# NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots

**NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots** are beneficial for students since it aids them in scoring high marks in the exam. The subject experts at BYJUโS outline the concepts in a distinct and well-defined fashion, keeping the IQ level of students in mind. These solution modules use various shortcut hints and practical examples to explain all the exercise questions in a simple and easily understandable language. Solving the NCERT Class 8 Solutions is a must to obtain an excellent score in the examination. Here, with the NCERT Solutions provided, the students will learn various techniques to determine whether a given natural number is a perfect square or not. The answers to all problems within this chapter in the NCERT books are presented here in a detailed and step by step way to help the students understand more effectively.

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Exercise 6.1 Page: 96

**1. What will be the unit digit of the squares of the following numbers?**

**i. 81**

**ii. 272**

**iii. 799**

**iv. 3853**

**v. 1234**

**vi. 26387**

**vii. 52698**

**viii. 99880**

**ix. 12796**

**x. 55555**

Solution:

The unit digit of square of a number having โaโ at its unit place ends with aรa.

i. The unit digit of the square of a number having digit 1 as unitโs place is 1.

โด Unit digit of the square of number 81 is equal to 1.

ii. The unit digit of the square of a number having digit 2 as unitโs place is 4.

โด Unit digit of the square of number 272 is equal to 4.

iii. The unit digit of the square of a number having digit 9 as unitโs place is 1.

โด Unit digit of the square of number 799 is equal to 1.

iv. The unit digit of the square of a number having digit 3 as unitโs place is 9.

โด Unit digit of the square of number 3853 is equal to 9.

v. The unit digit of the square of a number having digit 4 as unitโs place is 6.

โด Unit digit of the square of number 1234 is equal to 6.

vi. The unit digit of the square of a number having digit 7 as unitโs place is 9.

โด Unit digit of the square of number 26387 is equal to 9.

vii. The unit digit of the square of a number having digit 8 as unitโs place is 4.

โด Unit digit of the square of number 52698 is equal to 4.

viii. The unit digit of the square of a number having digit 0 as unitโs place is 01.

โด Unit digit of the square of number 99880 is equal to 0.

ix. The unit digit of the square of a number having digit 6 as unitโs place is 6.

โด Unit digit of the square of number 12796 is equal to 6.

x. The unit digit of the square of a number having digit 5 as unitโs place is 5.

โด Unit digit of the square of number 55555 is equal to 5.

**2. The following numbers are obviously not perfect squares. Give reason.**

**i. 1057**

**ii. 23453**

**iii. 7928**

**iv. 222222**

**v. 64000**

**vi. 89722**

**vii. 222000**

**viii. 505050**

Solution:

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

i. 1057 โน Ends with 7

ii. 23453 โน Ends with 3

iii. 7928 โน Ends with 8

iv. 222222 โน Ends with 2

v. 64000 โน Ends with 0

vi. 89722 โน Ends with 2

vii. 222000 โน Ends with 0

viii. 505050 โน Ends with 0

**3. The squares of which of the following would be odd numbers?**

**i. 431**

**ii. 2826**

**iii. 7779**

**iv. 82004**

Solution:

We know that the square of an odd number is odd and the square of an even number is even.

i. The square of 431 is an odd number.

ii. The square of 2826 is an even number.

iii. The square of 7779 is an odd number.

iv. The square of 82004 is an even number.

**4. Observe the following pattern and find the missing numbers. 11 ^{2} = 121**

**101 ^{2} = 10201**

**1001 ^{2} = 1002001**

**100001 ^{2} = 1 โฆโฆ.2โฆโฆโฆ1**

**10000001 ^{2} = โฆโฆโฆโฆโฆโฆโฆโฆ..**

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.

โด 100001^{2} = 10000200001

10000001^{2} = 100000020000001

**5. Observe the following pattern and supply the missing numbers. 112 = 121**

**1012 = 10201**

**101012 = 102030201**

**10101012 = โฆโฆโฆโฆโฆโฆโฆโฆโฆ**

**โฆโฆโฆโฆ2 = 10203040504030201**

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And, the square is symmetric about the middle digit. If the middle digit is 4, then the number to be squared is 10101 and its square is 102030201.

