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CHAPTER -1 UNIT 1 ADDITIONAL PROBLEMS ON “PLYING WITH NUMBERS” I. Choose the correct option 1. The general from of 456 is a) ( 4 x 100) + ( 5 x 10 ) + ( 6 x 1 ) b) ( 4 x 100) + ( 6 x 10 ) + ( 5 x 1 ) c) (5 x 100 ) + ( 4 x 10 ) + (6 x 1 ) d) ( 6 x 100 ) + ( 5 x 10 ) + ( 4 x 1) [a] 2. Computers use a) Decimal system b) binary system c)base 5 system d) base 6 system [b] 3. If abc is 3 digit number , then the number n = abc + acb + bac + cab + cba is always divisible by a) 8 b) 7 c) 6 d) 5 [c] 4. If abc is 3 digit number, then n = abc - acb - bac - bca - cab - cba is always divisible by a) 12 b) 15 c) 18 d) 21 [c] 5. If 1 K x K1 = K 2 K , the letter K stands for the digit a) 1 b) 2 c) 3 d) 4 [a] 6. The number 345111 is divisible by a) 15 b) 12 c) 9 d) 3 [d] 7. The number of integer of the form 3AB4 , where A, B denote some digits, which divisible by 11 is

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CHAPTER -1

UNIT – 1

ADDITIONAL PROBLEMS ON “PLYING WITH NUMBERS”

I. Choose the correct option

1. The general from of 456 is

a) ( 4 x 100) + ( 5 x 10 ) + ( 6 x 1 )

b) ( 4 x 100) + ( 6 x 10 ) + ( 5 x 1 )

c) (5 x 100 ) + ( 4 x 10 ) + (6 x 1 )

d) ( 6 x 100 ) + ( 5 x 10 ) + ( 4 x 1) [a]

2. Computers use

a) Decimal system b) binary system

c)base 5 system d) base 6 system [b]

3. If abc is 3 digit number , then the number n = abc + acb + bac + cab + cba is

always divisible by

a) 8 b) 7 c) 6 d) 5 [c]

4. If abc is 3 – digit number, then n = abc - acb - bac - bca - cab - cba is always

divisible by

a) 12 b) 15 c) 18 d) 21 [c]

5. If 1 K x K1 = K 2 K , the letter K stands for the digit

a) 1 b) 2 c) 3 d) 4 [a]

6. The number 345111 is divisible by

a) 15 b) 12 c) 9 d) 3 [d]

7. The number of integer of the form 3AB4 , where A, B denote some digits, which

divisible by 11 is

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a) 0 b) 4 c) 7 d) 9 [d]

2.What is the smallest 5 – digit number divisible by 11 and containing each of the

digits 2, 3, 4, 5, 6?

The smallest 5- digit number divisible by 11 which contain the digits 2, 3, 4, 5, 6 is 24365

[ ∵ 2 – 43 – 6 + 5 = 0 ]

3.How many 5 – digit number divisible by 11 are there containing each of the

digits 2,3, 4, 5, 6 ?

24365 , 26245, 24563, 26543

36542, 34562, 34265, 36245

54263, 56243, 54362, 56342

There are 12 5 digit numbers

4.If 49A and A49 where A > 0 , have a common factor , find all possible values of A .

If A = 2, 5, 7, 8 then the numbers formed by 49 A and have common factors.

Like

1. 492 and 249 common factor 3

2. 495 and 549 common factor 3

3. 497 and 749 common factor 7

4. 498 and 849 common factor 3

But if A = 1, 3, 4, 6 , 9 donot have common factor.

5. Write 1 to 10 using 3 and 5, eah al last once , and using addition and subtraction,

(For example , 7 = 5 + 5 – 3 )

1 = 3 + 3 – 5

2 = 5 – 3

3 = 5 + 3 – 5

4 = 3 + 3 + 3 – 5

5 = 3 -3 + 5

6 = 3 + 3 + 5 – 5

7 = 5 + 5 – 3

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8 = 5 + 3

9 = 5 + 5 + 5 – 3 – 3

10 = 3 – 3 + 5 + 5

6. Find all 2 – digit numbers each of which is divisible by the sum of digits.

2 – digit numbers which are divisible by the sum of its digits are

10, 12, 18 , 20 , 21 , 24 , 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80,

81, 84 , 90.

7. The page numbers of a book written in a row gives a 216 digit number. How many

pages are there in the book.

From Page No 1 to 9 there are 9 digits

From page No 10 to 99 there are 180 digits

From Page No 100 to 108 there are 27 digits

216

∴ Total No of pages 108 in a book when has 216 digit numbers.

8. Look at the following patters:

1

1

1

1

2

1

1

3

3

1

1

4

6

4

1

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1

1

1

1

2

1

1

3

3

1

1

4

6

4

1

1

5

10

10

5 1

1

6

15

20

15

6

1

1

7

21

35

35

21

7

1

1

8

20

56

70

56

28

8

1

9. This is called Pascal’s triangle. What is the middle number in the 9-th row?

The Middle number is the 9th Row is 70.

10. Complete the adjoining magic square. [ Hint: in a 3 x 3 magic square , ……., Magic

some is three times the central number?

Ans: The Number is 108

[ ∵ 108 = 1 + 0 + 8 = 9]

9 x 12 = 108

11. Find all 3- digit natural numbers which are 12 times as large as the ……. m of

their digits.

Find all digits x, y such that 34 x 5y is divisible by 36.

The no.s are divisible by 36 only when they are divisible by 4 and 9.

According to the divisibility condition of 4 and 9, then the numer is divisible by

4. If the sum of all the digits of a number is divisible 9, then the number s

divisible by 9.

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These condition one satisfied only

When x = 4, y = 2

Or x = 0, y = 6

Or x = 9 , y = 6

12.Can you divide the numbers 1,2,3,4,5,6,7,8,9,10 into two groups such that

the product of numbers in one group divides the product of numbers in the

other group and the quotient is minimum?

Group 1 1 x 2 x 3 x 4 x 5 x 6 7 = 5040

Group 2 8 x 9 x 10 = 720

5040 = 7

720

Minimum quotient i s 7

13.Find all 8 digit numbers 273A49B5 which are divisible by 11 as well as 25.

The number is divisible by 25

When the number ends with 00, 25 , 50 or 75

∴ In the number ‘B’ must be either 2 to 7.

The number is divisible by 11 if and only if ( Sum of odd placed digits) – (

Sum of even placed digits) is divisible by 11.

It is possible only when

i) If A = 1 and B = 2

Then the number is = 27314925

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Or

If A = 6 and B = 7

Then the number is 27364975

14.Suppose a, b are integers such that 2 + a and 25 –b are divisible by 11. Prove that

a + b is divisible by 11.

It two numbers are divisible by 11 then their sum and difference is also divisible by 11

Let us take the difference of the no. s

Then ( 2 + a ) – ( 35 – b ) is divisible by

= 2 + a – 35 + b

= 2 – 35 + a + b

= - 33 + a + b divisible by 11

Since – 33 is divisible by 11

a + b also must be divisible by 11

15.A list of numbers and corresponding codes are given in the adjacent table. Find the

Number L.

As per the given of numbers and their codes

2 corresponds to 8

0 corresponds to 0

1 corresponds to 6

∴ the number L = 8066

2011

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16. In the multiplication table A 8 x 3B = 2730 , A and B represent distinct digits different

from 0. Find A + B.

In A8 x 3B = 2740

To get 0 in unit place

8 is to be multiplied by 5

Ie is 8 x 5 = 40

A8 = 2730 = 78

35

AB = 78

⇒ A = 7

17. Find the least natural number which leaves the remainders 6 and when divided by 7 and 9

respectively.

LCM of 7 and 9 is 63

Least number which leaves the remainder 6 and 8

Go on finding the remainder by subtract in natural no. from 63

Ie 63 – 1 = 7 = 8

7) 62

56

6

II ly 63 – 1 = 6

9) 62

54

08

There is no other no. which gives 6 and 8 when divided by 7 and 9

Least number is 62.

