Squares and Square Roots ICSE Class-8th Concise Selina

Squares and Square Roots ICSE Class-8th Concise Selina Chapter-3. We provide step by step Solutions of Exercise / lesson-3 Squares and Square Roots ICSE Class-8th Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-3 A , Exe-3 B and Exe-3 C,  to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8.

Squares and Square Roots ICSE Class-8th Concise Selina Chapter-3


–: Select Topics :–

 

Exercise-3 A ,

Exercise-3 B

Exercise-3 C

 


Exercise – 3 A Squares and Square Roots ICSE Class-8th Concise Selina

Question 1.

Find the square of :
(i) 59
(ii) 63
(iii) 15

Answer 1:-

(i) Square of 59=59×59=3481

(ii) Square of 6.3=6.3×6.3=39.69

(iii) Square of 15=15×15=225

 

Question 2 :-

By splitting into prime factors, find the square root of :
(i) 11025
(ii) 396900
(iii) 194481

Answer 2 :-

(i) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 2.1

= 5 x 7 x 3 = 105

5 11025
5 2205
7 441
7 63
3 9
3
(ii)  Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 2.2

= 2 x 3 x 3 x 5 x 7 = 630

2 396900
2 198450
3 99225
3 33075
3 11025
3 3675
5 1225
5 245
7 49
7

Question 3 :-

(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be multiplied so that the product is a perfect square?

Answer 3 :-

 

(i) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 3.1

On grouping the prime factors of 2592 as shown; on factor i.e. 2 is left which cannot be paired with equal factor.

2 2592
2 1296
2 648
2 324
2 162
3 81

(ii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 3.2

On grouping the prime factors of 12748 as shown; one factor i.e. 3187 is left which cannot be paired with equal factor.

2 12748
2 6374
3187

The given number should be multiplied by 3187.

Question 4 :-

Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.

Answer 4 :-

10368 = Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 4

On grouping the prime factors of 10368 as shown; one factors i.e. 2 is left which cannot be paired with equal factor.

2 10368
2 5184
2 2592
2 1296
2 648
2 324
2 162
3 81
3 27
3 9
3

∴ The given number should be divided by 2.

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 4 .1

= 2 x 2 x 2 x 3 x 3 = 72

Question 5 :-

Find the square root of :
(i) 0.1764
(ii) 96\frac { 1 }{ 25 }
(iii) 0.0169

Answer 5 :-

(i) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 5.1

= 0.42

(ii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 5.2

(iii) 0.0169
Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 5.3

Question 6 :-

Evaluate :
(i) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.1

(ii)Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.2

(iii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.3

(iv) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.4
Answer 6:-

(i) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.1 a

= 0.8
(ii)  Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.2

17 28900
17 1700
10 100
10

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.2 a

(iii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.3

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.3 a

(iv)  Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.4

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.4 a

7 637
7 91
13
9 117
13

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 6.4 b

Question 7 :-
Evaluate:
(i)  Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.1
(ii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.2
(iii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.3

(iv) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.4

(v) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.5

Answer 7:-

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.1

(i) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.1 a

(ii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.2

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.2 a

(iii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.3

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.3 a

(iv) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.4

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.4 a

(v) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.5

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 7.5 a

Question 8 :-

A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all ₹ 1,296
Answer 8:-

Let the number of days he had spent = x
Number of rupees spent in each day = x
Total money spent = x x x = x2 = 1,296 (given)

x= √1296

4 1296
4 324
9 81
9

⇒ x = √4×4×9×9

x = 4 × 9

⇒ x = 36

Hence required number of days = 36

 

Question 9 :-

Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.

Answer 9 :-

Total number of students = 745
Students left after standing in arrangement = 16
No. of students who were to be arranged = 745 – 16 = 729
The number of rows = no. of students in each row
No. of rows = √729

3 729
3 243
3 81
3 27
3 9
3 3
1

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 9

= 3 x 3 x 3 = 27

Question 10 :-

13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.

