# Frank Learning Maths Class 8 Solutions Chapter 3

## Frank Learning Maths Class 8 Solutions Chapter 3 Squares and Square Roots

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 3 Squares and Square Roots. Here students can easily find step by step solutions of all the problems for Squares and Square Roots. Here students will find solutions for Exercise 3.1, 3.2 and 3.3 and also solution for chapter Check-up problems. 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 in this post all Mathematic solutions are based on the CBSE latest curriculum.

Squares and Square Roots Exercise 3.1 Solution

Question no – (1)

Solution :

(a) 192

= 19 × 19

= 361

(b) 472

= 47 × 47

= 2209

(c) 622

= 62 × 62

= 3844

(d) 1012

= 10201

Question no – (2)

Solution :

Let, Pythagorean triplets are 2m, m2 – 1, m2 + 1

(a) 2 m = 6

M = 3

M– 1 = 9 – 1 = 8

M2 + 1 = 9 + 1 = 10

(b) 2 m = 14

M = 7

M2 = 1 = 48

M2 + 1 = 50

(c) 2m = 18

M = 9

= M2 + 1 = 82

(d) 2 m = 16

M = 8

M2 – 1 = 63

M2 + 1 = 65

Question no – (3)

Solution :

(a) (29)2 – (28)2

=(29 + 28) (29 – 28)

= 57

(b) (68)2 – (67)2

= (68 – 67) (68 +67)

= 135

(c) (121)2 – (120)2

= (121) (121 – 120) (121 + 120)2

= 241

(d) (278)2 – (277)2

= (278 – 277) (278 + 277)

= 555

Question no – (4)

Solution :

LCM of 8, 9, 10 = 360

360 = 2 × 2 × 2 × 3 × 3 × 3 × 5

We should multiple the given number with (2 × 5) for perfect square.

360 × 10 = 36000

Therefore, 3600 is the perfect square.

Question no – (5)

Solution :

∴ LCM = 180

= 2 × 2 × 3 × 3 × 5

We should multiple the number by 5

180 × 5 = 900

Thus, the least square number is 900

Question no – (8)

Solution :

(i) 92 – 82

= 81 – 64

= 17

(ii) 132 – 122

= 25

Therefore, the Natural numbers between then 16 and 24

Question no – (9)

Solution :

(a) 502 – 512

= (51 + 50) (51 – 50)

= 101
Non-squares number between 512, 502 = (101 – 1)

= 100

(b) 902 – 912

= (91 + 90) (91 – 90)

= 181

Non-square number (181 – 1) = 180

(c) 10002 – 10012

= (1001 + 1000) ( 1001 – 1000)

= 2001

Non-square numbers (2001 – 1) = 2000

Question no – (10)

Solution :

(a) 13 × 15

= 13 (13 + 2)

= 132 + 13 x 2

= 169 + 26

= 195

(b) 28 × 30

= (30 – 2) x 30

= 302 – 2 x 30

= 900 – 60

= 840

(c) 53 × 55

= (55 – 2) x 55

= (55)2 – 2 x 55

= 3025 – 120

= 2905

(d) 100 × 102

= 100 (100 + 2)

= 1002 + 2 x 100

= 10000 + 200

= 10200

Squares and Square Roots Exercise 3.2 Solution

Question no – (3)

Solution :

(a) 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

(b) 64 – 1 → 63, 63 – 3 → 60, 60 – 5 → 55, 55 – 7 → 48, 48 – 9 → 39, 39 – 11 → 29, 29 – 13 → 15, 15 – 15 = 0.

√64 = 8

(c) 49 – 1 = 48, 48 – 3 = 45, 45 – 5 = 40, 40 – 7 = 33, 33 – 9 = 24, 24 – 11 = 13, 13 – 13 = 0.

√49 = 7

(d) 144 – 1 = 143, 143 – 3 = 140, 140 – 5 = 135, 135 – 7 = 128, 128 – 9 = 119, 119 – 11 = 108, 108 – 13 = 95, 95 – 15 = 80, 80 – 17 = 63, 63 – 19 = 44, 44 – 21 = 23, 23 – 23 = 0.