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18.Prove that the sum of cubes of three consecutive natural numbers always divisible by 3.

Let the 3 consecutive nature no.s be

X, x + 1 and x + 2

x³ + ( x + 1) ³ + ( x + 2 ) ³

x ³ + x ³+ 1 + 3x ( X + 1) + x ³ + 6 + 6x ( x + 2)

x³ + y³ + 1+ 3x³ + 3x + x³ + 6 + 6x + 12x

3x³ + 9x² + 15x + 9

3 ( x² + 3x + 5x + 3)

⇒ Sum of cubes of three consecutive natural numbers is always divisible by 3.

CHAPTER – 2

UNIT – 1

ADDITIONAL PROBLES ON ALGEBRIC EPXPRESSIONS

I. Choose the correct answer:

a. Terms having the same literal factors with same exponents are called

a) Exponents b) like terms

c)factors d) unlike terms [b]

b. The coefficient of ab in 2ab is:

a)ab b) 2 c) 2a d)2b [b]

c. The exponential from of a x a x a is :

a) 3a b) 3+a c) a³ d) 3-a [c]

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d. Sum of two negative integer is:

a) Negative b) positive

C)zero d) infinite [a]

e. What should be added to a² + 2ab to make it a complete square?

a)b² b) 2ab c) ab d)2a [a]

f. What is the product (x + 2) ( x – 3 )?

a) 2x – 6 b) 3x – 2

C)x² -x – 6 d) x² - 6x [c]

g. The value of (7.2)² is (use an identity to expand):

a) 49.4 b) 14.4 c) 51.84 d)49.04 [c]

h. The expansion of (2x – 3y)² is:

a) 2x² + 3y² + 6xy b)4x² + 9y² - 12 xy

c)2x² + 3y² - 6xy d) 4x² + 9y² + 12xy [b]

i. The product 57 x 62 is (use an identity):

a) 4596 b) 2596 c) 3596 d) 6596 [c]

2.Take away 8x – 8p + 10q from 10x + 10y – 7p + 9q.

10x + 10y - 7p + 9 q

8x - 7 y – 8 p + 10 q

( - ) (+) (+ ) ( - )

+2x + 17 y + p - q

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3.Expand:

(i) (4x + 3)² ii) (x + 2y)²

(iii) ( x + 1/x)² iv) ( x – 1 / x)²

i) (4x + 3) ² this is in the form

( a + b )² = a² + 2ab + b² Here a = 4x

(4x + 3) ² = (4x)² + 2 (4x) (3) + 3² b= 3

= 16x² + 24x + 9

ii) (x + 2y )² this is in the form

(a + b) ² = a² + 2ab + b² Here a= x

= x² + 2 (x) (2y) + ( 2y)² b = 2y

= x ²+ 4xy + 4y²

iii) ( x + 1 )² This is in the form x

( a+ b) = a + 2ab + b Here a = x

= x² + 2 . x . 1 + ( 1 )² b = 1 X x x

= x² + 2 + 1

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iv) ( x- 1 ) ² This is in the form

X

( a+ b) ² = a² – 2ab + b ² Here a = x

= x² + 2 , x , 1 + (1 ) ² b = 1

x x x

= x² + 2 + 1

3. Expand:

i) (2t + 5) ( 2t – 5) ii) ( xy + 8 ) ( xy – 8 )

iii)(2x + 3y ) ( 2x – 3y)

i) ( 2t + 5) ( 2t – 5) This is in the form

= ( a+ b) ( a- b) = a² – b²

= ( 2t + 5 ) ( 2t – 5) = (2t) ² – 5²

= 4t² - 25

ii) (xy + 8 ) ( cy – 8) This is in the form of

= ( a + b) ( a- b) = a² – b²

= ( xy + 8 ) ( xy – 8) = (xy) ² – 9²

= x² y² - 64

iii) ( 2x + 3y ) ( 2x – 3y ) This is in the form of

= (a + b) ( a- b) = a² – b²

= ( 2x + 3y ) ( 2x – 3y ) = (2x) ² - (3y) ²

= 4x² – 9y²

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5.Expand

i) ( n-1 ) ( n + 1) ( n² + 1)

ii) ( n – 1/n) ( n + 1 /n) ( n² +1/n²)

iii)(x -1 ) ( x + 1) ( x² + 1) ( x⁴ + 1)

iv)(2x – y ) ( 2x + y) ( 4x² + y²)

i) ( n – 1) ( n²+ 1) ( n -1) ( n + 1 ) = n² -1²

= ( n²- 1² ) ( n²+ 1²) This is in the form

= ( a- b) ( a + b) = a² - b²

= ( n² – 1²) ( n² + 1²) = (n²)² – ( 1²)²

= n⁴ - 1

ii) ( n – 1 ) ( n + 1 ) ( n² + 1 )

n n n²

= ( n² – 1 ) ( n² + 1 )

n² n² = n⁴ – 1 n⁴

iii) ( x – 1) ( x + 1) ( x² + 1) ( x⁴ + 1)

= (x² – 1² ) ( x² + 1²) ( x⁴ + 1)

=[ ( x² ) ² ) – ( 1²)²) ] [ x⁴ + 1⁴ )]

=(x⁴ – 1⁴ ) ( x⁴ + 1⁴ )

=( x⁴ )² – ( 1⁴ )²

= x ⁸- 1

iv) ( 2x – y) (2x + y) ( 4x² + y² )

=[(2x) ² – (y) ² ] ( 4x² + y² )

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= ( 4x² + y² ) ( 4x² + y²)

= (4x²)² – ( y²)²

=16x⁴ - y⁴

6. Use appropriate formula and compute:

i) ( 103) ² ii) (96) ² iii) 107 x 93

iv)1008 x 992 v) 185² - 115²

i) ( 103) ² = (100 + 3) ² This is in the form

( a + b) ² = a² + 2ab + b²

( 100 + 3 ) ² = ( 100 ) ² + 2 ( 100) (3) + 3²

= 10000 + 600 + 9

( 103 ) ² = 10609

ii) (96) ² = ( 100 – 4) ² This is in the form

( a – b) ² = a ² – 2ab + b²

(100 – 4) ² = ( 100) ² + 2 ( 100) (4) + 4²

= 10000 – 800 + 16

= 10016 – 800

( 96) ² = 9216

iii) 107 x 93

= ( 100 + 7) ( 100 – 7) This is the form of

( a + b) ( a- b) = a² – b ²

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( 100 + 7) ( 100 – 7) = ( 100) ² – 7²

= 1000 – 49

= 9951

iv) 1008 x 992

( 1000 + 8 ) ( 1000 – 8) This is in the form

( a+ b) ( a- b) = a² – b²

( 1000 + 8) ( 1000 – 8) = ( 1000) ² – 8²

= 100000 – 64

= 999936

v) 185² – 115² This is in the form

a² – b² = ( a + b) ( a- b)

( 185) ² – (115) ² = ( 185 + 115) ( 185 – 115)

= ( 300) (70)

(185) ² – (115) ² = (300) (70)

(185) ² – ( 115) ² = 21000

7.If x + y and xy = 12 find x² + y².

