RD Sharma Solutions for Class 9 Maths Chapter 3 Rationalisation

RD Sharma Solutions for Class 9 Maths Chapter 3 – Free PDF Download

RD Sharma Solutions for Class 9 Maths Chapter 3 – Rationalisation is one of the most important chapters in Class 9. Students learn about different algebraic identities and rationalisation of the denominator in Â RD Sharma Solutions for Class 9 Chapter 3. 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 the PDF of Chapter 3 click on the below link.

RD Sharma Solutions are useful for students as it helps them in scoring high marks in the examination. These solutions are prepared by subject matter experts at BYJUâ€™S, describing the complete method of solving problems. In order to excel in exams, students can view and download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 3.

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Exercise 3.1

Question 1: Simplify each of the following:

Solution:

(i)

(ii)

Question 2: Simplify the following expressions:

(i) (4 + âˆš7) (3 + âˆš2)

(ii) (3 + âˆš3)(5- âˆš2 )

(iii) (âˆš5 -2)( âˆš3 â€“ âˆš5)

Solution:

(i) (4 + âˆš7) (3 + âˆš2)

= 12 + 4âˆš2 + 3âˆš7 + âˆš14

(ii) (3 + âˆš3)(5- âˆš2 )

= 15 – 3âˆš2 + 5âˆš3 – âˆš6

(iii) (âˆš5 -2)( âˆš3 â€“ âˆš5)

= âˆš15 – âˆš25 – 2âˆš3 + 2âˆš5

= âˆš15 – 5 – 2âˆš3 + 2âˆš5

Question 3: Simplify the following expressions:

(i) (11 + âˆš11) (11 â€“ âˆš11)

(ii) (5 + âˆš7) (5 â€“âˆš7 )

(iii) (âˆš8 â€“ âˆš2 ) (âˆš8 + âˆš2 )

(iv) (3 + âˆš3) (3 – âˆš3)

(v) (âˆš5 – âˆš2) (âˆš5 + âˆš2)

Solution:

Using Identity: (a â€“ b)(a+b) = a2 â€“ b2

(i) (11 + âˆš11) (11 â€“ âˆš11)

= 112 â€“ (âˆš11)2

= 121 â€“ 11

= 110

(ii) (5 + âˆš7) (5 â€“âˆš7 )

= (52 â€“ (âˆš7)2 )

= 25 â€“ 7 = 18

(iii) (âˆš8 â€“ âˆš2 ) (âˆš8 + âˆš2 )

= (âˆš8)2 â€“ (âˆš2 ) 2

= 8 -2

= 6

(iv) (3 + âˆš3) (3 – âˆš3)

= (3)2 â€“ (âˆš3)2

= 9 â€“ 3

= 6

(v) (âˆš5 – âˆš2) (âˆš5 + âˆš2)

=(âˆš5)2 – (âˆš2)2

= 5 â€“ 2

= 3

Question 4: Simplify the following expressions:

(i) (âˆš3 + âˆš7)2

(ii) (âˆš5 â€“ âˆš3)2

(iii) (2âˆš5 + 3âˆš2 )2

Solution:

Using identities: (a â€“ b)2 = a2 + b2 â€“ 2ab and (a + b)2 = a2 + b2 + 2ab

(i) (âˆš3 + âˆš7)2

= (âˆš3)2 + (âˆš7)2 + 2(âˆš3)( âˆš7)

= 3 + 7 + 2âˆš21

= 10 + 2âˆš21

(ii) (âˆš5 â€“ âˆš3)2

= (âˆš5)2 + (âˆš3)2 – 2(âˆš5)( âˆš3)

= 5 + 3 – 2âˆš15

= 8 – 2âˆš15

(iii) (2âˆš5 + 3âˆš2 )2

= (2âˆš5)2 + (3âˆš2 )2 + 2(2âˆš5 )( 3âˆš2)

= 20 + 18 + 12âˆš10

= 38 + 12âˆš10

Exercise 3.2

Question 1: Rationalise the denominators of each of the following (i â€“ vii):

(i) 3/ âˆš5 (ii) 3/(2 âˆš5) (iii) 1/ âˆš12 (iv) âˆš2/ âˆš5

(v) (âˆš3 + 1)/ âˆš2 (vi) (âˆš2 + âˆš5)/ âˆš3 (vii) 3 âˆš2/ âˆš5

Solution:

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

= 3âˆš5/5

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

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

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

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

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

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

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

âˆš2 = 1.414, âˆš3 = 1.732, âˆš5 = 2.236 and âˆš10 = 3.162

Solution:

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

Solution:

Using identity: (a + b) (a â€“ b) = a2 â€“ b2

(i) Multiply and divide given number by 3âˆ’âˆš2

(ii) Multiply and divide given number by âˆš6 + âˆš5

(iii) Multiply and divide given number by âˆš41 + 5

(iv) Multiply and divide given number by 5âˆš3 + 3âˆš5

(v) Multiply and divide given number by 2âˆš5 + âˆš3

(vi) Multiply and divide given number by 2âˆš2 + âˆš3

(vii) Multiply and divide given number by 6 – 4âˆš2

(viii) Multiply and divide given number by 2âˆš5 + 3

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

Question 4: Rationales the denominator and simplify:

Solution:

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

(i) Multiply both numerator and denominator by âˆš3â€“âˆš2 to rationalise the denominator.

