# NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions

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

## NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions

Try These (Page No. 90)

Question 1.
Find the perfect square numbers between
(i) 30 and 40 and
(ii) 50 and 60.
Solution:
(i) Since,
1 × 1 = 1
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
6 × 6 = 4
7 × 7 = 49
Thus 36 is a perfect square number between 30 and 40.

(ii) Since, 7 × 7 = 49 and 8 × 8 = 64.
It means there is no perfect number between 49 and 64 and thus there is no perfect number between 50 and 60.

Question 2.
Can we say whether the following numbers are perfect squares? How do we know?
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 1069
(vi) 2061
Write five numbers that you can decide by looking at their ones digit that they are not square numbers.
Solution:
(i) 1057
∵ The ending digit is 7 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 1057 cannot be a square number.

(ii) 23453
∵ The ending digit is 3 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 23453 cannot be a square number.

(iii) 7928
∵ The ending digit is 8 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 7928 cannot be a square number.

(iv) 222222
∵ The ending digit is 2 (which is not one of 0, 1, 4, 5, 6, or 9).
∴ 222222 cannot be a square number.

(v) 1069
∵ The ending digit is 9
∴ It may or may not be a square number.
Also, 30 × 30 = 900
31 × 31 = 961
32 × 32 = 1024
33 × 33 = 1089
i.e., No natural number between 1024 and 1089 which is a square number.
1069 cannot be a square number.

(iv) 2061
∵ The ending digit is 1
∴ It may or may not be a square number.
∵ 45 × 45 = 2025 and 46 × 46 = 2116
i.e., No natural number between 2025 and 2116 which is a square number.
∴ 2061 is not a square number.
We can write many numbers which do not end with 0, 1, 4, 5, 6, or 9. (i.e., which are not square numbers)
Five such numbers can be: 1234, 4312, 5678, 87543, 1002007

Question 3.
Write five numbers that you cannot decide just by looking at their units digit (or one’s place) whether they are square numbers or not.
Solution:
Property 1.
Any natural number ending in 1, 4, 5, 6, or 9 can be or cannot be a square number.
Five such numbers are: 56790, 3671, 2454, 76555, 69209
Property 2.
If a number has 1 or 9 in the unit’s place, then its square ends in 1.
For example: (1)2 = 1, (9)2 = 81, (11)2 = 121, (9)2 = 361, (21)2 = 441

Try These (Page No. 91)

Question 4.
Which of 1232 , 772, 822, 1612, 1092 would end with the digit 1?
Solution:
The squares of those numbers end in 1 which end in either 1 or 9.
∴ The squares of 161 and 109 would end in 1.

Question 5.
Which of the following numbers would have digit 6 at the unit place.
(i) 192
(ii) 242
(iii) 262
(iv) 362
(v) 342
Solution:
(i) 192: Unit place digit = 9.
192 would not have unit digit as 6.

(ii) 242: Unit place digit = 4.
242 would not have unit digit as 6.

(iii) 262: Unit place digit = 6.
262 would have 6 as unit place.

(iv) 362: Unit place digit = 6.
362 would end in 6

(v) 342: Since the unit place digit is 4
342 would have unit place digit as 6.

Try These (Page No. 92)

Question 6.
What will be the “one’s digit” in the square of the following numbers?
(i) 1234
(ii) 26387
(iii) 52698
(iv) 99880
(v) 21222
(vi) 9106
Solution:
(i) 192: Ending digit = 4 and 42 = 16.
(1234)2 will have 6 as the ones digit.

(ii) 242: Ending digit is 7 and 72 = 49.
(26387)2 will have 9 as the ones digit.

(iii) 262: Ending digit is 8, and 82 = 64.
(52698)2 will end in 4.

(iv) 362: Ending digit is 0.
(99880)2 will end in 0.

(v) 22 = 4
Ending digit of (21222)2 is 4.

(v) 62 = 36
Ending digit of (9106)2 is 6.

Question 7.
The square of which of the following numbers would be an odd number/an even number? Why?
(i) 727
(ii) 158
(iii) 269
(iv) 1980
Solution:
(i) 727
Since 727 is an odd number.
Its square is also an odd number.

(ii) 158
Since 158 is an even number.
Its square is also an even number.

(iii) 269
Since 269 is an even number.
Its square is also an odd number.

(iv) 1980
Since 1980 is an even number.
Its square is also an even number.

Question 8.
What will be the number of zeros in the square of the following numbers?
(i) 60
(ii) 400
Solution:
(i) In 60, a number of zero is 1. Its square will have 2 zeros.
(ii) There are 2 zeros in 400. Its square will have 4 zeros.