( x + y ) ² = x² + y² + 2 xy ( x + y ) ² =

( x + y ) ² = 2xy = x² + y ² x² + y² – 2xy

7² = 2 ( 12) = x² + y² ∴ x² + y² =

49 – 24 = x ² + y² ( x + y) – 2xy

25 = x² + y²

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X² + y² = 25

8. If x + y = 12 and xy = 32 , find x² + y².

( x + y) ² = x² + y² + 2xy

( x + y ) ² – 2 xy = x² + y²

(12) ² – 2 ( 32)= x² + y²

144 – 64 = x² + y²

X² + y² = 144 – 64

X² + y² = 80

9. If 4x² + y² = 40 and xy = 10 , find x ² + y² .

4x² + y² = 40

= 2² x² + y² = 40

= ( 2x) ² + y² = 40

( 2x + y) ² = ( 2x )+ y² + 2 (2x) (y)

(2x + y) ² = 4x² + y² + 4 ( xy)

= 40 + 4 (6)

= 40 + 24

( 2x + y) ² = 64

2x + y =√ 64

2x + y =8

10. if xy = 3 and xy = 10 , find x² + y²

( x-y) ² = x² + y² – 2xy

3² = x² + y² – 2 (10)

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9 = x² + y² – 20

X² + y² = 9 + 20

X² + y² = 29

11. If x =( 1 )= 3, find x² + (1 ) and x³ + ( 1 )

x x² x³

case – 1 ( x + 1 ) = 3 x Squaring on both side

( x + 1 )² = 3² X

X² + 1 + 2 . x . 1 = 9

X ² x

X² + 1 = 9 – 2

x² + 1 = 7

x ²

Case – 2 x + 1 = 3

x

Cubing on both side

(x + 1)³ = 3³ x

x³ + 1 + 3 x . 1 ( x + 1 ) = 27

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x ³ x x

x³ + 1 + 3 ( 3) = 27

x ³

x³ + 1 + 9 = 27

x ³

x³ + 1 = 27

x ³

x³ + 1 = 18

x ³

12.If x + 1 = 6 , find x ²+ 1 and x⁴ + 1

X x ² x⁴

Case – 1 x + 1 = 6 x

Squaring on both side

( x + 1 )² = 6² X

X² + 1 + 2 . x. 1 = 36

X ² x

X² + 1 + 2 = 36 - 2

X ²

X² + 1 = 34

X ²

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Case – 2

Consider X² + 1 = 34

X ² Squaring on both side

( x + 1 )³ = 3³ X

( x² + 1 )² = ( 34)²

X⁴ + 1 + 2 . x² 1 = 1156 34² = 1156

X ⁴ x² X⁴ + 1 + 2 = 1156 X⁴ X⁴ + 1 = 1156 – 2 x⁴ ∴ X⁴ + 1 = 1154 x⁴

13.Simplify : (i) ( x + y)² + ( x – y)²

ii) ( x + y )² x ( x – y )²

i) ( x + y ) ² + ( x – y) ²

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= [x² + y² + 2 xy ] + [x² + y ²– 2xy]

=x² + y ²+ 2xy + x² + y² – 2xy

= 2x² + 2y ²

= 2 ( x² + y² )

ii) (x + y) ² x ( x- y) ²

= [ ( x + y ) ( x – y) ] ² ( a + b) ( a- b) = a² - b²

= [ x² – y ²]²

This is in the form

= ( a – b) ² = a² + b² – 2ab ( a- b) ² = a² + b² – 2ab

= ( x² – y² ) ² = x⁴ + y⁴ – 2 x² y² ( or)

(x² – y² ) = x⁴ – 2x²y ² + y⁴

14.Express the following as difference of two squares:

i) ( x + 2z) ( 2x + z)

ii) 4 (x + 2y) ( 2x + y)

iii) (x + 98) ( x + 102)

iv) 505 x 495 [Hint: 4ab = (a + b) ² – ( a- b) ²

i) ( x + 2z ) ( 2x + z)

= 2x² + xz + 4 xz + 2 z²

= 2x² + 2z ²+ 4 xz + xz

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= 2( x² + z² + 2xz ) + xz

= 2 ( x + z ) ² + xz

Multiply and divide by 4

= 8 ( x + z ) ² + 4xz ( x + z ) ² – (x – z) ² 4 4

= 8 ( x + z ) ² + 4 xz [x² + z² + 2xy ] – [x² + z² – 2zy]

4 4

= 8 ( x² + z ) ² + ( x + z) ² – ( x- z) ² 4

= 9 ( x + z ) ² - ( x - z ) ² 4 4

3(x – z) ² - x – z ² 4 2

ii) 4 (x + 2y) ( 2x + y)

4 [2x² + xy + 4xy + 2y² ]

4[2x² + 2y² + 4xy + xy ]

4[ 2 ( x² +y² + 2xy ) + xy]

4[2 (x +y) ² + xy]

8(x+ y ) 2 + 4xy

8 (x + y ) ² + ( x + y) ² – (x – y) ²

=9 ( x + y) ² – ( x- y) ²

=[3 (x + y)] ² – ( x- y ) ²

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iii) ( X + 98) ( x + 102)

= x² + 102 x + 98 x + 9996

= x ²+ 200x + ( 10000 – 4)

= x ²+ 200x + 10000 – 4

= x² + 200 x + (100) ² – 2²

=(x + 100) ² - 2²

iv) 505 x 495

= ( 500 + 5) ( 500 – 5)

= ( 500) ² – 5²

15. If a = 3x – 5y , b = 6x + 3y and c= 2y – 4x, find

i) a + b – c ii) 2a – 3b + 4c

i) a + b + c

= ( 3x – 5y) + ( 6x + 3y) – (2 y- 4x)

= 3x – 5y + 6x + 3y – 2y + 4x

= 13x – 4y

ii) 2a – 3b + 4c

=2(3x – 5y) – 3 ( 6x + 3y) + 4 ( 2y – 4x)

=6x – 10y -18x – 9y +8y -16 x

= 6x – 13x – 16x -10y- 9y + 8y

= 28x – 11y

16. The perimeter of a triangle is 15x² – 23 x + 9 and two of its side

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are 8x² + 8x – 1 and 6x² – 9x + 4 . Find the third side.

Perimeter of a triangle = sum of all the three sides of a triangle

Let the sides of the triangle be a, b and c then P = a + b + c

Let a = 5x² + 8x – 1 b = 6x² - 9x + 4 , and

P = 15x² – 23 x + 9

C = ?

C = P – ( a + b)

= ( 15x² – 23 + 9 ) – [( 5x² + 8x – 1 ) + (6x² – 9x + 4)

= ( 15x² + 23x + 9 ) – [11x² – x + 3]

= (15x² – 23x + 9)- 11x² + x -3

= 4x² – 22 x 6

17. The two adjacent sides of a rectangle are 2x² – 5xy + 3x² and

4xy – x² – x² . Find its perimeter.

Let a and b are the sides of the Rectangle

ℓ = 2x² - 5xy + 3x² b = 4xy - x² - x²

Then its perimeter P = 2 ( ℓ + b)

P = 2 [2x² - 5xy + 3z² + 4xy - x² - z² ]

P= 2 [ x² - xy + 2z²]

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P = 2x² - 52 xy + 4z²

18.The base and altitude of a triangle are ( 3x – 4y) and ( 6x + 5y)

respectively. Find its are.

Area of a ∆ = 1 b x h 2 Let b and h are the base and altitude of the triangle Then Area of triangle = 1 x b x h 2 A = 1 ( 3x – 4y ) ( 6x + 5y) 2 = 1 [ 18x + 15 xy - 2 xy – 20 y² ] 2 = 1 [ 18x² - 9xy – 20 y² ] 2 19.The sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of side length x + y is removed . What is the area of the remaining region? ( 2x + 3y)

( 3x + 2y)

Area of the remaining region =

X + y

X + y

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(Area of the Rectangle ) – (Area of the square)

(ℓ x b ) - ( side) ²

=[(2x + 3y) (3x + 2y)] – ( x + y) ²

= [ 16² + 4xy + 9xy + 6y² ] – ( x² + y² + 2xy)

= 6x² + 13xy + 6y² - x² - 2xy - y²

= 5x² + 11xy – 5y²

20. a, b, c are rational numbers such that a² + b² + c² -bc – ca = 0.

Prove that a = b = c.

a² + b² + c² - bc – ca = 0

Multiply both side by 2

2(a² + b² + c² - ab – bc –ca ) = 2 x 0

2a² + 2b² + 2c² - 2ab – 2bc – 2ca = 0

a² + a² + b² + c² + c² - 2ab – 2bc – 2ca = 0

a² + b² - 2ab + b² + c² - 2bc + c² +a² - 2ca = 0

( a- b) ² + ( b – c) ² + ( c- a) ² = 0

If the sum of squares of three terms is ‘0’

Then each term is equal to ‘0’

⇒ ( a- b) ² = 0 ( b – c ) ² = 0 ( c- a) ² = 0

a- b= 0 b – c = 0 c – a = 0

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a=b …. ( 1) b = c ….(2) c = a ….(3)

from equation ( 1) ( 2) and ( 3)

a = b = c

CHAPTER – 3

UNIT – 1

ADDITIONAL PROBLEMS ON “AXIOMS, POSTULATES”

1. Choose the correct option:

i) a= 60 and b = a , then b= 60 by ….

a. Axiom 1 b. Axiom 2

c.Axiom 3 d.Axiom 4 […..]

