Skip to content
Related Articles

Related Articles

Class 8 NCERT Solutions – Chapter 6 Squares and Square Roots – Exercise 6.3
  • Last Updated : 13 Nov, 2020

Question 1. What could be the possible one’s digits 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 × 3 × 3 × 3 × 3 × 3 × 3

729 = (3 × 3) × (3 × 3) × (3 × 3)

729 = (3 × 3 × 3) × (3 × 3 × 3)

729 = (3 × 3 × 3)2

Therefore, √729 = 3 × 3 × 3  = 27

ii. 400

Solution:

400 = 1 × 5 × 5 × 2 × 2 × 2 × 2

400 = (2 × 2) × (2 × 2) × (5 × 5)

400 = (2 × 2 × 5) × (2 × 2 × 5)

400 = (2 × 2 × 5)2

Therefore, √400 = 2 × 2 × 5 = 20

iii. 1764

Solution:

1764 = 2 × 2 × 3 × 3 × 7 × 7 × 1

1764 = (2 × 2) × (3 × 3) × (7 × 7)

1764 = (2 × 3 × 7) × (2 × 3 × 7)

1764 = (2 × 3 × 7)2

Therefore, √1764 = 2 × 3 × 7 = 42

iv. 4096

Solution:

4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 1

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

Therefore, √4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

v. 7744

Solution:

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

Therefore, √7744 = 2 × 2 × 2 × 11 = 88

vi. 9604

Solution:

9604 = 2 × 2 × 7 × 7 × 7 × 7× 1

9604 = (2 × 2) × (7 × 7) × (7 × 7)

9604 = (2 × 7 × 7) × (2 × 7 ×7)

9604 = (2 × 7 × 7)2

Therefore, √9604 = 2 × 7 × 7 = 98

vii. 5929

Solution:

5929 = 7 × 7 × 11 × 11

5929 = (7 × 7) × (11 × 11)

5929 = (7 × 11) × (7 × 11)

5929 = (7 × 11)2

Therefore, √5929 = 7 × 11 = 77

viii. 9216

Solution:

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

Therefore, √9216 = 96

ix. 529

Solution:

529 = 23 × 23 × 1

529 = (23)2

Therefore, √529 = 23

x. 8100

Solution:

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

Therefore, √8100 = 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 × 2 × 3 × 3 × 7

= (2 × 2) × (3 × 3) × 7

7 cannot be paired.

Therefore, multiply by 7 to get perfect square.

New number obtained = 252 × 7 = 1764

1764 = 2 × 2 × 3 × 3 × 7 × 7

1764 = (2 × 2) × (3 × 3) × (7 × 7)

1764 = (2 × 3 × 7)2

Therefore, √1764 = 2×3×7 = 42

ii. 180

Solution:

180 = 2 × 2 × 3 × 3 × 5

= (2 × 2) × (3 × 3) × 5

5 cannot be paired.

Therefore, multiply by 5 to get perfect square.

New number obtained = 180 × 5 = 900

900 = 2 × 2 × 3 × 3 × 5 × 5 × 1

900 = (2 × 2) × (3 × 3) × (5 × 5)

900 = (2 × 3 × 5)2

Therefore, √900 = 2 × 3 × 5 = 30

iii. 1008

Solution:

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

= (2 × 2) × (2 × 2) × (3 × 3) × 7

7 cannot be paired.

Therefore, multiply by 7 to get perfect square.

New number obtained = 1008 × 7 = 7056

7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

7056 = (2 × 2) × (2 × 2) × (3 × 3) × (7 × 7)

7056 = (2 × 2 × 3 × 7)2

Therefore, √7056 = 2 × 2 × 3 × 7 = 84

iv. 2028

Solution:

2028 = 2 × 2 × 3 × 13 × 13

= (2 × 2) × (13 × 13) × 3

3 cannot be paired.

Therefore, multiply by 3 to get perfect square. 

