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} ≡ a^{2} + 2ab + b^{2} 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} = 6^{2} = 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 = 4^{2} = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2

= 25 + 10 + 4 + 10

= 49 sq.units

[∵ (5 + 2)^{2} = 7^{2} = 49]

Question 2.

Verify the identity (a – b)^{2} ≡ a^{2} – 2ab+ b^{2} 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} = a^{2} – 2ab + b^{2} [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} = 2^{2} = 4]

ii)

∴ (a – b)^{2} = a^{2} – 2ab + b^{2}

Area of ABCD + Area of CYZS = a^{2} – 2ab + b^{2}

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) ≡ a^{2} – b^{2} geometrically by taking

(i) a = 3 units, b = 2 units

(ii) a = 2 units, b = 1 unit

Solution:

i)

a^{2} – b^{2} = Area of Fig I. + Area of Fig II

= a(a – b) + b(a – b)

= (a – b) (a + b)

= 3 × 3 – 2 × 2

a^{2} – b^{2} = 9 – 4= 5sq . units

[ ∵ 3^{2} – 2^{2 }= 9 – 4 = 5]

ii)

a^{2} – b^{2} = 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

a^{2} – b^{2} = 3 sq. units

[∵ (2^{2} – 1^{2}) = 4 – 1 = 3]