The DAV Class 8 Maths Book Solutions and **DAV Class 8 Maths Chapter 1 Worksheet 1** Solutions of Squares and Square Roots offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 1 WS 1 Solutions

Question 1.

Which of the following numbers are perfect squares?

11, 16, 32, 36, 50, 64, 75

Solution:

11 = 11 × 1 the prime factors are not in pair.

∴ 11 is not a perfect square number.

16 = 2 × 2 × 2 × 2, the prime factors are in pair.

∴ 16 is a perfect square number.

32 = 2 × 2 × 2 × 2 × 2, all the prime factors are not in pair.

∴ 32 is not a perfect square number.

36 = 2 × 2 × 3 × 3, all the prime factors are in pair.

∴ 36 is a perfect square number.

50 = 2 × 5 × 5, all the prime factors are not in pair.

∴ 50 is not a perfect square number.

64 = 2 × 2 × 2 × 2 × 2 × 2, all the prime factors are in pair.

∴ 64 is a perfect square number.

75 = 3 × 5 × 5, all the prime factors are not in pair.

∴ 75 is not a perfect square number.

Question 2.

Which of the following numbers are perfect squares of even numbers?

121, 225, 784, 841, 576, 6561

Solution:

As the numbers with their end digit even, their squares will also have their end digit even.

∴ 121 is not a perfect square of even number.

225 is not a perfect square of even number.

784 is a perfect square of even number.

841 is not a perfect square of even number.

576 is a perfect square of even number,

6561 is not a perfect square of even number.

Question 3.

Which of the following numbers are perfect squares?

100, 205000, 3610000, 212300000

Solution:

Any number ending with odd number of zeros is never a perfect square.

100 is a perfect square as it is ending with even zeros.

205000 is not a perfect square as it has odd number of zeros.

3610000 is a perfect square as it has even number of zeros.

212300000 is not a perfect square as it has odd number of zeros.

Question 4.

By just observing the digits at ones place, tell which of the following can be perfect squares?

1026, 1022, 1024, 1027

Solution:

The number ending with 2, 3, 7 and 8 are not perfect squares.

∴ 1026 and 1024 may be perfect squares whereas

1022 and 1027 may not be perfect squares.

Question 5.

How many non-square numbers lie between the following pairs of numbers?

(i) 7^{2} and 8^{2}

(ii) 10^{2} and 11^{2}

(iii) 40^{2} and 41^{2}

(iv) 80^{2} and 81^{2}

(v) 101^{2} and 102^{2}

(vi) 205^{2} and 206^{2}

Solution:

The non-perfect square numbers between the square of the numbers n and n + 1 is 2n.

(i) 7^{2} and 8^{2}

Here n = 7 and n + 1 = 8

∴ Number of non-perfect squares between 7^{2} and 8^{2}

= 2 × 7 = 14.

(ii) 10^{2} and 11^{2}

Here n = 10 and n + 1 = 11

∴ Non-perfect numbers between 10^{2} and 11^{2}

= 2 × 10 = 20.

(iii) 40^{2} and 41^{2}

Here n = 40 and n + 1 = 41

∴ Non-perfect square numbers between 40^{2} and 41^{2}

= 2 × 40 = 80.

(iv) 80^{2} and 81^{2}

Here n = 80 and n + 1= 81

∴ Non-perfect square numbers between 80^{2} and 81^{2}

= 2 × 80 = 160.

(v) 101^{2} and 102^{2}

Here n = 101 and n+ 1= 102

∴ Non-perfect square numbers between 101^{2} and 102^{2}

= 2 × 101 = 202.

(vi) 205^{2} and 206^{2}

Here n = 205 and n + 1 = 206

∴ Non-perfect square numbers between 205^{2} and 206^{2}

= 2 × 205 = 410.

Question 6.

Write down the correct number in the box:

(i) 100^{2} – 99^{2} = ______ = ______

(ii) 27^{2} – 26^{2} = ______ = ______

(iii) 569^{2} – 568^{2} = ______ = ______

Solution:

(i) 100^{2} – 99^{2} = 100 + 99 = 199

(ii) 27^{2} – 26^{2} = 27 + 26 = 53

(iii) 569^{2} – 568^{2} = 569 + 568 = 1137

Question 7.

Observe the pattern in the following and find the missing numbers:

121 = \(\frac{(22)^2}{1+2+1}\)

12321 = \(\frac{(333)^2}{1+2+3+2+1}\)

1234321 = ______

123454321 = ______

12345654321 = ______

Solution:

1234321 = \(\frac{(4444)^2}{1+2+3+4+3+2+1}\)

12345321 = \(\frac{(55555)^2}{1+2+3+4+5+4+3+2+1}\)

12345654321 = \(\frac{(666666)^2}{1+2+3+4+5+6+5+4+3+2+1}\)

Question 8.

