NCERT Solutions for class 8 Maths chapter 6 Square and square roots

Exercise 6.1  Exercise 6.2  Exercise 6.3  Exercise 6.4

Exercise 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

Sol :

We know that if a number has its unit’s place digit as a, then its square will end with the unit digit of the multiplication a × a.

(i) 81

Since the given number has its unit’s place digit as 1, its square will end with the unit digit of the multiplication (1 ×1 = 1) i.e., 1.

(ii) 272

Since the given number has its unit’s place digit as 2, its square will end with the unit digit of the multiplication (2 × 2 = 4) i.e., 4.

(iii) 799

Since the given number has its unit’s place digit as 9, its square will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.

(iv) 3853

Since the given number has its unit’s place digit as 3, its square will end with the unit digit of the multiplication (3 × 3 = 9) i.e., 9.

(v) 1234

Since the given number has its unit’s place digit as 4, its square will end with the unit digit of the multiplication (4 × 4 = 16) i.e., 6.

(vi) 26387

Since the given number has its unit’s place digit as 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.

(vii) 52698

Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.

(viii) 99880

Since the given number has its unit’s place digit as 0, its square will have two zeroes at the end. Therefore, the unit digit of the square of the given number is 0.

(xi) 12796

Since the given number has its unit’s place digit as 6, its square will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.

(x) 55555

Since the given number has its unit’s place digit as 5, its square will end with the unit digit of the multiplication (5 × 5 = 25) i.e., 5.

 

Question 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

Sol :

The square of numbers may end with any one of the digits 0, 1, 5, 6, or 9. Also, a perfect square has even number of zeros at the end of it.

(i) 1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.

(ii) 23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.

(iii) 7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.

(iv) 222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(v) 64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeros, it is not a perfect square.

(vi) 89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(vii) 222000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeros, it is not a perfect square.

(viii) 505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeros, it is not a perfect square.

 

Question 3

(i) 431 (ii) 2826

(iii) 7779 (iv) 82004

Sol :

The square of an odd number is odd and the square of an even number is even. Here, 431 and 7779 are odd numbers.

Thus, the square of 431 and 7779 will be an odd number.

 

Question 4

Observe the following pattern and find the missing digits.

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1…2…1

100000012 = …

Sol :

In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,

1000012 = 10000200001

100000012 = 100000020000001

 

Question 5

Observe the following pattern and supply the missing number.

112 = 121

1012 = 10201

101012 = 102030201

10101012 = …

2 = 10203040504030201

Sol :

By following the given pattern, we obtain

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

Sol :

From the given pattern, it can be observed that,

(i) The third number is the product of the first two numbers.

(ii) The fourth number can be obtained by adding 1 to the third number.

Thus, the missing numbers in the pattern will be as follows.

42 + 52 + = 212

52 + + 302 = 312

62 + 72 + =

 

Question 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

Sol :

We know that the sum of first n odd natural numbers is n2.

(i) Here, we have to find the sum of first five odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 = (5)2 = 25

(ii) Here, we have to find the sum of first ten odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100

(iii) Here, we have to find the sum of first twelve odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)2 = 144

 

Question 8

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

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

Sol :

We know that the sum of first n odd natural numbers is n2.

(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

 

Question 9

How many numbers lie between squares of the following numbers?

(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Sol :

We know that there will be 2n numbers in between the squares of the numbers n and (n + 1).

(i) Between 122 and 132, there will be 2 × 12 = 24 numbers

(ii) Between 252 and 262, there will be 2 × 25 = 50 numbers

(iii) Between 992 and 1002, there will be 2 × 99 = 198 numbers

Leave a Reply

Your email address will not be published. Required fields are marked *