Squares and Square Roots, Cubes and Cube Roots — Exercise 3(B)

Give reason to show that none of the numbers given below is a perfect square:

(i) 2162

(ii) 6843

(iii) 9637

(iv) 6598

(v) 640

(vi) 49000

(vii) 3600000

Answer

According to the properties of perfect squares, a number ending in 2, 3, 7 or 8 is never a perfect square and a number ending in an odd number of zeros is never a perfect square.

(i) The number 2162 ends in 2.

Hence, 2162 is not a perfect square as its unit digit is 2.

(ii) The number 6843 ends in 3.

Hence, 6843 is not a perfect square as its unit digit is 3.

(iii) The number 9637 ends in 7.

Hence, 9637 is not a perfect square as its unit digit is 7.

(iv) The number 6598 ends in 8.

Hence, 6598 is not a perfect square as its unit digit is 8.

(v) The number 640 ends in one zero.

Hence, 640 is not a perfect square as it has an odd number of zeros at the end.

(vi) The number 49000 ends in three zeros.

Hence, 49000 is not a perfect square as it has an odd number of zeros at the end.

(vii) The number 3600000 ends in five zeros.

Hence, 3600000 is not a perfect square as it has an odd number of zeros at the end.

Is the following number a square of an even number?

676

Answer

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 676 into prime factors, we get :

$\begin{array}{l|r} 2 & 676 \\ \hline 2 & 338 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

∴ 676 = (2 × 2) × (13 × 13)

$\Rightarrow \sqrt{676} = \sqrt{(2 \times 2) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{676} = 2 \times 13\\[1em] \Rightarrow \sqrt{676} = 26$

Since 26 is an even number.

Hence, 676 is the square of an even number.

Is the following number a square of an even number?

729

Answer

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 729 into prime factors, we get :

$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

∴ 729 = (3 × 3) × (3 × 3) × (3 × 3)

$\Rightarrow \sqrt{729} = \sqrt{(3 \times 3) \times (3 \times 3) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{729} = 3 \times 3 \times 3\\[1em] \Rightarrow \sqrt{729} = 27$

Since 27 is an odd number.

Hence, 729 is not the square of an even number.

Is the following number a square of an even number?

1089

Answer

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 1089 into prime factors, we get :

$\begin{array}{l|r} 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

∴ 1089 = (3 × 3) × (11 × 11)

$\Rightarrow \sqrt{1089} = \sqrt{(3 \times 3) \times (11 \times 11)} \\[1em] \Rightarrow \sqrt{1089} = 3 \times 11\\[1em] \Rightarrow \sqrt{1089} = 33$

Since 33 is an odd number.

Hence, 1089 is not the square of an even number.

Is the following number a square of an even number?

2304

Answer

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 2304 into prime factors, we get :

$\begin{array}{l|r} 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

∴ 2304 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)

$\Rightarrow \sqrt{2304} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{2304} = 48$

Since 48 is an even number.

Hence, 2304 is the square of an even number.

Is the following number a square of an even number?

5625

Answer

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 5625 into prime factors, we get :

$\begin{array}{l|r} 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

∴ 5625 = (3 × 3) × (5 × 5) × (5 × 5)

$\Rightarrow \sqrt{5625} = \sqrt{(3 \times 3) \times (5 \times 5) \times (5 \times 5)} \\[1em] \Rightarrow \sqrt{5625} = 3 \times 5 \times 5\\[1em] \Rightarrow \sqrt{5625} = 75$

Since 75 is an odd number.

Hence, 5625 is not the square of an even number.

Is the following number a square of an even number?

9216

Answer

According to the property of perfect squares, the square of an even number is always an even number and the square of an odd number is always an odd number.

Resolving 9216 into prime factors, we get :

$\begin{array}{l|r} 2 & 9216 \\ \hline 2 & 4608 \\ \hline 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

∴ 9216 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)

$\Rightarrow \sqrt{9216} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{9216} = 2 \times 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{9216} = 96$

Since 96 is an even number.

Hence, 9216 is the square of an even number.

Is the following number a square of an odd number?

