The document RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.2) Part - 1, Class 8 Math Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

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(i) 1547

(ii) 45743

(iii) 8948

(iv) 333333

(i) Its last digit is 7. Hence, 1547 cannot be a perfect square.

(ii) Its last digit is 3. Hence, 45743 cannot be a perfect square.

(iii) Its last digit is 8. Hence, 8948 cannot be a perfect square.

(iv) Its last digit is 3. Hence, 333333 cannot be a perfect square.

(i) 9327

(ii) 4058

(iii) 22453

(iv) 743522

(i) Its last digit is 7. Hence, 9327 is not a perfect square.

(ii) Its last digit is 8. Hence, 4058 is not a perfect square.

(iii) Its last digit is 3. Hence, 22453 is not a perfect square.

(iv) Its last digit is 2. Hence, 743522 is not a perfect square.

(i) 731

(ii) 3456

(iii) 5559

(iv) 42008

(i) 731 is an odd number. Hence, its square will be an odd number.

(ii) 3456 is an even number. Hence, its square will not be an odd number.

(iii) 5559 is an odd number. Hence, its square will not be an odd number.

(iv) 42008 is an even number. Hence, its square will not be an odd number.

Hence, only the squares of 731 and 5559 will be odd numbers.

(i) 52

(ii) 977

(iii) 4583

(iv) 78367

(v) 52698

(vi) 99880

(vii) 12796

(viii) 55555

(ix) 53924

(i) Its last digit is 2. Hence, the units digit is 2

(ii) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 7

(iii) Its last digit is 3. Hence, the units digit is 3

(iv) Its last digit is 7. Hence, the units digit is the last digit of 49 (49 = 7

(v) Its last digit is 8. Hence, the units digit is the last digit of 64 (64 = 8

(vi) Its last digit is 0. Hence, the units digit is 0

(vii) Its last digit is 6. Hence, the units digit is the last digit of 36 (36 = 6

(viii) Its last digit is 5. Hence, the units digit is the last digit of 25 (25 = 5

(ix) Its last digit is 4. Hence, the units digit is the last digit of 16 (16 = 4

1 + 3 = 2

1 + 3 + 5 = 3

1 + 3 Ã— 5 + 7 = 4

and write the value of 1 + 3 + 5 + 7 + 9 + ... upto

1 + 3 + 5 + 7 + ...

2

3

4

5

and find the value of

(i) 100

(ii) 111

(iii) 99

In a formula:

(n+1)2 âˆ’ (n)2 = (n+1) + n

Using this formula, we get:

(i) 100

= 199

(ii) 111

= (111 + 110) + (110 + 109)

= 440

(iii) 99

= 99 + 98 + 98 + 97 + 97 + 96

= 585

(i) (8, 15, 17)

(ii) (18, 80, 82)

(iii) (14, 48, 51)

(iv) (10, 24, 26)

(v) (16, 63, 65)

(vi) (12, 35, 38)

A triplet (

(i) The two smallest numbers are 8 and 15. The sum of their squares is:

8

Hence, (8, 15, 17) is a Pythagorean triplet.

(ii) The two smallest numbers are 18 and 80. The sum of their squares is:

18

Hence, (18, 80, 82) is a Pythagorean triplet.

(iii) The two smallest numbers are 14 and 48. The sum of their squares is:

14

Hence, (14, 48, 51) is not a Pythagorean triplet.

(iv) The two smallest numbers are 10 and 24. The sum of their squares is:

10

Hence, (10, 24, 26) is a Pythagorean triplet.

(v) The two smallest numbers are 16 and 63. The sum of their squares is:

16

Hence, (16, 63, 65) is a Pythagorean triplet.

(vi) The two smallest numbers are 12 and 35. The sum of their squares is:

12

Hence, (12, 35, 38) is not a Pythagorean triplet.