AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.5

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 – 2² = 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]