

DR AR

AR and the included angle

4
2
1

Math

ARC
ACR
Now, In

r
ARD [ight angle ]
AC AD
ADR . x, BC
ABC , ABC

b a, [ y construction ]

and BA AC BA AD BD s . Therefore, ABC is the required triangle.
Construction 2
The base of a triangle the base adjacent an acute angle and the difference of the other two sides are given. Construct the triangle. et the base a, a base adjacent angle x and
L
the difference d of the other two sides of a triangle ABC be given. It is required to construct the triangle.
Steps of Construction :
(1 rom any ray BE, cut the line segment BC ,
)F
equal to a. At B of the line segment BC draw an angle CBF = x.
(2 Cut a line segment BD equal to s from the
)
ray BE.
(3) J in C,D and at C, make an angle DCA o equal to EDC on the side of DC in which C lies. et the ray CA intersect BE at A.
L
Then ABC is the required triangle.
Proof :, In ACD , ADC
ACD [by construction]
AC
AD.
So, the difference of two sides AB AC AB AD BD d .
Now, In ABC , BC a, AB AC d and ABC
x. Therefore, required triangle.

ABC is the

Activity :
. If the given angle is not acute, the above construction is not possible. Why ?
1
Explore any way for the construction of the triangle under such circumstances.
. The base, the base adjacent angle and the difference of the other two sides of a
2
triangle are given. Construct the triangle in an alternate method.
Construction 3
The base adjacent two angles and the perimeter of a triangle are given.
Construct the triangle. et the base adjacent angles
L
x and y and the perimeter p be geven. It is required to construct the triangle.

5
2
1

Math

Steps of Construction :
(1 rom any ray DF, cut the part DE equal to the
) F perimeter p. Make angles EDL equal to x and
DEM equal to y on the same side of the line segment DE at D and E.
(2 Draw the bisectors BG and EH of the two angles.
)
(3) et these bisectors DG and EH intersect at a
L
point A. At the point A, draw ∠DAB equal to ∠ADE and ∠EAC equal to ∠AED.
(4) et AB intersect DE at B and AC intersect DE at
L
C.
Then, ABC is the required triangle.
Proof : In ADB , ADB b ,
AB
DAB [ y construction]
Again, in ACE , AEC
EAC ;
CA CE.
Therefore, in ABC , AB BC CA DB BC CE DE p.
1
1
ABC
ADB
DAB
x x x
2
2 and ACB

AEC

EAC

1 y 2

1 y 2

y. Therefore,

DB.

ABC is the required triangle.

Activity:
.
1 Two acute base adjacent angles and the perimeter of a triangle are given. Construct the triangle in an alternative way.
Example 1. Construct a triangle ABC, in which perimeter AB + BC + CA = 1 cm. H

B =60°
,

C =45° and the

G

Steps of Construction: ollow the steps below :
F
(1 Draw a line segment PQ =cm.
)
1
(2 At P, construct an angle of QPL = and at , an angle of
)
60°
Q
PQM = on
45°
the same side of PQ.
(3) Draw the bisectors PG and QH of the two angles. et the bisectors PG and QH
L
of these angles intersect at A.
(4) Draw perpendicular bisector of the segments PA of QA to intersect PQ at B and C.

6
2
1

Math

(5) J in A, B and A, C. o Then, ABC is the required triangle.
Activity : An adjacent side with the right angle and the difference of hypotenuse and the other side of a right- ngled tr iangle are given. Construct the triangle. a Exercise 7.1
1. Construct a triangle with the following data:
(a) The lengths of three sides are 3 cm, 3 5 cm, 2 8 cm. (b) The lengths of two sides are 4 cm , 3 cm and the included angle is 60 .
(c) Two angles are 60 and 45 and their included side is 5 cm.
(d) Two angles are 60 and 45 and the side opposite the angle 45 is 5 cm.
(e) The lengths of two sides are 4.5 cm and 3.5 respectively and the angle opposite to the second side is 4 cm.
(f) The lengths of the hypotenuse and a side are 6 cm and 4 cm respectively.
2. Construct a triangle ABC with the following data:
(a) Base 3.5 cm, base adjacent angle 60 and the sum of the two other sides 8 cm. (b) Base 4 cm, base adjacent angle 50 and the sum of the two other sides .5 cm.
7
(c) Base 4 cm, base adjacent angle 50 and the difference of the two other sides .5 cm.
1
(d) Base 5 cm, base adjacent angle 45 and the difference of the two other sides 1 cm. (e) Base adjacent angles 60 and 45 and the perimeter1
2
cm.
(f) Base adjacent angles 30 and 45 and the perimeter 0 cm.
1
3. Construct a triangle when the two base adjacent angles and the length of the perpendicular from the vertex to the base are given.
4. Construct a right- ngled triangle when the hypotenuse and the sum of the other a two sides are given.
5. Construct a triangle when a base adjacent angle, the altitude and the sum of the other two sides are given.
6. Construct an equilateral triangle whose perimeter is given.
. The base, an obtuse base adjacent angl e and the difference of the other two sides
7
of a trangle are given. Construcet the triangle.
7⋅2 Construction of Quadrilaterals
We have seen if three independent data are given, in many cases it is possible to construct a definite triangle. But with four given sides the construction of a definite quadrilateral is not possible. ive independent data are required for
F
construction of a definite quadrilateral. A definite quadrilateral can be constructed if any one of the following combinations of data is known :

Math

7
2
1

(a) our sides and an angle
F
(b) our sides and a diagonal
F
(c) Three sides and two diagonals
(d) Three sides and two included angles
(e) Two sides and three angles.
In class V the construction of quadrila terals with the above specified data has
III,
been discussed. If we closely look at the steps of construction, we see that in some cases it is possible to construct the quadrilaterals directly. In some cases, the construction is done by constructions of triangles. Since a diagonal divides the quadrilateral into two triangles, when one or two diagonals are included in data, construction of quadrilaterals is possible through construction of triangle.
(1 our sides and an angle
)F

(2 our sides and a diagonal
)F

(3) Three sides and two diagonals

(4) Three sides and two included angles (5) Two sides and three angles

8
2
1

Math

Sometimes special quadrilaterals can be constructed with fewer data. In such a cases, from the properties of quadrilaterals, we can retrieve five necessary data. or
F
example, a parallelogram can be constructed if only the two adjacent sides and the included angle are given. In this case, only three data are given. Again, a square can be constructed when only one side of the square is given. The four sides of a square are equal and an angle is a right angle; so five data are easily specified.
Construction 4
Two diagonals and an included angle between them of a parallelogram are given. Construct the parallelogram. et a and b be the diagonals of a parallelogram and x be
L
an angle included between them. The parallelogram is to be constructed.
Steps of construction: rom any ray AE, cut the line segment AC = a. Bisect the
F
line segment AC to find the mid- oint O, At O construct p the angle AOP = x and extend the ray PO to the opposite ray OQ. rom the rays OP and OQ cut two line
F
1 segments OB and OD equal to
b. J in A,B; A,D; C,B o 2 and C,D. Then ABCD is the required parallelogram.
Proof: In triangles AOB and COD ,
1
1 b OA OC a, OB OD b [ y construction]
2
2
=
[ and included AOB included COD opposite angle]
Therefore, AOB
COD .
So, AB CD and ABO
CDO ;but the two angles are alternate angles. AB and CD are parallel and equal.
Similarly, AD and BC are parallel and equal.
Therefore, ABCD is a parallelogram with diagonals
1
1
AC AO OC a a a
2
2
1
1 and BD BO OD b b b and the angle included
2
2 between the diagonals AOB
x.
Therefore, ABCD is the required parallelogram.

Math

9
2
1

Construction 5
Two diagonals and a side of a parallelogram are given. Construct the parallelogram. et a and b be the diagonals and c be a side of the
L
parallelogram. The parallelogram is to be constructed.
Steps of construction:
Bisect the diagonals a and b to locate their mid- oints. p rom any ray AX, cut the line segment AB = a. With
F
2 b centre at A and B draw two arcs with radii and a
2
respectively on the same side of AB. et the arcs
L
intersect at O. J in A, O and O, B. Extend AO and BO o 2 and OD = to AE and BF respectively. Now cut OC = a b from OE and OF respectively. J in A,D; D,C; C,B. o 2
Then ABCD is the required parallelogram.
Proof: In

AOB and COD , a b b OA OC
; OB OD
, [ y construction]
2
2 included COD opposite angle]
[
and included AOB =
AOB
COD .
AB CD and ABO
ODC ; but the angles are alternate angles.
AB and CD are parallel and equal.
Similarly, AD and BC are parallel and equal.
Therefore, ABCD is the required parallelogram.
Example 1. The parallel sides and two angles included with the larger side of a trapezium are given. Construct the trapezium. et a and b be the parallel sides of a trapezium where a > b and x and y be two
L
angles included with the side a. The trapezium is to be constructed.

3
10

Math

Steps of construction: rom any ray AX, cut the line segment AB = a. At A of the line segment AB,
F
construct the angle BAY = x and at B, construct the angle ABZ = y. rom the
F
line segment AB, cut a line segment AE=b. Now at E, construct BC || AY which cuts
BZ at C. Now construct CD || BA. The line segment CD intersects the ray AY at D.
Then ABCD is the required trapezium.
Proof : By construction, AB | CD and AD | EC. Therefore, AECD is a parallelogram and CD = AE = b. Now in the quadrilateral ABCD, AB=a, CD=b, AB || CD and
BAD = x, ABC = y (by construction). Therefore, ABCD is the required trapezium. Activity : The perimeter and an angle of a rhombus are given. Construct the rhombus. Exercise 7⋅2
. The two angles of a right angled triangle are given. Which one of the following
1
combination allows constructing the triangle?
a. 63 and 36
b. 30 and 0
7
c. 40 and 50
d. 8 and 0
0
2
. i. A rectangle is a parallelogram
2
ii. A square is a rectangle iii. A rhombus is a square
On the basis of the above information, which one of the following is true?
a. i and ii b. i and iii
c. ii and iii
d. i, ii and iii
In view of the given figure, answer the questions 3 and 4.

3. What is the area of AOB?
a. 6 sq. units
b. 7 units sq. c. 1 units
2
sq.
d.1 sq. units
4
4. The perimeter of the quadrilateral is
a. 2
1
units
b. 1 units
4
c. 2 units
0
d. 2
8
units
5. Construct a quadrilateral with the following data :
(a) The lengths of four sides are 3 cm, 3 5 cm, .5 cm, 3cm and an angle is 45 .
2
(b) The lengths of four sides are 3.5 cm, 4 cm, 2 5 cm 3.5 cm and a diagonal is 5 cm.
(c) The lengths of three sides are 3 2 3 cm, 3 5 cm and two diagonals are cm, 2 8 and 4.5 cm. cm, Math

3
11

6. Construct a parallelogram with the following data:
(a) The lengths of two diagonals are 4 cm, 6 5 cm and the included angle is 45 .
(b) The lengths of two diagonals are 5 cm, 6 5 cm and the included angle is 30 .
(c) The length of a side is 4 cm and the lengths of two diagonals are 5 cm and 6 5 cm.
(d) The length of a side is 5 cm and the lengths of two diagonals are 4.5 cm and 6 cm.
. The sides AB and BC and the angles B, C, D of the quadrilateral ABCD are
7
given. Construct the quadrilateral.
. The four segments made by the in tersecting points of the diagonals of a
8
4
5
parallelogram and an included angle between them are OA = cm. OB = cm.
4
0
8
respectively. Construct the
OC 3 5 cm OD = 5 cm and AOB = quadrilateral. 9. The length of a side of a rhombus and an angle are 3.5 cm. and 45 respectively; construct the rhombus.
0. The length of a side and a diag onal of a rhombus are given;construct the
1
rhombus.
. The length of two diagonals of a rh ombus are given. Construct the rhombus.
1
. The perimeter of a square is given. Construct the square.
2
1
3. The houses of Mr. oki and Mr. afrul
1
Z
Z
are in the same boundary and the area of their house is equal. But the house of Zoki is rectangular and the house of Mr. afrul is in shape of parallelogram.
Z
(a) Construct the boundary of each of their houses taking the length of base 1
0
units and height 8 units. (b) Show that the area of the house of Mr. oki is less than the area of the house
Z
of Mr. afrul.
Z
(c) If the ratio of the length and the breadth of the house of Mr. oki is 4:3 an d
Z
its area is 300 sq. units, find the ratio of the area of their houses.
4. The lengths of the hypotenuse and a side of right angled triangle are 7 and 4
1
cm cm. se the information to answer the following questions :
U
a. ind the length of the other side of the triangle.
F
b. Construct the triangle.
c. Construct a square whose perimeter is equal to the perimeter of the triangle.
0
8
95
5
1. AB = cm, BC = cm
4
5
A = o, B = o and C = o. of the quadrilateral
5
8
ABCD. se the information to answer the following questions:
U
a. Construct a rhombus and give the name.
b. se the above information to construct the quadrilateral ABCD.
U
c. Construct an equilateral triangle whose perimeter is equal to the perimeter of the quadrilateral ABCD.

Chapter Eight

Circle
We have already known that a circle is a geometrical figure in a plane consisting of points equidistant from a fixed point. Different concepts related to circles like centre, diameter, radius, chord etc has been discussed in previous class. In this chapter, the propositions related to arcs and tangents of a circle in the plane will be discussed.
At the end of the chapter, the students will be able to
Explain arcs, angle at the centre, angle in the circle, quadrilaterals inscribed in the cirlce
Prove theorems related to circle
State constructions related to circle.
8⋅1 Circle
A circle is a geometrical figure in a plane whose points are equidistant from a fixed point.
The fixed point is the centre of the circle. The closed path traced by a point that keeps it distance from the fixed centre is a circle. The distance from the centre is the radius of the circle. et O be a fixed point in a plane and r be a fixed measurement.
L
The set of points which are at a distance r from O is the circle with centre O and radius r. In the figure, O is the centre of the circle and A, B and C are three points on the circle. Each of OA,
OB and OC is a radius of the circle. Some coplanar points are called concylcic if a circle passes through these points, i.e. there is a circle on which all these points lie. In the above figure, the points A, B and C are concyclic.
Interior and Exterior of a Circle
If O is the centre of a circle and r is its radius, the set of all points on the plane whose distances from O are less than r, is called the interior region of the circle and the set of all points on the plane whose distances from O are greater than r, is called the exterior region of the circle. The line segment joining two points of a circle lies inside the circle.
The line segment drawn from an interior point to an exterior point of a circle intersects a circle at one and only one point. In the figure, P and Q are interior and exterior points of the circle respectively. The line segment PQ intersects the circle at
R
only.

3
13

Math

Chord and Diameter of a Circle
The line segment connecting two different points of a circle is a chord of the circle. If the chord passes through the centre it is known as diameter. That is, any chord forwarding to the centre of the circle is diameter. In the figure, AB and AC are two chords and O is the centre of the circle. The chord AC is a diameter, since it passes through the centre. OA and OC are two radii of the circle. Therefore, the centre of a circle is the midpoint of any diameter. The length of a diameter is 2r, where r is the radius of the circle.
Theorem 1
The line segment drawn from the centre of a circle to bisect a chord other than diameter is perpendicular to the chord. et AB be a chord (other than diameter) of a circle ABC with
L
centre O and M be the midpoint of the chord. J in O, M. It is o to be proved that the line segment OM is perpendicular to the chord AB.
Construction: J in O, A and O, B. o Proof:
Steps
(1 In
)
OAM and OBM,
OA = OB
AM = BM and OM = OM
Therefore, OAM
OBM
OMA= OMB
(2 Since the two angles are equal and together
)
maka a straight angle.

Justification
[M is the mid point of AB]
[adius of same circle] r [ ommon side] c [ SS theorem]
S

1
OMA
OMB =right angle.
Therefore, OM AB . (Proved).
Corollary 1: The perpendicular bisector of any chord passes through the centre of the circle.
Corollary 2: A straight line can not intersect a circle in more than two points.
Activity :
.
1 The theorem opposite of the theorem 1 ates that the perpendicular from st the centre of a circle to a chord bisects the chord. Prove the theorem.

3
14

Math

Theorem 2
All equal chords of a circle are equidistant from the centre. et AB and CD be two equal chords of a circle with
L
centre O. It is to be proved that the chords AB and CD are equidistant from the centre.
Construction: Draw from O the perpendiculars OE and OF to the chords AB and CD respectively. J in o O, A and O, C.
Proof:
Steps

J stification u (1 OE AB and OF CD
)
Therefore, AE = BE and CF = BF.
1
1
AB and CF =
CD
AE =
2
2

[ he perpendicular from
T
the centre bisects the chord] (2 But AB = DC
)
AE = CF

[supposition]

(3) Now in the right- ngled triangles OAE and a OCF hypotenuse OA =hypotenuse OC and AE = CF
OAE
OCF

[adius of same circle] r [ tep ]
S 2
[R S theorem]
H

OE = OF
(4) But OE and OF are the distances from O to the chords AB and CD respectively.
Therefore, the chords AB and CD are equidistant from the centre of the circle. (Proved)
Theorem 3
Chords equidistant from the centre of a circle are equal. et AB and CD be two chords of a circle with centre
L
O. OE and OF are the perpendiculars from O to the chords AB and CD respectively. Then OE and OF represent the distance from centre to the chords AB and CD respectively.
It is to be proved that if OE = OF, AB = CD.
Construction: J in O, A and O, C. o Math

3
15

Proof:
Steps

Justification

(1 Since OE AB and OF CD.
)
right angle
Therefore, OEA
OFC 1
(2 Now in the right angled triangles
)
OAE and OCF hypotenuse OA= hypotenuse OC and
OE OF
OAE
OCF
AE CF .

[right angles ]

(3) AE =

1
1
AB and CF = CD
2
2

(4) Therefore

[adius of same circle] r [R S theorem]
H
[ he perpendicular from the centre
T
bisects the chord]

1
1
AB
CD
2
2
CD (Proved)

AB
Corollary 1: The diameter is the greatest chord of a circle.
i.e.,

Exercise 8.1
. Prove that if two chords of a circle bisect each other, their point of intersection is
1
the centre of the circle.
.
2 Prove that the straight line joining the middle points of two parallel chords of a circle pass through the centre and is perpendicular to the chords.
3. Two chords AB and AC of a circle subtend equal angles with the radius passing through A. Prove that, AB = AC.
4. In the figure, O is the centre of the circle and chord AB = chord AC. Prove that
BAO
CAO.
5. A circle passes through the vertices of a right angled triangle. Show that, the centre of the circle is the middle point of the hypotenuse.
6. A chord AB of one of the two concentric circles intersects the other circle at points C and D. Prove that, AC = BD.
. If two equal chords of a circle inters ect each other, show that two segments of
7
one are equal to two segments of the other.
.
8 Prove that, the middle points of e qual chords of a circle are concyclic.
9. Show that, the two equal chords drawn from two ends of the diameter on its opposite sides are parallel.

3
16

Math

0. Show that, the two parallel chords of a circle drawn from two ends of a diameter
1
on its opposite sides are equal.
. Show that, of the two chords of a circle the bigger chord is nearer to the centre
1
than the smaller.
8.2 The arc of a circle
An arc is the piece of the circle between any two points of the circle. ook at the pieces of the circle between two points A and
L
B in the figure. We find that there are two pieces, one comparatively large and the other small. The large one is called the major arc and the small one is called the minor arc.
A and B are the terminal points of this arc and all other points are its internal points. With an internal fixed point C the arc is called arc ABC and is expressed by the symbol
ACB. Again, sometimes minor arc is expressed by the symbol AB. The two points A and B of the circle divide the circle into two arcs. The terminal points of both arcs are A and B and there is no other common point of the two arcs other than the terminal points.
Arc cut by an Angle
An angle is said to cut an arc of a circle if
(i) each terminal point of the arc lies on the sides of the angle (ii) each side of the angle contains at least one terminal point
(iii) Every interior point of the arc lies inside the angle. The angle shown in the figure cuts the APB arc of the circle with centre O.
Angle in a Circle
If the vertex of an angle is a point of a circle and each side of the angle contains a point of the circle, the angle is said to be an angle in the circle or an angle inscribed in the circle. The angles in the figure are all angles in a circle. Every angle in a circle cuts an arc of the circle. This arc may be a major or minor arc or a semi- ircle. c The angle in a circle cuts an arc of the circle and the angle is said to be standing on the cut off arc. The angle is also known as the angle inscribed in the conjugate arc. In the adjacent figure, the angle stands on the arc APB and is inscribed in the conjugate arc ACB. It is to be noted that APB and ACB are mutually conjugate.

Math

137

Remark: The angle inscribed in an arc of a circle is the angle with vertex in the arc and the sides passing through the terminal points of the arc. An angle standing on an arc is the angle inscribed in the conjugate arc.
Angle at the Centre
The angle with vertex at the centre of the circle is called an angle at the centre. An angle at the centre cuts an arc of the circle and is said to stand on the arc. In the adjacent figure,
AOB is an angle at the centre and it stands on the arc APB.
Every angle at the centre stands on a minor arc of the circle.
In the figure APB is the minor arc. So the vertex of an angle at the centre always lies at the centre and the sides pass through the two terminal points of the arc.
To consider an angle at the centre standing on a semi-circle the above description is not meaningful. In the case of semicircle, the angle at the centre BOC is a straight angle and the angle on the arc BAC is a right angle.
Theorem 4
The angle subtended by the same arc at the centre is double of the angle subtended by it at any point on the remaining part of the circle.
Given an arc BC of a circle subtending angles BOC at the centre O and BAC at a point A of the circle ABC. We need to prove that BOC = 2 BAC.
Construction: Suppose, the line segment AC does not pass through the centre O. In this case, draw a line segment AD at
A passing through the centre O.
Proof :
Steps

Justification

(1) In AOB, the external angle
BOD = BAO + ABO
(2) Also in AOB, OA = OB
Therefore, BAO = ABO
(3) From steps (1) and (2), BOD = 2 BAO.
(4) Similarly, COD = 2 CAO
(5) From steps (3) and (4),
BOD
COD 2 BAO 2 CAO
This is the same as BOC 2 BAC . [Proved]

[An exterior angle of a triangle is equal to the sum of the two interior opposite angles.] [Radius of a circle]
[Base angles of an isosceles triangle are equal]
[by adding]

3
18

Math

We can state the theorem in a different way. The angle standing on an arc of the circle is half the angle subtended by the arc at the centre.
Activity : Prove the theorem 4 when AC passes through the centre of the circle
ABC.
Theorem 5
Angles in a circle standing on the same arc are equal. et O be the centre of a circle and standing on the arc BD,
L
BAD and BED be the two angles in the circle. We need to prove that BAD = BED.
Construction : J in O, B and O, D. o Proof :
Steps

J stification u T
(1 The arc BD subtends an angle BOD at the centre O. [ he angle subtended
)
by an arc at the centre
Therefore, BOD =
2
BAD and BOD =
2
BED is double of the angle
2 BAD =
2
BED subtended on the or, BAD
BED (Proved) circle] Theorem 6
The angle in the semi- circle is a right angle. et AB be a diameter of circle with centre at O and ACB
L
is the angle subtended by a semi- ircle. It is to be proved c that ACB =right angle.
1
Construction: Take a point D on the circle on the opposite side of the circle where C is located.
Proof:
Steps
J stification u (1 The angle standing on the arc ADB
)
1
ACB =
(straight angle in the centre AOB)
2
(2 But the straight angle
)
AOB is equal to 2 right angles.
1
(2 right angles) =right angle. (Proved)
1
ACB =
2

[ he angle standing on
T
an arc at any point of the circle is half the angle at the centre]

Corollary 1. The circle drawn with hypotenuse of a right- ngled triangle as diameter a passes through the vertices of the triangle.
Corollary 2. The angle inscribed in the major arc of a circle is an acute angle.

3
19

Math

Activity :
. Prove that any angle inscribed in a minor arc is obtuse.
1

Exercise 8.2
1 ABCD is a quadrilateral inscribed in a circle with centre O. If the diagonals AB
.
and CD intersect at the point E, prove that AOB + COD = 2 AEB.
. Two chords AB and CD of the circle ABCD intersect at the point E. Show that,
2
AED and BEC are equiangular.
3. In the circle with centre O ADB + BDC =right angle. Prove that, A, B and
1
C lie in the same straight line.
4. Two chords AB and CD of a circle intersect at an interior point. Prove that, the sum of the angles subtended by the arcs AC and BD at the centre is twice AEC.
5. Show that, the oblique sides of a cyclic trapezium are equal.
6. AB and CD are the two chords of a circle ; P and Q are the middle points of the two minor arcs made by them. The chord PQ intersects the chords AB and AC at points D and E respectively. Show that, AD = AE.
8⋅3 Quadrilateral inscribed in a circle
An inscribed quadrilateral or a quadrilateral inscribed in a circle is a quadrilateral having all four vertices on the circle. Such quadrilaterals possess a special property.
The following avtivity helps us understand this property.
Activity:
Draw a few inscribed quadrilaterals ABCD. This can easily be accomplished by drawing circles with different radius and then by taking four arbitrary points on each of the circles. Measure the angles of the quadrilaterals and fill in the following table.
Serial No.
C
B+ D
A+ C
A
B
D
1
2
3
4
5
What to you infer from the table?
Circle related Theorems
Theorem 7
The sum of the two opposite angles of a quadrilateral inscribed in a circle is two right angles.

4
10

Math

e
Lt ABCD be a quadrilateral inscribed in a circle with centre O. It is required to prove that, ABC +
ADC =2right angles and
BAD + BCD =2 right angles.
Construction : J in O, A and O, C. o Proof :
Steps

J stification u (1 Standing on the same arc ADC, the angle at
)
centre AOC =(
2
ABC at the circumference) that is, AOC =
2
ABC.

[The angle subtended by an arc at the centre is double of the angle subtended by it at the circle]

(2 Again, standing on the same arc ABC, reflex
)
AOC at the centre = 2 (
ADC at the circumference) that is, reflex AOC =
2
ADC
AOC + reflex AOC =(
2
ABC + ADC)
But AOC + reflex AOC = right angles
4
2 ABC + ADC) = right angles
(
4
ABC + ADC =right angles.
2
In the same way, it can be proved that
BAD + BCD =right angles. (Proved)
2

[The angle subtended by an arc at the centre is double of the angle subtended by it at the circle]

Corollary 1: If one side of a cyclic quadrilateral is extended, the exterior angle formed is equal to the opposite interior angle.
Corollary 2: A parallelogram inscribed in a circle is a rectangle.
Theorem 8
If two opposite angles of a quadrilateral are supplementary, the four vertices of the quadrilateral are concyclic. e Lt ABCD be the quadrilateral with ABC + ADC =
2
right angles, inscribed in a circle with centre O. It is required to prove that the four points A, B, C, D are concyclic. Construction: Since the points A,B, C are not collinear, there exists a unique circle which passes through these three points. et the circle intersect AD at E. J in A,E.
L
o

4
11

Math

Proof :
Steps

J stification u (1 ABCE is a quadrilateral inscribed in the circle.
)
Therefore, ABC
2
AEC =right angles.
2
g
.
But ABC
ADC =right angles [ iven]
AEC = ADC
But this is impossible, since in CED , exterior AEC opposite interior ADC
Therefore, E and D points can not be different points.
So, E must coincide with the point D. Therefore, the points A, B, C, D are concyclic.

O

[ he sum of the two
T
opposite angles of an inscribed quadrilateral is two right angles.]
[ he exterior angle is
T
greater than any opposite interior angle.]

Exercise 8.3
. If the internal and extern al bisectors of the angles B and C of ABC meet at
1
P and Q respectively, prove that B, P, C, Q are concyclcic.
. Prove that, the bisector of any angle of a cyclic quadrilateral and the exterior
2
bisector of its opposite angle meet on the circumference of the circle.
3. ABCD is a circle. If the bisectors of CAB and CBA meet at the point P and the bisectors of DBA and DAB meet at Q, prove that, the four points A, Q, P, B are concyclic.
4. The chords AB and CD of a circle with centre D meet at right angles at some point within the circle, prove that, AOD + BOC =right angles.
2
5. If the vertical angles of two triangles standing on equal bases are supplementary, prove that their circum- ircles are equal. c 6. The opposite angles of the quadrilateral ABCD are supplementary to each other.
If the line AC is the bisector of BAD, prove that, BC = CD.
8⋅4 Secant and Tangent of the circle
Consider the relative position of a circle and a straight line in the plane. Three possible situations of the following given figures may arise in such a case:
(a) The circle and the straight line have no common points
(b) The straight line has cut the circle at two points
(c) The straight line has touched the circle at a point.

4
12

Math

A circle and a straight line in a plane may at best have two points of intersection. If a circle and a straight line in a plane have two points of intersection, the straight line is called a secant to the circle and if the point of intersection is one and only one, the straight line is called a tangent to the circle. In the latter case, the common point is called the point of contact of the tangent. In the above figure, the relative position of a circle and a straight line is shown. In figure (i) the circle and the straight line PQ have no common point; in figure (ii) the line PQ is a secant, since it intersects the circle at two points A and B and in figure (iii) the line PQ has touched the circle at A.
PQ is a tangent to the circle and A is the point of contact of the tangent.
Remark : All the points between two points of intersection of every secants of the circle lie interior of the circle.
Common tangent
If a straight line is a tangent to two circles, it is called a common tangent to the two circles. In the adjoining figures, AB is a common tangent to both the circles. In figure (a) and (b), the points of contact are different. In figure (c) and (d), the points of contact are the same.
If the two points of contact of the common tangent to two circles are different, the tangent is said to be
(a) direct common tangent if the two centres of the circles lie on the same side of the tangent and
(b) transverse common tangent, if the two centres lie on opposite sides of the tangent.
The tangent in figure (a) is a direct common one and in figure (b) it is a transverse common tangent.
If a common tangent to a circle touches both the circles at the same point, the two circles are said to touch each other at that point. In such a case, the two circles are said to have touched internally or externally according to their centres lie on the same side or opposite side of the tangent. In figure (c) the two circles have touched each other internally and in figure (d) externally.
Theorem 9
The tangent drawn at any point of a circle is perpendicular to the radius through the point of contact of the tangent. e Lt PT be a tangent at the point P to the circle with centre O and OP is the radius
OP.
throug0h the point of contact. It is required to prove that, PT

Math

4
13

e
Lt PT be a tangent at the point P to the circle with centre O and OP is the radius throug0h the point of contact. It is
OP.
required to prove that, PT
Construction: Take any point Q on PT and join O, Q.
Proof :
Since PT is a tangent to the circle at the point P, hence every point on it except P lies outside the circle. Therefore, the point Q is outside of the circle.
OQ is greater than OP that is, OQ > OP and it is true for every point Q on the tangent PT except P. So, OP is the shortest distance from the centre O to PT. Therefore, PT
OP. (Proved)
Corollary 1. At any point on a circle, only one tangent can be drawn.
Corollary 2. The perpendicular to a tangent at its point of contact passes through the centre of the circle.
Corollary 3. At any point of the circle the perpendicular to the radius is a tangent to the circle.
Theorem 10
If two tangents are drawn to a circle from an external point, the distances from that point to the points of contact are equal. et P be a point outside a circle ABC with centre O,
L
and two line segments PA and PB be two tangents to the circle at points A and B. It is required t o prove that. PA = PB.
Construction: et us join O, A; O, B and O, P.
L
Proof:
Steps

J stification u (1 Since PA is a tangent and OA is the radius
)
through the point of tangent PA OA.
=right angle
PAO =1
=right angle
Similarly, PBO =1 both PAO and PBO are right- ngled triangles. a (2 Now in the right angled triangles PAO and
)
PBO, hypotenuse PO = hypotenuse PO,

[The tangent is perpendicular to the radius through the point of contact of the tangent]

Math

4
14

Remarks:
. If two circles touch each other external ly, all the points of one excepting the point
1
of contact will lie outside the other circle.
. If two circles touch each other internally, all the points of the smaller circle
2
excepting the point of contact lie inside the greater circle.
Theorem 11
If two circles touch each other externally, the point of contact of the tangent and the centres are collinear. et the two circles with centres at A and B touch each
L
other externally at O. It is required to prove that the points A,O and B are collinear.
Construction: Since the given circles touch each other at O, they have a common tangent at the point
O. Now draw the common tangent POQ at O and join
O, A and O, B.
Proof: In the circles OA is the radius through the point of contact of the tangent and POQ is the tangent.
Therefore POA =right angle. Similarly
1
1
POB =right angle
1
1
2
Hence POA
POB =right angle +right angle =right angles or AOB =right angles i.e.
2
(Proved)

AOB is a straight angle.

A,O and B are collinear.

Corollary 1. If two circles touch each other externally, the distance between their centres is equal to the sum of their radii
Corollary 2. If two circles touch each other internally, the distance between their centres is equal to the difference of their radii.
Activity:
. Prove that, if two circles touch each othe r internally, the point of contact of the
1
tangent and the centres are collinear.

Exercise 8⋅4
. Two tangents are drawn from an external point P to the circle with centre O.
1
Prove that OP is the perpendicular bisector of the chord through the touch points.
. Given that tangents PA and PB touches the circle with centre O at A and B
2
respectively. Prove that PO bisects APB.

Math

4
15

3. Prove that, if two circles are concentric and if a chord of the greater circle touches the smaller, the chord is bisected at the point of contact.
4. AB is a diameter of a circle and BC is a chord equal to its radius. If the tangents drawn at A and C meet each other at the point D, prove that ACD is an equilateral triangle. 5. Prove that a circumscribed quadrilateral of a circle having the angles subtended by opposite sides at the centre are supplementary.
8⋅5 Constructions related to circles
Construction 1
To determine the centre of a circle or an arc of a circle.
Given a circle as in figure (a) or an arc of a circle as in figure (b). It is required to determine the centre of the circle or the arc.
Construction : In the given circle or the arc of the circle, three different points A, B,
C are taken. The perpendicular bisectors EF and GH of the chords AB and BC are drawn respectively. et the bisectors intersect at O. The O is the required centre of
L
the circle or of the arc of the circle.
Proof: By construction, the line segments EF and GH are the perpendicular bisectors of chords AB and BC respectively. But both EF and GH pass through the centre and their common point is O. Therefore, the point O is the centre of the circle or of the arc of the circle.
Tangents to a Circle
We have known that a tangent can not be drawn to a circle from a point internal to it. If the point is on the circle, a single tangent can be drawn at that point. The tangent is perpendicular to the radius drawn from the specified point. Therefore, in order to construct a tangent to a circle at a point on it, it is required to construct the radius from the point and then construct a perpendicular to it. Again, if the point is located outside the circle, two tangents to the circle can be constructed. Construction 2
To draw a tangent at any point of a circle. et A be a point of a circle whose centre is O. It is required to draw a tangent to the
L
circle at the point A.

