**NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and
Identities Ex 9.5**

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions
and Identities Ex 9.5 are the part of NCERT Solutions for Class 8 Maths. Here
you can find the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic
Expressions and Identities Ex 9.5**.**** **

**Ex 9.5 Class 8 Maths
Question 1.**

Use a suitable identity to get each of the following
products.(i) (x + 3) (x + 3)

(ii) (2y + 5) (2y + 5)

(iii) (2a – 7) (2a – 7)

(iv) (3a – ½) (3a – ½)

(v) (1.1m – 0.4) (1.1m + 0.4)

(vi) (a

^{2}+ b

^{2}) (-a

^{2}+ b

^{2})

(vii) (6x – 7) (6x + 7)

(viii) (-a + c) (-a + c)

(ix) (x/2 + 3y/4) (x/2 + 3y/4)

(x) (7a – 9b) (7a – 9b)

**Solution:**

**Ex 9.5 Class 8 Maths Question 2.**

Use the identity (x + a)(x + b) = x^{2}+ (a + b)x + ab to find the following products.

(i) (x + 3) (x + 7)

(ii) (4x + 5)(4x + 1)

(iii) (4x – 5) (4x – 1)

(iv) (4x + 5) (4x – 1)

(v) (2x + 5y) (2x + 3y)

(vi) (2a

^{2}+ 9) (2a

^{2}+ 5)

(vii) (xyz – 4) (xyz – 2)

**Solution:**

**Ex 9.5 Class 8
Maths Question 3.**

Find the following squares by using the identities.(i) (b – 7)

^{2}

(ii) (xy + 3z)

^{2}

(iii) (6x

^{2}– 5y)

^{2}

(iv) (2/3 m + 3/2 n)

^{2}

(v) (0.4p – 0.5q)

^{2}

(vi) (2xy + 5y)

^{2}

**Solution:**

**Ex 9.5 Class 8 Maths Question 4.**

Simplify.(i) (a

^{2}– b

^{2})

^{2}

(ii) (2x + 5)

^{2}– (2x – 5)

^{2}

(iii) (7m – 8n)

^{2}+ (7m + 8n)

^{2}

(iv) (4m + 5n)

^{2}+ (5m + 4n)

^{2}

(v) (2.5p – 1.5q)

^{2}– (1.5p – 2.5q)

^{2}

(vi) (ab + bc)

^{2}– 2ab

^{2}c

(vii) (m

^{2}– n

^{2}m)

^{2}+ 2m

^{3}n

^{2}

**Solution:**

**Ex 9.5 Class 8
Maths Question 5.**

Show that.(i) (3x + 7)

^{2}– 84x = (3x – 7)

^{2}

(ii) (9p – 5q)

^{2}+ 180pq = (9p + 5q)

^{2}

(iii) (4/3 m – 3/4 n)

^{2}+ 2mn = 16/9 m

^{2}+ 9/16 n

^{2}

(iv) (4pq + 3q)

^{2}– (4pq – 3q)

^{2}= 48pq

^{2}

(v) (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

**Solution:**

**Ex 9.5 Class 8 Maths Question 6.**

Using identities, evaluate.(i) 71

^{2}

(ii) 99

^{2}

(iii) 102

^{2}

(iv) 998

^{2}

(v) 5.2

^{2}

(vi) 297 × 303

(vii) 78 × 82

(viii) 8.9

^{2}

(ix) 1.05 × 9.5

**Solution:**

**Ex 9.5 Class 8 Maths Question 7.**

Using a^{2}– b

^{2}= (a + b) (a – b), find

(i) 51

^{2}– 49

^{2}

(ii) (1.02)

^{2}– (0.98)

^{2}

(iii) 153

^{2}– 147

^{2}

(iv) 12.1

^{2}– 7.9

^{2}

**Solution:
**(i) 51

^{2}– 49

^{2}= (51 + 49) (51 – 49) = 100 × 2 = 200

(ii) (1.02)

^{2}– (0.98)

^{2}= (1.02 + 0.98) (1.02 – 0.98) = 2.00 × 0.04 = 0.08

(iii) 153

^{2}– 147

^{2}= (153 + 147) (153 – 147) = 300 × 6 = 1800

(iv) 12.1

^{2}– 7.9

^{2}= (12.1 + 7.9) (12.1 – 7.9) = 20.0 × 4.2 = 84

**Ex 9.5 Class 8 Maths Question 8.**

Using (x + a) (x + b) = x^{2}+ (a + b)x + ab, find

(i) 103 × 104

(ii) 5.1 × 5.2

(iii) 103 × 98

(iv) 9.7 × 9.8

**Solution:
**(i) 103 × 104 = (100 + 3)(100 + 4) = (100)

^{2}+ (3 + 4) (100) + 3 × 4 = 10000 + 700 + 12 = 10712

(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) = (5)

^{2}+ (0.1 + 0.2) (5) + 0.1 × 0.2 = 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52

(iii) 103 × 98 = (100 + 3) (100 – 2) = (100)

^{2}+ (3 – 2) (100) + 3 × (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094

(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2) = (10)

^{2}– (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06

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