DAV Class 8 Maths Chapter 1 Worksheet 1 Solutions

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) 72 and 82
(ii) 102 and 112
(iii) 402 and 412
(iv) 802 and 812
(v) 1012 and 1022
(vi) 2052 and 2062
Solution:
The non-perfect square numbers between the square of the numbers n and n + 1 is 2n.
(i) 72 and 82
Here n = 7 and n + 1 = 8
∴ Number of non-perfect squares between 72 and 82
= 2 × 7 = 14.

(ii) 102 and 112
Here n = 10 and n + 1 = 11
∴ Non-perfect numbers between 102 and 112
= 2 × 10 = 20.

(iii) 402 and 412
Here n = 40 and n + 1 = 41
∴ Non-perfect square numbers between 402 and 412
= 2 × 40 = 80.

(iv) 802 and 812
Here n = 80 and n + 1= 81
∴ Non-perfect square numbers between 802 and 812
= 2 × 80 = 160.

(v) 1012 and 1022
Here n = 101 and n+ 1= 102
∴ Non-perfect square numbers between 1012 and 1022
= 2 × 101 = 202.

(vi) 2052 and 2062
Here n = 205 and n + 1 = 206
∴ Non-perfect square numbers between 2052 and 2062
= 2 × 205 = 410.

Question 6.
Write down the correct number in the box:
(i) 1002 – 992 = ______ = ______
(ii) 272 – 262 = ______ = ______
(iii) 5692 – 5682 = ______ = ______
Solution:
(i) 1002 – 992 = 100 + 99 = 199
(ii) 272 – 262 = 27 + 26 = 53
(iii) 5692 – 5682 = 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, m2 – 1, m2 + 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, m2 – 1 and m2 + 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, m2 – 1 and m2 + 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, m2 – 1 and m2 + 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, m2 – 1 and m2 + 1
= 2 × 1, (1)2 – 1 and (1)2 + 1 = 2, 0 and 2
Hence (2, 3, 4) are not Pythagorean triplet.

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?
1232, 772, 822, 562, 1612, 1092
Solution:
If a number has 1 or 9 in the unit’s place, then it is square ends in 1.
∴ only 1612 and 1092 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) 192
(ii) 242
(iii) 262
(iv) 362
(v) 342
Solution:
When a square number ends, the number whose square it is, will have either 4 or 6 in unit’s place.
(i) 192 is not a number which will have 6 in its unit’s place.
(ii) 242 is the number which will have 6 in its unit’s place.
(iii) 262 is the number which will have 6 in its units place.
(iv) 362 is the number which will have 6 in its unit’s place.
(v) 342 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) 82 and 92
(ii) 1102 and 1112
(iii) 2052 and 2062
Solution:
There are ‘2n’ non-perfect square numbers between the squares of ‘n’ and ‘(n + 1)’.

(i) 82 and 92
Here n – 8 and (n + 1) = 9
∴ Non-perfect square numbers between 82 and 92 are 2 × 8 = 16,

(ii) 1102 and 1112
Here n = 110 and n + 1 = 111
∴ Non-perfect square numbers between 1102 and 1112 are 2 × 110 = 220.

(iii) 2052 and 2062
Here n = 205 and n + 1 = 206
∴ Non-perfect square numbers between 2052 and 2062 are 2 × 205 = 410.

Question 7.
Fill in the blank boxes with correct numbers:
(i) 892 – 882 = ______ = ______
(ii) 562 – 552 = ______ = ______
(iii) 1052 – 1042 = ______ = ______
(iv) 5762 – 5752 = ______ = ______
(v) 1122 – 1112 = ______ = ______
Solution:
(i) 892 – 882 = 89 + 88 = 117
(ii) 562 – 552 = 56 + 55 = 111
(iii) 1052 – 1042 = 105 +104 = 209
(iv) 5762 – 5752 = 576 + 575 = 1151
(v) 1122 – 1112 = 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) ______ = ______ = 52
Solution:
(i) 1 = 1 = 12
(ii) 1 + 3 = 4 = 22
(iii) 1 + 3 + 5 = 0 = 32
(iv) 1 + 3 + 5 + 7 = 16 = 42
(v) 1 + 3 + 5 + 7 + 9 = 25 = 52

Question 9.
Express the following as the sum of two consecutive integers:
(i) 112
(ii) 132
(iii) 192
(iv) 212
(v) 332
Solution:
n2 = $$\frac{n^2-1}{2}+\frac{n^2+1}{2}$$

(i) 112 =
⇒ 60 + 61
Hence 112 = 60 + 61

(ii) 132 =
⇒ 84 + 85
Hence 132 = 84 + 85

(iii) 192 =
⇒ 180 + 181
Hence 192 = 180 + 181

(iv) 212 =
⇒ 220 + 221
Hence 212 = 220 + 221

(v) 332 =
⇒ 544 + 545
Hence 322 = 544 + 545

Question 10.
Fill in the blank boxes with correct numbers.
(i) 12 = ___
(ii) 112 = 1 ___ 1
(iii) 1112 = 12 ___ 21
(iv) 11112 = 123 ___ 321
(v) 111112 = 1234 ___ 4321
(vi) 1111112 = ______
Solution:
(i) 12 = 1
(ii) 112 = 121
(iii) 1112 = 12321
(iv) 11112 = 1234321
(v) 111112 = 123454321
(vi) 1111112 = 12345654321

