Q2. 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
Ans:
(i) 1057
Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)
∴ 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).
∴ 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).
∴ 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).
∴ 222222 is not a perfect square.
(v) 64000
Since the number of zeros is odd.
∴ 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).
∴ 89722 is not a perfect square.
(viii) 222000
Since the number of zeros is odd.
∴ 222000 is not a perfect square.
(viii) 505050
The unit’s digit is odd zero.
∴ 505050 can not be a perfect square.
Q3. The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Ans: Since the square of an odd natural number is odd and that of an even number is an even number.
(i) The square of 431 is an odd number.
[∵ 431 is an odd number.]
(ii) The square of 2826 is an even number.
[∵ 2826 is an even number.]
(iii) The square of 7779 is an odd number.
[∵ 7779 is an odd number.]
(iv) The square of 82004 is an even number.
[∵ 82004 is an even number.]
Q4. Observe the following pattern and find the missing digits.
11^{2} = 1 2 1
101^{2} = 10201
1001^{2} = 1002001
100001^{2} = 1 … 2 …1
10000001^{2} = …
Ans: We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.
∴ (100001)^{2} = 10000200001
(10000001)^{2} = 100000020000001
Q5. Observe the following pattern and supply the missing number.
11^{2} = 1 2 1
101^{2} = 1 0 2 0 1
10101^{2} = 102030201
1010101^{2} = ………….
………….^{2 }= 10203040504030201
Ans: Observing the above, we have:
(i) (1010101)^{2 }= 1020304030201
(ii) 10203040504030201 = (101010101)^{2}
Q6. 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} = __^{2}
Note: To find pattern:
Third number is related to first and second number. How?
Fourth number is related to third number. How?
Ans: Given, 1^{2} + 2^{2} + 2^{2} = 3^{2}
i.e 1^{2} + 2^{2} + (1×2 )^{2} = ( 1^{2} + 2^{2} 1 × 2 )^{2}
2^{2} + 3^{2} + 6^{2} =7^{2}
∴ 2^{2} + 3^{2} + (2×3 )^{2} = (2^{2} + 3^{2} 2 × 3)^{2}
3^{2 }+ 4^{2} + 12^{2} = 13^{2}
∴ 3^{2} + 4^{2} + (3×4 )^{2} = (3^{2} + 4^{2} – 3 × 4)^{2}
4^{2} + 5^{2} + (4×5 )^{2} = (4^{2} + 5^{2} – 4 × 5)^{2}
∴ 4^{2} + 5^{2} + 20^{2} = 21^{2}
5^{2} + 6^{2} + (5×6 )^{2} = (5^{2}+ 6^{2} – 5 × 6)^{2}
∴ 5^{2} + 6^{2} + 30^{2} = 31^{2}
6^{2} + 7^{2} + (6×7 )^{2} = (6^{2} + 7^{2} – 6 × 7)^{2}
∴ 6^{2} + 7^{2} + 42^{2} = 43^{2}
Q7. 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
Ans:
(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
Q8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Ans:
(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
Q9. 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 nonsquare numbers.
(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.
Q.1. Find the perfect square numbers between
(i) 30 and 40
(ii) 50 and 60.
Solution.
(i) Since,
⇒ 1 *1 = 1
⇒ 2 * 2 = 4
⇒ 3 * 3 = 9
⇒ 4 * 4 = 16
⇒ 5 * 5 = 25
⇒ 6 * 6 = 36
⇒ 7 * 7 = 49
Thus, 36 is a perfect square number between 30 and 40.
(ii) Since, 7 * 7 = 49 and 8 * 8 = 64.
It means there is no perfect number between 49 and 64, and thus there is no perfect number between 50 and 60.
Q.2. Can we say whether the following numbers are perfect squares? How do we know?
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 1069
(vi) 2061
Write five numbers which you can decide by looking at their one’s digit that they are not square numbers.
Solution.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 1069
Also,
i.e. No natural number between 1024 and 1089 is a square number.
∴ 1069 cannot be a square number.
(vi) 2061
45 * 45 = 2025
and 46 * 46 = 2116
i.e. No natural number between 2025 and 2116 is a square number.
∴ 2061 is not a square number.
We can write many numbers which do not end with 0, 1, 4, 5, 6 or 9. (i.e. which are not square number).
Five such numbers can be:
1234, 4312, 5678, 87543, 1002007.
Q.3. Write five numbers that you cannot decide just by looking at their unit’s digit (or one’s place) whether they are square numbers or not.
Solution. Any natural number ending in 0, 1, 4, 5, 6 or 9 can be or cannot be a squarenumber.
