NCERT Solutions(Part- 1)- Squares and Square Roots Class 8 Notes | EduRev

Mathematics (Maths) Class 8

Class 8 : NCERT Solutions(Part- 1)- Squares and Square Roots Class 8 Notes | EduRev

The document NCERT Solutions(Part- 1)- Squares and Square Roots Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.
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Q.1. Find the perfect square numbers between 
(i) 30 and 40
(ii) 50 and 60.
Solution.
(i) Since,
⇒ 1 *1 = 1        
⇒ 2 * 2 = 4    
⇒ 3 * 3 = 9
⇒ 4 * 4 = 16    
⇒ 5 * 5 = 25  
⇒ 6 * 6 = 36
⇒ 7 * 7 = 49
Thus, 36 is a perfect square number between 30 and 40.

(ii) Since, 7 * 7 = 49 and 8 * 8 = 64.
It means there is no perfect number between 49 and 64, and thus there is no perfect number between 50 and 60.

Q.2. Can we say whether the following numbers are perfect squares? How do we know?

(i) 1057 
(ii) 23453 
(iii) 7928 
(iv) 222222 
(v) 1069 
(vi) 2061

Write five numbers which you can decide by looking at their one’s digit that they are not square numbers.
Solution.
(i) 1057

The ending digit is 7 (which is not one of 0, 1, 4, 5, 6, or 9)

1057 cannot be a square number.

(ii) 23453

The ending digit is 3 (which is not one of 0, 1, 4, 5, 6, and 9).

∴  23453 cannot be a square number.

(iii) 7928

The ending digit is 8 (which is not one of 0, 1, 4, 5, 6 and 9).

 7928 cannot be a square number.

(iv) 222222

The ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).

222222 cannot be a square number.

(v) 1069

The ending digit is 9.

It may or may not be a square number.

Also,

►  30 * 30 = 900

► 31 * 31 = 961

► 32 * 32 = 1024

► 33 * 33 = 1089

i.e. No natural number between 1024 and 1089 is a square number.

∴ 1069 cannot be a square number.

(vi) 2061

∵ The ending digit is 1

∴ It may or may not be a square number.

∵ 45 * 45 = 2025

and 46 * 46 = 2116

i.e. No natural number between 2025 and 2116 is a square number.

∴ 2061 is not a square number.

We can write many numbers which do not end with 0, 1, 4, 5, 6 or 9. (i.e. which are not square number).
Five such numbers can be:

1234, 4312, 5678, 87543, 1002007.

Q.3. Write five numbers that you cannot decide just by looking at their unit’s digit (or one’s place) whether they are square numbers or not.
Solution. Any natural number ending in 0, 1, 4, 5, 6 or 9 can be or cannot be a square-number.
Five such numbers are:

56790, 3671, 2454, 76555, 69209

Property 1. If a number has 1 or 9 in the unit’s place, then its square ends in 1.
Example: (1)= 1, (9)2 = 81, (11)2 = 121, (19)2 = 361, (21)2 = 441.


Q.4. Which of 1232, 772, 822, 1612, 1092 would end with digit 1?
Solution. The squares of those numbers end in 1 which ends in either 1 or 9.

∴ The squares of 161 and 109 would end in 1.

Property 2. When a square number ends in 6, then the number whose square it will have 4 or 6 in its unit place.


Q.5. Which of the following numbers would have digit 6 at unit place.
(i) 19
(ii) 242 
(iii) 262 
(iv) 362 
(v) 342
Solution.
(i)192: Unit’s place digit = 9

∴ 19would not have unit’s digit as 6.

(ii) 242: Unit’s place digit = 4

∴ 242 would have unit’s digit as 6.

(iii) 262: Unit’s place digit = 6

∴ 262 would have 6 as unit’s place.

(iv) 362: Unit place digit = 6

∴ 362 would end in 6.

(v) 342: Since the unit place digit is 4

∴ 342 would have unit place digit as 6.

Q.6. What will be the “one’s digit” in the square of the following numbers?

(i) 1234 
(ii) 26387 
(iii) 52698 
(iv) 99880 
(v) 21222 
(vi) 9106
Solution.
(i) ∵ Ending digit = 4 and 42 = 16

∴ (1234)2 will have 6 as the one’s digit.

(ii) ∵ Ending digit is 7 and 7= 49

∴  (26387)2 will have 9 as the one’s digit.

(iii) ∵ Ending digit is 8, and 82 = 64

∴ (52692)2 will end in 4.

(iv) ∵ Ending digit is 0.

∴ (99880)2 will end in 0.

(v) ∵ 22 = 4

∴ Ending digit of (21222)2 is 4.