Answer :-

(13)2 = 169 and (31)2 = 961
Similarly, two such number can be 12 and 21
∴ (12)2 = 144 and (21)2 =441
and 102, 201
(102)2 = 102 x 102 =  10404
and (201)= 201 x 201 = 40401

 

Question 11 :-

Find the smallest perfect square divisible by 3, 4, 5 and 6.

Answer 11 :-

L.C.M. of 3, 4, 5, 6 = 2 x 2 x 3 x 5 = 60

2 3, 4, 5, 6
3 3, 2, 5, 3
1, 2, 5, 1

in which 3 and 5 are not in pairs L.C.M. = 2 x 3 x 2 x 5 = 60

We should multiple it by 3 x 5 i.e. by 15

Required perfect square = 60 x 15 = 900

Question 12 :-

If √784 = 28, find the value of:
(i) √7.84 + √78400
(ii) √0.0784 + √0.000784

Answer 12 :-

(i) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 12.1

(ii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions ans 12.2

 


 Exercise- 3 B Squares and Square Roots ICSE Class-8th Concise Selina

Question 1 :-

Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Answer 1 :-

(i) 4761

69
6 4761
36
129 1161
1161
x

Required square root = 69

(ii) 7744

88
8 7744
64
168 1344
1344
x

Required square root = 88

(iii) 15129

123
1 15129
1
22 51
44
243 729
729
x

Required Square root = 123

(iv) 0.2916

0.54
0.5 0.2916
0.25
0.104 416
416
x

Required square root = 0.54

(v) 0.001225

0.035
0.03 0.001225
9
0.065 325
325
x

Required square root = 0.035

(vi) 0.023104

0.152
0.1 0.023104
0.01
.25 131
125
302 604
604
x

Required square root = 0.152

(vii) 27.3529

5.23
5 27.3529
25
102 2.35
2.04
1043 3129
3129
x

Required square root = 5.23

Question 2 :-

Find the square root of:

(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Answer 2 :-

(i) 4.2025

2.05
2 4.2025
4
405 0.2025
0.2025
x

Required square root = 2.05

(ii) 531.7636

23.06
2 531.7636
4
43 131
129
4606 2.7636
2.7636
x

Required square root = 23.06

(iii) 0.007225

0.085
0.8 0.007225
64
0.165 825
825
x

Required square root = 0.085

Question 3 :-

Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Answer 3 :-

(i) 245

15.65
1 245
1
25 145
125
306 2000
1836
3125 16400
15625
775

Required square root = 15.65 up to two places of decimal.

(ii) 496

22.271
2 496
4
42 96
84
442 1200
884
4447 31600
31129
44541 47100
44541

Required square root = 22.2708 = 22.271 up to three places of decimals.

(iii) 82.6

9.088
9 82.60
81
1808 16000
14464
18168 153600
145324

Required square root = 9.088 = 9.09 up to two places of decimal.

(iv) 0.065

0.2549
0.2 0.0650
0.04
0.45 250
225
0.504 2500
2016
0.5089 48400
45801

Required square root = 0.255 up to three places of decimal.

(v) 5.2005

2.28
2 5.2005
4
42 120
84
448 3605
3584
456 2100

Required square root = 2.28 up to two places of decimal.

(vi) 0.602

0.775
0.7 .06020
.49
0.147 1120
1029
1545 9100
7725
1375

Required square root = 0.78 up to two places of decimals.

Question 4 :-

Find the square root of each of the following correct to two decimal places:
(i) 3  4/5
(ii) 6  7/8
Answer 4 :-

(i) 3  4/5

= 3.80

1.949
1 3.80
1
29 280
261
384 1900
1536
3889 36400
35001
1399

Required. square root = 1.949 = 1.95 up to two places of decimal.