4
16

Math

Construction :
O, A are joined. At the point A, a perpendicular AP is drawn to
OA. Then AP is the required tangent.
Proof: The line segment OA is the radius passing through A and AP is perpendicular to it. Hence, AP is the required tangent. Remark : At any point of a circle only one tangent can be drawn.
Construction 3
To draw a tangent to a circle from a point outside. et P be a point outside of a circle whose centre is O. A tangent is to be drawn to the
L
circle from the point P.
Construction :
(1 J in P, O. The middle point M of the line segment
)o
PO is determined.
(2 Now with M as centre and MO as radius, a circle is
)
drawn. et the new circle intersect the given circle at the
L
points A and B.
(3) A, P and B, P are joined.
Then both AP or BP are the required tangents.
Proof: A, O and B, O are joined. PO is the diameter of the circle APB.
PAO =right angle [the angle in the semi- ircle is a right angle]
1
c
.
So the line segment OA is perpendicular to AP. Therefore, the line segment AP is a tangent at A to the circle with centre at O. Similarly the line segment BP is also a tangent to the circle.
Remark: Two and only two tangents can be drawn to a circle from an external point.
Construction 4
To draw a circle circumscribing a given triangle. et ABC be a triangle. It is required to draw a circle
L
circumscribing it. That is, a circle which passes through the three vertices A, B and C of the triangle ABC is to be drawn.
Construction :
(1 EM and FN the perpendicular bisectors of AB and AC
)
respectively are drawn. et the line segments intersect
L
each other at O.

4
17

Math

(2 A, O are joined. With O as centre and radius equal to OA, a circle is drawn.
)
Then the circle will pass through the points A, B and C and this circle is the required circum- ircle of ABC. c Proof : B, O and C, O are joined. The point O stands on EM, the perpendicular bisector of AB.
OA= OB. Similarly, OA = OC
OA = OB = OC.
Hence, the circle drawn with O as the centre and OA as the radius passes through the three points A, B and C. This circle is the required circumcircle of ABC.
Activity:
In the above figure, the circumcircle of an acute angled triangle is constructed.
Construct the circumcircle of an obtuse and right- ngled triangles. a Notice that for in obtuse- ngled triangle, the circumcentre lies outside the triangle,in a acute- ngle triangle, the circumcentre lies within the triangle and in right- ngled a a triangle, the circumcentre lies on the hypotenuse of the triangle.
Construction 5
To draw a circle inscribed in a triangle. et ABC be a triangle. To inscribe a circle in it or to draw a
L
circle in it such that it touches each of the three sides BC, CA and AB of the triangle.
Construction : BL and CM, the bisectors respectively of the angles ABC and ACB are drawn. et the line segments
L
intersect at O. OD is drawn perpendicular to BC from O and let it intersect BC at D. With O as centre and OD as radius, a circle is drawn. Then, this circle is the required inscribed circle. Proof : rom O, OE and OF are drawn perpendiculars respectively to AC and AB.
F
et these two perpendiculars intersect the respective sides at E and F. The point O
L
lies on the bisector of ABC.
OF = OD.
Similarly, as O lies on bisector of
OD = OE = OF

ACB, OE = OD

4
18

Math

Hence, the circle drawn with centre as O and OD as radius passes through D, E and F.
Again, BC, AC and AB respectively are perpendiculars to OD, OE and OF at their extremities. Hence, the circle lying inside ABC touches its sides at the points D, E and F.
Hence, the circle DEF is the required inscribed circle of ABC.
Construction 6
To draw an ex-circle of a given triangle. et ABC be a triangle. It is required to draw its ex- ircle.
L
c
That is, to draw a circle which touches one side of ABC and the other two sides produced.
Construction: et AB and AC be produced to D and F
L
respectively. BM and CN, the bisectors of DBC and
FCB respectively are drawn. et E be their point of
L
intersection. rom E, perpendicular EH is drawn on BC
F
and let EH intersect BC at H. With E as centre and radius equal to EH, a circle is drawn.
The circle HGL is the ex- ircle of the triangle ABC. c Proof : rom E, perpendiculars EG and EL respectively are drawn to line
F
segments BD and CF. et the perpendicular intersect line segments G and L
L
respectively. Since E lies on the bisector of DBC
∴ EH = EG
Similarly, the point E lies on the bisector of

FCB, so EH = EL

EH = EG = EL
Hence, the circle drawn with E as centre and radius equal to EL passes through H,
G and L.
Again, the line segments BC, BD and CF respectively are perpendiculars at the extremities of EH, EG and EL. Hence, the circle touches the three line segments at the three points H, G and L respectively. Therefore, the circle HGL is the ex- ircle c of ABC.
Remark : Three ex- ircles can be drawn with any triangle. c Activity: Construct the two other ex- ircles of a triangle. c 4
19

Math

Exercise 8⋅5
. Observe the following information:
1
i. The tangent to a circle is perpendicular to the radius to the point of contact. ii. The angle subtended in a semi- ircle is a right angle. c iii. All equal chords of a circle are equidistant from the centre.
Which one of the following is correct ?
(a) i and ii
(b) i and iii (c) ii and iii
(d) i, ii and iii

se the above figure to answer questions 2 3:
U
and
2
.
BOD equals to
1
1
a.
BAC
b.
BAD c. 2 BAC
d. 2 BAD
2
2
3. The circle is of the triangle ABC
a. inscribed circle b. circumscribed circle
c. ex- ircle c d. ellipse
4. The angle inscribed in a major arc is
a. acute angle
b. right angle c. obtuse angle
d. complementary angle
5. Draw a tangent to a circle which is parallel to a given straight line.
6. Draw a tangent to a circle which is perpendicular to a given straight line.
. Draw two tangents to a circle such that the angle between them is 60°
7
.
. Draw the circum- ircle of the triangle whose sides are 3 cm, 4 cm and 4.5 cm
8
c and find the radius of this circle.
9. Draw an ex- ircle to an equilateral triangle ABC touching the side AC of the c triangle, the length of each side being 5 cm.
0. Draw the inscribed and the circumscribed circles of a square.
1
. Prove that two circles drawn on equal side s of an isosceles triangle as diameters
1
mutually intersect at mid point of its base.
1Prove that in a right- ngled triangle, the length of line segment joining mid point
2
. a of the hypotenuse to opposite vertex is half the hypotenuse.
1. ABC is a triangle. If the circle drawn with AB as diameter intersects BC at D,
3
prove that the circle drawn with AC as diameter also passes through D.

5
10

Math

5
1, If the chords AB and CD of a circles with centre O intersect at an internal point
1
E, prove that AEC =
( BOD + AOC).
2
6
1. AB is the common chord of two circles of equal radius. If a line segment meet through the circles at P and Q, prove that OAQ is an isosceles triangle.
. If the chord AB = x cm and OD AB, are in the circle ABC
7
1 with centre O use the adjoint figure to answer the following questions: a. ind the area of the circle.
F
b. Show that D is the mid point of AB. x 2
)
c. If OD = 2 – cm, determine x.
(
. The lengths of three sides of a triangle are 4 cm, 5 cm and 6 cm respectively. se
8
1
U
this information to answer the following questions:
a. Construct the triangle.
b. Draw the circumcircle of the triangle.
c. rom an exterior point of the circumcircle,, draw two tangents to it and show
F
that their lengths are equal.

Chapter Nine

Trigonometric Ratio
In our day to day life we make use of triangles, and in particular, right angled triangles. Many different examples from our surroundings can be drawn where right triangles can be imagined to be formed. In ancient times, with the help of geometry men learnt the technique of determining the width of a river by standing on its bank.
Without climbing the tree they knew how to measure the height of the tree accuratly by comparing its shadow with that of a stick. In all the situations given above, the disttances or heights can be found by using some mathematical technique which come under a special branch of mathematics called Trigonometry. The word
‘Trigonometry’ is derived from Greek words ‘tri’ (means three), ‘gon’ (means edge) and `metron’ (means measure). In fact, trigonometry is the study of relationship between the sides and angles of a triangle. There are evidence of using the
Trigonometry in Egyptian and Babilian civilization. It is believed that the Egyptians made its extensive use in land survey and engineering works. Early astrologer used it to determine the distances from the Earth to the farest planets and stars. At present trigonometry is in use in all branches of mathematics. There are wide usages of trigonometry for the solution of triangle related problems and in navigation etc. Now a days trigonometry is in wide use in Astronomy and Calculus.
At the end of the chapter, the students will be able to –
Describe the trigonometric ratios of acute angles
Determine the mutual relations among the trigonometric ratios of acute angle
Solve and prove the mathematical problems justifying the trigonometric ratios of acute angle
Determine and apply trigonometric ratios of acute angles 30o. 45 o, 60 o geometrically Determine and apply the value of meaningful trigonometric ratios of the angles 0o and
90o
Prove the trigonometric identities
Apply the trigonometric identities.
9⋅1 Naming of sides of a right angled triangle
We know that, the sides of right angles triangle are known as hypotenuse, base and height. This is successful for the horizontal position of triangle. Again, the naming of sides is based on the position of one of the two acute angles of right angled triangle.
As for example :
a. ‘Hypotenuse’, the side of a right angled triangle, which is the opposite side of the right angle.
b. ‘Opposite side’, which is the direct opposite side of a given angle.

Math

152

c. ‘ acent side’ which is a line segment constituting the given angle. dj A
,

For the angle OPN , OP is the
For the angle PON , OP is the a a hypotenuse, ON is the adj cent side hypotenuse, PN is the adj cent side and PN is the opposite side. and ON is the opposite side.
In the geometric figure, the capital letters are used to indicate the vertices and small letters are used to indicate the sides of a triangle. We often use the G letters to reek indicate angle. Widely used six letters of reek alphabet are :
G
beta β gamma γ theta θ phi φ omega ω alpha α reek letter are used in geometry and trigonometry through all the great
G
mathematician of ancient reek.
G
Example 1. Indicate the hypotenuse, the adj cent side and the opposite side for the a angle θ .

Solution:
(b) hypotenuse p
(a) hypotenuse 17 units opposite side r opposite side 8 units adj cent side q a adj cent side 15 units a Example 2. Find the lengths of hypotenuse, the adj cent side a for the angles α and β .

(a) For α angle, hypotenuse 25 units opposite side 24 units adj cent side 7 units. a (c) hypotenuse EF opposite side EG adj cent side FG a and the opposite side

(b) For β angle hypotenuse 25 units opposite side 7 units adj cent side 24 units. a Activity :
Indicate the hypotenuse , adj cent side and opposite for the angle θ and φ . a Math

153

9⋅2 Constantness of ratios of the sides of similar right-angled triangles
Activity : Measure the lengths of the sides of the following four similar triangles and complete the table below. What do you observe about the ratios of the triangles ?

BC

length of sides
AB
AC

BC/ AC

ratio (related to angle)
AB/ AC
BC/ AB

et,
L
XOA is an acute angle. We take a point P on the side OA . We draw a perpendicular from P to OX . s a result, a right angled triangle POM is formed.
A
The three ratios of the sides PM , OM and OP of POM do not depend on the position of the point P on the side OA.
If we draw the perpendiculars PM and P1 M 1 from two points P and P of OA to
1
the side OX , two similar triangles POM and POM1 are formed.
1
ow,
N
POM and POM1 are being similar,
1
PM
PM
OP
PM
= or, = 1 1 ..... (i)
P M1
OP
OP
OP
1
1
1
OM
OM
OP
OM1
=
or,
=
.... (ii)
OM1
OP
OP
OP
1
1
PM
OM
PM
PM
=
or,
= 1 1 .... (iii)
P M1
OM1
OM1
OM
1
That is, each of these ratios is constant. These ratios are called trigonometric ratios.
9⋅3 Trigonometric ratios of an acute angle et, L
XOA is an acute angle. We take any point P on OA .
We draw a perpendicular PM from the point P to OX . So, a right angled triangle POM is formed. The sixratios are obtained from the sides PM , OM and OP of POM which are called trigonometric ratios of the angle XOA and each of them are named particularly.
With respect to the XOA of right angled triangle POM , a PM is the opposite side. OM is the adj cent side and OP is the hypotenuse.
Denoting XOA θ , the obtained six ratios are described below for the angle

154

Math

From the figure :
PM
=
[ ine of angle θ ] s sin θ =
OP
OM
=
[ cos ine of angle θ ] cosθ =
OP
PM
=
[ tan gent of angle θ ] tan θ =
OM
nd opposite ratios of them are
A
1 cosec =
[ cosecant of angle θ ] sin θ
1
[ sec ant of angle θ ] secθ = cosθ 1
[ cot angent of angle θ ] cot θ = tan θ
We observe, the symbol sin θ means the ratio of sine of the angle θ , not the multiplication of sin and θ . sin is meaningless without θ . It is applicable for the other trigonometric ratios as well.
9⋅4

Relation among the trigonometric ratios

et,
L
XOA θ is an acute angle. from the adj cent figure, according to the definition a PM
1
OP
, cosec =
=
sin θ =
OP
sin θ
PM
OM
1
OP
, secθ =
=
cosθ =
OP
cosθ
OM
PM
1
OM
, cot θ =
=
tan θ =
OM
tan θ
PM
gain,
A

PM tan θ =
OM

= sin θ

cosθ

tan θ =

sin θ cosθ and similarly

cot θ = cosθ sin θ

PM
OP [ ividing the numerator and the denominator by OP ]
D
OM
OP

155

Math

9⋅5 Trigonometric identity
(i ) (sin θ )2 (cosθ )2 =

PM
OP

2

OM
OP

2

PM 2 OM 2
PM 2 OM 2 OP 2
=
=
2
2
OP 2
OP
OP
OP2
1 or, (sin θ )2 (cosθ )2 1
=

sin 2 θ

cos 2θ

[by the formula of Pthagoras ] y 1

Remark : For integer indices n we can write sin n θ for (sinθ )n and cosn θ for (cosθ ) n .
(ii ) sec 2 θ = (secθ ) 2 =

OP
OM

2

OP2
PM 2 OM 2
=
[OP is the hypotenuse of right angled
OM 2
OM 2
PM 2 OM 2
=
OM 2 OM 2
=

PM
OM

=1

sec2 θ or, or,

POM ]

2

= 1 (tan θ )2 = 1 tan 2 θ

1 tan 2 θ

sec 2 θ tan 2θ

tan 2 θ 1 sec 2θ 1

(iii ) cosec 2θ

(cosec θ ) 2 =

OP
PM

2

=

OP2
PM 2 OM 2
=
[ is the hypotenuse of right- ngled i a
PM 2
PM 2

=

PM 2
PM 2

OM 2
=1
PM 2

OM
PM

2

= 1 (cot θ ) 2 = 1 cot 2 θ

cosec 2θ cot 2 θ

1

and

cot 2θ

cosec2θ 1

POM ]

156

Math

Activity :
1. onstruct a table of the following trig onometric formulae for easy memorizing.
C
cosecθ

secθ tan θ

1 sin θ
1
cosθ
1
cot θ

tan θ

cot θ

sin θ cosθ cosθ sin θ

sin 2 θ cos 2θ 1 sec2 θ 1 tan 2 θ cosec 2θ 1 cot 2 θ

4
, find the other trigonometric ratios of the angle A .
3
4
.
Solution : iven that, tan A
G
3
So, opposite side of the angle A = adj cent side =
4, a
3
Example 1. If tan A

42

32

25 5
4
3
3
Therefore,, sin A = , cos A = , cot A =
5
4
5
5
5
cosec A = , sec A = .
4
3

hypotenuse =

Example 2.

B is the right angle of a right angled triangle ABC . If tan A

3
,
4

verify the truth of 2 sin A cos A =
1.
3
Solution : iven that, tan A
G
,
4
So, opposite side of the angle A = adj cent side =
4, a
3
hypotenuse = so, sin A

42

32

4
, cos A
5

= ence, 2 sin A cos A 2
H

5

5

3
4

4 3
24
=
5 5
25

1

1
Therefore, 2 sin A cos A = is a false statement.
Example 3.

B is a right angle of a right angled triangle ABC . If tan A 1 , verify

the j stification of 2 sin A cos A 1 . u 157

Math

Solution : iven that, tan A 1 .
G
So, opposite side of the A = adj cent side =
1, a
1
hypotenuse = 12 12
2
1
1
Therefore, sin A =
, cos A =
.
2
2
ow left hand side =
N
2

= sin A cos A 2

1
2

1
2
=
2

1
1
=
2

= right hand side.
= 2 sin A cos A 1 is a true statement.
Activity :
=
1. If C is a right angle of a right angled triangle ABC , AB 29 cm, cm and ABC θ , find the value of cos 2θ sin 2 θ .
Example 4. rove that, tan θ cot θ secθ , cos ecθ .
P
Proof : eft hand side = tanθ cotθ
L
sinθ cosθ
=
cosθ sinθ sin 2θ cos 2θ
=
sinθ cosθ
1
=
[ sin 2θ cos 2θ 1] sinθ cosθ
1
1
=
sinθ cosθ
= cosecθ secθ
= secθ cosecθ =
=.S. (proved)
R
.H
1
1
Example 5. rove that,
P
1
1 sin 2θ 1 cos ec 2θ
1
1
Proof : ..S. =
L
H
1 sin 2θ 1 cosec 2θ
1
1
=
2
1
1 sin θ 1 sin 2θ
1
sin 2θ
=
2
1 sin θ 1 sin 2θ

21
BC =

158

Math

1 sin 2θ
1 sin 2θ
R
.H
= 1 =.S. (proved)
1
Example 6. P that : rove 2 sin 2 A
=

Proof : ..S. =
L
H

1
2 sin 2 A
1
=
2 sin 2 A

1
2 sin 2 A
1
=
2 sin 2 A

=

1
2 tan 2 A

1

1
2 tan 2 A
1
sin 2 A
2
cos 2 A cos 2 A
2cos 2 A sin 2 A cos 2 A
2(1 sin 2 A) sin 2 A

1 cos 2 A
2
2 sin A 2 2 sin 2 A sin 2 A
1
1 sin 2 A
=
2 sin 2 A 2 sin 2 A
2 sin 2 A
=
2 sin 2 A
= 1 =H (proved)
RS.
.. tanA secA 1
Example 7. P that : rove 0 secA 1 tanA tanA secA 1 proof : ..S. =
L
H secA 1 tanA tan 2 A ( sec 2 A 1)
[ sec 2 θ 1 tan 2 θ ]
=
( secA 1)tanA tan 2 A tan 2 A
=
( secA 1)tanA
0
=
( secA 1)tanA
= 0 =.S. (proved)
R
.H
1 sinA secA tanA
Example 8. P that : rove 1 sinA
=

Math

159

1 sinA
1 sinA

Prove : ..S. =
L
H
=

(1 sinA)(1 sinA)
(1 sinA)(1 sinA)

(1 sinA)

2

=

(1 sinA)
1 sin 2 A

(1 sinA) 2 cos 2 A
1 sinA
=
cosA
1
sinA
=
Ñ cosA cosA
= secA tanA
=.S. (proved).
R
.H
Example 9. If tanA sinA a and tanA sinA b , prove that, a 2
Prove : H given that, tanA sinA a and tanA sinA b ere ..S. = a 2 b 2
L
H
= (tanA sinA) 2 (tanA sinA) 2
= 4 tanA sinA [ (a b) 2 (a b) 2 4ab]
=

= 4 tan 2 A sin 2 A
= 4 tan 2 A (1 cos 2 A)
= 4 tan 2 A tan 2 A cos 2 A
= 4 tan 2 A sin 2 A
= 4 (tanA sinA)(tanA sinA)
= 4 ab
=.S. (proved)
R
.H
Activity : 1. If cot 4 A cot 2 A 1 , prove that, cos 4 θ cos 2 A 1
2. If sin 2 A sin 4 A 1 , prove that, tan 4 A tan 2 A 1
5
Example 10. If secA tanA
, find the value of secA tanA .
2
5
Solution: ere given that, secA tanA
H
.............(i )
2
We know that, sec 2 A 1 tan 2 A

b2

4 ab .

Math

160

∴

or, sec 2 A tan 2 A 1 or, ( secA tanA)( secA tanA) 1
5
f or, ( secA tanA) 1 [rom (i ) ]
2
2 secA tanA
5

Exercise 9⋅1
1. erify whether each of the following math ematical statements is true or false.
V
ive argument in favour of your answer.
G
(a) The value of tan A is always less than 1.
(b) cot A is the multiplication of cot and A .
12
(c) For any value of A , sec A
.
5
(d) cos is the smallest form of cotangent .
3
, find the other trigonometric ratios of the angle A .
2. If sin A
4
3. G that 15 cot A 8 , find the values of sin A and sec A . iven 4. If C is the right angle of the right angled triangle ABC , AB = cm and BC
13
= cm. and
12
ABC θ , find the values of sin θ , cosθ and tan θ .
3 , verify the
B is the right angle of the right angled triangle ABC . If tan A truth of 3 sin A cos A 4 .
Prove (6 – 20) :
1
1
1
1
1
1
6. (i )
1 ; (ii )
1 ; (iii)
1;
sec 2 A cosec 2 A cos 2 A cot 2 A sin 2 A tan 2 A secA tanA
1
1 sinA cosA
7. (i )
1 . (iii)
1
1 ; (ii )
2
cosA cotA cosecA secA
1 sin A 1 cosec2 A cotA tanA
1
1
8. (i ) secA cosecA 1 ; (ii )
1
1 cotA 1 tanA
1 tan 2 A 1 cot 2 A cosA sinA
9.
sinA cosA. 10. tan A 1` sin 2 A sin A.
1 tanA 1 cotA secA tanA cosecA cotA cosecA cosecA
11.
12.
2sec 2 A. cosecA 1 cosecA 1 cosecA cotA secA tanA
1
1
1
1
14.
13.
2sec 2 A.
2 tan 2 A.
1 sinA 1 sinA cosecA 1 cosecA 1 tanA secA 1 sinA 1 cosA
15.
2 cosec A. 16.
0
secA 1 tanA 1 cosA sinA 5.

Math

161

1 sinθ
1 sinθ

cotA tanB cotA . tan B. cotB tanA
1 sinA secA 1 secA tan A. cotA cosec A.
20.
19.
1 sinA secA 1
21.
If cosA sinA
2 sin A
2cosA , prove that cos A sin A cosec 2 A sec 2 A
1
22. If tan A
, find the value of
.
cosec 2 A sec 2 A
3
4
, what is the value of cos ecA cotA ?
23. If cos ecA cotA
3
b a sinA b cosA
, find the value of
.
24. If cotA a a sinA b cosA

17. ( tanθ

secθ ) 2

18.

9⋅6 Trigonometric ratios of the angles 30 , 45 and 60
We have learnt to draw the angles having the measurement of 30 , 45 and 60 geometrically. The actual values of the trigonometric ratios for all these angles can be determined geometrically.
Trigonometric ratios of the angles 30 and 60 et, L
XOZ 30 and P is a point on the side OZ . tend PM upto Q
Draw PM OX and ex
A
tend such that MQ PM . dd O, Q and ex upto Z .
Now, between POM and QOM , PM QM ,
OM is the common side and included PMO
=
included
QMO 90
POM
QOM
Therefore, QOM
POM 30 and OQM
OPM 60
POQ
POM
QOM 30 30 60
OPQ is an equilateral triangle.
1
1
PQ
OP a [ ince POQ is an equilateral triangle] s If OP 2a , PM
2
2
From right- ngled OPM , we get, a gain,
A

OM
OP 2 PM 2
4a2 a2
3a .
We find the trigonometric ratios :
PM
a 1
OM
, cos 30 sin 30
OP 2a 2
OP

3a
2a

3
2

162

Math

PM a 1
.
OM
3a
3
OP 2a cosec 30
2 , sec 30
PM
a tan 30

cot 30

OM
PM

3a a OP
OM

2a
3a

2
3

3.

Similarly,
OM
3a
3
,
OP
2a
2
OM
3a
PM a 1
, tan 60 cos 60
OP 2a 2
PM
a
OP
2a
2
, cosec 60
OM
3a
3
OP 2a
PM
a sec 60
2 , cot 60
PM
a
OM
3a
Trigonometric ratio of the angle 45 et, L
XOZ 45 and P is a point on OZ .
Draw PM OX . In right angled triangle
OPM , POM 45
So, OPM 45
Therefore, PM OM = a (suppose) sin 60

2

2

3

1
.
3

2

Now, OP 2 OM 2 PM 2 = a + a 2 a
=
or, OP
2a
From the definition of trigonometric ratios, we get
PM
a
1
OM a 1
PM
, cos 45
, tan 45 sin 45
OP
OP
OM
2a
2
2a
2
1
1
1
1
cosec 45
2 , sec 45
2 , cot 45 tan 45 sin 45 cos 45
9⋅7 Trigonometric ratios of complementary angles
We know, if the sum of two acute angles is 90 , one of them is called complementary angle to the other. For exmple, 30 and 60 ; 15 and 75 are complementary a angles to each other.
In general, the angles θ and (90 θ ) are complementary angles to each other.

a
1
a

Math

163

Trigonometric ratios of complementary angles et, L
XOY θ and P is the point on the side OY of the angle. We draw PM OX .
Since the sum of the three angles of a triangle is two right angles therefore, in the right angled triangle POM , PMO 90 and OPM
POM one right angle = 90
OPM 90
POM 90 θ
[ ince POM
S
XOY θ ]
OM
= cos POM = cos θ sin (90 θ )
OP
PM
= sin POM = sin θ cos (90 θ )
OP
OM
= cot POM = cot θ tan (90 θ )
PM
PM
= tan POM = tan θ cot (90 θ )
OM
OP
= cosec POM = cosec θ sec (90 θ )
PM
OP
= sec POM = sec θ . cosec (90 θ )
OM
We can exress the above formulae in words below : p sine of complementary angle = cosine of angle cosine of complementary angle = sine of angle tangent of complementary angle = cotangent of angle etc.
5
, find the value of cosec θ cot θ .
Activity : 1. If sec (90 θ )
3
9⋅8 Trigonometric ratios of the angles 0 and 90
We have learnt how to determinal the trigonometric ratios for
N
the acute angle θ of a right angled triangle. ow, we see, if the angle is made gradually smaller, how the trigonometric ratios change. s θ get smallers the length of the side PN also gets
A
smaller. The point P closes to the point N and finally the angle θ comes closer to the angle 0 , OP is reconciled with imately. ON approx

164

Math

When the angle θ comes closer to 0 , the length of the line segment PN reduces to zero and in this case the value of sin θ PN is approx imately zero. Athe same time, t OP

the length of OP is equal to the length of ON and the value of cos

=

ON is 1
OP

approx imately. The angle, 0 is introduced for the convenience of discussion in trigonometry, and the edge line and the original line of the angle 0 are supposed the same ray.
Therefore, in line with the prior discussion, it is said that, cos 0 1, sin 0 0 .
If θ is the acute angle, we see sin θ cos θ
, cot θ
,
tan θ cosθ sin θ
1
1
, cosecθ
,
sec θ cos θ sin θ
We define the angle 0 in probable cases so that, those relations ex ists. sin 0
0
tan 0
0
1 cos 0
1
1
1.
sec 0 cos 0 1
Since division by 0 is not allowed, cosec 0 and cot 0 can not be defined.

imatel y gain, when the angle θ is very closed to 90 , hypotenuse OP is approx
A
equal to PN . So the value of sin θ is approx imately 1. n the other hand, if the
O
angle θ is is equal to 90 , ON is nearly zero;the value of cos θ is approx imately 0.
So, in agreement of formulae that are described above, we can say, cos 90 0 , sin 90 1 . cos 90
0
cot 90
0
1 sin 90
1
1
1
cosec 90 sin 90 1
Since one can not divided by 0, as before, tan 90 and sec 90 are not defined.
Observe : For convenience of using the values of trigonometric ratios of the angles
0 , 30 , 45 , 60 and 90 are shown in the following table :

165

Math

angle
R
atio sine 0
0

30
1
2

cosine

1

tangent

0

3
2
1
3
3

cotangent

undefined

45
1
2
1
2
1

60
3
2
1
2

90
1

3

undefined

1

1
3
2

0

0

2 undefined 2
3
2
1
cosecant undefined 1
2
3
2
Observe : asy method for remembering of the values of trigonometric ratios of
E
some fixd angles. e (i ) If we divide the numbers 0, 1, 2, 3 and 4 by 4 and take square root of the

secant

1

quotients, we get the values of sin 0 of sin 30 , sin 45 , sin 60 and sin 90 respectively. (ii ) If we divide the numbers 4, 3, 2, 1 and 0 by 4 and take square root of quotients, we get the values of cos 0 , cos 30 , cos 45 , cos 60 and cos 90 respectively. (iii ) If we divide the numbers 0, 1, 3 and 9 by 3 and take square root of quotients, we get the values of tan 0 , tan 30 , tan 45 and tan 60 , respectively (It is noted that tan 90 is undefined).
(iv ) If we divide the numbers 9, 3, 1 and 0 by 3 and take square root of quotients, we get the values of cot 30 , cot 45 , cot 60 , cot 90 respectively (It is noted that cot 0 is undefined).
Example 1. Find the values :
1 sin 2 45 tan 2 45
(a)
1 sin 2 45
(b)
cot 90 tan 0 sec 30 cosec 60
(c)
sin 60 cos 30 cos 60 sin 30
1 tan 2 60
(d)
sin 2 60
1 sin 2 60

166

Math

Solution :
(a)

1 sin 2 45
1 sin 2 45

iven exression =
G
p
1

=
1

tan 2 45

2

1
2
1
2

2

(1) 2

[ sin 45

1
I tan 45
2

1]

1
2
=
1
1
2
(b) iven exression =
G
p

1
4
1
1=2 1=
1=
3
3
3
2
cot 90 tan 0 sec 30 cosec 60
2 2
0
=0 0
3 3
2
, cosec 60
0, tan 0
0, sec 30
[ cot 90
3
iven exression =
G
p sin 60 cos 30 cos 60 sin 30
3 3 1 1
=
2 2 2 2
3
1
]
[ sin 60 cos 30
, sin 30 cos 60
2
2
4
3 1
=
=1
=
4
4 4
1 tan 2 60 iven exression =
G
p sin 2 60
1 sin 2 60
1

(c)

(d)

=

1

3

2
2

3
2

1
3
2
1 3 3
=
=
4
1 3 4
2 3 1
=
4
4

2
]
3

2

3
4

Example 2.
(a) If 2cos(A B) 1, 2sin (A B) values of A and B .

3 and A, B are acute angles, find the

Math

(b)
(c)
(d)

167

3
, find the value of A .
3
1 tan 2 A
, if A 45 .
P that, cos 2A rove 1 tan 2 A
Solve : 2cos 2θ 3sinθ 3 0 , where θ is an acute angle.

If

cosA sinA cosA sinA

Solution : (a)

1
1

2cos( A B ) 1

or, cos( A B ) or, cos( A B )

1
2

cos45 [

1
]
2

cos45

A B 45 ..................(i ) and 2 sin ( A B )
3
or, sin ( A B ) or, sin ( A

B)

3
2

sin 60

[

sin 60

3
]
2

A B 60 ....................(ii ) dding (i ) and (ii ) , we get,
A

2 A 105
1
105
A
52
2
2 gain, subtracting (i) from (ii), we get
A
2 B 15
15
or,
B
2
1
B 7
2
1
1
and B 7
R
equired A 52
2
2
3
cosA sinA 1
(b)
3 cosA sinA 1 cosA sinA cosA sinA 1 or, cosA sinA cosA sinA 1

3 1
3 1

3
3

Math

168

2cosA
2
2 sinA
2 3
1
cosA sinA 3 cotA cot 60
A 60 iven that, A 45
G

or, or, or,
(c)

we have to prove that, cos 2 A

1 tan 2 A
1 tan 2 A

..S. =
L
H cos 2 A
= cos(2 45 ) = cos 90 = 0
1 tan 2 A
1 tan 2 A
1 tan 2 45
1 (1) 2
=
=
1 tan 2 45
1 (1) 2
0
= =0
2
∴ ..S. =.S. (proved)
L R
H .H
(d)
G equation, 2cos 2θ 3sinθ 3 0 iven or, 2(1 sin 2θ 3(1 sinθ ) 0 or, 2(1 sinθ )(1 sinθ ) 3(1 sinθ ) 0 or, (1 sinθ ){(1 sinθ ) 3} 0
2
or, (1 sinθ ){2 sinθ 1} 0 or, 1 sinθ 0 or 2 sinθ 1 1 or, 2 sin θ 1 sin θ 1
1
or, sinθ or, sin θ sin 90
2
or, sin θ sin 30 θ 90 or, θ 30 θ is an acute angle, so θ 30 .

RS. =
.H
.

Exercise 9⋅2
1. If
(a)

cot θ

1
3

1
, which one is the value of cot
2

(b) 1

(c)

3

?
(d) 2

Math

169

2.
(i) sin 2θ 1 cos 2 θ
(ii) sec 2θ 1 tan 2 θ
(iii) cot 2θ 1 tan 2 θ
Which one of the followings is correct in accordance with the above statements.
(a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii

nswer of questions 3 and 4 on the basis of the figure :
A
3. What is the value of sin θ ?
3
4
3
4
(a)
(b)
(c)
(d)
4
3
5
5
4. What is the value of cot θ ?
3
3
4
4
(a)
(b)
(c)
(d)
4
5
5
3 v Ealuate (5 – :
8)
1 cot 2 60
5.
.
1 cot 2 60
6. tan 45 sin 2 60 tan 30 tan 60 .
1 cos 2 60
7.
sec 2 60 .
1 cos 2 60
8. cos 45 cot 2 60 cosec 2 30 . rove (9 – :
P
11)
9. cos 2 30 sin 2 30 cos 60 .
10. sin 60 cos 30 cos 60 sin 30 sin 90 .
11. cos 60 cos 30 sin 60 sin 30 cos 30 .
12. sin 3 A cos3 A , if` A 15 .
2 tanA
, if A 45 .
13. sin 2A
1 tan 2 A
2tanA
, if A 30 .
14. tan 2 A
1 tan 2 A
1 tan 2 A
, if A 60 .
15. cos 2A
1 tan 2 A
16. If 2cos(A B) 1 2sin(A B) and A, B are acute angles, show that A 45 ,
B 15 .

170

Math

17. If cos(A B) 1, 2sin(A B) and A, B are acute angle, find the values of
A and B . cosA sinA
3 1
18. Solve :
.
cosA sinA
3 1
19. If A, B are acute angle and cot(A B) 1, cot(A B)
3 , find the values of
A and B .
20. Show that, cos 3 A 4 cos 3 A 3cos A , when A 30 .
21. Solve : sinθ cosθ 1 , when 0 θ 90 .
22. Solve : cos 2θ sin 2θ 2 5cosθ , when θ is an acute angle.
23. Solve : 2sin 2θ 3cosθ 3 0 , θ is an acute angle.
24. Solve: tan 2θ (1
3 ) tanθ
3 0.
1
2
25. Find the value : 3 cot 60 cos ec 2 30
5 sin 2 45 4 cos 2 60 .
4
26. If B 90 , AB 5cm, BC 12cm. of ABC
(a) Find the length of AC .
(b) If C θ , find the value of sin θ cos θ .
(c) Show that, sec 2 θ cos 2 θ sec 2 θ cos ec 2 θ .
27.