(vi) ∵ 62 = 36

∴ Ending digit of (9106)is 6.

Property 3. A square number can only have even number of zeros at the end.

Property 4. The squares of odd numbers are odd and the squares of even numbers are even.


Q.7. The square of which of the following numbers would be an odd number/an even number? Why?

(i) 727 
(ii) 158 
(iii) 269 
(iv) 1980

Solution.

(i) 727 

Since 727 is an odd number.

∴ Its square is also an odd number.

(ii) 158

Since 158 is an even number.

∴ Its square is also an even number.

(iii) 269

Since 269 is an odd number.

∴ Its square is also an odd number.

(iv) 1980

Since 1980 is an even number.

∴ Its square is also an even number.

Q.8. What will be the number of zeros in the square of the following numbers?

(i) 60 
(ii) 400

Solution. 

(i) In 60, the number of zero is 1.

∴ Its square will have 2 zeros.

(ii) ∵ There are 2 zeros in 400.

∴ Its square will have 4 zeros.

Property 5. The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.

Property 6. There are 2n non-perfect square numbers between the squares of the numbers n and n + 1.


Q.9. How many natural numbers lie between 92 and 102? Between 112 and 122?

Solution.
(a) Between 92 and 102

Here, n = 9 and n + 1 = 10

∴ Natural number between 92 and 102 are (2 * n) or 2 * 9, i.e. 18.

(b) Between 112 and 122

Here, n = 11 and n + 1 = 12

∴ Natural numbers between 11and 122 are (2 * n) or 2 *11, i.e. 22.

Q.10. How many non-square numbers lie between the following pairs of numbers:

(i) 100and 1012       
(ii) 902 and 912      
(iii) 10002 and 10012

Solution. 
(i) Between 1002 and 1012

Here, n = 100
∴ n * 2 = 100 * 2 = 200

∴ 200 non-square numbers lie between 1002 and 1012.

(ii) Between 902 and 912

Here, n = 90
∴ 2 * n = 2 * 90 or 180

∴ 180 non-square numbers lie between 90 and 91.

(iii) Between 10002 and 10012

Here, n = 1000
∴ 2 * n = 2 * 1000 or 2000

∴  2000 non-square numbers lie between 10002 and 10012.

Property 7. The sum of first n odd natural numbers is n2.
OR
If there is a square number, it has to be the sum of the successive odd numbers starting from 1.


Q.11. Find whether each of the following numbers is a perfect square or not?

(i) 121  
(ii) 55  
(iii) 81  
(iv) 49  
(v)  69

Solution. Remember: If a natural number cannot be expressed as a sum of successive odd natural numbers starting from 1, then it is not a perfect square.

(i) 121

∵ 121 – 1 = 120
► 120 – 3 = 117
► 117 – 5 = 112
► 112 – 7 = 105
► 105 – 9 = 96
► 96 – 11 = 85
► 85 – 13 = 72
► 72 – 15 = 57
► 57 – 17 = 40
► 40 – 19 = 21
► 21 – 21 = 0

i.e. 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21. Thus, 121 is a perfect square.

(ii) 55

∵  55 – 1 = 54     30 – 11 = 19

► 54 – 3 = 51     19 – 13 = 6

► 51 – 5 = 46      6 – 15 = –9

► 46 – 7 = 39

► 39 – 9 = 30

Since, 55 cannot be expressed as the sum of successive odd numbers starting from 1.

∴ 55 is not a perfect square.

(iii) 81

Since,
► 81 – 1 = 80
► 80 – 3 = 77
► 77 – 5 = 72
► 72 – 7 = 65
► 65 – 9 = 56
► 56 – 11 = 45
► 45 – 13 = 32
► 32 – 15 = 17
► 17 – 17 = 0

∴ 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17.

Thus, 81 is a perfect square.

(iv) 49

Since, 49 – 1 = 48

          ► 48 – 3 = 45

          ► 45 – 5 = 40

          ► 40 – 7 = 33

          ► 33 – 9 = 24

          ► 24 – 11 = 13

          ► 13 – 13 = 0

∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

Thus, 49 is a perfect square.

(v) 69

Since
► 69 – 1 = 68
► 68 – 3 = 65
► 65 – 5 = 60
► 60 – 7 = 53
► 53 – 9 = 44
► 44 – 11 = 33
► 33 – 13 = 20
► 20 – 15 = 5
► 5 – 17 = –12

∴ 69 cannot be expressed as the sum of consecutive odd numbers starting from 1.

Thus, 69 is not a perfect square.

Property 8. The square of an odd number can be expressed as the sum of two consecutive natural numbers.

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