{"id":28039,"date":"2021-07-02T17:30:27","date_gmt":"2021-07-02T12:00:27","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28039"},"modified":"2021-07-02T17:30:27","modified_gmt":"2021-07-02T12:00:27","slug":"ncert-solutions-for-class-8-maths-chapter-9-ex-9-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-9-ex-9-5\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 9 Algebraic Expressions and Identities Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5<\/h2>\n

Question 1.
\nUse a suitable identity to get each of the following products.
\n(i) (x + 3) (x + 3)
\n(ii) (2y + 5) (2y + 5)
\n(iii) (2a – 7) (2a – 7)
\n(iv) (3a – \\(\\frac{1}{2}\\))(3a – \\(\\frac{1}{2}\\))
\n(v) (1.1m – 0.4) (1.1m + 0.4)
\n(vi) (a2<\/sup> + b2<\/sup>) (-a2<\/sup> + b2<\/sup>)
\n(vii) (6x – 7) (6x + 7)
\n(viii) (-a + c) (-a + c)
\n(ix) \\(\\left(\\frac{x}{2}+\\frac{3 y}{4}\\right)\\left(\\frac{x}{2}+\\frac{3 y}{5}\\right)\\)
\n(x) (7a – 9b) (7a – 9b)
\nAnswer:
\n(i) (x + 3) (x + 3) = (x + 3)2<\/sup>
\n= x2<\/sup> + 2 \u00d7 x \u00d7 3 + 32<\/sup>
\n= x2<\/sup> + 6x +9
\n[Using the identity (a + b)2<\/sup> = a2<\/sup>+ 2ab + b2<\/sup>]<\/p>\n

(ii) (2y + 5) (2y + 5) = (2y + 5)2<\/sup>
\n= (2y)2<\/sup> + 2 \u00d7 2y \u00d7 5 + 52<\/sup>
\n= 4y2<\/sup> + 20y + 25
\n[Using the identity (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]<\/p>\n

(iii) (2a – 7) (2a – 7) = (2a – 7)2<\/sup>
\n= (2a)2<\/sup> – 2 \u00d7 2a \u00d7 7 + (7)2<\/sup>
\n[using the identity (a – b)2<\/sup> = a2<\/sup> -2ab + b2<\/sup>]
\n= 4a2<\/sup> – 28a + 49<\/p>\n

(iv) (3a – \\(\\frac { 1 }{ 2 }\\)) (3a – \\(\\frac { 1 }{ 2 }\\)) = (3a – \\(\\frac { 1 }{ 2 }\\))2<\/sup>
\n= (3a)2<\/sup> – 2 \u00d7 3a \u00d7 \\(\\frac { 1 }{ 2 }\\) + \\(\\frac { 1 }{ 2 }\\)2<\/sup>
\n= 9a2<\/sup> – 3a + \\(\\frac { 1 }{ 4 }\\)
\n[using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n

(v) (1.1m – 0.4) (1.1m + 0.4)
\n= (1.1m)2<\/sup> – (0.4)2<\/sup> = 1.21m2<\/sup> – 0.16
\n[using the identity (a \u00d7 b)(a – b) = (a2<\/sup> – b2<\/sup>)]<\/p>\n

(vi) (a2<\/sup> + b2<\/sup>) (-a2<\/sup> + b2<\/sup>)
\n= (b2<\/sup> + a2<\/sup>) (b2<\/sup> – a2<\/sup>) = (b2<\/sup>)2<\/sup> – (a2<\/sup>)2<\/sup>
\n= b4<\/sup> – a4<\/sup>
\n[using identity (a + b) (a – b) = a2<\/sup> – b2<\/sup> ]<\/p>\n

(vii) (6x – 7) (6x + 7)
\n= (6x)2<\/sup> – 72<\/sup> = 36x2<\/sup> – 49
\n[using the identity (a + b)(a – b) = a2<\/sup> – b2<\/sup>]<\/p>\n

(viii) (- a + c) (- a + c) = (-a + c)2<\/sup>
\n= (-a)2<\/sup> -2 \u00d7 a \u00d7 c + c2<\/sup> = a2<\/sup> – 2ac + c2<\/sup>
\n[Using identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n