Question 11.
Observe the following following and fill in the blanks:
1 + 3 = 22
1 + 3 + 5 = 32
1 + 3 + 5 + 7 = 42
1 + 3 + 5 + 7 + 9 = ______
1 + 3 + 5 + 7 + 9 + 11 = ______
1 + 3 + 5 + 7 + ………. upto n terms = ______
Solution:
1 + 3 + 5 + 7 + 9 = 52
1 + 3 + 5 + 7 + 9 + 11 = 62
1 + 3 + 5 + 7 + …… upto n terms = n2

Question 12.
Fill in the blank boxes with correct numbers:
(i) 72 = 49
(ii) 672 = 4489
(iii) 6672 = 444889
(iv) 66672 = ______
(v) 666672 = ______
(vi) 6666672 = ______
Solution:
(iv) 66672 = 4444889
(v) 666672 = 4444488889
(vi) 6666672 = 444444888889

Question 13.
Observe the following pattern and supply the missing numbers:
(i) 112 = 121
(ii) 1012 = 10201
(iii) 101012 = 102030201
(iv) 10101012 = ___________
(v) 1010101012 = ___________
(vi) 101010101012 = ___________
Solution:
(iv) 1020304030201
(v) 10203040504030201
(vi) 102030405060504030201.

Question 14.
Using the given pattern, find the missing entries:
(i) 12 + 22 + 22 = 32
(ii) 22 + 32 + 62 = 72
(iii) 32 + 42 + ___ = 132
(iv) 42 + 52 + ___ = 212
(v) ___ + 62 + 302 = ___
(vi) 62 + 72 + ___ = 432
Solution:
(iii) 122
(iv) 52
(v) 202
(vi) 422

Question 15.
Find the squares of the following numbers without actual multiplications:
(i) 39
(ii) 42
(iii) 61
(iv) 75
(v) 85
Solution:
(i) 392 = (30 + 9)2
= 30(30 + 9) + 9(30 + 9)
= 302 + 30 × 9 + 9 × 30 + 92
= 900 + 270 + 270 + 81
= 900 + 540 + 81 = 1521

(ii) 422 = (40 + 2)2
= 40(40 + 2) + 2(40 + 2)
= 402 + 40 × 2 + 2 × 40 + 22
= 1600 + 80 + 80 + 4 = 1764

(iii) 612 = (60 + 1)2
= 60 (60 + 1) + 1(60 + 1)
= 602 + 60 × 1 + 1 × 60 + 12
= 3600 + 60 + 60 + 1 = 3721

(iv) 752 = (70 + 5)2
= 70(70 + 5) + 5(70 + 5)
= 702 + 70 × 5 + 5 × 70 + 52
= 4900 + 350 + 350 + 25 = 5625

(v) 852 = (80 + 5)2
= 80(80 + 5) + 5(80 + 5)
= 802 + 80 × 5 + 5 × 80 + 52
= 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, m2 – 1 and m2 + 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
m2 – 1 = 62 – 1 = 36 – 1 = 35
and m2 + 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
m2 – 1 = (4)2 – 1 = 16 – 1 = 15
and m2 + 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
m2 – 1 = (4)2 – 1 = 16 – 1 = 15
m2 + 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
m2 – 1 = 12 – 1 = 0
m2 + 1 = 12 + 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
m2 – 1 = (3)2 – 1 = 9 – 1 = 8 and
m2 + 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, m2 – 1 and m2 + 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 = 12
2 × 2 = 4 = 22
3 × 3 = 9 = 32
4 × 4 = 16 = 42
n × n = n2
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. 42 = 16,
62 = 36 and
82 = 64, etc.

(ii) Squares of odd numbers are always odd.
e.g. 92 = 81,
112 = 121 and
132 = 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. 52 – 42 = 25 – 16 = 9 = 5 + 4
72 – 62 = 49 – 36 = 13 = 7 + 6
92 – 82 = 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. 32 = 1 + 3 + 5 = 9
52 = 1 + 3 + 5 + 7 + 9 = 25
62 = 1 + 3 + 5 + 7 + 9 + 11 = 36

(vii) Squares of natural numbers composed of the only digit 1, follow a particular pattern.
e.g. 12 = 1 = 12
112 = 121 = 1 + 2 + 1 = 4 = 22
1112 = 12321 = 1 + 2 + 3 + 2 + 1 = 9 = 32
11112 = 1234321 = 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 = 42

(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. 32 = 9
42 = 16 = (15 + 1)
72 = 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. 42 = 16
52 = 25 = 24 + 1)
62 = 36
72 = 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 n2 and (n + 1)2 is 2n.

e.g.
(i) 12 = 1 and 22 = 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) 32 = 9 and 42 = 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.