So, 10101012 =1020304030201

1010101012 =10203040505030201

**6. Using the given pattern, find the missing numbers. 1 ^{2} + 2^{2} + 2^{2} = 3^{2}**

**2 ^{2} + 3^{2} + 6^{2} = 7^{2}**

**3 ^{2} + 4^{2} + 12^{2} = 13^{2}**

**4 ^{2} + 5^{2} + _2 = 21^{2}**

**5 + _ ^{2} + 30^{2} = 31^{2}**

**6 + 7 + _ ^{2} = _ ^{2}**

Solution:

Given, 1^{2} + 2^{2} + 2^{2} = 3^{2}

i.e 1^{2} + 2^{2} + (1ร2 )^{2} = ( 1^{2} + 2^{2} -1 ร 2 )^{2}

2^{2} + 3^{2} + 6^{2} =7^{2}

โด 2^{2} + 3^{2} + (2ร3 )^{2} = (2^{2} + 3^{2} -2 ร 3)^{2}

3^{2 }+ 4^{2} + 12^{2} = 13^{2}

โด 3^{2} + 4^{2} + (3ร4 )^{2} = (3^{2} + 4^{2} โ 3 ร 4)^{2}

4^{2} + 5^{2} + (4ร5 )^{2} = (4^{2} + 5^{2} โ 4 ร 5)^{2}

โด 4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + (5ร6 )^{2} = (5^{2}+ 6^{2} โ 5 ร 6)^{2}

โด 5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + (6ร7 )^{2} = (6^{2} + 7^{2} โ 6 ร 7)^{2}

โด 6^{2} + 7^{2} + 42^{2} = 43^{2}

**7. Without adding, find the sum.**

**i. 1 + 3 + 5 + 7 + 9**

Solution:

Sum of first five odd number = (5)^{2} = 25

**ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19**

Solution:

Sum of first ten odd number = (10)^{2} = 100

**iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23**

Solution:

Sum of first thirteen odd number = (12)^{2} = 144

**8. (i) Express 49 as the sum of 7 odd numbers.**

**Solution:**

We know, sum of first n odd natural numbers is n^{2} . Since,49 = 7^{2}

โด 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

**(ii) Express 121 as the sum of 11 odd numbers. **Solution:

Since, 121 = 11^{2}

โด 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

**9. How many numbers lie between squares of the following numbers?**

**i. 12 and 13**

**ii. 25 and 26**

**iii. 99 and 100**

Solution:

Between n^{2} and (n+1)^{2}, there are 2n nonโperfect square numbers.

i. 122 and 132 there are 2ร12 = 24 natural numbers.

ii. 252 and 262 there are 2ร25 = 50 natural numbers.

iii. 992 and 1002 there are 2ร99 =198 natural numbers.

Exercise 6.2 Page: 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} + 2ร30ร2 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 900 + 4 + 120

= 1024

ii. (35)^{2}

= (30+5 )^{2}

= (30)^{2} + (5)^{2} + 2ร30ร5 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 900 + 25 + 300

= 1225

iii. (86)^{2}

= (90 โ 4)^{2}

= (90)^{2} + (4)^{2} โ 2ร90ร4 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 8100 + 16 โ 720

= 8116 โ 720

= 7396

iv. (93)^{2}

= (90+3 )^{2}

= (90)^{2} + (3)^{2} + 2ร90ร3 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 8100 + 9 + 540

= 8649

v. (71)^{2}

= (70+1 )^{2}

= (70)^{2} + (1)^{2} +2ร70ร1 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 4900 + 1 + 140

= 5041

vi. (46)^{2}

= (50 -4 )^{2}

= (50)^{2} + (4)^{2} โ 2ร50ร4 [Since, (a+b)^{2} = a^{2}+b^{2} +2ab]

= 2500 + 16 โ 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, m2โ1, m2+1 is a Pythagorean triplet.

i. 2m = 6

โ m = 6/2 = 3

m2โ1= 32 โ 1 = 9โ1 = 8

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

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

ii. 2m = 14

โ m = 14/2 = 7

m2โ1= 72โ1 = 49โ1 = 48

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

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

iii. 2m = 16

โ m = 16/2 = 8

m2โ1 = 82โ1 = 64โ1 = 63

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

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

iv. 2m = 18

โ m = 18/2 = 9

m2โ1 = 92โ1 = 81โ1 = 80

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

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

Exercise 6.3 Page: 102

**1. What could be the possible โoneโsโ 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 unitโs digit of the square of a number having digit as unitโs

place 1 is 1 and also 9 is 1[9^{2}=81 whose unit place is 1].

