RD Sharma Solutions Class 9 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.

Download PDF of RD Sharma Solutions for Class 9 Chapter 3 Rationalisation

RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 01
RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 02
RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 03
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RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 05
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RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 08
RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 09
RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 10
RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 11
RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 12
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RD Sharma Solution class 9 Maths Chapter 3 Rationalisation 17

Access Answers to Maths RD Sharma 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 – 2(√5)( √3)

= 5 + 3 – 2√15

= 8 – 2√15

(iii) (2√5 + 3√2 )2

= (2√5)2 + (3√2 )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 – b2 and (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/a = 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

RD Sharma Solutions For Class 9 Chapter 3 Exercises:

Get detailed solutions for all the questions listed under below exercises:

Exercise 3.1 Solutions

Exercise 3.2 Solutions

Exercise VSAQs Solutions

RD Sharma Solutions for Chapter 3 Rationalisation

In this chapter students will study important concepts on Exponents of Real Numbers as listed below:

  • Rationalisation Introduction
  • Rationalisation of the denominator
  • Some important algebraic identities

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