RBSE Solutions for Class 7 Maths Chapter 3 Square and Square Root Ex 3.2 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 3.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Square and Square Root |

Exercise |
Ex 3.2 |

Number of Questions |
7 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 3 Square and Square Root Ex 3.2

Question 1

What can be the unit place digit (RBSESolutions.com) in the square root of the following numbers.

(i) 9604

(ii) 65536

(iii) 998001

(iv) 60481729

Solution:

(i) Unit place digit in the square root of 9604 = 2,8

(ii) Unit place digit in the square root of 65536 = 4,6

(iii) Unit place digit in die square root of 998001 = 1,9

(iv) Unit place digit in the square root of 60481729 = 3,7

Question 2

Estimate and tell which numbers (RBSESolutions.com) cannot be perfect squares?

(i) 48

(ii) 81

(iii) 102

(iv) 24636

Solution:

(i) 48

(iii) 102 and

(iv) 24636 may not be perfect square numbers.

Question 3

Find the square root by prime (RBSESolutions.com) factorization method.

(i) 1296

(ii) 729

(iii) 1764

(iv) 3969

(v) 4356

(vi) 1600

Solution:

(i) 1296

∴ 1296 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3

∴ \(\sqrt { 1296 }\) = 2 x 2 x 3 x 3

= 36

(ii) 729

∴ 729 = 3 x 3 x 3 x 3 x 3 x 3

∴ \(\sqrt { 729 }\) = 3 x 3 x 3

= 27

(iii) 1764

∴ 1764 = 2 x 2 x 3 x 3 x 7 x 7

∴ \(\sqrt { 1764 }\) = 2 x 3 x 7

= 42

(iv) 3969

∴ 3969 = 3 x 3 x 3 x 3 x 7 x 7

∴ \(\sqrt { 3969 }\) = 3 x 3 x 7

= 63

(v) 4356

∴ 4356 = 2 x 2 x 3 x 3 x 11 x 11

∴ \(\sqrt { 4356 }\) = 2 x 3 x 11

= 66

(vi) 1600

∴ 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

∴ \(\sqrt { 1600 }\) = 2 x 2 x 2 x 5

= 40

Question 4

Numbers given below are not perfect (RBSESolutions.com) squares. Find the smallest whole number to which when multiplied makes it perfect square.

(i) 252

(ii) 396

(iii) 1620

Solution:

(i) 252

∴ 252 = 2 x 2 x 3 x 3 x 7

There is no perfect pair of 7.

Therefore 7 is to (RBSESolutions.com) be multiplied.

(ii) 396

∴ 396 = 2 x 2 x 3 x 3 x 11

Here 11 has no pair so (RBSESolutions.com) required number is 11.

(iii) 1620

∴ 1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5

∴Here 5 has no pair. So the required number is 5.

Question 5

Numbers given below are not perfect (RBSESolutions.com) squares. Use prime factorization method to find the smallest number by which these can be divided to make them perfect squares.

(i) 1000

(ii) 867

(iii) 4375

Solution:

(i) 1000

∴ 1000 = 2 x 2 x 2 x 5 x 5 x 5

= 2 x 2 x 5 x 5 x 10

10 has no pair.

So 10 is (RBSESolutions.com) required to divide.

(ii) 867

∴ 867 = 3 x 17 x 17

∵ 3 has no pair.

∵3 is required (RBSESolutions.com) to divide.

(iii) 4375

∴ 4375 = 5 x 5 x 5 x 5 x 7

∵ 7 has no pair

∴ 7 is required to divide

Question 6

Rose plants to be planted in (RBSESolutions.com) a square garden. Each row has the number of roses same as the number of row. If there are 2401 plants, then find the number of rows.

Solution:

Let total number of roses = 2401. Let number of rows be x

∴ According to question

∴ x × x = 2401

⇒ x^{2} = 2401

= \(\sqrt { \underline { 7\times 7 } \times \underline { 7\times 7 } }\)

= 7 × 7

= 49

∴Number of rows = 49

Question 7

Find the smallest square number (RBSESolutions.com) which is completely divisible by 4, 9 and 10.

Solution:

L.C.M of 4, 9, 10 = 2 x 2 x 3 x 3 x 5 = 180

∵ There is no pair of 5.

∴ 5 is required to mulitply

∴ 5 x 180 = 900 = 2 x 2 x 3 x 3 x 5 x 5

∴ Smallest (RBSESolutions.com) required number = 900

We hope the RBSE Solutions for Class 7 Maths Chapter 3 Square and Square Root Ex 3.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 3 Square and Square Root Exercise 3.2, drop a comment below and we will get back to you at the earliest.