โด Unitโs digit of the square root of number 9801 is equal to 1 or 9.

ii. We know that the unitโs digit of the square of a number having digit as unitโs

place 6 is 6 and also 4 is 6 [6^{2}=36 and 4^{2}=16, both the squares have unit digit 6].

โด Unitโs digit of the square root of number 99856 is equal to 6.

iii. We know that the unitโs digit of the square of a number having digit as unitโs

place 1 is 1 and also 9 is 1[9^{2}=81 whose unit place is 1].

โด Unitโs digit of the square root of number 998001 is equal to 1 or 9.

iv. We know that the unitโs digit of the square of a number having digit as unitโs

place 5 is 5.

โด Unitโs 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. 153โน Ends with 3.

โด, 153 is not a perfect square

ii. 257โน Ends with 7

โด, 257 is not a perfect square

iii. 408โน Ends with 8

โด, 408 is not a perfect square

iv. 441โน Ends with 1

โด, 441 is a perfect square.

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

Solution:

100

100 โ 1 = 99

99 โ 3 = 96

96 โ 5 = 91

91 โ 7 = 84

84 โ 9 = 75

75 โ 11 = 64

64 โ 13 = 51

51 โ 15 = 36

36 โ 17 = 19

19 โ 19 = 0

Here, we have performed subtraction ten times.

โด โ100 = 10

169

169 โ 1 = 168

168 โ 3 = 165

165 โ 5 = 160

160 โ 7 = 153

153 โ 9 = 144

144 โ 11 = 133

133 โ 13 = 120

120 โ 15 = 105

105 โ 17 = 88

88 โ 19 = 69

69 โ 21 = 48

48 โ 23 = 25

25 โ 25 = 0

Here, we have performed subtraction thirteen times.

โด โ169 = 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.

729 = 3ร3ร3ร3ร3ร3ร1

โ 729 = (3ร3)ร(3ร3)ร(3ร3)

โ 729 = (3ร3ร3)ร(3ร3ร3)

โ 729 = (3ร3ร3)^{2}

โ โ729 = 3ร3ร3 = 27

ii.

400 = 2ร2ร2ร2ร5ร5ร1

โ 400 = (2ร2)ร(2ร2)ร(5ร5)

โ 400 = (2ร2ร5)ร(2ร2ร5)

โ 400 = (2ร2ร5)^{2}

โ โ400 = 2ร2ร5 = 20

iii.

1764 = 2ร2ร3ร3ร7ร7

โ 1764 = (2ร2)ร(3ร3)ร(7ร7)

โ 1764 = (2ร3ร7)ร(2ร3ร7)

โ 1764 = (2ร3ร7)^{2}

โ โ1764 = 2 ร3ร7 = 42

iv.

4096 = 2ร2ร2ร2ร2ร2ร2ร2ร2ร2ร2ร2

โ 4096 = (2ร2)ร(2ร2)ร(2ร2)ร(2ร2)ร(2ร2)ร(2ร2)

โ 4096 = (2ร2ร2ร2ร2ร2)ร(2ร2ร2ร2ร2ร2)

โ 4096 = (2ร2ร2ร2ร2ร2)^{2}

โ โ4096 = 2ร2ร2 ร2ร2ร2 = 64

v.

7744 = 2ร2ร2ร2ร2ร2ร11ร11ร1

โ 7744 = (2ร2)ร(2ร2)ร(2ร2)ร(11ร11)

โ 7744 = (2ร2ร2ร11)ร(2ร2ร2ร11)

โ 7744 = (2ร2ร2ร11)^{2}

โ โ7744 = 2ร2ร2ร11 = 88

vi.

