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

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 812 = 1
(ii) Unit’s digit of 2722 = 4
(iii) Unit’s digit of 7992 = 1
(iv) Unit’s digit of 38532 = 9
(v) Unit’s digit of 12342 = 6
(vi) Unit’s digit of 263872 = 9
(vii) Unit’s digit of 526982 = 4
(viii) Unit’s digit of 998802 = 0
(ix) Unit’s digit of 127962 = 6
(x) Unit’s digit of 555552 = 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.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1………..2……….1
100000012 = 1……………..
Solution:
(10001)2 = 100020001
(100001)2 = 10000200001
(10000001)2 = 100000020000001

Question 5.
Observe the following pattern and supply the missing numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ……………………..
……………..2 = 10203040504030201
Solution:
(1010101)2 = 1020304030201
(101010101)2 = 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
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:
42 + 52 + 202 = 212
52 + 62 + 302 = 312
52 + 72 + 42 = 432
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.
(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 = 52
(Sum of first five odd numbers is 52)
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 102 = 100
(Sum of first ten odd numbers is 102)
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144
(Sum of first ten odd number is 112)

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 = 72 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = 112 = 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 n2 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 252 and 262.
(iii) 2 × 99 = 198 numbers lies between 992 and 1002.