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ii) Given a point on the plane , one can draw …….

a. Unique b . two

c.finite number of

d.infinitely many …… through that point [d]

iii) Even two points in a plane, the number of lines which can be

drawn to pass through these two points is ….

a. Zero b. exactly one

c.at most one d. more than on [b]

iv) Two angles are supplementary , then their sum is …..

a. 90˚ b. 180˚ c. 270˚ d. 360˚ [b]

v)The measure of an angle which is 5 times its supplement is

a. 30˚ b. 60˚ c.120˚ d.150˚ [d]

2. What is the difference between a pair of supplementary angles and a pair of complementary angle? Supplementary angles : If the sum of two angles is 180˚ then they are supplementary angles. Complementary angles: If the sum of two angles is 90˚ then they are complementary angles.

3. What is the least number of on collinear points required to determine a plane? 3 non collinerar points.

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4. When do you say two angles are adjacent?

The angles are said to be adjacent if they have a common side and a common and point. A C D B

5. Let AB be a segment with C and D between them such that the order of

points on the segment is A, C , D, B, Suppose AD = BC . Prove that AC= DB.

AB is s line segment C and D are points on it such that AD = BC AC + CD = BD + CD AC = BD [ Axioms 3 ]

6. Let AB and CD be two straight lines intersecting at 0. Let 0X be the

dissector of BOD. Draw 0Y between 0D and 0A such that 0Y ⟘ 0X .

Prove that 0Y bisects DOA.

Y

A D

Q˚ P

X˚ X

0 X˚

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C B

AB and CD intersecting at ‘0’

OX bisects BOD

Let = BOX = DOX = X ̊

OY ⟘ OX

Let POY = P and YOA = Q

BOX + XOD + DOY + YOA = 180 ̊

x˚ + x˚ + p + q = 180˚ ……….. ( 1)

x + p = 90˚ …….. ( 2) given

∴ x + q = 90 ……… (3) ( Remaining angle of 180˚ )

From (2) + ( 3)

X + p = x + q = 90˚

P = q = 90˚

⇒ OY bisect DOA

7.Let AB and CD be two parallel lines and PQ be a transversal. Let PQ

intersect AB in L. Suppose the bisector of ALP intersect CD in R and

the bisector of PLP intersect CD in S. Prove that LRS + RSL = 90˚.

M

M M

X Y

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A X˚ Y˚ B

L

C S R P

Proof:

Let ALN = NLP = x˚ and

BLM = MLP = y˚

ALB = 180 ̊

ALN + NLP + PLM + MLB = 180˚

X + x + y + y = 180

2x + 2y = 180 ˚

X + y = 180 = 90˚ 2

⇒ NLM = 90 ̊

But NLM = RLS = 90˚ * V. O . A+

In the ∆ LRS

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LRS + RSL + LRS = 180 ̊

[Sum of all the angles of ∆le]

LRS + RSL + 90˚ = 180 ̊

LRS + RSL = 180˚ - 90 ̊

LRS + RSL = 90˚

8 In the adjoining figure. AB and CD are parallel lines. The transversals

PQ and RS interest at U on the line AB. Given that DWU = 110˚ and

CVP = 70 , find the measure of QUS.

S Q

X

U

A B

110 ̊

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C ( ( ) D

70˚ V W

P R

Ans: Given - DWU = 110˚ , CVP = 70˚

Proof-

UVP = KVP = 70˚ ( V.O. A)

VWU + UWD = 180˚ ( Adjacent Angles)

V + W + U = 180˚ ( Sum of angles of triangle is 180˚)

U = 180˚ - 70˚ - 70˚ = 40˚

X = U = 40˚ ( V. o . A)

∴ QUS = 40˚

9. What is the angle between the hour’s hand and minute’s hand

Of a clock at i) 1.40 hours , ii) 2.15 hours? ( Use 1˚ = 60 minute)

10. How much would hour’s hand moved from its position to 12 noon when the

time is 4.24 p.m?

11.Let AB be a line segment and let C be the midpoint of AB . Extend AB to D

such that B lines between A and D. Prove that AD + BD = 2CD.

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12.Let AB and CD be two lines intersecting BOD . Prove that the extension of 0X

to the left of 0 bisects AOC.

13.Let 0X be a ray and let 0A and 0B be two rays on the same side of 0X and 0B.

Let 0C be the bisector of AOB. Prove that AOX + XOB = 2 XOC.

14. Let 0A and 0B be two rays and let OX be a ray between 0A and 0B such that

AOX > XOB . Let 0C be the bisector of AOB. Prove that

AOX - XOB = 2 Cox.

15. Let 0A , 0B , 0C be the three rays such that 0C lies between 0A and 0B .

Suppose the bisectors of AOC and COB are perpendicular to each other. Prove

that B, O , A are collinear.

16. In the adjoining figure, AB DB . Prove that ABC + DCB + CDE = 180˚.

A B

D E

C

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17. Consider two parallel lines and a transversal . Among the measurement of

transversal. Among the measurement of triangles formed, how many distinct

numbers are there?

(Answer will be given in next issue)

CHAPTER - 1 UNIT1 – 2

EXERCISE 1. 2. 2

1. Express the following statement mathematically.

i) Square of 4 is 16

Ans: 4² = 16

ii) Square of 8 is 64

Ans: 8² = 64

iii) Square of 15 is 225

Ans : 15² = 225

2. Identify the perfect squares among the following numbers:

2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900,100, 1000, 100000.

Ans: Perfect squares : 6² = 36 , 7² = 49,

9² = 81 , 13² = 169, 25² = 625,

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30² = 900 , 10² = 100

3. Make a list of all perfect squares from 1 to 500.

Ans: Perfect squares: 1,4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225,

256, 289, 324, 361, 400, 441, 484.

4. Write 3 – digit number ending with 0, 1, 4, 5, 6,9 , one for each digit, but

none of them is a perfect square.

Ans: 110, 111, 114, 115, 116, 119 ( ending with 0, 1, 3, 5, 6, 9, but not a perfect

square) .

5. Find numbers from 100 to 400 that and with 0, 1, 4, 5, 6, or 9 which are

perfect squares.

Ans: Perfect squares from:

100 to 400 – 100 ( 10²)

144(12²), 169 ( 13²), 196 (14²),

225 (15²), 289 (17²), 324 (19²),

400(20²).

EXERCISE 1. 2. 3

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1. Find the sum 1 + 3 + 5 + …… + 51 ( the sum of all odd numbers from 1

to 51) without actually adding them.

Ans: 1 + 3 + 5 + …… +51 = 51 + 1 = 26 26² = 676

2

Sum of ‘n’ odd numbers = n²

Sum of 26 odd numbers = 26² = 676

2. Express 144 as a sum of 12 odd numbers.

Ans: 144 = 12²

Sum of ‘n’ odd numbers = n²

Sum of 12 odd numbers = 12²

1+ 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

12² = 144

3. Find the 15th

and 16th

triangular numbers and find their sum. Verify the

statement 8 for this sum.

Ans: 1+2

3 + 3

6+ 4

10 + 5

15+6

I II III IV V

21 + 7

28 +8

36+9

45+10

55+11

VI VII VIII IX X

66+12

78+13

91+14

105+15

XI XII XIII XIV

120+16

136+17

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XV XVI

n + ( n + 1) = ( n + 1 ) 2

15th

∆ no + 16th

∆ no = (16)2

120 + 136 = 256

4. What are the remainders of a perfect square when divided by 5.

Ex. Problem

Write 5th

and 6th

triangular numbers and verify the formula of statement 8.