9604 = 62 ร 2 ร 7 ร 7 ร 7 ร 7

โ 9604 = ( 2 ร 2 ) ร ( 7 ร 7 ) ร ( 7 ร 7 )

โ 9604 = ( 2 ร 7 ร7 ) ร ( 2 ร 7 ร7 )

โ 9604 = ( 2ร7ร7 )^{2}

โ โ9604 = 2ร7ร7 = 98

vii.

5929 = 7ร7ร11ร11

โ 5929 = (7ร7)ร(11ร11)

โ 5929 = (7ร11)ร(7ร11)

โ 5929 = (7ร11)^{2}

โ โ5929 = 7ร11 = 77

viii.

9216 = 2ร2ร2ร2ร2ร2ร2ร2ร2ร2ร3ร3ร1

โ 9216 = (2ร2)ร(2ร2) ร ( 2 ร 2 ) ร ( 2 ร 2 ) ร ( 2 ร 2 ) ร ( 3 ร 3 )

โ 9216 = ( 2 ร 2 ร 2 ร 2 ร 2 ร 3) ร ( 2 ร 2 ร 2 ร 2 ร 2 ร 3)

โ 9216 = 96 ร 96

โ 9216 = ( 96 )^{2}

โ โ9216 = 96

ix.

529 = 23ร23

529 = (23)^{2}

โ529 = 23

x.

8100 = 2ร2ร3ร3ร3ร3ร5ร5ร1

โ 8100 = (2ร2) ร(3ร3)ร(3ร3)ร(5ร5)

โ 8100 = (2ร3ร3ร5)ร(2ร3ร3ร5)

โ 8100 = 90ร90

โ 8100 = (90)^{2}

โ โ8100 = 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.

252 = 2ร2ร3ร3ร7

= (2ร2)ร(3ร3)ร7

Here, 7 cannot be paired.

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

New number = 252 ร 7 = 1764

1764 = 2ร2ร3ร3ร7ร7

โ 1764 = (2ร2)ร(3ร3)ร(7ร7)

โ 1764 = 2^{2}ร3^{2}ร7^{2}

โ 1764 = (2ร3ร7)^{2}

โ โ1764 = 2ร3ร7 = 42

ii.

180 = 2ร2ร3ร3ร5

= (2ร2)ร(3ร3)ร5

Here, 5 cannot be paired.

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

New number = 180 ร 5 = 900

900 = 2ร2ร3ร3ร5ร5ร1

โ 900 = (2ร2)ร(3ร3)ร(5ร5)

โ 900 = 2^{2}ร3^{2}ร5^{2}

โ 900 = (2ร3ร5)^{2}

โ โ900 = 2ร3ร5 = 30

iii.

1008 = 2ร2ร2ร2ร3ร3ร7

= (2ร2)ร(2ร2)ร(3ร3)ร7

Here, 7 cannot be paired.

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

New number = 1008ร7 = 7056

7056 = 2ร2ร2ร2ร3ร3ร7ร7

โ 7056 = (2ร2)ร(2ร2)ร(3ร3)ร(7ร7)

โ 7056 = 2^{2}ร2^{2}ร3^{2}ร7^{2}

โ 7056 = (2ร2ร3ร7)^{2}

โ โ7056 = 2ร2ร3ร7 = 84

iv.

2028 = 2ร2ร3ร13ร13

= (2ร2)ร(13ร13)ร3

Here, 3 cannot be paired.

โด We will multiply 2028 by 3 to get perfect square. New number = 2028ร3 = 6084

6084 = 2ร2ร3ร3ร13ร13

โ 6084 = (2ร2)ร(3ร3)ร(13ร13)

โ 6084 = 2^{2}ร3^{2}ร13^{2}

โ 6084 = (2ร3ร13)^{2}

โ โ6084 = 2ร3ร13 = 78

v.

1458 = 2ร3ร3ร3ร3ร3ร3

= (3ร3)ร(3ร3)ร(3ร3)ร2

Here, 2 cannot be paired.

