# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

## AP Board Class 8 Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Book Answers

 AP Board Class 8 Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Book Answers

## Andhra Pradesh State Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Books Solutions

 Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 8th Subject Maths Chapters Maths Chapter 11 Algebraic Expressions Ex 11.4 Provider Hsslive

2. Click on the Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Answers.
3. Look for your Andhra Pradesh Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks PDF.
4. Now download or read the Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook Solutions for PDF Free.

## AP Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

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

## Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks for Exam Preparations

Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 exam preparation. The AP Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Books State Board syllabus with maximum efficiency.

## FAQs Regarding Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook Solutions

#### Can we get a Andhra Pradesh State Board Book PDF for all Classes?

Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

## Important Terms

Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4, AP Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks, Andhra Pradesh State Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4, Andhra Pradesh State Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook solutions, AP Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks Solutions, Andhra Pradesh Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4, AP Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks, Andhra Pradesh State Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4, Andhra Pradesh State Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbook solutions, AP Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.4 Textbooks Solutions,
Share: