Square and Square Roots

Squares and square roots are fundamental concepts in mathematics.

Squares

A square of a number is the product of the number with itself.
If ‘a’ is a number, its square is a * a = a².
For example, the square of 5 is 5 * 5 = 25.

Square Roots

The square root of a number is the value which, when multiplied by itself, gives the original number.
It is denoted by the symbol √.
For example, the square root of 25 is √25 = 5.
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Relationship between Squares and Square Roots:

They are inverse operations. Finding the square root is the opposite of squaring a number.

Important Points

The square of a natural number is always a natural number.
A number ending with 2, 3, 7, or 8 cannot be a perfect square.
There are two square roots for a positive number (positive and negative).
The square root of a negative number is not a real number.

Methods to find square roots:

Prime factorization
Division method
Estimation

Exercise 5.1

1. What will be the unit digit of the squares of the following numbers?

(i) 81

(ii) 272

(iii) 799

(iv) 3853

(v) 1234

(vi) 20387

(vii) 52698

(viii) 99880

(ix) 12796

(x) 55555

Ans :

(i) 81: Unit digit is 1, so the unit digit of the square is 1^2 = 1.

(ii) 272: Unit digit is 2, so the unit digit of the square is 2^2 = 4.

(iii) 799: Unit digit is 9, so the unit digit of the square is 9^2 = 81, i.e., 1.

(iv) 3853: Unit digit is 3, so the unit digit of the square is 3^2 = 9.

(v) 1234: Unit digit is 4, so the unit digit of the square is 4^2 = 16, i.e., 6.

(vi) 20387: Unit digit is 7, so the unit digit of the square is 7^2 = 49, i.e., 9.

(vii) 52698: Unit digit is 8, so the unit digit of the square is 8^2 = 64, i.e., 4.

(viii) 99880: Unit digit is 0, so the unit digit of the square is 0^2 = 0.

(ix) 12796: Unit digit is 6, so the unit digit of the square is 6^2 = 36, i.e., 6.

(x) 55555: Unit digit is 5, so the unit digit of the square is 5^2 = 25, i.e., 5.

2. The following numbers are not perfect squares. Give reason.

(i) 1057

(ii) 23453

(iii) 7928

(iv) 222222

(v) 64000

(vi) 89722

(vii) 222000

(viii) 505050

Ans :

(i) 1057: The unit digit is 7, which is not a valid unit digit for a perfect square.

(ii) 23453: The unit digit is 3, which is not a valid unit digit for a perfect square.

(iii) 7928: The unit digit is 8, which is not a valid unit digit for a perfect square.

(iv) 222222: The unit digit is 2, which is not a valid unit digit for a perfect square.

(v) 64000: While the unit digit is 0, which is valid, the number of zeros at the end is odd, making it not a perfect square.

(vi) 89722: The unit digit is 2, which is not a valid unit digit for a perfect square.

(vii) 222000: The unit digit is 0, but the number of zeros at the end is odd, making it not a perfect square.

(viii) 505050: The unit digit is 0, but the number of zeros at the end is odd, making it not a perfect square.

3. The squares of which of the following would be odd numbers?

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Ans :

Analysis:

(i) 431: Odd number, so its square will be odd.
(ii) 2826: Even number, so its square will be even.
(iii) 7779: Odd number, so its square will be odd.
(iv) 82004: Even number, so its square will be even.

4. Observe the following pattern and find the missing digits.

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1…2…1

100000012 = ………

Ans :

Pattern:

11² = 121
101² = 10201
1001² = 1002001
100001² = 10000200001

Applying the pattern:

10000001² = 100000020000001

Therefore, the missing digits are: 100000020000001

5. Observe the following pattern and supply the missing numbers.

112 = 121

1012 = 10201

101012 = 102030201

10101012 = ……….

……….2 = 10203040504030201

Ans :

Applying the pattern:

1010101² = 1020304030201
101010101² = 10203040504030201

6. Using the given pattern, find the missing numbers.

1² + 2² + 2² = 3²

2² + 3² + 6² = 7²

3² + 4² + 12² = 13²

4² + 5² + ?² = 21²

5² + ?² + 30² = 31²

6² + 7² + ?² = ?²

Ans :

Applying the pattern

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

7. 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) 1 + 3 + 5 + 7 + 9

There are 5 odd numbers.
So, n = 5
Sum = n² = 5² = 25

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

There are 10 odd numbers.
So, n = 10
Sum = n² = 10² = 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

There are 12 odd numbers.
So, n = 12
Sum = n² = 12² = 144

8. (i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

Ans :

(i)

Here, n = 7. So, 49 = 7².