Five such numbers are:
56790, 3671, 2454, 76555, 69209
Property 1. If a number has 1 or 9 in the unit’s place, then its square ends in 1.
Example: (1)^{2 }= 1, (9)^{2} = 81, (11)^{2} = 121, (19)^{2} = 361, (21)^{2} = 441.
Q.4. Which of 123^{2}, 77^{2}, 82^{2}, 161^{2}, 109^{2} would end with digit 1?
Solution. The squares of those numbers end in 1 which ends in either 1 or 9.
∴ The squares of 161 and 109 would end in 1.
Property 2. When a square number ends in 6, then the number whose square it will have 4 or 6 in its unit place.
Q.5. Which of the following numbers would have digit 6 at unit place.
(i) 19^{2 }
(ii) 24^{2}
(iii) 26^{2}
(iv) 36^{2}
(v) 34^{2}
Solution.
(i) 19^{2}: Unit’s place digit = 9
∴ 19^{2 }would not have unit’s digit as 6.
(ii) 24^{2}: Unit’s place digit = 4
∴ 24^{2} would have unit’s digit as 6.
(iii) 26^{2}: Unit’s place digit = 6
∴ 26^{2} would have 6 as unit’s place.
(iv) 36^{2}: Unit place digit = 6
∴ 36^{2} would end in 6.
(v) 34^{2}: Since the unit place digit is 4
∴ 34^{2} would have unit place digit as 6.
Q.6. What will be the “one’s digit” in the square of the following numbers?
(i) 1234
(ii) 26387
(iii) 52698
(iv) 99880
(v) 21222
(vi) 9106
Solution.
(i) Since Ending digit = 4 and 4^{2} = 16
∴ (1234)^{2} will have 6 as the one’s digit.
(ii) Since Ending digit is 7 and 7^{2 }= 49
∴ (26387)^{2} will have 9 as the one’s digit.
(iii) Since Ending digit is 8, and 8^{2} = 64
∴ (52692)^{2} will end in 4.
(iv) Since Ending digit is 0.
∴ (99880)^{2} will end in 0.
(v) Since Ending digit = 2 and 2^{2} = 4
∴ Ending digit of (21222)^{2} is 4.
(vi) ∵ 6^{2} = 36
∴ Ending digit of (9106)^{2 }is 6.
Property 3. A square number can only have even number of zeros at the end.
Property 4. The squares of odd numbers are odd and the squares of even numbers are even.
Q.7. The square of which of the following numbers would be an odd number/an even number? Why?
(i) 727
(ii) 158
(iii) 269
(iv) 1980
Solution.
(i) 727
Since 727 is an odd number.
∴ Its square is also an odd number.
(ii) 158
Since 158 is an even number.
∴ Its square is also an even number.
(iii) 269
Since 269 is an odd number.
∴ Its square is also an odd number.
(iv) 1980
Since 1980 is an even number.
∴ Its square is also an even number.
Q.8. What will be the number of zeros in the square of the following numbers?
(i) 60
(ii) 400
Solution.
(i) In 60, the number of zero is 1.
∴ Its square will have 2 zeros.
(ii) Since there are 2 zeroes in 400.
∴ Its square will have 4 zeros.
Property 5. The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.
Property 6. There are 2n nonperfect square numbers between the squares of the numbers n and n + 1.
Q.9. How many natural numbers lie between 9^{2} and 10^{2}? Between 11^{2} and 12^{2}?
Solution.
(a) Between 9^{2} and 10^{2}
Here, n = 9 and n + 1 = 10
∴ Natural number between 9^{2} and 10^{2} are (2 * n) or 2 * 9, i.e. 18.
(b) Between 11^{2} and 12^{2}
Here, n = 11 and n + 1 = 12
∴ Natural numbers between 11^{2 }and 12^{2} are (2 * n) or 2 *11, i.e. 22.
Q.10. How many nonsquare numbers lie between the following pairs of numbers:
(i) 100^{2 }and 101^{2}
(ii) 90^{2} and 91^{2}
(iii) 1000^{2} and 1001^{2}
Solution.
(i) Between 100^{2} and 101^{2}
Here, n = 100
∴ n * 2 = 100 * 2 = 200
∴ 200 nonsquare numbers lie between 1002 and 1012.
(ii) Between 90^{2} and 91^{2}
Here, n = 90
∴ 2 * n = 2 * 90 or 180
∴ 180 nonsquare numbers lie between 90 and 91.
(iii) Between 1000^{2} and 1001^{2}
Here, n = 1000
∴ 2 * n = 2 * 1000 or 2000
∴ 2000 nonsquare numbers lie between 1000^{2} and 1001^{2}.
Property 7. The sum of first n odd natural numbers is n^{2}.
OR
If there is a square number, it has to be the sum of the successive odd numbers starting from 1.
Q.11. Find whether each of the following numbers is a perfect square or not?
(i) 121
(ii) 55
(iii) 81
(iv) 49
(v) 69
Solution. Remember: If a natural number cannot be expressed as a sum of successive odd natural numbers starting from 1, then it is not a perfect square.
(i) 121
Since,
121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
i.e. 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21. Thus, 121 is a perfect square.
(ii) 55
Since,
55 – 1 = 54
54 – 3 = 51
51 – 5 = 46
46 – 7 = 39
39 – 9 = 30
30 – 11 = 19
19 – 13 = 6
6 – 15 = –9
Since, 55 cannot be expressed as the sum of successive odd numbers starting from 1. Thus, 55 is not a perfect square.
(iii) 81
Since,
81 – 1 = 80
80 – 3 = 77
77 – 5 = 72
72 – 7 = 65
65 – 9 = 56
56 – 11 = 45
45 – 13 = 32
32 – 15 = 17
17 – 17 = 0
∴ 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17. Thus, 81 is a perfect square.
(iv) 49
Since,
49 – 1 = 48
48 – 3 = 45
45 – 5 = 40
40 – 7 = 33
33 – 9 = 24
24 – 11 = 13
13 – 13 = 0
∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13. Thus, 49 is a perfect square.
(v) 69
Since,
69 – 1 = 68
68 – 3 = 65
65 – 5 = 60
60 – 7 = 53
53 – 9 = 44
44 – 11 = 33
33 – 13 = 20
20 – 15 = 5
5 – 17 = –12
∴ 69 cannot be expressed as the sum of consecutive odd numbers starting from 1. Thus, 69 is not a perfect square.
Property 8. The square of an odd number can be expressed as the sum of two consecutive natural numbers.
or 21^{2} = 220 + 221 = 441
(ii) n = 13
∴ 132 = 85 + 84 = 169
Similarly,
(iii) 11^{2} = 60 + 61 = 121
(iv) 19^{2 }= 180 + 181 = 361
Q2. 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.
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} – 8^{2} = 81 – 64 = 17 = 9 + 8
10^{2} – 9^{2} = 100 – 81 = 19 = 10 + 9
15^{2 }– 14^{2} = 225 – 196 = 29 = 15 + 14
1012 – 1002 = 10201 – 10000 = 201 = 101 + 100
Property 11. If (n + 1) and (n – 1) are two consecutive even or odd natural numbers, then (n + 1) X (n – 1) = n^{2} – 1.
Example:
10 * 12 = (11 – 1) * (11 + 1) = 11^{2} – 1
11 * 13 = (12 – 1) * (12 + 1) = 12^{2} – 1
25 * 27 = (26 – 1) * (26 + 1) = 26^{2} – 1
Q3. Write the square, making use of the above pattern.
(i) 111111^{2}
(ii) 1111111^{2}
Solution: Using above pattern we can write:
(i) (111111)^{2} = 12345654321
(ii) (1111111)^{2} = 1234567654321
Q4. Can you find the square of the following numbers using the above pattern?
(i) 6666667^{2}
(ii) 66666667^{2}
Solution: Using the above pattern, we can have:
(i) 6666667^{2 }= 44444448888889
(ii) 66666667^{2} = 4444444488888889
(i) (32)^{2} = (30 + 2)^{2}
= 30^{2} + 2(30)(2) + (2)^{2}
= 900 + 120 + 4 = 1024
(ii) (35)^{2 }= (30 + 5)^{2}
= (30)^{2} + 2(30)(5) + (5)^{2}
= 900 + 300 + 25
= 1200 + 25 = 1225
Second method
35^{2} = 3 * (3 + 1) * 100 + 25
= 3 * 4 * 100 + 25
= 1200 + 25 = 1225
(iii) (86)^{2} = (80 + 6)^{2}
= (80)^{2} + 2(80)(6) + (6)^{2}
= 6400 + 960 + 36 = 7396
(iv) (93)^{2} = (90 + 3)^{2}
= (90)^{2} + 2(90)(3) + (3)^{2}
= 8100 + 540 + 9 = 8649
(v) (71)^{2} = (70 + 1)^{2}
= (70)^{2 }+ 2(70)(1) + (1)^{2}
= 4900 + 140 + 1 = 5041
(vi) (46)^{2} = (40 + 6)^{2}
= (40)^{2} + 2(40)(6) + (6)^{2}
= 1600 + 480 + 36 = 2116
Q2. Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Ans:
(i) Let 2n = 6 ∴n = 3
Now, n^{2} – 1 = 3^{2} – 1 = 8
and n^{2} + 1 = 3^{2} + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.
(ii) Let 2n = 14 ∴ n = 7
Now, n^{2} – 1 = 7^{2} – 1 = 48
and n^{2} + 1 = 7^{2} + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.
(iii) Let 2n = 16 ∴n = 8
Now, n^{2} – 1 = 8^{2} – 1
= 64 – 1 = 63
and n^{2} + 1 = 8^{2} + 1
= 64 + 1 = 65
Thus, the required Pythagorean triplet is 16, 63, 65.
(iv) Let 2n = 18 ∴n = 9
Now, n^{2} – 1 = 9^{2} – 1
= 81 – 1 = 80
and n^{2 }+ 1 = 9^{2} + 1
= 81 + 1 = 82
Thus, the required Pythagorean triple is 18, 80, 82.
Finding the Square of a number
Example. Find the square of 27.
Solution: (27)^{2 }= (20 + 7)^{2}
Using the formula (a + b)^{2} = a^{2 }+ 2ab + b^{2}, we have
(20 + 7)^{2} = (20)^{2} + 2 * (20) * (7) + (7)^{2}
= 400 + 280 + 49
= 729
Thus, (27)^{2} = 729
Note: For any number ending with 5, the square is a(a + 1) hundred + 25.
For example,
(25)^{2} = 2(2 + 1) * 100 + 25 = 625
(35)^{2} = 3(3 + 1) * 100 + 25 = 1225
(65)^{2} = 6(6 + 1) *100 + 25 = 4225
(125)^{2} = 12(12 + 1) * 100 + 25 = 15625
Pythagorean Triplets
If three numbers a, b and c are such that a^{2} + b^{2} = c^{2}, then they are called Pythagorean Triplets and they represent the sides of a right triangle.
Example:
(i) 3, 4, 5 form a Pythagorean triplet.
[∵ 3^{2} + 4^{2} = 5^{2}]
(ii) 8, 15, 17 form a Pythagorean triplet.
[∵ 8^{2} + 15^{2} = 17^{2}]
Note: For any natural number n, (n > 1), we have
(2n)^{2} + (n^{2} – 1)^{2} = (n^{2} + 1)^{2}
such that 2n, n^{2} – 1 and n^{2} + 1 are a Pythagorean triplet.
Example. Write a Pythagorean triplet whose one member is 15.
Solution: Since, a Pythagorean triplet is given by 2n, n^{2} – 1 and n^{2} + 1.
∴ 2n = 15 or n = 15/2 is not an integer.
So, let us assume that
n^{2} – 1 = 15
or n^{2} = 15 + 1 = 16
or n^{2} = 4^{2}, i.e. n = 4
Now, the required Pythagorean triplet is
2n, n^{2} – 1 and n^{2} + 1
or 2(4), 4^{2 }– 1 and 4^{2} + 1
or 8, 15 and 17
Remember
All Pythagorean triplets may not be obtained using the above form.
Question: Find the square of the following numbers containing 5 in unit’s place.
(i) 15 (ii) 95 (iii) 105 (iv) 205
Solution:
(i) (15)^{2} = 1 * (1 + 1) * 100 + 25
= 1 * 2 * 100 + 25
= 200 + 25 = 225
(ii) (95)^{2} = 9(9 + 1) * 100 + 25
= 9 * 10 * 100 + 25
= 9000 + 25 = 9025
(iii) (105)^{2} = 10 * (10 + 1) * 100 + 25
= 10 *11 * 100 + 25
= 11000 + 25 = 11025
(iv) (205)^{2} = 20 * (20 + 1) * 100 = 25
= 20 * 21 * 100 + 25
= 42000 + 25 = 42025
Q. Write all the square numbers between 100 and 300.
Ans: 121, 144, 169, 196, 225, 256 and 289
Q. Which of the following would end with digit 9:
123^{2}, 77^{2}, 84^{2}, 161^{2}, and 10^{2}.
Ans: (123)^{2} and (77)^{2}
Q. Write all the nonsquare numbers between 42 and 52.
Ans: 17, 18, 19, 20, 21, 22, 23 and 24
Q. Fill in the blanks:
(i) 1 + 3 + 5 + 7 + 9 + … = 5^{2}
(ii) 1 + 3 + 5 + 7 + 9 + … = 6^{2}
Ans: (i) 11 (ii) 11 + 13
Q. Fill in the blanks:
(i) 11^{2} = …… + ……
(ii) (…..)^{2} = 112 + 113
Ans: (i) 60 + 61 (ii) (15)^{2}
Q. Find the square of the following numbers actual multiplication:
(i) 39 (ii) 42
Ans: (i) 1521 (ii) 1764
Q. Find the square of:
(i) 35 (ii) 105
Ans: (i) 1225 (ii) 11025
Q. Write the Pythagorean triplet whose smallest number is 8.
Ans: 8, 15, 17
Q. Find a Pythagorean triplet in which one member is 12.
Ans: 12, 35, 37
Square Roots
Finding the square root of a number is just the opposite operation of squaring it. For example, the square of 5 is 25.
∴ Square root of 25 is 5.
Note: We kwon that (–2) * (–2) = 4, then we say that (–2) is also the square root of 4. Similarly, square roots of 100 are 10 and (–10).
But in this chapter, we shall be studying about positive square roots only
Finding Square Root Through Prime Factorisation
We know that a factor that occurs once in the prime factorisation of a number, occurs twice in the prime factorisation of its square. Thus, we can use this fact of prime factorisation of a number to find the square root of a perfect square.
Note: A perfect square has complete pairs of its prime factors.
Example. Is 1008 a perfect square? If not, find the smallest multiply of 1008 which is a perfect square end then find the square root of the new number.
Solution: We have
1008 = 2 * 2 * 2 * 2 * 3 * 3 * 7
As the prime factor 7 has no pair.
∴1008 is not a perfect square. Obviously, if 7 gets a pair, then the number will become a perfect square.
∴1008 * 7 = [2 * 2 * 2 * 2 * 3 * 3 * 7] * 7
or 7056 = 2 * 2 * 2 * 2 * 3 * 3 * 7
Thus, 7056 is the required multiple of 1008 which is a perfect square.
Now, = 2 * 2 * 3 * 7 = 84
Question: (i) 11^{2} = 121. What is the square root of 121?
(ii) 14^{2} = 196. What is the square root of 196?
Solution: (i) The square root of 121 is 11.
(ii) The square root of 196 is 14.
Think, Discuss and Write
Question:
(–1)^{2} = 1. Is –1, a square root of 1?
(–2)^{2} = 4. Is –2, a square root of 4?
(–9)^{2} = 81. Is –9, a square root of 81?
Solution:
(i) Since (–1) * (–1) = 1
i.e. (–1)^{2} = 1
∴ Square root of 1 can also be –1.
Similarly,
(ii) Yes (–2) is a square root of 4.
(iii) Yes (–9) is a square root of 81.
Since, we have to consider the positive square roots only and the symbol for a positive square
root is √
∴ = 4 [and not (–4)]
Similarly, means, the positive square root of 25, i.e. 5.
Note: We can also find the square root by subtracting successive odd numbers starting from 1.
Question: By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square, then find its square root.
(i) 121 (ii) 55 (iii) 36 (iv) 49 (v) 90
Solution:
(i) Subtracting the successive odd numbers from 121, we have
121 – 1 = 120 120 – 3 = 117
117 – 5 = 112 112 – 7 = 105
105 – 9 = 96 96 – 11 = 85
85 – 13 = 72 72 – 15 = 57
57 – 17 = 40 40 – 19 = 21
21 – 21 = 0
∴ = 11. [∵ We had to subtract the first 11 odd numbers.]
(ii)
∵ 55 – 1 = 54 54 – 3 = 51
51 – 5 = 46 46 – 7 = 39
39 – 9 = 30 30 – 11 = 19
19 – 13 = 6 6 – 15 = –9
and we do not reach to 0. ∴ 55 is not a perfect square.
(iii)
∵ 36 – 1 = 35 35 – 3 = 32
32 – 5 = 27 27 – 7 = 20
20 – 9 = 11 11 – 11 = 0
and we have obtained 0 after subtracting 6 successive odd numbers.
∴ 36 is a perfect square.
Thus, = 6.
(iv) We have
49 – 1 = 48 48 – 3 = 45
45 – 5 = 40 40 – 7 = 33
33 – 9 = 24 24 – 11 = 13
13 – 13 = 0
∵ We have obtained 0 after successive subtraction of 7 odd numbers.
∴ 49 is a perfect square,
Thus, = 7.
(v) We have:
90 – 1 = 89 89 – 3 = 86
86 – 5 = 81 81 – 7 = 74
74 – 9 = 65 65 – 11 = 54
54 – 13 = 41 41 – 15 = 26
26 – 17 = 9 9 – 19 = –10
Since, we can not reach to 0 after subtracting successive odd numbers.
∴ 90 is not a perfect square.
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