NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

NCERT Solutions Class 8 Mathematics 
Chapter – 6 (Squares and Square) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 6 Squares and Square Exercise 6.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter 6: Squares and Square Roots

• NCERT Solution Class 8 Maths Ex – 6.2
• NCERT Solution Class 8 Maths Ex – 6.3
• NCERT Solution Class 8 Maths Ex – 6.4

Exercise – 6.1 

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

(i) If a number has 1 or 9 in its unit digit, then its square ends with 1.
Since 81 has 1 as its unit digit, 1 will be the unit digit of its square. (1 × 1 = 1)

(ii) If a number has either 2 or 8 as its unit digit, then its square ends with 4.
Since 272 has 2 as its unit digit, 4 will be the unit digit of its square. (2 × 2 = 4)

(iii) If a number has 1 or 9 in its unit digit, then its square ends with 1.
Since 799 has 9 as its unit digit, 1 will be the unit digit of its square. (9 × 9 = 81)

(iv) If a number has either 3 or 7 as its unit digit, then its square number ends with 9.
Since 3853 has 3 as its unit digit, 9 will be the unit digit of its square. (3 × 3 = 9)

(v) If a number has either 4 or 6 as its unit digit, then its square ends with 6.
Since 1234 has 4 as its unit digit, 6 will be the unit digit of its square. (4 × 4 = 16)

(vi) If a number has either 3 or 7 as its unit digit, then its square number ends with 9.
Since 26387 has 7 as its unit digit, 9 will be the unit digit of its square. (7 × 7 = 49)

(vii) If a number has either 2 or 8 as its unit digit, then its square ends with 4.
Since 52698 has 8 as its unit digit, 4 will be the unit digit of its square. (8 × 8 = 64)

(viii) If a number has 0 as its unit digit, then its square ends with 0.
Since 99880 has 0 as its unit digit, 0 will be the unit digit of its square.

(ix) If a number has either 4 or 6 as its unit digit, then its square ends with 6.
Since 12796 has 6 as its unit digit, 6 will be the unit digit of its square. (6 × 6 = 36)

(x) If a number has 5 as its unit digit, then its square ends with 5.
Since 55555 has 5 as its unit digit, 5 will be the unit digit of its square. (5 × 5 = 25)

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
The number 1057 is not a perfect square because it ends with 7 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(ii) 23453
The number 23453 is not a perfect square because it ends with 7 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(iii) 7928
The number 7928 is not a perfect square because it ends with 8 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(iv) 222222
The number 222222 is not a perfect square because it ends with 2 at unit’s place whereas the Square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(v) 64000
The number 64000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

(vi) 89722
The number 89722 is not a square number because it ends in 2 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(vii) 222000
The number 222000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

(viii) 505050
The number 505050 is not a square number because it has 1 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

3. The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(v) 82004

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

(i) 431
431 is an odd number
∴ Its square will also be an odd number.

(ii) 2826
2826 is an even number
∴ Its square will not be an odd number.

(iii) 7779
7779 is an odd number
∴ Its square will be an odd number.

(v) 82004
82004 is an even number
∴ Its square will not be an odd number.

4. Observe the following pattern and find the missing numbers. 11² = 121
101² = 10201
1001² = 1002001
100001² = 1____2____1
10000001² = ________

Solution –
101² = 10201
1001² = 1002001
100001² = 10000200001
10000001² = 100000020000001

5. Observe the following pattern and supply the missing numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ________
______2 = 10203040504030201

Solution –
112 = 121
1012 = 10201
101012 = 102030201
10101012 = 1020304030201
1010101012 = 10203040504030201

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
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
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19
(iii) 1+3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23

Solution –

(i) 1 + 3 + 5 + 7 + 9
Sum of first five odd number = (5)² = 25

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19
Sum of first ten odd number = (10)² = 100

(iii) 1+3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23
Sum of first thirteen odd number = (12)² = 144

8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.

Solution –

(i) 49 = (7)²
Therefore, 49 is the sum of first 7 odd natural numbers
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 121 = (11)²
Therefore, 121 is the sum of first 11 odd natural numbers
121 = 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 – ‘2n’ numbers lie between the square of two consecutive numbers n and (n+1).

(i) 12 and 13
Here, n = 12.
Thus, 2n = 2 × 12 = 24,
i.e., 24 numbers lie between (12)² and (13)²

(ii) 25 and 26
Here, n = 25
Thus, 2n = 2 × 25 = 50,
i.e., 50 numbers lie between (25)² and (26)²

(iii) 99 and 100
Here, n = 99
Thus, 2n = 2 × 99 = 198,
i.e., 198 numbers between (99)² and (100)²