You can Download KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1

Question 1.

What will be the unit’s 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) Unit’s digit of 81^{2} = 1

(ii) Unit’s digit of 272^{2} = 4

(iii) Unit’s digit of 799^{2} = 1

(iv) Unit’s digit of 3853^{2} = 9

(v) Unit’s digit of 1234^{2} = 6

(vi) Unit’s digit of 26387^{2} = 9

(vii) Unit’s digit of 52698^{2} = 4

(viii) Unit’s digit of 99880^{2} = 0

(ix) Unit’s digit of 12796^{2} = 6

(x) Unit’s digit of 55555^{2} = 5

Question 2.

The following numbers are obviously not perfect squares. Give reason.

(i) 1057

(ii) 23453

(iv) 7928

(iv) 222222

(v) 64000

(vi) 89722

(vii) 222000

(viii) 505050

Solution:

A perfect square does not have its unit digit 2, 3, 7, or 8.

(i) 1057 – Its unit digit is 7, hence, not a perfect square.

(ii) 23453 – Its unit digit is 3, hence, not a perfect square.

(iii) 7928 – Unit digit is 8, hence, not a perfect square.

(iv) 222222 – Unit digit is 2, hence, not a perfect square.

(v) 64000 – Its unit digit is 0, which can make a number, perfect square but these zeros are in odd number. Hence, not a perfect square.

(vi) 89722 – Unit digit is 2, hence, not a perfect square.

(vii) 222000 – Not perfect square, because zeros are in odd numbers.

(viii) 505050 – Not perfect square, because zeros, in the end, are in odd numbers.

Question 3.

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

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Solution:

431 and 7779 are odd, hence, the square of these are odd.

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} = 1……………..

Solution:

(10001)^{2} = 100020001

(100001)^{2} = 10000200001

(10000001)^{2} = 100000020000001

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:

(1010101)^{2} = 1020304030201

(101010101)^{2} = 10203040504030201

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}

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?

Solution:

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

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

5^{2} + 7^{2} + 4^{2} = 43^{2}

Pattern

1 × 2 = 2, 2 + 1 = 3

2 × 3 = 6, 6 + 1 = 7

3 × 4 = 12, 12 + 1 = 13

4 × 5 = 20, 20 + 1 = 21

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) 1 + 3 + 5 + 7 + 9 = 5^{2}

(Sum of first five odd numbers is 5^{2})

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10^{2} = 100

(Sum of first ten odd numbers is 10^{2})

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12^{2} = 144

(Sum of first ten odd number is 11^{2})

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 = 7^{2} = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 121 = 11^{2} = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

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) n = 12 and n + 1 = 12 + 1 = 13

So, 2n are the numbers lies between n^{2} and (n + 1)^{2}.

Hence, 2 × 12 = 24 numbers lie between the squares of 12 and 13.

(ii) Similarly 2 × 25 = 50 numbers lies between 25^{2} and 26^{2}.

(iii) 2 × 99 = 198 numbers lies between 99^{2} and 100^{2}.