**Frank Learning Maths Class 8 Chapter 3 Squares and Square Roots Exercise 3.2 Solutions**

This post is created to help all the CBSE Class 8 students for the Solutions of Frank Learning Maths Class 8 Mathematics Book, Chapter 3 Squares and Square Roots Exercise 3.1. Here students can easily find step by step solutions of all the problems for Squares and Square Roots Exercise 3.1. Step by Step proper solutions for every problems. All the problem are solved with easily understandable methods with proper guidance so that all the students can understand easily.

**Chapter 3 – Squares and Square Roots**

**Squares and Square Roots – Exercise 3.2 all Questions Solution**

**(3) Find the square root of the following by successive subtraction
**

**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

**(4) Find the square root of the following by finding the un its and the tens digits.**

**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

**(5) Find the square root of the following by finding the units and the tens digits**

**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 58

^{2}< 3450 < 59

^{2}

∴ perfect square should be – 59^{2}

**(b) 7895
**∴ We have 89

^{2}< 1895 < 90

^{2}

∴ Perfect square should be – 90^{2}

**(c) 54725
**∴ We have 233

^{2}< 54725 < 234

^{2 }

∴ Perfect square should be – 234^{2}

**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

X^{2} = 1000000

X = 1000

∴ The numbers are – 1000, 2000