(a) what is the measurement of AC .
(b) Find the value of tan A tan C.
(c) Find the values of x and y.

Chapter Ten

Distance and Height
From very ancient times trigonometrical ratios are applied to find the distance and height of any distant obj ct. t pres ent trigonometrcal ratios are of boundless e A importance because of its increasing usage. The heights of the hills, mountains and trees and the widths of those rivers which cannot be measured in ordinary method are measured the heights and widths with the help of trigonometry. In this condition it is necessary to know the trigonometrical ratios values of acute angle.
Athe end of this chapter, the students will be able to – t xlain the geoline, vertical plane a nd angles of elevation and declination
E
p
Solve mathematical problem related to distance and height with the help of trigonometry Measure practically different types of distances and heights with the help of trigonometry. Horizontal line, Vertical line and Vertical plane :
The horizontal line is any straight line on the plane. A straight line parallel to horizon is also called a horizontal line. The vertical line is any line perpendicular to the horizontal plane. It is also called normal line.
A
horizontal line and a vertical line intersected at right angles on the plane define a plane. It is known as vertical plane.
In the figure : A with height of AB is standing vertically tree H at a distance of CB from a point C on the plane. ere, CB is the horizontal line. BA is the vertical line and the plane ABC is perpendicular to the horizontal plane which is a vertical plane. Angle of Elevation and Angle of Depression : bserve the figure, AB is a straight line parallel to the horizon.
O
The points P, O, B lie on the same vertical plane. The point P on the straight AB makes angle POB with the line AB . ere at O , the angle of elevation of P is POB .
H
So, the angle at any point above the plane with the straight line parallel to horizon is called the angle of elevation.

gain the point Q, O, B lie on the same vertical plane and point Q lines at lower
A
H side of the straight line AB parallel to horizon. ere, the angle of depression at O of
Q is ∠QOB. So, the angle at any point below the straight line parallel to the plane is called the angle of depression.

172

Math

Activity : o Pint the figure and show the horizontal line, vertical line, vertical plane, angle of elevation and angle of and L depression. N.B. : For solving the problems in this chapter approx imately right figure is needed.
While drawing the figure, the following techniques are to be applied. perpendicular. (1) While drawing 30 angle, it is needed base >
(2) While drawing 45 angle, it is needed base = perpendicular. (3) While drawing 60 angle, it is needed base < perpendicular. Example 1. The angle of elevation at the top of a tower at a point on the ground is
30 at a distance of 75 metre from the foot. Find the height of the tower.
Solution : et, the height of the tower is AB h metre.
L
The angle of elevation at C from the foot of the tower
BC 75 metre of A on the ground is ACB 30
From

ABC we get, tan ACB

or, tan 30

h
1
or,
75
3

h
75

AB
BC
tan 30

1
3

or,

3h

75 or, h

75
3

75 3
3]
or,
[ ultiplying the numerator and denominator by m 3 h 25 3 h 43.301 metre (app.). equired height of the tower is 43.301 metre (app.).
R
Example 2. The height of a tree is 105 metre. If the angle of elevation of the tree at a point from its foot on the ground is 60 , find the distance of the point on the ground from the foot of the tree.
Solution : et, the distance of the point on the ground
L
from the foot of tree is BC x metre. H eight of the tree
AB 105 metre and at C the angle of elevation of the vertex tree is ACB 60 of From ABC we get, or, h

173

Math

tan ACB

AB or, tan 60 0
BC

105 x tan 60 0

3

105
105
105 3 or, 3 x 105 or, x or, x or, x 35 3
3
x
3
x 60.622 (app.)
The required distance of the point on the ground from the foot of the tree is
60.622 metre (app.).

3

or,

Activity:
In the picture, AB is a tree. Information from the picture –
1. Find the height of the tree.
2. Find the distance of the point C on the ground from the foot of the tree.
Example 3. A ladder of 18 metres. long touche s the roof of a wall and makes an angle of 45o with the horizon. Find the height of the wall.
Solution : et, the height of the wall AB is = h metre,
L
length of ladder AC is 18 m. and makes angles with the ground ACB 45 .

or, sin 45
1
2

or,

2h 18

or, h

h
18

h
18

or,

or, h

AB
AC

ABC we get, sin ACB

From

18 2
2
9 2

sin 450

or, h

1
2

or,

2h 18

or, h

18
2

18
2

[ ultiplying the numerator and denominator by m 2]

h 12.728 (app.)
Therefore, required height of the wall is 12.728 m. (app.).
Example 4. A leaned due to storm. The stic k with height of 7 metre from its foot tree was learned against the tree to make it straight. If the angle of depression at the point of contacting the stick on the ground is 30 , find the length of the stick ?

174

Math

Solution : et, the height of the stick from the
L
foot learned against the tree of AB 7 metre and angle of depression is DBC 30 a ACB
DBC 30 [ lternate angle]
From ABC we get,
7
AB or, sin 30 0 sin ACB
BC
BC
1
7
1
0 or, sin 30
2 BC
2
BC 14 equired height of the stick is 14 metre.
R
Activity :
In the figure, if depression angle CAE 60 , elevation angle
ADB 30 , AC 36 metre and B, C , D lie on the same straight line, find the lengths of the sides AB, AD and CD .
Example 5. The angle of elevation at a point of the roof of a building is 60 in any point on the ground. Moving back 42 metres from the angle of elevation of the point of the place of the building becomes 45 . Find the height of the building.
Solution : et, the height of the building is AB h
L
metres. The angle of elevation at the top ACB 60 .
The angle of elevation becomes ADB 45 moving back from C by CD 42 metres. et, BC x metre
L
BD BC CD x 42 metre
From ABC we get,
AB
h or, 3 tan 60 0 tan 60 0
3
x
BC
h x ............ i
3

ABD we get, tan 450

gain, from
A
or, 1 or, h

h x 42 h 3

tan 450

1 or, h

42; by equation (i)

AB
BD
x 42

Math

or,

175

3h

h

42 3 or,

3h h

42 3 or,

3 1h

42 3 or, h

42 3
3 1

h 99.373 (app.) eight of the building is 99.373 metres (app.)
H
`
Example 6. A pole is broken such that the broken part makes an angle of 30 with the other and touches the ground at a distance of 10 metres form its foot. Find the lengths of the pole.
Solution : et, the total height of the pole is AB h
L
metre. Breaks at the height of BC x matre without separation and makes an angle with the other, and touches the ground at a
BCD 30 distance BD 10 metres from the foot. ere, CD AC AB BC h x metre
H
From BCD we get,
1 10
BD
x 10 3 tan 30 or, =
3 x
BC
1
10
BD gain, sin 30
A
or,
2 h x
CD
[ or, h x 20 or, h 20 x or, h 20 10 3; putting the value of x ] h 37.321 (app.) eight of the pole is 37.321 metres (app.).
H
Activity :
A
balloon is flying above any point be tween two mile posts. t the point of the
A
balloon the angle of depression of the two posts are 30 and 60 respectively. Find the height of the balloon.

Exercise 10
1. (a) Find the measurement of CAD
(b) Find the lengths of AB and BC.
(c) Find the distance between A and D.
2. From a helicopter above a point O between two kilometre posts, the angles of depression of the two points A and B are 60o and 30o respectively.
(a) Draw a figure with short description.
(b) Find the height of the helicopter from the ground.
(c) Find the direct distance from the point A of the helicopter.
3. What is the elevation angle of point P from the point O ?
(b) POA
(c) QOA
(d) POB
(a) QOB
4. (i) The horizontal line is any straight line lying on the plane.
(ii) ertical line is any line perpendicular to the plane.
V
(iii) A horizontal line and a vertical plane define a plane. It is known as vertical plane.

176

Math

which one is right of the above speech ?
(a) i and ii (b) i and iii
(c) ii and iii
(d) i, ii and iii nswer the questions 5 – from the adj cent figure :
A
6 a 5. The length of BC be – will 4 m (b) 4m (c) 4 2 m
(d) 4 3 m
(a)
3
6. The length of A will be –
B
4 m (b) 4m (c) 4 2 m
(d) 4 3 m
(a)
3
7. If the angle of the elevation of the top of the miner is 30 at a point on the ground and the height is 26 metres, then find the distance of the plane from the Miner.
8. If the top of a tree is 20 metres distance from the foot on the ground at any point and the angle of elevation of 60 , find the height of the tree.
18
9. Forming 45 angle with ground A metres long ladder touches the top of the wall, find the height of the wall.
10. If the of depression of a point on the ground 20 metres from the top of the house is 30 then, find the height of the house.
11. The angle of elevation of a tower at any point on the ground is 60 . If moved back 25 metre, the angle of elevation becomes 30 , find the height of the tower.
12. The angle of elevation of a tower 60 moving 60 metres towards a minar. Find the height of the minar.
13. A man standing at a place on the bank of a river observesd that the angle of elevation of a tower exctly opposite to him on the other bank was 60 o. Moving a 32 metres back he observed that the angle of elevation of the tower was 30o. Find the height of the tower and the width of the river.
14. A of 64 metre long breaks into two parts without complete separation and makes pole an angle 60 with the ground. Find the length of the broken part of the pole.
15. A is broken by a storm such that the broken part makes an angle of 30 with tree the other and touches the ground at a distance of 10 metres from it. Find the length of the whole tree.
16. Standing any where on the bank of a river, a man observed a tree exctly straight a to him on the other bank that the angle j urney by boat of elevation of he top of o the tree of 150 metres length is 30 . The man started for the tree. But he reached at 10 metres away from the tree due to current.
(a) Show the above description by a figure.
(b) Find the width of the river.
(c) Find the distance from the starting point to the destination.

Chapter Eleven

Algebraic Ratio and Proportion
It is important for us to have a clear concept of ratio and proportion. A rithmetical ratio and proportion have been elaborately discussed in class V In this chapter,
II.
algebraic ratio and proportion will be discussed. We regularly use the concept of ratio and proportion in construction materials and in the production of food staff, in consumers production, in using fertilizer in land, in making the shapes and design of many things attractive and good tooking and in many areas of our daily activities.
Many problems of daily lives can be solved by using ratio and proportion.
At the end of this chapter, the students will be able to : xlain algebraic ratio and proportion.
E
p se different types of rules of tr ansformation related to proportion.
U
Describe successive proportion. se ratio, proportion, successive propor tion in solving real lives problem.
U
11⋅1 Ratio f two quantities of same kind and unit, how many times or parts of other can be
O
exressed by a fraction. This fraction is called the ratio of two quantities. p p
. The quantities p and q
The ratio of two quantities p and q is written in p : q q are to be of same kind and same unit. p is called antecedent and q is called subsequent of the ratio.
Some times we use ratio in approx imate mea sure. Such as, the number of cars on the road at 8 .M. doubles the number at 10 .M. In this case, it is not necessary to
A
A know the exct number of cars to determine the ratio. A a gain, in many occasions, we say that the area of your house is three times the area of mine. ere, also it is not
H
necessary to know the exct area of the house. We use the concept of ratio in cases a of practical life.
11 2 Proportion
If four quantities are such that the ratio of first and second quantities is equal to the ratio of third and fourth quantities, those four quantities form a proportion. If a, b, c, d are four such quantities, we write a : b = c : d . The four quantities of proportion need not to be of same kinds. The two quantities of the ratio are to be of the same kind only.

178

Math

In the above figure, let the base of two triangles be a and b respectively and their height is h unit. If the areas of the triangle be A and B square units, we can write,
1
ah
A 2 a or A : B a : b
B 1 b bh
2
i.e. ratio of two areas is equal to ratio of two bases.
Ordered proportional
By ordered proportional of a, b, c it is meant that a : b = b : c. a, b, c will be ordered proportional if and only if b2 = ac. In case of ordered proportional, all the quantities are to be of same kinds. In this case, c is called third proportional of a and b and b is called mid- roportional of a and c. p Example 1. A and B traverses fixd distance in t1 and t2 minutes. Find the ratio of e average velocity of A and B. m m
Solution : et the average velocities of A and B be v1 sec/ etre and v 2 sec/ etre
L
respectively. So, in time t1 minutes A traverses v1t1 metres and in t2 minutes B traverses the distance v2t2 meters. ccording to the problem, v
A

1t1

=v 2t2

v1 v2 t2 t1 ere, ratio of the velocities is invers ely proportional to the ratio of time.
H
Activity: 1. Eress 3.5 : 5.6 into 1 : a and b : 1 x p
2. If x : y = : 6, 3 x : 5y =What ?
5
11 3 Transformation of Ratio ere, the quantities of ratios are positive numbers.
H
I
(1) If a : b c : d then b : a d : c [nvertendo]
Proof : G that, iven a b c d m ad bc [ ultiplying both the sides by bd] ad bc or, [ ividing both the sides by ac where a d ac ac d b or, c a

0, c

0]

179

Math

(2)

i.e., b : a d : c
If a : b c : d then a : c

Proof : iven that,
G

a b a b : d [ lternendo]

c d m ad bc [ ultiplying both the sides by bd] ad bc or, [ ividing both the sides by cd where c d cd cd a b or, c d
i.e., a : c b : d a b c d
[ omponendo] c (3) If a : b c : d then b d a c
Proof : iven that,
G
b d a 1 b i.e.,
(4)

a b b If a : b

c d 0, d

d
A
1 [ ding 1 to both the sides]

c d d c : d then

a b b c d d [dividendo]

Proof : a : b

c:d a c s 1
1 [ ubtracting 1 from both the sides] b d a b c d
i.e.,
b d (5)

If a : b

c : d then

Proof : iven that,
G

a b a b a b

c d c d

[ omponendo – c dividendo]

c d a b c d
..............(i)
b d a b c d b d

By componendo, gain A by dividendo, or, b a b

d
[ y invertendo]............. ( ii) b c d

0]

180

Math

a b b c d d [ ultiplying (i) and (ii)] m b a b d c d a b c d
i.e.,
. [here a b and c d ] a b c d a c e g a c e g
(6) If then each of the ratio =
.
b d f h b d f h a c e g
Proof : et,
L
k. b d f h a bk , c dk , e fk , g hk a c e g bk dk fk hk k (b d f h) k, b d f h b d f h b d f h
But k is equal to each of the ratio. a c e g a c e g
.
b d f h b d f h
Activity : 1. Sum of ages of mother and sister is s years. Before t years, the ratio of their ages was r : p. What will be the ratio of their ages after x years ?
2. The shadow of a man of height r metre, standing at p metre from a lightpost is s metre. If the height of the lightpost be h metre, what was the distance of the man from the lightpost ?
Example 2. The ratio of present ages of father and son is 7 : 2 , and the ratio will be
8 : 3 after 5 years. What are their present ages ?
Solution : et the present age of father be a and that of son is b . So, by the
L
conditions of first and second of the problems, we have,
Therefore,

a 7
.......... .(i) b 2 a 5 8
.......... .(ii ) b 5 3
From equation (i), we get,
7b
a
...........(iii)
2
From equation (ii), we get,

3(a 5) 8(b 5) or, 3a 15 8b 40 or, 3a 8b 25
7b
8b 25 [ y using ( iii)] or, 3 b 2

181

Math

21b 16b
25
2 or, 5b 50 b 10
Ptting b 10 in equation (iii), we get, a u or,

35

The present age of father is 35 years, and that of son is 10 years.
Example 3. If a : b

b : c , prove that

Solution : iven that,
G

a b b c

2

a2 b2

a:b b:c b 2 ac

a b b c

ow,
N

2

( a b) 2
(b c)

and

2

a2 b2

k;

(bk ) 2

b2

b 2 (k 2 1)

k2 1

(bk ) 2 b 2

b 2 (k 2 1)

k2 1

ac bd ac bd

bk dk bd bk dk bd

bd (k 2 1)

k2 1

bd (k 2 1)

k2 1

a2

b2

2

2

ac ac a2

b2

a2 b2 and a

b

a b ac bd
.
ac bd

c d ow,
N

2

a c b2 c2
2
2 a c
Example 4. If
, show that a b b d a2 b2
Solution : et,
L

2

2bc c 2
2ab ac

ac 2bc c 2 a (a 2b c) c(a 2b c)
2

b2

2ab b 2

b2 a2 a b b c

a2 b a2

bd
.
bd

.

b2 c2

a bk and c dk

c

a2

ac

ac c 2 a(a c) c(a c) a c

182

Math

1 ax 1
1 ax 1
1 ax
Solution : iven that,
G
1 ax
1 bx 1 ax
1 bx 1 ax

Example 5. Solve :

[ quaring both the sides] s (1 ax) 2

1 2ax a 2 x 2

1 bx
1 bx

or,

(1 ax) 2

1 bx
1 bx

or,

bx
1, 0 b 2a 2b. bx 1 bx
1
1 bx

1 2ax a 2 x 2

or,

or,

2
2bx

1 2ax a 2 x 2 1 2ax a 2 x 2

1 bx 1 bx
1 bx 1 bx

[ y componendo and dividendo] b 1 2ax a 2 x 2 1 2ax a 2 x 2

2(1 a 2 x 2 )
4ax

1 1 a2 x2 bx 2ax or, 2ax bx(1 a 2 x 2 ) or, x{2a b(1 a 2 x 2 )} 0 ither x 0 or 2a b(a a 2 x 2 ) 0
E
or, b(1 a 2 x 2 ) 2a or, or, 1 a 2 x 2 or, a 2 x 2 or, x 2 x 2a b 2a
1
b
1 2a
1
a2 b
1 2a
1
a b

Example 6. If

6 x 1 2a
1.
a b
1
x 3a x 3b
, show that b x 3a x 3b

x 0, x

equired solution
R

1 a 2, a b.

Math

183

6 1 1 x a b m 6ab ( a b) x [ ultiplying both the sides by abx]
6ab
i.e., x
( a b) x 2b or, 3a a b x 3a 2b a b
[ y componendo and dividendo] b x 3a 2b a b
3a a 3b
3a b a x 2a
3b a b
3b 2a a b
[ y componendo and dividendo] b 3b 2a a b
3b 3a b
3b a b x 3a x 3b a 3b 3a b x 3a x 3b b a a b a 3b 3a b a 3b 3a b 2(b a)
2.
b a b a b a b a x 3a x 3b
2.
x 3a x 3b

Solution : iven that,
G

x x or, gain, A or, x x x x Now,

Example 7. If

1 x
1 x

1 x
1 x

p , prove that,

p2

2p
1 0. x 1 x
1 x p 1 x
1 x
1 x
1 x
1 x p 1
[ y componendo -dividendo] b 1 x
1 x
1 x p 1

Solution : G that, iven 1 x
1 x or, 2 1 x
2 1 x

or,

1 x
1 x

p 1 p 1

or,

1 x
1 x

( p 1) 2

p2

( p 1) 2

p2 2 p 1

p 1 p 1

2 p 1 [ quaring both sides] s Math

184

or, or, p2

1 x 1 x
1 x 1 x
1
x p2 2 p 1 p2

2p 1

2

2

2p 1

p

2p 1 p

p2 1 or, p2 1
2p
2p
1 0. x [ y componendo -dividendo] b 2p x a3 b3 a(a b) , prove that a, b, c are ordered proportional. a b c a3 b3 a ( a b)
Solution : iven that,
G
a b c

Example 8. If

or,

a 3 b3 a b c

or,

(a b)(a 2 ab b 2 ) a b c

a(a b) a(a b)

a 2 ab b 2
[ ividing both sides by ( a + b)]
D
a a b c or, a 2 ab b 2 a 2 ab ac or, b 2 ac a, b, c are ordered proportional.

Example 9. If

a b b c

c d a b Solution : iven that,
G

or, or, or,

or, or, a b c 1 b c d a b b c b c a c c a b c d a a c a c b c d a
1
(a c) b c

d
, prove that c a b c d c d a

a or a b c d 0 .

d s 1 [ ubtracting 1 from both the sides] a c d d a d a

0
1
d

a

0

Math

185

d a b c
0
b c d a or, (a c)(d a b c) 0 or, (a c)

E ither a c 0 i.e., a or, a b c d 0.
Example 10. If

c

x y y z z

z x x

and x, y, z are not mutually equal, prove

y

1 or equal

that the value of each ratio is either equal

x

Solution : et,
L

y

z

k

y z z x x y x k ( y z )................(i) y k ( z x)...................(ii z k ( x y )...................(iii)

Subtracting (ii) from (i), we get, x y k ( y x) or, k ( y x) y x k 1 gain, adding (i ) , (ii ) and (iii ) , we get,
A
x y z k ( y z z x x y ) 2k ( x

1( x y z )
2( x y z )

k

by

y

z)

1
2

1 or

Therefore, the value of each of the ratio is
Example 11. If ax

1
.
2

cz , show that

x2 yz y2 zx 1
.
2 z2 xy

bc

ca

ab

2

2

c2

a

b

Solution : et, ax by cz k
L
k k k x , y
, z a b c Now,

i.e.,

x2 yz y2 zx z2 xy k2

bc

k2

ca

k2

ab

bc

ca

2

2

2

2

2

2

2

2

x2 yz y2 zx z2 xy bc

ca

2

2

a

a

k

b

b

ab c2 k

.

c

k

a

b

ab c2 .

186

Math

Exercise 11⋅1
1.
2.
3.
4.

5.
6.
7.

If the sides of two squares be a and b metres respectively, what will be the ratio of their areas ?
If the area of a circle is equal to the area of a square, find the ratio of their paremetre. If the ratio of two numbers is 3 : 4 and their ..M. is 180, find the two
L
C numbers. The ratio of absent and present students of a day in your class is found to be
1:4. xress the number in percentage of absent students in terms of total
E
p students. A thing is bought and sold at the loss of 28%. Find the ratio of buying and selling cost.
Sum of the ages of father and son is 70 years. 7 years ago, the ratio of their ages were 5 : 2. What will the ratio of their ages be after 5 years.
If a : b b : c , prove that,

a2 b2 b2 c2 abc(a b c) 3
(iii )
(ab bc ca) 3

(i )

a c 8. Solve : (i )

(iii )
(iv )

a2

a x

a b (ii )
10. If

a b (ii )

x2

2

(iv ) a 2b c

1
3

a x

2

a

x 1 x 1

(vi ) 81
9. If

1

1 x
1 x

1
1

1 x
1 x

x

x 6 x 6
3

1 a3 (ii ) a 2 b 2 c 2

a a x x b
, 2a b 0 and x x 5

(v )

1 b3 ac bd c 2 d 2 ac bd c 2 d 2 b c
, show that, c d

ab b 2 ab b 2

a a x x (b c) 2 c b

0.

ax b ax b

ax b ax b

c2 c2 d2 d2 cd cd a3 b3 c3

( a b) 2 a 1 x
1 x

a2 c , show that, (i ) 2 d a

1 c3 c

187

Math

a 3 b3 b3 c3
(ii ) (a 2 b 2 c 2 )(b 2 c 2 d 2 ) (ab bc cd ) 2 b3 c3 c3 d 3
4 ab x 2a x 2b
, show that,
11. If x
2, a b. a b x 2a x 2b
(i )

3

12. If x
13. If x

16.
17.

18.

19.

2a
2a

3
3

m 1
3
, prove that, x m 1

3b
3b

2a
2a

3mx 2

3b , show that, 3bx 2
3b

3x m

0

4ax 3b 0.

( a b)2
2 , prove that, a, b, c are ordered proportional. b 2 c 2 (a c) a b c x y z
If
, prove that,
.
y z x z x y x y z b c c a a b x y z bz cy cx az ay bx
If
, prove that,
.
a b c a b c a b c b c a c a b
If
and a b c 0 , prove that, a b b c c a a b c. x y z and x y z 0 , show
If
xa yb zc ya zb xc za xb yc
1
. that, each of the ratio is a b c
If (a b c) p (b c a)q (c a b)r (a b c) s ,
1 1 1 1 prove that,
.
q r s p

14. If
15.

3

m 1 m 1

a2

20. If lx
21. If

p q b2

my

nz , show that,

a2 a and
2
b b x2 yz y2 zx z2 xy mn l2 a q p q
, show that, a a q

nl m2 lm
.
n2

p q
.
q

11⋅4. Successive Ratio et oni’ earning be Tk. 1000, Soni’ earning be Tk. 1500 and Somi’ earning be
LR s s s
Tk. 2500. ere, oni’ earning : Soni’ earning =
H R s s 1000 : 1500 = : 3 ;Soni’
2
s

188

Math

earning : Sami’ earning =1500 : 2500 =3 : 5. ence, Rni’ : Soni’ : Sami’ s H o s s s earning = : 3 : 5.
2
If two ratios are of the form a : b and b : c, they can be put in the form a : b : c. This is called successive ratio. ny two or more than two ratios can be put in this form. It
A
is to be noted that if two ratios are to be put in the form a : b c, antecedent of the first ratio and subsequent of the second ratio are to be equal. Such as, if two ratios 2 : 3 and 4 : 3 are to be put in the form a : b : c the subsequent quantity of first ratio is to be made equal to antecedent quantity of the second ratio. That is those quantities are to be made equal to their ..M.
L
C
4 4 3 12
2 2 4 8 gain, ere, 2 : 3
H
4:3
12 : 9
8 : 12 A
3 3 3 9
3 3 4 12
Therefore, if the ratios 2 : 3 and 4 : 3 are put in the form, a : b : c will be 8 : 12 : 9.
It is to be noted that if the earning of Sami in the above exmple is 1125, the ratio of a their earnings will be 8 : 12 : 9.
Example 12. If a, b, c are quantities of same kind and a : b = : 4, b : c = : 7, what
3
6 will be a : b : c ?
Solution :

a b 3
4

b
3 3 9 and c
4 3 12

6
7

6 2
7 2

12
[.C of 4 and 6 is 12]
LM.
.
14

a : b : c = : 12 : 14.
7
Example 13. The ratio of angles of a triangle is 3 : 4 : 5. xress the angles i p E degree. º

Solution : Sum of three angles =
180

et the angles, according to given ratio, be 3 x, 4x and 5x.
L
ccording to the problem, 3 x + x + x =
A
4 5
180
Therefore, the angles are 3x = ×
3 15
4x = ×
4 15

º

=
60

º

=
75

=
45

or, 12x =
180

º

or, x =
15

º

º

º

and 5x = ×
5 15

º

º

º

Example 14. If the sides of a square increase by 10%, how much will the area be increased in percentage ?

n

189

Math

Solution : et each side of the square be
L

a metre

rea of the square be a 2 square metre
A
If the side increases by 10%, each side will be (a 10% of a ) metre or 1 10a metre. In this case, the area of the square will be (1 10a ) 2 square metre or, 1 21a 2 square metre. rea increases by (1 21a 2 a 2 ) 0.21a 2 square metre
A
The percentage of increment of the area will be

0 21a 2
100% 21% a2 Activity :
1.

There are 35 male and 25 female students in your class. The ratios of rice and pulse are 3 : 1 and 5 : 2 given by each of the male and female students for taking khisuri in a picnic. Find the ratio of total rice and total pulse.

11⋅5 Proportional Division
Division of a quantity into fixd ratio is called proportional division. If S is to be e divided into a : b : c : d , dividing S by (a + b + c +d) the parts a, b, c and d are to be taken.
Therefore,

a

of S

Sa a b c d

of S

Sb a b c d

3rd part

c of S a b c d

Sc a b c d

4th part

d of S a b c d

Sd a b c d

1st part

2nd part

a b c d b a b c d

In this way, any quantity may be divided into any fixd ratio. e 190

Math

Example 15. Divide Tk. 5100 among 3 persons in such a way that 1st person’ part : s 2nd person’ part : 3rd person’ part are s s

Solution : ere,
H

1 1 1
: : .
2 3 9

1
1
1
[.M.
L
18 .C of 2, 3, 9 is 18]
18 :
18 :
9
3
2

1 1 1
: :
2 3 9

9:6:2
Sum of the quantities of ratio = + + =
9 6 2 17.

Tk. 5100

1st person’ part s 9
= 2700
Tk.
17

2nd person’ part s Tk. 5100

6
= 1800
Tk.
17

3rd person’ part s Tk. 5100

2
= 600
Tk.
17

Therefore, three persons will have Tk. 2700, Tk. 1800 and Tk. 600 respectively.

Exercise 11⋅2
1.

If a,b,c are ordered proportional, which one is correct of the followings ?
(a) a 2

2.

(b) b 2

bc

ac

(c) ab

The ratio of ages of A and kib is 5 rif A

bc

(d) a

b c

: 3 ;if rif is of 20 years old, how many
A

years later the ratio of their ages will be 7 : 5. ?
(a) 5 years
3.

(b) 6 years

(c) 8 years

(d) 10 years

ote the following information :
N
(i) ll the four quantities need not to be of same kind in proportion.
A
(ii) The ratio of areas of two triangles is equal to the ratio of areas of their bases.
(iii) If

a b c d e f g a c e g
, value of each ratio will be
.
h b d f h

Math

191

n
Othe basis of above information, which one of the followings is correct?
(a) (i) and (ii)

(b) (ii) and (iii)

(c) (i) and (iii)

(d) (i), (ii) and (iii)

The ratio of angles of

ABC is 2 : 3 : 5 and the ratio of angles of the

quadrilateral ABCD is 3 : 4 : 5 : 6. Based on this information, answer questions number 4 and 5.
4.

If the sides of a square double, how much will the area of a square be increased?
(a) 2 times

(b) 4 times

(c) 8 times

(d) 6 times

5.

If x : y 7 : 5, y : z

6.

The estimated cost for the construction of a wooden bridge is Tk. 90,000. But

how much ?
5:7 , z : x =

Tk. 21,000 has been spent more. What is the percentage of the exess cost ? c 7.

The ratio of rice and husk in paddy to 7 : 3. What is the percentage of rice in it ?

8.

The weight ot 1 cubic cm. wood is 7 decigram. What is the percentage of the weight of wood to the equivalent volume of water ?

9.

Distribute Tk. 300 among a, b, c, d in such a way that the raios are a’ part : b’ s s part = 2 : 3, b’ part : c’ part = : 2 and s s
1

c’ part : d’ part 3 : 2. s s

10. Three fishermen have caught 690 pieces of fishes. The ratios of their parts are

2 4
5
, and . ow many fishes each of them will get. ?
H
3 5
6
11.

The parametre of a triangle is 45 cm. The ratio of the lengths of the sides is
3 : 5 : 7. Find the length of each sides.

3 4 6
: : .
4 5 7

12.

Distribute Tk. 1011 in the ratio

13.

If the ratio of two numbers is 5 : 7 and their ..F. is 4, what is ..M. of the
H
C
L
C numbers. 192

Math

14.

In a cricket game, the total runs scored by Sakib, Mushfique and Mashrafi were
171. The ratio of runs scored by Sakib and Mushfiques, and Mushfique and
Mashrafi was 3 : 2. What were the runs scored by them individually.

15.

In a office, there were 2 officers, 7 clarks and 3 bearers. If a bearer gets Tk. 1, a clerk gets Tk. 2 and an officer gets Tk. 4. Their total salary is Tk. 15,000. What is their individual salary ?

16.

In selecting the leader of a society, Mr. Denal won in ratio of 4 : 3 votes of the two contestants. If total numbers of members were 581 and 91 members did not cast their votes, what was the difference of votes by which opposite of Mr.
Denal had been defeated ?

17.

If the sides of a square are increased by 20%, what is percentage of increment of the area of the square ?

18.

If the length of a rectangle is increased by 10% and the breadth is decreased by
10%, what is the percentage of increase or decrease of the area of the rectangle? 19.

In a field, the ratio of production is 4 : 7 before and after irrigation. In that field, the production of paddy in a land previously was 304 quintal. What would be the production of paddy after irrigation ?

20.

If the ratio of paddy and rice produced from paddy is 3 : 2 and the ratio of wheat and suzi produced from wheat is 4 : 3, find the ratio of rice and suzi produced from equal quantity of rice and wheat.

21.

The are of a land is 432 square metre. If the ratios of lengths and breadths of that land and that of another land be 3 : 4 and 2 : 5 respectively, that what is the area of another land ?

22.

Zami and Simi take loans of different amounts at the rate of 10% simple profit on the same day from same Bank. The amount on capital and profit which Zimi refunds after two years, the same amount Simi refunds after three years on capital and profit. Find the ratio of their loan.
The ratio of sides of a triangle is 5 : 12 : 13 and parametre is 30 cm.

23.

(a) Draw the triangle and write what type of triangle in respect of angles.

Math

193

(b) Determine the area of a square drawn with the diagonal of a rectangle taking greater side as length and smaller side as breadth as the sides of a square. (c) If the length is increased by 10% breadth is increased by 20%wha and ,

t

will be percentage of increase of the area ?
24.

The ratio of present and absent of students of a day in a class is 1 : 4.
(a) xress the percentage of absent students against total students.
E
p
(b) The ratio of present and absent students would be 1 : 9 if 10 more students were present. What was the total number of students ?
(c) Othe total number of students, the f number of female students is less than

male students by 20. Find the ratio of male and female students.

Chapter Twelve

Simple Simultaneous Equations with Two Variables
For solv the mathematical problems, th e most important topic of Algebra is ing equation. In classes Vand V we hav got the idea of simple equation and hav
I
II, e e known how to solv the simple equation with one vriable. In class V we hav e a
III,
e solvd the simple simultaneous equations by the methods of substitution and e elimination and by graphs. We hav also learnt how to form and solv simple e e simultaneous equations related to reallife, problems. In this chapter, the idea of simple simultaneous equations hav been exanded and new methods of solution e p hav been discussed. Besides, in this ch apter, solution by graphs and formation of e simultaneous equations related to real life problems and their solutions hav been e discussed in detail.
At the end of the chapter, the students will be able to − erify the consistency of simple s imultaneous equations with two ariables.
V
v erify the mutual dependence of two simple simultaneous equations with two
V
ariables v Elain the method of cross- ultiplication x p m Form and solv simultaneous equations related to real life mathematical e problems olv the simultaneous equations with two ariables by graphs.
Se
v
12⋅1 Simple simultaneous equations.
S
imple simultaneous equations means two s imple equations with two ariables when v they are presented together and the two vriables are of same characteristics. Sch a u two equations together are also called system of simple equations. In class III, we
V
hav solvd such system of equations and learnt to form and solv simultaneous e e e equations related to real life problems. In this chapter, these hav been discussed in e more details.
First, we consider the equation 2 x y 12 . This is a simple equation with two ariables. v
In the equation, can we get such vlues of x and y on the left hand side for which a the sum of twice the first with the second will be equal to 12 of the right hand side ; that is, the equation will be satisfied by those two alues ? v ow,
N we fill in the following chart from the equation 2 x y 12 :

195

Math

alue of
V

2
0
3
5
....

x

alue of
V

16
12
6
2
.....

y

alue of .H.S(
V
L .
4 16
0 12
6 6
10 2
.....

