Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1

Question 1.

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

- 81
- 272
- 799
- 3853
- 1234
- 26387
- 52698
- 99880
- 12796
- 55555

Solution:

1. āµ 1 Ć 1 = 1

ā“ The unit digit of (81)^{2} will be 1.

2. āµ 2 Ć 2 = 4

The unit digits of (272)^{2} will be 4.

3. Since, 9 Ć 9 = 81

ā“ The unit digit of (799)^{2} will be 1.

4. Since, 3 Ć 3 = 9

ā“ The unit digit of (3853)^{2} will be 9.

5. Since, 4 Ć 4 = 16

ā“ The unit digit of (1234)^{2} will be 6.

6. Since 7 Ć 7 = 49

ā“ The unit digit of (26387)^{2} will be 9.

7. Since, 8 Ć 8 = 64

ā“ The unit digit of (52698)^{2} will be 4.

8. Since 0 Ć 0 = 0

ā“ The unit digit of (99880)^{2} will be 0.

9. Since 6 Ć 6 = 36

ā“ The unit digit of (127969)^{2} will be 6.

10. Since, 5 Ć 5 = 25

ā“ The unit digit of (555559)^{2} will be 5.

Question 2.

The fĆ²llowing numbers are obviously not perfect squares. Give reason?

- 1057
- 23453
- 7928
- 222222
- 64000
- 89722
- 222000
- 505050

Solution:

1. 1057

Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)

ā“ 1057 is not a perfect square.

2. 23453

Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9).

ā“ 23453 is not a perfect square.

3. 7928

Since, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9).

ā“ 7928 is not a perfect square.

4. 222222

Since, the ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).

ā“ 222222 is not a perfect square.

5. 64000

Since, the number of zeros is odd.

ā“ 64000 is not a perfect square.

6. 89722

Since, the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9).

ā“ 89722 is not a perfect square.

7. 222000

Since, the number of zeros is odd.

ā“ 222000 is not a perfect square.

8. 505050

The units digit is odd zero.

ā“ 505050 can not be a perfect square.

Question 3.

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

- 431
- 2826
- 7779
- 82004

Solution:

Since the square of an odd natural number is odd and that of an even number is an even number.

1. ā“ The square of 431 is an odd number.

[āµ431 is an odd number]

2. The square of 2826 is an even number.

[āµ 2826 is an even number]

3. The square of 7779 is an odd number.

[āµ 7779 is an odd number]

4. The square of 82004 is an even number.

[āµ 82004 is an even number]

Question 4.

Observe the following pattern and find the missing digits?

11^{2} = 121

101^{2} = 10201

10101^{2} = 1002001

1010101^{2} = I …………….. 2 …………………. 1

10000001^{2} = …………………………

Solution:

Observing the above pattern, we have

- (100001)
^{2}= 10000200001 - (10000001)
^{2}= 100000020000001

Question 5.

Observe the following pattern and supply the missing number?

11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} = ………………………..

…………………… ^{2} = 10203040504030201

Solution:

Observing the above, we have

- (1010101)
^{2}= 1020304030201 - 10203040504030201 = (101010101)
^{2}

Question 6.

Using the given pattern, find the missing numbers?

1^{2} + 2^{2} + 3^{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}

Note:

To find pattern:

Third number is related to first and second number. How?

Fourth number is related to third number. How?

Soution:

The missing numbers are

- 4
^{2}+ 5^{2}+ 20^{2}= 21^{2} - 5
^{2}+ 6^{2}+ 30^{2}= 31^{2} - 6
^{2}+ 7^{2}+ 42^{2}= 43^{2}

Question 7.

Without adding, find the sum,

- 1 + 3 + 5 + 7 + 9
- 1 + 3 + 5 + 7 + 9 + 11 + 13 + 13 + 17
- 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution:

- The sum of first 5 odd numbers = 5
^{2}= 25 - The sum of first 10 odd numbers = 10
^{2}= 100 - The sum of first 12 odd numbers = 12
^{2}= 144

Question 8.

- Express 49 as she suns of 7 odd numbers.
- Express 121 as the suns oft! odd numbers.

Solution:

1. 49 = 7^{2} Sum of first 7 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13

2. 121 = 11^{2} = Sum of first 11 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Question 9.

How mars numbers lie between squares of the following numbers?

- 12 and 13
- 25 and 26
- 99 and 100

Solution:

Since between n^{2} and (n + 1)^{2}, there are 2n, non-square numbers.

- Between 12
^{2}and 13^{2}, there are 2 Ć 12, ie; 24 numbers - Between 25
^{2}and 26^{2}, there are 2 Ć 25, i.e; 50 numbers - Between 99
^{2}and 100^{2}, there arc 2 Ć 99. i.e; 198 numbers