RD Sharma Solutions Class 9 Chapter 3 Rationalisation | - GMS - Learning Simply
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  • RD Sharma Solutions Class 9 Maths Chapter 3 Rationalisation

    In Class 9, Rationalisation is one of the most important chapters. RD Sharma solutions for class 9 Chapter 3 is about different algebraic identities and rationalisation of the denominator. A rationalisation is a process by which radicals in the denominator of a fraction are eliminated. In this chapter, students will learn to simplify algebraic expressions using identities. Students are advised to memorize all the identities before they attempt any rationalisation question. To download pdf of chapter 3 click on the below link.


    Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 3 Rationalisation

    Exercise 3.1

    Question 1: Simplify each of the following:

    Class 9 Maths Chapter 3 Rationalisation

    Solution:

    (i)

    Class 9 Maths Chapter 3 Rationalisation

    (ii)

    Class 9 Maths Chapter 3 Rationalisation

    Question 2: Simplify the following expressions:

    (i) (4 + √7) (3 + √2)

    (ii) (3 + √3)(5- √2 )

    (iii) (√5 -2)( √3 – √5)

    Solution:

    (i) (4 + √7) (3 + √2)

    = 12 + 4√2 + 3√7 + √14

    (ii) (3 + √3)(5- √2 )

    = 15 – 3√2 + 5√3 – √6

    (iii) (√5 -2)( √3 – √5)

    = √15 – √25 – 2√3 + 2√5

    = √15 – 5 – 2√3 + 2√5

    Question 3: Simplify the following expressions:

    (i) (11 + √11) (11 – √11)

    (ii) (5 + √7) (5 –√7 )

    (iii) (√8 – √2 ) (√8 + √2 )

    (iv) (3 + √3) (3 – √3)

    (v) (√5 – √2) (√5 + √2)

    Solution:

    Using Identity: (a – b)(a+b) = a2 – b2

    (i) (11 + √11) (11 – √11)

    = 112 – (√11)2

    = 121 – 11

    = 110

    (ii) (5 + √7) (5 –√7 )

    = (52 – (√7)2 )

    = 25 – 7 = 18

    (iii) (√8 – √2 ) (√8 + √2 )

    = (√8)2 – (√2 ) 2

    = 8 -2

    = 6

    (iv) (3 + √3) (3 – √3)

    = (3)2 – (√3)2

    = 9 – 3

    = 6

    (v) (√5 – √2) (√5 + √2)

    =(√5)2 – (√2)2

    = 5 – 2

    = 3

    Question 4: Simplify the following expressions:

    (i) (√3 + √7)2

    (ii) (√5 – √3)2

    (iii) (2√5 + 3√2 )2

    Solution:

    Using identities: (a – b)2 = a2 + b2 – 2ab and (a + b)2 = a2 + b2 + 2ab

    (i) (√3 + √7)2

    = (√3)2 + (√7)2 + 2(√3)( √7)

    = 3 + 7 + 2√21

    = 10 + 2√21

     

    (ii) (√5 – √3)2

    = (√5)2 + (√3)– 2(√5)( √3)

    = 5 + 3 – 2√15

    = 8 – 2√15

    (iii) (2√5 + 3√2 )2

    = (2√5)2 + (3√2 )+ 2(2√5 )( 3√2)

    = 20 + 18 + 12√10

    = 38 + 12√10

    Exercise 3.2

    Question 1: Rationalise the denominators of each of the following (i – vii):

    (i) 3/ √5 (ii) 3/(2 √5) (iii) 1/ √12 (iv) √2/ √5

    (v) (√3 + 1)/ √2 (vi) (√2 + √5)/ √3 (vii) 3 √2/ √5

    Solution:

    (i) Multiply both numerator and denominator to with same number to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 part 1

    = 3√5/5

    (ii) Multiply both numerator and denominator to with same number to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 part 2

    (iii) Multiply both numerator and denominator to with same number to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 part 3

    (iv) Multiply both numerator and denominator to with same number to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 part 4

    (v) Multiply both numerator and denominator to with same number to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 part 5

    (vi) Multiply both numerator and denominator to with same number to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 part 6

    (vii) Multiply both numerator and denominator to with same number to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 part 7

    Question 2: Find the value to three places of decimals of each of the following. It is given that

    √2 = 1.414, √3 = 1.732, √5 = 2.236 and √10 = 3.162

    RD sharma class 9 maths chapter 3 ex 3.2 question 2

    Solution:

    RD sharma class 9 maths chapter 3 ex 3.2 question 1 solution
    RD sharma class 9 maths chapter 3 ex 3.2 question 2 solution
    RD sharma class 9 maths chapter 3 ex 3.2 question 2 solutions

    Question 3: Express each one of the following with rational denominator:

