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

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.1

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:
(i) The unit digit of (81)2 is 1 (1 × 1 = 1)
(ii) The unit digit of (272)2 is 4 (2 × 2 = 4)
(iii) The unit digit of (799)2 is 1 (9 × 9 = 81)
(iv) The unit digit of (3853)2 is 9 (3 × 3 = 9)
(v) The unit digit of (1234)2 is 6 (4 × 4 = 16)
(vi) The unit digit of (26387)2 is 9 (7 × 7 = 49)
(vii) The unit digit of (52698)2 is 4 (8 × 8 = 64)
(viii) The unit digit of (99880)2 is 0 (0 × 0 = 0)
(ix) The unit digit of (12796)2 is 6 (6 × 6 = 36)
(x) The unit digit of (55555)2 is 5 (5 × 5 = 25)

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:
(i) 1057 is not a perfect square,
As the last digit is 7 [It is not one of 0, 1, 4, 5, 6, and 9].

(ii) 23453 is not a perfect square,
As the last digit is 3 [It is not one of 0, 1, 4, 5, 6, and 9].

(iii) 7928 is not a perfect square,
As the last digit is 8 [It is not one of 0, 1, 4, 5, 6, and 9].

(iv) 222222 is not a perfect square,
As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(v) 64000 is not a perfect square,
As the number of zeros is odd.

(vi) 89722 is not a perfect square,
As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(vii) 222000 is not a perfect square,
As the number of zeros is odd.

(viii) 505050 is not a perfect square,
As the number of zeros is odd.

Question 3.
The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Note: The square of an odd natural number is always odd and that of an even number always an even number.
Solution:
(i) The square of 431 is an odd number.
(ii) The square of2826 is an even number.
(iii) The square of 7779 is an odd number.
(iv) The square of82004 is an 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:
By observing the above pattern, we get
(i) 1000012 = 10000200001
(ii) 100000012 = 100000020000001

Question 5.
Observe the following pattern and supply the missing number.
(a) 112 = 121
1012 = 10201
101012 = 102030201
10101012 = …………….
………..2 = 1020304030201
Solution:
By observing the above pattern, we get
(I) (1010101)2 = 1020304030201
(ii) 10203040504030201 = (101010101)2

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:
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432
Note: 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?

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) Sum of the first 5 odd numbers = 52 = 25
(ii) Sum of the first 10 odd numbers = 102 = 100
(iii) Sum of the first 12 odd numbers = 122 = 144

Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as to the sum of 11 odd numbers.
Solution:
(i) 72 = 49
1 + 3 + 5 + 7 + 9 + 11 + 13 (First 7 odd numbers)
(ii) 112 = 121
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (First 11 odd numbers)

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:
Note: Since between n and (n + 1) there are 2n non-square numbers
(i) Between 122 and 132, there are 2 × 12 numbers i.e. 24 numbers
(ii) Between 252 and 262, there are 2 × 25 numbers i.e. 50 numbers
(iii) Between 992 and 1002, there are 2 × 99 numbers i.e. 198 numbers
Note: For any number ending with 5, the square is a (a + 1) × 100 + 25
(i) 352
= 3(3 + 1) × 100 + 25
= 1200 + 25
= 1225

(ii) 552
= 5(5 + 1) × 100 + 25
= 3000 + 25
= 3025

(iii) 1252
= 12(12 + 1) × 100 + 25
= 12 × 13 × 100 + 25
= 15600 + 25
= 15625

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