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