โด We will multiply 1458 by 2 to get perfect square. New number = 1458 ร 2 = 2916

2916 = 2ร2ร3ร3ร3ร3ร3ร3

โ 2916 = (3ร3)ร(3ร3)ร(3ร3)ร(2ร2)

โ 2916 = 3^{2}ร3^{2}ร3^{2}ร2^{2}

โ 2916 = (3ร3ร3ร2)^{2}

โ โ2916 = 3ร3ร3ร2 = 54

vi.

768 = 2ร2ร2ร2ร2ร2ร2ร2ร3

= (2ร2)ร(2ร2)ร(2ร2)ร(2ร2)ร3

Here, 3 cannot be paired.

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

New number = 768ร3 = 2304

2304 = 2ร2ร2ร2ร2ร2ร2ร2ร3ร3

โ 2304 = (2ร2)ร(2ร2)ร(2ร2)ร(2ร2)ร(3ร3)

โ 2304 = 2^{2}ร2^{2}ร2^{2}ร2^{2}ร3^{2}

โ 2304 = (2ร2ร2ร2ร3)^{2}

โ โ2304 = 2ร2ร2ร2ร3 = 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.

252 = 2ร2ร3ร3ร7

= (2ร2)ร(3ร3)ร7

Here, 7 cannot be paired.

โด We will divide 252 by 7 to get perfect square. New number = 252 รท 7 = 36

36 = 2ร2ร3ร3

โ 36 = (2ร2)ร(3ร3)

โ 36 = 2^{2}ร3^{2}

โ 36 = (2ร3)^{2}

โ โ36 = 2ร3 = 6

ii.

2925 = 3ร3ร5ร5ร13

= (3ร3)ร(5ร5)ร13

Here, 13 cannot be paired.

โด We will divide 2925 by 13 to get perfect square. New number = 2925 รท 13 = 225

225 = 3ร3ร5ร5

โ 225 = (3ร3)ร(5ร5)

โ 225 = 3^{2}ร5^{2}

โ 225 = (3ร5)^{2}

โ โ36 = 3ร5 = 15

iii.

396 = 2ร2ร3ร3ร11

= (2ร2)ร(3ร3)ร11

Here, 11 cannot be paired.

โด We will divide 396 by 11 to get perfect square. New number = 396 รท 11 = 36

36 = 2ร2ร3ร3

โ 36 = (2ร2)ร(3ร3)

โ 36 = 2^{2}ร3^{2}

โ 36 = (2ร3)^{2}

โ โ36 = 2ร3 = 6

iv.

2645 = 5ร23ร23

โ 2645 = (23ร23)ร5

Here, 5 cannot be paired.

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

New number = 2645 รท 5 = 529

529 = 23ร23

โ 529 = (23)^{2}

โ โ529 = 23

v.

2800 = 2ร2ร2ร2ร5ร5ร7

= (2ร2)ร(2ร2)ร(5ร5)ร7

Here, 7 cannot be paired.

โด We will divide 2800 by 7 to get perfect square. New number = 2800 รท 7 = 400

400 = 2ร2ร2ร2ร5ร5

โ 400 = (2ร2)ร(2ร2)ร(5ร5)

โ 400 = (2ร2ร5)^{2}

โ โ400 = 20

vi.

1620 = 2ร2ร3ร3ร3ร3ร5

= (2ร2)ร(3ร3)ร(3ร3)ร5

Here, 5 cannot be paired.

โด We will divide 1620 by 5 to get perfect square. New number = 1620 รท 5 = 324

324 = 2ร2ร3ร3ร3ร3

โ 324 = (2ร2)ร(3ร3)ร(3ร3)

โ 324 = (2ร3ร3)^{2}

โ โ324 = 18

**7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Ministerโs 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.

โด Each student donate Rs.x .

Total many contributed by all the students= xรx=x^{2} Given, x^{2} = Rs.2401

x^{2} = 7ร7ร7ร7

โ x^{2} = (7ร7)ร(7ร7)

โ x^{2 }= 49ร49

โ x = โ(49ร49)

โ x = 49

โด 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.