Therefore, 49 can be expressed as the sum of the first 7 odd natural numbers:

49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii)

Similarly, here n = 11. So, 121 = 11².

Therefore, 121 can be expressed as the sum of the first 11 odd natural numbers:

121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100.

Ans :

The number of numbers between the squares of two consecutive natural numbers, n and (n+1), is 2n.

Applying the Pattern

(i) Between 12² and 13², there are 2 * 12 = 24 numbers.

(ii) Between 25² and 26², there are 2 * 25 = 50 numbers.

(iii) Between 99² and 100², there are 2 * 99 = 198 numbers.

Exercise 5.2

1. Find the square of the following numbers.

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

Ans :

(i) 32² = 1024

(ii) 35² = 1225

(iii) 86² = 7396

(iv) 93² = 8649

(v) 71² = 5041

(vi) 46² = 2116

2. Write a Pythagorean triplet whose one member is

(i) 6

(ii) 14

(iii) 16

(iv) 18

Ans :

If one member of a Pythagorean triplet is 2m, then the other two members are m² – 1 and m² + 1.

Applying the Pattern

(i) 6

Let 2m = 6
- m = 3
- m² – 1 = 3² – 1 = 8
- m² + 1 = 3² + 1 = 10
Pythagorean triplet: 6, 8, 10

(ii) 14

Let 2m = 14
- m = 7
- m² – 1 = 7² – 1 = 48
- m² + 1 = 7² + 1 = 50
Pythagorean triplet: 14, 48, 50

(iii) 16

Let 2m = 16
- m = 8
- m² – 1 = 8² – 1 = 63
- m² + 1 = 8² + 1 = 65
Pythagorean triplet: 16, 63, 65

(iv) 18

Let 2m = 18
- m = 9
- m² – 1 = 9² – 1 = 80
- m² + 1 = 9² + 1 = 82
Pythagorean triplet: 18, 80, 82

Exercise 5.3

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

(i) 9801

(ii) 99856

(iii) 998001

(iv) 657666025

Ans :

(i) 9801: The unit digit is 1, so the square root ends in 1 or 9.

(ii) 99856: The unit digit is 6, so the square root ends in 4 or 6.

(iii) 998001: The unit digit is 1, so the square root ends in 1 or 9.

(iv) 657666025: The unit digit is 5, so the square root ends in 5.

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

(i) 153

(ii) 257

(iii) 408

(iv) 441

Ans :

Analyzing the given numbers:

153: The unit digit is 3, which is not in the list of possible unit digits for a perfect square.
257: The unit digit is 7, which is not in the list of possible unit digits for a perfect square.
408: The unit digit is 8, which is not in the list of possible unit digits for a perfect square.
441: The unit digit is 1, which is a possible unit digit for a perfect square. However, we cannot definitively say if it is a perfect square without further calculation.

Therefore, the numbers that are surely not perfect squares are 153, 257, and 408.

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

Ans :

Finding the Square Root of 100

100 – 1 = 99
99 – 3 = 96
96 – 5 = 91
91 – 7 = 84
84 – 9 = 75
75 – 11 = 64
64 – 13 = 51
51 – 15 = 36
36 – 17 = 19
19 – 19 = 0

We subtracted 10 odd numbers to reach 0. Therefore, the square root of 100 is 10.

Finding the Square Root of 169

169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0

We subtracted 13 odd numbers to reach 0. Therefore, the square root of 169 is 13.

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

(x) 8100

Ans :

i) 729

Prime factorization of 729
= 3 * 3 * 3 * 3 * 3 * 3
Grouping pairs: (3 * 3) * (3 * 3) * (3 * 3)
Square root of 729 = 3 * 3 * 3 = 27

ii) 400

Prime factorization of 400 = 2 * 2 * 2 * 2 * 5 * 5
Grouping pairs: (2 * 2) * (2 * 2) * (5 * 5)
Square root of 400 = 2 * 2 * 5 = 20

iii) 1764

Prime factorization of 1764
= 2 * 2 * 3 * 3 * 7 * 7
Grouping pairs: (2 * 2) * (3 * 3) * (7 * 7)
Square root of 1764
= 2 * 3 * 7 = 42

iv) 4096

Prime factorization of 4096
= 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
Grouping pairs: (2 * 2) * (2 * 2) * (2 * 2) * (2 * 2) * (2 * 2) * (2 * 2)
Square root of 4096 = 2 * 2 * 2 * 2 * 2 * 2 = 64

v) 7744

Prime factorization of 7744 = 2 * 2 * 2 * 2 * 2 * 2 * 11 * 11
Grouping pairs: (2 * 2) * (2 * 2) * (2 * 2) * (11 * 11)
Square root of 7744 = 2 * 2 * 2 * 11 = 88

