RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 8 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 8. Students can also read RBSE Class 8 Maths Important Questions for exam preparation. Students can also go through RBSE Class 8 Maths Notes to understand and remember the concepts easily. Practicing the class 8 maths chapter 6 try these solutions will help students analyse their level of preparation.

RBSE Class 8 Maths Solutions Chapter 6 Square and Square Roots Ex 6.1

Question 1.
What will be the unit digit of the squares of the following numbers?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Answer:
The unit digit of the squares of the given numbers is shown in the table :
RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 1

RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1

Question 2.
The following numbers are obviously not perfect squares. Give reason,
(i) 1057
Answer:
A number that ends either with 2, 3, 7 or 8 cannot be a perfect square. Also, a number that ends with odd number of zero(s) cannot be a perfect square.
∵ The given number 1057 ends with 7, so it cannot be a perfect square.

(ii) 23453
Answer:
∵ 23453 ends with 3, so it cannot be a perfect square.

(iii) 7928
Answer:
∵ 7928 ends with 8, so it cannot be a perfect square.

(iv) 222222
Answer:
∵ 222222 ends with 2, so it cannot be a perfect square.

(v) 64000
Answer:
∵ 64000 ends with an odd number of zeros, so it cannot be a perfect square.

(vi) 89722
Answer:
∵ 89722 ends with 2, so it cannot be a perfect square.

(vii) 222000
Answer:
∵ 222000 ends with an odd number of zeros, so it cannot be a perfect square.

(viii) 505050
Answer:
∵ 505050 ends with an odd number of zeros, so it cannot be a perfect square.

Question 3.
The squares of which of the following would be odd numbers?
(i) 431
Answer:
431 is an odd, so its square must be odd.

(ii) 2826
Answer:
2826 in an even, so its square must be even.

(iii) 7779
Answer:
7779 is an odd, so its square must be odd.

(iv) 82004
Answer:
82004 is an even, so its square must be even.

RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1

Question 4.
Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 ............... 2 ................... 1
100000012 = .............................
Answer:
1000012 = 10000200001
100000012= 100000020000001

Question 5.
Observe the following pattern and supply the missing numbers.
1122 = 121
1012 = 10201
101012 = 102030201
10101012 = _________
__________2 = 10203040504030201
Answer:
The missing numbers are
10101012 = 1020304030201
1010101012 = 10203040504030201

Question 6.
Using the given pattern, find the missing numbers.
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + ....2 = 212
52 + ....2 + 302 = 312
62 + 72 + .........2 = ......2
Answer:
The missing numbers are:
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432

Question 7.
Without adding, find the sum.
(i) 1+ 3 + 5 + 7 + 9
Answer:
1 + 3 + 5+ 7 + 9
= Sum of first 5 odd numbers = 52 = 25

(ii) 1 + 3 + 5+ 7 + 9 + 11 + 13 + 15 + 17 + 19
Answer:
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
= Sum of first 10 odd numbers = 102 = 100

(iii) 1+3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Answer:
1 + 3 + 5 + 7 + 9+11 + 13 + 15 + 17 + 19 + 21 + 23
= Sum of first 12 odd numbers = 122 = 144

Question 8.
(i) Express 49 as the sum of 7 odd numbers.
Answer:
49 = 72 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers.
Answer:
121 = 112
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

RBSE Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1

Question 9.
How many numbers lie between squares of the following numbers?
(i) 12 and 13
Answer:
There are 24 (i.e. 2 × 12) numbers lie between 122 and 132.

(ii) 25 and 26
Answer:
There are 50 (i.e. 2 × 25) numbers lie between 252 and 262.

(iii) 99 and 100
Answer:
There are 198 (i.e. 2 × 99) numbers lie between 992 and 1002.

Prasanna
Last Updated on May 17, 2022, 12:14 p.m.
Published May 17, 2022