# NCERT Solutions for Class 8 Maths Chapter 6 :Square roots CBSE Part 1

In this page we have NCERT Solutions for Class 8 Maths Chapter 6 :Square roots for EXERCISE 1 . Hope you like them and do not forget to like , social share and comment at the end of the page.
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
The unit digit square decide the number for the square of any number
 1 1 Explanation: Since, 12 =1 2 4 Explanation: Since, 22 = 4, 3 1 Explanation: Since, 92 = 81 4 9 Explanation: Since 32 = 9 5 6 Explanation: Since, 42 = 16 6 9 Explanation: Since, 72 = 49 7 4 Explanation: Since, 82 = 64. So 8 0 Since, 02 = 0. 9 6 Explanation: Since, 62 = 36 10 5 Explanation: Since, 52 = 25
Question 2
The following numbers are obviously not perfect squares. Give reason.
1. 1057
2. 23453
3. 7928
4. 222222
5. 64000
6. 89722
7. 222000
8. 505050
The square of any number will have 0,1,4,5,6 or 9 at its unit place
So  (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 1, 4, 5, 6, or 9 at unit’s place, so they are not be perfect squares.
The square of Zeros will be even always
So (v), (vii) and (viii) don’t have even number of zeroes at the end so they are not perfect squares.
Question 3
The squares of which of the following would be odd numbers?
1. 431
2. 2826
3. 7779
4. 82004
1. 431  square will end in 1,So odd number
2.  2826  square will end in 6 ,so even number
3. 779 square will end in 1,So odd number
4. 82004 square will end in 6 ,so even number
Question 4
Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1.........2.......1
100000012 = ...............
Solution:
1000012 = 10000200001
100000012 = 100000020000001
Reasoning Start with 1 followed as many zeroes as there are between the first and the last one, followed by two again followed by as many zeroes and end with 1.
Question 5
Observe the following pattern and supply the missing numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ..................
..............2 =10203040504030201
10101012 = 1020304030201
1010101012 =10203040504030201
Reasoning Start with 1 followed by a zero and go up to as many number as there are number of 1s given, follow the same pattern in reverse order.
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 + _+ 302 = 312
62 + 72 + _2 = _2
We can see following pattern the above series
Relation among first, second and third number - Third number is the product of first and second number
Relation between third and fourth number - Fourth number is 1 more than the third number
So
42 + 52 + 202 = 212
52 + 6+ 302 = 312
62 + 72 + 422 = 432

Question 7
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Explanation:
1 + 3 = 22 = 4
1 + 3 + 5 = 32 = 9
1 + 3 + 5 + 7 = 42 =16
1 + 3 + 5 + 7 + 9 = 52 = 25
So Sum of n odd numbers starting from 1 = n2
From the above derivation we can answer the above questions
1. Since, there are 5 consecutive odd numbers, Thus, their sum = 52 = 25
2. Since, there are 10 consecutive odd numbers, Thus, their sum = 102 = 100
3. Since, there are 12 consecutive odd numbers, Thus, their sum = 122 = 144
Question 8
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Explanation:
1 + 3 = 22 = 4
1 + 3 + 5 = 32 = 9
1 + 3 + 5 + 7 = 42 =16
1 + 3 + 5 + 7 + 9 = 52 = 25
So Sum of n odd numbers starting from 1 = n2
1) Since, 49 = 72
So, 72 can be expressed as follows:
1 + 3 + 5 + 7 + 9 + 11 + 13
2) Since, 121 = 112
Therefore, 121 = 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