2x + y)
12
12
12
12
12

R .
.H.S

12
12
12
12
12

The equation has infinite number of solutions. Among those, four solutions are
( 2, 16), (0, 12), (3, 6) and (5, 2) .
Again, we fill in the following chart from another equation x y 3 : alues of y
V
alue of .H.S
V
L .
R .
.H.S
alue of x
V
( x y)
2 5 3
2
5
3
0 3
3
0
3
3
3 0
3
3
0
3
5 2
3
5
2
3
.....
3

....

.....

3

The equation has infinite number of solutions. Among those, four solutions are
( 2, 5), (0, 3), (3, 0) and (5, 2) .
If the two equations discussed abov are considered together a system, both the e equations will be satisfied simultaneously only by (5, 2). Both the equations will not be satisfied simultaneously by any other alues. v Therefore, the solution of the system of equations 2 x y 12 and x y 3 is

( x, y) (5, 2)
Activity : Write down fiv solutions for each of the two equations x 2 y 1 0 e ists. and 2 x y 3 0 so that among the solutions, the common solutions also ex
12⋅2 Conformability for the solution of simple simultaneous equations with two variables. 2 x y 12 hav unique (only e (a) As discussed earlier, the system of equations x y 3 one) solution. uch system of equations are called consistent. omparing the
S
C coefficient of x and y (taking the ratio of the coefficients) of the two equations, we
2 1 get, ; any equation of the system of equations cannot be exressed in terms of p 1
1
the other. That is why, such system of equations are called mutually independent. In the case of consistent and mutually independent system of equations, the ratios are not equal. In this case, the constant terms need not to be compared.

196

Math

(b) ow we shall consider the system of equations
N

2x

y

6

4 x 2 y 12

. Will this two

equations be solvd ? e Here, if both sides of first equation are multiplied by 2, we shall get the second equation. Again, if both sides of second equation are div ided by 2, we shall get the first equation. That is, the two equations are mutually dependent.
We know, first equation has infinite number of solutions. o, 2nd equation has also
S
the same infinite number of solutions. uch system of equations are called consistent
S
and mutually dependent. uch system of equations hav infinite number of solutions.
S
e
Here, comparing the coefficients of x, y and the constant terms of the two
2
1 6
1
. equations, we get,
4
2 12
2
That is, in the case of the system of such simultaneous equations, the ratios become equal.
(c) ow, we shall try to solv the system of equations
N
e

2 x y 12
.
4x 2 y 5

Here, multiplying both sides of first equation by 2, we get, 4 x 2 y 24 second equation is 4 x 2 y 5 subtracting, 0 19 , which is impossible. o, we can say, such system of equations cannot be solvd. uch system of equations are
S
e S inconsistent and mutually independent. uch system of equations hav no solution.
S
e
Here, comparing the coefficients of x , y and constant terms from the two equations,
2 1 12 we get,
. That is, in case of the system of inconsistent and mutually
4 2 5 independent equations ratios of the coefficients of the ariables are not equal to the ratio v of the constant terms. enerally, conditions for comformability of two simple
G
simultaneous equations, such as, system of equations (i )

a1 x b1 y a2 x b2 y

(ii )

a1 x b1 y

c1 c2 c1

a2 x b2 y c2
(iii ) a1 x b1 y c1 a2 x b2 y c2

a1 x b1 y

c1

a2 x b2 y

c2

Comparison of coeff.and const. terms a1 b1 a2 b2

are givn in the chart below : e consistent/ incon sistent consistent has solution mutually dependent/ (how many)
/ no. independent independent es
Y
(only one) dependent

a1 a2 b1 b2 c1 c2 consistent

a1 a2 b1 b2 c1 c2 inconsistent independent

es (infinite
Y
numbers) o N

197

Math

Now, if there is no constant terms in both the equations of a system of equations ; a1 b1 e the
i.e., c1 c2 0 , if with reference to the abov discussion from (i), if a2 b2 system of equations are always consistent and independent of each other. In that case, there will be only one (unique) solution. a b1
From (ii ) and (iii ) if 1
, the system of equations are consistent and a2 b2 dependent of each other. In that case, there will be infinite number of solutions.
Example : xlain whether the following sy stem of equations are consistent /
E
p inconsistent, dependent/ independent of each other and indicate the number of solutions in each case.
(b) 2 x 5 y 3
(c) 3 x 5 y 7
(a) x 3 y 1
2x 6 y 2 x 3y 1
6 x 10 y 15
Solution : x 3y 1
(a) ien system of equations are :
G
v
2x 6 y 2 atio of the coefficients of
R

1
2
3 y is
6

x is

atio of the coefficients of
R
atio of constant terms is
R

or

1
2

1
2

1 3 1
2 6 2
Therefore, the system of equations are consistent and mutually dependent. The system of equations hav infinite number of solutions. e 2x 5 y 3
(b) ien system of equations are :
G
v x 3y 1
R of the coefficients of atio x is

atio of the coefficients of
R

y is

2
1
5
3

2
5
1
3
Therefore, the system of equations are consistent and mutually independent. The system of equations hav only one (unique) solution. e we hav, e 198

Math

(c) ien system of equations are :
G
v

3x 5 y 7
6 x 10 y 15
3
1 or R of the coefficients of x is atio 6
2
5
1
or
R of the coefficients of y is atio 10
2
7 ratio of the constant terms is
15
3
5
7 we get,
6
10 15
Therefore, the system of equations are inconsistent and mutually independent. The system of equations hav no solution. e Activity : erify whether the system of equations x 2 y 1 0, 2 x y 3 0
V
are consistent and dependent and indicate how many solutions the system of equations may hav. e Exercise 12⋅1
Mention with arguments, whether the following simple simultaneous equations are consistent/inconsistent, mutually dependen t/independent and indicate the number of solutions :
2. 2 x y 3
3. x y 4 0
1. x y 4 x y 10
3 x 3 y 10 0
4x 2 y 6
5. 3 x 2 y 0
6. 5 x 2 y 16 0
4. 3 x 2 y 0
6
3x y 2
9x 6 y 0
6x 4 y 0
5
1
1
1
7.
x y
1
8. x y 0
9.
x y
1
2
2
2 x 2y 2 x y 5 x 2y 0
10. ax cy 0 cx ay c 2 a 2 .
12⋅3 Solution of simple sumultaneous equations
We shall discuss the solutions of only the consistent and independent simple simultaneous equations. uch system of equation has only one (unique) solution.
S
Here, four methods of solutions are discussed :
(1) Method of substitution, (2) Method of elimination
(3) Method of crossmultiplication (4) G raphical method.

Math

199

In class V we hav known how to solv by the methods of substitution and
III,
e e elimination. Here, exmples of one for each of these two methods are givn. a e
Example 1. Slv by the method of substitution : oe 2x y 8
3x 2 y 5
Solution : ien equations are :
G
v
2 x y 8..............(1)
3 x 2 y 5............( 2)
From equation (1), y 8 2 x.........(3)
Putting the alue of y form equation (3) in equation (2), we get v 3 x 2(8 2 x) 5
Putting the vlue of x in equation (3) a y 8 2 3 or, 3 x 16 4 x 5 or, 3 x 4 x 5 16
8 6 or, 7 x 21
2
or, x 3
Slution ( x, y) (3, 2) o Solution by the method of substitution : onvniently, from any of the two equations,
Ce
vlue of one ariable is exressed in a v p terms of the other vriable and putting the obtained alue in the other equation, we a v shall get an equation with one ariable. olv this equation, alue of the ariable v S ing v v can be found. This alue can be put in any of the equations. But, if it is put in the v equation in which one ariable has been exr essed in terms of the other vriable, the v p a solution will be easier. From this equation, alue of the other ariable will be found. v v
Example 2. Slv by the method of elimination : oe 2x y 8
3x 2 y 5
[N.B. : To show the difference between the methods of substitution and elimination, same equations of exmple 1 hav been taken in this exmple 2] a e a Solution : ien equations are
G
v
2 x y 8..............(1)
3 x 2 y 5............( 2)
Multiplying both sides of equation (1) by 2, 4 x 2 y 16..............(3) equation (2) is 3 x 2 y 5............( 2)
Adding (3) and (2), 7 x 21 or x 3 .
Putting the alue of x in equation (1), we get v 2 3 y 8 or, y 8 6 or, y 2
Slution ( x, y) (3, 2) o Math

200

Solution by the method of elimination : o e
Cnvniently, one equation or both equations are multiplied by such a number so that after multiplication, absolute alue of the coefficients of the same ariable v v become equal. Then as per need, if the equations are added or subtracted, the a vriable with equal coefficient will be eliminated. Then, solv the obtained ing equation, the vlue of the ex a isting ariable will be found. If that vlue is put v a convniently in any of the givn equations, alue of the other ariable will be found. e e v v
(3) Method of cross-multiplication :
We consider the following two equations : a1 x b1 y c1 0...........(1) a2 x b2 y c2 0..........( 2)
Multiplying equation (1) by b2 and equation (2) by b1 , we get, a1b2 x b1b2 y b2 c1 0.........(3)

a2b1 x b1b2 y b1c2 0.........( 4) ubtracting equation (4) from equation (3), we get
S

(a1b2 or, (a1b2

a2b1 ) x b2 c1 b1c2 0 a2b1 ) x b1c2 b2 c1
1
x or, ..........(5) b1c2 b2 c1 a1b2 a2b1
Again, multiplying equation (1) by a2 and equation (2) by a1 , we get, a1a2 x a2b1 y c1a2 0.........(6) a1a2 x a1b2 y c2 a1 0.........(7) ubtracting equation (7) from equation (6), we get
S

(a2b1 a1b2 ) y c1a2 c2 a1 0
(a1b2 a2b1 ) y
(c1a2 c2 a1 )
1
y or, ..........(8) c1a2 c2 a1 a1b2 a2b1 or, From (5) and (8) we get, x y b1c2 b2 c1 c1a2 c2 a1

1 a1b2 a2b1

From such relation between x and y , the technique of finding their vlues is called a the method of cross- ultiplication. m From the abov relation between x and y , we get, e Math

201

x

1

b1c2

b2 c1

Again,

a1b2

a2 b1

y c1a2 1 c2 a1

b1c2 a1b2 b2c1 a2b1 , or y

c1a2 a1b2 , or x

a1b2

a2 b1

c2 a1 a2b1 b1c2 a1b2 The solution of the givn equations : ( x, y ) e b2 c1 c1a2
,
a2b1 a1b2

c2 a1 a2b1 We observe : quations E

x and y

elation between
R

a1 x

b1 y

c1

a2 x

b2 y

c2

x

0
0

b1c2

y b2 c1

c1a 2

Illustration

c2 a1

a1b2

y

x

1 a 2 b1

a1

b1

c1

a2

b2

1

c2

a1

b1

a2

b2

[N.B. : The method of cross- ultiplication can also be applied by keeping the m constant terms of both equations on the right hand side. In that case, changes of sign will occur ; but the solution will remain same.]
Activity : If the system of equations a1 x b1 y c1 0
4x y 7 0 are exressed as the system of equations p ,
3x y
0
a2 x b2 y c2 0 find the alues of a1 , b1 , c1 , a2 , b2 , c2 . v Example 3. Slv by the method of cross- ultiplication : oe m

6x y 1
3 x 2 y 13
Solution : Making the right hand side of the equations 0 (zro) by transposition, we get, e 6x y 1 0 comparing the equations with

3 x 2 y 13

a1 x b1 y2 c1 0 respectivly, e a2 x b2 y c2 0

0

a1

6, b1

1, c1

1

a2

we get,

3, b2

2, c2

13
Illustration : y x
1

By the method of cross- ultiplication, we get, m x b1c2 y b2 c1

c1a2

1 c2 a1

a1b2

a2b1

a1

b1

c1

a2

b2

c2

a1 a2 b1 b2 202

Math

or

x
( 1) ( 13)

or

x
13 2

y
3 78

x
15
x
15

1
15

y
75
1
,
15 y Again,
75

or

2 ( 1)

( 1) 3

y
( 13) 6

6 2

1
3 ( 1)

1
12 3

x

y

1
2

15
1
or x
15
75
1
, or y
15
15
( x, y ) (1,5)

1

1
6

13

3

5

Slution o Example 4. Slv by the method of cross- ultiplication : oe m

3x 4 y 0
2x 3y

1

Solution : ien equations are :
G
v

3x 4 y
2x 3y

0

or

1

3x 4 y 0 0
2x 3y 1 0

By the method of cross- ultiplication, we get, m x y 1
4 1 ( 3) 0 0 2 1 3 3 ( 3) 2 ( 4) or x
4 0

y
0 3

x

y

3

4

0

3

4

2

3

1 2

3

1

1
9 8

x y 1 or 4
3
1 x y 1 or 4 3 1 x 1
, or x 4
4 1 y 1
Again,
, or y 3
3 1
Slution ( x, y ) ( 4,3) . o Example 5. Slv by the method of cross- ultiplication : oe m

x
2

y
3

8

5x
3y
4

Solution : Arranging the givn equations in the form e ax by c

3

0 , we get,

1
2

203

Math

x y
8
2 3
3x 2 y or 8
6
or 3 x 2 y 48 0

5x
3y
3
4
5 x 12 y or 3
4
or 5 x 12 y 12 0
Again,

the givn equations are : e 3 x 2 y 48 0
5 x 12 y 12 0
By the method of cross- ultiplication, we get, m x
2 12 ( 12) ( 48)

y
( 48) 5 12 3

1
3 ( 12) 5 2

x

y

1

3

2

48

3

2

5

12

12

5

12

x y 1
24 576
240 36
36 10 x y
1
or
552
276
46
1 x y or 552 276 46
552
x
1
or, x
12
552 46
46
276 y 1
, or y
6
Again,
46
276 46
Slution ( x, y ) (12,6) o or

Verification of the correctness of the solution : e x and y in givn equations, we get, x y 12 6
In 1st equation, .H.S=
L .
6 2
2 3 2 3
.H.S
8 R .
5 12
5x
In 2nd equation, .H.S=
L .
3 6
3y
4
4
R .
.
15 18
3 =H.S
Putting the vlues of a the solution is correct.
Example 6. Slv by the method of cross- ultiplication : oe m
Solution : ien equations are
G
v ax by ab ax by ab 0 or, bx ay ab bx ay ab 0
By the method of cross- ultiplication, we get, m ax by

ab

bx ay.

204

Math

x

y

( b ) ( ab ) ( a )( ab )

( ab ) b

x

1
( ab ) a

a ( a ) b ( b)

y

1

y
1
ab 2 a 2 b a2 b2 x y
1
or ab( a b) ab( a b)
( a b)( a b) x y
1
or ab( a b) ab( a b) ( a b)( a b) ab( a b) x 1
, or x
( a b)( a b) ab( a b) ( a b)( a b) y 1 ab( a
, or y
Again,
ab( a b) ( a b)( a b)
( a b)( a ab ab
( x, y )
,
a b a b

b

ab

a

b

b

or

a

a

ab

b

a

x

ab 2

a 2b

Exercise 12⋅2 oe Slv by the method of substitution (1 -3) :
1. 7 x 3 y 31 x y
1
2.

9x 5 y

41

2 x 3

3 y 2

1

3.

ab a b
b)
ab
b) a b

x y
2
a b ax by a 2

b2

oe
Slv by the method of elimination (4 -6) :
4. 7 x 3 y

31
41

5. 7 x 8 y
9
5x 4 y
3
olv by the method of cross- ultiplication (7 -15) :
Se
m

9x 5 y

7. 2 x

3y 5 0
4x 7 y 6 0
10. 4 x 3 y
12
2x 5
13. ax by

a2

b2

2bx ay ab
7)( y 3) 7

15. ( x

6. ax

by c a2 x b2 y c2

8. 3 x

9. x 2 y 7
5y 9 0
5x 3 y 1 0
2x 3y 0
11. 7 x 8 y 9
12. 3 x y 7 0
5x 4 y
3
14. y(3 x) x(6 y )

3(3

x)

5( y 1)

( y 3)( x 1) 5

2x

y 3

Math

205

12⋅4 Solution by graphical method
In a simple equation with two vriab les, the relation of ex a isting ariables v x and y can be exressed by picture. This picture is called the graphs of that relation. In the p graph of such equation, there ex infin ite number of points. Plotting a few such ist points, if they are j ined with each other, we shall get the graph. o ach of a simple simultaneous equations ha s infinite number of solutions. raph of
E
G each equation is a straight line. oordinates of each point of the straght line satisfies
C
the equation. To indicate a graph, two or more than two points are necessary. ow N we shall try to solv graphically the following system of equations : e 2 x y 3..........(1) x 3
1 0
4 x 2 y 6........(2)
From equation (1), we get, y 3 2 x . y 5
3
3
Taking some vlues of x in the equation, we find the a corresponding vlues of y and make the adj ining table : a o three points on the graph of the equation are : ( 1, 5), (0, 3) and (3, 3) |
6 4x
Again, from equation (2) , we get, 2 y 6 4 x or, y x 6
2 0
2
Taking some vlues of x in the equations, we find the a y
7
3
9
the corresponding vlues of y and make the adj ining table : a o three points on the graph of the equation are : ( 2, 7), (0, 3) and (6, 9)
In a graph paper let XOX and YOY be x axs and y axs and O is the
-i
-i respectivly the e origin.
We take each side of smallest squares of the graph paper as unit along with both axs. ow, we plot the e N points ( 1, 5), (0, 3) and (3, 3) obtained from equation (1) and oin them each other. The graph is j a straight line.
Again, we plot the points ( 2, 7), (0, 3) and (6, 9) obtained from equation (2) and j in them each o other. In this case also the graph is a straight line.
But we observ that the two straight lines coin cide and they hav turned into the one e e straight line. Again, if both sides of equation (2) are div by 2, we get he equation ided (1). That is why the graphs of the two equations coincide.
2 x y 3..........(1) are consistent and mutually
Here, the system of equations,
4 x 2 y 6........(2) dependent. uch system of equations hav infinite number of solutions and its graph
S
e is a straight line.

206

Math

ow,
N we shall try to solv the system of equations : e 2 x y 4..........(1)
4 x 2 y 12......(2)

From equation (1), we get, y 2 x 4 .
Taking some vlues of x in the equation, we find the corresponding a o alues of y and make the adj ining table : v x
1 0
4
three points on the graph of the equation are : y ( 1, 6), (0, 4), (4, 4) .
6
4 4
Again, from equation (2), we get,
4 x 2 y 12 , or 2 x y 6 x or y 2 x 6 y Taking some vlues of x in the equation, we find the a 6 the corresponding vlues of y and make the adj ining table : a o three points on the graph of the equation are : (0, 6), (3, 0), (6, 6) |
In graph paper let XOX and YOY be respectivly x ax and y ax and O is the e - is
- is origin. Taking each side of smallest squares in the graph paper as unit, we plot the points ( 1, 6), (0, 4) and (4, 4) obtained from equation (1) and oin j them each other. The graph is a straight line.
Again, we plot the points (0, 6), (3, 0), (6, 6) obtained from equation (2) and j in them each o other. In this case also the graph is a straight line.
We observ in the graph, though each of the givn e e equations has separately infinite number of solutions, they hav no common solution as e system of simultaneous equations. Further, we observ that the graphs of the two equations are straight lines parallel to each other. e That is, the lines will nevr intersect each other. Therefore, there will be no common e point of the lines. In this case we say, such system of equations hav no solution. We e know, such system of equations are inconsistent and independent of each other. ow, we shall solv the system of tw o consistent and independent equations by
N
e graphs. G raphs of two such equations with two vriables intersect at a point. Both a the equations will be satisfied by the coordinates of that point. The ery coordinates v of the point of intersection will be the solution of the two equations.
Example 7. olv and show the solution in graph :
Se
2x y 8

3x 2 y y 8 0...........(1)
3 x 2 y 5 0.............( 2)
By the method of cross- ultiplication, we get, m Solution : : ien two equations are :
G
v

2x

5

Math

207

x
1 ( 5) ( 2) ( 8) or x
5 16

y
( 8) 3 ( 5) 2

y
24 10

1
2( 2) 3 1

1
4 3

x y 1 or 21
14
7 x y 1 or 21 14 7 x 1
21
, or x
3
21 7
7
y 1
14
Again,
, or y
2
14 7
7
solution is ( x, y ) (3,2)
- is
- is e O, the origin. et XOX and YOY be x ax and y ax respectivly and
L
Taking each two sides of the smallest squares along with both axs of the graph e paper as one unit, we plot the point (3, 2).

x

3x
5x
Solution : : ien equations are :
G
v

y 3 y 21
3x
5x

y y 3..........(1)
21........(2)

1

y

Example 8. Slv with the help of graphs : oe 6

x

0

3

3

3

4

From equation (1), we get, 3 x y 3 , or y 3 x 3 y 6
Taking some vlues of x in equation (1), we get a corresponding vlues of y and make the table beside : a three points on the graph of the equation are : ( 1, 6), (0, 3), (3,6)
Again, from equation (2), we get, 5 x y 21 , or y 21 5 x
Taking some alues of x in equation (2), we find v the corresponding alues of y and make the v adj ining table : o three points on the graph of the equation are :
(3,6), (4,1), (5, 4) . e -x et XOX and YOY be respectivly x ais and
L
-i y axs and O be the origin.
We take each side of the smallest square of the graph paper as unit.

1

6
5

4

Math

208

ow, we plot the points ( 1, 6), (0, 3), (3,6) obtained from equation (1) and oin
N
j them successivly. The graph is a straigh t line. S e imilarly, we plot the points j e
(3,6), (4,1), (5, 4) obtained from equation (2) and oin them successivly. In this case also the graph is a straight line. et the two straight lines intersect each other at
L
P . It is seen from the picture that the coordinates of P are (3,6) . solution is ( x, y ) (3,6)
Example 9. Slv by graphical method : 2 x 5 y oe 14

4 x 5 y 17 x 2x 5 y
14...........(1)
4 x 5 y 17........(2) y 2 x 14
From equation (1), we get, 5 y
14 2 x , or y
5
Taking some convnient alues of e v x in the equation, we find the corresponding vlues of y and make the adj ining table : a o three points on the graph of the equation are : x 1
(3, 4), , 3), ( 2, 2) .
2
4 x 17 y Again, from equation (2), 5 y 4 x 17 , or y
Solution : ien equations are :
G
v

3

4

3

1

1
2
3

1
2
3

2

2

2

5

5

Taking some convnient alues of x in the equation (2), e v we find the corresponding alues of y and make the adj ining table : v o three points on the graph of the equation are :
(3, 1),

1
, 3 , ( 2, 5) .
2

et
L
be x ax and
- is
-i
y axs
XOX and YOY respectivly and O, the origin. e We take each two sides of the smallest squares as unit along with both axs. e ow, we plot the points (3, 4), 1 , 3 and ( 2, 2)
N
2 obtained from equation (1) in the graph paper and oin them each other. The graph is j a straight line. S imilarly, we plot the points (3, 1), 1 , 3 , ( 2, 5) obtained from
2

equation (2) and oin them each other. The graph is a straight line. j et the straight lines intersect at P . It is seen from the graph, coordinates of P are 1 , 3 .
L
2

solution is ( x, y )

1
, 3
2

Math

209

Example 10. olv with the help of graphs :
Se
Solution : ien equation is
G
v

3

3 x 2

3

3 x 2

x

8 4x

y

0

2

y

6

3

0

x

1

2

3

y

4

0

8 4x

3 x 8 4x
2
3 y 3
x.........(1)
2
And, y 8 4 x..........( 2) ow, Taking some alues of
N
v x in equation (1), we find the corresponding alues of y and make the adj ining v o table : three points on the graph of the equation (1) are :

et,
L

2

3

4

( 2,6), (0,3), (2,0)

Again, taking some vlues of x in equation (2), we a find the corresponding alues of y and make the v adj ining table : o three points on the graph of the equation (2) are :

(1,4), (2,0), (3, 4)

et XOX and YOY be x ax
L
- is, y ax respectivly
- is e and O, the origin. We take each side of the smallest squares along with both axs as e unit. N we plot the points ( 2,6), (0,3), ( 2,0) , obtained from equation (1) on the ow, graph paper and oin them each other. The graph is a straight line. In the same way, j j we plot the points (1,4), ( 2,0), (3, 4) obtained from equation (2) and oin them each other. This graph is also a straight line. et the two straight lines intersect at P . It is
L
seen from the picture that the coordinates of the point of intersection are (2,0) . solution is x 2 , or solution is 2
Activity : Find four points on the graph of the equation 2 x y 3 0 in terms of a table. Then, taking unit of a fixd length on the graph paper, plot the points and e oin them each other. Is the graph a straight line ? j Exercise 12⋅3
Solve by graphs :
1. 3 x 4 y 14

4x 3y
4. 3 x 2 y

5x 3 y

2. 2 x

5x

2
2

5

5.

y 1 y 13

x y
2
2 3
2 x 3 y 13

3. 2 x

5y 1 x 3y 2

6. 3 x

y

6

5 x 3 y 12

210

Math

3x 2 y

8.

4

3x 4 y 1

x
2

y
3
x

y
6

9. 3 x

3

2

x 2

3

10. 3 x 7 3 2 x
12⋅5 Formation of simultaneous equations from real life problems and solution. In evryday life, there occur some such ma thematical problems which are easier to e solv by forming equations. For this, from the condition or conditions of the e problem, two mathematical symbols, mostly the ariables x, y are assumed for two v unknown exressions. Two equations are to be formed for determining the alues of p v those unknown exressions. If the two equations thus formed are solvd, vlues of he p e a unknown quantities will be found.
Example 11. If 5 is added to the sum of the two digits of a number consisting of two digits, the sum will be three times the digits of the tens place. Moreovr, if the places e of the digits are interchanged, the number thus found will be 9 less than the original number. Find the number.
Solution : et the digit of the tens pl ace of the required number be x and its digits
L
of the units place is y. Therefore, the number is 10 x y. by the 1st condition, x y 5 3 x.........(1) and by the 2nd condition, 10 y x (10 x y ) 9.......( 2)
From equation (1), we get, y 3 x x 5 , or y 2 x 5........(3)
Again from equations (2), we get,
10 y y x 10 x 9 0 putting the alue of x in (3), we get, v or or or or 9y y 2x x 9x 9 0 x 1 0
5 x 1 0
4

y

2 4 5
8 5
3
the number will be

[

10 x y y 10 4 3
40 3
43

the number is 43
Example 12. 8 years ago, father’s age was eight times the age of his son. After 10 years, father’s age will be twice the age of the son. What are their present ages ?
Solution : et the present age of father be x year and age of son is y year.
L
by 1st condition x 8 8( y 8)........(1)

Math

211

and by 2nd condition, x 10 2( y 10).......( 2)
From (1) , we get, x 8 8 y 64 or x 8 y 64 8 or x 8 y 56.............(3)
From (2) , we get, x 10 2 y 20 or p a 8 y 56 10 2 y 20 [ utting the vlue of x from (3)] or 8 y 2 y 20 56 10 or 6 y 66 or y 11 from (3) , we get, x 8 11 56 88 56 32 at present, Father’s age is 32 years and son’s age is 11 years.
Example 13. Twice the breadth of a rectangular garden is 10 metres more than its length and perimeter of the garden is 100 metre.
a. Assuming the length of the garden to be x metre and its breadth to be y metre, form system of simultaneous equations.
b. Find the length and breadth of the garden.
c. There is a path of width 2 metres around the outside boundary of the garden. To make the path by bricks, it costs 110 00 per square metre. What will be the total cost ?
Solution : a. ength of the rectangular garden is x metre and its breadth is y metre.
L
by 1st condition, 2 y x 10........(1) and by 2nd condition, 2( x y ) 100.......( 2)
b. From equation (1), we get, 2 y x 10........(1)
From equation (2), we get, 2 x 2 y 100.......(2) p v or 2 x x 10 100 [ utting the alue of 2 y from (1)] or 3 x 90 or x 30 p a from (1), we get, 2 y 30 10 [ utting the vlue of x ] or, 2 y 40 or, y 20 length of the garden is 30 metres and its breadth is 20 metres.
c. ength of the garden with the path is (30+) metres.
L
4
= metres
34

212

Math

and breadth is (20+) metres = metres.
4
24
Area of the path =
Area of the graden with the path area of the garden
(34 24 30 20) square metre
(816 600) square metre
216 square metre cost for making the path by bricks
=
Tk. 216 110
=
Tk. 23760
Activity : If in triangle ABC ,
B 2 x degree, C v B
C , find the alue of x and y .
A y degree and A

x

degree,

Exercise 12⋅4
1.

For which of the following conditions are the system of equations ax by c 0 and px qy r 0 consistent and mutually independent ? a b a b c a b a b c
b.
c.
d.
a. p q r p q p q r p q

If x y 4, x y 2 , which one of the following is the alue of ( x, y ) ? v a. (2, 4)
b. (4, 2)
c. (3, 1)
d. (1, 3) v 3. If x y 6 and 2 x 4 , what is the alue of y ?
b. 4
c. 6
d. 8
a. 2
4. For which one of the following equation is the adj ining chart correct? o b. y 8 x
a. y x 4
c. y 4 2 x
d. y 2 x 4
5. If 2 x y 8 and x 2 y 4 , what is x y ?
a. 0
b. 4
c. 8
d. 12
6. bserv the following information : :
O e
i. The equations 2 x y 0 and x 2 y 0 are mutually dependent.
G
ii. raph of the equation x 2 y 3 0 passes through the point ( 3, 0) .
G
iii. raph of the equation 3 x 4 y 1 is a straight line. n the basis of information abov, which one of the following is correct ?
O
e
a. i and ii
b. ii and iii
c. i and iii
d. i, ii and iii
7. ength of the floor of a rectangular r oom is 2 metres more than its breadth and
L
perimeter of the floor is 20 metres.
2.

Math

213

Answer the following questions :
(1) What is the length of the floor of the room in metre ?
a. 10
b. 8
c. 6
d. 4
(2) What is the area of the floor of the room in square metre ?
b. 32
c. 48
d. 80
a. 24
(3) How much taka will be the total cost for decorating the floor with mosaic at
Tk. 900 per square metre ?
a. 72000
b. 43200
c. 28800
d. 21600
Solve by forming simultaneous equations (8 - 15) :
8. If 1 is added to each of numerator and denominator of a fraction, the fraction
4
will be . Again, if 5 is subtracted from each of numerator and denominator,
5
1 the fraction will be . Find the fraction.
2
9. If 1 is subtracted from numerator and 2 is added to denaminator of a fraction,
1
the fraction will be . Again, if 7 is subtracted from numerator and 2 is
2
1 subtracted from denominator, the fraction will be . Find the fraction.
3
10. The digit of the units place of a number consisting of two digits is 1 more than three times the digit of tens place. But if the places of the digits are interchanged, the number thus found will be equal to eight times the sum of the digits. What is the number ?
11. Difference of the digits of a number consisting of two digits is 4. If the places of the digits are interchanged, sum of the numbers so found and the original number will be 110 ; find the number.
12. Present age of mother is four times the sum of the ages of her two daughters.
After 5 years, mother’s age will be twice the sum of the ages of the two daughters. What is the present age of the mother ?
13. If the length of a rectangular region is decreased by 5 metres and breadth is increased by 3 metres, the area will be less by 9 square metres. Again, if the length is increased by 3 metres and breadth is increased by 2 metres, the area will be increasd by 67 square metres. Find the length and breadth of the rectangle.

21
4

Math

4. A boat, rowing in favour of current, goes 1 per hour and rowing against the
1
5 km current goes 5 per hour. ind the speed of the boat and current. km F
. A labourer of a garments serves on the basis of monthly salary. At the end o f
5
1 every year she gets a fixd increment. Her monthly salary beomes Tk. 45 e 0 after 4 years and Tk. 0
5
after 8 years. ind the salary at the beginning of he
F
service and amount of annual increment of salary.
.
6
1 A system of simple equations are x y 1
0

3x 2 y

r

0

a. Sow that the equations are consistent. How many solutions do they have ? h b. olving the system of equations, find ( x, y) .
S
c. ind the area of the triangle formed by the straight lines indicated by the
F
equations with the x ax
- is.
. If 7 added to the num. of a fraction, the value of the fraction is the integer 2. Again
7
1 is ,

if 2 is subtracted from the denominator, value of the fraction is the integer .
1
a. orm a system of equations by the fraction
F

x
.
y

b. ind
F
( x, y ) by solving the system of equations by the method of crossmultiplication. W is the fraction ? hat c. raw the graphs of the system of equations and verify the correctness of the
D
obtained values of ( x, y ) .

Chapter Thirteen

Finite Series
The term order’is widely used in our day to day life. Sch as, the concept of order
‘
u is used to arrange the commodities in the shops, to arrange the incidents of drama and ceremony, to keep the commodities in attractive way in the godown. Again, to make many works easier and attractive, we use large to small, child to old, light to originated heavy etc. types of order. Mathematical series have been of all these concepts of order. In this chapter, the relation between sequence and series and contents related to them have been presented.
At the end of this chapter, the students will be able to – escribe the sequence and series and determine the difference between them
D
Exlain finite series p orm formulae for determining the fix d term of the series and the sum of
F
e fixd numbers of terms and solve math ematical problems by applying the e formulae etermine the sum of squares and cubes of natural numbers
D
o
Slve mathematical problems by appl ying different formulae of series
Cnstruct formulae to find the fixd o e term of a geometrical progression and sum of fixd numbers of terms and solv e mathematical problems by applying e the formulae.
Sequence
et us note the following relation :
L
1
2
4
3
5 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

2
4
6
8
1 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 2 n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
0
Here, every natural number n is related to twice the number 2 n . That is, the set of positive even numbers {2, 4, 6, 8,.......} is obtained by a method from the set of natural numbers N {1, 2, 3,.......} . This arranged set of even number is a sequence.
Hence, some quantities are arranged in a particular way such that the antecedent and subsequent terms becomes related. The set of arranged quantities is called a sequence. The aforesaid relation is called a function and defined as f (n) 2n . The general term of this sequence is 2n . The terms of any sequence are infinite. The way of writing the sequence with the help of general term is 2n , n 1, 2, 3, ... ... or,
2n n 1 or, < .
2n >

Math

21
6

The first quantity of the sequence is called the first term, the second quantity is called second term, the third quantity is called the third term etc. The first term of the sequence , 3, , , … 1the second term is 2 etc.
1 57 …, is ollowings are the four exmples of sequence :
F
a
1 , 2 , 3 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , n , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
1 , 3 , 5 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , (2n 1) , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
1 , 4 , 9 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , n 2 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
1 , 2 , 3 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , n , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
2

3

n 1

4

Activity : 1 G
. eneral terms of the six sequences sequences :
(i) 1

(ii) n 1

n

n 1

(iii) 1n
2

(iv) 1 1 n 2

are given below. W down the rite (v) ( 1) n 1 n

n 1

(vi) ( 1) n 1 n

.

