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

Question 1.

Find the perfect square numbers between 40 and 50.

Solution:

Perfect square numbers between 40 and 50 = 49.

Question 2.

Which of the following 24^{2}, 49^{2}, 77^{2}, 131^{2} or 189^{2} end with digit 1?

Solution:

Only 49^{2}, 131^{2} and 189^{2} end with digit 1.

Question 3.

Find the value of each of the following without calculating squares.

(i) 27^{2} – 26^{2}

(ii) 118^{2} – 117^{2}

Solution:

(i) 27^{2} – 26^{2} = 27 + 26 = 53

(ii) 118^{2} – 117^{2} = 118 + 117 = 235

Question 4.

Write each of the following numbers as difference of the square of two consecutive natural numbers.

(i) 49

(ii) 75

(iii) 125

Solution:

(i) 49 = 2 × 24 + 1

49 = 25^{2} – 24^{2}

(ii) 75 = 2 × 37 + 1

75 = 38^{2} – 37^{2}

(iii) 125 = 2 × 62 + 1

125 = 63^{2} – 62^{2}

Question 5.

Write down the following as sum of odd numbers.

(i) 7^{2}

(ii) 9^{2}

Solution:

(i) 7^{2} = Sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 9^{2} = Sum of first 9 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

## NCERT Solutions for Class 8 Maths

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

Ex 6.1 Class 8 Maths Question 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) 20387

(vii) 52698

(viii) 99880

(ix) 12796

(x) 55555

Solution:

(i) Unit digit of 81^{2} = 1

(ii) Unit digit of 272^{2} = 4

(iii) Unit digit of 799^{2} = 1

(iv) Unit digit of 3853^{2} = 9

(v) Unit digit of 1234^{2} = 6

(vi) Unit digit of 26387^{2} = 9

(vii) Unit digit of 52698^{2} = 4

(viii) Unit digit of 99880^{2} = 0

(ix) Unit digit of 12796^{2} = 6

(x) Unit digit of 55555^{2} = 5

Ex 6.1 Class 8 Maths Question 2.

The following numbers are not perfect squares. Give reason.

(i) 1057

(ii) 23453

(iii) 7928

(iv) 222222

(v) 64000

(vi) 89722

(vii) 222000

(viii) 505050

Solution:

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

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

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

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

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

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

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

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

Ex 6.1 Class 8 Maths Question 3.

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

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Solution:

(i) 431^{2} is an odd number.

(ii) 2826^{2} is an even number.

(iii) 7779^{2} is an odd number.

(iv) 82004^{2} is an even number.

Ex 6.1 Class 8 Maths Question 4.

Observe the following pattern and find the missing digits.

11^{2} = 121

101^{2} = 10201

1001^{2} = 1002001

100001^{2} = 1…2…1

10000001^{2} = ………

Solution:

According to the above pattern, we have

100001^{2} = 10000200001

10000001^{2} = 100000020000001

Ex 6.1 Class 8 Maths Question 5.

Observe the following pattern and supply the missing numbers.

11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} = ……….

……….^{2} = 10203040504030201

Solution:

According to the above pattern, we have

1010101^{2} = 1020304030201

101010101^{2} = 10203040504030201

Ex 6.1 Class 8 Maths Question 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} + ….^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + …..^{2} = ……^{2}

Solution:

According to the given pattern, we have

4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + 42^{2} = 43^{2}

Ex 6.1 Class 8 Maths 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 = n^{2}

(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]

Ex 6.1 Class 8 Maths Question 8.

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

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

Solution:

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

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

Ex 6.1 Class 8 Maths Question 9.

How many numbers lie between squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100.

Solution:

(i) We know that numbers between n^{2} and (n + 1)^{2} = 2n

Numbers between 12^{2} and 13^{2} = (2n) = 2 × 12 = 24

(ii) Numbers between 25^{2} and 26^{2} = 2 × 25 = 50 (∵ n = 25)

(iii) Numbers between 99^{2} and 100^{2} = 2 × 99 = 198 (∵ n = 99)