NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.3

NCERT Solutions have been carefully compiled and developed as per the latest CBSE syllabus. NCERT solutions for Class 8 maths, Chapter 6 Exercise 6.3, exercise questions and answers helps students to understand the topics and concepts related to Squares and Square Roots.

NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.3 are prepared using step by step approach, with an aim to improving problem-solving skills.

Download free NCERT Solutions for Maths Chapter 6 prepared by the subject experts at BYJU’S, from the link below.

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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 6.2 Solutions 2 Questions

Exercise 6.4 Solutions 9 Questions

Access Answers of Maths NCERT Class 8 Chapter 6- Squares and Square Roots Exercise 6.3 Page Number 102

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

i. 9801

ii. 99856

iii. 998001

iv. 657666025

Solution:

i. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[92=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 9801 is equal to 1 or 9.

ii. We know that the unit’s digit of the square of a number having digit as unit’s

place 6 is 6 and also 4 is 6 [62=36 and 42=16, both the squares have unit digit 6].

∴ Unit’s digit of the square root of number 99856 is equal to 6.

iii. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[92=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 998001 is equal to 1 or 9.

iv. We know that the unit’s digit of the square of a number having digit as unit’s

place 5 is 5.

∴ Unit’s digit of the square root of number 657666025 is equal to 5.

2. Without doing any calculation, find the numbers which are surely not perfect squares.

i. 153

ii. 257

iii. 408

iv. 441

Solution:

We know that natural numbers ending with the digits 0, 2, 3, 7 and 8 are not perfect square.

i. 153⟹ Ends with 3.

∴, 153 is not a perfect square

ii. 257⟹ Ends with 7

∴, 257 is not a perfect square

iii. 408⟹ Ends with 8

∴, 408 is not a perfect square

iv. 441⟹ Ends with 1

∴, 441 is a perfect square.

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Solution:

100

100 – 1 = 99

99 – 3 = 96

96 – 5 = 91

91 – 7 = 84

84 – 9 = 75

75 – 11 = 64

64 – 13 = 51

51 – 15 = 36

36 – 17 = 19

19 – 19 = 0

Here, we have performed subtraction ten times.

∴ √100 = 10

169

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 88

88 – 19 = 69

69 – 21 = 48

48 – 23 = 25

25 – 25 = 0

Here, we have performed subtraction thirteen times.

∴ √169 = 13

4. Find the square roots of the following numbers by the Prime Factorisation Method.

i. 729

ii. 400

iii. 1764

iv. 4096

v. 7744

vi. 9604

vii. 5929

viii. 9216

ix. 529

x. 8100

Solution:

i.

NCERT Solution For Class 8 Maths Chapter 6 Image 1

729 = 3×3×3×3×3×3×1

⇒ 729 = (3×3)×(3×3)×(3×3)

⇒ 729 = (3×3×3)×(3×3×3)

⇒ 729 = (3×3×3)2

⇒ √729 = 3×3×3 = 27

ii.

NCERT Solution For Class 8 Maths Chapter 6 Image 2

400 = 2×2×2×2×5×5×1

⇒ 400 = (2×2)×(2×2)×(5×5)

⇒ 400 = (2×2×5)×(2×2×5)

⇒ 400 = (2×2×5)2

⇒ √400 = 2×2×5 = 20

iii.

NCERT Solution For Class 8 Maths Chapter 6 Image 3

1764 = 2×2×3×3×7×7

⇒ 1764 = (2×2)×(3×3)×(7×7)

⇒ 1764 = (2×3×7)×(2×3×7)

⇒ 1764 = (2×3×7)2

⇒ √1764 = 2 ×3×7 = 42

iv.

