Joy of Mathematics Class 8 Solutions Chapter 4

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Joy of Mathematics Class 8 Solutions Chapter 4 Squares and Square Roots

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 4 Squares and Square Roots. Here students can easily find step by step solutions of all the problems for Squares and Square Roots. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 4.1, 4.2 and 4.3

Squares and Square Roots Exercise 4.1 Solution

Question no – (1)

Solution :

(a) (51)²

= 51 × 51

= 2601

(b) (- 42)²

= (-1)² 42 × 42

= 1 × 42 × 42

= 1764

(c) (- 4/17)²

= (-1)² 4/17 × 4/17

= 16/289

(d) (0.65)²

= 0.65 × 0.65

= 0.4225

(e) (3 3/5)²

= 18/5 × 18/5

= 324/25

(f) (0.006)²

= 0.006 × 0.006

= 000036

Question no – (2)

Solution :

(a) Given,

=> 1250 = 5 × 5 × 5 × 5 × 2

∴ in this factorization; factors 2 are left unpaired.

∴ 1250 is not perfect square number.

(b) Given, 9801

∴ 9801 = 3 × 3 × 3 × 3 × 3 × 11 × 11

∴ In this factorization, all the factors are grouped into pairs of identical factors.

(c) Given, 2025

∴ 2025 = 5 × 5 × 3 × 3 × 3 × 3

∴ In this factorization, all the factors are grouped into pairs of identical factors.

∴ 2025 is a perfect square.

(d) Given, 62500

∴ 62500 = 5 × 5 × 5 × 5 × 5 × 5 × 2 × 2

∴ In this factorization, all the factors are grouped into pairs of identical factors.

∴ 62500 is a perfect square.

(e) Given, 16641

∴ 16641 = 3 × 3 × 43 × 43

∴ In this factorization, all the factors are grouped into pairs of identical factors.

∴ 16641 is a perfect square.

(f) Given, 152100

∴ 152100 = 3 × 5 × 5 × 2 × 2 × 3 × 13 × 13

∴ In this factorization, all the factors are grouped into pairs of identical factors.

∴ 152100 is a perfect square.

Question no – (3) 

Solution :

(a) Given, 40368

∴ 40368 = 2 × 2 × 2 × 2 × 3 × 29 × 29

Since, factor 3 is left unpaired, so we multiply 40368 by 3 then no factor will be left unpaired and the number obtained is = 40368 × 3 = 121,104 which is a perfect square number.

(b) Given, 19044

∴ 15044 = 2 × 2 × 3 × 3 × 23 × 23

In this factorization, all the factors are grouped into pairs.

So, we multiply 19044 by 1

∴ 19044 × 1 = 19044, which is a perfect square.

(c) Given, 268912

∴ 268912 = 2 × 2 × 2 × 2 × 7 × 7 × 7 × 7 × 7

∴ Since, factor 7 is left unpaired, so we multiply 268912 by 7 then no factor will be left unpaired and the number obtained is = 268912 × 7 = 1882384 which is a perfect square number.

Question no – (4)

Solution :

(a) Given, 3675

∴ 3675 = 5 × 5 × 3 × 7 × 7

∴ Since factor 3 is left unpaired, so we divide 3675 by 3 then no factor will be left unpaired and the number obtained is = 3675 ÷ 3 = 1225 which is a perfect square number.

(b) Given, 20184

∴ 20184 = 2 × 2 × 2 × 3 × 29 × 29

Since, factor 3 is left unpaired so we divided 20184 by 3, then no factor will be left unpaired and the number obtained is 20184 ÷ 3 = 6728, which is a perfect square number.

(c) Given, 253125

∴ 253125 = 5 × 5 × 5 × 5 × 5 × 3 × 3 × 3 × 3

Since, factor 5 is left unpaired so we divided 253125 by 5, then no factor will be left unpaired and the number obtained is 253125 ÷ 5 = 50625 which is a perfect square number.