\"NCERT<\/p>\n

(x) (7a – 9b) (7a – 9b) = (7a – 9b)2<\/sup>
\n= (7a)2<\/sup> – 2 \u00d7 7a \u00d7 9b \u00d7 (-9b)2<\/sup>
\n[using identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 49a2<\/sup> – 126ab + 81b2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 2.
\nUse the identity (x + a) (x + b) = x2<\/sup> (a + b) x + ab to find the following products.
\n(i) (x + 3) (x + 7)
\n(ii) (4x + 5) (4x + 1)
\n(iii) (4x – 5) (4x – 1)
\n(iv) (4x + 5) (4x – 1)
\n(v) (2x + 5y) (2x + 3y)
\n(vi) (2a2<\/sup> + 9) (2a2<\/sup> + 5)
\n(vii) (xyz – 4) (xyz – 2)
\nAnswer:
\n(i) (x + 3)(x + 7)
\n= x2<\/sup> + (3 + 7) x + 3 \u00d7 7 = x2<\/sup> + 10x + 21<\/p>\n

(ii) (4x + 5)(4x + 1)
\n= (4x)2<\/sup> + (5 + 1) 4x + 5 \u00d7 1
\n= 16x2<\/sup> + 6 \u00d7 4x + 5
\n= 16x2<\/sup> + 24x + 5<\/p>\n

(iii) (4x – 5) (4x – 1)
\n= (4x)2<\/sup> + (-5 -1) 4x + (-5) (-1)
\n= 16x2<\/sup> + (-6) 4x + 5
\n= 16x2<\/sup> – 24x + 5<\/p>\n

(iv) (4x + 5) (4x – 1)
\n= (4x)2<\/sup> + (5 – 1) 4x + 5 x (-1)
\n= 16x2<\/sup> + (4) 4x – 5
\n= 16x2<\/sup> + 16x – 5<\/p>\n

(v) (2x + 5y) (2x + 3y)
\n= (2x)2<\/sup> + (5y + 3y) 2x + 5y \u00d7 3y
\n= 4x2<\/sup> + (8y) 2x + 15y2<\/sup>
\n= 4x2<\/sup> + 16xy + 15y2<\/sup><\/p>\n

(vi) (2a2<\/sup> + 9) (2a2<\/sup> + 5)
\n= (2a2<\/sup>)2<\/sup> + (9 + 5) 2a2<\/sup> + 9 \u00d7 5
\n= 4a4<\/sup> + (14) 2a2<\/sup> + 45
\n= 4a4<\/sup> + 28a2<\/sup> + 45<\/p>\n

(vii) (xyz – 4) (xyz – 2)
\n= (xyz)2<\/sup> + (-4 -2) xyz + (-4) (-2)
\n= x2<\/sup>y2<\/sup>z2<\/sup> + (- 6) xyz + 8
\n= x2<\/sup>y2<\/sup>z2<\/sup> – 6xyz + 8<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind the following squares by using the identities.
\n(i) (b – 7)2<\/sup>
\n(ii) (xy + 3z)2<\/sup>
\n(iii) (6x2<\/sup> – 5y)2<\/sup>
\n(iv)(\\(\\frac { 2 }{ 3 }\\)m + \\(\\frac { 3 }{ 2 }\\)n)2<\/sup>
\n(v) (0.4p – 0.5q)2<\/sup>
\n(vi) (2xy + 5y)2<\/sup>
\nAnswer:
\n(i) (b – 7)2<\/sup> = b2<\/sup> – 2 b 7 + (-7)2<\/sup>
\n= b2<\/sup> – 14b + 49
\n[Using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup> ]<\/p>\n

(ii) (xy + 3z)2<\/sup> = (xy)2<\/sup> + 2 (xy) 3z + (3z)2<\/sup>
\n= x2<\/sup>y2<\/sup> + 6xyz + 9z2<\/sup>
\n[using the identity (a x b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]<\/p>\n

(iii) (6x2<\/sup> – 5y)2<\/sup>
\n= (6x2<\/sup>)2<\/sup> – 2 \u00d7 6x2<\/sup> \u00d7 5y + (5y)2<\/sup>
\n= 36x2<\/sup> – 60x2<\/sup>y + 25y2<\/sup>
\n[Using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup><\/p>\n