NCERT Solution For Class 8 Maths Chapter 6 Image 4

4096 = 2×2×2×2×2×2×2×2×2×2×2×2

⇒ 4096 = (2×2)×(2×2)×(2×2)×(2×2)×(2×2)×(2×2)

⇒ 4096 = (2×2×2×2×2×2)×(2×2×2×2×2×2)

⇒ 4096 = (2×2×2×2×2×2)2

⇒ √4096 = 2×2×2 ×2×2×2 = 64

v.

NCERT Solution For Class 8 Maths Chapter 6 Image 5

7744 = 2×2×2×2×2×2×11×11×1

⇒ 7744 = (2×2)×(2×2)×(2×2)×(11×11)

⇒ 7744 = (2×2×2×11)×(2×2×2×11)

⇒ 7744 = (2×2×2×11)2

⇒ √7744 = 2×2×2×11 = 88

vi.

NCERT Solution For Class 8 Maths Chapter 6 Image 6

9604 = 62 × 2 × 7 × 7 × 7 × 7

⇒ 9604 = ( 2 × 2 ) × ( 7 × 7 ) × ( 7 × 7 )

⇒ 9604 = ( 2 × 7 ×7 ) × ( 2 × 7 ×7 )

⇒ 9604 = ( 2×7×7 )2

⇒ √9604 = 2×7×7 = 98

NCERT Solution For Class 8 Maths Chapter 6 Image 7vii.

5929 = 7×7×11×11

⇒ 5929 = (7×7)×(11×11)

⇒ 5929 = (7×11)×(7×11)

⇒ 5929 = (7×11)2

⇒ √5929 = 7×11 = 77

viii.

NCERT Solution For Class 8 Maths Chapter 6 Image 7

9216 = 2×2×2×2×2×2×2×2×2×2×3×3×1

⇒ 9216 = (2×2)×(2×2) × ( 2 × 2 ) × ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 )

⇒ 9216 = ( 2 × 2 × 2 × 2 × 2 × 3) × ( 2 × 2 × 2 × 2 × 2 × 3)

⇒ 9216 = 96 × 96

⇒ 9216 = ( 96 )2

⇒ √9216 = 96

ix.

NCERT Solution For Class 8 Maths Chapter 6 Image 8

529 = 23×23

529 = (23)2

√529 = 23

x.

NCERT Solution For Class 8 Maths Chapter 6 Image 9

8100 = 2×2×3×3×3×3×5×5×1

⇒ 8100 = (2×2) ×(3×3)×(3×3)×(5×5)

⇒ 8100 = (2×3×3×5)×(2×3×3×5)

⇒ 8100 = 90×90

⇒ 8100 = (90)2

⇒ √8100 = 90

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

i. 252

ii. 180

iii. 1008

iv. 2028

v. 1458

vi. 768

Solution:

i.

NCERT Solution For Class 8 Maths Chapter 6 Image 10

252 = 2×2×3×3×7

= (2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will multiply 252 by 7 to get perfect square.

New number = 252 × 7 = 1764

NCERT Solution For Class 8 Maths Chapter 6 Image 11

1764 = 2×2×3×3×7×7

⇒ 1764 = (2×2)×(3×3)×(7×7)

⇒ 1764 = 22×32×72

⇒ 1764 = (2×3×7)2

⇒ √1764 = 2×3×7 = 42

ii.

NCERT Solution For Class 8 Maths Chapter 6 Image 12

180 = 2×2×3×3×5

= (2×2)×(3×3)×5

Here, 5 cannot be paired.

∴ We will multiply 180 by 5 to get perfect square.

New number = 180 × 5 = 900

NCERT Solution For Class 8 Maths Chapter 6 Image 13

900 = 2×2×3×3×5×5×1

⇒ 900 = (2×2)×(3×3)×(5×5)

⇒ 900 = 22×32×52

⇒ 900 = (2×3×5)2

⇒ √900 = 2×3×5 = 30

iii.

NCERT Solution For Class 8 Maths Chapter 6 Image 14

1008 = 2×2×2×2×3×3×7

= (2×2)×(2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will multiply 1008 by 7 to get perfect square.

