RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.2 are available here in PDF for easy access and download. In the Class 9 RD Sharma textbook, students will explore the various topics in more detail. In Chapter 4, students will learn about the identity of the square of a trinomial. Refer to the RD Sharma Class 9 Maths Chapter 4 Exercise 4.2 PDF given below.
RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.2
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 4 Algebraic Identities Exercise 4.2 Page Number 4.11
Exercise 4.2 Page No: 4.11
Question 1: Write the following in the expanded form:
(i) (a + 2b + c)2
(ii) (2a − 3b − c)2
(iii) (−3x+y+z)2
(iv) (m+2n−5p)2
(v) (2+x−2y)2
(vi) (a2 +b2 +c2) 2
(vii) (ab+bc+ca) 2
(viii) (x/y+y/z+z/x)2
(ix) (a/bc + b/ac + c/ab) 2
(x) (x+2y+4z) 2
(xi) (2x−y+z) 2
(xii) (−2x+3y+2z) 2
Solution:
Using identities:
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz
(i) (a + 2b + c)2
= a2 + (2b) 2 + c2 + 2a(2b) + 2ac + 2(2b)c
= a2 + 4b2 + c2 + 4ab + 2ac + 4bc
(ii) (2a − 3b − c)2
= [(2a) + (−3b) + (−c)]2
= (2a) 2 + (−3b) 2 + (−c) 2 + 2(2a)(−3b) + 2(−3b)(−c) + 2(2a)(−c)
= 4a2 + 9b2 + c2 − 12ab + 6bc − 4ca
(iii) (−3x+y+z)2
= [(−3x) 2 + y2 + z2 + 2(−3x)y + 2yz + 2(−3x)z
= 9x2 + y2 + z2 − 6xy + 2yz − 6xz
(iv) (m+2n−5p)2
= m2 + (2n) 2 + (−5p) 2 + 2m × 2n + (2×2n×−5p) + 2m × −5p
= m2 + 4n2 + 25p2 + 4mn − 20np − 10pm
(v) (2+x−2y)2
= 22 + x2 + (−2y) 2 + 2(2)(x) + 2(x)(−2y) + 2(2)(−2y)
= 4 + x2 + 4y2 + 4 x − 4xy − 8y
(vi) (a2 +b2 +c2) 2
= (a2) 2 + (b2) 2 + (c2 ) 2 + 2a2 b2 + 2b2c2 + 2a2c2
= a4 + b4 + c4 + 2a2 b2 + 2b2 c2 + 2c2 a2
(vii) (ab+bc+ca) 2
= (ab)2 + (bc) 2 + (ca) 2 + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)
= a2 b2 + b2c2 + c2 a2 + 2(ac)b2 + 2(ab)(c) 2 + 2(bc)(a) 2
(viii) (x/y+y/z+z/x)2
(ix) (a/bc + b/ac + c/ab) 2
(x) (x+2y+4z) 2
= x2 + (2y) 2 + (4z) 2 + (2x)(2y) + 2(2y)(4z) + 2x(4z)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
(xi) (2x−y+z) 2
= (2x) 2 + (−y) 2 + (z) 2 + 2(2x)(−y) + 2(−y)(z) + 2(2x)(z)
= 4x2 + y2 + z2 − 4xy−2yz+4xz
(xii) (−2x+3y+2z) 2
= (−2x) 2 + (3y) 2 + ( 2z) 2 + 2(−2x)(3y)+2(3y)(2z)+2(−2x)(2z)
= 4x2 + 9y2 + 4z2 −12xy+12yz−8xz
Question 2: Simplify
(i) (a + b + c)2 + (a − b + c) 2
(ii) (a + b + c)2 − (a − b + c) 2
(iii) (a + b + c)2 + (a – b + c) 2 + (a + b − c) 2
(iv) (2x + p − c)2 − (2x − p + c) 2
(v) (x2 + y2 − z2) 2 − (x2 − y2 + z2) 2
Solution:
(i) (a + b + c)2 + (a − b + c) 2
= (a2 + b2 + c2 + 2ab+2bc+2ca) + (a2 + (−b) 2 + c2 −2ab−2bc+2ca)
= 2a2 + 2 b2 + 2c2 + 4ca
(ii) (a + b + c)2 − (a − b + c) 2
= (a2 + b2 + c2 + 2ab+2bc+2ca) − (a2 + (−b) 2 + c2 −2ab−2bc+2ca)
= a2 + b2 + c2 + 2ab + 2bc + 2ca − a2 − b2 − c2 + 2ab + 2bc − 2ca
= 4ab + 4bc
(iii) (a + b + c)2 + (a – b + c) 2 + (a + b − c) 2
= a2 + b2 + c2 + 2ab + 2bc + 2ca + (a2 + b2 + (c) 2 − 2ab − 2cb + 2ca) + (a2 + b2 + c2 + 2ab − 2bc – 2ca)
= 3 a2 + 3b2 + 3c2 + 2ab − 2bc + 2ca
(iv) (2x + p − c)2 − (2x − p + c) 2
= [4x2 + p2 + c2 + 4xp − 2pc − 4xc] − [4x2 + p2 + c2 − 4xp− 2pc + 4xc]
= 4x2 + p2 + c2 + 4xp − 2pc − 4cx − 4x2 − p2 − c2 + 4xp + 2pc− 4cx
= 8xp − 8xc
= 8(xp − xc)
(v) (x2 + y2 − z2) 2 − (x2 − y2 + z2) 2
= (x2 + y2 + (−z) 2) 2 − (x2 − y2 + z2) 2
= [x4 + y4 + z4 + 2x2y2 – 2y2z 2 – 2x2z2 − [x4 + y4 + z4 − 2x2y2 − 2y2z2 + 2x2z2]
= 4x2y2 – 4z2x2
Question 3: If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca.
Solution:
a + b + c = 0 and a2 + b2 + c2 = 16 (given)
Choose a + b + c = 0
Squaring both sides,
(a + b + c)2 = 0
a2 + b2 + c2 + 2(ab + bc + ca) = 0
16 + 2(ab + bc + c) = 0
2(ab + bc + ca) = -16
ab + bc + ca = -16/2 = -8
or ab + bc + ca = -8
RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.2
RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.2 are based on the identity for the square of a trinomial.
Some special cases:
- (a + b – c)2 = a2 + b2 + c2 + 2ab – 2bc – 2ca
- (a – b +c)2 = a2 + b2 + c2 – 2ab – 2bc + 2ca
- (-a + b +c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
- (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
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