Square Roots

Exercise 6.3 Part 1

Question 1: What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801

Answer: 1 and 9.
Explanation: Since 1² and 9² give 1 at unit’s place, so these are the possible values of unit digit of the square root.

(ii) 99856

Answer: 4 and 6
Explanation: Since, 4²= 16 and 6²= 36, hence, 4 and 6 are possible digits

(iii) 998001

Answer: 1 and 9

(iv) 657666025

Answer: 5
Explanation: Since, 5²= 25, hence 5 is possible.

Question 2: Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Answer: (i) 153 (ii) 257 (iii) 408
Explanation: Since, (i), (ii) and (iii) are surely not be perfect square as these numbers end with 3, 7 and 8. A number can be a perfect square if it ends with 0, 1, 4, 5, 6, 9 only

Question 3: Find the square roots of 100 and 169 by the method of repeated subtraction.

Answer: Square root of 100 by Repeated subtraction:
1. 100 - 1 = 99
2. 99 - 3 = 96
3. 96 - 5 = 91
4. 91 -7 = 84
5. 84 - 9 = 75
6. 75 - 11 = 64
7. 64 - 13 = 51
8. 51 - 15 = 3610. 19 - 19 = 0
We get 0 at 10th step. Thus, 10 is the square root of 100.

Square root of 169 by Repeated subtraction:
1. 169 - 1 = 168
2. 168 - 3 = 165
3. 165 - 5 = 160
4. 160 - 7 = 153
5. 153 - 9 = 144
6. 144 - 11 = 133
7. 133 - 13 = 120
8. 120 - 15 = 105
9. 105 - 17 = 88
10. 88 - 19 = 69
11. 69 - 21 = 48
12. 48 - 23 = 25
13. 25 - 25 = 0
We get 0 at 13th step. Thus 13 is the square root of 169