โด the number of plants in each rows = x.

Total many contributed by all the students = x ร x =x^{2}

Given,

x_{2} = Rs.2025

x^{2} = 3ร3ร3ร3ร5ร5

โ x^{2} = (3ร3)ร(3ร3)ร(5ร5)

โ x2 = (3ร3ร5)ร(3ร3ร5)

โ x2 = 45ร45

โ x = โ45ร45

โ x = 45

โด 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:

L.C.M of 4, 9 and 10 is (2ร2ร9ร5) 180.

180 = 2ร2ร9ร5

= (2ร2)ร3ร3ร5

= (2ร2)ร(3ร3)ร5

Here, 5 cannot be paired.

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

Hence, the smallest square number divisible by 4, 9 and 10 = 180ร5 = 900

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

Solution:

L.C.M of 8, 15 and 20 is (2ร2ร5ร2ร3) 120.

120 = 2ร2ร3ร5ร2

= (2ร2)ร3ร5ร2

Here, 3, 5 and 2 cannot be paired.

โด We will multiply 120 by (3ร5ร2) 30 to get perfect square.

Hence, the smallest square number divisible by 8, 15 and 20 =120ร30 = 3600

Exercise 6.4 Page: 107

**1. Find the square root of each of the following numbers by Division method.**

**i. 2304**

**ii. 4489**

**iii. 3481**

**iv. 529**

**v. 3249**

**vi. 1369**

**vii. 5776**

**viii. 7921**

**ix. 576**

**x. 1024**

**xi. 3136**

**xii. 900**

Solution:

i.

โด โ2304 = 48

ii.

โด โ4489 = 67

iii.

โด โ3481 = 59

iv.

โด โ529 = 23

v.

โด โ3249 = 57

vi.

โด โ1369 = 37

vii.

โด โ5776 = 76

viii.

โด โ7921 = 89

ix.

โด โ576 = 24

x.

โด โ1024 = 32

xi.

โด โ3136 = 56

xii.

โด โ900 = 30

**2. Find the number of digits in the square root of each of the following numbers (without any**

**calculation).64**

**i. 144**

**ii. 4489**

**iii. 27225**

**iv. 390625**

Solution:

i.

โด โ144 = 12

Hence, the square root of the number 144 has 2 digits.

ii.

โด โ4489 = 67

Hence, the square root of the number 4489 has 2 digits.

iii.

โ27225 = 165

Hence, the square root of the number 27225 has 3 digits.

iv.

โด โ390625 = 625

Hence, the square root of the number 390625 has 3 digits.

**3. Find the square root of the following decimal numbers.**

**i. 2.56**

**ii. 7.29**

**iii. 51.84**

**iv. 42.25**

**v. 31.36**

Solution:

**i.**

โด โ2.56 = 1.6

ii.

โด โ7.29 = 2.7

iii.

โด โ51.84 = 7.2

iv.

โด โ42.25 = 6.5

v.

โด โ31.36 = 5.6

**4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.**

**i. 402**

**ii. 1989**

**iii. 3250**

**iv. 825**

**v. 4000**

Solution:

i.

โด โ400 = 20

โด We must subtracted 2 from 402 to get a perfect square.

New number = 402 โ 2 = 400

โด โ400 = 20

ii.

โด We must subtracted 53 from 1989 to get a perfect square. New number = 1989 โ 53 = 1936

โด โ1936 = 44

iii.

โด We must subtracted 1 from 3250 to get a perfect square.

New number = 3250 โ 1 = 3249

โด โ3249 = 57

iv.

โด We must subtracted 41 from 825 to get a perfect square.

New number = 825 โ 41 = 784

โด โ784 = 28

โด We must subtracted 31 from 4000 to get a perfect square. New number = 4000 โ 31 = 3969

โด โ3969 = 63

**5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.**

**(i) 525**

**(ii) 1750**

**(iii) 252**

**(iv)1825**

**(v)6412**

Solution:

(i)

Here, (22)2 < 525 > (23)2

We can say 525 is ( 129 โ 125 ) 4 less than (23)2.