Try These (Page No. 94)

Question 9.
How many natural numbers lie between 92 and 102 Between 112 and 122?
Solution:
(a) Between 92 and 102
Here, n = 9 and n + 1 = 10
Natural numbers between 92 and 102 are (2 × n) or 2 × 9, i.e. 18.

(b) Between 112 and 122
Here, n = 11 and n + 1 = 12
Natural numbers between 112 and 122 are (2 × n) or 2 × 11, i.e. 22.

Question 10.
How many non-square numbers lie between the following pairs of numbers:
(i) 1002 and 1012
(ii) 902 and 912
(iii) 10002 and 10012
Solution:
(i) Between 1002 and 1012
Here, n = 100
n × 2 = 100 × 2 = 200
200 non-square numbers lie between 1002 and 1012

(ii) Between 902 and 912
Here, n = 90
2 × n = 2 × 90 = 180
180 non-square numbers lie between 902 and 912.

(iii) Between 10002 and 10012
Here, n = 1000
2 × n = 2 × 1000 = 2000
2000 non-square numbers lie between 10002 and 10012.

Question 11.
Find whether each of the following numbers is a perfect square or not?
(i) 121
(ii) 55
(iii) 81
(iv) 49
(v) 69
Solution:
If a natural number cannit be expressed as a sum of successive off natural numbers starting from 1, then it is not a perfect square.
(i) 121
Since, 121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
i.e., 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.
Thus, 121 is a square.

(ii) 55
Since, 55 – 1 = 54
54 – 3 = 51
51 – 5 = 46
46 – 7 = 39
39 – 9 = 30
30 – 11 = 19
19 – 13 = 6
6 – 15 = -9
Since 55 cannot be expressed as the sum of successive odd numbers starting from 1.
55 is not a perfect square.

(iii) 81
Since, 81 – 1 = 80
80 – 3 = 77
77 – 5 = 72
72 – 7 = 65
65 – 9 = 56
56 – 11 = 45
45 – 13 = 32
32 – 15 = 17
17 – 17 = 0
81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
Thus, 81 is a perfect square.

(iv) 49
Since, 49 – 1 = 48
48 – 3 = 45
45 – 5 = 40
40 – 7 = 33
33 – 9 = 24
24 – 11 = 13
13 – 13 = 0
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
Thus, 49 is a perfect square.

(iv) 69
Since, 69 – 1 = 68
68 – 3 = 65
65 – 5 = 60
60 – 7 = 53
53 – 9 = 44
44 – 11 = 33
33 – 13 = 20
20 – 15 = 5
5 – 17 = -12
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
69 cannot be expressed as the sum of consecutive odd numbers starting from 1.
Thus, 69 is not a perfect square.

Try These (Page No. 95)

Question 12.
Express the following as the sum of twp consecutive integers.
(i) 212
(ii) 132
(iii) 112
(iv) 192
Solution:
(i) n = 21
(n2 – 1)/2 = (441 – 1)/2 = 440/2 = 220
(n2 + 1)/2 = (441 + 1)/2 = 442/2 = 221
n2 = (n2 – 1)/2 + (n2 + 1)/2
212 = 220 + 221 = 441

(ii) n = 13
(n2 + 1)/2 = (132 + 1)/2 = (169 + 1)/2 = 85
(n2 – 1)/2 = (132 – 1)/2 = (169 – 1)/2 = 84
132 = 85 + 84 = 169

(iii) n = 11
(n2 + 1)/2 = (112 + 1)/2 = 61
(n2 – 1)/2 = (112 – 1)/2 = 60
112 = 60 + 61 = 121

(iv) n = 19
(n2 – 1)/2 = (192 – 1)/2 = 180
(n2 + 1)/2 = (192 + 1)/2 = 181
192 = 180 + 181 = 361

Question 13.
Do you think the reverse is also true, i.e., is the sum of any two consecutive positive integers is a perfect square of a number? Give example to support your answer.
Solution:
No, it is not always true.
For example:
(i) 5 + 6 = 11, 11 is not a perfect square.
(ii) 21 + 22 = 43, 43 is not a perfect square.
Examples:
92 – 82 = 81 – 64 = 17 = 9 + 8
102 – 92 = 100 – 81 = 19 = 10 + 9
152 – 142 = 225 – 196 = 29 = 15 + 14
1012 – 1002 = 10201 – 10000 = 201 = 101 + 100
For Example,
10 × 12 = (11 – 1) × (11 + 1) = 112 – 1
11 × 12 = (12 – 1) × (12 + 1) = 112 – 1
25 × 27 = (26 – 1) × (26 + 1) = 262 – 1

Question 14.
Write the square, making use of the above pattern.
(i) 1111112
(ii) 11111112
Solution:
using the above pattern we can write
(i) (111111)2 = 12345654321
(ii) (1111111)2 = 1234567654321

