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- Sqare Roots : Solution of NCERT Exercise 6.1

Question: 1 - What will be the unit digit of the squares of the following numbers?

(i) 81

Answer: 1

Explanation: Since, 1^{2} ends up having 1 as the digit at unit’s place so 81^{2} will have 1 at unit’s place.

(ii) 272

Asnwer: 4

Explanation: Since, 2^{2} = 4, therefore, square of 272 will have 2 at it's unit place.

So, 272^{2} will have 4 at unit’s place

(iii) 799

Answer: 1

Explanation: Since, 9^{2} = 81, So, 799^{2} will have 1 at unit’s place

(iv) 3853

Answer: 9

Explantion: Since 3^{2} = 9, so, 3853^{2} will have 9 at unit’s place.

(v) 1234

Answer: 6

Explanation: Since, 4^{2} = 16, so, 1234^{2} will have 6 at unit’s place

(vi) 26387

Answer: 9.

Explanation: Since, 7^{2} = 49. Therefore, 26387^{2} will have 9 at unit’s place

(vii) 52698

Answer: 4

Explanation: Since, 8^{2} = 64. So, 52698^{2} will have 4 at unit’s place

(viii) 99880

Answer: 0

Since, 0^{2} = 0. So, 99880^{2} will have 0 at unit’s place

(ix) 12796

Answer: 6

Explanation: Since, 6^{2} = 36. So, 12796^{2} will have 6 at unit’s place

(x) 55555

Answer: 5

Explanation:Since, 5^{2} = 25, therefore, 55555^{2} will have 5 at unit’s place

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

Answer: (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 1, 4, 5, 6, or 9 at unit’s place, so they are not be perfect squares.

(v), (vii) and (viii) don’t have even number of zeroes at the end so they are not perfect squares.

Question: 3. The squares of which of the following would be odd numbers?

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

Answer: (i) 431 and (iii) 779.

Explanation: (i) and (ii) have odd numbers as their square, because an odd number multiplied by another odd number always results in an odd number.

Question: 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} = ...............

Solution:

100001^{2} = 10000200001

10000001^{2} = 100000020000001

Explanation: Start with 1 followed as many zeroes as there are between the first and the last one, followed by two again followed by as many zeroes and end with 1.

Question: 5. Observe the following pattern and supply the missing numbers.

11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} = ..................

..............^{2} =10203040504030201

Solution:

1010101^{2} = 1020304030201

101010101^{2} =10203040504030201

Explanation: Start with 1 followed by a zero and go up to as many number as there are number of 1s given, follow the same pattern in reverse order.

Question: 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:

4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2 }+ 30^{2} = 31^{2}

6^{2} + 7^{2} + 42^{2} = 43^{2}

Relation among first, second and third number - Third number is the product of first and second number

Relation between third and fourth number - Fourth number is 1 more than the third number

Question: 7. Without adding, find the sum.

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

Answer:Since, there are 5 consecutive odd numbers, Thus, their sum = 5^{2} = 25

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

Answer:Since, there are 10 consecutive odd numbers, Thus, their sum = 10^{2} = 100

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

Answer:Since, there are 12 consecutive odd numbers, Thus, their sum = 12^{2} = 144

Explanation:

1 + 3 = 2^{2} = 4

1 + 3 + 5 = 3^{2} = 9

1 + 3 + 5 + 7 = 4^{2} =16

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

In other words this is a way of finding the sum of n odd numbers starting from 1.

Therefore, Sum of n odd numbers starting from 1 = n^{2}

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

Solution:

Since, 49 = 7^{2}

So, 7^{2} can be expressed as follows:

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

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

Solution: Since, 121 = 11^{2}

Therefore, 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

Solution:

12^{2} = 144

13^{2} = 169

Now, 169 - 144 = 25

So, there are 25 - 1 = 24 numbers lying between 12^{2} and 13^{2}

(ii) 25 and 26

Solution:

We know that, 25^{2} = 625

And, 26^{2} = 676

Now, 676 - 625 = 51

So, there are 51 - 1 = 50 numbers lying between 25^{2} and 26^{2}

(iii) 99 and 100

Solution:

We know that, 99^{2} = 9801

And, 100^{2} = 10000

Now, 10000 - 9801 = 199

So, there are 199 - 1 = 198 numbers lying between 99^{2} and 100^{2}

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