2n 1

2. Each of you write a general term and then write the sequence.
Series
If the terms of a sequence are connected successively by + sign, a series is obtained.
Sch as, 1 3 5 7 ....... is a series. The difference between two successive u terms of the series is equal. Again, 2 4 8 6
1 ........ is a series. The ratio of two successive terms is equal. Hence, the characterstic of any series depends upon the relation between its two successive terms. Among the series, two important series are arithmetic series and geometric series.
Arithmetic series
If the difference between any term and its antecedent term is always equal, the series is called arithmetic series.
Example : 1 3 5 7 9 1 is a series. The first term of the series is , the
1
second term is 3, the third term is , etc.
5
second term = 5 3 2,
Here, second term – first term = 3 1 2, third term – fourth term – third term = 7 5 2, fifth term – fourth term = 9 7 2, six term – th fifth term = 1 9 2.
Hence the series is an arithmetic series. In this series, the difference between two terms is called common difference. The common difference of the mentioned series is 2. The numbers of terms of the series are fixd. That is why the series is finite e series. It is to be noted that if the terms of the series are not fixd, the series is called e infinite series, such as, 1 4 7 0 ... ... is an infinite series. In an arithmetic
1
series, the first term and the common difference are generally denoted by a and d respectively. Then by definition, if the first term is a , the second term is a d , the third term is a 2d , etc. Hence, the series will be a (a d ) (a 2d ) ...
Determination of common term of an arithmetic series et the first term of arithmetic series be a and the common difference be d ,
L
terms of the series are :

Math

21
7

first term a a (1 1) d second term a d a (2 1) d third term a 2d a (3 1) d forth term a 3d a (4 1) d
.... ....
.... ....
.... ....
....
.... ....
.... ....
.... ....
....
n th term = a (n 1)d
This n th term is called common term of arithmetic series. If the first term of an arithmetic series in a and common difference is d, all the terms of the series are determined successively by putting n 1, 2, 3, 4,... ... in the n th term. et the first term of an arithmetic series be 3 and the common difference be 2. Then
L
second term of the series
3 2 5, third term 3 2 2 7, forth term
3 3 2 9 etc.
Therefore, n th term of the series 3 n 1 2 2n 1 .
Activity : If the first term of an arithmetic series is 5 common difference is , and 7 find the first six terms, 22nd term, r th term and (2p+ term.
)th
1
Example 1. f the series, 5 8 1 4
O
which term is 383 ?
1
Solution : The first term of the series a 5, common difference

d

8 5 1

8 3

It is an arithmetic series. et, n th term of the series = 383
L
W e know that, n th term = a (n 1)d .

(n 1)d 383
(n 1)3 383
3n 3 383
383 5 3
381
381 or, n
3
n 17
2

or, or, or, or, a
5
5
3n
3n

17 th term of the given series = 383 .
2
Sum of n terms of an Arithmetic series et the first term of any arithmetic series be a , last term be p, common difference be
L
d , number of terms be n and sum of n numbers of terms be S n . riting from the first term and conversely from the last term of the series we get,
W

Sn

a ( a d ) ( a 2d ) p 2d a p a

and S n p p d
Adding, 2S n a p

p

p 2d p d p a 2d ( a d ) a
... ... a p a p a

(i)
(ii ) p 21
8

Math

or, 2 S n

na

n a 2

Sn

p

[ number of terms of the series is n]

p

(iii )

Again, n th term

Sn
i.e., S n

p a (n 1)d . Putting this value of p in (iii) we get, n [a {a (n 1)d }
]
2 n 2a n 1 d
(iv)
2

If the first term of arithmetic series a, last term p and number of terms n are known, the sum of the series can be determined by the formula (iii ) . ut if first term
B
the a , common difference d , number of terms n are known, the sum of the series are determined by the formula (iv) .
Determination of the sum of first n terms of natural numbers et Sn be the sum of n numbers of natural numbers i.e.
L

Sn 1 2 3

(n 1) n

(i)

riting
W from the first term and conversely fr om the last term of the series we get,

Sn 1 2 3

n 2

and S n n n 1 n 2
Adding, 2S n (n 1) (n 1) (n 1) or, 2Sn n(n 1)

Sn

(n 1) n
(i)
3
2 1
(ii )
(n 1) [ n number of terms]
-

n(n 1)
2

(iii )

Example 2. ind the sum total of first 0
F
5 terms of natural numbers.
Solution : sing formula ( iii) we get,
U

S0
5

0 (0 1)
5 5
2

25 1
5

25
17

The sum total of first 0
5
natural numbers is 25
1.
7 what ?
Example 3. 1 2 3 4
9
Solution : The first term of the series a 1, common difference d 2 1 1 and the last term p =
.
9
It is an arithmetic series. et the n th term of the series = 9
L
Alternative method : n th term of an arithmetic S
W know, e ince progression = a (n 1)d n Sn a p, a (n 1)d 9
2
or, 1 (n 1)1 9
9
S9
1 9 or, 1 n 1 9
2

21
9

Math

n

9

9

0
1
2

49
0
5

rom ( iv) formula, the sum of first n -erms of an arithmetic series
F
t

Sn

n
{2a (n 1)d }
.
2

9
9
(2 8 )
9
{2 1 (9 1) 1}
2
2
9 0
1
9 0
5
49
0
5
2
Example 4. W is the sum of 30 hat terms of the series
7 1 1
2 7
Solution : irst term of the series a 7 , common difference d 1
F
2 7 5
It is an arithmetic series. Here, number of terms n 30.
W
e know that the sum of n terms of an arithmetic series n Sn
{2a (n 1)d}
.
2
30
o, the sum of 30
S
terms S30 = {2.7 (30 1)5} 1 (1 29 5)
5 4
2
5 5
9
1 (1 15 ) 1 1
5 4 4
2385
Hence, the sum 9 terms of the series

S9

Example 5. A deposits Tk. 20from his salary in the first month and in every
1
month of subsequent months, he deposits Tk. 0
1
more than the previous months.
( i ) How much does he deposit in n th month ? p ( ii ) Exress the aforesaid problem in series upto n terms.
( iii ) How much does he deposit in first n - onths ? m ( iv ) How much does he deposit in a year ?
Solution : ( i ) In the first month, he deposits Tk. 2.
10
Tk. 10
In the second month, he deposits Tk. (10
2
0
1 )=
3
In third month, he deposits Tk. (30
Tk. 10
1
0 )=
1
4
In forth month, he deposits Tk. (10
Tk. 1
4
0
1 )=
5
0
Hence, it is an arithmetic series whose first term is a 20 , common difference
1

d

30
1
20
1
0 .
1
n th term of the series a (n 1)d
20
1
(n 1)0
1
20
1
0 n 0
1
1
0 n. 0
1
1
Therefore, he deposits Tk. (0 n 0
1
1 ) in n th month.
( i i) The series, in this case upto n numbers of terms will be
-

220

Math

10
2

10
3

10
4

(1 n 1
0
0

) n ( iii ) In n numbers of months, he deposits Tk. {2a (n 1)d }
2
n
=
Tk.
{2 20
1
(n 1)0 }
1
2 n n
=
Tk.
Tk.
(240 0 n 0 ) =
0 1
1
2(0
5
1
2
2
=
Tk. n(5 n 1
0
5 )
0
( iv ) W e know that 1 year =
2
2 months. Here n 1 .
1
Therefore, A deposits in 1 year =
Tk. 1 (5 1 1
2 0 2 5 )
0
=
Tk. 1 (6
2 0
1 )
5
0
=
Tk. 2 0
1 5
7
1
.
=
Tk. 21
0

0 n)
5

Exercise 13.1
. ind the common difference and the 1th terms of the series
1 F
2

2 5 1
2

1
9

2. W term of the series 8 1 1 1 hich is 39 ?
2
4 7
3. W term of the series 4 7 1 1 hich is 30 ?
1
0 3
4. If the p th term of an arithmetic series is p2 and q th term is q2 , what is
( p q) th term of the series ?
. If the m th tem of an arithmetic series is n and n th term is m , what is
5
( m n )th term of the series ?
. W is the number of
6 hat
?
n terms of the series 1 3 5 7 n? . W is the sum of first 9
7 hat terms of the series
8 1 24
6
8. 5 1 7 23
W ? hat 1
9
5 hat 9 29 25 21
.
23 W ?
. The 2th term of an arithmetic series is
0
1
1
. hat
7W is the sum of the first 23 terms?
. If the th term of an arithmetic series is
1
1
6
20 , what will be the sum of first 31 terms ? is 44 . ind
F
2
1. The total sum of first n terms of the series 9 7 5
1
the value of n .
3
1. If the sum of first n terms of the series 2 4 6 8 is 25find the
,
0
5
value of n.
4
1. If the sum of first n terms of the series is n(n 1) , find the series.
. If the sum of first n terms of the series is n(n 1) , what is the sum of first 0
5
1
1
terms ?
.
6If the sum of 2 terms of an arithmetic
1
1 series is 44 and first 20
1
terms is 6find
5
,
0
the sum of first 6 terms. 221

Math

. The sum of the first m terms of an arithmetic series is n and the first n terms is
7
1
m. ind the sum of first ( m+n) terms.
F
8. If the p th, q th and r th terms of an arithmetic series are a, b, c , respectively,
1
show that a(q r) b(r p) c( p q) 0.
1Sow that,
9h
.
1 3 5 7
25 9
1
6
1
1
7
3
7
1
20 .
9
20A man agrees to refund the loan of Tk. 20in some parts. Each part is Tk. 2
.
0 more than the previous part. If the first part is Tk. 1 in how many parts will the
,
man be able to refund that amount ?
Determinaton of the sum of Squares of the first n numbers of Natural Numbers et Sn be the number of squares of the first n numbers of natural numbers
L
i.e., S n 12 2 2 32 n2 W e know,

r 3 3r 2 3r 1 (r 1) 3 or, r 3 (r 1) 3 3r 2 3r 1
In the above identity, putting, r 1, 2, 3,
13 03 3.12 3.1 1
23 13
3

3

3.2 2

3.2 1

2

3.3 1

3

2

n3

3.3

(n 1) 3

3.n 2

Adding, we get, n 3 03 3(12 2 2 32 or, or,

2

2n n 1)
2

or,

3.n 1 n 2 ) 3(1 2 3

n(n 1) n 3 3S n 3. n 1 2 3
2
3n(n 1)
3Sn n3 n 2
2n 3 3n 2 3n 2n 2n 3 3n 2 n n( 2n 2

3S n
Sn

, n we get,

n(n 1)(2n 1)
2
n(n 1)(2n 1)
6

n) (1 1 1

n

n ( 2n 2

2 n{2n(n 1) 1(n 1)}
2

n(n 1)

3n 1)
2

2

1)

222

Math

The sum of cubes of the first n numbers of Natural Numbers et Sn be the sum of cubes of the first n numbers of natural numbers.
L
That is, S n 13 2 3 33 n3 (r 1)2 (r 1)2 (r 2 2r 1) (r 2 2r 1) 4r. or, (r 1) 2 r 2 r 2 (r 1) 2 4r.r 2 4r 3 [ ultiplying both the sides by r2]
M

W e know that,

In the above identity, putting r 1, 2, 3,
W
e get,
2 2.12 12.0 2 4.13
32.2 2 2 2.12 4.23
4 2.32 32.32 4.33
(n 1) 2 n 2

n 2 (n 1) 2

4n 3

Adding, we get, (n 1) 2 .n 2 12.0 2 or, (n 1)2.n2

,n

4(13

23 33

n3 )

4Sn

or, Sn

n2(n 1)2
4

Sn

n(n 1)
2

2

ecessary formulae :
N

n(n 1)
2

n

1 2 3
12

22

32

n2

13

23 33

n3

N.B: 13
.

23

33

n(n 1)(2n 1)
6
2 n(n 1)
2

n3

(1 2 3

n) 2 .

Activity : . ind the sum of natural even numbers of the first n- umbers.
1F
n
2. ind the sum of squares of natural odd numbers of the first n numbers.
F

Math

223

Geometric series
If the ratio of any term and its antecedent term of any series is always equal i.e., if any term divided by its antecedent term, the quotient is always equal, the series is called a geometric series and the quotient is called common ratio. uch as, of the series 2 4 8 1 32 ,
S
6 the first term is 2, the second term is 4, the third term is 8, the fourth term is 6 the
1
and fifth term is 32. Here, the ratio of the second term to the first term = third term to the second term =

6
1
8

8
4

4
2

2 , the ratio of the

2 , the ratio of the fourth term to the third term =

2 , the ratio of the fifth term to the fourth term =

32
6
1

2.

In this series, the ratio of any term to its antecedent term is always equal. The common ratio of the mentioned series is 2. The numbers of terms of the series are fixd. That is why the series is finite geomet ric series. The geometric series is widely e used in different areas of physical and biological science, in organiztions like B a ank and ife Insurance etc, and in different b ranches of technology. If the numbers of
L
terms are not fixd in a geometric series, it is called an infinite geometric series. e The first term of a geometric series is generally exressed by a and common ratios p o by r . S by definition, if the first term is a , the second term is ar, the third term is ar 2 , etc. Hence the series will be a ar ar 2 ar 3
Activity : W down the geometric series in the following cases : rite (i) The first term 4, common ratio 0
1

( ii) The first term , common ratio 1
9
3

(iii) The first term , common ratio 1
7
0
1
1
(v) The first term , common ratio
1
2

(iv) The first term 3, common ratio 1
(vi) The first term 3, common ratio 1
.

General term of a Geometric series et the first term of a geometric series be a, and common ratio be r. Then, of the
L
series, second term ar ar 2 1 first term a ar 1 1 ,
2
3 1 third term ar ar , fourth term ar3 ar 4 1
... ... ...
...
... ...
... ... ...
...
... ... n th term = ar n 1
This n th term is called the general term of the geometric series. If the first term of a geometric series a and the common ratio r are known, any term of the series can be determined by putting r 1, 2, 3, etc. successively in the n th term.
Example 6. W is the th term of the series hat 0
1
?
2 4 8 1
6

Math

224

4
2

2, common ratio r

Solution : The first term of the series a

The given series is a geometric series. n W e know that the n th term of geometric series = ar th 0 1
1
0
1 term of the series = 2 2
= 2 2 9 24
0
1
Example 7. W is the general term of the series 18 hat 2

1

6
4

32

28 , common ratio r
1

Solution : The first term of the series a

It is a geometric series.
W
e know that the general term of the series =

ar n

2.

?

4
6
28
1

1
.
2

1

n 1

1
27
1
1
. n 1 n 1 7
2
2
2n 8
2
Example 8. The first and the second terms of a geometric series are 27and . ind
7F
the th and the th terms of the series.
5
0
1
Solution : The first term of the given series a 27, the second term is 9 .

Hence, the general term of the series = 18
2

Then the common ratio r

9
27

The th term = ar 5
5

1
3

1

27

4

1
.
3
1
3

27 1
27 3

9

1
1
33
1
.
3
3 3 6 3 6 79
3
2
Determination of the sum of a Geometric series et the first term of th e geometric series be a, common ratio r and number of terms
L
n. If S n is the sum of n terms,

and the th term =
0
1

Sn

0 ar 1

1

a ar ar 2

27

ar n

and r.S n ar ar 2 ar 3 u Sbtracting, S n rS n a ar n or, S n (1 r ) a(1 r n )

Sn

a(1 r n )
, when r
1 r

2

ar n

1

(i)
[
ar n 1 ar n multiplying ( i) by r] ( ii)

1

Again, subtracting (ii) form (i) we get, rS n S n ar n a or, S n (r 1) a(r n 1)

a(r n 1)
, when r 1 .
(r 1)
Observe : If common ratio is r 1 , each term
i.e., S n

a

Math

225

Hence, in this case S n

a a a an. upto n.

Activity : A employed a man from the first April for taking his son to school and taking back home for a month. His wages were fixd to be – paisa in first day, e one twice of the first day in second day i.e. two paisa, twice of the second day in the third day i.e. four paisa. If the wages were paid in this way, how much would he get after one month including holidays of the week ?
Example 9. W is the sum of the series hat 2
1

24 48

2 , common ratio r
1

Solution : The first term of the series is a the series is a geometric series. et the nth term of the series 7
L
6
8
n 1
W
e know, n th term = ar or, or, or, or,

8 ?
6
7
24
2
2
1

ar n 1 7
8
6
1 2n 1 7
2
6
8
2n

7
6
8
1
2

1

6
4

2 n 1 26 n 1 6 n 7. a(r n 1) when r
,
(r 1)

Therefore, the sum of the series

1

2
1 (2 7 1)
2
1 (18 1) 1 17 14 .
2
2 2
2
5
2 1
Example 10. ind the sum of first eight terms of the series 1 1 1 1
F
2 4 8 st Solution : The 1 term of the series is a

1
2
1

1, common ratio r

It is a geomeric series.
Here the number of terms n 8.
W
e know, sum of n terms of a geometric series
Sn

a(1 r n ) when r
,
1 r

1.
1

1

Hence, sum of eight terms of the series is S
8
1
2

25 1
6
25
6

25
5
18
2

17
2
1
18
2

1
2
1
2

8

1

1
25
6
1
2

1
1
2

1.

226

Math

Exercise 13.2
.
1 If a, b, c, d are cosecutive four terms of a arithmetic series, which one is correct
?
c d b d b c a c
(a) b
(b) a
(c) c
(d) d
2
2
2
2
2. (i) If a (a d ) a 2d ) ........ the sum of first n is terms of the series is n { (n )
1
2 2a + – d} n(n 1)(2n 1)
(ii) 12 + +
+ 3 .......... + n 6
(iii) + ++
1 5........... +
2
(2n 1) n 2 hich one of the followings is correct according to the above statements.
W
(a) i and ii
(b) i and iii
(c) ii and iiii
(d) i, ii and iii
Answer the questions 3 and 4 on the basis of following series : log 2 + 4 + 8 + log log ........
3. W one is the common difference of the series ? hich (a) 2
(b) 4
(c) log 2
(d) 2 log 2
4. W one is the 7 term of the series hich th
(a) log 32 (b) log 4 (c) log 28 (d) log 25
6
1
6
. etermine the 8th term of the s eries 4 + ++ +
5 D
6 32 68 .............
1
. etermine the sum of first fourteen te rms of the series 3 ++ +
6 D
927..............
1 of the series 28 + + +
1 4 32 ...........
6
. W
7 hich of the term is
2
8. If the th terms of a geometric series
5

2 3 and the th term are
0
1
9

8 2
, find the
8

3rd term of the series.
. W
9 hich of the term is

8 2 of the sequence

1
, 1, 2 ,
2

?

1If 5 x y 15 is geometric series, find the value of x and y .
0
.
3
.
1 If 3 x y z 243 is geometric series, find the value of x, y and z.
2. W is the sum of first seven terms of the series
1 hat
?
2 4 8 6
1
3. ind the sum of (2n 1) terms of the series 1 1 1 1
1 F
4. W is the sum of first ten terms of the series
1 hat
?
log 2 log 4 log 8
. ind the sum of first twelve terms of the series
5F
1
?
log 2 log 1 log 5
6
1
2
. If the sum of n terms of the series 2 4 8 6
6
1 is 25, what is the
4
1 value of n ?

227

Math

1W is sum of
7hat
.
?
(2n 2) terms of the series 2 2 2 2
8. If the sum of cubes of n natural numbers is 441find the value of n and find the
1
, sum of those terms. n natural numbers is 225 , find the value of n
. If the sum of cubes of the first
9
1 and find the sum of square of those terms ?
20Sow that 13 23 33 ..............1 3 (1 2 3 ..... 1 ) 2 .
.h
0
0
21If
.

13 23 33
1 2 3

n3 n 21 , what is the value of n ?
0

22. A iron- ar with length one metre is divided into ten- ieces such that the lengths b p of the pieces form a geometric progression. If the largest pieces is ten times of the smallest one, find the length in appox imate millimetre of the smallest piece.
23. The first term of a geometric series is a, common ratio is r , the fourth term of the series is 2 and the th term is 8 2 .
–
9
(a) Exress the above information by two equations. p (b) ind the 2th term of the series.
F
1
(c) ind the series and then determine th e sum of the first seven terms of the
F
series.
24. The nth term of The a series is 2n –
4.
(a) ind the series.
F
(b) ind the th term of the series
F
0
1
and determine the sum of first 20 terms. (c) Cnsidering the first term of the obta ined series as 1t term and the common o s difference as common ratio, construct a new series and find the sum of first 8 terms of the series by applying the formula.

Chapter Fourteen

Ratio, Similarity and Symmetry or comparing two quantities, their ratios are to be considered. Again, for
F
determining ratios, the two quantities are to be measured in the same units. In algebra we have discussed this in detail.
At the end of this chapter, the students will be able to
Exlain geometric ratios p Exlain the internal division of a line segment p Verify and prove theorems related to ratios
Verify and prove theorems related to similarity
Exlain the concepts of symmetry p Verify line and rotational symmetry of real obj cts practically. e 14⋅1 Properties of ratio and proportion
(i ) If a : b = x : y and c: d = x : y, it follows that a: b = c : d.
(ii ) If a : b = b : a, it follows that a b
(iii ) If a : b = x : y, it follows that b t a y t x (inversendo)
(iv ) If a : b = x : y, it follows that a t x b t y (alternendo) bc (cross multiplication)
(v ) If a : b = c : d, it follows that ad
(vi ) If a : b = x : y, it follows that a b t b x y t y (componendo) and a b t b x y t y (dividendo) a c a b c d
, it follows that
(componendo- ividendo) d (vii ) If b d a b c d
Geometrical Proportion
Earlier we have learnt to find the area of a triangular region. Two necessary concept of ratio are to be formed from this.
(1 If the heights of two triangles are equal, their bases and areas are
)
proportional.

et the bases of the triangles ABC and DEF be BC = a, EF = d respectively and the
L
height in both cases be h.

229

Math

Hence, the area of the triangle ABC =

1 a h
2

and the area of the triangle

1
DEF= d h .
2

1
1
Therefore, area of the triangle ABC :area of the triangle DEF = a h : d h
2
2
= a : d = BC : EF that is, the areas and bases ate proportional.
(2) If the bases of two triangles are equal, their heights and areas are proportional. et the heights of the triangles ABC and DEF be AP = h, DQ = k respectively and
L
1 the base in both cases be b. Hence, the area of the triangle ABC = b h and the
2
1 area of the triangle DEF= b k
2
1
1
Therefore, area of the triangle ABC :area of the triangle DEF = b h : b k
2
2
= h: k = AP: DQ
Theorem 1
A straight line drawn parallel to one side of a triangle intersects the other two sides or those sides produced proportionally.
Proposition : In the figure, the straight line DE is parallel to the side BC of the triangle ABC. DE intersects AB and AC or their produced sections at D and E respectively. It is required to prove that,
AD : DB = AE : EC. igF2 Construction: J in B, E and C, D. o igF1
Proof:
J stificaltin u S teps T
(1 The heights of ADE and BDE are [ he bases of the triangles of equal height
)
are proportional] equal. 230

Math

ADE
AD
[ he bases of the triangles of equal
T
BDE
DB
height are proportional]
(2) Again, The heights of ADE and
[ the same base and between same n O
DEC are equal. pair of lines]
ADE
AE
DEC
EC
(3) ut
B
BDE
DEC
ADE
ADE
BDE
DEC
AD
AE
(4) Therefore,
DB
EC
i.e., AD t DB AE t EC.
Corollary 1. If the line parallel to BC of the triangle AB
C
intersects the sides
AB and
AB AC
AB AC and .
AC at D and E respectively,
AD AE
BD CE
Corollary 2. The line through the mid point of a side of a triangle parallel to another side bisects the third line.
The proposition opposite of theorem 1 al so true. That is, if a line segment divides is the two sides of a triangle or the line produced proportionally it is parallel to the third side. Here follows the proof of the theorem.
Theorem 2
If a line segment divides the two sides or their produced sections of a triangle proportionally, it is parallel to the third side.
Proposition : In the triangle ABC the line segment DE divides the two sides
AB and AC or their produced sections proportionally. That is, AD : DB = AE :
EC. It is required to prove that DE and
BC are proportional.
Construction: J in B, E and C, D. o Proof:
S
teps
J stificaltin u ADE
AD
(1
)
[Triangles with equal height ]
BDE
DB
[Triangles with equal height]
[ iven] g ADE
AE
and
[om (i) and (ii)] r F
DEC
EC

Math

231

AE
EC
ADE
ADE
(3) Therefore,
DEC
BDE
BDE
DEC
(4) ut
B
BDE and DEC are on the same side of the common base DE. S o they lie between a pair of parallel lines.
Hence BC and DE are parallel.
(2) ut
B

AD
DB

Theorem 3
The internal bisector of an angle of a triangle divides its opposite side in the ratio of the sides constituting to the angle.
Proposition : In ABC the line segment AD bisects the internal angle A and intersects the side BC at D.
It is required to prove that BD : DC = BA : AC.
Construction: D the line segment CE parallel to raw DA, so that it intersects the side BA produced at E.
Proof:
S teps J stificaltin u b
(1 ince
)S
DA ll CE and both BC and AC are their [ y construction]
[ orresponding angles] c transversal
[ lternate angles] a and AEC
BAD
[ upposition] s and ACE
CAD
[Theorem ]
1
(2) ut
B
BAD
CAD
[ tep (2)] s AEC
ACE ;
AC AE
(3) Again, since DA ll CE ,
BD
BA
DC
AE
(4) ut AE AC
B
BD
BA
DC
AC
Theorem 4
If any side of a triangle is divided internally, the line segment from the point of division to the opposite vertex bisects the angle at the vertex.

Math

232

Proposition : et ABC be a triangle and the line segment AD
L
from vertex A divides the side BC at D such that BD : DC =
BA : AC. It is required to prove that AD bisects BAC,
i.e. BAD = CAD.
Construction : raw at C the line segment CE parallel to
D
DA, so that it intersects the side BA produced at E.
Proof:
S teps J stificaltin u [ y construction] b (1 In
)
BCE , DA ll CE
[Theorem ]
1
BA t AE BD t DC
[supposition ]
(2) ut BD t DC BA t AC
B
[rom steps (1 and (2) ] f )
BA t AE BA t AC
[ se angles of isosceles are equal] a B
AE AC
Therefore, ACE
[ rresponding angles ]
C
o
AEC
(4) ut
B
[ lternate angles ] a AEC
BAD
and ACE
[rom step (2) ] f CAD
Therefore, BAD
CAD
i.e., the line segment AD bisects BAC .

Exercise 14⋅1
.
1

The bisectors of two base angles of a triangle intersect the opposite sides at X and Y respectively. If Y parallel to
X
is the base, prove that the triangle is an isosceles triangle.
2. Prove that if two lines intersect a few parallel lines, the matching sides are proportional. 3. Prove that the diagonals of a trapezu m are divided in the same ratio at their i point of intersection.
4. Prove that the line segment oining the mid points of oblique sides of a j trapezum and two parallel sides are parallel. i . The medians AD and BE of the triangle ABC intersects each other at G. A line
5
segment is drawn through G parallel to DE which intersects AC at F. Prove that
AC = EF.
6

.
6
.
7

In the triangle ABC, X is any point on BC and O is a point on AX. Prove that
AOB t AOC BX t XC
In the triangle ABC, the bisector of A intersects BC at D. A line segment drawn parallel to BC intersects AB and AC at E and F respectively. Prove that
BD : DC = BE : CF.

Math

8.

233

If the heights of the equiangular triangles ABC and DEF are AM and DN respectively, prove that AM : DN = AB : DE.

14⋅2 Similarity
The congruence and similarity of triangles have been discussed earlier in class VII. In general, congruence is a special case of similarity. If two figures are congruent, they are similar ; but two similar triangles are not always congruent.
Equiangular Polygons:
If the angles of two polygons with equal number of sides are sequentially equal, the polygons are known as equiangular polygons.

Equiangular triangle

Equiangular quadrilateral

Similar Polygons:
If the vertices of two polygons with equal number of sides can be matched in such a sequential way that
(i) The matching angles are equal
(ii) The ratios of matching sides are equal, the two polygons are called similar polygons. In the above figures, the rectangle ABCD and the square PQRS are equiangular since the number of sides in both the figures is 4 and the angles of the rectangle are sequentially equal to the angles of the square (all right angles). Though the similar angles of the figure are equal, the ratios of the matching sides are not the same. Hence the figures are not similar. In case of triangles, situation like this does not arise. As a result of matching the vertices of triangles, if one of the conditions of similarity is true, the other condition automatically becomes true and the triangles are similar. That is, similar triangles are always equiangular and equiangular triangles are always similar.
If two triangles are equiangular and one of their matching pairs is equal, the triangles are congruent. The ratio of the matching sides of two equiangular triangles is a constant. Proofs of the related theorems are given below.

234

Theorem 5
If two triangles are equiangular, their matching sides are proportional.
Proposition : et ABC and DEF be triangles with
L
A = D, B = E and C = F. We need to
AB AC BC prove that
DE DF EF
Construction: onsider the matching sides of the
C
triangles ABC and DEF unequal. Take two points P and Q on AB and AC respectively so that AP = DE and AQ = DF. J in P and Q and complete the o construction.
Proof:
Steps
J stificaltin u (1) In the triangles APQ EF and D
D
AP DE , AQ DF , A
[SAS theorem]
Therefore, APQ
DEF
ence,
H
APQ
DEF
ABC and AQP
DFE
ACB.
That is, the corresponding angles produced as a result of intersections of AB and AC by the line segment PQ are equal.
AB AC
Therefore,
or,
PQ ll BC ;
AP AQ
[heorem 1] t AB AC
.
DE DF
(2) Similarly, cutting line segments EDand EF from A and C
B
B respectively, it can be shown that
BA BCEF
i.e.,
ED
[heorem 1] t AB BC
AB AC
BC
i.e.,
.
;
DE EF
DE DF
EF
The proposition opposite of theorem 5 also true. is Theorem 6
If the sides of two triangles are proportional, the opposite angles of their matching sides are equal.

Math

235

Math

Proposition : et in ABC and DEF ,
L
AB AC BC
.
DE DF EF
It is to prove that,
A
D, B
E, C
F.
Construction: onsider the matching sides of the
C
triangles ABC and DEF unequal. Take two points
P and Q on AB and AC respectively so that AP =
DE and AQ = DF. J in P and Q. o Proof:
Steps
AB AC
AB AC
(1) Since
, so,
.
DE DF
AP AQ
Therefore, PQ ll BC
ABC
APQ and ACB
AQP
Triangles ABC and APQ are equiangular.
AB BC
AB BC
Therefore,
, so,
.
DE AQ
AP PQ
AB BC
BC BC
[ upposition]; s DE EF
EF PQ
EF PQ
Therefore, APQ and DEF are congruent.

J stificaltin u [ heorem 2]
T
[ rresponding angles made
C
o by the transversal AB
]
[ rresponding angles made
C
o by the transversal AC
]
[ heorem ]
T
5

[SSS Theorem]
PAQ
EDF , APQ
DEF . AQP
DFE ,
APQ
ABC and AQP
ACB
A
D, B
E, C
F.
Theorem 7
If one angle of a triangle is equal to an angle of the other and the sides adjacent to the equal angles are proportional, the triangles are similar.
Proposition : et in
L
ABC and DEF, A = D
AB AC
.
and
DE DF
It is to be proved that the triangles ABC and DEF are similar.
Construction: onsider the matching sides
C
of ABC and DEF unequal. Take two points P and

236

Math

Q on AB and AC respectively so that AP = DE and AQ = DF. J in P and Q. o Proof:
Steps
(1) In APQ and DEF , AP DE , AQ DF and included A included D
ABC
DEF
A
D, APQ
E , AQP
F.
(2) Again,
AB AC
AB AC
DE DF , so AP AQ since J stificaltin u [ AS Theorem]
S

[ heorem 2]
T

PQ ll BC
APQ and ACB
AQP
Therefore, ABC
A
D, B
E and C
F
i.e., triangles ABC and DEF are equiangular.
Therefore ABC and DEF are similar.
Theorem 8
The ratio of the areas of two similar triangles is equal to the ratio of squares on any two matching sides.
Proposition : et the triangles ABC and DEF
L
be similar and BC and EF be their matching sides respectively. It is required to prove that
ABC t DEF BC 2 t EF 2
Construction: raw perpendiculars AG and
D
DH on BC and EF respectively. et AG h,
L
DH p.
Proof:
Steps
J stificaltin u 1
1
(1) ABC
BC.h and DEF
EF . p
2
2
ABC
DEF

1
2
1
2

BC.h
EF . p

h.BC h
=
p.EF p BC
EF

(1) ut in the triangles ABG and DEG,
B
E,

AGB

1
DHE (= right angle)

B

237

Math

BAG
EDH
(3) because ABC and DEF are similar. h AB BC p DE EF

ABC
DEF

h p BC
EF

BC
EF

BC
EF

[ riangles AB
T
C and EF
D
are similar]

BC 2
EF 2

14⋅1 Internal Division of a Line Segment in definite ratio
If A and B are two different points in a plane and m and n are two natural numbers, we acknowledge that there exits a unique point X lying between A and B and AX : XA
= m : n.
In the above figure, the line segment AB is divided at X internally in the ratio m : n,
i.e. AX : XB = m : n.
Construction 1
To divide a given line segment internally in a given ratio. et the line segment AB be divided internally in the
L
ration m : n.
Construction: et an angle
L
BAX be drawn at A. From
AX cut the lengths AE = m and EC = n sequentially. o J in B, C. At E, draw line segment ED parallel to CB which intersects AB at D. Then the line segment AB is divided at D internally in the ratio m : n.
Proof: Since the line segment DE is parallel to a side
BC of the triangle ABC
AD : DB = AE : EC = m : n.
Activity :
1. ivide a given line segment in definite ra tio internally by an alternative method.
D
Example 1. ivide a line segment of length 7 internally in the ratio 3.2.
D
cm
Solution: raw any ray AG. From AG, cut a
D
line segment AB =7cm. raw an angle
D
BAX at A.
From AX, cut the lengths AE =3 cm and EC =
2 cm. from EX. J in B, C. At E, draw an o angle AED equal to ACB whose side intersects AB at D. Then the line segment AB is divided at D internally in the ratio 3:2.

238

Math

Exercise 14⋅2
1.

onsider the following information:
C
i. ratios are considered to compare two expressions ii. to find ratio, expressions are measured in the same unit iii. to find ratio, expressions must be of the same type.
Which case of the following is true?
a. i and ii
b. ii and iii
c. i and iii
d. i, ii and iii

se the information from the above figure to answer the questions 2 and 3.
U
What is the ratio of the height and base of the triangle ABC?
4
2
5
1
b.
c.
d.
a.
5
5
4
2
3. What is the area of triangle ABD in sq. units?
a. 6
b. 20
c. 40
d. 5
0
4. In triangle ABC, if PQ | BC, which of the following is true?
a. AP : PB = AQ : QC
b. AB : PQ = AC : PQ
c. AB : AC = PQ : BC
d. PQ : BC = BP : BQ
. In a square how many lines of symmetry are there?
5
a. 10
b. 8
c. 6
d. 4
. Prove that if each of the two triangle s is similar to a third triangle, they are
6
congruent to each other.
. Prove that, if one acute angle of a right angled triangle is equal to an acute
7
angle of another right angled triangle, the triangles are similar.
8 Prove that the two right angled triangles formed by the perpendicular from the
.
vertex containing the right angle are similar to each other and also to the original triangle.
.
9 In the adjacent figure,
B = D and
CD = AB. Prove that BD = BL.
4
5
2.

