AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4

Question 1.

Select a suitable identity and find the following products

(i) (3k + 4l)(3k + 4l)

(ii) (ax^{2} + by^{2})(ax^{2} + by^{2})

(iii) (7d – 9e)(7d – 9e)

(iv) (m^{2} – n^{2})(m^{2} + n^{2})

(v) (3t + 9s) (3t – 9s)

(vi) (kl – mn) (kl + mn)

(vii) (6x + 5)(6x + 6)

(viii) (2b – a)(2b +c)

Solution:

(3k + 4l) (3k 4l) = (3k + 4l)^{2} is in the form of (a + b)^{2}.

=(3k)^{2} + 2 × 3k × 4l+ (4l)^{2} [ (a+ b)^{2} = a^{2} + 2ab + b^{2}

= 3k × 3k + 24kl + 4l × 4l

= 9k^{2} + 24kl + 16l^{2}

ii) (ax^{2} + by^{2}) (ax^{2} + by^{2}) = (ax^{2} + by^{2})^{2} is in the form of (a + b)^{2}.

= (ax^{2})^{2} + 2 × ax^{2} × by^{2} + (by^{2})^{2} [ ∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= ax^{2} × ax^{2} + 2abx^{2}y^{2} + by^{2} × by^{2}

= a^{2}x^{4} + 2ab x^{2}y^{2} + b^{2}y^{4}

iii) (7d – 9e) (7d – 9e)

= (7d – 9e)^{2} is in the form of (a – b)^{2}.

= (7d)^{2} – 2 × 7d × 9e + (9e)^{2} [ ∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 7d × 7d – 126de + 9e × 9e

= 49d^{2} – 126de + 81e^{2}

iv) (m^{2} – n^{2}) (m^{2} + n^{2}) is in the form of (a + b) (a – b).

∴ (a + b) (a – b) = a^{2} – b^{2}

∴ (m^{2} + n^{2}) (m^{2} – n^{2}) = (m^{2})^{2} – (n^{2})^{2} = m^{4} – n^{4}

v) (3t + 9s) (3t – 9s) = (3t)^{2} – (9s)^{2} [ ∵ (a + b) (a – b) = a^{2} – b^{2} ]

= 3t × 3t – 9s × 9s

= 9t^{2} – 81s^{2}

vi) (kl – mn) (kl + mn) = (kl)^{2} – (mn)^{2} [ ∵(a + b) (a – b) = a^{2} – b^{2} ]

= kl × kl – mn × mn

= k^{2}l^{2} – m^{2}n^{2}

vii) (6x + 5) (6x + 6) is in the form of

(ax + b) (ax + c).

(ax + b) (ax + c) = a^{2}x^{2} + ax(b + c) + bc

(6x + 5) (6x + 6) = (6)^{2}x^{2} + 6x (5 + 6) + 5 × 6

= 36x^{2} + 6x × 11 + 30

= 36x^{2} + 66x + 30

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).

(ax – b) (ax + c) = a^{2}x^{2} + ax(c – b) – cb

(2b – a) (2b + c) = (2)^{2}(b)^{2} + 2b (c – a) – ca

= 4b^{2} + 2bc – 2ab – ca

Question 2.

Evaluate the following by using suitable identities:

(i) 304^{2}

(ii) 509^{2}

(iii) 992^{2}

(iv) 799^{2}

(v) 304 × 296

(vi) 83 × 77

(vii) 109 × 108

(viii) 204 × 206

Solution:

i) 304^{2} = (300 + 4)^{2} is in the form of (a + b)^{2}.

∵ (a+b)^{2} = a^{2} + 2ab + b^{2}

a = 300, b = 4

(300 + 4)^{2} = (300)^{2} + 2 × 300 × 4 + (4)^{2}

= 300 × 300+ 2400 + 4 × 4

= 90,000 + 2400 + 16

= 92,416

ii) 509^{2} = (500 + 9)^{2}

a = 500, b = 9

= (500)^{2} + 2 × 500 × 9 + (9)^{2}

[ ∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 500 × 500 + 9000 + 9 × 9

= 2,50,000 + 9000 + 81

= 2,59,081

iii) 992^{2} = (1000 – 8)^{2}

a = 1000, b = 8

= (1000)^{2} – 2 × 1000 × 8 + (8)^{2} [∵ (a-b)^{2} = a^{2} – 2ab + b^{2}]

= 1000 × 1000 – 16,000 + 8 × 8

= 10,00,000 – 16000 + 64

= 10,00,064 – 1600

= 9,98,464

iv) 799^{2} = (800 – 1)^{2}

a = 800, b = 1

= (800)^{2} – 2 × 800 × 1 + (1)^{2}

= 800 × 800 – 1600 + 1

= 6,40,000 – 1600 + 1

= 6,40,001 – 1600

= 6,38,401

v) 304 × 296 = (300 + 4) (300 – 4) is in the form of (a + b) (a – b).

(a + b) (a – b) = a^{2} – b^{2}

∴ (300 + 4) (300 – 4) = (300)^{2} – (4)^{2}

= 300 × 300 – 4 × 4

= 90,000 – 16

= 89,984

vi) 83 × 77 = (80 + 3) (80 – 3)

= (80)^{2} – (3)^{2} [ ∵ (a + b) (a – b) = a^{2} – b^{2}]

= 80 × 80 – 3 × 3

= 6400 – 9

= 6391

vii) 109 × 108 = (100 + 9) (100 + 8)

= (100)^{2} + (9 + 8)100 + 9 × 8

= 10,000 + 1700 + 72

= 11,772

viii) 204 × 206 = (205 – 1) (205 + 1)

= (205)^{2} – (1)^{2} [∵ (a + b)(a-b) = a^{2} – b^{2}]

= 205 × 205 – 1 × 1

= 42,025 -1

= 42,024