484

Answer

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 484 into prime factors, we get :

$\begin{array}{l|r} 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

∴ 484 = (2 × 2) × (11 × 11)

$\Rightarrow \sqrt{484} = \sqrt{(2 × 2) × (11 × 11)} \\[1em] \Rightarrow \sqrt{484} = 2 \times 11\\[1em] \Rightarrow \sqrt{484} = 22$

Since 22 is an even number.

Hence, 484 is not the square of an odd number.

Is the following number a square of an odd number?

961

Answer

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 961 into prime factors, we get :

$\begin{array}{l|r} 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$

∴ 961 = (31 × 31)

$\Rightarrow \sqrt{961} = \sqrt{31 \times 31} \\[1em] \Rightarrow \sqrt{961} = 31$

Since 31 is an odd number.

Hence, 961 is the square of an odd number.

Is the following number a square of an odd number?

4225

Answer

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 4225 into prime factors, we get :

$\begin{array}{l|r} 5 & 4225 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

∴ 4225 = (5 × 5) × (13 × 13)

$\Rightarrow \sqrt{4225} = \sqrt{(5 \times 5) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{4225} = 5 \times 13\\[1em] \Rightarrow \sqrt{4225} = 65$

Since 65 is an odd number.

Hence, 4225 is the square of an odd number.

Is the following number a square of an odd number?

7396

Answer

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 7396 into prime factors, we get :

$\begin{array}{l|r} 2 & 7396 \\ \hline 2 & 3698 \\ \hline 43 & 1849 \\ \hline 43 & 43 \\ \hline & 1 \\ \end{array}$

∴ 7396 = (2 × 2) × (43 × 43)

$\Rightarrow \sqrt{7396} = \sqrt{(2 \times 2) \times (43 \times 43)} \\[1em] \Rightarrow \sqrt{7396} = 2 \times 43\\[1em] \Rightarrow \sqrt{7396} = 86$

Since 86 is an even number.

Hence, 7396 is not the square of an odd number.

Is the following number a square of an odd number?

6241

Answer

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 6241 into prime factors, we get :

$\begin{array}{l|r} 79 & 6241 \\ \hline 79 & 79 \\ \hline & 1 \\ \end{array}$

∴ 6241 = (79 × 79)

$\Rightarrow \sqrt{6241} = \sqrt{79 \times 79} \\[1em] \Rightarrow \sqrt{6241} = 79$

Since 79 is an odd number.

Hence, 6241 is the square of an odd number.

Is the following number a square of an odd number?

8649

Answer

According to the property of perfect squares, the square of an odd number is always an odd number and the square of an even number is always an even number.

Resolving 8649 into prime factors, we get :

$\begin{array}{l|r} 3 & 8649 \\ \hline 3 & 2883 \\ \hline 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$

∴ 8649 = (3 × 3) × (31 × 31)

$\Rightarrow \sqrt{8649} = \sqrt{(3 \times 3) \times (31 \times 31)} \\[1em] \Rightarrow \sqrt{8649} = 3 \times 31\\[1em] \Rightarrow \sqrt{8649} = 93$

Since 93 is an odd number.

Hence, 8649 is the square of an odd number.

Without adding find the sum:

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

Answer

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, the given numbers are the first 11 odd natural numbers.

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

⇒ 11²

⇒ 121

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

Without adding find the sum:

1 + 3 + 5 + 7 + ..................... + 47 + 49

Answer

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, the last odd number is 49.

$\Rightarrow 2n - 1 = 49 \\[1em] \Rightarrow 2n = 50 \\[1em] \Rightarrow n = 25$

So, the given numbers are the first 25 odd natural numbers.

⇒ 1 + 3 + 5 + 7 + ..... + 47 + 49

⇒ 25²

⇒ 625

Hence, 1 + 3 + 5 + 7 + ..... + 47 + 49 = 625.

Without adding find the sum:

1 + 3 + 5 + 7 + ................. + 97 + 99.

Answer

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, the last odd number is 99.

