The document NCERT Solutions(Part- 2)- Squares and Square Roots Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.

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**Q.1.** **Express the following as the sum of two consecutive integers.****(i) 21 ^{2} **

or 21^{2} = 220 + 221 = 441

**(ii)** n = 13

∴ 132 = 85 + 84 = 169

Similarly,

**(iii) **11^{2} = 60 + 61 = 121

**(iv)** 19^{2 }= 180 + 181 = 361

**Q.2. ****Do you think the reverse is also true, i.e. is the sum of any two consecutive positive integers is perfect square of a number? Give example to support your answer.****Solution. **No, it is not always true.**Example:****(i)** 5 + 6 = 11, 11 is not a perfect square.**(ii) **21 + 22 = 43, 43 is not a perfect square.

Property 10.The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.

**Examples:**

9^{2} – 8^{2} = 81 – 64 = 17 = 9 + 8

10^{2} – 9^{2} = 100 – 81 = 19 = 10 + 9

15^{2 }– 14^{2} = 225 – 196 = 29 = 15 + 14

1012 – 1002 = 10201 – 10000 = 201 = 101 + 100

Property 11.If (n + 1) and (n – 1) are two consecutive even or odd natural numbers, then (n + 1) X (n – 1) = n^{2}– 1.

**Example:**

10 * 12 = (11 – 1) * (11 + 1) = 11^{2} – 1

11 * 13 = (12 – 1) * (12 + 1) = 12^{2} – 1

25 * 27 = (26 – 1) * (26 + 1) = 26^{2} – 1

**Q.1.** **Write the square, making use of the above pattern.**

**(i) 111111 ^{2} **

**Solution. **

**(i)** (111111)^{2} = 12345654321

**(ii)** (1111111)^{2} = 1234567654321

**Q.2. ****Can you find the square of the following numbers using the above pattern?****(i) 6666667 ^{2} **

**Exercise 6.1****Q.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****Solution.****(i) **∵ 1 * 1 = 1

∴ The unit’s digit of (81)^{2} will be 1.

**(ii) **∵ 2 * 2 = 4

∴ The unit’s digits of (272)^{2 }will be 4.

**(iii)** Since, 9 * 9 = 81

∴ The unit’s digit of (799)^{2} will be 1.

**(iv) **Since, 3 * 3 = 9

∴ The unit’s digit of (3853)^{2 }will be 9.

**(v)** Since, 4 * 4 = 16

∴ The unit’s digit of (1234)^{2} will be 6.

**(vi)** Since 7 * 7 = 49

∴ The unit’s digit of (26387)^{2} will be 9.

**(vii) **Since, 8 * 8 = 64

∴ The unit’s digit of (52698)^{2} will be 4.

**(viii)** Since 0 * 0 = 0

∴ The unit’s digit of (99880)^{2} will be 0.

**(ix)** Since 6 * 6 = 36

∴ The unit’s digit of (12796)^{2} will be 6.

**(x) **Since, 5 * 5 = 25

∴ The unit’s digit of (55555)^{2 }will be 5.**Q.2.** **The following numbers are obviously not perfect squares. Give reason.****(i) 1057 **

**(ii) 23453 **

**(iii) 7928 **

**(iv) 222222(v) 64000 **

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

**∴ 1057 is not a perfect square**.

**(ii)** 23453

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

**∴ 23453 is not a perfect square**.

**(iii)** 7928

Since, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9).

**∴ 7928 is not a perfect square.**

**(iv) **222222

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

**∴ 222222 is not a perfect square.**

**(v) **64000

Since the number of zeros is odd.

**∴ 64000 is not a perfect square.**

**(vi) **89722

Since, the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9).

**∴ 89722 is not a perfect square.**

**(viii)** 222000

Since the number of zeros is odd.

**∴ 222000 is not a perfect square.**

**(viii) **505050

The unit’s digit is odd zero.

**∴ 505050 can not be a perfect square.****Q.3. ****The squares of which of the following would be odd numbers?****(i) 431 ****(ii) 2826 ****(iii) 7779 ****(iv) 82004**

**Solution. **Since the square of an odd natural number is odd and that of an even number is an even number.

**(i)** The square of 431 is an odd number.

[∵ 431 is an odd number.]**(ii)** The square of 2826 is an even number.

[∵ 2826 is an even number.]

**(iii) **The square of 7779 is an odd number.

[∵ 7779 is an odd number.]

**(iv)** The square of 82004 is an even number.

[∵ 82004 is an even 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} = …**

**(ii) **(10000001)^{2} = 100000020000001**Q.5.** **Observe the following pattern and supply the missing number.**

**11 ^{2} = 121**

**101 ^{2} = 10201**

**10101 ^{2} = 102030201**

**1010101 ^{2} = ………….**

**…………. ^{2 }= 10203040504030201**

**(ii)** 10203040504030201 = (101010101)^{2}**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} = __**

**Note: **__To find pattern:__

Third number is related to first and second number. How?

Fourth number is related to third number. How?**Solution.**__The missing numbers are:__

**(i)** 4^{2} + 5^{2} + 20^{2} = 21^{2}

**(ii)** 5^{2} + 6^{2} + 30^{2} = 31^{2}

**(iii)** 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.****(i)**** **The sum of first 5 odd numbers = 5^{2 }= 25

**(ii)**** **The sum of first 10 odd numbers = 10^{2 }= 100

**(iii)** The sum of first 12 odd numbers = 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.****(i)** 49 =7^{2 }= Sum of first 7 odd numbers

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

**(ii) **121 = 11^{2} = Sum of first 11 odd numbers

= 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. **Since between **n ^{2} and (n + 1)^{2}, there are 2n non-square numbers**.

**(ii)** Between 25^{2} and 26^{2}, there are 2 * 25, i.e.** 50 numbers.**

**(iii)** Between 99^{2} and 100^{2}, there are 2 * 99, i.e. **198 numbers.**

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