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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**Question 1.** **Express the following as the sum of two consecutive integers.**

**(i) 21 ^{2} (ii) 13^{2} (iii) 11^{2} (iv) 19**

**Solution:**

**(i) **n = 21

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

**(ii)** n = 13

âˆ´ 132 = 85 + 84 = 169

Similarly,

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

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

**Question 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.

For 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} â€“ 8^{2} = 81 â€“ 64 = 17 = 9 + 8

10^{2} â€“ 9^{2} = 100 â€“ 81 = 19 = 10 + 9

15^{2 }â€“ 14^{2} = 225 â€“ 196 = 29 = 15 + 14

1012 â€“ 1002 = 10201 â€“ 10000 = 201 = 101 + 100

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

**For example,**

10 * 12 = (11 â€“ 1) * (11 + 1) = 11^{2} â€“ 1

11 * 13 = (12 â€“ 1) * (12 + 1) = 12^{2} â€“ 1

25 * 27 = (26 â€“ 1) * (26 + 1) = 26^{2} â€“ 1

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

**(i) 111111 ^{2} (ii) 1111111**

**Solution:** Using above pattern we can write,

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

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

**Question 2: ****Can you find the square of the following numbers using the above pattern?**

**(i) 6666667 ^{2} (ii) 66666667**

**Solution:** Using the above pattern, we can have,

**(i) **6666667^{2 }= 44444448888889

**(ii)** 66666667^{2} = 4444444488888889

**Exercise 6.1**

**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**

**Solution: **

**(i) **âˆµ 1 * 1 = 1

âˆ´ The unitâ€™s digit of (81)^{2} will be 1.

**(ii) **âˆµ 2 * 2 = 4

âˆ´ The unitâ€™s digits of (272)^{2 }will be 4.

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

âˆ´ The unitâ€™s digit of (799)^{2} will be 1.

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

âˆ´ The unitâ€™s digit of (3853)^{2 }will be 9.

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

âˆ´ The unitâ€™s digit of (1234)^{2} will be 6.

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

âˆ´ The unitâ€™s digit of (26387)^{2} will be 9.

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

âˆ´ The unitâ€™s digit of (52698)^{2} will be 4.

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

âˆ´ The unitâ€™s digit of (99880)^{2} will be 0.

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

âˆ´ The unitâ€™s digit of (12796)^{2} will be 6.

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

âˆ´ The unitâ€™s digit of (55555)^{2 }will be 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**

**Solution: **

**(i) **1057

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

âˆ´ 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).

âˆ´ 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).

âˆ´ 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).

âˆ´ 222222 is not a perfect square.

**(v) **64000

Since, the number of zeros is odd.

âˆ´ 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).

âˆ´ 89722 is not a perfect square.

**(viii)** 222000

Since, the number of zeros is odd.

âˆ´ 222000 is not a perfect square.

**(viii) **505050

The unitâ€™s digit is odd zero.

âˆ´ 505050 can not be a perfect square.

**Question 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.

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

[âˆµ 431 is an odd number.]

** (ii)** The square of 2826 is an even number.

[âˆµ 2826 is an even number.]

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

[âˆµ 7779 is an odd number.]

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

[âˆµ 82004 is an even number.]

**Question 4. ****Observe the following pattern and find the missing digits.**

**11 ^{2} = 121**

**101 ^{2} = 10201**

**1001 ^{2} = 1002001**

**100001 ^{2} = 1 â€¦ 2 â€¦1**

**10000001 ^{2} = â€¦**

**Solution:** Observing the above pattern, we have

**(i) **(100001)^{2} = 10000200001

**(ii) **(10000001)^{2} = 100000020000001

**Question 5.** **Observe the following pattern and supply the missing number.**

**11 ^{2} = 121**

**101 ^{2} = 10201**

**10101 ^{2} = 102030201**

**1010101 ^{2} = â€¦â€¦â€¦â€¦.**

**â€¦â€¦â€¦â€¦. ^{2 }= 10203040504030201**

**Solution:** Observing the above, we have

(i) (1010101)^{2 }= 1020304030201

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

**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} = __**

**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}

**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**

**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

**Question 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

**Question 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.

âˆ´ (i) Between 12^{2 }and 13^{2}, there are 2 *12, i.e. 24 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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