Which of the following triplets are Pythagorean?

(3, 4, 5), (6, 7, 8), (10, 24, 26), (2, 3, 4)

[Hint: Let the smallest even number be 2m and find m for it. Then, fmd (2m, m^{2} – 1, m^{2} + 1). If you get the triplet, it is Pythagorean).

Solution:

(i) Taking the triplet (3, 4, 5)

Here, the smallest even number = 4

∴ 2m = 4 ⇒ m = 2

∴ Pythagorean triplets are 2m, m^{2} – 1 and m^{2} + 1

= 2 × 2, (2)^{2} – 1 and (2)^{2} + 1 = 4, 3, 5

Hence (3, 4, 5) is a Pythagorean triplet.

(ii) Taking the triplet (6, 7, 8)

Here, the smallest even number = 6

∴ 2m = 6 ⇒ m = 3

∴ Pythagorean triplets are 2m, m^{2} – 1 and m^{2} + 1

= 2 × 3, (3)^{2} – 1 and (3)^{2} + 1 = 6, 8, 10

Hence (6, 7, 8) is not a Pythagorean triplet.

(iii) Taking the triplet (10, 24, 26)

Here, the smallest natural number = 10

∴ 2m = 10 ⇒ m = 5

∴ Pythagorean triplets are 2m, m^{2} – 1 and m^{2} + 1

⇒ 2 × 5, (5)^{2} – 1 and (5)^{2} + 1 ⇒ 10, 24 and 26

Hence (10, 24, 26) is a Pythagorean triplet.

(iv) Taking the triplet (2, 3, 4)

Here, the smallest even number = 2 2m = 2 m = 1

Pythagorean triplets are 2m, m^{2} – 1 and m^{2} + 1

= 2 × 1, (1)^{2} – 1 and (1)^{2} + 1 = 2, 0 and 2

Hence (2, 3, 4) are not Pythagorean triplet.

Additional Questions:

Question 1.

Which of the following numbers are not perfect squares?

(i) 1057

(ii) 7928

(iii) 22222

(iv) 23453

(v) 2061

(vi) 1069

(vii) 256

(viii) 36124

Solution:

The numbers ending with 2, 3, 7 or 8 can never be perfect squares.

(i) 1057 = ending with 7, so it is not a perfect square.

(ii) 7928 = ending with 8, so it is not a perfect square.

(iii) 22222 = ending with 2, so it is not a perfect square.

(iv) 23453 = ending with 3, so it is not a perfect square.

(v) 2061 = It may be a perfect square as it is ending with 1 (other than 2, 3, 7, or 8).

(vi) 1069 = It may be a perfect square number as it is ending with 9 (other than 2, 3, 7, or 8).

(vii) 256 = It may be a perfect square number as it is ending with 6.

(viii) 36124 = It may be a perfect square number as it is ending with 4 (other than 2, 3, 7, 8).

Question 2.

Which of the following squares would end with digit 1?

123^{2}, 77^{2}, 82^{2}, 56^{2}, 161^{2}, 109^{2}

Solution:

If a number has 1 or 9 in the unit’s place, then it is square ends in 1.

∴ only 161^{2} and 109^{2} would end with 1 as they have 1 and 9 respectively in the unit’s place.

Question 3.

Which of the following numbers would have digit 6 at unit place?

(i) 19^{2}

(ii) 24^{2}

(iii) 26^{2}

(iv) 36^{2}

(v) 34^{2}

Solution:

When a square number ends, the number whose square it is, will have either 4 or 6 in unit’s place.

(i) 19^{2} is not a number which will have 6 in its unit’s place.

(ii) 24^{2} is the number which will have 6 in its unit’s place.

(iii) 26^{2} is the number which will have 6 in its units place.

(iv) 36^{2} is the number which will have 6 in its unit’s place.

(v) 34^{2} is the number which will have 6 in its unit’s place.

Question 4.

Which of the following numbers are perfect squares of even numbers?

225, 121, 1024, 1296, 361, 289, 576

Solution:

A number having even number digits at its unit place, will have even numbers at the unit place of its square.

∴ Only 1024, 1296 and 576 are the perfect squares of even numbers.

Question 5.

What will be the number of zeros in the square of the following numbers?