(ii) Multiply both numerator and denominator by 7â€“4âˆš3 to rationalise the denominator.

(iii) Multiply both numerator and denominator by 3+2âˆš2 to rationalise the denominator.

(iv) Multiply both numerator and denominator by 3âˆš5+2âˆš6 to rationalise the denominator.

(v) Multiply both numerator and denominator by âˆš48â€“âˆš18 to rationalise the denominator.

(vi) Multiply both numerator and denominator by 2âˆš2 â€“ 3âˆš3 to rationalise the denominator.

Exercise VSAQs

Question 1: Write the value of (2 + âˆš3) (2 â€“ âˆš3).

Solution:

(2 + âˆš3) (2 â€“ âˆš3)

= (2)2 â€“ (âˆš3)2

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

= 4 – 3

= 1

Question 2: Write the reciprocal of 5 + âˆš2.

Solution:

Question 3: Write the rationalisation factor of 7 â€“ 3âˆš5.

Solution:

Rationalisation factor of 7 â€“ 3âˆš5 is 7 + 3âˆš5

Question 4: If

Find the values of x and y.

Solution:

[Using identities : (a + b)(a – b) = a2 – b2 and (a – b)2 = a2 + b2 â€“ 2ab]

Question 5: If x = âˆš2 â€“ 1, then write the value of 1/x.

Solution:

x = âˆš2 â€“ 1

or 1/x = 1/(âˆš2 â€“ 1)

Rationalising denominator, we have

= 1/(âˆš2 â€“ 1) x (âˆš2 + 1)/(âˆš2 + 1)

= (âˆš2 + 1)/(2-1)

= âˆš2 + 1

Question 6: Simplify

Solution:

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

Question 7: Simplify

Solution:

[ Because: (a – b)2 = a2 + b2 – 2ab ]

Question 8: If a = âˆš2 +1, then find the value of a â€“ 1/a.

Solution:

Given: a = âˆš2 + 1

1/a = 1/(âˆš2 + 1)

= 1/(âˆš2 + 1) x (âˆš2 – 1)/(âˆš2 – 1)

= (âˆš2 – 1)/ ((âˆš2)2 – (1)2)

= (âˆš2 – 1)/1

= âˆš2 – 1

Now,

a – 1/a = (âˆš2 + 1) – (âˆš2 – 1)

= 2

Question 9: If x = 2 + âˆš3, find the value of x + 1/x.

Solution:

Given: x = 2 + âˆš3

1/x = 1/(2 + âˆš3)

= 1/(2 + âˆš3) x (2 – âˆš3)/(2 – âˆš3)

= (2 – âˆš3)/ ((2)2 – (âˆš3)2)

= (2 – âˆš3)/(4-3)

= (2 – âˆš3)

Now,

x + 1/x = (2 + âˆš3) + (2 – âˆš3)

= 4

Question 10: Write the rationalisation factor of âˆš5 â€“ 2.

Solution:

Rationalisation factor of âˆš5 â€“ 2 is âˆš5 + 2

Question 11: If x = 3 + 2âˆš2, then find the value of âˆšx â€“ 1/âˆšx.

Solution:

RD Sharma Solutions for Class 9 Maths Chapter 3 Rationalisation

In the 3rd chapter of Class 9 RD Sharma Solutions students will study important concepts on Rationalisation as listed below:

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

Students can get acquainted with various methods of solving problems with ease and face exams more confidently. Meanwhile students are suggested to solve these solutions without fail to procure great marks in their board exams.

Frequently Asked Questions on RD Sharma Solutions for Class 9 Maths Chapter 3

How many exercises are present in RD Sharma Solutions for Class 9 Maths Chapter 3?

RD Sharma Solutions for Class 9 Maths Chapter 3 has 2 exercises. The topics discussed in these exercises are 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. Practice is an essential task to learn and score well in Mathematics. Hence the solutions are designed by BYJUâ€™S experts to boost confidence among students in solving the complex problems effortlessly.

Why should I opt for RD Sharma Solutions for Class 9 Maths Chapter 3?

The concepts present in RD Sharma Solutions for Class 9 Maths Chapter 3 are explained in simple language which help students to understand the subject better. Solutions are prepared by a set of experts at BYJUâ€™S with the aim of helping students boost their exam preparation. Students can download the solutions in PDF format for effective learning as per their needs.

Is RD Sharma Solutions for Class 9 Maths Chapter 3 difficult to learn?

No, practising RD Sharma Solutions for Class 9 Maths Chapter 3 diligently helps students to understand the concepts easily and score high marks in exams. These solutions are formulated by a set of experts at BYJUâ€™S to enhance studentâ€™s skills in Mathematics. Students are highly recommended to follow RD Sharma Solutions for effective exam preparation.