**Frank Learning Maths Class 8 Chapter 3 Squares and Square Roots Exercise 3.1 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 3.1 all Questions Solution**

**(1) Find the Square of the following by multiplication**

**Solution : **

**(a)** 19^{2}^{}

= 19 × 19

= 361

**(b)** 47^{2}

= 47 × 47

= 2209

**(c)** 62^{2}^{}

= 62 × 62

= 3844

**(d)** 101^{2}

= 10201

**(2) Find the Pythagorean triplets of the numbers, the smallest of which is given below**

**Solution : **

Let, Pythagorean triplets are 2m, m^{2} – 1, m^{2} + 1

**(a)** 2 m = 6

M = 3

M^{2 }– 1 = 9 – 1 = 8

M^{2} + 1 = 9 + 1 = 10

**(b)** 2 m = 14

M = 7

M^{2} = 1 = 48

M^{2} + 1 = 50

**(c)** 2m = 18

M = 9

= M^{2} + 1 = 82

**(d)** 2 m = 16

M = 8

M^{2} – 1 = 63

M^{2} + 1 = 65

**(3) Evaluate the following using properties of perfect squares
**

**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)** 9^{2} – 8^{2}

= 81 – 64

= 17

**(ii)** 13^{2} – 12^{2}

= 25

Therefore, the Natural numbers between then 16 and 24

**Question no – (9) **

**Solution : **

**(a)** 50^{2} – 51^{2}

= (51 + 50) (51 – 50)

= 101

**∴ **Non-squares number between 51^{2}, 50^{2} = (101 – 1)

= 100

**(b)** 90^{2} – 91^{2}

= (91 + 90) (91 – 90)

= 181

**∴ **Non-square number (181 – 1) = 180

**(c)** 1000^{2} – 1001^{2}

= (1001 + 1000) ( 1001 – 1000)

= 2001

**∴ **Non-square numbers (2001 – 1) = 2000

**Question no – (10) **

**Solution : **

**(a)** 13 × 15

= 13 (13 + 2)

= 13^{2} + 13 x 2

= 169 + 26

= 195

**(b)** 28 × 30

= (30 – 2) x 30

= 30^{2} – 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)

= 100^{2} + 2 x 100

= 10000 + 200

**=** 10200