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

Last Updated : 11 Sep, 2024

In Chapter 6 of Class 8 Mathematics, students explore Squares and Square Roots which are fundamental concepts in number theory. Exercise 6.3 specifically focuses on the calculation and application of square roots including using the methods like prime factorization and long division. Understanding these concepts is crucial for solving problems involving areas, powers, and various algebraic expressions.

What are Squares and Square Roots?

  • Squares: The square of a number is the product of the number multiplied by itself. For example 42=16. Mathematically for any number x its square is π‘₯2.
  • Square Roots: The square root of a number is the value that when multiplied by itself gives the original number. For example, the square root of the 16 is 4 as 4Γ—4=16.

Question 1. What could be the possible one digit of the square root of each of the following numbers?

i. 9801

Solution:

Unit place digit of the number is 1

And we all know 12 = 1 & 92 = 81 whose unit place is 1

Therefore, one’s digit of the square root of 9801 should equal to 1 or 9.

ii. 99856

Solution:

Unit place digit of the number is 6

And we all know 62 = 36 & 42 = 16, both the squares have unit place 6.

Therefore, one’s digit of the square root of 99856 is equal to 6 or 4.

iii. 998001

Solution:

Unit place digit of the number is 1

And we all know 12 = 1 & 92 = 81 whose unit place is 1

Therefore, one’s digit of the square root of 998001 should equal to 1 or 9.

iv. 657666025

Solution:

Unit place digit of the number is 5

And we all know 52 = 25 whose unit place is 5

Therefore, one’s digit of the square root of 657666025 should equal to 5.

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

i. 153

Solution:

Unit place digit of the number is 3.

Therefore, 153 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].

ii. 257

Solution:

Unit place digit of the number is 7.

Therefore, 257 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].

iii. 408

Solution:

Unit place digit of the number is 8.

Therefore, 408 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].

iv. 441

Solution:

Unit place digit of the number is 1.

Therefore, 441 is a perfect square

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

Solution:

For 100

100 - 1 = 99                 [1]

99 - 3 = 96                   [2]

96 - 5 = 91                   [3]

91 - 7 = 84                   [4]

84 - 9 = 75                   [5]

75 - 11 = 64                 [6]

64 - 13 = 51                 [7]

51 - 15 = 36                 [8]

36 - 17 = 19                 [9]

19 -19 = 0                 [10]

Here, subtraction has been performed for ten times.

Therefore,  √100 = 10

For 169

169 - 1 = 168                 [1]

168 - 3 = 165                 [2]

165 - 5 = 160                 [3]

160 - 7 = 153                 [4]

153 - 9 = 144                 [5]

144 - 11 = 133                 [6]

133 - 13 = 120                 [7]

120 - 15 = 105                 [8]

105 - 17 = 88                 [9]

88 - 19 = 69                 [10]

69 - 21 = 48                 [11]

48 - 23 = 25                 [12]

25 - 25 = 0                 [13]

Here, subtraction has been performed for thirteen times.

Therefore, √169 = 13

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

i. 729

Solution:

729 = 1 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 3

729 = (3 Γ— 3) Γ— (3 Γ— 3) Γ— (3 Γ— 3)

729 = (3 Γ— 3 Γ— 3) Γ— (3 Γ— 3 Γ— 3)

729 = (3 Γ— 3 Γ— 3)2

Therefore, βˆš729 = 3 Γ— 3 Γ— 3  = 27

ii. 400

Solution:

400 = 1 Γ— 5 Γ— 5 Γ— 2 Γ— 2 Γ— 2 Γ— 2

400 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (5 Γ— 5)

400 = (2 Γ— 2 Γ— 5) Γ— (2 Γ— 2 Γ— 5)

400 = (2 Γ— 2 Γ— 5)2

Therefore, βˆš400 = 2 Γ— 2 Γ— 5 = 20

iii. 1764

Solution:

1764 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 7 Γ— 7 Γ— 1

1764 = (2 Γ— 2) Γ— (3 Γ— 3) Γ— (7 Γ— 7)

