NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.2

NCERT exercise solutions are helpful to improve the hold of the students in the problems related to square and square roots.. All the questions of this exercise have been solved by subject experts. NCERT solutions for Class 8 maths helps students in enhancing their skills. Download free Maths NCERT Solutions for Chapter 6 of class 8 and get going with the homework as well as the exam preparation.

Download PDF of NCERT Solutions for class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.2

 

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Access other exercise solutions of class 8 Maths Chapter 6- Squares and Square Roots

Exercise 6.1 Solutions 9 Questions

Exercise 6.3 Solutions 10 Questions

Exercise 6.4 Solutions 9 Questions

Access Answers of Maths NCERT class 8 Chapter 6- Squares and Square Roots Exercise 6.2 Page Number 98

1. Find the square of the following numbers.

i. 32

ii. 35

iii. 86

iv. 93

v. 71

vi. 46

Solution:

i. (32)2

= (30 +2)2

= (30)2 + (2)2 + 2×30×2 [Since, (a+b)2 = a2+b2 +2ab]

= 900 + 4 + 120

= 1024

ii. (35)2

= (30+5 )2

= (30)2 + (5)2 + 2×30×5 [Since, (a+b)2 = a2+b2 +2ab]

= 900 + 25 + 300

= 1225

iii. (86)2

= (90 – 4)2

= (90)2 + (4)2 – 2×90×4 [Since, (a+b)2 = a2+b2 +2ab]

= 8100 + 16 – 720

= 8116 – 720

= 7396

iv. (93)2

= (90+3 )2

= (90)2 + (3)2 + 2×90×3 [Since, (a+b)2 = a2+b2 +2ab]

= 8100 + 9 + 540

= 8649

v. (71)2

= (70+1 )2

= (70)2 + (1)2 +2×70×1 [Since, (a+b)2 = a2+b2 +2ab]

= 4900 + 1 + 140

= 5041

vi. (46)2

= (50 -4 )2

= (50)2 + (4)2 – 2×50×4 [Since, (a+b)2 = a2+b2 +2ab]

= 2500 + 16 – 400

= 2116

2. Write a Pythagorean triplet whose one member is.

i. 6

ii. 14

iii. 16

iv. 18

Solution:

For any natural number m, we know that 2m, m2–1, m2+1 is a Pythagorean triplet.

i. 2m = 6

⇒ m = 6/2 = 3

m2–1= 32 – 1 = 9–1 = 8

m2+1= 32+1 = 9+1 = 10

∴ (6, 8, 10) is a Pythagorean triplet.

ii. 2m = 14

⇒ m = 14/2 = 7

m2–1= 72–1 = 49–1 = 48

m2+1 = 72+1 = 49+1 = 50

∴ (14, 48, 50) is not a Pythagorean triplet.

iii. 2m = 16

⇒ m = 16/2 = 8

m2–1 = 82–1 = 64–1 = 63

m2+ 1 = 82+1 = 64+1 = 65

∴ (16, 63, 65) is a Pythagorean triplet.

iv. 2m = 18

⇒ m = 18/2 = 9

m2–1 = 92–1 = 81–1 = 80

m2+1 = 92+1 = 81+1 = 82

∴ (18, 80, 82) is a Pythagorean triplet.


The exercise 6.2 of NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots is based on the following topics:

  1. Finding the square of a number
    • Patterns in squares
    • Pythagorean triplets

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