(d) Give, 188771

∴ 188771 = 11 × 131 × 131

Since, factor 11 is left unpaired so, we divided 188771 by 11, then no factor will be left unpaired and the number obtained is 188771 ÷ 11 = 17161, which is a perfect square number.

Question no – (5) 

Solution :

(a) The number of zeros at the end of a perfect square is never not perfect square.

(b) The square of an even number is always perfect square.

(c) The square of an odd number is always perfect square.

(d) If the ones digit of a number is 4, its square will end with perfect square.

(e) 425² – 424² = Perfect square.

(f) The square of a number of (other than 1) is either a multiple of 3 or exceeds a multiple of 3 by perfect square.

Square and square roots Exercise 4.2 Solution : 

Question no – (1) 

Solution :

(a) 784

∴ 784 = 2 × 2 × 2 × 2 × 7 × 7

∴ √784 = 2 × 2 × 7

= 28

(b) 2304

∴ 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

∴ √2304 = 2 × 2 × 2 × 2 × 3

= 48

(c) 1296

∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

∴ √1296 = 2 × 2 × 3 × 3

= 36

(d) 3969

∴ 3969 = 3 × 3 × 3 × 3 × 7 × 7

∴ √3969 = 3 × 3 × 7

= 21 × 3

= 63

(e) 196/81

= 13 × 13/3 × 3 × 3 × 3

=> √196/√81

= 13/9

(f) 121/10000

= 11 × 11/2 × 5 × 5 × 5 × 5 × 2 × 2 × 2

∴ √121/√10000 = 11/2 × 2 × 5 × 5

= 11/4 × 25

= 11/100

(g) 225/289

= 5 × 5 × 3 × 3/17 × 17

∴ √225/√289 = 5 × 3/17

= 15/17

(h) 625/1296

∴ 5 × 5 × 5 × 5/3 × 3 × 3 × 3 × 2 × 2 × 2 × 2

∴ √625/√1296 = 5 × 5/3 × 3 × 2 × 2

= 25/36

(i) 38 11/0.5

= 961/25

= 31 × 31/5 × 5

∴ √961/√25

= 31/5

(j) 23 26/121

= 2809/121

= 53 × 53/11 × 11

∴ √2809/√121

= 53/11

(k) 10 74/169 = 1764/169

∴ 1764/169 = 2 × 2 × 3 × 3 × 7 × 7/13 × 13

∴ √1764/√169 = 2 × 3 × 7/13

= 42/13

(l) 7 393/529

= 4096/529

∴ 4096/529 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2/23 × 23

∴ √4096/√529 = 2 × 2 × 2 × 2 × 2 × 2/23

= 64/23

(m) 0.09

= 0.09/100

∴ 9/100 = 3 × 3/2 × 2 × 5 × 5

∴ √9/√100 = 3/10

= 0.3

(n) 0.0004/10000

∴ 4/10000 = 2 × 2/5 × 5 × 5 × 5 × 2 × 2 × 2 × 2

∴ √4/√10000 = 2/5 × 5 × 2 × 2

= 2/100

= 0.02

(o) 0.0121

= 0.0121/10000

= 11 × 11/ 5 × 5 × 5 × 5 × 2 × 2 × 2 × 2

∴ √121/√10000 = 11/5 × 5 × 2 × 2

= 11/100

= 0.11

(p) 0.0324

= 0.0324/10000

= 2 × 2 × 3 × 3 × 3 × 3/5 × 5 × 5 × 5 × 2 × 2 × 2 × 2

∴ √324/1000 = 2 × 3 × 3/5 × 5 × 2 × 2

= 18/100

= 0.18

Question no – (3)

Solution :