√144 = 12

Question no – (4)

Solution :

(a) 324

The units digit of the square root may be 2 or 8

Tens digit – 1

probably 12 , 18

√324 = 18

(b) 4225

The unit digit of the square root is – 5

tens digit – 6

√4225 = 65

(c) 9801

The unit digit of the square root may be – 1 , 9

Tens digit – 9

probably – 91 , 99

√9801 = 99

(d) 5041

The unit digits may be – 1, 9

Tens digit – 7

probably – 71, 79

√5041 = 71

Question no – (5)

Solution :

(a) 4931

We should subtracted by 31 ten perfect square,

= 4900 = (70)2

(b) 18255

We should subtracted by 30 ten perfect square,

= 18225 = (135)2

(c) 27285

We should subtracted by 160 ten perfect square,

27285 = (165)2

Question no – (6)

Solution :

(a) 3450

We have 582 < 3450 < 592

perfect square should be – 592

(b) 7895

We have 892 < 1895 < 902

Perfect square should be – 902

(c) 54725

We have 2332 < 54725 < 2342

Perfect square should be – 2342

Question no – (7)

Solution :

We should multiplied by -11

The square,

= (2 × 11 × 13)

= 286

Question no – (8)

Solution :

We should divided by 6
The square = 5 × 7 = 35

Question no – (9)

Solution :

The least number of 6 digit that is a perfect square =  (317)2

Question no – (10)

Solution :

The perfect square is = (316)2

Question no – (11)

Solution :

(a) 6, 11, 12

The smallest square,

= 2 × 2 × 3 × 11 × 3 × 11

= (2 × 3 × 11)2

= (66)2

(b) 5, 14, 21

The smallest square,

= 2 × 2 × 3 × 3 × 5 × 5 × 7 × 7

= (2 × 3 × 5 × 7)2

= (210)2

Question no – (12)

Solution :

There are 52 number of students.

Question no – (13)

Solution :

Therefore, 11 students are left out.

Question no – (14)

Solution :

Area of the square park,

= 96 × 54

= 5184

Therefore, the Length of the side will be 72 m

Question no – (15)

Solution :

Let the number one x, 2x

2x × x = 2000000

X2 = 1000000

X = 1000

The numbers are – 1000, 2000

Squares and Square Roots Exercise 3.3 Solution

Question no – (1)

Solution :

(a) √121/625

= √121/√625

= √11 × 11/ √25 × 25

= 11/25

(b) √225/729

= √225/729

= 15/27

(c) √64/√441

= 8/21

(d) √81/576

= √81/576

= 9/24

Question no – (2)

Solution :

(a) √31 × √98

= √32 98

= √4 × 4 × 2 × 2 × 7 × 7

= 4 × 2 × 7

= 56

(b) √12 × 27

= √3 × 4 × 3 × 9

= 3 × 2 × 3

= 18

(c) √48 × 108

= √4 × 4 × 3 × 2 × 2 × 3 × 9

= 4 × 3 × 2 × 3

= 72

(d) √104 × 234

= √2 × 2 × 2 × 13 × 2 × 13 × 9

= 2 × 2 × 13 × 3

= 156

Question no – (3)

Solution :

(a) Given, √5 19/25

= √144/25

= 12/5

(b) Given, √96/294

= √16/49

= 4/7

(c) Given, √84 37/121

= √10201/121

= 101/11

(d) Given, √6 30/289

= √1764/289

= 42/17

Question no – (4)

Solution :

(a) Given, 33.64

(b) Given, 0.002809

(c) Given, 15.625

(d) Given, 44.1

(e) Given, 3

Question no – (5)

Solution :

(a) 350

182 < 350 < 192

√350 to the nearest integer = 19

(b) 150

122 < 150 < 132

√150 to the nearest integer = 13

(c) 5000

702 < 5000 < 72

√5000 to the nearest integer 71

(d) 720

262 < 720 < 272

√720 to the nearest integer 27

Question no – (6)

Solution :

(a) √21609.0 = 147.0

(b) √216.09 = 14.7

(c) √2.1609 = 1.47

(d) √0.021609 = 0.147

(e) √0.00021609 = 0.0147

(f) √2160900 = 1470

Question no – (7)

Solution :

Length of the side 11.5 km

4 side of the square 4 × 11.5 = 46 km

He travelled 3 times = (46 × 3) = 138 km

Therefore, the distance he travelled is 138 km.

