NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1.

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

Ex 6.1 Class 8 Maths 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

Solution:
The unit digits of the squares of the given numbers are shown against the numbers in the table given below:

Ex 6.1 Class 8 Maths Question 2.

The following numbers are obviously not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050

Solution:
A number which ends with 2, 3, 7 or 8 cannot be a perfect square. Again, a number which ends with odd number of zero(s) cannot be a perfect square.
(i) Since the number 1057 ends with 7, so it cannot be a perfect square.
(ii) Since the number 23453 ends with 3, so it cannot be a perfect square.
(iii) Since the number 7928 ends with 8, so it cannot be a perfect square.
(iv) Since the number 222222 ends with 2, so it cannot be a perfect square.
(v) Since the number 64000 ends with an odd number of zeros, so it cannot be a perfect square.
(vi) Since the number 89722 ends with 2, so it cannot be a perfect square.
(vii) Since the number 222000 ends with an odd number of zeros, so it cannot be a perfect square.
(viii) Since the number 505050 ends with an odd number of zeros, so it cannot be a perfect square.

Ex 6.1 Class 8 Maths Question 3.

The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004

Solution:
(i) The number 431 is an odd number, so its square would be odd.
(ii) The number 2826 is an even number, so its square would be even.
(iii) The number 7779 is an odd number, so its square would be odd.
(iv) The number 82004 is an even number, so its square would be even.
Hence, the squares of the numbers 431 and 7779 would be odd numbers.

Ex 6.1 Class 8 Maths Question 4.

Observe the following pattern and find the missing digits:
                              11² = 121
                            101² = 10201
                          1001² = 1002001
                      100001² = 1 ………….. 2 ……….. 1
                  10000001² = ………………..

Solution:
The missing digits are as follows:
100001² = 10000200001
10000001² = 100000020000001

Ex 6.1 Class 8 Maths Question 5.

Observe the following pattern and supply the missing numbers:
                                    11² = 121
                                  101² = 10201
                              10101² = 102030201
                          1010101² = ………..
                        ……………. ² = 10203040504030201

Solution:
The missing numbers are as follows:
1010101² = 1020304030201
101010101² = 10203040504030201

Ex 6.1 Class 8 Maths 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

Solution:
The missing numbers are as follows:
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²

Ex 6.1 Class 8 Maths 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 = Sum of first 5 odd numbers
                                  = 5² = 25
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
                                  = Sum of first 10 odd numbers
                                  = 10² = 100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
                                  = Sum of first 12 odd numbers
                                  = 12² = 144

Ex 6.1 Class 8 Maths 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

Ex 6.1 Class 8 Maths 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:

We know that there are 2n numbers lie between n² and (n + 1)².
(i) Therefore, between 12² and 13², there are 24 (i.e., 2 × 12) numbers.
(ii) Therefore, between 25² and 26², there are 50 (i.e., 2 × 25) numbers.
(iii) Therefore, between 99² and 100², there are 198 (i.e., 99 × 2) numbers.