    RD sharma class 9 maths chapter 3 ex 3.2 question 3

    Solution:

    Using identity: (a + b) (a – b) = a2 – b2

    (i) Multiply and divide given number by 3−√2

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 1

    (ii) Multiply and divide given number by √6 + √5

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 2

    (iii) Multiply and divide given number by √41 + 5

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 3

    (iv) Multiply and divide given number by 5√3 + 3√5

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 4

    (v) Multiply and divide given number by 2√5 + √3

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 5

    (vi) Multiply and divide given number by 2√2 + √3

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 6

    (vii) Multiply and divide given number by 6 – 4√2

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 7

    (viii) Multiply and divide given number by 2√5 + 3

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 8

    (ix) Multiply and divide given number by √(a2+b2) – a

    RD sharma class 9 maths chapter 3 ex 3.2 question 3 part 9

    Question 4: Rationales the denominator and simplify:

    RD sharma class 9 maths chapter 3 ex 3.2 question 4

    Solution:

    [Use identities: (a + b) (a – b) = a2 – b2 ; (a + b)2 = (a2 + 2ab + b2 and (a – b)2 = (a2 – 2ab + b2 ]

    (i) Multiply both numerator and denominator by √3–√2 to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 4 part 1

    (ii) Multiply both numerator and denominator by 7–4√3 to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 4 part 2

    (iii) Multiply both numerator and denominator by 3+2√2 to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 4 part 3

    (iv) Multiply both numerator and denominator by 3√5+2√6 to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 4 part 4

    (v) Multiply both numerator and denominator by √48–√18 to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 4 part 5

    (vi) Multiply both numerator and denominator by 2√2 – 3√3 to rationalise the denominator.

    RD sharma class 9 maths chapter 3 ex 3.2 question 4 part 6

    Exercise VSAQs

    Question 1: Write the value of (2 + √3) (2 – √3).

    Solution:

    (2 + √3) (2 – √3)

    = (2)2 – (√3)2

    [Using identity : (a + b)(a – b) = a2 – b2]

    = 4 – 3

    = 1

    Question 2: Write the reciprocal of 5 + √2.

    Solution:

    rd sharma class 9 maths chapter 3 Short answers question 2

    Question 3: Write the rationalisation factor of 7 – 3√5.

    Solution:

    Rationalisation factor of 7 – 3√5 is 7 + 3√5

    Question 4: If

    rd sharma class 9 maths chapter 3 Short answers question 4

    Find the values of x and y.

    Solution:

    [Using identities : (a + b)(a – b) = a2 – band (a – b)2 = a2 + b2 – 2ab]

    rd sharma class 9 maths chapter 3 Short answers question 4 solution

    Question 5: If x = √2 – 1, then write the value of 1/x.

    Solution:

    x = √2 – 1

    or 1/x = 1/(√2 – 1)

    Rationalising denominator, we have

    = 1/(√2 – 1) x (√2 + 1)/(√2 + 1)

    = (√2 + 1)/(2-1)

    = √2 + 1

    Question 6: Simplify

    rd sharma class 9 maths chapter 3 Short answers question 6

    Solution:

    rd sharma class 9 maths chapter 3 Short answers question 6 solution

    [ Because: (a + b)2 = a2 + b2 + 2ab ]

    Question 7: Simplify

    rd sharma class 9 maths chapter 3 Short answers question 7

    Solution:

    rd sharma class 9 maths chapter 3 Short answers question 7 solution

    [ Because: (a – b)2 = a2 + b2 – 2ab ]

    Question 8: If a = √2 +1, then find the value of a – 1/a.

    Solution:

    Given: a = √2 + 1

    1/a = 1/(√2 + 1)

    = 1/(√2 + 1) x (√2 – 1)/(√2 – 1)

    = (√2 – 1)/ ((√2)2 – (1)2)

    = (√2 – 1)/1

    = √2 – 1

    Now,

    a – 1/a = (√2 + 1) – (√2 – 1)

    = 2

    Question 9: If x = 2 + √3, find the value of x + 1/x.

    Solution:

    Given: x = 2 + √3

    1/x = 1/(2 + √3)

    = 1/(2 + √3) x (2 – √3)/(2 – √3)

    = (2 – √3)/ ((2)2 – (√3)2)

    = (2 – √3)/(4-3)

    = (2 – √3)

    Now,

    x + 1/x = (2 + √3) + (2 – √3)

    = 4

    Question 10: Write the rationalisation factor of √5 – 2.

    Solution:

    Rationalisation factor of √5 – 2 is √5 + 2

    Question 11: If x = 3 + 2√2, then find the value of √x – 1/√x.

    Solution:

    rd sharma class 9 maths chapter 3 Short answers question 11 solution

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