# Squares And Square Roots : Exercise 6.1 (Mathematics NCERT Class 8th)

**Q.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**

**(i) 81**

*Sol.*The digit 81 ends with â€˜1â€™. And square of 1 is 1.

Hence, the unit digit of square of 81 is 1.

(ii) 272

The digit 272 ends with â€˜2â€™. And square of 2 is 4.

Hence, the unit digit of square of 272 is 4.

(iii) 799

The digit 799 ends with â€˜9â€™. And square of 9 is 81.

Hence, the unit digit of square of 799 is 1.

(iv) 3853

The digit 3853 ends with â€˜3â€™. And square of 3 is 9.

Hence, the unit digit of square of 3583 is 9.

(v) 1234

The digit 1234 ends with â€˜4â€™. And square of 4 is 16.

Hence, the unit digit of square of 1234 is 6.

(vi) 26387

The digit 26387 ends with â€˜7â€™. And square of 7 is 49.

Hence, the unit digit of square of 26387 is 9.

(vii) 52698

The digit 52698 ends with â€˜8â€™. And square of 8 is 64.

Hence, the unit digit of square of 52698 is 4.

(viii) 99880

The digit 99880 ends with â€˜0â€™. And square of 0 is 0.

Hence, the unit digit of square of 99880 is 0.

(ix) 12796

The digit 12796 ends with â€˜6â€™. And square of 6 is 36.

Hence, the unit digit of square of 12796 is 6.

(x) 55555

The digit 55555 ends with â€˜5â€™. And square of 5 is 25.

Hence, the unit digit of square of 55555 is 5.

**Q.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**

**(i) 1057**

*Sol.*It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(ii) 23453

It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(iii) 7928

It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(iv) 222222

It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(v) 64000

It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(vi) 89722

It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(vii) 222000

It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(viii) 505050

It is not a perfect square since its unitâ€™s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

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

**(i) 431 (ii) 2826 (iii) 7779 (iv) 82004**

**(i) 431**

*Sol.*The digit 431 ends with â€˜1â€™. And square of 1 is 1.

Hence, the square of 431 would be an odd number.

(ii)2826

The digit 2826 ends with â€˜6â€™. And square of 6 is 36.

Hence, the square of 2826 would not be an odd number.

(iii)7779

The digit 7779 ends with â€˜9â€™. And square of 9 is 81.

Hence, the square of 7779 would be an odd number.

(iv)82004

The digit 82004 ends with â€˜4â€™. And square of 4 is 16.

Hence, the square of 82004 would not be an odd number.

**Q.4 Observe the following pattern and find the missing digits.
**

**11**

^{2}= 121**101**

^{2}= 10201**1001**

^{2}= 1002001**100001**

^{2}= 1 ......... 2 ......... 1**10000001**

^{2}= ...........................**100001**

*Sol.*

^{2}= 10000200001

10000001

^{2}= 100000020000001

**Q.5 Observe the following pattern and supply the missing numbers.
**

**11**

^{2}= 1 2 1**101**

^{2}= 1 0 2 0 1**10101**

^{2}= 102030201**1010101**

^{2}= ...........................**............**

^{2}= 10203040504030201**1010101**

*Sol.*

^{2}= 1020304030201

101010101

^{2}= 10203040504030201

**Q.6 Using the given pattern, find the missing numbers.
**

**1**

^{2}+ 2^{2}+ 2^{2}= 3^{2}**2**

^{2}+ 3^{2}+ 6^{2}= 7^{2}**3**

^{2}+ 4^{2}+ 12^{2}= 13^{2}**4**

^{2}+ 5^{2}+ _^{2}= 21^{2}**5**

^{2}+ _^{2}+ 30^{2}= 31^{2}**6**

^{2}+ 7^{2}+ _^{2}= __^{2 }**4**

*Sol.*

^{2}+ 5

^{2}+ 20

^{2}= 21

^{2 }5

^{2}+ 6

^{2}+ 30

^{2}= 31

^{2 }6

^{2}+ 7

^{2}+42

^{2}= 43

^{2}

**Q.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**

**We know that the sum of first**

*Sol.**n*odd natural numbers is

*n*

^{2}.

(i) 1 + 3 + 5 + 7 + 9

Here, there are five odd numbers.

Therefore, 1 + 3 + 5 + 7 + 9 = 5

^{2}= 25.

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

Here, there are ten odd numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 = 10^{2} = 100

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

Here, there are twelve odd numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23= 12^{2} = 144

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

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

**We know that the sum of first**

*Sol.**n*odd natural numbers is

*n*

^{2}.

(i) 49 = 7^{2}. Therefore, it is the sum of 7 odd numbers.

Hence, 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 121 = 11^{2}. Therefore, it is the sum of 11 odd numbers.

Hence, 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

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

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

**We know that there will be 2**

*Sol.**n*numbers in between the squares of the numbers

*n*and (

*n*+1).

(i) 12 and 13

Here,

*n*= 12. Therefore, there will 2

*n*= 2 x 12 = 24 square numbers between 12 and 13.

(ii) 25 and 26

Here, *n* = 25. Therefore, there will 2*n* = 2 x 25 = 50 square numbers between 25 and 26.

(iii) 99 and 100

Here, *n* = 99. Therefore, there will 2*n* = 2 x 99 = 198 square numbers between 99 and 100.