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.
We hope the RBSE Solutions for Class 7 Maths Chapter 3 Square and Square Root Ex 3.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 3 Square and Square Root Exercise 3.1, drop a comment below and we will get back to you at the earliest.