 # 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, $1^2=1$ 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.
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.
$11^2= 121$
$101^2= 10201$
$1001^2= 1002001$
$100001^2$= 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=$10203040504030201$
$1010101^2= 1020304030201$
$101010101^2=10203040504030201$
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.
$1^2+ 2^2+ 2^2= 3^2$
$2^2+ 3^2+ 6^2= 7^2$
$3^2+ 4^2+ 12^2= 13^2$
42+ 52+ _2= 212
52+ _2+ 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
$4^2+ 5^2+ 20^2= 21^2$
$5^2+ 6^2+ 30^2= 31^2$
$6^2+ 7^2+ 42^2= 43^2$

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) $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23$
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
1. Since, there are 5 consecutive odd numbers, Thus, their sum = $5^2= 25$
2. Since, there are 10 consecutive odd numbers, Thus, their sum = $10^2= 100$
3. Since, there are 12 consecutive odd numbers, Thus, their sum = $12^2= 144$
Question 8
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Explanation:
$1 + 3 = 2^2= 4$
$1 + 3 + 5 = 3^2= 9$
$1 + 3 + 5 + 7 = 4^2=16$
$1 + 3 + 5 + 7 + 9 = 5^2= 25$
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
i. $12^2= 144$
$13^2= 169$
Now, 169 - 144 = 25
So, there are 25 - 1 = 24 numbers lying between $12^2$ and $13^2$
ii. 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 $100^2$

Reference Books for class 8 Math

Given below are the links of some of the reference books for class 8 Math.

1. Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
2. Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
3. Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
4. Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

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