# 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

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

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

(iii) 998001

(iv) 657666025

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

Question 4: Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729

(ii) 400

(iii) 1764

(iv) 4096

(v) 7744

(vi) 9604

(vii) 5929

(viii) 9216

(ix) 529