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

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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