(iv) ( \\(\\frac { 2 }{ 3 }\\)m + \\(\\frac { 3 }{ 2 }\\)n)2<\/sup>
\n\"NCERT
\n[Using the identity (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]<\/p>\n

(v) (0.4p – 0.5q)2<\/sup>
\n= (0.4p)2<\/sup> – 2 \u00d7 0.4p \u00d7 0.5q + (0.5q)2<\/sup>
\n= 0.16p2<\/sup> – 0.4pq + 0.25q2<\/sup>
\n[Using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n

(vi) (2xy + 5y)2<\/sup>
\n= (2xy)2<\/sup> + 2 \u00d7 2xy \u00d7 5y + (5y)2<\/sup>
\n= 4x2<\/sup>y2<\/sup> + 20xy2<\/sup> + 25y2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 4.
\nSimplify:
\n(i) (a2<\/sup> – b2<\/sup>)2<\/sup>
\n(ii) (2x + 5)2<\/sup> – (2x – 5)2<\/sup>
\n(iii) (7m – 8n)2<\/sup> + (7m + 8n)2<\/sup>
\n(iv) (4m + 5n)2<\/sup> + (5m + 4n)2<\/sup>
\n(v) (2.5p – 1.5q)2<\/sup> – (1.5p -2.5q)2<\/sup>
\n(vi) (ab + bc)2<\/sup> – 2ab2<\/sup>c
\n(vii) (m2<\/sup> – n2<\/sup>m2<\/sup>)2<\/sup> + 2m3<\/sup>n2<\/sup>
\nAnswer:
\n(i) (a2<\/sup> – b2<\/sup> )2<\/sup> = (a2<\/sup>)2<\/sup> – 2 \u00d7 a2<\/sup> \u00d7 b2<\/sup> + (b2<\/sup>)2<\/sup>
\n= a4<\/sup> – 2a2<\/sup> b2<\/sup> + b4<\/sup>
\n[using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n

(ii) (2x + 5)2<\/sup> – (2x – 5)2<\/sup>
\n= (2x)2<\/sup> + 2 \u00d7 2x \u00d7 5 + (5)2<\/sup> – [(2x)2<\/sup> – 2 \u00d7 2x \u00d7 5 + 52<\/sup>]
\n[using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]
\n(a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>
\n= 4x2<\/sup> + 20x + 25 – (4x2<\/sup> – 20x + 25)
\n= 4x2<\/sup> + 20x + 25 – 4x2<\/sup> + 20x – 25
\n= 4x2<\/sup> – 4x2<\/sup> + 20x + 20x + 25 – 25
\n= 0 + 40x + 0
\n= 40x<\/p>\n

(iii) (7m – 8n)2<\/sup> + (7m + 8n)2<\/sup>
\n= (7m)2<\/sup> – 2 \u00d7 7m \u00d7 8n + (8n)2<\/sup> + (7m)2<\/sup> + 2 \u00d7 7m \u00d7 8n + (8n)2<\/sup>
\n[Using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup> and (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]
\n= 49m2<\/sup> – 112mn + 64n2<\/sup> + 49m2<\/sup> + 112mn + 64n2<\/sup>
\n= 49m2<\/sup> + 49m2<\/sup> – 112mn + 112 mn + 64n2<\/sup> + 64n2<\/sup>
\n= 98m2<\/sup> + 0 + 128n2<\/sup>
\n= 98m2<\/sup> + 128n2<\/sup><\/p>\n

(iv) (4m + 5n)2<\/sup> + (5m + 4n)2<\/sup>
\n= (4m)2<\/sup> + 2 \u00d7 4m \u00d7 5n + (5n)2<\/sup> + (5m)2<\/sup> + 2 \u00d7 5m \u00d7 4n + (4n)2
\n[using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]
\n= 16m2<\/sup> + 40mn + 25n2<\/sup> + 25m2<\/sup> + 40mn + 16n2<\/sup>
\n= 16m2<\/sup> + 25m2<\/sup> + 40mn + 40mn + 25n2<\/sup> + 16n2<\/sup>
\n= 41m2<\/sup> + 80mn + 41n2<\/sup><\/p>\n

