**Download NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.1 – Square and Square roots**

**NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1 – Square and Square roots, has been designed by the NCERT to test the knowledge of the student on the following topics : **

- Introduction
- Properties of Square Numbers
- Some More Interesting Patterns

1. Adding triangular numbers

2. Numbers between square numbers

3. Adding odd numbers

4. A sum of consecutive natural numbers

5. Product of two consecutive even or odd natural numbers

6. Some more patterns in square numbers

**Q.1 What will be the unit digit of the squares of the following numbers?**

**Solution:**

**(i) 81**

Unit’s place of 81 is 1. since square of 1 ie., 1×1 is 1, unit’s place of square of 81 will be 1.

**(ii) 272**

Unit’s place of 272 is 2. since square of 2 ie., 2×2 is 4, unit’s place of square of 272 will be 4.

**(iii) 799**

Unit’s place of 799 is 9. since square of 9 ie., 9×9 is 81, unit’s place of square of 799 will be 1. [as the square of unit’s place of 799 is ending with 1]

**(iv) 3853**

Unit’s place of 3853 is 3. since square of 3 ie., 3×3 is 9, unit’s place of square of 3853 will be 9.

**(v) 1234**

Unit’s place of 1234 is 4. since square of 4 ie., 4×4 is 16, unit’s place of square of 1234 will be 6. [as the square of unit’s place of 1234 is ending with 6]

**(vi) 26387**

Unit’s place of 26387 is 7. since square of 7 ie., 7×7 is 49, unit’s place of square of 26387 will be 9. [as the square of unit’s place of 26387 is ending with 9]

**(vii) 52698**

Unit’s place of 52698 is 8. since square of 8 ie., 8×8 is 64, unit’s place of square of 52698 will be 4. [as the square of unit’s place of 52698 is ending with 4]

**(viii) 99880**

Unit’s place of 99880 is 0. since square of 0 ie., 0×0 is 0, unit’s place of square of 99880 will be 0.

**(ix) 2796**

Unit’s place of 12796 is 6. since square of 6 ie., 6×6 is 36, unit’s place of square of 12796 will be 6. [as the square of unit’s place of 12796 is ending with 6]

**(x) 55555**

Unit’s place of 55555 is 5. since square of 5 ie., 5×5 is 25, unit’s place of square of 55555 will be 5. [as the square of unit’s place of 55555 is ending with 5]

**Q.2 The following numbers are obviously not perfect squares. Give reason.**

**Solution: **Trick: As we know that the square of numbers will end with the digits 0, 1, 4, 5, 6 or 9. Also, a perfect square will end with even number of zeroes.

**(i) 1057**

Unit’s place of 1057 is 7. Therefore, 1057 is not a perfect square.

**(ii) 23453**

Unit’s place of 23453 is 3. Therefore, 23453 is not a perfect square.

**(iii) 7928**

Unit’s place of 7928 is 8. Therefore, 7928 is not a perfect square.

**(iv) 222222**

Unit’s place of 222222 is 2. Therefore, 222222 is not a perfect square.

**(v) 64000**

64000 has three zeros at the end. Since, a perfect square cannot end with odd number of zeroes, 64000 is not a perfect square.

**(vi) 89722**

Unit’s place of 89722 is 2. Therefore, 89722 is not a perfect square.

**(vii) 222000**

222000 has three zeroes at the end. since, a perfect square cannot end with odd number of zeroes, 222000 is not a perfect square.

**(viii) 505050**

505050 has one (ie., odd number of) zero at the end. Since a perfect square cannot end with odd number of zeroes, 505050 is not a perfect square.

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

**Solution: **

We know that, odd numbers end with 1, 3, 5, 7 and 9. Unit’s place of square of a number will be the Unit’s place of square of unit’s digit of the number

**(i) 431**

Unit’s digit of square of 431 will be 1. Therefore, square of 431 would be an odd number

**(ii) 2826**

Unit’s digit of square of 2826 will be 6. Therefore, square of 2826 would not be an odd number

**(iii) 7779**

Unit’s digit of square of 7779 will be 1. Therefore, square of 7779 would be an odd number

**(iv) 82004**

Unit’s digit of square of 82004 will be 6. Therefore, 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}=1___2___1**

**Solution:**

11^{2}=121

101^{2}=10201

1001^{2}=1002001

100001^{2}=10000200001

10000001^{2}=100000020000001

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

**11 ^{2}=121**

**101 ^{2}=10201**

**10101 ^{2}=102030201**

**1010101 ^{2}=__________**

**_______ ^{2}=10203040504030201**

**Solution:**

11^{2}=121

101^{2}=10201

10101^{2}=102030201

1010101^{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}**

**Solution:**

**To find pattern: **

If we keenly observe the given pattern, third number is the product of first and second numbers whereas the fourth number is the successor of third number.

Therefore,

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} + 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**

**Solution:**

We know that, Sum of first n odd numbers is equal to square of n ie., n^{2}

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

Sum of first five odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 = 5^{2} = 25

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

Sum of first ten odd natural 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

Sum of first twelve odd natural 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.**

**Solution:**

We know that, Sum of first n odd numbers is equal to square of n ie., n^{2}

**(i)** 49

49 = 7^{2}; Therefore, 49 is the sum of first 7 odd natural numbers.

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

**(ii)** 121

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

** **

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

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

**Solution:**

We know that, there will be 2n non perfect square numbers between the squares of the numbers n and (n + 1)

Therefore, between

**(i)** 12^{2} and 13^{2}, there will be 2×12 = 24 numbers

**(ii)** 25^{2} and 26^{2}, there will be 2×25 = 50 numbers

**(iii)** 99^{2} and 100^{2}, there will be 2×99 = 198 numbers

**The next Exercise for** **NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2 – Square and Square roots can be accessed by clicking here**

** Maths – NCERT Solutions Class 8**

**Download NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.1 – Square and Square roots**

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