In this page we have NCERT book Solutions for Class 8th Maths: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
Answer
The unit digit square decide the number for the square of any number
1 
1 Explanation: Since, 1^{2} =1 
2 
4 Explanation: Since, 2^{2} = 4, 
3 
1 Explanation: Since, 9^{2} = 81 
4 
9 Explanation: Since 3^{2} = 9

5 
6 Explanation: Since, 4^{2} = 16 
6 
9 Explanation: Since, 7^{2} = 49 
7 
4 Explanation: Since, 8^{2} = 64. So 
8 
0 Since, 0^{2} = 0. 
9 
6 Explanation: Since, 6^{2} = 36

10 
5 Explanation: Since, 5^{2} = 25 
Question 2
The following numbers are obviously not perfect squares. Give reason.
Answer
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?
Answer
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:
100001^{2} = 10000200001
10000001^{2} = 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.
11^{2} = 121
101^{2} = 10201
10101^{2} = 102030201
1010101^{2} = ..................
..............^{2} =10203040504030201
Answer
1010101^{2} = 1020304030201
101010101^{2} =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.
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}
Answer
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
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.
(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
Answer
Explanation:
1 + 3 = 2^{2} = 4
1 + 3 + 5 = 3^{2} = 9
1 + 3 + 5 + 7 = 4^{2} =16
1 + 3 + 5 + 7 + 9 = 5^{2} = 25
So Sum of n odd numbers starting from 1 = n^{2}
From the above derivation we can answer the above questions
Question 8
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer
Explanation:
1 + 3 = 2^{2} = 4
1 + 3 + 5 = 3^{2} = 9
1 + 3 + 5 + 7 = 4^{2} =16
1 + 3 + 5 + 7 + 9 = 5^{2} = 25
So Sum of n odd numbers starting from 1 = n^{2}
1) Since, 49 = 7^{2}
So, 7^{2} can be expressed as follows:
1 + 3 + 5 + 7 + 9 + 11 + 13
2) Since, 121 = 11^{2}
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
Answer
1) 12^{2} = 144
13^{2} = 169
Now, 169  144 = 25
So, there are 25  1 = 24 numbers lying between 12^{2} and 13^{2}
2) We know that, 25^{2} = 625
And, 26^{2} = 676
Now, 676  625 = 51
So, there are 51  1 = 50 numbers lying between 25^{2} and 26^{2}
3) We know that, 99^{2} = 9801
And, 100^{2} = 10000
Now, 10000  9801 = 199
So, there are 199  1 = 198 numbers lying between 99^{2} and 100^{2}