Exercise 6.1 - Chapter 6 Square & Square Roots class 8 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 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

Answer:

We know that if a number has its unit’s place digit as a, then its square will end with the unit digit of the multiplication a × a.

(i) 81

Since the given number has its unit’s place digit as 1, its square will end with the unit digit of the multiplication (1 ×1 = 1) i.e., 1.

(ii) 272

Since the given number has its unit’s place digit as 2, its square will end with the unit digit of the multiplication (2 × 2 = 4) i.e., 4.

(iii) 799

Since the given number has its unit’s place digit as 9, its square will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.

(iv) 3853

Since the given number has its unit’s place digit as 3, its square will end with the unit digit of the multiplication (3 × 3 = 9) i.e., 9.

(v) 1234

Since the given number has its unit’s place digit as 4, its square will end with the unit digit of the multiplication (4 × 4 = 16) i.e., 6.

(vi) 26387

Since the given number has its unit’s place digit as 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.

(vii) 52698

Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.

(viii) 99880

Since the given number has its unit’s place digit as 0, its square will have two zeroes at the end. Therefore, the unit digit of the square of the given number is 0.

(xi) 12796

Since the given number has its unit’s place digit as 6, its square will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.

(x) 55555

Since the given number has its unit’s place digit as 5, its square will end with the unit digit of the multiplication (5 × 5 = 25) i.e., 5.

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:

The square of numbers may end with any one of the digits 0, 1, 5, 6, or 9. Also, a perfect square has even number of zeroes at the end of it.

(i) 1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.

(ii) 23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.

(iii) 7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.

(iv) 222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.

(v) 64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

(vi) 89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.

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

(viii) 505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

Question 3:

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

(i) 431 (ii) 2826

(iii) 7779 (iv) 82004

Answer:

The square of an odd number is odd and the square of an even number is even. Here, 431 and 7779 are odd numbers.

Thus, the square of 431 and 7779 will be an odd number.

Question 4:

Observe the following pattern and find the missing digits.

11² = 121

101² = 10201

1001² = 1002001

100001² = 1…2…1

10000001² = …

Answer:

In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,

100001² = 10000200001

10000001² = 100000020000001

Question 5:

Observe the following pattern and supply the missing number.

11² = 121

101² = 10201

10101² = 102030201

1010101² = …

…² = 10203040504030201

Answer:

By following the given pattern, we obtain

1010101² = 1020304030201

101010101² = 10203040504030201

Question 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

5² + _ ² + 30² = 31²

6² + 7² + _ ² = __²

Answer:

From the given pattern, it can be observed that,

(i) The third number is the product of the first two numbers.

(ii) The fourth number can be obtained by adding 1 to the third number.

Thus, the missing numbers in the pattern will be as follows.

4² + 5² +   = 21²

5² +   + 30² = 31²

6² + 7² +   = 

Question 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

Answer:

We know that the sum of first n odd natural numbers is n².

(i) Here, we have to find the sum of first five odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 = (5)² = 25

(ii) Here, we have to find the sum of first ten odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)² = 100

(iii) Here, we have to find the sum of first twelve odd natural numbers.

Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)² = 144

Question 8:

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

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

Answer:

We know that the sum of first n odd natural numbers is n².

(i) 49 = (7)²

Therefore, 49 is the sum of first 7 odd natural numbers.

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

(ii) 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

Question 9:

How many numbers lie between squares of the following numbers?

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

Answer:

We know that there will be 2n numbers in between the squares of the numbers n and (n + 1).

(i) Between 12² and 13², there will be 2 × 12 = 24 numbers

(ii) Between 25² and 26², there will be 2 × 25 = 50 numbers

(iii) Between 99² and 100², there will be 2 × 99 = 198 numbers