1764 = (2 Γ— 3 Γ— 7) Γ— (2 Γ— 3 Γ— 7)

1764 = (2 Γ— 3 Γ— 7)2

Therefore, βˆš1764 = 2 Γ— 3 Γ— 7 = 42

iv. 4096

Solution:

4096 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 1

4096 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2)

4096 = (2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2) Γ— (2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ—2)

4096 = (2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2)2

Therefore, βˆš4096 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 = 64

v. 7744

Solution:

7744 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 11 Γ— 11 Γ— 1

7744 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (11 Γ— 11)

7744 = (2 Γ— 2 Γ— 2 Γ— 11) Γ—( 2 Γ— 2 Γ— 2 Γ— 11)

7744 = (2 Γ— 2 Γ— 2 Γ— 11)2

Therefore, βˆš7744 = 2 Γ— 2 Γ— 2 Γ— 11 = 88

vi. 9604

Solution:

9604 = 2 Γ— 2 Γ— 7 Γ— 7 Γ— 7 Γ— 7Γ— 1

9604 = (2 Γ— 2) Γ— (7 Γ— 7) Γ— (7 Γ— 7)

9604 = (2 Γ— 7 Γ— 7) Γ— (2 Γ— 7 Γ—7)

9604 = (2 Γ— 7 Γ— 7)2

Therefore, βˆš9604 = 2 Γ— 7 Γ— 7 = 98

vii. 5929

Solution:

5929 = 7 Γ— 7 Γ— 11 Γ— 11

5929 = (7 Γ— 7) Γ— (11 Γ— 11)

5929 = (7 Γ— 11) Γ— (7 Γ— 11)

5929 = (7 Γ— 11)2

Therefore, βˆš5929 = 7 Γ— 11 = 77

viii. 9216

Solution:

9216 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3 Γ— 3 Γ— 1

9216 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (3 Γ— 3)

9216 = (2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3) Γ— (2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3)

9216 = 96 Γ— 96

9216 = (96)2

Therefore, βˆš9216 = 96

ix. 529

Solution:

529 = 23 Γ— 23 Γ— 1

529 = (23)2

Therefore, βˆš529 = 23

x. 8100

Solution:

8100 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 5 Γ— 5 Γ— 1

8100 = (2 Γ— 2) Γ— (3 Γ— 3) Γ— (3 Γ— 3) Γ— (5 Γ— 5)

8100 = (2 Γ— 3 Γ— 3 Γ— 5) Γ— (2 Γ— 3 Γ— 3 Γ— 5)

8100 = 90 Γ— 90

8100 = (90)2

Therefore, βˆš8100 = 90

Question 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

Solution:

252 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 7

= (2 Γ— 2) Γ— (3 Γ— 3) Γ— 7

7 cannot be paired.

Therefore, multiply by 7 to get perfect square.

New number obtained = 252 Γ— 7 = 1764

1764 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 7 Γ— 7

1764 = (2 Γ— 2) Γ— (3 Γ— 3) Γ— (7 Γ— 7)

1764 = (2 Γ— 3 Γ— 7)2

Therefore, βˆš1764 = 2Γ—3Γ—7 = 42

ii. 180

Solution:

180 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 5

= (2 Γ— 2) Γ— (3 Γ— 3) Γ— 5

5 cannot be paired.

Therefore, multiply by 5 to get perfect square.

New number obtained = 180 Γ— 5 = 900

900 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 5 Γ— 5 Γ— 1

900 = (2 Γ— 2) Γ— (3 Γ— 3) Γ— (5 Γ— 5)

900 = (2 Γ— 3 Γ— 5)2

Therefore, βˆš900 = 2 Γ— 3 Γ— 5 = 30

iii. 1008

Solution:

1008 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3 Γ— 3 Γ— 7

= (2 Γ— 2) Γ— (2 Γ— 2) Γ— (3 Γ— 3) Γ— 7

7 cannot be paired.

