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. RD Sharma Solutions for Class 9Â 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.
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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 + √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.
= 3√5/5
(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
√2 = 1.414, √3 = 1.732, √5 = 2.236 and √10 = 3.162
Solution:
Question 3: Express each one of the following with rational denominator:
Solution:
Using identity: (a + b) (a – b) = a2 – b2
(i) Multiply and divide given number by 3−√2
(ii) Multiply and divide given number by √6 + √5
(iii) Multiply and divide given number by √41 + 5
(iv) Multiply and divide given number by 5√3 + 3√5
(v) Multiply and divide given number by 2√5 + √3
(vi) Multiply and divide given number by 2√2 + √3
(vii) Multiply and divide given number by 6 – 4√2
(viii) Multiply and divide given number by 2√5 + 3
(ix) Multiply and divide given number by √(a2+b2) – a
Question 4: Rationales the denominator and simplify:
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.
(ii) Multiply both numerator and denominator by 7–4√3 to rationalise the denominator.
(iii) Multiply both numerator and denominator by 3+2√2 to rationalise the denominator.
(iv) Multiply both numerator and denominator by 3√5+2√6 to rationalise the denominator.
(v) Multiply both numerator and denominator by √48–√18 to rationalise the denominator.
(vi) Multiply both numerator and denominator by 2√2 – 3√3 to rationalise the denominator.
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:
Question 3: Write the rationalisation factor of 7 – 3√5.
Solution:
Rationalisation factor of 7 – 3√5 is 7 + 3√5
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 – 2ab]
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
Solution:
Question 7: Simplify
Solution:
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 Solutions for Class 9 Maths Chapter 3 Rationalisation
In the 3rd chapter of Class 9 RD Sharma Solutions students will study important concepts on Exponents of Real Numbers as listed below:
- Rationalisation Introduction
- Rationalisation of the denominator
- Some important algebraic identities
Students can get acquainted with writing exams and will be able to face exams more confidently. Meanwhile students are suggested to solve these solutions without fail to procure great marks in their board exams.