A line segment drawn through the vertex A of the parallelogram ABCD intersects the
BC and DC at M and N respectively. Prove that BM DN is a constant.
11. In the adjacent figure, BD AC and
1
DQ BA 2 AQ
QC . BD 5 BL. Prove that, DA
DC .
2

10.

Math

239

12. In the triangles ABC and DEF A = D. Prove that,
ABC t DEF AB. AC t DE .DF .
13. The bisector AD of A of the triangle ABC intersects BC at D. The line segment CE parallel to DA intersects the line segment BA extended.
a. raw the specified figure.
D
b. Prove that ,BD : DC = BA : AC.
c. If a line segment parallel to BC intersect AB and AC at P and Q respectively, prove that BD : DC = BP : CQ.
14. In the figure, ABC and DEF are two similar triangles.
a. ame the matching sides and
N
matching angles of the triangles.
b. Prove that,
ABC
DEF

AB 2
DE 2

AC 2
DF 2

BC 2
EF 2

BC 3 and ABC = sq cm,
3
AB 2 draw the triangle DEF and find its area.
14.4 Symmetry
Symmetry is an important geometrical concept, commonly exhibited in nature and is used almost in every field of our activity. Artists, designers, architects, carpenters always make use of the idea of symmetry. The tree-eaves, the flowers, the beehives, l houses, tables, chairs - everywhere we find symmetrical designs. A figure has line symmetry, if there is a line about which the figure may be folded so that the two parts of the figure will coincide.

c. If BC

3 cm , EF

8 cm , B

0
6 ,

Each of the above figures has the line of symmetry. The last figure has two lines of symmetry
Activity:
1. Sumi has made some paper- ut design as shown in the c adjacent figure. In the figure, mark the lines of symmetry. ow H many lines of symmetry does the figure have?
2. Write and identify the letters in English alphabet having line symmetry. Also mark their line of sysmmetry.
14.5 Lines of Symmetry of Regular Polygons
A polygon is a closed figure made of several line segments. A polygon is said to be regular if all its sides are of equal length and all its angles are equal. The triangle is a

Math

240

polygon made up of the least number of line segments. An equilateral triangle is a regular polygon of three sides. An equilateral triangle is regular because its sides as well as angles are equal. A square is the regular polygon of four sides. The sides of a square are equal and each of the angles is equal to one right angle. Similarly, in regular pentagons and hexagons, the sides are equal and the angles are equal as well.

Each regular polygons is a figure of symmetry. Therefore, it is necessary to know their lines of symmetry. Each regular polygon has many lines of symmetry as it has many sides.
Three lines of
Four lines of
Five lines of
Six lines of symmetry symmetry symmetry symmetry

Equilateral triangle

Square

Regular pentagon

Regular hexagon

The concept of line symmetry is closely related to mirror reflection. A geometrical figure, has line symmetry when one half of it is the mirror image of the other half.
So, the line of symmetry is also called the reflection symmetry.

Exercise 14.3
1. Which of the following figures have line symmetry?
(a) A house (b) A mosque (c) A temple (d) A church
(f) Parliament house (g) The Tajmahal.
2. The line of symmetry is given, find the other hole:

(e) A pagoda

3. In the following figures, the line of symmetry is given; complete and identify the figure. Math

241

4. Identify the lines of symmetry in the following geometrical figures:

5. Complete each of the following incomplete geometrical shapes to be symmetric about the mirror line:

6. Find the number of lines of symmetry of the following geometrical figures:
(a) An isosceles triangle
(b) A scalene triangle
(c) A square
(d) A rhombus
(e) A pentagon
(f) A regular hexagon
(g) A circle
7. Draw the letters of the English alphabet which have reflection symmetry with respect to
(a) a vertical mirror
(b) a horizontal mirror
(c) both horizontal and vertical mirrors.
8. Draw three examples of shapes with no line of symmetry.
14.6 Rotational Symmetry
When an object rotates around any fixed point, its shape and size do not change. But the different parts of the object change their position. If the new position of the object after rotation becomes identical to the original position, we say the object has a rotational symmetry. The wheels of a bicycle, ceiling fan, square are examples of objects having rotational symmetry etc.. As a result of rotation the blades of the fan looks exactly the same as the original position more than once. The blades of a fan may rotate in the clockwise direction or in the anticlockwise direction. The wheels of a bicycle may rotate in the clockwise direction or in the anticlockwise direction. The rotation in the anti clockwise direction is considered the positive direction of rotation.
This fixed point around which the object rotates is the centre of rotation. The angle of turning during rotation is called the angle of rotation. A full-turn means rotation by 360°; a half-turn is rotation by 180°.
In the figure below, a fan with four blades rotating by 90° is shown in different positions. It is noted that us a fall turn of the four positions (rotating about the angle by 90°, 180°, 270° and 360°), the fan looks exactly the same. For this reason, it is said that the rotational symmetry of the fan is order 4.

24
2

ath
M

Here is one more example for rotational symmetry. onsider the intersection of two
C
diagonals of a square the centre of rotation. In the quarter turn about the centre of the square, any diagonal position will be as like as the second figure. In this way, when you complete four quarter-urns, the square re aches its original position. It is said t that a square has a rotational symmetry of order 4.

bserve also that every object occupies sa me position after one complete revolution.
O
S every geometrical object has a rotationa l symmetry of order 1. uch cases have o S no interest for us. For finding the rotational symmetry of an object, one need to consider the following matter.
(a) The centre of rotation
(b) The angle of rotation
(c) The direction of rotation
(d) The order of rotational symmetry.
Activity:
1.G examples of 5 plane objects from your surroundings which have ive rotational symmetry.
2. Find the order of rotational symmetry of the following figures.

14⋅7 Line Symmetry and Rotational Symmetry
Whave seen that some geometrical shapes have only line symmetry, some have e only rotational symmetry and some have both line symmetry and rotational symmetry. For example, the square has four lines of symmetry as well rotational symmetry of order .
4
The circle is the most symmetrical figure, because it can be rotated around its centre through any angle. Therefore, it has unlimited order of rotational of symmetry. At the same time, every line through the centre forms a line of reflection symmetry and so it has unlimited number of lines of symmetry.

24
3

ath
M

Activity:
1. Determine the line of symmetry and the rotational symmetry of the given alphabet and complete the table below:
Letter
Line of
Number of
Rotational
Order of symmetry lines of symmetry rotational symmetry symmetry
Z
o
N
0 es Y
2
H
O
E
C

Exercise 14⋅4
1. Find the rotational symmetry of the following figures:
(c)
(d)
(a)
(b)

(e

(f)

2. W you slice a lemon the cross- ection looks as shown in hen s the figure. Determine the rotational symmetry of the figure.

.
3 Fill in the blanks:
Shape

Centre of
Rotation

Order of
Rotation

Angle of
Rotation

quare
S
R ectangle Rombus h quilateral
E
triangle
Smi- ircle e c egular pentagon
R
4 N
. ame the quadrilaterals which have line of symmetry and rotational symmetry of order more than 1.
5. an we have a rotational symmetry of a body of order more than 1 whose angle
C
of rotation is 18 stify your answer.
?J
°u

Chapter Fifteen

Area Related Theorems and Constructions
Wknow that bounded plane figures may have different shapes. If the region is e bounded by four sides, it is known as quadrilateral. Quadrilaterals have classification and they are also named after their shapes and properties. Apart from these, there are regions bounded by more than four sides. These are polygonal regions or simply polygons. The measurement of a closed region in a plane is known as area of the region. For measurement of areas usually the area of a square with sides of 1 unit of length is used as the unit area and their areas are expressed in square units. For example, the area of B angladesh is 1.4
4
lacs square kilometres (approximately).
Thus, in our day to day life we need to know and measure areas of polygons for meeting the necessity of life. S, it is important for the learners to have a o comprehensive knowledge about areas of polygons. Areas of polygons and related theorems and constructions are presented here.
At the end of the chapter, the students will be able to xplain the area of polygons
E
erify and prove theorems related to areas
V
Cnstruct polygons and justify construction by using given data o onstruct a quadrilateral with area equal to the area of a triangle
C
onstruct a triangle with area equal to the area of a quadrilateral
C
15⋅1 Area of a Plane Region very closed plane region has definite area. In order to measure such area, usually
E
the area of a square having sides of unit length is taken as the unit. For example, the area of a square with a side of length 1 cm. is 1 square centimetre.
W
e know that,
(a) In the rectangular region ABCD if the length AB = a units (say, metre), breadth
BC = b units (say, metre), the area of the region ABCD = ab square units (say, square metres).
(b) In the square region ABCD if the length of a side AB a units (say, metre), the
=
area of the region ABCD = a2 square units (say, square metres).

Math

245

When the area of two regions are equal, the sign ‘=’ is used between them. For example, in the figure the area of the rectangular region ABCD = Area of the triangular region
AED.
It is noted that if ABC and DEF are congruent, we write ABC
DEF . In this case, the area of the triangular region ABC = area of the triangular region DEF.
But, two triangles are not necessarily congruent when they have equal areas. For example, in the figure, area of ABC = area of DBC but ABC and DBC are not congruent. Theorem 1
Areas of all the triangular regions having same base and lying between the same pair of parallel lines are equal to one another.

Let the triangular regions ABC and DBC stand on the same base BC and lie between the pair of parallel lines BC and AD. It is required to prove that, region ABC = region DBC.
Construction : At the points B and C of the line segment BC, draw perpendiculars
BE and CF respectively. They intersect the line AD or AD produced at the points E and F respectively. As a result a rectangular region EBCF is formed.
Proof : According to the construction, EBCF is a rectangular region. The triangular region ABC and rectangular region EBCF stand on the same base BC and lie between
1
the two parallel line segments BC and ED. Hence, region ABC =
(rectangular
2 region EBCF)
1
Similarly, region DBC = (rectangular region EBCF)
2
region ABC = region DBC (proved).

24
6

ath
M

Theorem 2
Parallelograms lying on the same base and between the same pair of parallel lines are of equal area.

Let the parallelograms regions ABCD and EFGH stand on the same base and lie between the pair of parallel lines AF and DG and AB = EF. It is required to prove that, area of the parallelogram ABCD = area of the parallelogram EFGH.
Construction :
The base EF of EFGH is equal. J in AC and EG. From the points C and G, draw o perpendiculars CL and GK to the base AF respectively.
1
1
Proof: The area of ABC = 2 AB CL and the area of EFG is EF GK .
2
AB EF and CL = GK (by construction)
Therefore, area of ABC = area of the triangle EFG
1
1 area of the parallelogram ABCD = area of the parallelogram EFGH
2
2
Area of the parallelogram ABCD = area of the parallelogram EFGH. (P roved) Theorem 3 (Pthagoras Theorem) y In a right angles triangle, the square of the hypotenuse is equal to the sum of squares of other two sides.
Proposition: Let ABC be a right angled triangle in which
ACB is a right angle and hypotenuse is AB. It is to be proved that AB 2 BC 2 AC 2 .
Construction: Draw three squares ABED, ACGF and
BCHK on the external sides of AB, AC and BC respectively. Through C, draw the line segment CL parallel to AD which intersects AB and DE at M and L respectively. J in C, D and B, F. o Proof: teps S
J stification u (1) In CAD and FAB , CA = AF , AD = AB and included CAD
CAB
BAD
[ BAD =
CAF = 1
= CAB right angle]
CAF
= included BAF

ath
M

24
7

Therefore, CAD
FAB
(2) Triangular region CAD and rectangular
[S
S theorem]
A
region ADLM lie on the same base AD and between the parallel lines AD and CL .
Therefore, ectangular region ADLM = 2(triangular
R
region CAD )
[ heorem 1]
T
(3 Triangular region BAF and the square ACGF lie on the
)
same base AF and between the parallel lines
AF and BG .
Hence quare region ACGF = 2(triangular region FAB ) = [ heorem 1]
S
T
2(triangular region CAD )
(4 R
) ectangular region
ADLM = square region ACGF
[ rom (2) and(3]
F
)
(5) S imilarly joining C, E and A, K , it can be proved that rectangular region BELM = square region BCHK
(6 R
) ectangular region ( ADLM + BELM )= square
[ rom (4 and (5)]
F
) square region BCHK region ACGF + or, square region ABED = square region ACGF + square region BCHK
That is, AB 2 BC 2 AC 2 [oved]
P
r
Construction 1
Construct a parallelogram with an angle equal to a definite angle and area equal to that of a triangular region.

Let ABC be a triangular region and x be a definite angle. It is required to construct a parallelogram with angle equal to x and area equal to the area of the triangular region ABC.
Construction:
isect the line segment BC at E. At the point E of the line segment EC, construct
B
CEF equal to x . Through A, construct AG parallel to BC which intersects the ray EF at F. Again, through C, construct the ray CG parallel to EF which intersects the ray AG at G. Hence, ECGF is the required parallelogram.
Proof: J in A, E. ow, area of the triangular region ABE = area of the triangular o N region AEC [since base BE = base EC and heights of both the triangles are equal]
.
area of the triangular region ABC = 2 (area of the triangular region AEC).

24
8

ath
M

Again, area of the parallelogram region ECGF is 2 (area of the triangular region
AEC) [since both lie on the same base and EC ll AG ). area of the parallelogram region ECGF = area of the triangular region ABC.
Again CEF = x [ ince EF|| CB by construction] s .
S the parallelogram C is the required parallelogram. o E
F
G
Construction 2
Construct a triangle with area of the triangular region equal to that of a quadrilateral region.

Let ABCD be a quadrilateral region. To construct a triangle such that area of the is triangular region equal to that of a rectangular region ABCD.
Construction :
J in D, B. Through C, construct CE parallel to DB which intersects the side AB o extended at E. J in D, E. Then, DAE is the required triangle. o Proof: The triangles BDC and BDE lie on the same base BD and DB ll CE (by construction). area of the triangular region BDC = area of the triangular region BDE. area of the triangular region BDC + area of the triangular region ABD = area of the triangular region BDE + area of the triangular region ABD area of the quadrilateral region ABCD = area of the triangular region ADE .
Therefore, ADE is the required triangle.
N.B. Applying the above mentioned method innumerable numbers of triangles can be drawn whose area is equal to the area of a given quadrilateral region.
Construction 3
Construct a parallelogram, with a given an angle and the area of the bounded region equal to that of a quadrilateral region.

Let ABCD be a quadrilateral region and x be a definite angle. It is required to x and the area equal to area of the construct a parallelogram with angle quadrilateral region ABCD.

ath
M

24
9

Construction:
J in B, D. Through C construct CF parallel to DB which intersects the side AB o extended at F. Find the midpoint G of the line segment AF. At A of the line segment
AG, construct GAK equal to x and draw GH ll AK through G. Again draw
KDH||AG through D which intersects AK and GH at K and H respectively.
Hence AGHK is the required parallelogram.
Proof: J in D, E. B construction AGHK is a parallelogram. o y where GAK x . Again, area of the triangular region DAF = area of the rectangular region ABCD and area of the parallelogram AGHK = area of the triangular region DAF. Therefore, AGHK is the required parallelogram.

Exercise 15
1. The lengths of three sides of a triangle are given, in which case below the construction of the right angled triangle is not possible ?
(a) 3 4 5 cm cm, cm,
(b) 6 8 10 cm, cm, cm
(c) 5 cm, 7 9 cm, cm
(d) 5 cm, 12 cm, 13 cm 2. O bserve the following information :
i. ach of the bounded plane has definite area.
E
ii. If the area of two triangles is equal, the two angles are congruent. iii. If the two angles are congruent, their area is equal.
W one of the following is correct ? hich (a) i and ii
(b) i and iii
(c) ii and iii
(d) i, ii and iii
Answer question no. 3 and 4 the basis of the information in the figure below, on ABC is equilateral, AD BC and AB = 2 :

BD = W ? hat (c) 2
(d) 4
(a) 1
(b) 2
. W is the height of the triangle ?
4 hat
2
4
(a)
sq. unit (b) 3 sq. unit
(c)
sq. unit
(d) 2 3 sq. unit.
3
3
5. P rove that the diagonals of a parallelo gram divide the parallelogram into four equal triangular regions.
. rove that the area of a square is half the area of the square drawn on its
6 P diagonal. 3
.

250

ath
M

. rove that any median of a triangle divides the triangular region into two regions
7 P of equal area.
. A parallelogram and a rectangular region of equal area lie on the same side of the
8
bases. how that the perimeter of the pa rallelogram is greater than that of the
S
rectangle.
. X and Y are the mid points of the sides AB and AC of the triangle ABC. rove tha t
9
P
1
the area of the triangular region AXY = (area of the triangular region ABC)
4
10 In the figure, ABCD is a trapezum with sides AB and CD parallel. Find the area
.
i of the region bounded by the trapezum ABCD. i 11. P is any point interior to the parallelogram ABCD. P rove that the area of the
1
triangular region PAB + area of the triangular region PCD = (area of the the 2 parallelogram ABCD).
12. A line parallel to BC of the triangle ABC intersects AB and AC at D and E respectively. P rove that the area of the triangular region DBC = area of the triangular region EBC and area of the triangular region DBF = area of the triangular region CDE.
13 A 1 right angle of the triangle ABC. D is a point on AC. rove tha t
.
P
BC 2 AD 2 BD 2 AC 2 .
14 ABC is an equilateral triangle and AD is perpendicular to BC. rove tha t
.
P
4 AD 2 3 AB 2 .
15. ABC is an isosceles triangle. BC is its hypotenuse and P is any point on BC. rove that PB 2 PC 2 2PA 2 .
P
16 C is an obtuse angle of ABC; AD is a perpendicular to BC. Sow that.
.
h
AB2 = AC2 +
C 2 +C
B
2B
.C
D
17 C is an acute angle of ABC ; AD is a perpendicular to BC. Sow that
.
h
AB 2 AC 2 BC 2 2 BC.CD .
18 AD is a median of ABC. Sow that, AB 2 AC 2 2( BD 2 AD 2 ) .
.
h

Chapter Sixteen

Mensuration
The length of a line, the area of a place, the volume of a solid etc. are determined for practical purposes. In the case of measuring any such quantity, another quantity of the same kind having some definite magnitude is taken as unit. The ratio of the quantity measured and the unit defined in the above process is the amount of the quantity.
Quantity measured
i.e. magnitude = Unit quantity
In the case of a fixed unit, every magnitude is a number which denotes how many times of the unit of the magnitude is the magnitude of the quantity measured. For example, the bench is 5 metre long. Here metre is a definite length which is taken as a unit and in comparison to that the bench is 5 times in length.
At the end of the Chapter, the students will be able to −
Determine the area of polygonal region by applying the laws of area of triangle and quadrilateral and solve allied problems.
Determine the circumference of the circle and a length of the chord of a circle.
Determine the area of circle.
Determie the area of circle. Determining the area of a circle and its part, solve the alied problems.
Determine the area of solid rectangles, cubes and cylinder and solve the allied problems. Determine the area of uniform and non uniform polygonal regions.
16⋅1 Area of Triangular region
In the previous class, we learned that area of triagular region =

1 base 2

height.

(1) Right Angled Triangle :
Let in the right angled triangle ABC , BC a and AB b are the adjacent sides of the right angle. Here if we consider BC the base and AB the height,
1
Area of ABC = base height
2
1
= ab
2
(2) Two sides of a triangular region and the angle included between them are given. 252

Math

et in
L
ABC the sides are :
BC a , CA b , AB c .
AD is drawn perpendicular from A to BC . et altitude height) AD h .
L
(
AD
o
Cnsidering the angle Cwe get,
,
sin C
CA
h or, sin C or, h b sin C b 1
Area of ABC = BC AD
2
1
= a b sin C
2
1
= absin C
2
1
S
imilarly, area of
ABC = bc sin A
2
1
= ca sin B
2
(3) Three sides of a triangle are given. et in
L
ABC , BC a , CA b and AB c . erimeter of the triangle 2 s a b c
P
W e draw AD BC et, BD x , so CD a x
L
In right angled ABD and ACD
AD2 AB2 BD2 and AD2 AC 2 CD2
AB2 BD2 AC 2 CD2 or, c2 x2 b2 (a x)2 or, c 2 x 2 b 2 a 2 2ax x 2 or, 2ax c2 a2 b2 c 2 a 2 b2 x 2a
Again, AD2 c2 x2
=c

2

= c

c 2 a 2 b2
2a

c2

a 2 b2
2a

2

c

c2

a 2 b2
2a

253

Math

2ac c2 a2 b2 2ac c2 a2 b2
2a
2a
{ c a)2 b2} b2 (c a)2}
(
{
=
4a2
(a b c( a b c 2b( a b c 2a( a b c 2c)
)
)
)
=
4a2
2s(2s 2b( 2s 2a( 2s 2c)
)
)
=
2
4a
4s(s a( s b( s c)
)
)
=
a2
=

2 s(s a( s b( s c)
)
) a 1
Area of ABC = BC AD
2
1
2
= a s ( s a ( s b( b c )
)
) a 2
= s ( s a ) s b) b c )
(
(
AD

(4) Equilateral Triangle : et the length of each side of the equilateral triangular region ABC be a .
L
a
Draw AD BC
BD CD
2
In right angled ABD
BD2 AD2 AB2 or, AD2
AD

Area of

AB2 BD2

a2

a
2

2

a2

a2
4

3a 2
4

3a
2
ABC =

1
BC AD
2

1
3a
3 2 or, a a 2
2
4
(5) Isosceles triangle : et ABC be an isosceles triangle in which AB
L
and BC b

=

AC

a

254

Math

Draw AD
In

BC .

b
2

BD CD

ABD right angled

AD2

AD

AB 2

BD2

a2

b
2

2

a2

b2
4

4a 2 b2
4

4a2 b2
2

Area of isosceles

ABC =

1
BC AD
2

1
4a2 b2 b 2
2
b
2
=
4a b2
4
Example 1. The lengths of the two sides of a right angled triangle, adj cent to right a angle are 6 cm. and 8 cm. respectively. Find the area of the triangle.
Solution : et, the sides adj cent to right angle are
L
a a 8 cm. and b 6 cm. respectively. 1
Its area = ab
2
1
=
8 6 square cm. = 24 square cm.
2
equired area 24 square cm.
R
Example 2. The lengths of the two sides of a triangle are 9 cm. and 10 cm. respectively and the angle included between them is 60 . Find the area.
Solution : et, the sides of triangle are a 9 cm. and b 10 cm. respectively.
L
Their included angle θ 60 .
1
Area of the triangle = ab sin 60o
2
=

3
1
sq. cm.
9 10
2
2
(
= 38 97 sq. cm. approx) equired area 38 97 sq. cm. approx)
R
(
Exmple 3. The lengths of the three sides of a triangle are 7 cm., 8 cm. and 9 cm. respectively. Find its area.

=

Math

255

Solution : et, the lengths of the sides of the triangle are a
L
c 9 cm. a b c 7 8 9 cm. = 12 cm. emi perimeter s
S
2
2
Its area
= s(s a) s b) s c)
(
(

7 cm., b 8 cm. and

= 12(12 7) 12 8) 12 9) sq. cm.
(
(
(
= 12 5 6 7 sq. cm. = 50 2 sq. cm. approx)
(
The area of the triangle is 50 2 sq. cm. approx)
Eample 4. The area of an equilateral triangle increases by 3 3 sq. metre when the x length of each side increases by 1 metre. Find the length of the side of the triangle.
Solution : et, the length of each side of the equilateral triangle is a metre.
L
Its area =

3 2 a sq. m.
4

The area of the triangle when the length of each side increases by 1m. =

3
(a 1)2
4

sq. metre.
According to the question,

3
(a 1)2
4

3 2 a 4

3 3

3
]
4 or, a2 2a 1 a2 12 or, 2a 11 or, a 5 5
The required length is 5 5 metre.
Example 5. The length of the base of an isosceles triangle is 60 cm. If its area is
1200 sq. metre, find the length of equal sides.
Solution : et the base of the isosceles triangle be b 60 cm. and the length of
L
equal sides be a . b Area of the triangle =
4a2 b2
4
b
According to the question,
4a2 b2 1200
4
60 or, 4a2 (60)2 1200
4

or, (a 1)2

a2

d
12 ;[ ividee by

or, 15 4a2 3600 or, 4a2 3600

1200
80

Math

256

or, 4a2 3600 6400 ;[ y squaring] b or, 4a2 10000 or, a2 2500 a 50
The length of equal sides of the triangle is 50 cm.
Example 6. From a definite place two roads run in two directions mak are angle ing 120o . From that definite place, persons move in the two directions with speed of 10 m per hour and 8 m per hour respectiv ely. hat will be the direct distance k k
W
between them after 5 hours ? k/ m/
Solution : et two men start from A with velocities 10 mhour and 8 k hour
L
respectively and reach B and C after 5 hours. Then after 5 hours, the direct distance between them is BC . From C perpendicular CD is drawn on BA produced. k k k k
AB 5 10 m = 50 m, AC 5 8 m = 40 m. and BAC 120o
DAC = 180 120 = 60
From the right angled triangle ACD o CD
3
o sin 60o or, CD AC sin 60o 40
20 3
2
AC
1
AD and 20 cos 60o or, AD AC cos 60o 40
AC
2
Again, we get from right angled BCD ,
BC 2 BD2 CD2 (BA AD)2 CD2
= (50 20)2 (20 3)2 4900 1200 6100
(
BC 78 1 app.) k ( equired distance is 78 1 m. approx)
R
Example 7. The lengths of the sides of a triangle are 25, 20, 15 units respectively.
Find the areas of the triangles in which it is divided by the perpendicular drawn from the vertex opposite of the greatest side.
Solution : et in triangle ABC , BC 25 units, AC 20 units, AB 15 units.
L
The drawn perpendicular AD from vertex A on side BC divides the triangular region into ABD and ACD . et BD x and AD h
L
A
CD BC BD 25 x
In right angle ABD
BD2 AD2 AB2 or, x2 h2 (15)2 x2 h2 225..........(i) and ACD is right angled
CD2 AD2 AC 2 or, (25 x)2 h2 (20)2

Math

257

or, 625 50x x2 h2 400 or, 625 50x 225 400 ;[ ith the help of equation (i) ] w or, 50x 450 ; x 9 uting the value of x in equation (i) , we get,
P
81 h2 225 or, h2 144 h 12
1
1
Area of ABD = BD AD
9 12 square units = 36 square units
2
2
1
1 and area of ACD = BD AD
(25 9) 12 square units
2
2
1
=
16 12 square units = 96 square units
2
equired area is 36 square units and 96 square units.
R

Exercise 16⋅1
1. The hypotenuse of a right angled triangle is 25 m. If one of its sides is

3 th of the
4

other, find the length of the two sides.
2. A ladder with length 20m. stands vertically against a wall. How much further should the lower end of the end of the ladder be moved so that its upper end descends 4 metre?
5
3. The perimeter of an isosceles triangle is 16 m. If the length of equal sides is th
6
of base, find the area of the triangle.
4. The lengths of the two sides of a triangle are 25 cm., 27 cm. and perimeter is 84 cm. Find the area of the triangle.
5. W the length of each side of an eq uilateral triangle is increased by 2 metre, its hen area is increased by 6 3 square metre. Find the length of side of the triangle.
6. The lengths of the two sides of a triangle are 26 m., 28 m. respectively and its area is 182 square metre. Find the angle between the two sides.
11
7. The perpendicular of a right angled triangle is 6cm less than times of the
12
4 base, and the hypotenuse is 3 cm less than times of the base. Find the length
3
of the base of the triangle.
8. The length of equal sides of an isosceles triangle is 10m and area 48 square metre. Find the length of the base.

258

Math

9. Two roads run from a definite place with an angle of 135o in two dinrections.
Two persons move from the definite place in two directions with the speed of 7 m per hour and 5 m per hour respectiv ely. W will be the direct distance k k hat between them after 4 hours?
10. If the lengths of the perpendiculars from a point interior of an equilateral triangle to three sides are 6 cm., 7 cm., 8 cm. respectively;find the length of sides of the triangle and the area of the triangular region.
16⋅2 Area of Quadrilateral Region
(1) Area of rectangular region et, the length of AB a , breadth
L
BC b and diagonal AC d , of rectangle ABCD .
W ow, the diagonal of a rectangle divides. en k the rectangle into two equal triangular regions.
1
Area of the rectangle ABCD 2 area of ABC = 2 a b ab
2
perimeter of the rectangular region, s 2(a b) the ABC is right angled. d a 2 b2
AC 2 AB2 BC 2 or, d 2 a 2 b 2 ;
(1) Area of square region et the length of each side of a square ABCD
L
be a and diagonal d. The diagonal AC divides the square region into two equal triangular regions.
Area of square region ABCD = 2 area of ABC
1
=2 a a = a2
2
Observe that, the perimeter of the square region s 4a a2 a2
2a 2
2a
and diagonal d
(3) Area of parallelogram region.
(a) Base and height are given. et, the base AB b
L
and height DE h of parallelogram ABCD
The diagonal BD divides the parallelogram into two equal triangular regions.
The area of parallelogram ABCD = 2 area of

= 2 1b h
2

= bh

ABD

Math

259

(b) The length of a diagonal and the length of a perpendicular drawn from the opposite angular point on that diagonal are given. et, the diagonal be AC d and the perpendicular from opposite angular point D
L
on AC be DE h of a parallelogram ABCD . Diagonal AC divides the parallelogram into two equal triangular regions.
The area of parallelogram region ABCD = 2 area of ACD
1
=2 d h
2
= dh
(4) Area of Rhombus Region
Two diagonals of a rhombus region are given et the diagonals be AC d1 , BD = d2 of the rhombus ABCD and the diagonals
L
intersect each other at O . Diagonal AC divides the rhombus region into two equal triangular regions.
W ow that the diagonals of a rhombu s bisect each other at right angles. en k d Height of ACD = 2
2
Area of the rhombus region ABCD = 2 area of ACD d 1
=2
d1 2
2
2
1
= d1d2
2
(5) Area of trapezium region
Two parallel sides of trapezium region and the distance of perpendicular between them are given. et ABCD be a trapezium whose lengths of parallel sides are AB a unit, CD b
L
unit and distance between them be CE h . Diagonal AC divides the trapezium region ABCD into ABC and ACD .
Area of trapezium region ABCD
Area of the triangular region ACD .
= Area of the triangular region ABC +
1
1
= AB CE
CD AF
2
2
1
1
1
= ah bh = h(a b)
2
2
2

3 times of breadth. If the area is 384
2
square metre, find the perimeter and length of the diagonal.
Example 1. ength of a rectangular room is
L

260

Solution : et breath of the rectangular room is x metre
L
3x ength of the room is
L
metre
2
3x2
3x
square metre.
Area
x or,
2
2
3x2
According to the question,
384 or, 3x2 768 or, x2 256 ;
2
3 ength of the rectangular room =
L
16 metre = 24 metre
2
and breath = 16 metre.
It’ perimeter = 2 24 + metre = 80 metre. s ( 16)

Math

x 16 metre

and length of the diagonal = (24)2 (16)2 metre = 832 metre = 28 84 metre app.) ( .
The required perimetre is 80 metre and length of diagonal is 28 84 metre approx)
(
.
Example 2. The area of a rectangular region is 2000 square metre. If the length is reduced by 10 metre, it becomes a square region. Find the length and breadth of the rectangular region.
Solution : et length of the rectangular region be x metre and breadth y metre.
L
Area of the rectangular region = xy square metre
According to the question xy 2000................( 1) and x 10 y.............(2)
W from equation ( , e get,
2)
y x 10.............(3)
From equations 1)and ( , we get,
(
3) x( x 10) 2000 or, x2 10x 2000 0 or, x2 50x 40x 2000 0 or, ( x 50) x 40) 0
(
x 50 0 or, x 40 0 or, x 50 or, x
40
ut length can never be negative.
B
x 50
Now puting the value of x in equation 3, we get
()
y 50 10 40 length = 50 m. and breadth = 40 m.
Example 3. There is a road of 4 metre width inside around a square field. If the area of the road is 1 hector, determine the area of the field excluding the road.
Solution : et, the length of the square field is x metre.
L
Its area is x2 square metre.