New number = 1008×7 = 7056

NCERT Solution For Class 8 Maths Chapter 6 Image 15

7056 = 2×2×2×2×3×3×7×7

⇒ 7056 = (2×2)×(2×2)×(3×3)×(7×7)

⇒ 7056 = 22×22×32×72

⇒ 7056 = (2×2×3×7)2

⇒ √7056 = 2×2×3×7 = 84

iv.

NCERT Solution For Class 8 Maths Chapter 6 Image 16

2028 = 2×2×3×13×13

= (2×2)×(13×13)×3

Here, 3 cannot be paired.

∴ We will multiply 2028 by 3 to get perfect square. New number = 2028×3 = 6084

NCERT Solution For Class 8 Maths Chapter 6 Image 17

6084 = 2×2×3×3×13×13

⇒ 6084 = (2×2)×(3×3)×(13×13)

⇒ 6084 = 22×32×132

⇒ 6084 = (2×3×13)2

⇒ √6084 = 2×3×13 = 78

v.

NCERT Solution For Class 8 Maths Chapter 6 Image 18

1458 = 2×3×3×3×3×3×3

= (3×3)×(3×3)×(3×3)×2

Here, 2 cannot be paired.

∴ We will multiply 1458 by 2 to get perfect square. New number = 1458 × 2 = 2916

NCERT Solution For Class 8 Maths Chapter 6 Image 19

2916 = 2×2×3×3×3×3×3×3

⇒ 2916 = (3×3)×(3×3)×(3×3)×(2×2)

⇒ 2916 = 32×32×32×22

⇒ 2916 = (3×3×3×2)2

⇒ √2916 = 3×3×3×2 = 54

vi.

NCERT Solution For Class 8 Maths Chapter 6 Image 20

768 = 2×2×2×2×2×2×2×2×3

= (2×2)×(2×2)×(2×2)×(2×2)×3

Here, 3 cannot be paired.

∴ We will multiply 768 by 3 to get perfect square.

New number = 768×3 = 2304

NCERT Solution For Class 8 Maths Chapter 6 Image 21

2304 = 2×2×2×2×2×2×2×2×3×3

⇒ 2304 = (2×2)×(2×2)×(2×2)×(2×2)×(3×3)

⇒ 2304 = 22×22×22×22×32

⇒ 2304 = (2×2×2×2×3)2

⇒ √2304 = 2×2×2×2×3 = 48

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

i. 252

ii. 2925

iii. 396

iv. 2645

v. 2800

vi. 1620

Solution:

i.

NCERT Solution For Class 8 Maths Chapter 6 Image 22

252 = 2×2×3×3×7

= (2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will divide 252 by 7 to get perfect square. New number = 252 ÷ 7 = 36

NCERT Solution For Class 8 Maths Chapter 6 Image 23

36 = 2×2×3×3

⇒ 36 = (2×2)×(3×3)

⇒ 36 = 22×32

⇒ 36 = (2×3)2

⇒ √36 = 2×3 = 6

ii.

NCERT Solution For Class 8 Maths Chapter 6 Image 24

2925 = 3×3×5×5×13

= (3×3)×(5×5)×13

Here, 13 cannot be paired.

∴ We will divide 2925 by 13 to get perfect square. New number = 2925 ÷ 13 = 225

NCERT Solution For Class 8 Maths Chapter 6 Image 24

225 = 3×3×5×5

⇒ 225 = (3×3)×(5×5)

⇒ 225 = 32×52

⇒ 225 = (3×5)2

⇒ √36 = 3×5 = 15

iii.

NCERT Solution For Class 8 Maths Chapter 6 Image 25

396 = 2×2×3×3×11

= (2×2)×(3×3)×11

Here, 11 cannot be paired.

