NCERT Solutions for Class 8 Maths Chapter 6 - Squares and Square Roots Exercise 6.1

The NCERT solutions for Class 8 Maths enhance topics with frequent, focused, engaging maths challenges and activities that strengthen maths concepts. Each question of exercise 6.1 in NCERT Class 8 Maths Solutions has been carefully solved for the students to understand, keeping the examination point of view in mind.

Class 8 Maths Chapter 6 – Squares and Square Roots Exercise 6.1 Questions and answers help students to understand the difference between squares and square roots, as well as how to find them out. These NCERT Solutions are prepared by subject experts at BYJU’S using a step-by-step approach.

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Access other exercise solutions of Class 8 Maths Chapter 6 – Squares and Square Roots

Exercise 6.2 Solutions 2 Questions

Exercise 6.3 Solutions 10 Questions

Exercise 6.4 Solutions 9 Questions

Access Answers of Maths NCERT Class 8 Chapter 6 – Squares and Square Roots Exercise 6.1 Page Number 96

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

Solution:

The unit digit of the square of a number having ‘a’ at its unit place ends with a×a.

i. The unit digit of the square of a number having digit 1 as the unit’s place is 1.

∴ The unit digit of the square of the number 81 is equal to 1.

ii. The unit digit of the square of a number having the digit 2 as the unit’s place is 4.

∴ The unit digit of the square of the number 272 is equal to 4.

iii. The unit digit of the square of a number having the digit 9 as the unit’s place is 1.

∴ The unit digit of the square of the number 799 is equal to 1.

iv. The unit digit of the square of a number having the digit 3 as the unit’s place is 9.

∴ The unit digit of the square of the number 3853 is equal to 9.

v. The unit digit of the square of a number having the digit 4 as the unit’s place is 6.

∴ The unit digit of the square of the number 1234 is equal to 6.

vi. The unit digit of the square of a number having the digit 7 as the unit’s place is 9.

∴ The unit digit of the square of the number 26387 is equal to 9.

vii. The unit digit of the square of a number having the digit 8 as the unit’s place is 4.

∴ The unit digit of the square of the number 52698 is equal to 4.

viii. The unit digit of the square of a number having the digit 0 as the unit’s place is 01.

∴ The unit digit of the square of the number 99880 is equal to 0.

ix. The unit digit of the square of a number having the digit 6 as the unit’s place is 6.

∴ The unit digit of the square of the number 12796 is equal to 6.

x. The unit digit of the square of a number having the digit 5 as the unit’s place is 5.

∴ The unit digit of the square of the number 55555 is equal to 5.

2. The following numbers are obviously not perfect squares. Give reason.

i. 1057

ii. 23453

iii. 7928

iv. 222222

v. 64000

vi. 89722

vii. 222000

viii. 505050

Solution:

We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.

i. 1057 ⟹ Ends with 7

ii. 23453 ⟹ Ends with 3

iii. 7928 ⟹ Ends with 8

iv. 222222 ⟹ Ends with 2

v. 64000 ⟹ Ends with 0

vi. 89722 ⟹ Ends with 2

vii. 222000 ⟹ Ends with 0

viii. 505050 ⟹ Ends with 0

3. The squares of which of the following would be odd numbers?

i. 431

ii. 2826

iii. 7779

iv. 82004

Solution:

We know that the square of an odd number is odd, and the square of an even number is even.

i. The square of 431 is an odd number.

ii. The square of 2826 is an even number.

iii. The square of 7779 is an odd number.

iv. The square of 82004 is an even number.

4. Observe the following pattern and find the missing numbers. 11² = 121

101² = 10201

1001² = 1002001

100001² = 1 …….2………1

10000001² = ……………………..

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and the middle digit is 2. And the number of zeros between the left-most digit 1 and the middle digit 2 and the right-most digit 1 and the middle digit 2 is the same as the number of zeros in the given number.

∴ 100001² = 10000200001

10000001² = 100000020000001

5. Observe the following pattern and supply the missing numbers. 112 = 121

1012 = 10201

101012 = 102030201

10101012 = ………………………

…………2 = 10203040504030201

Solution:

We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And the square is symmetric about the middle digit. If the middle digit is 4, the number to be squared is 10101, and its square is 102030201.

So, 10101012 =1020304030201

1010101012 =10203040505030201

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

5 + _ ² + 30² = 31²

6 + 7 + _ ² = _ ²

Solution:

Given, 1² + 2² + 2² = 3²

i.e 1² + 2² + (1×2 )² = ( 1² + 2² -1 × 2 )²

2² + 3² + 6² =7²

∴ 2² + 3² + (2×3 )² = (2² + 3² -2 × 3)²

3² + 4² + 12² = 13²

∴ 3² + 4² + (3×4 )² = (3² + 4² – 3 × 4)²

4² + 5² + (4×5 )² = (4² + 5² – 4 × 5)²

∴ 4² + 5² + 20² = 21²

5² + 6² + (5×6 )² = (5²+ 6² – 5 × 6)²

∴ 5² + 6² + 30² = 31²

6² + 7² + (6×7 )² = (6² + 7² – 6 × 7)²

∴ 6² + 7² + 42² = 43²

7. Without adding, find the sum.

i. 1 + 3 + 5 + 7 + 9

Solution:

Sum of first five odd numbers = (5)² = 25

ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

Solution:

Sum of first ten odd numbers = (10)² = 100

iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution:

Sum of first thirteen odd numbers = (12)² = 144

8. (i) Express 49 as the sum of 7 odd numbers.

Solution:

We know that the sum of the first n odd natural numbers is n² . Since,49 = 7²

∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers. Solution:

Since, 121 = 11²

∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?

i. 12 and 13

ii. 25 and 26

iii. 99 and 100

Solution:

Between n² and (n+1)², there are 2n non–perfect square numbers.

i. 122 and 132, there are 2×12 = 24 natural numbers.

ii. 252 and 262, there are 2×25 = 50 natural numbers.

iii. 992 and 1002, there are 2×99 =198 natural numbers.

Exercise 6.1 of NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots is based on the following topics:

1. Introduction to squares and square roots
2. What are square numbers?
3. Properties of Square Numbers
4. Some interesting patterns
   1. Adding triangular numbers
   2. Numbers between square numbers
   3. Adding odd numbers
   4. A sum of consecutive natural numbers

Also, explore – 

NCERT Solutions for Class 8