(v) (2.5p – 1.5q)2<\/sup> – (1.5p – 2.5q)2<\/sup>
\n= (2.5p)2<\/sup> – 2 \u00d7 2.5p \u00d7 1.5q + (1.5q)2<\/sup> – [(1.5p)2<\/sup> – 2 \u00d7 1.5p \u00d7 2.5q + (2.5q)2<\/sup>]
\n[Using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 6.25p2<\/sup> – 7.5pq + 2.25q2<\/sup> – (2.25p2<\/sup> – 7.5pq + 6.25q2<\/sup>)
\n= 6.25p2<\/sup> – 7.5pq + 2.25q2<\/sup> – 2.25p2<\/sup> + 7.5pq – 6.25q2<\/sup>
\n= 6.25p2<\/sup> – 2.25p2<\/sup> + 2.25q2<\/sup> – 6.25q2<\/sup> – 7.5pq + 7.5pq
\n= 4p2<\/sup> – 4q2<\/sup> + 0 = 4p2<\/sup> – 4q2<\/sup><\/p>\n

(vi) (ab + bc)2<\/sup> – 2ab2<\/sup>c
\n= (ab)2<\/sup> + 2 \u00d7 ab \u00d7 bc + (bc)2<\/sup> – 2ab2<\/sup>c
\n[using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]
\n= a2<\/sup>b2<\/sup> + 2ab2<\/sup>c + b2<\/sup>c2<\/sup> – 2ab2<\/sup>c
\n= a2<\/sup>b2<\/sup> + b2<\/sup>c2<\/sup> + 2ab2<\/sup>c – 2ab2<\/sup>c
\n= a2<\/sup>b2<\/sup> +b2<\/sup>c2<\/sup> + 2ab2<\/sup>c – 2ab2<\/sup>c
\n= a2<\/sup>b2<\/sup> + b2<\/sup>c2<\/sup> + 0
\n= a2<\/sup>b2<\/sup> + b2<\/sup>c2<\/sup><\/p>\n

(vii) (m2<\/sup> – n2<\/sup>m)2<\/sup> + 2m3<\/sup>n2<\/sup>
\n= (m2<\/sup>)2<\/sup> – 2 \u00d7 m2<\/sup> \u00d7 n2<\/sup>m + (n2<\/sup>m)2<\/sup> + 2m3n2
\n= m4<\/sup> – 2m3<\/sup>n2<\/sup> + n4<\/sup>m2<\/sup> + 2m3<\/sup>n2<\/sup>
\n= m4<\/sup> + n4<\/sup>m2<\/sup> – 2m3<\/sup>n2<\/sup> + 2m3<\/sup>n2<\/sup>
\n= m4<\/sup> + n4<\/sup>m2<\/sup> + 0
\n= m4<\/sup> + n4<\/sup>m2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 5.
\nShow that:
\n(i) (3x + 7)2<\/sup> – 84x = (3x – 7)2<\/sup>
\n(ii) (9p – 5q)2<\/sup> + 180pq = (9p + 5q)2<\/sup>
\n(iii) (\\(\\frac { 4 }{ 3 }\\) m – \\(\\frac { 3 }{ 4 }\\) n) + 2mn = \\(\\frac { 16 }{ 9 }\\)m2<\/sup> + \\(\\frac { 16 }{ 9 }\\)n2<\/sup>
\n(iv) (4pq + 3q)2<\/sup> – (4pq – 3q)2<\/sup> = 48pq2<\/sup>
\n(v) (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0
\nAnswer:
\n(i) L.H.S. = (3x + 7)2<\/sup> – 84x
\n= (3x)2<\/sup> + 2 \u00d7 3x \u00d7 7 + 72<\/sup> – 84x
\n[using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]
\n= 9x2<\/sup> + 42x + 49 – 84x
\n= 9x2<\/sup> + 42x – 84x + 49
\n= 9x2<\/sup> – 42x + 49
\nR.H.S. = (3x – 7)2<\/sup>
\n= (3x)2<\/sup> – 2 \u00d7 3x \u00d7 7 + 72<\/sup> = 9x2<\/sup> – 42x + 49
\nR.H.S = L.H.S. Hence, proved<\/p>\n

