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)^{2 }= 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 ^{2 }**

∴ 19^{2 }would not have unit’s digit as 6.

**(ii)** 24^{2}: Unit’s place digit = 4

∴ 24^{2} would have unit’s digit as 6.

**(iii)** 26^{2}: Unit’s place digit = 6

∴ 26^{2} would have 6 as unit’s place.

**(iv)** 36^{2}: Unit place digit = 6

∴ 36^{2} would end in 6.

**(v) **34^{2}: Since the unit place digit is 4

∴ 34^{2} 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 4^{2} = 16

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

**(ii)** ∵ Ending digit is 7 and 7^{2 }= 49

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

**(iii) **∵ Ending digit is 8, and 8^{2} = 64

∴ (52692)^{2} will end in 4.

**(iv) **∵ Ending digit is 0.

∴ (99880)^{2} will end in 0.

**(v)** ∵ 2^{2} = 4

∴ Ending digit of (21222)^{2} is 4.

**(vi) **∵ 6^{2} = 36

∴ Ending digit of (9106)^{2 }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 9 ^{2} and 10^{2}? Between 11^{2} and 12^{2}?**

**Solution.****(a)** Between 9^{2} and 10^{2}

Here, n = 9 and n + 1 = 10

**∴ Natural number between 9 ^{2} and 10^{2} are (2 * n) or 2 * 9, i.e. 18.**

**(b)** Between 11^{2} and 12^{2}

Here, n = 11 and n + 1 = 12

**∴ Natural numbers between 11 ^{2 }and 12^{2} are (2 * n) or 2 *11, i.e. 22**.

**(i) 100 ^{2 }and 101^{2} **

**Solution.** **(i)** Between 100^{2} and 101^{2}

Here, n = 100

∴ n * 2 = 100 * 2 = 200

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

**(ii)** Between 90^{2} and 91^{2}

Here, n = 90

∴ 2 * n = 2 * 90 or 180

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

**(iii) **Between 1000^{2} and 1001^{2}

Here, n = 1000

∴ 2 * n = 2 * 1000 or 2000

**∴ 2000 non-square numbers lie between 1000 ^{2} and 1001**

Property 7.The sum of first n odd natural numbers is n^{2}.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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