# 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 **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)

= 5 x 7 x 3 = 105

5 | 11025 |

5 | 2205 |

7 | 441 |

7 | 63 |

3 | 9 |

3 |

= 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 |

= 3 x 3 x 7 x 7 = 441

3 | 194481 |

3 | 64827 |

3 | 21609 |

3 | 7203 |

7 | 2401 |

7 | 343 |

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 :-**

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.

**Answer 4 :-**

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)

(iii) 0.0169

**Answer 5 :-**

(i)

= 0.42

(ii)

(iii) 0.0169

**Question 6 :-**

**Evaluate :**

(i)

(ii)

(iii)

(iv)

**Answer 6:-**

(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)

**Answer 7:-**

(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

**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 = x^{2} = 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 |

= 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)^{2 }= 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) **

**(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

**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) ^{2 }**

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.

**Answer 7 :-**

**√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.

**Answer 8 :-**

**√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) ^{1764}⁄2809

(ii) ^{507}⁄4107

(iii)

(iv) 0.01 + √0.0064

**Answer 9 :-**

(i) ^{1764}⁄2809

4 | 1764 16 |

82 | 164 164 |

x |

53 | |

5 | 2809 25 |

103 | 309 309 |

x |

Hence, square root of √**^{1764}⁄2809 **=

^{42}⁄53

^{507}⁄4107= ^{507+3}⁄4107+3

= ^{169}⁄1369

13 | |

1 | 169 1 |

23 | 69 69 |

x |

37 | |

3 | 1369 9 |

67 | 469 469 |

x |

Hence, square root of √**^{169}⁄1369 **=

^{13}⁄37** **

= √219024

468 | |

4 | 219024 16 |

86 | 590 516 |

928 | 7424 7424 |

x |

Hence, = 468

OR

2 | 108 |

2 | 54 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

2 | 2028 |

2 | 1014 |

3 | 507 |

13 | 169 |

13 | 13 |

1 |

= 2 x 2 x 3 x 13 = 468

(iv) **0.01 + √0.0064**

0.08 | |

8 | 0.0064 64 |

x |

**= 0.01 + 0.08 = 0.09**

**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 34^{2}

34 | |

3 | 1205 9 |

64 | 305 256 |

49 |

∴ On adding the required number to 1205, we shall be getting 35^{2} 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 |

**Question 14 :-**

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 74^{2}

74 | |

9 | 5483 49 |

144 | 583 576 |

7 |

∴ On adding the required number to 5483, we shall be getting 75^{2} 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) 37^{2} – 36^{2}

(ii) 85^{2} – 84^{2}

(iii) 101^{2} – 100^{2}

**Answer 12 :-**

(i) 37^{2} – 36^{2}

Using property, for any natural number n,

(n + 1)^{2} – n^{2} = (n + 1) + n

⇒ (36 + 1)^{2} – 36^{2} = (36 + 1) + 36

⇒ 37^{2} – 36^{2 }= 37 + 36

⇒ 37^{2 }– 36^{2} = 73

^{2}– 84

^{2}

(n + 1)

^{2}– n

^{2}= (n + 1) + n

⇒ (84 + 1)

^{2}– 84

^{2}= (84 + 1) + 84

⇒ 85

^{2}– 84

^{2 }= 85 + 84

⇒ 85

^{2}– 84

^{2}= 169

^{2}– 100

^{2}

(n + 1)

^{2}– n

^{2}= (n + 1) + n

⇒ (100 + 1)

^{2}– 100

^{2}= (100 + 1) + 100

⇒ 101

^{2}– 100

^{2}= 101 + 100

⇒ 101

^{2}– 100

^{2}= 201

**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

= Sum of first 27 odd natural number = 272 = 729

**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

3^{2} + 4^{2} = 5^{2}

⇒ 9 + 16 = 25

⇒ 25 = 25

In 6, 8 and 10

6^{2} + 8^{2} = 10^{2}

⇒ 36 + 64 = 100

⇒ 100 = 100

In 5, 12, and 13

5^{2 }+ 12^{2} = 13^{2}

⇒ 25 + 144 = 169

⇒ 169 = 169

— End of **Squares and Square Roots** Solutions :–

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