Question no – (8)

Solution :

Given, √59.29 – √5.29 / √59.29 + √5.29

∴ (√59.29 – √5.29) (√59.29 + √5.29) / (√59.29 + √5.29)2

= 59.29 – 5.29 / 59.29 + 2 √59.29 – 5.29 + 5.29

= 54.00 / 64. 58 + 2√5929/100 × 529/100

= 54/ 6458/100 + 2√77 × 77 × 23 × 23/100

= 5400/6458 + 2 × 77 × 23

= 5400/6458 + 3542

= 5400/10000

= 0.54…(Simplified)

Question no – (9)

Solution :

(a) As per the question,

AB = 6cm

BC = 8cm

AC = √62 + 82

= √36 + 64

= √100

= 10 cm

(b) As per the question,

AC = 13 cm,

BC = 5 cm

AB = √132 – 52

= √169 – 25

= √144

= 12 cm

Squares and Square Roots Chapter Check-up Solution :

Question no – (1)

Solution :

(a) The value of √176 + v2401 = 15

Explanation :

= √176 + √2401

= √176 + 49

= √225

= 15

(b) There are 10 natural numbers lying between 52 and 62

(c) One’s digit in the square of 14 and 36 is 6.

(d) If one number of a Pythagorean triplet is 2 m, the other two one (m2 – 1) and (m2 + 1).

(e) The sum of first n odd natural numbers is n2.

(f) The perfect square will never end with 2, 3, 7, and 8.

(g) The value of √248 + √52 + √144 is 16.

Explanation :
= √248 + √52 + √144
= √248 + √52 + 12
= √248 + √64
= √248 + 8
= √256
= 16

(h) A number having 7 at its units place will have 9 at the unit place of its square.

Question no – (2)

Solution :

∴ 176 = 2 × 2 × 4 × 11

Therefore, we should multiplied by 11 for perfect square.

Question no – (3)

Solution :

Given, √(35)2 + (12)2

= √1225 + 144

= √1369

= 37 m

Therefore, he walk 37 m while returning.

Question no – (4)

Solution :

Two Pythagorean triplets each having one of the numbers as 5 are 3, 4, 5 and 5, 5.25, 6.25

Question no – (5)

Solution :

(a) 1369

(b) 5625

(c) 27.04

(d) 1.44

Question no – (6)

Solution :

Given number, 1385

∴ Subtracted by 16

∴ Square root be 37

Question no – (7)

Solution :

Given number, 6200

∴ 782 < 6200 < 792

∴ 792 = 6241

Therefore, we must be added 41 for a perfect square.

Question no – (8)

Solution :

We know, The greatest three digits is 999

∴ The greatest number of three digit is (31)2 = 961

Question no – (9)

Solution :

∴ LCM = 120

= 120 = 4 × 2 × 3 × 5

Therefore, we need 4 × 2 × 2 × 3 × 3 × 5 × 5 for divisible by the given number 3600

Question no – (10)

Solution :

As per the given question,

A decimal number is multiplied by itself.

The product is = 51.84,

The number = ?

Hence, the number will be 7.2

Question no – (11)

Solution :

4a = 96 m

a = 24

∴ Length = 24 m

∴ Area = (24)2 = 576 m2

Thus, the area of a square field will be 576 m2

Question no – (12)

Solution :

Given, √(35)2 + (12)2

= √1225 + 144

= √1369

= 37 m

Therefore, he walk 37 m while returning.

Question no – (13)

Solution :

(2x)2 + (3x)2 + (5x)2 = 608

4x2 + 9x2 + 25x2 = 608

38x= 608

x2 = 16

x = 4

Therefore, the numbers are 8, 12 and 20

Question no – (14)

Solution :

√101 1/400 m2

= 201/20 m

Therefore, Length of one side will be 10 1/20 m.

Previous Chapter Solution :

Updated: June 5, 2023 — 6:32 am