∴ We will divide 396 by 11 to get perfect square. New number = 396 ÷ 11 = 36

NCERT Solution For Class 8 Maths Chapter 6 Image 26

36 = 2×2×3×3

⇒ 36 = (2×2)×(3×3)

⇒ 36 = 22×32

⇒ 36 = (2×3)2

⇒ √36 = 2×3 = 6

iv.

NCERT Solution For Class 8 Maths Chapter 6 Image 27

2645 = 5×23×23

⇒ 2645 = (23×23)×5

Here, 5 cannot be paired.

∴ We will divide 2645 by 5 to get perfect square.

New number = 2645 ÷ 5 = 529

NCERT Solution For Class 8 Maths Chapter 6 Image 28

529 = 23×23

⇒ 529 = (23)2

⇒ √529 = 23

v.

NCERT Solution For Class 8 Maths Chapter 6 Image 29

2800 = 2×2×2×2×5×5×7

= (2×2)×(2×2)×(5×5)×7

Here, 7 cannot be paired.

∴ We will divide 2800 by 7 to get perfect square. New number = 2800 ÷ 7 = 400

NCERT Solution For Class 8 Maths Chapter 6 Image 30

400 = 2×2×2×2×5×5

⇒ 400 = (2×2)×(2×2)×(5×5)

⇒ 400 = (2×2×5)2

⇒ √400 = 20

vi.

NCERT Solution For Class 8 Maths Chapter 6 Image 31

1620 = 2×2×3×3×3×3×5

= (2×2)×(3×3)×(3×3)×5

Here, 5 cannot be paired.

∴ We will divide 1620 by 5 to get perfect square. New number = 1620 ÷ 5 = 324

NCERT Solution For Class 8 Maths Chapter 6 Image 32

324 = 2×2×3×3×3×3

⇒ 324 = (2×2)×(3×3)×(3×3)

⇒ 324 = (2×3×3)2

⇒ √324 = 18

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Solution:

Let the number of students in the school be, x.

∴ Each student donate Rs.x .

Total many contributed by all the students= x×x=x2 Given, x2 = Rs.2401

NCERT Solution For Class 8 Maths Chapter 6 Image 33

x2 = 7×7×7×7

⇒ x2 = (7×7)×(7×7)

⇒ x2 = 49×49

⇒ x = √(49×49)

⇒ x = 49

∴ The number of students = 49

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Solution

Let the number of rows be, x.

∴ the number of plants in each rows = x.

Total many contributed by all the students = x × x =x2

Given,

x2 = Rs.2025

NCERT Solution For Class 8 Maths Chapter 6 Image 34

x2 = 3×3×3×3×5×5

⇒ x2 = (3×3)×(3×3)×(5×5)

⇒ x2 = (3×3×5)×(3×3×5)

⇒ x2 = 45×45

⇒ x = √45×45

⇒ x = 45

∴ The number of rows = 45 and the number of plants in each rows = 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution:

NCERT Solution For Class 8 Maths Chapter 6 Image 35

L.C.M of 4, 9 and 10 is (2×2×9×5) 180.

180 = 2×2×9×5

= (2×2)×3×3×5

= (2×2)×(3×3)×5

Here, 5 cannot be paired.

∴ we will multiply 180 by 5 to get perfect square.

Hence, the smallest square number divisible by 4, 9 and 10 = 180×5 = 900

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution:

NCERT Solution For Class 8 Maths Chapter 6 Image 36

L.C.M of 8, 15 and 20 is (2×2×5×2×3) 120.

120 = 2×2×3×5×2

= (2×2)×3×5×2

Here, 3, 5 and 2 cannot be paired.

∴ We will multiply 120 by (3×5×2) 30 to get perfect square.

Hence, the smallest square number divisible by 8, 15 and 20 =120×30 = 3600


Exercise 6.3 of NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots is based on the following topics:

  1. Square Roots
    • Finding square roots
    • Finding square root through repeated subtraction
    • Finding square root through prime factorisation

1 Comment

  1. Square root is good and best chepter I will read all exercise of this chapter

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