(ii) 6  7/8

 = 6.875

2.62
2 6.8750
4
46 287
276
522 1150
1044
106

Required. square root = 2.62

Question 5 :-

For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Answer 5 :-
(i) 796

Taking square root of 796, we find that 12 has been left

28
2 7 96
4
48 396
384
12

∴ Least number to be subtracted = 12

(ii) 1886

Taking square root of 1886, we find that 37 has been left

43
4 1886 16
83 286
249
37

∴ Least number to be subtracted = 37

(iii) 23497

Taking square root of 23497, we find that 88 has been left

153
1 23497
1
25 134
125
303 997
909
88

∴ Least number to be subtracted = 88

Question 6 :-

For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Answer 6 :-

(i) 511

Taking square root of 511, we find that 27 has been left We see that 511 is greater than (22)

22
2 5 11
4
42 111
84
27

On adding the required number to 511, we get (23)2 i.e., 529
So, the required number = 529 – 511 = 18

(ii) 7172

Taking square root of 7172, we find that 116 has been left
We see that 7172 is greater than (84)2

84
8 71 72
64
164 772
656
116

∴ On adding the required number to 7172, we get (85)2 i.e., 7225
Required number = 7225 – 7172 = 53

(iii) 55078

234
2 55078
4
43 150
129
464 2178
1856
322

Taking square root of 55078, we find that 322 has been left
We see that 55078 is greater than (234)2
On adding the required number to 55078, we get (235)2 i.e., 55225
Required number = 55225 – 55078 = 147

Question 7 :-

Find the square root of 7 correct to two decimal places; then use it to find the value of Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions exercise b ans 7 correct to three significant digits.
Answer 7 :-

√7 = 2.645 = 2.65

2.645
2 7.000000
4
46 300
276
524 2400
2096
5285 30400
26425
3975

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions exercise b ans 7 a

= 2.22

Question 8 :-

Find the value of √5 correct to 2 decimal places; then use it to find the square root of  Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions exercise b ans 8 correct to 2 significant digits.
Answer 8 :-

√5 = 2.236 = 2.24

2.236
2 5.000000
4
42 100
84
443 1600
1329
4466 27100
26796
304

Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions exercise b ans 8 a

 

Question 9 :-

Find the square root of:
(i) 17642809
(ii) 5074107
(iii) Squre and Squres RootsICSE Class-8th Concise Selina Power Solutions exercise b ans 9
(iv) 0.01 + √0.0064

Answer 9 :-

(i) 17642809

4 1764
16
82 164
164
x

 

53
5 2809
25
103 309
309
x

Hence, square root of √17642809 = 4253

Question 10 :-

Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.

Answer 10 :-

(i)

Square root of 7.832

2.7985
4 7.832000
4
49 383
329
549 5420
4941
5588 47900
44704
5596 319600
279825
39775

√7.832= 2.80 up to two decimal places

(ii)

Square root of 7.832

2.7985
4 7.832000
4
49 383
329
549 5420
4941
5588 47900
44704
5596 319600
279825
39775

= 2.8 up to two significant digits

 

Question 11 :-

Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.

Answer 11 :-

Clearly, if 49 is subtracted from 1205, the number will be a perfect square.

34
3 1205
9
64 305
256
49

 

∴ 1205 – 49 = 1156 and √1156 = 34

 

Question 12 :-

Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Answer 12 :-

Clearly, 1205 is greater than 342

34
3 1205
9
64 305
256
49

∴ On adding the required number to 1205, we shall be getting 352 i.e., 1225
∴ The required number = 1225 – 1225 = 20

 

Question 13 :-

Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Answer 13 :-

Clearly; if 12 is subtracted from 2037, the remainder will be a perfect square.

