RBSE Solutions for Class 7 Maths Chapter 3 Square and Square Root Ex 3.3 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 3 Square and Square Root Exercise 3.3.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Square and Square Root |

Exercise |
Ex 3.3 |

Number of Questions |
8 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 3 Square and Square Root Ex 3.3

Question 1

Find the smallest root of the (RBSESolutions.com) following number using division method

(i) 441

(ii) 576

(iii) 1225

(iv) 2916

(v) 4624

(vi) 7921

Solution:

Make the pair of 2 – 2 digits from (RBSESolutions.com) unit place in left

(i) 441

So \(\sqrt { 441 }\) = 21

(ii) 576

So \(\sqrt { 576 }\) = 24

(iii) 1225

So \(\sqrt { 1225 }\) = 35

(iv) 2916

So \(\sqrt { 2916 }\) = 54

(v) 4624

So \(\sqrt { 4624 }\) = 68

(vi) 7921

So \(\sqrt { 7921 }\) = 89

Question 2

Find the square root of the following (RBSESolutions.com) numbers without really calculating

(i) 121

(ii) 256

(iii) 4489

(iv) 60025

Solution:

First we will prepare a table by using unit digit of given number to find the possible unit digit in their squares.

Unit place digit number of given number |
Possible unit digit in square |

1 | 1 or 9 |

4 | 2 or 8 |

5 | 5 |

6 | 4 or 6 |

9 | 3 or 7 |

∵ Unit place digit number = 1

∴ Unit place digit (RBSESolutions.com) in square root = 1 or 9

leaving 21 remaining digit = 1

∵ 12 = 1 < 2^{2}

∴ 121 value of 11^{2}

∴ \(\sqrt { 121 }\) = 11

(ii) 256

∵ Unit place digit = 6

∴ Unit place digit in square root number = 4 or 6

Leaving 56 remaining digit = 2

∵1^{2} < 2^{2} < 2^{2}

Value of 256 = 14^{2} or 16^{2}

∵ 15^{2} = 225

∴ \(\sqrt { 256 }\) = 16

(iii) 4489

∵ Unit place digit = 9

∴ Unit place digit (RBSESolutions.com) in square root = 3 or 7

Leaving 89 remaining digits = 44

∵ 6^{2} < 44 <7^{2}

∴ Value of 4489 will be 63^{2} or 67^{2}

∵ 65^{2} = 4225

∴ \(\sqrt { 4489 }\) = 67

(iv) 60025

∵ Unit place of digit = 5

∴ Unit place digit in square root = 5

Leaving 25 remaining number = 600

∵ 24^{2} < 600 < 25^{2}

∴value of 60025 will be 2452

∴ \(\sqrt { 60025 }\) = 245

Question 3

Find the square root of (RBSESolutions.com) following decimal numbers

(i) 6.25

(ii) 2.89

(iii) 32.49

(iv) 31.36

(v) 57.76

Solution:

(i) 6.25.

∴ \(\sqrt { 6.25 }\) = 2.5

(ii) 2.89

∴ \(\sqrt { 2.89 }\) = 1.7

(iii) 32.49

∴ \(\sqrt { 32.49 }\) = 5.7

(iv) 31.36

∴ \(\sqrt { 31.36 }\) = 5.6

(v) 57.76

∴ \(\sqrt { 57.76 }\) = 7.6

Question 4

What is to be added in the following (RBSESolutions.com) numbers so that they become perfect squares ?

(i) 420

(ii) 2000

(iii) 837

(iv) 3500

Solution:

(i) … 420 is greater than 20 from the square number 400.

∴ 420 – 20 = 400 ⇒ \(\sqrt { 400 }\) = 20

∵ Something is to (RBSESolutions.com) be added to 420.

So we will think from 20 to 21.

∵ (21)^{2} = 441

∴ 441 – 420 = 21 should be added.

(ii) we see that 2000 is more than 64 from a square numbers 1936

∴ 2000 – 64= 1936

\(\sqrt { 1936 }\) = 44

∴ To get a perfect (RBSESolutions.com) square 44 + 1 = 45

∵ (45)^{2} = 2025

∴ 2025 – 2000 = 25 should be added.

(iii) We see that 837 is more than 53 from a perfect square number.

∴ 837 – 53 = 784

∵ \(\sqrt { 784 }\) = 28

To get a perfect (RBSESolutions.com) square = 28 + 1 = 29

∵ (29)^{2} = 841

∴ 841 – 837 = 4 should be added

(iv) We see that 3500 is more than 19 from a perfect square number

∴ 3500 – 19 = 3481

∵\(\sqrt { 3481 }\) = 59

To get a perfect square = 59 + 1 = 60,

∵ (60)^{2} = 3600

∴ 3600 – 3500 = 100 should be added.

Question 5

What is to be subtracted from the (RBSESolutions.com) following numbers so that they become perfect square numbers ?

(i) 555

(ii) 252

(iii) 1650

(iv) 6410

Solution:

(i) 555

∴ Number to be (RBSESolutions.com) subtracted = 26

(ii) 252

∴ Number to be subtracted = 27

(iii) 1650

∴ Number to be (RBSESolutions.com) subtracted = 50

(iv) 6410

∴ Number to be subtracted = 10

Question 6

Chair arc tos be arranged for a wedding function is square configuration. 1000 chairs (RBSESolutions.com) are available. How many additional chairs will be requried for square configuration? At the same time, also find the number of chairs in each row.

Solution:

Total chairs = 1000

1000 from a square number

39 are more than

∴ 1000 – 39 = 961

∴ \(\sqrt { 961 }\) = 31

∴ Number of more (RBSESolutions.com) chairs required

= (32)^{2} – 1000 = 1024 – 1000 = 24

Number of chairs in each row = 31 + 1 = 32

Question 7

Area of a square farm is 361 m^{2}. How much wire will be required for fencing the four sides ?

Solution:

Area of square field = 361 m^{2}

∴ Side of square field = \(\sqrt { 361 }\) = 19 m

∴ Required length of wire = 4 x side = 4 x 19 = 76 m

Question 8

Find the smallest number that (RBSESolutions.com) can divide 2352 to make it a perfect square.

Solution:

On factorisation of 2352

∴ 2352 = 2 x 2 x 2 x 2 x 7 x 7 x 3

∴ On dividing by 3.

Remainder (RBSESolutions.com) will be a perfect square number.

We hope the RBSE Solutions for Class 7 Maths Chapter 3 Square and Square Root Ex 3.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 3 Square and Square Root Exercise 3.3, drop a comment below and we will get back to you at the earliest.

## Leave a Reply