New number obtained = 2028 × 3 = 6084

6084 = 2 × 2 × 3 × 3 × 13 ×13

 6084 = (2 × 2) × (3 × 3) × (13 × 13)

6084 = (2 × 3 × 13)2

Therefore, √6084 = 2×3×13 = 78

v. 1458

Solution:

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

= (3 × 3) × (3 × 3) × (3 × 3) × 2

2 cannot be paired.

Therefore, multiply by 2 to get perfect square. 

New number obtained = 1458 × 2 = 2916

2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

2916 = (3 × 3) × (3 × 3) × (3 × 3) × (2 × 2)

2916 = (3×3×3×2)2

Therefore, √2916 = 3×3×3×2 = 54

vi. 768

Solution:

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × 3

3 cannot be paired.

Therefore, multiply 768 by 3 to get perfect square.

New number obtained  = 768×3 = 2304

2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

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

2304 = (2 × 2 × 2 × 2 × 3)2

√2304 = 2 × 2 × 2 × 2 × 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 × 2 × 3 × 3 × 7

= (2 × 2) × (3 × 3) × 7

7 cannot be paired.

Divide 252 by 7 to get perfect square. 

Therefore, New number obtained = 252 ÷ 7 = 36

36 = 2 × 2 × 3 × 3

36 = (2 × 2) × (3 × 3)

36 = (2 × 3)2

Therefore, √36 = 2 × 3 = 6

ii. 2925

Solution:

252 = 2 × 2 × 3 × 3 × 7

= (2 × 2) × (3 × 3) × 7

7 cannot be paired.

Divide by 7 to get perfect square. 

Therefore, New number obtained = 252 ÷ 7 = 36

36 = 2 × 2 × 3 × 3

36 = (2 × 2) × (3 × 3)

36 = (2 × 3)2

Therefore, √36 = 2 × 3 = 6

iii. 396

Solution:

396 = 2 × 2 × 3 × 3 × 11

= (2 × 2) × (3 × 3) × 11

11 cannot be paired.

Divide by 11 to get perfect square. 

Therefore, New number obtained = 396 ÷ 11 = 36

36 = 2 × 2 × 3 × 3

36 = (2 × 2) × (3 × 3)

36 = (2 × 3)2

Therefore, √36 = 2 × 3 = 6

iv. 2645

Solution:

2645 = 5 × 23 × 23

2645 = (23 × 23) × 5

5 cannot be paired.

Divide  by 5 to get perfect square.

Therefore, New number obtained = 2645 ÷ 5 = 529

529 = 23 × 23

529 = (23)2

Therefore, √529 = 23

v. 2800

Solution:

2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

= (2 × 2) × (2 × 2) × (5 × 5) × 7

7 cannot be paired.

Divide by 7 to get perfect square. 

Therefore, New number obtained = 2800 ÷ 7 = 400

400 = 2 × 2 × 2 × 2 × 5 × 5

400 = (2 × 2) × (2 × 2) × (5 × 5)

400 = (2 × 2 × 5)2

Therefore, √400 = 20

vi. 1620

Solution:

1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5

= (2 × 2) × (3 × 3) × (3 × 3) × 5

5 cannot be paired.

Divide by 5 to get perfect square. 

Therefore, New number obtained = 1620 ÷ 5 = 324

324 = 2 × 2 × 3 × 3 × 3 × 3

324 = (2 × 2) × (3 × 3) × (3 × 3)

324 = (2 × 3 × 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 × 7 × 7 × 7

a2 = (7 × 7) × (7 × 7)

a2 = 49 × 49

a = √(49 × 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 × 3 × 3 × 3 × 5 × 5

a2 = (3 × 3) × (3 × 3) × (5 × 5)

a2 = (3 × 3 × 5) × (3 × 3 × 5)

a2 = 45 × 45

a = √(45 × 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 × 2 × 3 x 3 × 5) = 180.

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

180 = 2 × 2 × 9 × 5

= (2 × 2)× 3 × 3 × 5

= (2 × 2) × (3 × 3) × 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 × 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 × 2 × 5 × 2 × 3) = 120.

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

120 = 2 × 2 × 3 × 5 × 2

= (2 × 2) × 3 × 5 × 2

3, 5 and 2 cannot be paired.

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

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

= 3600

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :