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 = 36
10. 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
Thus, Answer = 27
(ii) 400
Thus, Answer = 20
(iii) 1764
Answer:
Thus, Answer = 42
(iv) 4096
Answer:
Thus, Answer = 64
(v) 7744
Answer:
Thus, Answer = 88
(vi) 9604
Answer:
Thus, Answer = 98
(vii) 5929
Answer:
Thus, Answer = 77
(viii) 9216
Answer:
Thus, Answer = 96
(ix) 529
Answer:
Thus, Answer = 23
(x) 8100
Answer:
Thus, Answer = 90
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