Question 15.
Can you find the square of the following numbers using the above pattern?
(i) 66666672
(ii) 666666672
Solution:
using the above pattern we can write
(i) (6666667)2 = 44444448888889
(ii) (66666667)2 = 4444444488888889

Try These (Page No. 97)

Question 16.
Find the square of the following numbers containing 5 in unit place.
(i) 15
(ii) 95
(iii) 105
(iv) 205
Solution:
(i) (15)2 = 1 × (1 + 1) × 100 + 25
= 1 × 2 × 100 + 25
= 200 + 25
= 225

(ii) (95)2 = 9 × (1 + 1) × 100 + 25
= 9 × 10 × 100 + 25
= 9000 + 25
= 9025

(iii) (105)2 = 10 × (10 + 1) × 100 + 25
= 10 × 11 × 100 + 25
= 1100 + 25
= 11025

(iv) (205)2 = 20 × (20 + 1) × 100 + 25
= 20 × 21 × 100 + 25
= 4200 + 25
= 42025

Try These (Page No. 99)

Question 17.
(i) 112 = 121. What are the square roots of 121?
(ii) 142 = 196. What are the square roots of 196?
Solution:
(i) The square root of 121 is 11.
(ii) The square root of 196 is 14.

Try These (Page No. 100)

Question 18.
By repeated subtraction of numbers starting from 1, find whether the following numbers are perfect squares the number is a perfect square, then square root.
(i) 121
(ii) 55
(iii) 36
(iv) 49
(v) 90
Solution:
(i) Subtracting the successive odd number from 121,
We have 121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
√121 = 11 [we had to subtract the first 11 odd numbers.]

(ii) 55 – 1 = 54
54 – 3 = 51
51 – 5 = 46
46 – 7 = 39
39 – 9 = 30
30 – 11 = 19
19 – 13 = 6
6 – 15 = -9
and we do not reach to 0.
55 is not a perfect square.

(iii) 36 – 1 = 35
35 – 3 = 32
32 – 5 = 27
27 – 7 = 20
20 – 9 = 11
11 – 11 = 0
and we do have obtained 0 after subtracting 6 successive odd numbers.
36 is a perfect square.
Thus, √36 = 6.

(iv) We have
49 – 1 = 48
48 – 3 = 45
45 – 5 = 40
40 – 7 = 33
33 – 9 = 24
24 – 11 = 13
13 – 13 = 0
We have obtained 0 after successive subtraction of 7 odd numbers.
49 is a perfect square Thus, √49 = 7.

(v) We have
90 – 1 = 89
89 – 3 = 86
86 – 5 = 81
81 – 7 = 74
74 – 9 = 65
65 – 11 = 54
54 – 13 = 41
41 – 15 = 26
26 – 17 = 9
9 – 19 = -10
Since we can not reach 0 after subtracting successive odd numbers.
90 is not a perfect square.

Try These (Page No. 105)

Question 19.
Without calculating square roots, find the number of digits in the square root of the following numbers.
(i) 25600
(ii) 100000000
(iii) 36864
Solution:
(i) 25600
n = 5 [an odd number]
Its square root will have $$\frac{n+1}{2}$$
i.e. $$\frac{5+1}{2}=\frac{6}{2}$$ = 3 digits

(ii) 100000000
n = 9 (odd number)
Number of digits of its square root
$$\frac{n+1}{2}=\frac{9+1}{2}=\frac{10}{2}=5$$ digits

(iii) 36864
n = 5 (odd number)
Number of digits in its square root
$$\frac{n+1}{2}=\frac{5+1}{2}=\frac{6}{2}=3$$ digits

Try These (Page No. 107)

Question 20.
Estimate the value of the following the nearest whole number.
(i) √180
(ii) √100
(iii) √350
(iv) √500
Solution:
(i) √80
102 = 100, 92 = 81, 82 = 64
and 80 is between 64 and 81.
i.e., 64 < 80 < 81
or 82 < 80 < 92
or 8 < √80 < 9
Thus, √80 lies between 8 and 9.

(ii) √1000
We know that
302 = 900, 312 = 961, 322 = 1024
1000 lies between 961 and 1024.
i.e., 916 < 1000 < 1024
or 312 < 1000 < 322
or 312 < √1000 < 32.
Thus, √1000 lies between 31 and 32.

(iii) √350
We have 182 = 324, 192 = 361
Since 350 lies between 324 and 316.
i.e., 324 < 350 < 361
or 182 < 350 < 192
or 18 < √350 < 19.
Thus, √350 lies between 18 and 19.

(iv) √500
222 = 484 and 232 = 529
Since, 500 lies between 484 and 529.
or 484 < 500 < 529
or 222 < 500 < 232
or 22 < √350 < 23.
Thus, √500 lies between 22 and 23.

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