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

## AP Board Class 8 Maths Chapter 11 Algebraic Expressions Ex 11.5 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Book Answers

 AP Board Class 8 Maths Chapter 11 Algebraic Expressions Ex 11.5 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Book Answers

## Andhra Pradesh State Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Books Solutions

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## AP Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

Question 1.
Verify the identity (a + b)2 ≡ a2 + 2ab + b2 geometrically by taking
(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit
Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)2 = 62 = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 42 = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2
= 25 + 10 + 4 + 10
= 49 sq.units
[∵ (5 + 2)2 = 72 = 49]

Question 2.
Verify the identity (a – b)2 ≡ a2 – 2ab+ b2 geometrically by taking
(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units
Solution:
(i)

Area of AIFE + Area of FGCH = (a – b)2 = a2 – 2ab + b2 [area of AIFE – area of IBGF – area of EFHD + area of FGCH]
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
= 9 – 3 – 3 + 1 = 4
∴ (a – b)2 = 4 sq. units
[∵ (3 – 1 )2 = 22 = 4]

ii)

∴ (a – b)2 = a2 – 2ab + b2
Area of ABCD + Area of CYZS = a2 – 2ab + b2
area of ABCD – area of BXYC – area of DCST + area of CYZS
=5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)2 = (3)2 = 9]

Question 3.
Verify the identity(a + b)(a – b) ≡ a2 – b2 geometrically by taking
(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit
Solution:
i)

a2 – b2 = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a – b) (a + b)
= 3 × 3 – 2 × 2
a2 – b2 = 9 – 4= 5sq . units
[ ∵ 32 – 2= 9 – 4 = 5]

ii)

a2 – b2 = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a + b) (a – b)
=(2 + 1)(2 – 1)
= 3 × 1 = 3
a2 – b2 = 3 sq. units
[∵ (22 – 12) = 4 – 1 = 3]

## Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Textbooks for Exam Preparations

Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 exam preparation. The AP Board STD 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 Books State Board syllabus with maximum efficiency.

## FAQs Regarding Andhra Pradesh Board Class 8th Maths Chapter 11 Algebraic Expressions Ex 11.5 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

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