โด If we add 4 to 525, it will be perfect square. New number = 525 + 4 = 529

โด โ529 = 23

(ii)

Here, (41)2 < 1750 > (42)^{2}

We can say 1750 is ( 164 โ 150 ) 14 less than (42)^{2}.

โด If we add 14 to 1750, it will be perfect square.

New number = 1750 + 14 = 1764

โดโ1764 = 42

(iii)

Here, (15)2 < 252 > (16)2

We can say 252 is ( 156 โ 152 ) 4 less than (16)2.

โด If we add 4 to 252, it will be perfect square.

New number = 252 + 4 = 256

โด โ256 = 16

(iv)

Here, (42)2 < 1825 > (43)2

We can say 1825 is ( 249 โ 225 ) 24 less than (43)2.

โด If we add 24 to 1825, it will be perfect square.

New number = 1825 + 24 = 1849

โด โ1849 = 43

(v)

Here, (80)2 < 6412 > (81)2

We can say 6412 is ( 161 โ 12 ) 149 less than (81)2.

โด If we add 149 to 6412, it will be perfect square.

New number = 6412 + 149 = 656

โด โ6561 = 81

**6. Find the length of the side of a square whose area is 441 m2.**

Solution:

Let the length of each side of the field = a Then, area of the field = 441 m2

โ a2 = 441 m2

โa = โ441 m

โด The length of each side of the field = a m = 21 m.

**7. In a right triangle ABC, โ B = 90ยฐ.**

**a. If AB = 6 cm, BC = 8 cm, find AC**

**b. If AC = 13 cm, BC = 5 cm, find AB**

Solution:

a.

Given, AB = 6 cm, BC = 8 cm

Let AC be x cm.

โด AC2 = AB2 + BC2

Hence, AC = 10 cm.

b.

Given, AC = 13 cm, BC = 5 cm

Let AB be x cm.

โด AC2 = AB2 + BC2

โ AC2 โ BC2 = AB2

Hence, AB = 12 cm

**8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows**

**and the number of columns remain same. Find the minimum number of plants he needs more for this.**

Solution:

Let the number of rows and column be, x.

โด Total number of row and column= xร x = x2 As per question, x2 = 1000

โ x = โ1000

Here, (31)2 < 1000 > (32)2

We can say 1000 is ( 124 โ 100 ) 24 less than (32)2.

โด 24 more plants are needed.

**9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.**

Solution:

Let the number of rows and column be, x.

โด Total number of row and column= x ร x = x2 As per question, x2 = 500

x = โ500

Hence, 16 children would be left out in the arrangement

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

The NCERT Solutions Chapter 8 deals with the concept of ratios and percentages, along with other main topics and concepts.

The major concepts covered in this chapter include:

6.1 Introduction

6.2 Properties of Square Numbers

6.3 Some interesting patterns

6.4 Finding the square of a number

6.4.1 Patterns in squares

6.4.2 Pythagorean triplets

6.5 Square Roots

6.5.1 Finding square roots

6.5.2 Finding square root through repeated subtraction

6.5.3 Finding square root through prime factorisation

6.5.4 Finding square root by division method

6.6 Square Roots of Decimals

6.7 Estimating Square Root

Exercise 6.1 Solutions 9 Questions

Exercise 6.2 Solutions 2 Questions

Exercise 6.3 Solutions 10 Questions

Exercise 6.4 Solutions 9 Questions

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

Chapter 6 of NCERT Solutions for Class 8 Maths discusses the following:

- If a natural number m can be expressed as n
^{2}, where n is a natural number, then m is a square number. - All square numbers end with 0, 1, 4, 5, 6 or 9 at units place.
- Square numbers can only have an even number of zeros at the end.
- Square root is the inverse operation of square.
- There are two integral square roots of a perfect square number

Learning the chapter Squares and Square Roots enables the students to understand:

- Square and Square roots
- Square roots using factor method and division method for numbers containing(a) no more than a total of 4 digits and (b) no more than 2 decimal places
- Estimating square roots
- Learning the process of moving nearer to the required number