(i) 60

(ii) 400

(iii) 2000

(iv) 150000

Solution:

(i) Square of 60 will have two zeros in the end.

(ii) Square of 400 will have four zeros in the end.

(iii) Square of 2000 will have six zeros in the end.

(iv) Square of 150000 will have eight zeros in the end.

Question 6.

How many non-square numbers lie between the following pairs of numbers?

(i) 8^{2} and 9^{2}

(ii) 110^{2} and 111^{2}

(iii) 205^{2} and 206^{2}

Solution:

There are ‘2n’ non-perfect square numbers between the squares of ‘n’ and ‘(n + 1)’.

(i) 8^{2} and 9^{2}

Here n – 8 and (n + 1) = 9

∴ Non-perfect square numbers between 8^{2} and 9^{2} are 2 × 8 = 16,

(ii) 110^{2} and 111^{2}

Here n = 110 and n + 1 = 111

∴ Non-perfect square numbers between 110^{2} and 111^{2} are 2 × 110 = 220.

(iii) 205^{2} and 206^{2}

Here n = 205 and n + 1 = 206

∴ Non-perfect square numbers between 205^{2} and 206^{2} are 2 × 205 = 410.

Question 7.

Fill in the blank boxes with correct numbers:

(i) 89^{2} – 88^{2} = ______ = ______

(ii) 56^{2} – 55^{2} = ______ = ______

(iii) 105^{2} – 104^{2} = ______ = ______

(iv) 576^{2} – 575^{2} = ______ = ______

(v) 112^{2} – 111^{2} = ______ = ______

Solution:

(i) 89^{2} – 88^{2} = 89 + 88 = 117

(ii) 56^{2} – 55^{2} = 56 + 55 = 111

(iii) 105^{2} – 104^{2} = 105 +104 = 209

(iv) 576^{2} – 575^{2} = 576 + 575 = 1151

(v) 112^{2} – 111^{2} = 112 + 111 = 223

Question 8.

Observe the following pattern in the following and fill in the blanks:

(i) 1 = ______ = ______

(ii) 1 + 3 = ______ = ______

(iii) ______ = 9 = ______

(iv) 1+ 3 + 5 + 7 = ______ = ______

(v) ______ = ______ = 5^{2}

Solution:

(i) 1 = 1 = 1^{2}

(ii) 1 + 3 = 4 = 2^{2}

(iii) 1 + 3 + 5 = 0 = 3^{2}

(iv) 1 + 3 + 5 + 7 = 16 = 4^{2}

(v) 1 + 3 + 5 + 7 + 9 = 25 = 5^{2}

Question 9.

Express the following as the sum of two consecutive integers:

(i) 11^{2}

(ii) 13^{2}

(iii) 19^{2}

(iv) 21^{2}

(v) 33^{2}

Solution:

n^{2} = \(\frac{n^2-1}{2}+\frac{n^2+1}{2}\)

(i) 11^{2 }=

⇒ 60 + 61

Hence 11^{2} = 60 + 61

(ii) 13^{2} =

⇒ 84 + 85

Hence 13^{2} = 84 + 85

(iii) 19^{2} =

⇒ 180 + 181

Hence 19^{2} = 180 + 181

(iv) 21^{2} =

⇒ 220 + 221

Hence 21^{2} = 220 + 221

(v) 33^{2} =

⇒ 544 + 545

Hence 32^{2} = 544 + 545

Question 10.

Fill in the blank boxes with correct numbers.

(i) 1^{2} = ___

(ii) 11^{2} = 1 ___ 1

(iii) 111^{2} = 12 ___ 21

(iv) 1111^{2} = 123 ___ 321

(v) 11111^{2} = 1234 ___ 4321

(vi) 111111^{2} = ______

Solution:

(i) 1^{2} = 1

(ii) 11^{2} = 121

(iii) 111^{2} = 12321

(iv) 1111^{2} = 1234321

(v) 11111^{2} = 123454321

(vi) 111111^{2} = 12345654321

Question 11.

Observe the following following and fill in the blanks:

1 + 3 = 2^{2}

1 + 3 + 5 = 3^{2}

1 + 3 + 5 + 7 = 4^{2}

1 + 3 + 5 + 7 + 9 = ______

1 + 3 + 5 + 7 + 9 + 11 = ______

1 + 3 + 5 + 7 + ………. upto n terms = ______

Solution:

1 + 3 + 5 + 7 + 9 = 5^{2}

1 + 3 + 5 + 7 + 9 + 11 = 6^{2}

1 + 3 + 5 + 7 + …… upto n terms = n^{2}

Question 12.

