These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

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

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) 26387

(vii) 52698

(viii) 99880

(ix) 12796

(x) 55555

Solution:

(i) The unit digit of (81)^{2} is 1 (1 × 1 = 1)

(ii) The unit digit of (272)^{2} is 4 (2 × 2 = 4)

(iii) The unit digit of (799)^{2} is 1 (9 × 9 = 81)

(iv) The unit digit of (3853)^{2} is 9 (3 × 3 = 9)

(v) The unit digit of (1234)^{2} is 6 (4 × 4 = 16)

(vi) The unit digit of (26387)^{2} is 9 (7 × 7 = 49)

(vii) The unit digit of (52698)^{2} is 4 (8 × 8 = 64)

(viii) The unit digit of (99880)^{2} is 0 (0 × 0 = 0)

(ix) The unit digit of (12796)^{2} is 6 (6 × 6 = 36)

(x) The unit digit of (55555)^{2} is 5 (5 × 5 = 25)

Question 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:

(i) 1057 is not a perfect square,

As the last digit is 7 [It is not one of 0, 1, 4, 5, 6, and 9].

(ii) 23453 is not a perfect square,

As the last digit is 3 [It is not one of 0, 1, 4, 5, 6, and 9].

(iii) 7928 is not a perfect square,

As the last digit is 8 [It is not one of 0, 1, 4, 5, 6, and 9].

(iv) 222222 is not a perfect square,

As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(v) 64000 is not a perfect square,

As the number of zeros is odd.

(vi) 89722 is not a perfect square,

As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(vii) 222000 is not a perfect square,

As the number of zeros is odd.

(viii) 505050 is not a perfect square,

As the number of zeros is odd.

Question 3.

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

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Note: The square of an odd natural number is always odd and that of an even number always an even number.

Solution:

(i) The square of 431 is an odd number.

(ii) The square of2826 is an even number.

(iii) The square of 7779 is an odd number.

(iv) The square of82004 is an even number.

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:

By observing the above pattern, we get

(i) 100001^{2} = 10000200001

(ii) 10000001^{2} = 100000020000001

Question 5.

Observe the following pattern and supply the missing number.

(a) 11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} = …………….

………..^{2} = 1020304030201

Solution:

By observing the above pattern, we get

(I) (1010101)^{2} = 1020304030201

(ii) 10203040504030201 = (101010101)^{2}

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:

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

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

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

Note: To find pattern

The third number is related to the first and second numbers. How?

The fourth number is related to the third number. How?

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:

(i) Sum of the first 5 odd numbers = 5^{2} = 25

(ii) Sum of the first 10 odd numbers = 10^{2} = 100

(iii) Sum of the first 12 odd numbers = 12^{2} = 144

Question 8.

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

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

Solution:

(i) 7^{2} = 49

1 + 3 + 5 + 7 + 9 + 11 + 13 (First 7 odd numbers)

(ii) 11^{2} = 121

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (First 11 odd numbers)

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:

Note: Since between n and (n + 1) there are 2n non-square numbers

(i) Between 12^{2} and 13^{2}, there are 2 × 12 numbers i.e. 24 numbers

(ii) Between 25^{2} and 26^{2}, there are 2 × 25 numbers i.e. 50 numbers

(iii) Between 99^{2} and 100^{2}, there are 2 × 99 numbers i.e. 198 numbers

Note: For any number ending with 5, the square is a (a + 1) × 100 + 25

(i) 35^{2}

= 3(3 + 1) × 100 + 25

= 1200 + 25

= 1225

(ii) 55^{2}

= 5(5 + 1) × 100 + 25

= 3000 + 25

= 3025

(iii) 125^{2}

= 12(12 + 1) × 100 + 25

= 12 × 13 × 100 + 25

= 15600 + 25

= 15625