vi) 9604

Prime factorization of 9604 = 2 * 2 * 7 * 7 * 7 * 7
Grouping pairs: (2 * 2) * (7 * 7) * (7 * 7)
Square root of 9604 = 2 * 7 * 7 = 98

vii) 5929

Prime factorization of 5929 = 7 * 7 * 11 * 11
Grouping pairs: (7 * 7) * (11 * 11)
Square root of 5929
= 7 * 11 = 77

viii) 9216

Prime factorization of 9216 = 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 3
Grouping pairs: (2 * 2) * (2 * 2) * (2 * 2) * (3 * 3) * (3 * 3)
Square root of 9216 = 2 * 2 * 2 * 3 * 3 = 96

ix) 529

Prime factorization of 529 = 23 * 23
Grouping pairs: (23 * 23)
Square root of 529 = 23

x) 8100

Prime factorization of 8100
= 2 * 2 * 3 * 3 * 3 * 3 * 5 * 5
Grouping pairs: (2 * 2) * (3 * 3) * (3 * 3) * (5 * 5)
Square root of 8100 = 2 * 3 * 3 * 5 = 90

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.

(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

Ans :

i) 252

Prime factorization:
2 * 2 * 3 * 3 * 7
The factor 7 doesn’t have a pair.
multiply by 7.
252 * 7 = 1764, which is the square of 42.

ii) 180

Prime factorization:
2 * 2 * 3 * 3 * 5
The factor 5 doesn’t have a pair.
multiply by 5.
180 * 5 = 900, which is the square of 30.

iii) 1008

Prime factorization: 2 * 2 * 2 * 2 * 3 * 3 * 7
The factor 7 doesn’t have a pair.
multiply by 7.
1008 * 7 = 7056, which is the square of 84.

iv) 2028

Prime factorization: 2 * 2 * 3 * 13 * 13
The factor 3 doesn’t have a pair.
To make it a perfect square, multiply by 3.
2028 * 3 = 6084, which is the square of 78.

v) 1458

Prime factorization: 2 * 3 * 3 * 3 * 3 * 3
The factor 2 doesn’t have a pair.
multiply by 2.
1458 * 2 = 2916, which is the square of 54.

vi) 768

Prime factorization: 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3
The factor 3 doesn’t have a pair.
To make it a perfect square, multiply by 3.
768 * 3 = 2304, which is the square of 48.

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained.

(i) 252

(ii) 2925

(iii) 396

(iv) 2645

(v) 2800

(vi) 1620

Ans :

Number	Divisor	Perfect Square	Square Root
252	None	252 (already perfect square)	16
2925	None	2925 (already perfect square)	54
396	None	396 (already perfect square)	19
2645	None	2645 (already perfect square)	51
2800	None	2800 (already perfect square)	50
1620	None	1620 (already perfect square)	40

7. The students of class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Ans :

Each student donated Rs x.

Therefore, total donation = x * x = x²

But, it is given that the total donation is Rs 2401.

Therefore, x² = 2401

=> x = √2401 = 49

Hence, the answer is 49.

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Ans :

Let the number of rows be x. Since each row has the same number of plants as the number of rows, there are also x plants in each row.

Total plants = Number of rows * Number of plants per row

2025 = x * x

x² = 2025

To find x, we need to find the square root of 2025.

x = √2025 = 45

Therefore, there are 45 rows and 45 plants in each row.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Ans :

Step 1: Find the LCM of 4, 9, and 10

Prime factorization of 4 = 2 * 2
Prime factorization of 9
= 3 * 3
Prime factorization of 10 = 2 * 5

LCM (Least Common Multiple) = 2 * 2 * 3 * 3 * 5 = 180

Step 2:

180 is not a perfect square because the prime factor 5 does not have a pair.

Step 3: Make the LCM a perfect square To make 180 a perfect square, we need to multiply it by 5.

180 * 5 = 900

Step 4: Verify if the new number is a perfect square 900 is a perfect square as its square root is 30.

10. Find the smallest number that is divisible by each of the numbers 8, 15 and 20.

Ans :

Find the LCM (Least Common Multiple):
- Prime factorization of 8 = 2 * 2 * 2
- Prime factorization of 15 = 3 * 5
- Prime factorization of 20 = 2 * 2 * 5
- LCM = 2 * 2 * 2 * 3 * 5
- = 120
Check if the LCM is a perfect square:
- 120 is not a perfect square as the prime factors 2, 3, and 5 do not have pairs.
Make the LCM a perfect square:
- To make 120 a perfect square, we need to multiply it by 2 * 3 * 5 = 30.
Find the smallest square number:
- 120 * 30 = 3600
- 3600 is a perfect square as its square root is 60.