NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.1

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

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.1:–

Question 1.

What will be the unit digit of the squares of the following numbers?

(i) 81

Solu:

Unit digit of 81² = 1

(ii) 272

Solution:

Unit digit of 272² = 4

(iii) 799

Solution:

Unit digit of 799² = 1

(iv) 3853

Solution:

 Unit digit of 3853² = 9

(v) 1234

Solution: 

Unit digit of 1234² = 6

(vi) 20387

Solution:

 Unit digit of 26387² = 9

(vii) 52698

Solution:

Unit digit of 52698² = 4

(viii) 99880

Solution:

Unit digit of 99880² = 0

(ix) 12796

Solution:

Unit digit of 12796² = 6

(x) 55555

Solution:

Unit digit of 55555² = 5

Question 2.

The following numbers are not perfect squares. Give reason.

(i) 1057

Solution:

1057 ends with 7 at unit place. So it is not a perfect square number.

(ii) 23453

Solution:

23453 ends with 3 at unit place. So it is not a perfect square number.

(iii) 7928

Solution:

7928 ends with 8 at unit place. So it is not a perfect square number.

(iv) 222222

Solution:

222222 ends with 2 at unit place. So it is not a perfect square number.

(v) 64000

Solution:

64000 ends with 3 zeros. So it cannot a perfect square number.

(vi) 89722

Solution:

89722 ends with 2 at unit place. So it is not a perfect square number.

(vii) 222000

Solution:

22000 ends with 3 zeros. So it can not be a perfect square number.

(viii) 505050

Solution:

505050 ends with 1 zero. So it is not a perfect square number.

Question 3. 

The squares of which of the following would be odd numbers?

(i) 431

Solution:

431² is an odd number.

(ii) 2826

Solution:

2826² is an even number.

(iii) 7779

Solution:

 7779² is an odd number.

(iv) 82004

Solution:

82004² is an even number.

Question 4. 

Observe the following pattern and find the missing digits.

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1…2…1

100000012 = ………

Solution:

According to the above pattern, we have

1000012 = 10000200001

100000012 = 100000020000001

Question 5.

 Observe the following pattern and supply the missing numbers.

112 = 121

1012 = 10201

101012 = 102030201

10101012 = ……….

……….2 = 10203040504030201

Solution:

According to the above pattern, we have

10101012 = 1020304030201

1010101012 = 10203040504030201

Question 6.

 Using the given pattern, find the missing numbers.

12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + ….2 = 212

52 + ….2 + 302 = 312

62 + 72 + …..2 = ……2

Solution:

According to the given pattern, we have

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

Question 7. 

Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution:

We know that the sum of n odd numbers = n2

(i) 1 + 3 + 5 + 7 + 9 = (5)2 = 25 [∵ n = 5]

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100 [∵ n = 10]

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12)2 = 144 [∵ n = 12]

Question 8.

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

Solution:

49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)

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

Solution:

121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)

Question 9. 

How many numbers lie between squares of the following numbers?

(i) 12 and 13

Solution:

 We know that numbers between n2 and (n + 1)2 = 2n

Numbers between 12² and 13² = (2n) = 2 × 12 = 24

(ii) 25 and 26

Solution:

Numbers between 25² and 26² = 2 × 25 = 50 (∵ n = 25)

(iii) 99 and 100.

Solution:

 Numbers between 99² and 100² = 2 × 99 = 198 (∵ n = 99)