Ans: 1+2

3+3

6+4

10+5

I II III IV

15+6

21+ 7

V VI

n+ ( n+1) = ( n + 1)2

5th

∆ no + 6th ∆ no = ( 6)2

15 + 21 = 36 ∴ Statement 8 is true

Methods for squaring a number:

1842 = (180 + 4)2 *( Using identity)

=180 2 + 4

2 + 2 x 180 x 4

= 32,400 + 16 + 1,440

= 33,856

*No. ending with 5 –

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1.9 x 10

5 2 2. 7

x8 5 2 3. 5

x6 5

2

9025 5625 3025

EERSICE 1.2.4

1. Find the squares of :

i) 31 = (30 + 1)2

(a + b)2 = a

2 + b

2 + 2ab

(30+1)2 = ( 30)

2 + (1)

2 + 2 x 30 x 1

= 900 + 1 + 60

= 961

ii) (72)2 = (70 + 2)

2

(70 + 2 ) 2 = (70)

2 + (2)

2 + 2 x 70 x 2

= 4,900 + 4 + 280

= 5184

iii) 37 = (30 + 7)2

(30 + 7)2 = (30)

2 + (7)

2 + 2 x 30 x 7

= 900 + 49 + 420

= 1369

iv) 166 = ( 160 + 6)2

(160 + 6)2 = ( 160)

2 + (6)

2 + 2 x 60 x 6

= 25,600 + 36 + 1920

= 27,556

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2. Find the squares of :

i) 8 x 9

5 = 7225

ii) 11 x12

5 = 13225

iii) 16 x 17

5 = 27225

3. Find the square of 1468 by writing this as 1465 + 3

Ans: 1465 + 3

( a + b)2 = a

2 + b

2 + 2ab

(1465 + 3 )2 = ( 1465)

2 + (3)

2 + 2 x 1465 x 3

= 21,46,225 + 9 + 8790

= 2146234 + 8790

= 21,55,024

EXERCISE 1.2.5

1. Find the squares root of the following numbers by

factorization:

i) 196

2 196

2 98

7 49

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7 7 √ 196 = √ 2 2 x 7

2 = 2 x 7

1

√ 196 = 14

ii) 256

2 256

2 128

2 64

2 32

2 16

2 8

2 4

2 2

1

√256 = √ 2 2 x 2

2 x 2

2 x 2

2 =

2x 2x2x2

√256 = 16

iii) 10404

2 10404

2 5202

3 2607

3 867

17 287

17 17

1

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√10404 = √ 2 2 x 3

2 x 17

2 =

2 x 3 17

= 102

iv) 1156

2 1156

2 578

17 289

17 17

1

√1156 = √2 2 x 17

2 = 2 x 17

√1156 = 34

v) 13225

5 13225

5 2645

23 529

23 23

1

√13225 = √5 2 x 23

2 =

5 x 23

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√13225 = 115

2. Find the squares of :

i) 100 + 36 = 10 + 6 = 16

2 100

2 50

5 25

5 5

1

√100 = √2 2 x 5 2 = 2 x 5

√100 = 10

3 36

3 12

2 4

2 2 √36 = √3 2 x 2

2 =

3 x 2 = 6

1

ii) √1360 + 9

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√1369 = 37

37 1369 √1369 = √ 372 = 37

37 37

1

iii) √2704 + √144 +√ 289

52 + 12 + 17

64 + 17

= 81

2 2704

2 1352

2 676

2 338

2 169

13 13

1

√2704 =√ 22 x 3

2 x 13

2 =

2 x 2 x 13 =

52

2 144

2 72

2 36

2 18

3 9

3 3

1

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√144 = √2 2 x 2

2 x 3

2 =

2 x 2 x 3 = 12

17 289

17 17 √ 289 = √17 2

= 17

1

iv) 225 - 25 = 15 – 5 = 10

5 225

4 45

3 9

3 3 √225 = √5 2 x 3

2 =

2 x 2 = 15

1

5 25

5 5 √25 = √ 5 2 = 5

1

v) 1764 - 1444 = 42 - 38 = 4

2 1764

2 882

3 441

3 147

7 49

7 7

1

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√1764 = √ 2 2 x 3

2 x 7

2 = 2 x 3 x 7

√1764 = 42

2 1444

2 722

19 361

19 19 √1444= √ 2 2 x 19

2

= 2 x 9

1

√1444 = 38

vi) 169 x 361

213 x 19

117

13x

247

13 169

13 13 √ 169 = √13 2 = 13

1

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19 361

19 361

1 √ 361 = √ 19 2 19

3. Square yard has area 1764m2 . From a corner of this

yard, a square part of area 784 m2 is taken out for

public utility. The remaining portion is divided in to

5 equal square parts. what is the parameter of these

equal parts?

Ans. Area of square = a2

Area of given square yard = 1764 m2

Area used for public utility = 784 m2

∴ Area of remaining portion = 1764 –

784

980 m2

If the area is divided into 5 parts, then the area of each

Square = 980 = 196 m2

5

Length of each side = √ 196

a- = 14m

Perimeter of square = 4a

= 4 x 14

= 56m

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4. Find the smallest positive interger with which one has to

multiply each of the following numbers to get a perfect

square.

Ans i) 847

11 847

11 77

7 7 847 = 11 2 x 7

1

∴ 7 is the smallest positive number to be added to 847 to get a

perfect square.

ii)450

2 450

5 225

5 45

3 9 450 = 5 2 x 3

2 x 2

3 3

1

∴ 2 is the smallest positive interger to be multiplied with 450 to get a perfect

square.

iii) 1445

5 1445

17 289 1445 = 17 2 x 5

17 17

1

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∴ 5 is the smallest positive interger to be multiplied with 1445 to get a perfect

square.

iv) 1352

2 1352

2 676

2 338

13 169 1352 = 2 2 x 13

2 x 2

13 13

1

∴ 2 is the smallest positive interger to be multiplied with 1352 to get a perfect

square.

5. Find the largest perfect square factor of each of the following numbers:

i) 48

2 48

2 24

2 12

2 6

3 3

1

2 x 2 x 2 x 2 x 3

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4 x 4 x 3

16 x 3

∴ 16 is the largest perfect square factor.

ii) 11280

2 11280

2 5640

2 2820

2 1410

5 705

3 141

47 47 2 x 2 x 2 x 2 x 5 x 3 x 47

1 4 x 4 x 5 x 3x 47

16 x 5 x 3 x 47

∴ 16 is the largest perfect square factor.

iii) 729

3 729

3 243

3 81

3 27

3 9

3 3

1

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3 x 3 x 3 x 3 x 3

9 x 9 x 9

81 x 9

729

∴ 729 is the largest perfect square factor.

iv) 1352

2 1352

2 676

2 338

13 169

13 13

1

2 x 2 x 2 x 13 x 13

4 x 2 x 169

676 x 2

∴ 676 is the largest perfect square factor.

6. Find the proper positive factor of 48 and a proper positive multiple of

48 which add up to a perfect square. Can you prove that there are infinitely

many such pairs?

Ans: The factor of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

Multiples of 48 are = 48, 96 , 144, 192 , 240, 280 ….

1) 48 + 1 2 ) 192 + 4

49 = 7 2 192 = 14

2

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3)288 + 1 4 ) 96 + 4

289 = 11 2 100 = 10

2

5)96 + 48 6) 48 + 16

144 = 12 2 64 = 8

2

7)240 + 16

256 = 16 2

Yes, we can prove that there are infinitely many such fairs.

EXERCISE 1.2.6

1. Find the nearest integer to the square of the following numbers.

Ans: i) 232

225 < 232 < 256

15 2 16

2

∴ Square root of 232 is nearest to 15

ii)600

576 < 600 < 625

242 25

2

∴ Square root of 600 s nearest to 24.

iii)728

678 < 728 < 729

262 27

2

∴ Square root of 728 is nearest to 27.

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iv)824

784 < 824 < 841

282 29

2

∴Square root of 824 is nearest to 27

v)1729

1684 < 1729 < 1764

412 42

2

∴Square root of 1729 is nearest to 42.