∴ 2037 – 12 = 2025 and √2025 = 45

45
4 2037
49
85 437
425
12

Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Answer 14 :-

Clearly, 5483 is greater than 742

74
9 5483
49
144 583
576
7

∴ On adding the required number to 5483, we shall be getting 752 i.e. 5625.
Hence, the required number = 5625 – 5483
= 142

 


Exercise – 3 C Concise Selina solutions of Squares and Square Roots for ICSE Class-8th 

Question 1 :-

Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699

Answer 1 :-

We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9
So, the following numbers can be squares: 3051, 5684, and 50699 i.e., (i), (iii), and (v)

 

Question 2 :-

Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64

Answer 2 :-

The square of the following numbers will have 1 at their units place as (1)2 = 1, (9)2 = 81
81 and 139 i.e., (i) and (iii)

 

Question 3 :-

Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Answer 3 :-

The square of the following numbers will not have 1 at their units place: as only (1)2 = 1, (9)2 = 81 have 1 at then units place
322, 572, 2652 i.e., (i), (ii) and (v)

 

Question 4 :-

Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98

Answer 4 :-

The squares of the following numbers, Will not have 6 at their units place as only (4)2 = 16, (6)2 = 36 has but its units place 35, 23 and 98 i.e., (i), (ii), and (v)

 

Question 5 :-

Which of the following numbers will have 6 at their unit’s place :
(i) 262
(ii) 492
(iii) 342
(iv) 432
(v) 2442

Answer 5 :-

The following numbers have 6 at their units place as (4)2 = 16, (6)2 = 36 has 6 at their units place 262, 342, 2442 i.e., (i), (iii) and (v)

 

Question 6 :-

If a number ends with 3 zeroes, how many zeroes will its square have?
Answer 6 :-

We know that if a number ends with n zeros, then its square will have 2n zeroes at their ends
A number ends with 3 zeroes, then its square will have 3 x 2 = 6 zeroes

 

Question 7 :-

If the square of a number ends with 10 zeroes, how many zeroes will the number have?
Answer 7 :-

We know that if a number ends with n zeros Then its square will have 2n zeroes Conversely, if square of a number have 2n zeros at their ends then the number will have n zeroes
The square of a number ends 10 zeroes, then the number will have 10/2 = 5 zeroes

 

Question 8 :-

Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Answer 8 :-

No, it is not possible for the square of a number, to have 5 zeroes which is odd because the number of zeros of the square must be 2n zeroes i.e., even number of zeroes

 

Question 9 :-

Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Answer 9 :-

A number having 2,3,7 or 8 at the unit place is never a perfect square.

 

Question 10 :-

State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(ii) 76
(iv) 75
Answer 10 :-

(i) 23 – odd

(ii) 54 – even

(ii) 76 – odd

(iv) 75 – even

 

Question 11 :-

Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Answer 11 :-

No, number has an even number of zeroes.

 

Question 12 :-

Evaluate:
(i) 372 – 362
(ii) 852 – 842
(iii) 1012 – 1002

Answer 12 :-

(i) 372 – 362
Using property, for any natural number n,
(n + 1)2 – n2 = (n + 1) + n
⇒ (36 + 1)2 –  362 = (36 + 1) + 36
⇒ 372 –  36= 37 + 36
⇒ 37–  362 = 73

Question 13 :-

Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53
Answer 13 :-

(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23

= Sum of first 12 odd natural numbers = 122 = 144

(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41

= Sum of first 21 odd natural numbers = 212 = 441

 

Question 14 :-

Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Answer 14 :-

The three sets of Pythagorean triplets such that each set has numbers less than 30 are 3, 4 and 5; 6, 8 and 10; 5, 12 and 13
Proof:
In 3, 4 and 5
32 + 42 = 52
⇒ 9 + 16 = 25
⇒ 25 = 25
In 6, 8 and 10
62 + 82 = 102
⇒ 36 + 64 = 100
⇒ 100 = 100
In 5, 12, and 13
5+ 122 = 132
⇒ 25 + 144 = 169
⇒ 169 = 169

 

 

— End of Squares and Square Roots Solutions :–


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