NCERT Solutions For Class 8 Maths Chapter – 6 Squares and square roots Exercise – 6.1

Q1. What will be the unit digit of the squares of the following numbers ?

 

• 81

Solution:-

 

The unit  digit of a square of a number having digit 1 as unit place is 1.

 

Unit digit of the square of number is 81 is equal to 1.

 

• 272

Solution:-

 

The unit digit of the square of a number having digit 2 as unit place is 4

 

Unit digit of the square of number 272 is equal to 4.

 

• 799

Solution:-

 

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

 

Unit digit of the square of number 799 is equal to 1

 

• 3853

Solution:-

 

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

 

Unit digit of the square of number 3853 is equal to 9.

 

• 1234

Solution:-

 

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

 

Unit digit of the square of number 1234 is equal to 6.

 

• 26387 

• Also Read

• NCERT Class 7 English Book pdf Download Honeycomb

Solution:-

 

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

 

Unit digit of the  square of number 26387 is equal to 9.

 

• 52698

Solution:-

 

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

 

Unity digit of the square of number 52698 is equal to 4.

 

• 99880

Solution:-

 

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

 

 

Unit digit of the square of number 99880 is equal to 0.

 

• 12796

Solution:-

 

The unit digit of the square of a number having digit 5 as unit place is 6.

 

Unit digit of the square  of number 12796 is equal to 6.

 

• 55555

Solution:-

 

The unit digit of the square of number having digit 5 as unit place is 5.

 

Unit digit of the square of number 55555  is equal to 5.

 

Q2. The following numbers are obviously not perfect squares. Give reason.

 

• 1057

Solution:-

 

Ends with 7.

 

• 23453

Solution:-

 

Ends with 3.

 

• 7928

Solution:-

 

Ends with 8.

 

• 222222

Solution:-

 

Ends with 2.

 

• 64000

Solution:-

 

Ends with 0

 

• 89722

Solution:-

 

Ends with 2.

 

• 222000

Solution:-

 

Ends with 0.

 

• 505050

Solution:-

 

Ends with 0.

 

Q3. The squares of which of the following would be odd numbers ?

• 431

Solution:-

 

The squares of 431 is an odd number.

 

• 2826

Solution:-

 

The squares of 2826 is an even number.

 

• 7779

Solution:-

 

The squares of 7779 is an odd number.

 

• 82004

Solution:-

 

The squares of 82004 is an even number.

 

Q4. 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 squares 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 , then the number to be squared is 10101  and it s square is 102030201.

So, 10101012 = 1020304030201

1010101012 = 10203040505030201

 

Q6. 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

1. e 1² + 2² + ( 1 x 2 )² = ( 1² + 2² – 1 x 2 )²

2² + 3² + 6² = 7²

ATQ    2² + 3² + (2 x 3)² = (2² + 3² -2 x 3)²

3² + 4² + ( 3 x 4 )² = ( 3² + 4² -3 x 4)²

4² + 5² + 20² = 21²

5² + 6² + 30² = 31²

6²  + 7²  + 42²  = 43²

 

Q7. without adding , find the sum .

1. 1 + 3 + 5 + 7 + 9

Solution:-

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

 

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

Solution:-

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

 

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

Solution:-

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

 

Q8. (i.) express 49 as the sum of 7 odd numbers.

Solution:-

We know sum of first n odd naturals numbers is n². since  , 49 = 7²

ATQ: 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

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

 

Q9. how many numbers lie between squares of the following numbers ?

• 12 and 13

Solution:-

 

12 and 13 there are  2 x 12 = 24 natural numbers

 

• 25 and 26

Solution:-

 

25 and 26 there are 2 x 25 = 50 natural numbers.

 

• 99 and 100

Solution:-

 

99 and 100 there are 2 x 99 = 198  natural numbers

एकवचन बहुवचन singular – plural -in – Hindi