Therefore, multiply by 7 to get perfect square.

New number obtained = 1008 Γ— 7 = 7056

7056 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3 Γ— 3 Γ— 7 Γ— 7

7056 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (3 Γ— 3) Γ— (7 Γ— 7)

7056 = (2 Γ— 2 Γ— 3 Γ— 7)2

Therefore, βˆš7056 = 2 Γ— 2 Γ— 3 Γ— 7 = 84

iv. 2028

Solution:

2028 = 2 Γ— 2 Γ— 3 Γ— 13 Γ— 13

= (2 Γ— 2) Γ— (13 Γ— 13) Γ— 3

3 cannot be paired.

Therefore, multiply by 3 to get perfect square. 

New number obtained = 2028 Γ— 3 = 6084

6084 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 13 Γ—13

 6084 = (2 Γ— 2) Γ— (3 Γ— 3) Γ— (13 Γ— 13)

6084 = (2 Γ— 3 Γ— 13)2

Therefore, βˆš6084 = 2Γ—3Γ—13 = 78

v. 1458

Solution:

1458 = 2 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 3

= (3 Γ— 3) Γ— (3 Γ— 3) Γ— (3 Γ— 3) Γ— 2

2 cannot be paired.

Therefore, multiply by 2 to get perfect square. 

New number obtained = 1458 Γ— 2 = 2916

2916 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 3

2916 = (3 Γ— 3) Γ— (3 Γ— 3) Γ— (3 Γ— 3) Γ— (2 Γ— 2)

2916 = (3Γ—3Γ—3Γ—2)2

Therefore, βˆš2916 = 3Γ—3Γ—3Γ—2 = 54

vi. 768

Solution:

768 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3

= (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— 3

3 cannot be paired.

Therefore, multiply 768 by 3 to get perfect square.

New number obtained  = 768Γ—3 = 2304

2304 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3 Γ— 3

2304 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (2 Γ— 2) Γ— (3 Γ— 3)

2304 = (2 Γ— 2 Γ— 2 Γ— 2 Γ— 3)2

√2304 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 3 = 48

Question 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

Solution:

252 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 7

= (2 Γ— 2) Γ— (3 Γ— 3) Γ— 7

7 cannot be paired.

Divide 252 by 7 to get perfect square. 

Therefore, New number obtained = 252 Γ· 7 = 36

36 = 2 Γ— 2 Γ— 3 Γ— 3

36 = (2 Γ— 2) Γ— (3 Γ— 3)

36 = (2 Γ— 3)2

Therefore, βˆš36 = 2 Γ— 3 = 6

ii. 2925

Solution:

252 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 7

= (2 Γ— 2) Γ— (3 Γ— 3) Γ— 7

7 cannot be paired.

Divide by 7 to get perfect square. 

Therefore, New number obtained = 252 Γ· 7 = 36

36 = 2 Γ— 2 Γ— 3 Γ— 3

36 = (2 Γ— 2) Γ— (3 Γ— 3)

36 = (2 Γ— 3)2

Therefore, βˆš36 = 2 Γ— 3 = 6

iii. 396

Solution:

396 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 11

= (2 Γ— 2) Γ— (3 Γ— 3) Γ— 11

11 cannot be paired.

Divide by 11 to get perfect square. 

Therefore, New number obtained = 396 Γ· 11 = 36

36 = 2 Γ— 2 Γ— 3 Γ— 3

36 = (2 Γ— 2) Γ— (3 Γ— 3)

36 = (2 Γ— 3)2

Therefore, βˆš36 = 2 Γ— 3 = 6

iv. 2645

Solution:

2645 = 5 Γ— 23 Γ— 23

2645 = (23 Γ— 23) Γ— 5

5 cannot be paired.

Divide  by 5 to get perfect square.