2. A piece of land is in the shape of a square and its area is 1000m2. This

has to be fenced using barbed wire. The barbed wire is available only in

integral lengths. What is the minimum length of the barbed wire that

has to be bought for this purpose.

Ans: Area of the land = 1000m2

Area = a2 = 1000

Perimeter = 4a

Squaring , (Perimeter)2 = ( 4a)

2

= 16a2

= 16 x 1000

P2 = 16,000

P = 16000

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√15876 < √16000 < √16129

126 127

∴ Nearest number is 126

But this is not enough to cover the hard.

∴Length of barbed wire required = 127 m.

3. A student was asked to find√ 961 . He read it wrongly and found √691

to the nearest integer. How much small was his number form the

correct answer?

Ans. √ 961 = 31

√691

√676 < √691 < √729

26 27

26 is nearest number to √ 691

∴ difference = 31 – 26 = 5

EXERCISE 1.2.7

1. Looking at the pattern, fill in the gaps in the following:

2 3 4 -5 - 8 -

23=8 3

3= - - = 64 - = - 6

3 = - - = - - = - 729

Ans: 33 = 27 , 4

3 = 64

(-5)3 = -125 , 6, 6

3 = 216

83 = 512 -9, (-9)

3 = -729.

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2. Find the cubed of the first five odd natural numbers and the cubes of

the first five even natural numbers. What can you say about parity of

the odd cubes and even cubes?

Ans:

Odd Cubes Even Cubes

13 = 1

33 = 27

53 = 125

73 = 343

93 = 729

23 = 8

43 = 64

63 = 216

83 = 512

103 = 1000

Parity

Cubes of odd numbers is always odd.

Cubes of even numbers is always even.

3. How many perfect cubes you can find from 1 to 100 ? How many from –

100to 100?

Ans: 1 1to 100 - 4 perfect cubes

-100 to 100 - 4 perfect cubes

4. How many perfect cubes are there from 1 to 500? How many are perfect

square among cubes?

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Ans: There are 7 perfect cubes from 1 to 500. Only 13

and 43 are the squares

among these cubes.

5. Find the cubes of 10, 30 , 100, 1000. What can you say about the zeros

at the end?

Ans: 103 = 1000

303 = 27000

1003 = 1000000

10003 = 10000000000

The number of zero of a cube are 3 times, the no. of zero of numbers.

6. What are the digits in the unit’s place of the cubes 1, 2, 3, 4, 5, 6, 7, 8 , 9,

10? Is it possible to say that a number is not a perfect cube by looking at

the digit in unit’s place of the given number, just like you did for

squares?

Ans:

Numbers The digit in the unit’s place

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1

2

3

4

5

6

7

8

9

10

1

8

7

4

5

6

3

2

9

0

EXCERSISE 1.2.8

1. Find the cube root by prime factorization.

i) 10648

2 10648

2 5324

2 2662

11 1331

11 11

1

3 √ 10648 =

3√ 2 2 x 11

2 =

2x11 = 22

ii) 46656

2 46656

2 23328

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2 11664

2 5832

2 2916

2 1458

9 729

9 81

9 9

1

3 √15625 =

3√ 53 x 5

3 =

5x5 = 25

2.Find the cube root of the following by looking at the last digit

and using estimation.

i) 9112 5

43 < 91 < 5

3

403 < 91125 < 50

3

45

ii) 16637 5

53 < 166 < 6

3

503 < 1166375 < 60

3

55

iii) 70496 9

83 < 704 < 9

3

803 < 704969 < 90

3

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89

3.Find the nearest integer to the cube root of each of the following.

i) 331776

216000 < 331776 < 343000

603 40

3

3 √331776 lies between 60 and 70.

ii) 46656

27000 < 46656 < 64000

303 40

3

3 √466656 lies between 30 and 40.

iii) 373248

216000 < 373248 < 343000

3√ 373248 lies between 60 and 70.

UNIT 2

CHAPTER – 2

EXERCISE 2.2.2

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1. Resolve into factors:

i) x2 + xy ii)3x

2 – 6x

x(x+y) 3x(x-2)

iii)(1.6)a2 – (0.8)a iv) 5-10m – 20m

0.8a (2a – 1) 5(1-2m – 4n)

II. Factories.

i) a2+ ax + ab + bx

a(a + x) (a + b)

(a + x) ( a + b)

ii) 3ac + 7bc – 3ad - 7bd

3ac + 7bc – 3ad – 7bd

c(3a + 7b) – d(3a + 7b)

(3a + 7b) (c – d)

iii) 3xy – 6yz – 3xt + 6zt

3y ( x-2z) – 3t(x-2z)

(x-2z) (3y- 3t)

iv) y3 -3y

3 + 2y – 6 – xy + 3x

y2 (y-3) + 2(y -3) – x(y-3)

(y-3) ( y2

+ 2 – x)

3.Factorise

i) 4a2 – 25

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(2a)2 – ( 5)2

(2a + 5) (2a – 5)

ii) x2 – 9

16

(x)2 - ( 3 )

2

4

( x + 3 ) ( x – 3)

4 4

iii) x4 – y

4

(x2)2- (y

2)

2

(x2 + y

2) ( x

2 – y

2)

iv) 7 3 2 - 2 1

2

10 10

7 3 + 2 1 7 3 – 2 1

10 10 10 10

= 94 52 10 10

= 1222

25

v) (0.7)2 – (0.3)

2

(0.7 + 0.3) (0.7 – 0.3)

(1) (0.4) = 0.4

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vi) (5a – 2b)2 – (2a – b)

2

[5a – 2b + ( 2a – b )] [ 5a – 2b – ( 2a – b)]

(5a – 2b + 2a – b ) ( 5a – 2b – 2a + b)

(7a – 3b) ( 3a – b)

EXERCISE 2.2.3

1. In the following you are given the product pq and the

sum p+ q. Determine p and q.

i) pq= 18 and p + q = 11

+9

Product = 18

+2

Sum = 11

∴ p = 9, q = 2

ii) pq= 32 and p + q = - 12

-8

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Product = 32

-4

Sum= -12

∴ p = - 8 , q = -4

iii) pq = - 24 and p + q = 2

+6

Product = 24

-4

Sum = 2

iv)pq = -12 and p + q = 11

+12

Product = - 12

-1

Sum = 11

∴p = + 12 , q = -1

v)pq = -6 and p + q = -5

-6

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Product = - 6

+1

Sum= - 7

∴ p= - 11 , q = + 4

II. Factorise

i) x2 + 6x + 8 +4x

x2 + 4x + 2x + 8

x(x + 4) +2 (x + 4) Product = 8x2

x(x+4) +2 (x + 4)

(x + 4) ( x + 2) +2x

Sum = 6x

ii) x2 +4x + 3

x2 + 3x + 1x + 3 +3x

x( x + 3) + 1 ( x + 3) Product = 3x2

(x + 3) ( x + 1) + 1x

Sum = + 4x

iii) a2 + 5a + 6 +3x

a2 + 3a + 2a + 6

a(a + 3) + 1 ( a + 3) Product =6a2

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(a + 3) (a + 2)

+2a

Sum = +5a

iv) a2 – 5a + 6 -2a

a2 – 2a – 3a + 6

a(a – 2) – 3( a- 2) Product = 6a2

(a – 2) ( a – 3)

-3a

Sum = - 5a

v) a2 – 3a – 40 -8a

a2 – 8a + 5a – 40

a(a – 8) + 5(a -8) Product = -40a2

(a – 8) ( a+ 5)

+5a

Sum = - 3a

vi) x2 – x – 72 -9x

x2 – 8x + 5x – 40

x(x – 9) + 8 ( x – 9) Product = - 72x2

( x – 9) ( x + 8 )

+8x

Sum= - x

ADDITIONAL PROBLEM ON “ FACTORISATION”