(ii) L.H.S. = (9p – 5q)2<\/sup> + 180pq
\n= (9p)2<\/sup> – 2 x 9p x 5q + (5q)2<\/sup> + 180pq
\n[Using the formula (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 81p2<\/sup> – 90pq + 25q2<\/sup> + 180pq
\n= 81p2<\/sup> – 90pq + 180pq + 25q2<\/sup>
\n= 81p2<\/sup> + 90pq + 25q2<\/sup>
\nR.H.S. = (9p + 5q)2<\/sup>
\n[Using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]
\n= (9p)2<\/sup> + 2 x 9p x 5q + (5q)2<\/sup>
\n= 81p2<\/sup> + 90pq + 25q2<\/sup>
\nR.H.S. = L.H.S.
\n\u2234 Hence, proved.<\/p>\n

(iii) L.H.S = ( \\(\\frac{4}{3}\\)m – \\(\\frac{3}{4}\\)n)2<\/sup> + 2mn
\n[Using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n\"NCERT<\/p>\n

= \\(\\frac{16}{9}\\)m2<\/sup> + \\(\\frac{9}{16}\\)n2<\/sup> = RHS
\nL.H.S. = R.H.S.
\nHence, proved.<\/p>\n

(iv) L.H.S. = (4pq + 3q)2<\/sup> – (4pq – 3q)2<\/sup>
\n= (4pq)2<\/sup> + 2 \u00d7 4pq \u00d7 3q + (3q)2<\/sup> – [(4pq)2<\/sup> – 2 \u00d7 4pq \u00d7 3q + (3q)2<\/sup>]
\n[Using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> and (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 16p2<\/sup>q2<\/sup> + 24pq2<\/sup> + 9q2<\/sup> – [ 16p2<\/sup>q2<\/sup> – 24pq2<\/sup> + 9q2<\/sup>]
\n= 16p2<\/sup>q2<\/sup> + 24pq2<\/sup> + 9q2<\/sup> – 16p2<\/sup>q2<\/sup> + 24pq2<\/sup> – 9q2<\/sup>
\n= (16p2<\/sup>q2<\/sup> – 16p2<\/sup>q2<\/sup>) + 24pq2<\/sup>+24pq2<\/sup> + 9q2<\/sup> – 9q2<\/sup>
\n= 0 + 48pq2<\/sup> + 0s
\n= 48pq2<\/sup>
\nL.H.S. = R.H.S.
\nHence, proved.<\/p>\n

(v) L.H.S.
\n=(a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a)
\n= a2<\/sup> – b2<\/sup> + b2<\/sup> – c2<\/sup> + c2<\/sup> – a2<\/sup>
\n[Using the identity (a + b) (a – b) = a2 – b2]
\n= a2<\/sup> – a2<\/sup> + b2<\/sup> – b2<\/sup> + c2<\/sup> – c2<\/sup>
\n= 0 + 0 + 0 = 0
\nL.H.S. = R.H.S.
\nHence, proved.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nUsing identities, evaluate.
\n(i) 712<\/sup>
\n(ii) 992<\/sup>
\n(iii) 1022<\/sup>
\n(iv) 9982<\/sup>
\n(v) 5.22<\/sup>
\n(vi) 297 \u00d7 303
\n(vii) 78 \u00d7 82
\n(viii) 8.92<\/sup>
\n(ix) 1.05 \u00d7 9.5
\nAnswer:
\n(i) 712<\/sup> = (70 + 1)2 = 702<\/sup> + 2 \u00d7 70 \u00d7 1 + 12<\/sup> [Usingtheidentity(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]
\n= 4900 + 140 + 1 = 5041<\/p>\n

(ii) 992<\/sup> = (100 – 1)2<\/sup> = 1002<\/sup> – 2 \u00d7 100 \u00d7 1 + 12<\/sup>
\n[Usingtheidentity(a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 10000 – 200 + 1 = 9801<\/p>\n

(iii) (102)2<\/sup> = (100 + 2)2<\/sup> = 1002<\/sup> + 2 \u00d7 100 \u00d7 2 + 22<\/sup>
\n[Using (a + b)2 = a2<\/sup> + 2ab + b2<\/sup>]
\n= 10000 + 400 + 4 = 10404<\/p>\n

(iv) (998)2<\/sup> = (1000 – 2)2<\/sup>
\n= 10002<\/sup> – 2 \u00d7 1000 \u00d7 2 + 22<\/sup>
\n[Using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 1000000 – 4000 + 4 = 996004<\/p>\n