261

Math

There is a road around the field with width 4 m. ength of the square field excluding the road = ( x 2 4) , or ( x 8) m
L
Area of the square field excluding the road is ( x 8)2 square m.
Area of the road = {x 2 ( x 8) 2 } square m.
W ow, 1 hector = en k
10000 square m.
According the question, x2 (x 8)2 = 10000 or, x 2 x 2 16 x 64 10000 or, 16 x 10064 x 629
Area of the square field excluding the road = (629 8)2 square m.
= 385641 square m.
= 38 56 hector approx.)
(
(
.
The required area is 38 56 hector approx.)
Example 4. The area of a parallelogram is 120 sq. cm. and length of one of its diagonal is 24 cm. Determine the length of the perpendicular drawn on that diagonal from the opposite vertex.
Solution : e a diagonal of a parallelogram be d 24 cm. and the length of the
L
perpendicular drawn on the diagonal from the opposite vertex be h cm.
Area of the parallelogram = dh square cm.
120 120
As per question, dh 120 or, h
5
d
24
The required length of the perpendicular is 5 cm.
Example 5. If the length of the sides of a parallelogram are 12 m. 8 m. If the length of the smaller diagonal is 10 m, determine the length of the other diagonal.
Solution : et, in the parallelogram ABCD ; AB a 12 m. and AD c 8 m. and
L
diagonal BD b 10 m. et us draw the perpendiculars DF and C
L
E from D and C on the extended part of AB , respectively. J in A, C and B, D . o 12 10 8
Smi perrimeter of e m. = 15 m.
ABD is s
2
Area of the triangular region ABD = s ( s a( s 6( s c)
)
)
= 15(15 12( 15 10( 15 8) sq. m.
)
)
= 1575 sq. m.
= 39 68 sq. m. approx.)
(
1
Again, area of the triangular region ABD = AB DF
2
1 or, 39 68
DF
12 DF or, 6 DF 39 68 ;
2

6 61

262

Math

N in right angled triangle ow, BCE ,
BE 2 BC 2 CE 2 AD2 DF 2 82 (6 61)2 20 31
BE 4 5
Therefore, AE AB BE 12 4 5 16 5
From right angled triangle BCE , we get,
AC 2 AE 2 CE 2 (16 5)2 (6 61)2 315 94
(
AC 17 77 approx.)
(
The required length of the diagonal is 17 77 m. approx.)
Example 6. The length of a diagonal of a rhombus is 10m. and its area is 120 sq. m.
Determine the length of the other diagonal and its perimeter.
Solution : et, the length of a diagonal of rhombus ABCD is BD d1 10 metre
L
and another diagonal AC d2 metre.
1
Area of the rhombus = d1d2 sq. m.
2
1
120 2 120 2
As per question, d1d2 120 or, d2
24
2
10
10
Wnow, the diagonals of rhombus bisect e k each other at right angles. et the
L
diagonals interset at the point 0.
10
24
m. = 5 m. and OA OC
m. = 12 m.
OD OB
2
2 and in right angled triangle AOD , we get
AD 13
AD 2 OA 2 OD 2 5 2 (12) 2 169 ;
The length of each sides of the rhombus is 13 m.
The perimeter of the rhombus = 4 13 m. = 52 m..
The required length of the diagonal is 24 m. and perimeter is 52 m.
Example 7. The lengths of two parallel sides of a trapezium are 91 cm. and 51 cm. and the lengths of two other sides are 37 cm and 13 cm respectively. Determine the area of the trapezium.
Solution : et, in trapezium ABCD ; AB 91 cm. CD 51 cm. et us draw the
L
L perpendiculars DF and CF on AB from D and C respectively.
CDEF is a rectangle.
EF CD 51 cm. et, AE x and DE CF h
L
BF AB AF 91 ( AE EF ) 91 ( x 51) 40 x
From right angled triangle ADE , we get,
AE 2 DE 2 AD 2 or, x2 h2 (13)2 or, x2 h2 169.........(i)
Again, from right angled triangle BCF , we get,
BF 2 CF 2 BC 2 or, (40 x) 2 h 2 (37) 2

Math

263

or, 1600 80x x2 h2 1369
[
() or, 1600 80x 169 1396 with the help of 1] or, 1600 169 1396 80x ;or, 80x 400 ; x 5
Now putting the value of x in equation 1, we get
()
52 h2 163 or, h 169 25 144 ; h 12
1
Area of ABCD = ( AB CD) h
2
1
= (91 51) 12 square cm.
2
= 852 square cm.
The required area is 852 square cm.
16⋅3 Area of regular polygon
The lengths of all sides of a regular polygon are equal. Again, the angles are also equal. egular polygon with n sides produces n isocsceles triangles by adding centre
R
to the vertices.
Area of the regular polygon = n area of one triangular region. et ABCDEF be a regular polygon whose centre is O.
L
It has n sides and the length of each side is a. e oin
W
j
O, A ; θ O, B.
M
et in
L
AOB height OM h and OAB θ
The angle produced at each of the vertices of regular polygon = 2θ .
Angle produced by n number of vertices in the polygon = 2θ n
Angle produced in the polygon at the centre = 4 right angles.
The sum of angles of n number of triangles = 2θ (n 4) right angles. um of 3 angles of
S
OAB = 2 right angles.
The wise, summation of the angles of n numbers of triangles = n 2 right angles
2θ (n 4) right angles = n 2 right angles or, 2θ n (2n 4) right angles
2n 4 right angles or, θ
2n
or, θ

1

or, θ

1

Now, tan θ

h a 2

2 right angles n 2 n 2h ; a 90 = 90 h a tan θ
2

180 n Math

264

Area of

1 ah 2
1
a tan θ
= a
2
2 a2 180
=
tan 90 n 4

OAB =

=

180 a2 cot
4
n

Area of a regular polygon having n sides =

180 a2 cot
.
4 n Example 8. If the length of each side of a regular pentagon is 4 cm, determine its area. Solution : et, length of each side of a regular pentagon is a 4 cm.
L
and number of sides n 5 a2 180 cot W ow, area of a regular polygon = en k n 4
2
4
180
cot
Area of the pentagon =
S. cm. q 4
5
= 4 cot 36o sq. cm.
= 4 1 376 sq. cm. [ ith the help of calculator] w = 5 506 sq. cm. approx.)
(
(
The required area = 5 506 sq. cm. approx.)
Example 9. The distance of the centre to the vertex of a regular hexagon is 4 m.
Determine its area. o Solution : et, ABCDEF is a regular hexagon whose centre is O , O is j ined to
L
each of the vertex and thus 6 triangles of equal area are formed.
360
COD
60
6 et the distance of centre O to its vertex is a m.
L
a = 4.
Area of
=

COD

=

1 a a sin 60
2

3 2
4 sq. m. = 4 3 sq. m.
4
Area of the regular hexagon

1 2 3 a 2
2

3 2 a 4

Math

265

= 6 4 3 sq, m.
= 24 3 sq. m.

Exercise 16⋅2
1. The length of a rectangular region is twice its width. If its area is 512 sq. m., determine its perimeter.
2. The length of a plot is 80 m. and the breadth is 60 m. A rectangular pond was excavated in the plot. If the width of each side of the border around the pond is 4 metre, determine the area of the border of the pond.
3. The length of a garden is 40 metre and its breadth is 30 metre. There is a pond in
1
side the garden with around border of equal width. If the area of the pond is of 2 that of the garden, find the length and breadth of the pond.
4. Outside a square garden, there is a path 5 metre width around it. If the area of the path is 500 square metre, find the area of the garden.
5. The perimeter of a square region is equal to the perimeter of a rectangular region.
The length of the rectangular region is thrice its breadth and the area is 768 sq.
m. How many stones will be required to cover the square region with square stones of 40 cm each?
6. Area of a rectangular region is 160 sq. m. If the length is reduced by 6 m., it becomes a square region. Determine the length and the breadth of the rectangle.
3
7. The base of a parallelogram is th of the height and area is 363 square inches.
4
Determine the base and the height of the parallelogram.
8. The area of a parallelogram is equal to the area of a square region. If the base of the parallelogram is 125m. and the height is 5 m, find the length of the diagonal the square.
9. The length of two sides of a parallelogram are 30 cm and 26 cm, If the smaller diagonal is 28 cm, find the length of the other diagonal.
10. The perimeter of a rhombus is 180 cm. and the smaller diagonal is 54 cm. Find the other diagonal and the area.
11. Deference of the length of two parallel sides of a trapezium is 8 cm. and their perpendicular distance is 24 cm. Find the lengths of the two parallel sides of the trapezium. 12. The lengths of two parallel sides of a trapezium are 31 cm. and 11 cm. respectively and two other sides are 10 and 12 cm. respectively. Find the area of the trapezium.

266

Math

13. The distance from the centre to the vertex of a regular octagon is 1.5 m. Find the area of the regular octagon.
14. The length of a rectangular flower garden is 150 m. and breadth is 100 m. For nursing the garden, there is a path with 3 m. width all along its length and breadth right at the middle of the garden.
(a) Describe the above information with figure.
(b) Determine the area of the path.
(c) ow many bricks of 25 cm. length and 12 5 cm. width will be required to
H
make the path metalled ?
15. From the figure of the polygon determine its area.
12cm

16. From the information given below determine the area of the figures :

6.4
Measurement regarding circle
(1) Circumference of a Circle
The length of a circle is called its circumference. et r be the radius
L
of a circle, its circumference c 2π r , where π 3 14159265.......... which is an irrational number value of π = 3 1416 is used as the actual value.
Therefore, if the radius of a circle is known, we can find the approximate value of the circumference of the circle by using the value of π .
Example 1. The diameter of a circle is 26 cm. Find its circumference.
Solution : et, the radius of the circle is r .
L
diameter of the circle = 2r and circumference = 2π r
26
As per question, 2r 26 or, r r 13
2
circumference of the circle = 2π r = 2 3 1416 13 cm.
= 3 1416 26 cm (approx.)
The required circumference of the circle is 81 68 cm. (approx.) θ 267

Math

(2) Length of arc of a circle et O the centre of a circle whose radius is
L be r and are AB s , which produces θ o angle at the centre.
C
ircumference of the circle = 2π r
Total angle produced at the centre of the circle = 360o and arc s produces angle θ at the centre. Wknow, any interior angle at the centre of a circle produced by any e arc is proportional to the arc. π rθ θ s or, s
180
360
2π r
(3) Area of circular region and circular segment
The subset of the plane formed by the union of a circle and its interior is called a circular region and the circle is called the boundary of the such circular region.
Circular segment: The area formed by an arc and the radius related to the oining points of that arc is called circular segment. j If A and B are two points on a circle with centre , the subset of
O
the plane formed by the union of the intersection of AOB and the interior of the circle with the line segment OA , OB and the arc
AB , is called a circular segment.
In previous class, we have learnt that if the radius of a circle is r , the area is = πr 2 e know, any angle produced by an arc at the centre of a circle is
W
proportional to the arc. θ S, at this stage we can accept that the area of two circular o segments of the same circle are proportional to the two arcs on which they stand. et us draw a radius r with centre O.
L
The circular segment AOB stands on the arc APB whose measurement is θ . Draw a perpendicular OC on OA .
Measurement of
Area of circular segment AOB
=
=
Area of circular segment AOC
Measurement of

AOB
AOC

θ
Area of circular segment AOB
; [ AOC 90 o ]
=
Area of circular segment AOC
90
θ or, Area of circular segment AOB = area of circular segment AOC
90
θ
1
area of the circle
=
90 4

or,

268

Math

1 θ πr 2
90 4 θ = πr 2
360
θ
S, area of circular segment = o πr 2
360
Example 2. The radius of a circle is 8 cm. and a circular segment substends an angle
56 o at the centre. Find the length of the arc and area of the circular segment.
Solution : et, radius of the circle, r 8 cm, length of arc is s and the angle
L
subtended by the circular segment is 56o .

=

πrθ
180

3 1416 8 56 cm.. = 7 82 cm. (approx)
180
Area of circular segment = θ πr 2
360

W e know,

s

56
= 360

3.1416

82 sq.cm.

= 62 55 sq. cm. (approx)
Example 3. If the difference between the radius and circumference of a circle is 90 cm., find the radius of the circle.
Solution : et the radius of the circle be r
L
Diameter of the circler is 2r and circumference = 2πr
As per question, 2πr 2r 90
90
45 or, 2r (π 1) 90 or, r
21 01 (approx.)
2(π 1) 3 1416 1
The required radius of the circle is 21 01 cm. (approx.).
Example 4. The diameter of a circular field is 124 m. There is a path with 6 m. width all around the field. Find the area of the path.
Solution : et the radius of the circular field be r and radius of the
L
field with the path be R .
124
m. = 62 m. and R (62 6) m. = 68 m. r 2
Area of the circular field = πr 2 and area of the circular field with the path = πR 2
Area of the path = Area of field with path –
Area of the field
= (πR 2 πr 2 ) = π ( R 2 r 2 )
= 3 1416{ 68) 2 (62) 2 } sq. m.
(
= 3 1416(4624 3844)

Math

269

= 3 1416 780 sq. m.
= 2450 45 sq. m. (approx)
The required area of the path is 2450 44 square m. (approx.)
Activity : ircumference of a circle is 440 m. Determine the length of the sides of
C
the inscribed square in it.
Example 5. The radius of a circle is 12 cm. and the length of an arc is 14 cm.
Determine the angle subtended by the circular segment at its centre.
Solution : et, radius of the circle is r = 12 cm., the length of the arc is s = 14 cm.
L
and the angle subtended at the centre is θ . πsθ W e know, s
180
or, πrθ 180 s
180 s
180 14 or, θ
66 85 (approx) πr 3 1416 12
The required angle is 66 85 (approx)
Example 6. Diameter of a wheel is 4.5 m. for traversing a distance of 360 m.; how many times the wheel will revolve ?
Solution : iven that, the diameter of the wheel is 4.5 m.
G
4 5
m. and circumference = 2πr
The radius of the wheel, r
2
et, for traversing 360 m, the wheel will revolve n times
L
As per question, n 2πr = 360
360
360 2 or, n
= 25.46 (approx)
2πr 2 3 1416 4 5
The wheel will revolve 25 times (approx) for traversing 360 m..
Example 7. Two wheels revolve 32 and 48 times respectively to cover a distance of
211 m. 20 cm. Determine the difference of their radii.
Solution : 211 m. 20 cm. = 21120 cm. et, the radii of two wheels are R and r respectively ;where R r.
L
C ircumferences of two wheels are 2πR and 2πr respectively and the difference of radii is ( R r )
As per question, 32 2πR 21120
21120
21120 or, R
105 04 (approx)
32 2π 32 2 3 1416 and 48 2πr 21120
21120
21120 or, r
70 03 (approx)
48 2π 48 2 3 1416

270

Math

R r (105 04 70 03) cm.. = 35 01 cm. = 35 m. (approx)
The difference of radii of the two wheels is 35 m (approx)
Example 8. The radius of a circle is 14 cm. The area of a square is equal to the area of the circle. Determine the length of the square.
Solution : et the radius of the circle, r 14 cm. and the length of the square is a
L
Area of the square region = a 2 and the area of the circle = πr 2
According to the question, a 2 = πr 2
22
m. = 11 m. adius of the half circle r
R
2 or, a πr 3 1416 14 24 81 (approx)
The required length is 24 81 cm. (approx.)
Example 9 : In the figure, ABCD is a square whose length of each side is 22 m. and
AED region is a half circle. Determine the area of the whole region.
Solution : et, the length of each side of the square ABCD be a.
L
Area of square region = a 2
Again, AED is a half circle. et ;be its radius
L
1 2
Area of the half circle, AED πr 2
Area of the whole region = Area of the square ABCD + area of the half circle AED.
1 2
= a2 πr 2
1
= { 22) 2 sq. (
3 1416 (11) 2 } metre
2
22
[ = 22, r = 2 = 11] a = 674.07 sq. m (app.)
The required area is 674.07 square metre (approx)
Example 10. In the figure, ABCD is a rectangle whose the length is 12 m., the breadth is 10 m. and DAE is a circular region.
Determine the length of the arc DE and the area of the whole region.
Solution : et the radius of the circular segment, r AD 12 m. and the angle
L
subtended at centre θ 30 πrθ length of the arc DE
180
3 1416 12 30
=
m. = 6.28 m. (approx.)
180

Math

271

Area of the circular segment DAE =

θ
360

πr 2

30
3 1416 (12) 2 sq. m. =
360

37.7 sq.cm (approx)
The length of the rectangle ABCD is 12 m. and the breadth is 10 m.
Area of the rectangle = length breadth = 12 10 sq. m = 120 sq. m.
Area of the whole region = (37.7 + 120) sq. m. = 157.7 square metre.
The required area is 157.7 square metre. (approx)
Activity : Determine the area of the dark marked region in the figure.

Exercise 16.3
1. Angle subtended by a circular segment at the centre is 30o. If the diameter of the circle is 126 cm., determine the length of the arc.
1
2. A horse turned around a circular field with a speed of 66 m per minute in 1
2
minute. Determine the diameter of the field.
3. Area of a circular segment is 77 sq. m. and the radius is 21 m. Determine the angle subtended at the centre by the circular segment.
4. The radius of a circle is 14 cm. and an arc subtends an angle 76o at its centre.
Determine the area of the circular segment.
5. There is a road around a circular field. The outer circumference of the road is greater than the inner circumference by 44 metres. Find the width of the road.
6. The diameter of a circular park is 26 m. There is a road of 2 m. width around the park outside. Determine the area of the road.
7. The diameter of the front wheel of a car is 28 cm. and the back wheel is 35 cm.
To cover a distance of 88 m, how many times more the front wheel will revolve than the back one ?
8. The circumference of a circle is 220 m. Determine the length of the side of the inscribed square in the circle.
9. The circumference of a circle is equal to the perimeter of an equilateral triangle.
Determine the ratio of their areas ?
10. Determine the area of the dark marked region with the help of the information given below :

Math

272

6⋅5 Rectangular solid
The region surrounded by three pairs of parallel rectangular planes is known as rectangular solid. et, ABCDEFGH is a rectangular solid, whose length AB a , and breadth
L
BC b and height AH c
(1) Determining the diagonal: AF is the diagonal of the rectangular solid
ABCDEFGH
In ABC , BC AB and AC is hypotenuse
AC 2 AB2 BC 2 a2 b2
Again, in ACF , FC AC and AF is hypotenuse
AF 2 AC 2 CF 2 a2 b2 c2

AF

a 2 b2 c 2

the diagonal of the rectangular solid = a2 b2 c2
(2) Determination of area of the whole surface :
There are 6 surfaces of the rectangular solid where
1
the opposite surfaces are equal figure
2
Area of the whole surface of the rectangular solid area of the
= (area of the surface of ABCD +
2
1
3
3 surface of ABGH +area of the surface of
BCFG )
2
= 2( AB AD AB AH BC BG)
= 2(ab ac bc)
= 2(ab bc ca) width height
(3) olume of the rectangular solid =
V
length
= abc
Example 1. The length, width and height of a rectangular solid are 25 cm., 20 cm. and 15 cm. respectively. Determine its area of the whole surface, volume and the length of the diagonal.
Solution : et, the length of the rectangular solid is a 25 cm., width b 20 cm.
L
and height c 15 cm.
Area of the whole surface of the rectangular solid = 2(ab bc ca)
= 2(25 20 20 15 15 25) sq. cm.
= 2350 square cm. olume = abc
V
= 25 20 15 cube cm.
= 7500 cube cm.
And the length of its diagonal =

a2

b2

c2

Math

273

=

(25)2 (20)2 (15)2 cm.

=

625 400 225 cm.

= 1250 cm.
= 35 353 cm. (approx.)
The required area of the whole surface is 2350 cm2., volume 7500 cm3. and the length of the diagonal is 35 353 cm. (approx.).
Activity : Determine the volume, area of the whole surface and the length of the diagonal of your mathematics book calculating its length, width and height.
6⋅6 Cube
If the length, width and height of a rectangular solid are equal, it is called a cube. et, ABCDEFGH is a cube.
L
Its height = width = height = a (1) The length of diagonal of the cube = a2 a2 a2 = 3a2 = 3a
(2) The area of the whole surface of the cube = 2(a a a a a a)
= 2(a 2 a 2 a 2 ) = 6a 2
(3) The volume of the cube = a a a = a3
Example 2. The area of the whole surface of a cube is 96 m2. Determine the length of its diagonal.
Solution : et, the sides of the cube is a
L
The area of its whole surface = 6a2 and the length of diagonal = 3a a 4
As per question, 6a2 96 or, a 2 16 ;
The length of diagonal of the cube = 3a = 3 4 = 6 928 m. (approx.)
The required length of the diagonal is 6 928 m. (approx.)
Activity : The sides of 3 metal cube are 3 cm., 4 cm. and 5 cm. respectively. A new cube is formed by melting the 3 cubes. Determine the area of the whole surface and the length of the diagonal of new cube.
6⋅7 Cylinder :
The solid formed by a complete revolution of any rectangle about one of its sides as axis is called a cylinder or a right circular Clinder. The two ends of a right circular cylinder y are circles. The curved face is called curved surface and the total plane is called whole surface. The side of the rectangle which is parallel to the axis and revolves about the axis is called the generator line of the cylinder. et, figure (a) is a right circular cylinder, whose radius is r and height h
L
(1) Area of the base = πr 2

274

Math

(2) Area of the curved surface
=
perimeter of the base height = 2πrh
(3) Area of the whole surface
= (πr 2 2πrh πr 2 ) = 2πr(r h)
(4) The volume = area of the base height = πr 2h
Example 3. If the height of a right circular cylinder is 10 cm. and radius of the base is 7 cm, determine its volume and the area of the whole surface.
Solution : et, the height of the right circular cylinder is h 10 cm. and radius of
L
the base is r .
Its volume = πr 2h = 3 1416 72 10
= 1539 38 cube cm. (approx.)
And the area of the whole surface = 2πr (r h)
= 2 3 1416 7(7 10) sq. cm. (approx.)
= 747 7 sq. cm. (approx.)
Activity : Make a right circular cylinder using a rectangular paper. Determine the area of its whole surface and the volume.
Example 4. The outer measurements of a box with its top are 10 cm., 9 cm. and
7cm. respectively and the area of the whole inner surface is 262 cm2. Find the thickness of its wall if it is uniform on all sides.
Solution : et, the thickness of the box is x cm.
L
The outer measurements of the box with top are 10 cm., 9 cm. and 7 cm. respectively. The inside measurement of the box are respectively a (10 2x) cm., b (9 2x) cm. and c (7 2 x) cm.
The area of the whole surface of the inner side of the box = 2(ab bc ca)
As per question, 2(ab bc ca) = 262 or, (10 2 x)(9 2 x) (9 2 x)(7 2 x) (7 2 x)(10 2 x) 131 or, 90 38x 4x2 63 32x 4x2 70 34x 4x2 131 0 or, 12x2 104x 92 0 or, 3x2 26x 23 0 or, 3x2 3x 23x 23 0 or, 3x( x 1) 23( x 1) 0 or, ( x 1)(3x 23) 0 or, x 1 0 or, 3x 23 0
23
or, x 1 or, x
7 67 (approx.)
3

275

Math

ut the thickness of a box cannot be greater than or equal to the length or width or height
B
x 1
The required thickness of the box is 1 cm.
Example 5. If the length of diagonal of the surface of a cube is 8 2 cm., determine the length of its diagonal and volume.
Solution : et, the side of the cube is a .
L
The length of diagonal of the surface = 2a
3a
ength of diagonal =
L
And the volume = a3
As per question, 2a = 8 2 ; a 8
The length of the cube’ diagonal = 3 8 cm. = s 8
3 cm.
And the volume = 83 cm3. = 512 cm3.
The required length of the diagonal is 8 3 cm. (approx) and the volume is 512 cm3.
Example 6. The length of a rectangle is 12 cm. and width 5 cm. If it is revolved around the greater side, a solid is formed. Determine the area of its whole surface and the volume.
Solution : iven that, the length of a rectangle is 12 cm. and width 5 cm. If it is
G
revolved around the greater side, a circle based right cylindrical solid is formed with height h 12 cm. and radius of the base r 5 cm.
The whole surface of the produced solid = 2πr(r h)
= 2 3 1416 5(5 12) sq. sm.
= 534 071 sq. cm. (approx.)
And the volume = πr 2h
= 3 1416 52 12 cm3.
= 942 48 cube cm3. (approx.)
The required area of whole surface is 534 071 sq. cm. (approx.) and the volume is
942 48 cm3. (approx.)

Exercise 16⋅4
1. The length and witdth of two adj cent sides of a parallelogram are 7 cm., and 5 a cm. respectively. What is the half of its perimeter in cm.. ?
(a) 12
(b) 20
(c) 24
(d) 28
2. The length of the side of an equilateral triangle is 6 cm. What is its area (cm2) ?
(a) 3 3
(b) 4 3
(c) 6 3
(d) 9 3
3. If the height of a trapezum is 8 cm. and the lengths of the parallel sides are 9 cm. i and 7 cm. respectively, what is its area (cm2) ?
(b) 64
(c) 96
(d) 504
(a) 24

276

Math

4. Follow the information given below :
(i) A square stone with the side of 4 cm. has 16 cm. perimeter.
(ii) The area of circular sheet with the radius 3 cm.is 3 cm2.
(iii) The volume of a cylinder with height of 5cm. and the radius of 2 cm. is 20π cm3.
According to the information above, which one of the following is correct ?
(b) i and iii
(c) ii and iii
(d) i , ii and iii
(a) i and ii
Answer the following questions (5 – 7) as per information from the picture below:

5. What is the length of the diagonal of the rectangle ABCD in cm. ?
(b) 14
(c) 14 4 (app.)
(d) 15
(a) 13
6. What is the area of the circular segment ADF in sq. cm.?
(a) 16
(b) 32
(c) 64
(d) 128
7. What is the circumference of the half circle AGB in cm. ?
(b) 18 85 (app.)
(c) 37 7 (app.)
(d) 96
(a) 18
8. The length, width and height of a rectangular solid are 16 m. 12 m. and 4 5 m. respectively. Determine the area of its whole surface, length of the diagonal and the volume.
9. The ratios of the length, width and height of a rectangular solid are 21 : 16 : 12 and the length of diagonal is 87 cm. Determine the area of the whole surface of the solid.
10. A rectangular solid is standing on a base of area 48 m2 Its height is 3m and diagonal is 13 m. Determine the length and width of the rectangular solid.
11. The outer measurements of a rectangular wooden box are 8 cm., 6 cm. and 4 cm., respectively and the area of the whole inner surface is 88 cm2. Find the thickness of the wood of the box.
12. The length of a wall is 25 m, height is 6 m. and breadth is 30 cm. The length, breadth and height of a brick is 10 cm. 5 cm. and 3 cm. respectively. Determine the number of bricks to build the wall with the bricks.
13. The area of the surface of a cube is 2400 sq. cm. Determine the diagonal of the cube. 14. The radius and the height of a right circular cylinder are 12 cm. and 5 cm. respectively. Find the area of the curved surface and the volume of the cylinder.
15. The area of a curved surface of a right circular cylinder is 100 sq. cm. and its volume is 150 cubic cm. Find the height and the radius of the cylinder.

Math

277

16. The area of the curved surface of a right circular cylinder is 4400 sq. cn. If its height is 30 cm., find the area of its whole surface.
17. The inner and outer diameter of a iron pipe is 12 cm. and 14 cm. respectively. If the height of the pipe is 5 m., find the weight of the iron pipe where weight of 7.2 gm. iron = cm 3.
1
18. The length and the breadth of a rectangular region are 12 m. and 5m. respectively. There is a circular region j st around the rectangle. The places u which is are not occupied by the rectangle, are planted with grass.
(a) Describe the information above with a figure.
(b) Find the diameter of the circular region.
(c) If the cost of planting grass per sq. m. is Tk. 50, find the total cost.
19. ABC and BCD are on the same base BC and on the same parallel lines BC and AD .
(a) Draw a figure as per the description above. region ABC = region BCD .
(b) P that roof (c) Draw a parallelogram whose area is equal to the area of ABC and whose one of the angles is equal to a given angle (construction and description of construction is must).
20. ABCD is a parallelogram and BCEF is a rectangle and BC is the base of both of them.
(a) Draw a figure of the rectangle and the parallelogram assuming the same height. (b) Sow that the perimeter of ABCD is greater than the perimeter of BCEF . h (c) Ratio of length and width of the rectangle is 5 : 3 and its perimeter is 48
m. Determine the area of the parallelogram

Chapter Seventeen

Statistics
Oing to the contribution of information an d data, the world has become a global w village for the rapid advancement of science and information. G lobaliztion has been a made possible due to rapid transformation and expansion of information and data.
S, to keep the continuity of development and for participating and contribute in o globaliztions, it is essential for the studen ts at this stage to have clear knowledge a about information and data. In the context, to meet the demands of students in acquiring knowledge, information and data have been discussed from class Vand
I
class-wise contents have been arranged step by step. In continuation of this, the students of this class will know and learn cumulative frequency, frequency polygon, give curve in measuring of central tenden cy mean, median, mode etc. in short-cut
O
method.
At the end of this chapter, the students will be able to Explain cumulative frequency, frequency polygon and ogive curve;
Explain data by the frequency polygon, and ogive curve ;
Explain the method of measuring of central tendency ;
Explain the necessity of short-cut method in the measurement of central tendency ;
Find the mean, median and mode by the short-cut method ;
Explain the diagram of frequency polygon and ogive curve.
Presentation of Data : We know that numerical information which are not qualitative are the data of statistics. The data under investigation are the raw materials of statistics. They are in unorganizd form and it is not possible to take e necessary decision directly from the unorganiz d data. It is necessary to organiz and e e tabulate the data. nd the tabulation of da ta is the presentation of the data. In
A
previous class we have learnt how to organiz the data in tabulation. We determine e know that it is required to the range of data for tabulation. Then determining the class interval and the number of classes by using tally marks, the frequency distribution table is made. ere, the metho ds of making frequency distribution table
H
are to be re-discussed through example for convenient understanding.
Example 1. In a winter season, the temperature (in celsius) of the month of J nuary a in the district of rimangal is placed belo w. Find the frequency distribution table of
S
the temperature.
14 , 14 , 14 , 13 , 12 , 13 , 10 , 10 , 11 , 12 , 11 , 10 , 9 , 8 , 9 ,
11 , 10 , 10 , 8 , 9 , 7 , 6 , 6 , 6 , 6 , 7 , 8 , 9 , 9 , 8 , 7 .

Math

279

Solution : ere the minimum and maximum numerical values of the data of
H
temperature are 6 and 14 respectively. ence the range =
H
14
6+ =
1 9.
9
If the class interval is considered to be 3, the numbers of class will be or, 3.
3
Considering 3 to be the class interval, if the data are arranged in 3 classes, the frequency table will be :
Temperature (in celcius)
6 8
9 11
12 14

Tally llll llll l llll llll lll llll ll

Frequency
11
13
7
Total =
31

Activity : Form two groups of all the students studying in your class. Find the frequency distribution table of the weights (in K of all the members of the gs) groups.
Cumulative Frequency :
In example1, considering 3 the class interval and determining the number of classes, the frequency distribution table has been made. The numbers of classes of the mentioned data are 3. The limit of the first class is 6
8 . The lowest range of the class is 6o and the highest range is 8oC. The frequency of this class is 11.
The frequency of the second class is 13. ow if the frequency 11 of first class is
N
added to the frequency 13 of the second class, we get 24. This 24 will be the cumulative frequency of the second class and the cumulative frequency of first class as begins with the class will be 11. A gain, if the cumulative frequency 24 of the second class is added to the frequency of the third class, we get 24 + = which is
7 31 the cumulative frequency of the third class. Thus cumulative frequency distribution table is made. In the context of the above discussion, the cumulative frequency distribution of temperature in example 1 is as follow :
Temperature (in celsius)
Frequency
Cumulative Frequency
6 8
11
11
9 11
13
(11 + =
13) 24
12 14
7
(24 + =
7) 31
Example 2. The marks obtained in English by 40 students in an annual examination are given below. Make a cumulative frequency table of the marks obtained.
70, 40, 35, 60, 55, 58, 45, 60, 65, 80, 70, 46, 50, 60, 65, 70, 58, 69, 48, 70, 36, 85,
60, 50, 46, 65, 55, 61, 72, 85, 90, 68, 65, 50, 40, 56, 60, 65, 46, 76.

Math

280

Solution : Range of the data =
(highest numerical value – lowest numerical value) +
1
=
90
35 1
+
= +
55 1
=
56 et the class interval be 5, the number of classes =
L

56
5

= 2 or 12
11
ence the cumulative frequency distribution table at a class interval of 5 will be as follow :
H
Obtained Frequency Cumulative Obtained Frequency Cumulative marks frequency marks frequency
35 39
2
2
70 74
4
4+ =
31 35
40 44
2
2+ =
2 4
75
79
1
1+ =
35 36
45 49
5
5+ =
4 9
80
84
1
1+ =
36 37
50 54
3
3+ =
9 12
85
89
2
2+ +
37 39
55 59
5
5+ =
12 17
90
94
1
1+ =
39 40
60 64
8
8+ =
17 25
95
99
0
0+ =
40 40
65 69
6
6+ =
25 31
Variable : We know that the numerical information is the data of statistics. The numbers used in data are variable. uch as, the numbers indicating temperatures are variable.
S
imilarly, in example 2, the secured marks used in the data are the variables.
S
Discrete and Indiscrete Variables : The variables used in statistics are of two types. Sch as, discrete and indiscrete vari ables. The variables whose values are only u integers, are discrete variables. The marks obtained in example 2 are discrete variables. imilarly, only integers are used in population indicated data. That is why,
S
the variables of data used for population are discrete variables. nd the variables
A
whose numerical values can be any real number are indiscrete variables. Sch as, in u example 1, the temperature indicated data which can be any real number. esides,
B
any real number can be used for the data related to age, height, weight etc. That is why, the variables used for those are indiscrete variables. The number between two indiscrete variables can be the value of those variables. Sme times it becomes o necessary to make class interval indiscrete. To make the class interval indiscrete, the actual higher limit of a class and the lower limit of the next class are determined by fixing mid-point of a higher limit of any class and the lower limit of the next class.
Sch as, in example 1 the actual higher-lower limits of the first class are 8 5 and u 5 5 respectively and that of the second class are 11 5o and 8 5o etc.
Activity : Form a group of maximum 40 students of your class. Form frequency distribution table and cumulative frequency table of the group with the weights/ eights of the members. h Math

281

Diagram of Data : We have seen that the collected data under investigation are the raw materials of the statistics. If the frequency distribution and cumulative frequency distribution table are made with them, it becomes clear to comprehend and to draw a conclusion. If that tabulated data are presented through diagram, they become easier to understand as well as attractive. That is why, presentation of statistical data in tabulation and diagram is widely and frequently used method. In class V different
III,
types of diagram in the form of line graph and histogram have been discussed elaborately and the students have been taught how to draw them. ere, how
H
frequency polygon, pie-chart, ogive curve drawn from frequency distribution and cumulative frequency table will be discussed.
Frequency Polygon : In class V we have learnt how to draw the histogram of
III,
discrete data. ere how to draw frequency polygon from histogram of indiscrete data
H
will be put for discussion through example.
Example 3. The frequency distribution table of the weights (in kg) of 60 students of class X of a school are is follows :
Weight (in kg)
46 50
51 55 56 60 61 65
66 70
Frequency
5
10
20
15
10
(N of tudents)
o. S
(a) Draw the histogram of frequency distribution.
(b) Draw frequency polygon of the histogram.
Solution : The class interval of the data in the table is discrete. If the class interval are made indiscrete, the table will be :
Class interval of the weight (in Discrete class Mid point of
Frequency
kg) interval class
46 50
45 5 50 5
48
5
51 55
50 5 55 5
53
10
56 60
55 5 60 5
58
20
61 65
60 5 65 5
63
15
66 70
65 5 70 5
68
10
(a) istogram has been drawn taking each square of graph paper as unit of class
H
interval along with x -axis and frequency along with y -axis. The class interval along with x -axis has started from. The broken segments have been used to show the presence of previous squares starting from from origin to 45 5.

282

Math

(b) The mid-points of the opposite sides parallel to the base of rectangle of the histogram have been fixed for drawing frequency polygon from histogram. The mid-points have been j ined by o line segments to draw the frequency polygon
(shown in the adj cent figure). The mid-points a of the first and the last rectangles have been j ined with x -axis representing the class interval by the end points of line segments o to show the frequency polygon attractive.
Frequency Polygon : The diagram drawn by oining frequency indicated points j opposite to the class interval of indiscrete data by line segments successively is frequency polygon.
Example 4. Draw polygon of the following frequency distribution table :
Class interval
Mid-point
Frequency

10 20

20 30

30 40

40 50

50 60

60 70

70 80

80 90

15

25

35

45

55

65

75

85

8

10

15

30

45

41

15

7

Solution : istogram of frequency distribution is
H
drawn taking two squares of graph paper as 5 units of class interval along with x -axis and 2 squares of graph paper as 5 units of frequency along with y axis. The mid-points of the sides opposite to the base of rectangle of histogram are identified which are the mid-points of the class. ow the fixed mid-points are
N
oined. The end-points of the first and the last classes j are j ined to x -axis representing the class interval to o draw frequency polygon.
Activity : Draw frequency polygon from the marks obtained in angla by the
B
students of your class in first terminal examination.
Example 5. The frequency distribution table of the marks obtained by 50 students of class X in science are given. Draw the frequency polygon of the data (without using histogram) :
Class interval of marks obtained
Frequency

31 40

41 50

51 60

61 70

71 80

81 90

91 100

6

8

10

12

5

7

2

Solution : ere the given data are discrete. In this case, it is convenient to draw
H
frequency polygon directly by finding the mid-point of class interval.