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1. Choose the correct answer :

a. 4a + 12 b is equal to

a. 4a b. 12b

c. 4(a + 3b) d. 3a (c )

b. The perfect of two numbers is positive and their sum negative only

when

a. Both are positive

b. both are negative

c. one positive the other negative

d. one of them equal to zero ( b)

c. F a c t 2 + 6x + 8 , we get

a. ( x + 1) ( x + 8) b. ( x + 6) ( x + 2)

c. ( x + 10 ) ( x + 2) d. ( x + 4) ( x + 2) (d)

d. The denominator of an algebraic fraction should not be

a. 1 b. 0 c. 4 d.7 (b)

e. If the sum of two integers is – 2 d their product is – 24 the number

are

a. 6 and 4 b. 6 and 4

c. – 6 and – 4 d. 6 and – 4 ( b)

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f . The difference ( 0.7) 2 – ( 0.3)2 simplifies to

a. 0.4 b. 0.04 c. 0.49 d. 0.56 (a)

2. Factorise the following:

i) x2 + 6x + 9

x2 + 3x + 3x + 9

x( x+ 3) + 3 ( x + 3)

(x + 3) ( x + 3)

= ( x + 3)2

ii) 1- 8x + 16 x2

16x2 – 8x + 1

16x2 – 4x – 4x +1

4x (4x – 1) – 1(4x – 1))

=(4x – 1) 4x – 1)

iii) 4x2 – 81y

2

(2x)2 – ( 9y)

2

(2x + 9y) ( 2x – 9y)

iv) 4a2 + 4ab + b

2

4a2 + 2ab + 2ab + b

2

2a(2a + b) + b(2a + b)

(2a + b) ( 2a + b)

v) A2 b

2 + c

2 d

2 – a

2 c

2 – b

2 d

2

A2 b

2 – a

2 c

2 + c

2 d

2 – b

2 d

2

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A2 (b

2 –c

2) + a

2 (c

2 – b

2)

A2 ( b

2 – c

2 ) – d

2 (b

2 – c

2)

=( b2 – c

2) ( a

2 – d

2)

3. Factories the following:

i) x2 + 7x + 12 +4x

x2 + 4x + 3x + 12

x( x + 4 ) + 3( x + 4) P=12 x 2

(x + 4) ( x + 3)

+3x

Sum = 7x

ii)x2 + x – 12 +4x

x2 – 3x + 4x – 12 Product =12x

x(x-3) + 4(x – 3) - 3

( x – 3 ) ( x + 4 ) Sum = +x

iii)x2 – 3x – 18 -6x

x2 – 6x + 3x - 18 Product = 18x2

x( x – 6 )+ 3 (x – 6) +3x

( x – 3 ) ( x – 7) Sum = -3x

iv)x2 + 4x – 192 +7x

x2 – 3x + 7x – 21 P= +21x2

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x( x – 3) + 7 ( x – 3) -3x

( x – 3 ) ( x – 7) Sum = +4x

v)x2 – 4x – 192 +12x

x2 – 16x + 12x – 192 P = + 192 x

x( x – 16 ) + 12 ( x – 16) -16x

( x – 16) ( x + 12 ) Sum = -4x

vi)x4 - 5x

2 + 4

x4 – 4x

2 – 1x

2 + 4 -4x

2

x2 ( x

2 – 4) – 1 ( x

2 – 4) P = + 4x4

(x2 – 4) ( x

2 – 1) -1 x

2

= [x2 – (2)

2 ] [ x

2 – ( 1)

2] Sum = - 5x

2

(x + 2) ( x – 2) ( x + 1) ( x -1)

vii)x4 – 13x

2y

2 + 36y

4 -9x

2y

2

x4 – 9x

2y

2 – 4x

2 y

2 + 36y

2 P=+36x

4y

4

x2(x

2 – 9y) – 4y

2 (x

2 – 9y

2) -4x

2y

2

(x2 – 9y

2) (x

2 – 4y

2)

[x2 – (3y)

2] [x

2 – (2y)

2]

( x + 3y) ( x – 3y) ( x + 2y) ( x – 2y)

4. Factorise the following:

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i) 2x2 + 7x +6 24x

2x2 + 4x + 3x + 6 Product = 12x

2

2x( x+ 2) + 3 ( x + 2) +3x

( x + 2) ( 2x + 3) Sum = 7x

ii) 3x2 – 17x + 20 -22x

3x2 – 12 x + 5x + 20 P = + 60x2

3x ( x – 4) – 5(x – 4) + 5x

( x – 4) ( 3x – 5) Sum = - 17x

iii) 6x2 – 5x – 14 -12xy

6x2 – 12x + 7x -14 P= x

2y

2

6x (x-2) + 7 ( x-2) + 7x

(x – 2) (6x – 7) Sum = -5x

iv) 4x2 + 12xy + 5y2

4x2 + 2xy + 10xy + 5y2 +2xy

2x ( 2x+y) 5y (2x + y) P=20x2y2

(2x + y) (2x + 5y) +10xy

Sum= 12xy

v) 4x4 – 5x

2 +1

4x4 – 4x

2 – 1x

2 + 1 -4x

2

X2(x

2 – 1) -1 ( +1x

2 -1 ) P=4x

4

(x2 – 1) (4x

4 – 1) -1x

2

[(x2) – (1)

2] [(2x)

2 – (1)

2] Sum= -5x

2

(x+1) (x -1)(2x+1) (2x-1)

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CHAPTER – 3

UNIT – 2

EXERCISE 3.2.1

1. Match the following

i)

(a) Equilateral triangle

ii)

(b) Acute angled triangle

iii)

( c) Right angled triangle

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iv)

(c) Obtuse angled triangle

Ans: i. c ii. d. iii. a iv. b.

2. Based on the sides , classify the following triangles ( figures not

drawn to the scales)

i) ii)

3cm 5cm 7cm 5cm

4cm 4cm

Scalene Triangle Scalene Triangle

iii). iv)

6.5 cm 3.5 cm 6.5cm 6.5cm

6cm 6cm

Scalene Triangle Isosceles Triangle

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v). vi)

3cm 5.6cm 2.5cm 4.1cm

4.3 cm 3.2 cm

Scalene Triangle Scalene Triangle

vii). viii). 3cm

9cm 5cm 3cm 3cm

6cm

Scalene Triangle Equilatereal Triangle

ix). x).

6cm 6cm

5cm 5cm

8cm

3.5cm

Isosceles Triangle Isosceles Triangle

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EXERCISE 3.2.2

1. In a triangle ABC , if A = 55˚ and B = 40˚, find C

Ans: In ∆ ABC

A + B + C = 180˚ ( Theorem 3)

55˚ + 40˚ + C = 180˚

C = 180˚ - ( 55˚ + 40˚ )

C = 85˚

2. In a right angled triangle, if one of the other two angle is 35˚ , find the

remaining angle.

Ans: A = 35˚ , B = 90˚ , ( In a right angled triangle, one angle should be

90˚), in a right angled triangle

A + B + C = 180˚ ( Theorem )

35˚ + 90˚ + C = 180˚

C = 180˚ - ( 35˚ + 90˚ )

C = 180˚ – 125˚

C = 55˚

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3. If the vertex angle of an isosceles triangle is 50, find the other angles.

Ans: A = 50˚ , Let B and C be x ( Because in isosceles ∆ two angles are

equal)

A + B + C = 180˚ ( Theorem)

50˚ + x + x = 180˚

2x = 180˚ – 50˚

X = 130˚ 2

X = 65˚

∴ A = 50˚ , B = 65˚ and C = 65˚

4. The angles of triangle are in the ratio 1:2:3 Determine the three angles.

Ans: 1:2:3

Let the angles be 1x, 2x and 3x

1x + 2 x + 3x = 180˚ ( Theorem 3)

6x = 180˚

x = 180˚ 6

x = 30˚, 2x = 2 x 30 = 60˚

3x = 3 x 30 = 90˚

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5. In the adjacent triangle ABC, find the value of x and calculate the

measure of all the angles of the triangle.

Ans: In∆ ABC

A + B + C = 180˚

( Theorem 3)

x+ 15 + x + 15 + x + 30˚ = 180˚

30x + 30 = 180˚

3x = 180˚ - 30

x = 150

3

x = 50 ̊

∴ A = x + 15 + 50 + 15 = 65˚

∴ B = x – 15 = 50 – 15 = 35˚

∴ C = x + 30 = 50 + 30 = 80˚

6. The angles of triangle are arranged in ascending order of their

magnitude. If the difference between two consecutive angles is 10˚ , find

the three angles.

