RBSE Solutions for Class 7 Maths Chapter 3 Square and Square Root Ex 3.1 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 3.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Square and Square Root |

Exercise |
Ex 3.1 |

Number of Questions |
7 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 3 Square and Square Root Ex 3.1

Question 1

What will be the unit (RBSESolutions.com) place squares of following numbers?

(i) 24

(ii) 17

(iii) 100

(iv) 55

(v) 111

(vi) 1023

(vii) 5678

(viii) 12796

(ix) 2412

Solution:

S.No |
Number |
Square of number |
Unit place to digit in square number |

(i) | 24 | 24 x 24 | 4 x 4 = 1⑥ |

(ii) | 17 | 17 x 17 | 7 x 7 = 4⓪ |

(iii) | 100 | 100 x 100 | 0 x 0 = ⓪ |

(iv) | 55 | 55 x 55 | 5 x 5 = 2 |

(v) | 111 | 111 x 111 | 1 x 1 = 2⑤ |

(vi) | 1023 | 1023 x 1023 | 3 x 3 = ① |

(vii) | 5678 | 5678 x 5678 | 8 x 8 = 6④ |

(viii) | 12796 | 12796 x 12796 | 6 x 6 = 3⑥ |

(ix) | 2412 | 2412 x 2412 | 2 x 2 = 4 |

Question 2

Find the squares of (RBSESolutions.com) numbers given below :

(i) 18

(ii) 11

(iii) 107

(iv) 15

(v) 200

(i) 27

Solution:

(i) 18 = (18)^{2} = 18 × 18 = 324

(ii) 11 = (11)^{2} = 11 × 11 = 121

(iii) 107 = (107)^{2} = 107 × 107 = 11449

(iv) 15 = (15)^{2} = 15 × 15 = 225

(v) 200 = (200)^{2} = 200 × 200 = 40000

(vi) 27 = (27)^{2} = 27 × 27 = 729

Question 3

Which of the following numbers (RBSESolutions.com) has their square as even number?

(i) 235

(ii) 396

(iii) 5508

(i) 2001

(v) 82003

(vi) 10224

Solution:

Only even numbers have their squares as even (RBSESolutions.com) number so the numbers having even square will be

(ii) 396,

(iii) 5508,

(vi) 10224

Question 4

Find the following sums without operating:

(i) 1 + 3 + 5 + 7

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

Solution:

Sum of the numbers given is (RBSESolutions.com) equal to the square of their number

(i) 1 + 3 + 5 + 7 = 4 sum of numbers = (4)^{2} = 16

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13

= 7 sum of numbers = (7)^{2} = 49

(iii) 1+3 + 5 + 7 + 9+11 + 13 + 15 + 17 + 19

= 10 sum of numbers = (10)^{2} = 100

Question 5

Write down 64 in the form (RBSESolutions.com) of sum of eight odd numbers.

Solution:

64 = 1+ 3 + 5 + 7 + 9 + 11 +13 + 15

Question 6

How many numbers are there in between the following numbers?

(i) 10 and 11

(ii) 17 and 18

(iii) 30 and 31

Solution:

(i) 10 + 11 =21 numbers

(ii) 17 + 18 = 35 numbers

(iii) 30 +31 =61 numbers

Question 7

Check whether the given set (RBSESolutions.com) of three numbers either a Pythagorean triplets or not?

(i) 9,12, 15

(ii) 7, 11, 13

(iii) 10, 24, 26

Solution:

(i) (9)^{2} + (12)^{2} = 81 + 144 = 225 = (15)^{2}

So (9)^{2} + (12)^{2}= (15)^{2}

or 9, 12, 15 Pythagorean triplets.

(ii) (7)^{2} + (11)^{2} = 49 + 121 = 170

∴ (7)^{2} + (11)^{2} ≠ (13)^{2}

or 7, 11, 13 Pythagorean (RBSESolutions.com) triplets.

(iii) (10)^{2} + (24)^{2} = 100 + 576 = 676 = (26)^{2}

∴ (10)^{2} + (24)^{2} = (26)^{2}

or 10, 24, 26 Pythagorean triplets.

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