Fill in the blank boxes with correct numbers:

(i) 7^{2} = 49

(ii) 67^{2} = 4489

(iii) 667^{2} = 444889

(iv) 6667^{2} = ______

(v) 66667^{2} = ______

(vi) 666667^{2} = ______

Solution:

(iv) 6667^{2} = 4444889

(v) 66667^{2} = 4444488889

(vi) 666667^{2} = 444444888889

Question 13.

Observe the following pattern and supply the missing numbers:

(i) 11^{2} = 121

(ii) 101^{2} = 10201

(iii) 10101^{2} = 102030201

(iv) 1010101^{2} = ___________

(v) 101010101^{2} = ___________

(vi) 10101010101^{2} = ___________

Solution:

(iv) 1020304030201

(v) 10203040504030201

(vi) 102030405060504030201.

Question 14.

Using the given pattern, find the missing entries:

(i) 1^{2} + 2^{2} + 2^{2} = 3^{2}

(ii) 2^{2} + 3^{2} + 6^{2} = 7^{2}

(iii) 3^{2} + 4^{2} + ___ = 13^{2}

(iv) 4^{2} + 5^{2} + ___ = 21^{2}

(v) ___ + 6^{2} + 30^{2} = ___

(vi) 6^{2} + 7^{2} + ___ = 43^{2}

Solution:

(iii) 12^{2}

(iv) 5^{2}

(v) 20^{2}

(vi) 42^{2}

Question 15.

Find the squares of the following numbers without actual multiplications:

(i) 39

(ii) 42

(iii) 61

(iv) 75

(v) 85

Solution:

(i) 39^{2} = (30 + 9)^{2}

= 30(30 + 9) + 9(30 + 9)

= 30^{2} + 30 × 9 + 9 × 30 + 9^{2}

= 900 + 270 + 270 + 81

= 900 + 540 + 81 = 1521

(ii) 42^{2} = (40 + 2)^{2}

= 40(40 + 2) + 2(40 + 2)

= 402 + 40 × 2 + 2 × 40 + 2^{2}

= 1600 + 80 + 80 + 4 = 1764

(iii) 61^{2} = (60 + 1)^{2}

= 60 (60 + 1) + 1(60 + 1)

= 60^{2} + 60 × 1 + 1 × 60 + 1^{2}

= 3600 + 60 + 60 + 1 = 3721

(iv) 75^{2} = (70 + 5)^{2}

= 70(70 + 5) + 5(70 + 5)

= 70^{2} + 70 × 5 + 5 × 70 + 5^{2}

= 4900 + 350 + 350 + 25 = 5625

(v) 85^{2} = (80 + 5)^{2}

= 80(80 + 5) + 5(80 + 5)

= 80^{2} + 80 × 5 + 5 × 80 + 5^{2}

= 6400 + 400 + 400 + 25 = 7225.

Question 16.

Which of the triplets are Pythagoreans?

(i) (12, 35, 37)

(ii) (8, 15, 17)

(iii) (7, 8, 12)

(iv) (2, 3, 5)

(v) (6, 8, 10)

Solution:

The Pythagorean triplets are 2 m, m^{2} – 1 and m^{2} + 1

if 2 m is the smallest even number of the triplet.

(i) (12, 35, 37)

Here 12 is the smallest even number

∴ 2m = 12 ⇒ m = 6

∴ triplets are 2m = 2 × 6 = 12

m^{2} – 1 = 6^{2} – 1 = 36 – 1 = 35

and m^{2} + 1 = (6)^{2} + 1 = 36 + 1 = 37

Hence (12, 35, 37) is the Pythagorean triplet.

(ii) (8, 15, 17)

Here the smallest even number is 8

2m = 8

∴ m = 4

m^{2} – 1 = (4)^{2} – 1 = 16 – 1 = 15

and m^{2} + 1 = (4)^{2} + 1 = 16 + 1 = 17

Hence (8, 15, 17) is a Pythagorean triplet.

(iii) (7, 8, 12)

Here the smallest even number is 8

2m = 8

∴ m = 4

m^{2} – 1 = (4)^{2} – 1 = 16 – 1 = 15

m^{2} + 1 = (4)^{2} + 1 = 16 + 1 = 17

So, (7, 8, 12) is not a Pythagorean triplet.

(iv) (2, 3, 5)

Here smallest even number is 2

2m = 2

∴ m = 1

m^{2} – 1 = 1^{2} – 1 = 0

m^{2} + 1 = 1^{2} + 1 = 2

Hence (2, 3, 5) is not a Pythagorean triplet.