Ans: Let the angles be x , x + 10 and x + 20

x + x +10 + 20 = 180˚ ( Theorem 3)

3x + 30 = 180˚

3x = 180˚ – 30

x = 150

3 x = 50

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∴ x + 10 = 50 + 10 = 60 ̊

∴ x + 20 = 50 + 20 = 70 ̊

∴ The angles are 50˚ , 60˚ and 70˚

EXERCISE 3.2.3

1. The exterior angles obtained on producing the base of a triangle both

ways are 104 ̊and 136˚. Find the angels of the triangle.

Ans: A

136 104

B C T M

Given – ACT is triangle with the lines

BC and TM extending both ways.

BCA = 136˚ , ATM = 104˚

Proof : BCA + ACT = 180˚

ACT = 180˚ – 136˚ = 44˚

CTA + ATM = 180˚ ( Pro.1)

CTA = 180˚ – 104˚

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CTA = 76 ̊

In∆ ACT, A + C + T = 180˚

( Interior angle theorem)

A + 76˚ = 180˚

A + 180˚ – ( 44˚ + 76˚ )

A + 180˚ – 120˚

A = 60˚

∴The Angles in the∆ ACT are,

A = 60˚ , C = 44˚ , T = 76˚

2. Sides BC, CA and AB of a triangle ABC are produced in an order,

forming exterior angles ACD, DAE and CBF. Show that ACD +

DAE + CBF = 360˚.

Ans:

E E

A

B D C

F

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ACB + ACD = 180˚ ( Proposition 1)

BAC + CAE = 180˚ ( Proposition 1)

FBC + ABC = 180˚ ( Proposition 1)

Adding above equations.

ACB + ACD + BAC + CAE + FBC + ABC = 360˚ + 160˚

ACD + CAE + CBF + { ACB + BAC + ABC } = 360˚ + 180˚

ACD + BAE + CBF + 180˚ ( Interior angle theorem ) = 360˚ + 180˚

ACD + BAE + CBF = 360˚ ( Axiom 3)

3. Compute the value of x in each of the following figures:

Ans:

i) Given – ABC is an isosceles single and CD is extended

ABC = 50˚ A

To prove - ACD

Proof : In∆ ABC 50˚ x

C = B = 50˚ ( AB = AC ) B C D

BCA + ACD = 180˚ ( Proposition 1) P

ACD = 180˚- 50˚ T 180˚

ACD = 130˚

∴ x = 130˚ 160˚ x

S Q R

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ii) Given – PTR = 130˚ , SQT = 106˚

To prove – QRT

Proof : SQT + TQR = 180˚ ( Prop . 1)

TQR = = 180˚ - 106˚

TQR = 74˚

PTR + PTQ = 180˚ ( Prop .1 )

PTR = 180˚ - 130˚

PTR = 50 ̊

In ∆PTQ , R + T + Q = 180˚

( Interior angle theorem)

R = 180˚ - ( 50˚ + 74˚ )

R = 56 ̊

QRT = 56˚ , x = 56 ̊

iii) Given – XRY = 65˚ , RST = 100˚

To Prove – RZS

Proof : ZRS = XRY = 65 ̊

(Vertically opposite angle)

ZSR + RST = 180˚ ( Prop . 1)

ZSR = 180˚ – 100˚

ZSR = 80˚

In ∆RZS , R + Z + S = 180˚ ( Interior angle theorem)

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Z =180˚ - ( 65˚ + 80˚ ) x y

Z = 180˚ – 145˚ 65˚ R

Z = 35 ̊

RZS = 35˚ x 100˚

∴ X = 35˚ z s T

iv) Given – ABE = 120˚ , BED = 112˚

To prove – BEC A

Proof : ABE + EBC = 180˚ ( Proo.1) 120˚ B

EBC = 180˚ - 120˚

EBC = 60˚

EGB + BCD = 180˚ ( Prop.1) x 112˚

ECB = 180˚ – 120˚ E C D

ECB = 68˚

In ∆ EBC , E + B + C = 180˚ ( Interior angle theorem)

E = 180˚ - ( 60˚ + 68˚ )

E = 52˚ BEC = 52 ̊

∴ x = 52˚

v) Given – ACT is an isosceler triangle.

ATC = 20˚

To prove - ACN

Proof : In ∆ ACT , A = T = 20˚ ( AC = CT )

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In ∆ ACT , A + C + T = 180˚ ( Interior angle theorem)

ACN = 180˚ - ( 20˚ + 20˚ ) A

ACN = 140˚

ACN = 40˚

∴ x = 40˚ X 20˚

N C T

4. In the figure , QT ⏊ PR , TQR = 40˚ and SPR = 30˚ , Find TRS

and PSQ

Ans: In ∆ TRQ,

QTR = 90˚ ( data)

TQR = 40˚ (data)

TQR = 180˚ - ( 90˚ + 40˚ ) (Remaining angle)

= 180˚ - 130˚

TRQ = 50˚

In ∆ PSR,

Ext PSQ = SPR + PRS ( Exterior angle theorem)

= 30˚ + 50˚

∴Ext PSQ = 80˚

5. An exterior angle of triangle is 120 and one of the interior opposite

angles is 30. Find the other angles of the triangles.

Ans:

ACB + ACD = 180˚ ( Proposition 1)

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ACB = 180˚ - 120˚

ACB = 60˚ ( Third angle)

In ABC, A

A + B + C = 180˚

( Interior angle theorem)

A = 180˚ ( 30˚ + 60˚ ) 30˚ 120˚

A = 180˚ - 90˚ B C D

A = 90˚ ( Other interior opposite angle)

6. Find the sum of all the angles at the five vertices of the adjoining

star.

Ans: 2 + 4 + 6 = 180˚

(Interior angle theorem) 3

1 + 8 + 10 = 180˚ ( Th .3) 2 8 9 4

2 + 9 + 5 = 180˚ ( Th.3) 7 10

1 + 8 + 4 = 180˚ ( Th.3) 6

3 + 7 + 5 = 180˚ ( Th.3) 1 5

Adding above equations,

2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 6 + 7 + 8 + 9 + 10

= 5 x 180

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2 ( 1 + 2 + 3 + 4 + 5 ) + 6 + 7 + 8 + 9 + 10 = 900

2 ( 1 + 2 + 3 + 4 + 5 ) + 540˚ ( Sum of angles of a

pertragon is 540˚) = 900

2 ( 1 + 2 + 3 + 4 + 5 ) = 900 - 540

1 + 2 + 3 + 4 + 5 = 360 2

1 + 2 + 3 + 4 + 5 = 180˚

ADDITIONAL PROBLEM ON THEOREMS

1. Fill in the blanks to make the following statements true.

a. The sum of the angles of a triangle is ( 180˚ )

b. An exterior angle of triangle is equal to the sum of ____ opposite

angles. ( Interior )

c. An exterior angle of a triangle is always ____ than either of the interior

opposite angle. ( Larger)

d. A triangle cannot have more than _____ right angle. (1)

2. Choose the correct answer from the given alternatives:

1. In a triangle , ABC , A = 80 ̊ and AB = AC , then B is

a. 50˚ b. 60˚ c. 40˚ d.70˚ (a)

2. In right angled triangle, A is right angle and B = 35 ̊ , then C is __

a. 65˚ b. 55˚ c. 75˚ d.45˚ (d)

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3. In a triangle ABC , A = C = 45 ̊, then the triangle is ___

a. right angle

b. Acute angled triangle

c. Obtuse angle triangle

d. Equilateral triangle (a)

4. In an equilateral triangle, each exterior angle is _____

a. 60˚ b . 90˚ c.120˚ d. 150˚ ( c )

5. Sum of three exterior angles of a triangle is _____

a. Two right angles

b. Three right angles

c. One right angle

d. Four right angles (d)