283

Math

Class interval
Mid-point

31 40

40 31
2
35 5

41 50
45 5

51 60
55 5

61 70
65 5

Frequency
6
8
10
12
The polygon is drawn by taking 2 squares of graph paper as 10 units of mid-points of class interval along with x -axis and taking two squares of graph paper as one units of frequency along with y -axis.

71 80
75 5

81 90
85 4

91 100
95 5

5

7

2

Activity : Draw frequency polygon from the frequency distribution table of heights of 100 students of a college. eights (in
H
cm.)
Frequency

141 150

151 160

161 170

171 180

181 190

5

16

56

11

4

Cumulative Frequency Diagram or Ogive curve : Cumulative frequency diagram or give curve is drawn by taking the upper limit of class interval along with x -axis
O
and cumulative frequency along with y -axis after classification of a data
Example 6. The frequency distribution table of the marks obtained by 50 students out of 60 students is as follow :
Class interval of
1 10 11 20 21 30
31 40
41 50 marks obtained
Frequency
8
12
15
18
7
Draw the O curve of this frequency distribution. give Solution : The cumulative frequency table of frequency distribution of the given data is :
Class interval of
1
11 20
21 30
31 40
41 50 marks obtained
10
Frequency
8
12
15
18
7
Cumulative
8
8+ =
12 20
15 + =
20 35
18 + =
35 53 7 + =
53 60 frequency give curve of cumulative frequency of data is drawn
O
taking two squares of graph paper as unit of upper limit of class interval along with x -axis and one square of graph paper as 5 units of cumulative frequency along with y -axis.

284

Math

Activity : Make cumulative frequency table of the marks obtained 50 and above in Mathematics by the students of your class in an examination and draw an O curve. give Central Tendency : Central tendency and its measurement have been discussed in class V and V We have seen if the da ta under investigation are arranged in order
II
III. of values, the data cluster round near any central value. A if the disorganizd gain e data are placed in frequency distribution table, the frequency is found to be abundant in a middle class i.e. frequency is maximum in middle class. In fact, the tendency of data to be clustered around the central value is number and it represents the data. The central tendency is measured by this number. enerally, the measurement of central
G
tendency is of three types (1) A rithmetic means (2) Median (3) Mode :
Arithmetic Mean : We know if the sum of data is divided by the numbers of the data, we get the arithmetic mean. ut this method is complex, time consuming and
B
there is every possibility of committing mistake for large numbers of data. In such cases, the data are tabulated through classification and the arithmetic mean is determined by short-cut method.
Example 7. The frequency distribution table of the marks obtained by the students of a class is as follows. Find the arithmetic mean of the marks.
Class interval
25 34 35 44 45 54 55 64 65 74 75 84 85 94
Frequency
5
10
15
20
30
16
4
Solution : ere class interval is given and that is why it is not possible to know the
H
individual marks of the students. In such case, it becomes necessary to know the mid-value of the class.
Class upper value class lower value
Mid-value of the class =
2
If the class mid-value is x1 (i 1, ......., k ) , the mid-value related table will be as follows:
Frequency f i f i xi
Class interval
Class mid-value xi
25 34
29 5
5
147 5
35 44
39 5
10
395 0
45 54
49 5
15
742 5
55 64
59 5
20
1190 0
65 74
69 5
30
2085 0
75 84
79 5
16
1272 0
85 94
89 5
4
358 0
Total
100
6190 0
1 k
The required mean = f i xi ni1 =

1
6190
100

= 9.
61

285

Math

rithmetic mean of classified data (short-cut method)
A
The short-cut method is easy for determining arithmetic mean of classified data.
The steps to determine mean by short-cut method are :
1. To find the mid-value of classes.
2. To take convenient approximated mean (a) from the mid-values.
3. To determine steps deviation, the difference between class mid-values and approximate mean are divided by the class interval i.e. mid value approximate mean u= class interval
4. To multiply the steps deviation by the corresponding class frequency.
5. To determine the mean of the deviation and to add this mean with approximate mean to find the required mean.
Short-cut method : The formula used for determining the mean of the data by this
1
method is x a f i ui h where x is required mean, a is approximate n mean, The fi is class frequency of i th class, ui f i is the product of step deviation with class intervals of i th class and h is class interval.
Example 8. The production cost (in hundred taka) of a commodity at different stages is shown in the following table. Find the mean of the expenditure by short-cut method.
P
roduction cost
2-6 6-10 10-14 14-18 18-22 22-26 26-30 30-34
(in hundred taka)
Frequency
1
9
21
47
52
36
19
3
Solution : To determine mean in the light of followed steps in short-cut method, the table will be :
Frequency f i Step deviation Frequency and class
Class
Midxi a interval f i ui interval value xi ui h
2 6
4
1
4
4
6 10
8
9
2
27
10 14
12
21
3
42
14 18
16
47
1
47
18 22
20 a
52
0
0
22 26
24
36
1
36
26 30
28
19
2
38
30 34
32
3
3
9
Total
188
37
Mean x

a

f i ui n h

286

Math

37
4
188

= 20

=
20
79
= 22
19
Mean production cost is Tk. 19 22 hundred.
Weighted mean : In many cases the numerical values x1 , x2 ,......., xn of statistical data under investigation may be influenced by different reasons /importance weight. In such
/
i
/
case, the values of the data x1 , x2 ,....... xn along with their reasons/mportance weight w1 , w2 ,....... wn are considered to find the arithmetic mean.
If the values of n numbers of data are x1 , x2 ,....... xn and their weights are w1 , w2 ,....... wn , the weighted mean will be n xi wi i 1 n xw

wi i 1

Example 9. The rate of passing in degree ono ur’ class and the number of students
H
s of some department of a niversity are pres ented in the table below. Find the mean
U
rate of passing in degree honour’ class of those departments of the university. s Math tatistics English
S
B angla oology
Z
Pl. o ame of the
N
Sience c department
Rate of passing (in
70
80
50
90
60
85 percentage) umber of S
N
tudents
80
120
100
225
135
300
Solution : ere, the rate of passing and the numb er of students are given. The weight
H
of rate of passing is the number of students. If the variables of rate of passing are x and numerical variable of students is w, the table for determining the arithmetic mean of given weight will be as follows :
Department
xi wi xi wi
Math
70
80
5600
S
tatistics
80
120
9600
English
50
100
5000
angali
B
90
225
20250 oology Z
60
135
8100
ol. cience
P S
85
300
25500
Total
960
74050
6
xi wi xw i 1
6

wi

74050
960

i 2

Mean rate of passing is 77 14

77 14

Math

287

Activity : Collect the rate of passing students and their numbers in .S
SC.
. examination of some schools in your pazlla and find mean rate of passing.
Ui
Median
We have already learnt in class V the va lue of the data which divide the data when
III
arranged in ascending order into two equal parts are median of the data. We have also learnt if the numbers of data are n and n is an odd number, the median will be n 1 the value of th term. ut if n is an even number, the median will be numerical
B
2 n and n 1 th terms. ere we present through example how
H
mean of the value of
2
2 mean is determined with or without the help formulae.
Example 10. The frequency distribution table of 51 students is placed below. Find the median. eight (in cm.)
H
150
155
160
165
170
175
Frequency
4
6
12
16
8
5
Solution : Frequency distribution table for finding mean is an follows : eight (in cm.)
H
150
155
160
165
170
175
Frequency
4
6
12
16
8
5
Cumulative Frequency
4
10
22
38
46
51 ere, n = which is an odd number.
H
51
Median = value of the 51 1 th term
2

= value of 26 th term = the 165
Required median is 165 c.m.
Note : The value of the terms from 23th to 38th is 165.
Example 11. The frequency distribution table of marks obtained in mathematics of
60 students is as follows. Find the median :
Marks obtained 40
45 50
55
60 70 80 85 90 95 100
Frequency
2
4
4
3
7
10 16 6
4
3
1
Solution : Cumulative frequency distribution table for determining median is :
Marks obtained 40 45 50 55 60 70 80 85 90 95 100
Frequency
2
4
4
3
7
10 16
6
4
3
1
Cumulative
2
6
10 13 20 30 46 52 56 59 60 frequency ere, n = which is an even number.
H
60
Median =

The sum of values of

60
60
1th terms th and
2
2
2

Math

288

The sum of values of 30th and 31th terms
2
70 80 150
=
75
2
2
Required Median is 75.
Activity : 1. Make frequency distribution table of the heights (in cm.) of 49 students of your class and find the mean without using any formula.
2. From the above problem, deduct the heights of 9 students and then find the median of heights (in cm.) of 40 students.
Determining Median of Classified Data
=

n th term of classified data is
2
n median. nd the formula used to determine the median or the value of
A
th term is :
2
If the number of classified data is n , the value of

n h , where
L is the lower limit of the median class, n is the
Fc
2 fm frequency, Fc is the cumulative frequency of previous class to median class, f m is the frequency of median class and h is the class interval.
Example 12. Determine median from the following frequency distribution table :

Median = L

Time (in sec.) 30 35 36 41 42 47 48 53 54 59 60 65
Frequency
3
10
18
25
8
6
Solution : Frequency distribution table for determining median :
Time (in sec.)
Frequency
Cumulative
(class interval)
Frequency
30 35
3
3
36 41
10
13
42 47
18
31
48 53
25
56
54 59
8
64
60 65
6
70 n= 70 n 70 ere, n 70 and
H
or 35.
2 2
Therefore, median is the value of 35th term. 35th term lies in the class (48 –53). ence the median class is (48 –
H
53).
25 and h 6 .
Therefore, L 48, F 31, F c Median

= L

m

n
Fc
2

h fm Math

289

= +
48 (35

31)

6 = +
48 4
25

6
25

= +
48 0 96
= 96
48
Required median is 48 96
Activity : Make two groups with all the students of your class. (a) Make a frequency distribution table of the time taken by each of you to solve a problem,
(b) find the median from the table.
Mode :
In class V we have learned that th e number which appears maximum times in a
III,
data is the mode of the data. In a data, there may be one or more than one mode. If there is no repetition of a member in a data, data will have no mode. N we shall ow discuss how to determine the mode of classified data using formula.
Determining Mode of Classified Data
The formula used to determine the mode of classified data is : f 1
Mode = L h , where L is the lower limit of mode-class i.e. the class f f
1

2

where the mode lies, f1 = frequency of mode-class – frequency of the class previous to mode class, f 2 frequency of mode class –frequency of next class of mode class interval. class and h =
Example 13. Find the mode from the following frequency distribution table.
Class
Frequency
31 40
4
41 50
51 60
61 70
71 80
81 90
91 100
Solution
Mode = L

6
8
12
9
7
4
f

f

1

1

f

h
2

ere, the maximum numbers of repetition of frequency is 12 which lies in the class
H
(61 –
70). ence,
H
L 61 f 12 8 4
2
f 12 9 3
2
h 10

290

Math

4
4
10
10 61
7
4 3

Mode = +
61

40
61 5 7 66 7
7
Therefore, the required mode is 66 714
Example 14. Find the mode from the frequency distribution table below :
Solution : ere, maximum numbers
H
Class
Frequency
of frequency are 25 which lie in the
41 50
25
class (41–0). S it is evident that
5 o
51
60
20
mode is in this class. We know that
61 70
15
f
1
Mode = L h 71 80
8
f f
= +
61

1

2

ere, L 41 [f the frequency is maximum in the first class, the frequency of
H
I previous class is zro] e f 25 0
1

f2

5

25 20

25
10
25 5
25
10 51 8 33
30

Mode = +
41
= +
41

= 33
49
Therefore, required mode is 49 33
In classified data, if the first class is mode class the frequency of previous class is considered to be zro. e Example 15. Determine the mode of the following frequency distribution table :
Solution : The maximum numbers of
Class
Frequency frequency are 25 which lie in the class
10 20
4
(41 –
50). S it is obvious that this class is the o 21
30
16 class of mode. We know that,
31 40
20
f
41 50
25
Mode =
1
L

ere,
H

f

1

f

h

2

L

41

f1

25 20 5

291

Math

h 10
5
10
25
= + =
41 2 43
The required mode is 43.

Therefore, mode = +
41

Exercise 17
Put tick (√) mark in the correct answer :
1. f the following, which one is class interval ?
O
(a) The difference between the highest and the lowest data
(b) The difference between the first and the last data
(c) The difference between the highest and the lowest number of each class
(d) The sum of the highest and the lowest numbers of each class.
2. Which one indicates the data included in each class when the data are classified?
(a) Class limit
(b ) Mid-point of the class
(c) umbers of classes
N
(d) Class frequency
3. If the disorganizd data of statistics are arranged according to the value, the data e cluster round near any central value. This tendency of data is called
(a) mode (b) cetral tendency
(c) mean
(d) median
In winter, the statistics of temperatures (in celsius) of a region in angladesh is
B
10 , 9 , 8 , 6 , 11 , 12 , 7 , 13 , 14 , 5 . In the context of this statistics, answer the questions from (4 –
6).
4. Which is the mode of the above numerical data ?
(a) 12
(b) 5
(c) 14
(d) no mode
5. Which one is the mean of temperature of the above numerical data ?
(a) 8
(b) 8.5
(c) 9.5
(d) 9
6. Which one is the median of the data ?
(a) 9.5
(b) 9
(c) 8.5
(d) 8
7. The number of classified data included in the table is n , the lower limit of median class is L , the cumulative data of previous class to median class is Fc , the frequency of median class is fm and class interval is h . In the light of these information, which one is the formula for determining the median ? n h n h
(a) L
(b) L fm Fc
2
Fm fm 2
(c) L

n
Fc
2

h fm (d) L

n
2

fn

h
Fm

292

Class Interval
Frequency
Cumulative Frequency

Math

31 40
6
6

41 50
12
18

51 60
16
34

61 80
24
58

71 80
12
70

81 90
8
78

91 100
2
80

8. In how many classes have the data been arranged ?
(a) 6
(b) 7
(c) 8
(d) 9
9. What is the class interval of the data presented in the table ?
(a) 5
(b) 9
(c) 10
(d) 15
10. What is the mid value of the 4th class ?
(a) 71.5
(b) 61.5
(c) 70.5
(d) 75.6
11. Which one is the median class of the data ?
(a) 41 50
(b) 51 60
(c) 61 70
(d) 71 80
12. What is the cumulative frequency of the previous class to the median class ?
(a) 18
(b) 34
(c) 58
(d) 70
13. What is the lower limit of median class ?
(a) 41
(b) 51
(c) 61
(d) 71
14. What is the frequency of median class ?
(a) 16
(b) 24
(c) 34
(d) 58
15. What is the median of the presented data?
(a) 63
(b) 63.5
(c) 65
(d) 65.5
16. What is the mode of the presented data ?
(a) 61.4
(b) 61
(c) 70
(d) 70.4
17. The weights (in kg) of 50 students of class X of a school are :
45, 50, 55, 51, 56, 57, 56, 60, 58, 60, 61, 60, 62, 60, 63, 64, 60,
61, 63, 66, 67, 61, 70, 70, 68, 60, 63, 61, 50, 55, 57, 56, 63, 60,
62, 56, 67, 70, 69, 70, 69, 68, 70, 60, 56, 58, 61, 63, 64.
(a) Make frequency distribution table considering 5 as a class interval.
(b) Find the mean from the table in short-cut method.
(c) Draw frequency polygon of the presented data in frequency distribution table.
18. Frequency distribution table of the marks obtained in mathematics of 50 students of class X are provided. Draw the frequency polygon of the provided data.
Class interval 31 40 41 50 51 60 61 70 71 80 81 90 91 100
Frequency
6
8
10
12
5
7
2
19. The frequency distribution table of a terminal examination in 50 marks of 60 students of a class is as follows :
Marks obtained
1 10 11 20
21 30
31 40
41 50
Frequency
7
10
16
18
9
Draw an O curve of the data. give 20. The frequency distribution table of weights (in kg) are provided below.
Determine the median.
Weight (kg)
45
50
55
60
65
70
Frequency
2
6
8
16
12
6

Math

293

21. The frequency distribution table of weights (in kg) of 60 students of a class are:
Interval
45-49 50-54 55-59 60-64 65-69 70-74
Frequency
4
8
10
20
12
6
Cumulative Frequency
4
12
22
42
54
60
(a) Find the median of the data.
(b) Find the mode of the data.
22. In case of data, Mode is(i) Measures of central tendency
(ii) Represented value which is mostly occured
(iii) May not unique in all respect
Which is correct on the basis of above information?
a) i and ii
b) i and iii
c) ii and iii
d) i, ii, and iii
23. The following are the marks obtained in Mathematics of fifty students of class IX in a school :

(a) What is the type of the given information? What indicate frequency in a class of distribution?
(b) Make frequency table taking appropiate class enterval.
(c) Determine the mean of the given number by shortcut method.
24.

(a) In the above figure, what is class midvalue?
(b) Express by data of information demonstrated in the figure (b).
(c) Find the median of frequency obtained from (b).

Answer
Exercise 1
4. (a) 0 16 (b) 0 63 (c) 3 2 (d) 3 53 5. (a)
(e) 6

769
3330

71
2
35
2
(b)
(c)
(d) 3
9
99
15
90

6 (a) 2 333, 5 235 (b) 7 266, 4 237 (c) 5 777777, 8 343434 ,

6 245245 (d) 12 3200, 2 1999, 4 3256 7. (a) 0 5 (b) 0 589 (c) 17 1179

(d) 1 92631 8.
(a) 1 31 (b) 1 665 (c) 3 1334 (d) 6 11062 9. (a) 0 2 (b) 2 (c) 0 2074
(d) 12 185 10. (a) 0 5 (b) 0 2 (c) 5 21951 (d) 4 8 11. (a) 3 4641, 3 464
(b) 0 5025, 0 503 (c) 1 1595, 1 160 (d) 2 2650, 2 265
12.(a) Rational (b) Rational (c) Irrational (d) Irrational (e) Irrational (f)
Irrational (g) Rational ` (h) Rational
13. (a) 9 (b) 5 (c) 8

Exercise 2.1
1. (a) {4, 5} (b) { 3, 4, 5, 6} (c) {6, 12, 18, 36}
2. (a) {x

N : x is odd number and 1

(d) {3, 4}

13} (c) {x

N : x, in the multiple

of 36 }
(c) {x

N : x, is the multiple of 4 and x 40} (d) {x Z : x2 16

and x3

216}

3. (a) {1} (L) {1, 2, 3, 4, a}

(b) {2}

(c) {2, 3, 4, a}

(d) {2}

5. {{x, y), ( x), ( y}, φ} , {{m, n, l), (m, n), (m, l}, {n, l}, {m}, {n}, {l}, φ}
7. (a) 2, 3 (L) (a, c) (b) (1, 5)
8. (a) {(a, b), (a, c)}, {(b, a), (c, a)} (b) {(4, x), (4, y), (5, x), (5, y)}
(c) {3, 3), (5, 3), (7, 3)}
9. {1, 3, 5, 7, 9, 15, 35, 45} and {1, 5} 10. {35, 105} 11. 5 persons

Math

295

Exercise 2⋅2
4. {(3, 2), (4, 2)} 5. {(2, 4), (2, 6)} 6.

7, 23,

7
4
7. 2 8. 1 or 2 or 3 9.
16
x

11. (a) {2}, {1, 2, 3} (b) { 2, 1, 0, 1, 2), (2, 1)}
(c)

1
5
, 1,
, {0, 1, 1, 2,
2
2

2}

12. (a) {( 1, 2), (0, 1) , (1, 0), (2, 1)} , { 1, 0, 1, 2} , {2, 1, 0, 1}
(b) {( 1, 2), (0, 0) , (1, 2)} , { 1, 0, 1} , { 2, 0, 2}
13. (a)

41 (b) 5 (c) 13

Exercise 3⋅1
4x2
y2

1. (a) 4a2 12ab 9b2 (b) 4a2b2 12ab2c 9b2c2 (c) x4
1
a2

(d) a2 2

(i) 9 p

2

16q

2

(f) a2b2 2abc c2

(e) 16 y 2 40xy 25x2

(g) 25x4 10x2 y

y 2 (h) x2

25r

2

4 y 2 16z 2

4 y4 4xy 16 yz 8zx

24 pq 40qr 30 pr (j) 9b2

25c2

4a2 30bc 20ca 12ab

(k) a2 x2 b2 y 2 c2 z 2 2abxy 2bcyz 2cazx
(l) a2 b2 c2 d 2 2ab 2ac 2ad 2bc 2bd 2cd
(m) 4a2 9x2 4 y 2 25z 2 12ax 8ay 20az 12xy 30xz 20 yz (n) 10201
(o) 994009
2.

(p) 10140491

(a) 16a2 (b) 36x2 (c) p2 49r 2 14rp (d) 36n2 24 pn 4 p2 (e) 100
(f) 4410000

3.

16

4.

1 5.

(g) 10 (h) 3104
3m 6. 130 8.

1
4

11. 19 12. 25 13. 6 14. 138

15. 9 17. (2a b c)2 (b a c)2 18. ( x 1)2 82 19. (x 5)2 12 20. (i) 3

Math

296

Exercise 3⋅2
1.

(a) 8x3 60x2 150x 125

(b) 8x6 36x4 y 2 54x2 y 4 27 y6
(d) 343m6 294m4n 84m2n2 8n3

(c) 64a3 240a2 x2 300ax4 125x6
(e) 65450827

(P) 994011992

(f) 8a3 b3 27c3 12a2b 36a2c 6ab2 54ac2 9b2c 27bc2 36abc
(g) 8 x 3 27 y 3 z 3 36 x 2 y 12 x 2 z 54 xy 2 27 y 2 z 6 xz 2 9 yz 2 36 xyz
2.

(a) 8a3 (b) 64x3 (c) 8x3 (d) 1 (e) 8(b c)3 (f) 64m3n3 (g) 2( x 3

y3

z3)

(h) 64x3
3.

665 4. 54 5. 8 6. 42880

7. 1728 10. (a) 3 (b) 9 11. (a) 133

(b) 665
12. a 3 3a 13.

p3 3 p 13. 46 5

Exercise 3⋅3
1.

(a b)(a c)

2.

3.

2( x y)(x

4. b(x y)(a c)

5.

(3x 4)2

7.

( x2

9.

(2a 3b 2c)(2a 3b 2c)

2xy

y z)

(b 1)(a 1)

6. y 2)(x2

2xy

y2)

11. (a y 2)(a y 4)

(a2 5a 1)(a2 5a 1)

8.

(ax by ay by )(ax bx ay bx)

10. 9( x a)(x a)(x 2a)(x 2a)
12. (4x 5 y)(4x 5 y 2z)

13. (a b c)(b c a)(c a b)(a b c) 14. ( x 4)(x 9)
15. (x 2)(x 2)(x2 5)

16. (a 18)(a 12)

17. ( x3 y3 3)(x3 y3 2)

18. (a4 2)(a4 1)

19. (ab 7)(ab 15)

20. ( x 13)(x 15)

21. ( x 2)(x 2)(2x 3)(2x 3)

22. (2x 5)(6x 4)

23.

y 2 ( x 1)(9x 14)

24. (x 3)(x 3)(4x2 9)

Math

297

25. (x a)(ax 1)

26. (a2 2a 4)(3a2 6a 10)

27. (2 z 3x 5)(10 x 7 z 3)

28.

29. ( x ay y)(ax x y)

30. 3x(2x 1)(4x2 2x 1)

31. (a b)2(a4 2a3b 6a2b2 2ab3 b4)

(3a 17b)(9a 7b)

32. ( x 2)(x2 x 1)

33. (a 3)(a2 3a 3)

34. (a b)(2a2 5ab 8b2)

35. (2x 3)(4x2 12x 21)

36.

37.
39.

1
(2a 1)(4a2
8
2a

1
2a

2a

1
(6a b)(36a 2 6ab b 2 )
27

a2
3

b2

a4
9

a 2b 2
3

2a 1)

38.

1
2a

b4

40. (a 4)(19a2 13a 7)

2

41. (x 6)(x 10)

42. ( x2 7 x 4)(x2 7 x 18)

43. ( x2 8x 20)(x2 8x 2)

Exercise 3⋅4
(a 1)(3a 2 3a 5)

1.

(6x 1)(x 1)

2.

3.

(x

4. (x 6)(x 1)

5.

(2x 3)(x 1)

6.

( x 3)(3x 2)

7.

( x 2)( x 1)( x 3)

8.

( x 1)( x 2)( x 3)

9.

(a 3)(a 2

10. (a 1)(a 1)(a2 2a 3)

y)(x 3 y)(x 2 y)

3a 12)

11. (a 1)(a 4)(a 2)

12. (x 2)(x2 x 2)

13. (a b)(a2 6ab b2)

14. ( x 3)( x2 3x 8)

15. ( x y )( x 3 y )( x 2 y )

16. (x 2)(2x 1)(x2 1)

17. (2x 1)(x 1)(x 2)(2x 1)

18. x( x 1)(x2 x 1)(x2 x 1)

19. (4x 1)(x2 x 1)

20. (2x 1)(3x 2)(3x 1)

Math

298

Exercise 3⋅5
1.
5.
9.
13.
17.

(c)
(d)
(a)
(a)
(a)

2.
6.
10.
14.
18.

21. (2) (b)
24.

xy x y

(d)
(c)
(c)
(b)
(b)

3. (b)
7. (d)
11. (d)
15. (c)
19. (c)

21. (3) (d)

days

22.

4.
8.
12.
16.
20.

(b)
(d)
(b)
(b)
(d)

2
( p r ) days 23. 5 hours
3

25. 95 persons

26. Speed of current is

d 1
2 q

1 d 1 km per hour and speed of boat is
2 p p 1 q km. per hour.
27. The speed of the oar is 8 km/hour and the speed of current is 2 km/hour t1t 2 minutes t 2 t1

28

29. 240 liter

30. Tk. 10

31. Tk. 48

Tk. 120 , (b) Tk. 80 , (c) Tk. 60 33. Purchase value Tk. 450
625 36. Tk. 5%
39. Tk. 61
37. Tk. 522 37 (approx.) 38. Tk. 780
40. VAT is Tk.

32.

(a)

34. 4% 35. Tk.

px
; the amount of VAT is Tk. 300.
100 x

Exercise 4⋅1
1.

10
7

7. 4

ab
3a 2b
1
8.
9
2.

3. 27
17.

3
2

4.

a2 b 18. 3

Exercise 4⋅2
1. (a) 4 (b)

1
1
5
(c)
(c) 4 (e)
3
2
6

5. 343

6. 1

19. 5

20. 0, 1

299

Math

2. (a) 125 (b) 5

(c) 4
13
(c) 0
4. (a) og 2 (b)
15

Exercise 4⋅3
1. b 2. d 3. c 4. a 5. c 7. d 8. (1) d (2) c (3) a
9. (a) 6 530 103 (b) 6 0831 103 (c) 2 45 10

4

(d) 3 75 10 7

(e) 1 4 10 7
10. (a) 100000 (b) 0 000001 (c) 25300 (d) 0 009813 (e) 0 0000312
11. (a) 3 (b) 1 (c) 0 (d) 2 (e) 5
12. (a) characteristics 1 , Mantissa 43136
80035
(c) characteristics 0 , Mantissa 14765
65896

(b) characteristics 1 , Mantissa
(d) characteristics 2 , Mantissa

(e) characteristics 4 , Mantissa 82802
13. (a) 1 66706 (b) 1 64562 (c) 0 81358 (d) 3 78888
14. (a) 0 95424 (b) 1 44710 (c) 1 62325
15. a.2 a 3 53 b. 6 25 10

c. characteristics 1, Mantissa 79588

Exercise 5⋅1
1. 1

2. ab 3.
7.

a b
2
13. { a}

9.

15.

1
3

6

a b
2

4. 1

5.

3
5

6.

5
2

8. a b

10. 3

11. {2}

12. {4(1

2 )}

14. φ
16.

20. 28, 70 21.

m n
2

17.

3
22. 72
4

7
2
23. 72

18. 6
24. 18

19.
25. 9

(a 2

b2

c2 )

300

Math

26. Number of coin of twenty five and fifty paisa are 100 and 20 respectively.
22. 120 km.

Exercise 5⋅2
1. c
9.

2. b

3. b

3

10.

2,

5d

6. b

3 2 2 3
,
2
3

7. a

11. 1, 6

8. (1) d (2) c (3) a
12.

7 13.

3
2

6,

3
20

14. 1,
15.

4. c

1
, 2
2

16. 0,

2
3

17.

ab

18. 0, a b 19. 3,

1
2

2
, 2
3

20.

21. { a,

b}

22. {1,

23. {1} 24. {0, 2a} 25.

1}

or 87 27. Length 16 metre, breadth 12 metre

1
,1
3

28. 9 cm., 12 cm.

26.
29.

78
27

cm.

30. 21 persons, Tk. 20 31. 70 32. a. 70 9 x, 9 x 7 b. 34 c. 5 cm. 5 2 cm.
33. b. 5 cm.. c. 2 : 5 : 8

Exercise 9.1
2. cos A

7
, tan A
4

3
, cot A
7

3. sin A

15
8
, cos A
17
17

4. sin θ

5
, cos nθ
13

7
, sec A
3

22.

1
3
23.
4
2

24.

12
, tan θ
13
4
3

25.

5
12
a2 b2 a2 b2

4
, cos ecA
7

4
3

301

Math

Exercise 9.2

5.

1
2

6.

19. A 37

3
4 2
1
,B
2

7.
7

23
2 2
8.
5
3

1
2

21. θ

17. A 30 , B 30

90

22. θ

60 23. θ

18. A 30
60 24. θ

45

25. 3

Exercise 10
7. 45.033 metre (app.) 8. 34.641 metre (app.) 9. 12.728 metre (app.)
10. 10 metres
11. 21.651 metre (app.) 12. 141.962 metre(app.) 13. 83.138 metre (app.) and
48 metre
14. 34.298 metre (app.) 15. 44.785 metre (app.) 16. (b) 259.808 metre.

Exercise 11⋅1
1. a2 : b2, 2. π : 2 , 3. 45, 60, 4. 20%, 5. 18 : 25, 6. 13 : 7, 8. (i)
(iii) x

2ab b 2 , (iv) 10, (v)

3
2ab
, (ii) 2 ,
4
b 1

1 b 1
, (vi) , 2, 22. 3 c 2
2a
c

Exercise 11⋅2
1. a 2. c 3. c 4. b 5. b 6. 24%, 7. 70%, 8. 70%, 9. a Tk. 40, b Tk.60, c Tk.
120, d Tk. 80, 10. 200, 240, 250, 11. 9 cm. 15cm., 21cm., 12. Tk. 315, Tk.
336, Tk. 360, 13. 140, 14. 81 runs 54 runs, 36 runs, 15. Officer Tk. 24000,
Clark Tk. 12000 bearer Tk. 6000 16. 70, 17. 44%, 18. 1% 19. 532 quintal,
20. 8 : 9, 21. 1440 sq.metre, 22. 13 : 12.

Exercise 12.1
1. Consistent, not dependent, single solution
2. Consistent, dependent, innumerable solution
3. inconsistent not dependent, has no solution
4.
consistent, dependent, innumerable solution 5. consistent, not dependent,

302

Math

single solution 6. inconsistent, not dependent, has no solution 7. Consistent, dependent, innumerable solution 8. Consistent, dependent, innumerable solution 9. Consistent, dependent, single solution 10. Consistent, not dependent, single solution .

Exercise 12.2

Exercise 12.3

Exercise 12.4
1. a

2. c

3. b

4. b

15
9.
26

7
7(3) d 8.
9

5. b

10. 27

6. b

11. 37 or 73

7(1) . c

7(2). d

12. 30 years

13.

length 17 m. breadth 9 metre 14. spread of boat 10 km. per hour, spead of corrent 5 km. per hour 15. starting salary Tk. 4000, yearly increment Tk. 25
16. a. one

b. (4, 6)

c. sq.unit 17. a.

x 7 y 2,

x y 2

1,

b. (3, 5),

Exercise 13.1
1. 7 Ges 75, 2. 129 Zg, 3. 100 Zg, 4. p2

pq q2 , 5. 0 , 6. n2 , 7. 360 , 8. 320 ,

9. 42 , 10. 1771 , 11. 620 , 12. 18, 13. 50, 14. 2 4 6 ........,, 15. 110 , 16. 0,
17. (m+n), 20. 50 .

Exercise 13.2
5.

1
1
, 2. 3 6. (314 1), 7. 9th term, 8.
, 9. 9th term, 10. x 15, y
2
3

45 ,

3
5

Math

11. x 9, y

303

27, z 81 , 12. 86, 13. 1, 14. 55log 2, 15. 650log 2, 16. n

7,

17. 0 , 18. n 6, S 21 , 19 n 5, S 165 , 20. 3069 , 21. 20, 22. 24.47mm
.
512

app.)
(

Exercise 16⋅1
1. 20 m., 15 m.

2. 12 m. 3. 12 sq. m. 4. 327 26 sq. m. app.) 5. 5 m.
(

6. 30o 7. 36 or 12 cm. a ,
10. 24 249 cm. ( pp.)

8. 12 or 16 m.

9 44 44 km. ( pp.)
.
a

a
254 611 sq. cm. ( pp.)

Exercise 16⋅2
1. 9 m.
6

2. 1056 sq. m.

3. 30 m. and 20 m.

4. 400 m.

5. 6400 6. 16 m. and 10 m. 7. 16 5 m. and 22 m.8. 35 35 m. ( pp.) a .
9 48 66 cm. ( pp.) 10. 72 cm., 194 sq. cm. a 4
12. 9 75 sq. cm. app.)
5
(

11. 17 cm. and 9 cm.

13. 6 36 sq. m. ( pp.) a Exercise 16.3
1. 32 97 cm. ( pp.) 2. 31 513 m. ( pp.) 3. 20 008 ( pp.) 4. 128 282 sq. a a a 8 cm. ( pp.) 5. 7 003 m. ( pp.) 6. 175 9 m. app.) 7. 20 times a a
(
3
(
.
8. 49 517 m. app.) 9

3 3 :π

Exercise 16⋅4
8. 636 sq. m., 20 5 m., 864 cubic metre. . 14040 sq. m.
9
10. 12 m., 4 m.
11. 1 cm. 12. 300000 13. 34 641 sq. cm. 14. 534 071 sq. cm. app.)
(
(
4
92 48 cubic cm. app.) 15. 5 305 sq. cm. 3 cm.. 16. 6111 8 sq. cm.
17. 147 027 kg. ( pp.) a Exercise 17
1. ( ) 2. ( ) 3. ( ) 4. ( ) 5. ( ) 6. ( ) 7. a) 8. ( ) 9( ) c b b d c a
(
b .c
10. ( ) 11. ( ) 12. ( ) 13. ( ) 14. ( ) 15. ( ) 16. ( ) 20. Median c c c c b b a 60 21. a)62 K , ( )62.8 g.
(
g b K

2013
Academic Year

9-10 Math