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.
[âˆµ 3^{2} + 4^{2} = 5^{2}]
(ii) 8, 15, 17 form a Pythagorean triplet.
[âˆµ 8^{2} + 15^{2} = 17^{2}]
Note: For any natural number n, (n > 1), we have
(2n)^{2} + (n^{2} â€“ 1)^{2} = (n^{2} + 1)^{2}
such that 2n, n^{2} â€“ 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} â€“ 1 and n^{2} + 1.
âˆ´ 2n = 15 or n = 15/2 is not an integer.
So, let us assume that
n^{2} â€“ 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} â€“ 1 and n^{2} + 1
or 2(4), 4^{2 }â€“ 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 unitâ€™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
Exercise 6.2
Question 1. Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
Solution:
(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
Question 2. Write a Pythagorean triplet whose one member is
(i) 6 (ii) 14 (iii) 16 (iv) 18
Solution:
(i) Let 2n = 6 âˆ´n = 3
Now, n^{2} â€“ 1 = 3^{2} â€“ 1 = 8
and n^{2} + 1 = 3^{2} + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.
(ii) Let 2n = 14 âˆ´ n = 7
Now, n^{2} â€“ 1 = 7^{2} â€“ 1 = 48
and n^{2} + 1 = 7^{2} + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.
(iii) Let 2n = 16 âˆ´n = 8
Now, n^{2} â€“ 1 = 8^{2} â€“ 1
= 64 â€“ 1 = 63
and n^{2} + 1 = 8^{2} + 1
= 64 + 1 = 65
âˆ´ The required Pythagorean triplet is 16, 63, 65.
(iv) Let 2n = 18 âˆ´n = 9
Now, n^{2} â€“ 1 = 9^{2} â€“ 1
= 81 â€“ 1 = 80
and n^{2 }+ 1 = 9^{2} + 1
= 81 + 1 = 82
âˆ´ The required Pythagorean triple is 18, 80, 82.
Q. Write all the square numbers between 100 and 300. Ans: 121, 144, 169, 196, 225, 256 and 289 Q. Write all the non-square numbers between 42 and 52. Ans: 17, 18, 19, 20, 21, 22, 23 and 24 Q. Fill in the blanks: Ans: (i) 11 (ii) 11 + 13 Q. Fill in the blanks: Q. Find the square of the following numbers actual multiplication: Ans: (i) 1521 (ii) 1764 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.
âˆ´ Square root of 25 is 5.
Note: We kwon that (â€“2) * (â€“2) = 4, then we say that (â€“2) is also the square root of 4. Similarly, square roots of 100 are 10 and (â€“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.
âˆ´1008 is not a perfect square. Obviously, if 7 gets a pair, then the number will become a perfect square.
âˆ´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:
(â€“1)^{2} = 1. Is â€“1, a square root of 1?
(â€“2)^{2} = 4. Is â€“2, a square root of 4?
(â€“9)^{2} = 81. Is â€“9, a square root of 81?
Solution:
(i) Since (â€“1) * (â€“1) = 1
i.e. (â€“1)^{2} = 1
âˆ´ Square root of 1 can also be â€“1.
Similarly,
(ii) Yes (â€“2) is a square root of 4.
(iii) Yes (â€“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 âˆš
âˆ´ = 4 [and not (â€“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 â€“ 1 = 120 120 â€“ 3 = 117
117 â€“ 5 = 112 112 â€“ 7 = 105
105 â€“ 9 = 96 96 â€“ 11 = 85
85 â€“ 13 = 72 72 â€“ 15 = 57
57 â€“ 17 = 40 40 â€“ 19 = 21
21 â€“ 21 = 0
âˆ´ = 11. [âˆµ We had to subtract the first 11 odd numbers.]
(ii)
âˆµ 55 â€“ 1 = 54 54 â€“ 3 = 51
51 â€“ 5 = 46 46 â€“ 7 = 39
39 â€“ 9 = 30 30 â€“ 11 = 19
19 â€“ 13 = 6 6 â€“ 15 = â€“9
and we do not reach to 0. âˆ´ 55 is not a perfect square.
(iii)
âˆµ 36 â€“ 1 = 35 35 â€“ 3 = 32
32 â€“ 5 = 27 27 â€“ 7 = 20
20 â€“ 9 = 11 11 â€“ 11 = 0
and we have obtained 0 after subtracting 6 successive odd numbers.
âˆ´ 36 is a perfect square.
Thus, = 6.
(iv) We have
49 â€“ 1 = 48 48 â€“ 3 = 45
45 â€“ 5 = 40 40 â€“ 7 = 33
33 â€“ 9 = 24 24 â€“ 11 = 13
13 â€“ 13 = 0
âˆµ We have obtained 0 after successive subtraction of 7 odd numbers.
âˆ´ 49 is a perfect square,
Thus, = 7.
(v) We have:
90 â€“ 1 = 89 89 â€“ 3 = 86
86 â€“ 5 = 81 81 â€“ 7 = 74
74 â€“ 9 = 65 65 â€“ 11 = 54
54 â€“ 13 = 41 41 â€“ 15 = 26
26 â€“ 17 = 9 9 â€“ 19 = â€“10
Since, we can not reach to 0 after subtracting successive odd numbers.
âˆ´ 90 is not a perfect square.