(v) 5.22<\/sup> = (5 + 0.2)2<\/sup> = 52<\/sup> + 2 \u00d7 5 \u00d7 0.2 + (0.2)2<\/sup>
\n[Using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]
\n= 25 + 2 + 0.04 = 27.04<\/p>\n

(vi) 297 \u00d7 303 = (300 – 3) (300 + 3) = 3002<\/sup> \u00d7 32<\/sup>
\n[Using (a + b) (a – b) = a2<\/sup> – b2<\/sup> ]
\n= 90000 – 9 = 89991<\/p>\n

(vii) 78 \u00d7 82 = (80 – 2) (80 + 2) = 802<\/sup> – 22<\/sup>
\n[Using (a + b) (a – b) = a2<\/sup> – b2<\/sup>]
\n= 6400 – 4 = 6396<\/p>\n

(viii)(8.9)2<\/sup> = (9 – 0.1)2<\/sup> = 92<\/sup> – 2 \u00d7 9 \u00d7 0.1 + (0.1)2<\/sup>
\n[Using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 81 – 1.8 + 0.01 = 79.21<\/p>\n

(ix) 1.05 \u00d7 9.5 = \\(\\frac{1}{10}\\) \u00d7 10.5 \u00d7 9.5
\n= \\(\\frac{1}{10}\\)[(10 + 0.5)(10 – 0.5)]
\n= \\(\\frac{1}{10}\\) [102<\/sup> – 0.52<\/sup>]
\n[Using (a + b) (a – b) = a2<\/sup> – b2<\/sup>]
\n= \\(\\frac{1}{10}\\)[100 – 0.25] = \\(\\frac{1}{10}\\) \u00d7 99.75
\n= 9.975<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nUsing a2<\/sup> – b2<\/sup> = (a + b) (a – b) find
\n(i) 512<\/sup> – 492<\/sup>
\n(ii) (1.02)2<\/sup> – (0.98)2<\/sup>
\n(iii) 1532<\/sup> – 1472<\/sup>
\n(iv) 12.12<\/sup> – 7.92<\/sup>
\nAnswer:
\n(i) 512<\/sup> – 492<\/sup> = (51 + 49)(51 – 49)
\n= (100) \u00d7 2 = 200<\/p>\n

(ii) (1.02)2<\/sup> – (0.98)2<\/sup>
\n= (1.02 + 0.98) (1.02 – 0.98)
\n= 2 \u00d7 0.04 = 0.08<\/p>\n

(iii) 1532<\/sup> – 1472<\/sup> = (153 + 147) (153 – 147)
\n= 300 \u00d7 6 = 1800<\/p>\n

(iv) (12.1)2<\/sup> – (7.9)2<\/sup>
\n= (12.1 + 7.9) (12.1 – 7.9)
\n= 20 \u00d7 4.2 = 84<\/p>\n

Question 8.
\nUsing (x + a) (x + b) = x2<\/sup> + (a + b) x + ab, find
\n(i) 103 \u00d7 104
\n(ii) 5.1 \u00d7 5.2
\n(iii) 103 \u00d7 98
\n(iv) 9.7 \u00d7 9.8
\nAnswer:
\n(i) 103 \u00d7 104 = (100+ 3) (100 + 4)
\n= 1002<\/sup> + (3 + 4) 100 + 3 \u00d7 4
\n= 10000 + 700 + 12 = 10712<\/p>\n

(ii) 5.1 \u00d7 5.2= (5 + 0.1) (5 + 0.2)
\n= 52<\/sup> + (0.1 +0.2) 5+ 0.1 \u00d7 0.2
\n= 25 + 1.5 + 0.02 = 26.52<\/p>\n

(iii) 103 \u00d7 98 = (100 + 3) (100 – 2)
\n= 1002<\/sup> + (3 – 2) 100 + (3) (- 2)
\n= 10000 + 100 – 6 = 10094<\/p>\n

(iv) 9.7 \u00d7 9.8 = (10 – 0.3) (10 – 0.2)
\n= 102<\/sup> + (- 0.3 – 0.2) 10 + (- 0.3) (- 0.2)
\n= 100 + (- 0.5) \u00d7 10 + 0.06
\n= 100 – 5 + 0.06 = 95.06.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Question 1. Use a suitable identity to get each of the following products. (i) (x …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-9-ex-9-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts. 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