Class 8 | NCERT Solution Maths Chapter 6 | Squares and Square Roots | Exercise 6.1

 Exercise 6.1 Page: 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

∴ Unit digit of the square of 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

we. 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 middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is 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, then 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 number = (5)2 = 25

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

Solution:

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

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

Solution:

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

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

Solution:

We know, sum of 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.

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