(a) √196 + √0.0064 – √100

= √14 × 14 + √8 × 8/100 × 100 – √10 × 10

= 14 + 8/100 – 10

= 14 + 0.08 – 10

= 4 + 0.08

= 4.08

(b) √20² – 16²

= √20 × 20 – 16 × 16

= √400 – 256

= √144

= √12 × 12

= 12

(c) √0.0361/576

= √361/576 × 10000

= √19 × 19/24 × 24 × 100 × 100

= 19/24 × 100

= 19/2400

(d) (- √144/576) × (-16/√64)

= – (√12 × 12/24 × 24) × (- 16/√8 × 8)

= – 12/24 × (- 16/8)

= – 1/2 × (- 2)

= 1/2 × 2

= 1

(e) [(- 15)² × √81] + 5²

= [225 × √9 × 9] + 5²

= (225 × 9) + 5²

= 2025 + 25

= 2050

(f) √(37 + 2 1/16) × 0.0144/0.25

= √(37 + 33/16) × 144 × 100/25 × 10000

= √ (592 + 33/16) × 144 × 100/25 × 10000

= √625/16 × 144 × 100/25 × 10000

= √25 × 9 × 100/10000

= 5 × 3 × 10/100

= 15/10

= 1.5

Question no – (4)

Solution :

Let, the length of one side of the square field = x

∴ Area of square = x² (length)²

= x²

A/Q, x² = 334.89

=> x = √334.89

=> = √33489/100

=> x = √183 × 183/10 × 10

= 183/10

= 18.3

∴ The length of one side of the field 18.3m.

Question no – (5)

Solution :

Let, the length of one side of a square = x

∴ Area of a square = x²

∴ x² = 306 1/4

=> x² = 1225/4

=> x = √1225/4

=> x = √35 × 35/2 × 2

=> x = 35/2

= 17.5

∴ Length of the square = 17.5m.

∴ Perimeter of the square

= 4 × x m

= 4 × 35/2 m

= 70m

∴ Perimeter of the square = 70m.

Question no – (6)

Solution :

∴ From the division we conclude that if 212 is subtracted from the given number, will become a perfect square.

Question no – (7)

Solution :

We observe that the given number is greater than (212)² but less than (213)²

∴ The number to be added is

= (213)² – 45156

= 45369 – 45156

= 213

Question no – (8)

Solution :

The least number of three digits 100, which is a perfect square.

Question no – (9)

Solution :

The least number of four digit 1000 finding the √1000, we observe that,

(31)² < 1000 < (32)²

∴ The least four digit number which is a perfect square is (32)² = 1024

Question no – (10)

Solution :

The greatest number of three digit = 999

Finding √999, we observe that (31)² is less than 999 by 38

∴ The required number is = 999 – 38

= 961

Question no – (11)

Solution :

The greatest number of five digit = 99999

Finding √99999, we observe that, (316)² is less than 99999 by 143

∴ The required number is

= 99999 – 143

= 99856

Question no – (12)

Solution :

Total number of students = 8175

∴ 75 students left out then = 8175 – 75

= 8100

∴ Number of students in each row

= √8100

= √90 × 90

= 90

Question no – (13)

Solution :

Here,

L.C.M. of 12, 15 and 24, 120

∴ 120 = 2 × 2 × 2 × 3 × 5

= 2³ × 3 × 5

= 2² × 2 × 3 × 5

∴ In order to get the required number, we should multiply 120 by 2 × 3 × 5 = 30

∴ Hence the required number is

= 120 × 30

= 3600

Question no – 14

Solution :

Let, the number of the temples in the city be x

∴ Number of fruits offered in each temple is = x

So, total number of fruits offered = x × x = x²

∴ x² = 4913 × 17

= 83521

∴ The number of temples x = √83521

= √289 × 289

= 289

Question no – (15)

Solution :

A garden wanted to planet = 580

∴ He find 4 plants are left = 580 – 4

= 576

∴ Number of plants he planted in each row

= √576

= √24 × 24

= 24

Previous Chapter Solution : 

👉 Chapter 3