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) (ax2 + by2)(ax2 + by2)
(iii) (7d – 9e)(7d – 9e)
(iv) (m2 – n2)(m2 + n2)
(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 = a2 + 2ab + b2
= 3k × 3k + 24kl + 4l × 4l
= 9k2 + 24kl + 16l2

ii) (ax2 + by2) (ax2 + by2) = (ax2 + by2)2 is in the form of (a + b)2.
= (ax2)2 + 2 × ax2 × by2 + (by2)2 [ ∵ (a + b)2 = a2 + 2ab + b2]
= ax2 × ax2 + 2abx2y2 + by2 × by2
= a2x4 + 2ab x2y2 + b2y4

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 = a2 – 2ab + b2]
= 7d × 7d – 126de + 9e × 9e
= 49d2 – 126de + 81e2

iv) (m2 – n2) (m2 + n2) is in the form of (a + b) (a – b).
∴ (a + b) (a – b) = a2 – b2
∴ (m2 + n2) (m2 – n2) = (m2)2 – (n2)2 = m4 – n4

v) (3t + 9s) (3t – 9s) = (3t)2 – (9s)2 [ ∵ (a + b) (a – b) = a2 – b2 ]
= 3t × 3t – 9s × 9s
= 9t2 – 81s2

vi) (kl – mn) (kl + mn) = (kl)2 – (mn)2 [ ∵(a + b) (a – b) = a2 – b2 ]
= kl × kl – mn × mn
= k2l2 – m2n2

vii) (6x + 5) (6x + 6) is in the form of
(ax + b) (ax + c).
(ax + b) (ax + c) = a2x2 + ax(b + c) + bc
(6x + 5) (6x + 6) = (6)2x2 + 6x (5 + 6) + 5 × 6
= 36x2 + 6x × 11 + 30
= 36x2 + 66x + 30

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).
(ax – b) (ax + c) = a2x2 + ax(c – b) – cb
(2b – a) (2b + c) = (2)2(b)2 + 2b (c – a) – ca
= 4b2 + 2bc – 2ab – ca

Question 2.
Evaluate the following by using suitable identities:
(i) 3042
(ii) 5092
(iii) 9922
(iv) 7992
(v) 304 × 296
(vi) 83 × 77
(vii) 109 × 108
(viii) 204 × 206
Solution:
i) 3042 = (300 + 4)2 is in the form of (a + b)2.
∵ (a+b)2 = a2 + 2ab + b2
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) 5092 = (500 + 9)2
a  = 500, b = 9
= (500)2 + 2 × 500 × 9 + (9)2
[ ∵ (a + b)2 = a2 + 2ab + b2]
= 500 × 500 + 9000 + 9 × 9
= 2,50,000 + 9000 + 81
= 2,59,081

iii) 9922 = (1000 – 8)2
a = 1000, b = 8
= (1000)2 – 2 × 1000 × 8 + (8)2 [∵ (a-b)2 = a2 – 2ab + b2]
= 1000 × 1000 – 16,000 + 8 × 8
= 10,00,000 – 16000 + 64
= 10,00,064 – 1600
= 9,98,464

iv) 7992 = (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) = a2 – b2
∴ (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) = a2 – b2]
= 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) = a2 – b2]
= 205 × 205 – 1 × 1
= 42,025 -1
= 42,024