# NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

**Algebraic Expressions and Identities Exercise 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 – 12) (3a – 12)

(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) (x2 + 3y4) (x2 + 3y4)

(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) (23 m + 32 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) (43 m – 34 n)^{2} + 2mn = 169 m^{2} + 916 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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