Therefore, New number obtained = 2645 Γ· 5 = 529

529 = 23 Γ— 23

529 = (23)2

Therefore, βˆš529 = 23

v. 2800

Solution:

2800 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 5 Γ— 5 Γ— 7

= (2 Γ— 2) Γ— (2 Γ— 2) Γ— (5 Γ— 5) Γ— 7

7 cannot be paired.

Divide by 7 to get perfect square. 

Therefore, New number obtained = 2800 Γ· 7 = 400

400 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 5 Γ— 5

400 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (5 Γ— 5)

400 = (2 Γ— 2 Γ— 5)2

Therefore, βˆš400 = 20

vi. 1620

Solution:

1620 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 3 Γ— 3 Γ— 5

= (2 Γ— 2) Γ— (3 Γ— 3) Γ— (3 Γ— 3) Γ— 5

5 cannot be paired.

Divide by 5 to get perfect square. 

Therefore, New number obtained = 1620 Γ· 5 = 324

324 = 2 Γ— 2 Γ— 3 Γ— 3 Γ— 3 Γ— 3

324 = (2 Γ— 2) Γ— (3 Γ— 3) Γ— (3 Γ— 3)

324 = (2 Γ— 3 Γ— 3)2

√324 = 18

Question 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 as assume number of students be, a

So, Each Student has donated Rs a.

Therefore, Total amount donated = a x a

That mean's a x a = 2401

a2 = 2401

a2 = 7 Γ— 7 Γ— 7 Γ— 7

a2 = (7 Γ— 7) Γ— (7 Γ— 7)

a2 = 49 Γ— 49

a = √(49 Γ— 49)

a = 49

Therefore, The number of students = 49

Question 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 as assume number of rows be, a

So, Each row has number of plants = a.

Therefore, Total number of plants = a x a

That mean's a x a = 2025

a2 = 3 Γ— 3 Γ— 3 Γ— 3 Γ— 5 Γ— 5

a2 = (3 Γ— 3) Γ— (3 Γ— 3) Γ— (5 Γ— 5)

a2 = (3 Γ— 3 Γ— 5) Γ— (3 Γ— 3 Γ— 5)

a2 = 45 Γ— 45

a = √(45 Γ— 45)

a = 45

Therefore, The number of rows = 45 and also number of plants in each rows = 45.

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

Solution:

First, we have to find L.C.M of 4, 9 and 10

4 = 2 x 2 x 1

9 = 3 x 3 x 1

5 = 1 x 5

Therefore,  L.C.M = (2 Γ— 2 Γ— 3 x 3 Γ— 5) = 180.

Now we have to find  the smallest whole number divisible by 180

180 = 2 Γ— 2 Γ— 9 Γ— 5

= (2 Γ— 2)Γ— 3 Γ— 3 Γ— 5

= (2 Γ— 2) Γ— (3 Γ— 3) Γ— 5

5 cannot be paired.

Therefore,  multiply 180 by 5 to get perfect square.

The smallest square number divisible by 180 and also by  4, 9 and 10 = 180 Γ— 5 

= 900

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

Solution:

First, we have to find L.C.M of 8, 15 and 20

8 = 1 x 2 x 2 x 2

15 = 1 x 5 x 3

20 = 1 x 2 x 5 x 2

Therefore,  L.C.M = (2 Γ— 2 Γ— 5 Γ— 2 Γ— 3) = 120.

Now we have to find  the smallest whole number divisible by 120

120 = 2 Γ— 2 Γ— 3 Γ— 5 Γ— 2

= (2 Γ— 2) Γ— 3 Γ— 5 Γ— 2

3, 5 and 2 cannot be paired.

Therefore, multiply 120 by (3 Γ— 5 Γ— 2) i.e 30 to get perfect square.

The smallest square number divisible by 120 and also by 8, 15 and 20 = 120 Γ— 30 

= 3600

Conclusion

Chapter 6 of the Class 8 NCERT textbook helps students solidify their understanding of the Squares and Square Roots emphasizing methods for the calculating square roots efficiently. Exercise 6.3 enhances these skills by providing the practical problems that help in the grasping these concepts for the future mathematical applications.

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