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

(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

(i) = 5 x 7 x 3 = 105

 5 11025 5 2205 7 441 7 63 3 9 3
(ii) = 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?

(i) 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) 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.

10368 = 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. = 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

(i) = 0.42

(ii) (iii) 0.0169 #### Question 6 :-

Evaluate :
(i) (ii) (iii) (iv) (i) = 0.8
(ii) 17 28900 17 1700 10 100 10 (iii)  (iv)  7 637 7 91 13
 9 117 13 Question 7 :-
Evaluate:
(i) (ii) (iii) (iv) (v)  (i) (ii)  (iii)  (iv)  (v)  #### 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

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.

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 = 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.

(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.

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

(i) (ii) ### 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

(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

(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

(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

(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
(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

(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 correct to three significant digits.

√7 = 2.645 = 2.65

 2.645 2 7.000000 4 46 300 276 524 2400 2096 5285 30400 26425 3975 = 2.22

#### Question 8 :-

Find the value of √5 correct to 2 decimal places; then use it to find the square root of correct to 2 significant digits.

√5 = 2.236 = 2.24

 2.236 2 5.000000 4 42 100 84 443 1600 1329 4466 27100 26796 304 #### Question 9 :-

Find the square root of:
(i) 17642809
(ii) 5074107
(iii) (iv) 0.01 + √0.0064

(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.

(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.

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.

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.

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.

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

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

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

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

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

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?

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?

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.

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

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

(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.

No, number has an even number of zeroes.

#### Question 12 :-

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

(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

(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.

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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