(v) (6, 8, 10)

Here 6 is the smallest even number

2m = 6

∴ m = 3

m^{2} – 1 = (3)^{2} – 1 = 9 – 1 = 8 and

m^{2} + 1 = (3)^{2} + 1 = 9 + 1 = 10

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

Note: [All Pythagorean triplets may not be obtained using this form of 2 m, m^{2} – 1 and m^{2} + 1. example: (5, 12, 13) and (7, 24, 25). Although they have 12 and 24 as their due of the members.]

Question 17.

Fill in the following blanks:

(i) 1 + 3 + 5 = (___)^{2}

(ii) 1 + 3 + 5 + 7 = (___)^{2}

(iii) 1 + 3 + 5 + 7 + 9 = (___)^{2}

(iv) 1 + 3 + 5 + 7 + 9 + 11 = (___)^{2}

(u) 1 + 3 + 5 + 7 + 9 + 11+ 13 = (___)^{2}

Solution:

(i) 3

(ii) 4

(iii) 5

(iv) 6

(v) 7.

### DAV Class 8 Maths Chapter 1 Worksheet 1 Notes

The number obtained by multiplying any number by itself is called the square of the number.

For example:

1 × 1 = 1 = 1^{2}

2 × 2 = 4 = 2^{2}

3 × 3 = 9 = 3^{2}

4 × 4 = 16 = 4^{2}

n × n = n^{2}

Numbers, such as 1, 4, 9, 16, 25 … are called perfect squares.

**Significance of a perfect square:**

(i) Squares of even numbers are also even.

e.g. 4^{2} = 16,

6^{2} = 36 and

8^{2} = 64, etc.

(ii) Squares of odd numbers are always odd.

e.g. 9^{2} = 81,

11^{2} = 121 and

13^{2} = 169, etc.

(iii) The numbers ending with an odd number of zeros are never perfect squares.

e.g. 180, 36000, etc.

(iv) The difference between the squares of two consecutive natural numbers is equal to their sum.

e.g. 5^{2} – 4^{2} = 25 – 16 = 9 = 5 + 4

7^{2} – 6^{2} = 49 – 36 = 13 = 7 + 6

9^{2} – 8^{2} = 81 – 64 = 17 = 9 + 8

(v) The numbers ending with 2, 3, 7 and 8 are not perfect squares.

e.g. 12, 23, 67, 238 and 353, etc.

(vi) The square of a natural number n’ is equal to the sum of the first n odd natural numbers.

e.g. 3^{2} = 1 + 3 + 5 = 9

5^{2} = 1 + 3 + 5 + 7 + 9 = 25

6^{2} = 1 + 3 + 5 + 7 + 9 + 11 = 36

(vii) Squares of natural numbers composed of the only digit 1, follow a particular pattern.

e.g. 1^{2} = 1 = 1^{2}

11^{2} = 121 = 1 + 2 + 1 = 4 = 2^{2}

111^{2} = 12321 = 1 + 2 + 3 + 2 + 1 = 9 = 3^{2}

1111^{2} = 1234321 = 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 = 4^{2}

(viii) The square of a number other than 0 and 1 is either a multiple of 3 or exceeds the multiple of 3 by 1,

e.g. 3^{2} = 9

4^{2} = 16 = (15 + 1)

7^{2} = 49 = (48 + 1)

(ix) The square of a number other than 0 and 1 is either a multiple of 4 or exceeds a multiple of 4 by 1.

e.g. 4^{2} = 16

5^{2} = 25 = 24 + 1)

6^{2} = 36

7^{2} = 49 = (48 + 1)

**Numbers between two square**** numbers:**

Let us take any two consecutive numbers n and (n + 1). The number of non-squares between n^{2} and (n + 1)^{2} is 2n.

e.g.

(i) 1^{2} = 1 and 2^{2} = 4

∴ Non-square numbers between 1 and 4 are 2, 3.

1, 2, 3, 4 → 2 non-square numbers i.e. 2 × 1

(ii) 2 = 4 and 32 = 9

∴ Non-square numbers between 4 and 9 are 5, 6, 7, 8.

4, 5, 6, 7, 8, 9 → 4 non-square numbers i.e. 2 × 2

(iii) 3^{2} = 9 and 4^{2} = 16

∴ Non-square numbers between 9 and 16 are 10, 11, 12, 13, 14, 15.

9, 10, 11, 12, 13, 14, 15, 16 → 6 non-square number i.e. 2 × 3

Hence, there are 2n non-square natural numbers between the two consecutive perfect square numbers.