$\Rightarrow 2n - 1 = 99 \\[1em] \Rightarrow 2n = 100 \\[1em] \Rightarrow n = 50$

So, the given numbers are the first 50 odd natural numbers.

⇒ 1 + 3 + 5 + 7 + ..... + 97 + 99

⇒ 50²

⇒ 2500

Hence, 1 + 3 + 5 + 7 + ..... + 97 + 99 = 2500.

Express 144 as the sum of 12 odd numbers.

Answer

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, 144 = 12², so 144 is the sum of the first 12 odd natural numbers.

⇒ 144

⇒ 12²

⇒ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Hence, 144 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23.

Express 225 as the sum of 15 odd numbers.

Answer

We know that,

For every natural number $n$, the sum of first $n$ odd natural numbers = $n^2$.

Here, 225 = 15², so 225 is the sum of the first 15 odd natural numbers.

⇒ 225

⇒ 15²

⇒ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

Hence, 225 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29.

Write a Pythagorean triplet whose smallest member is 6.

Answer

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$\Rightarrow 2m = 6\\[1em] \Rightarrow m = 3$

∴ $m^2 - 1 = 3^2 - 1 = 9 - 1 = 8$ and $m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$.

Hence, (6, 8, 10) is a Pythagorean triplet whose smallest member is 6.

Write a Pythagorean triplet whose smallest member is 14.

Answer

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$\Rightarrow 2m = 14\\[1em] \Rightarrow m = 7$

∴ $m^2 - 1 = 7^2 - 1 = 49 - 1 = 48$ and $m^2 + 1 = 7^2 + 1 = 49 + 1 = 50$.

Hence, (14, 48, 50) is a Pythagorean triplet whose smallest member is 14.

Write a Pythagorean triplet whose smallest member is 16.

Answer

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$\Rightarrow 2m = 16\\[1em] \Rightarrow m = 8$

∴ $m^2 - 1 = 8^2 - 1 = 64 - 1 = 63$ and $m^2 + 1 = 8^2 + 1 = 64 + 1 = 65$.

Hence, (16, 63, 65) is a Pythagorean triplet whose smallest member is 16.

Write a Pythagorean triplet whose smallest member is 20.

Answer

We know that,

For every natural number $m \gt 1, (2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

Putting the given value as $2m$,

$\Rightarrow 2m = 20\\[1em] \Rightarrow m = 10$

∴ $m^2 - 1 = 10^2 - 1 = 100 - 1 = 99$ and $m^2 + 1 = 10^2 + 1 = 100 + 1 = 101$.

Hence, (20, 99, 101) is a Pythagorean triplet whose smallest member is 20.

Evaluate:

(29)² - (28)²

Answer

Solving,

⇒ (29)² - (28)²

⇒ (29 + 28) × (29 - 28)

⇒ 57 × 1

⇒ 57

Hence, (29)² - (28)² = 57.

Evaluate:

(56)² - (55)²

Answer

Solving,

⇒ (56)² - (55)²

⇒ (56 + 55) × (56 - 55)

⇒ 111 × 1

⇒ 111

Hence, (56)² - (55)² = 111.

Evaluate:

(82)² - (81)²

Answer

Solving,

⇒ (82)² - (81)²

⇒ (82 + 81) × (82 - 81)

⇒ 163 × 1

⇒ 163

Hence, (82)² - (81)² = 163.

Evaluate:

(108)² - (107)²

Answer

Solving,

⇒ (108)² - (107)²

⇒ (108 + 107) × (108 - 107)

⇒ 215 × 1

⇒ 215

Hence, (108)² - (107)² = 215.

Evaluate:

(221)² - (220)²

Answer

Solving,

⇒ (221)² - (220)²

⇒ (221 + 220) × (221 - 220)

⇒ 441 × 1

⇒ 441

Hence, (221)² - (220)² = 441.

Using the identity $(a + b)^2 = (a^2 + 2ab + b^2)$, evaluate:

(204)²

Answer

Solving,

⇒ (204)²

⇒ (200 + 4)²

⇒ (200)² + 2 × 200 × 4 + (4)²

⇒ 40000 + 1600 + 16

⇒ 41616

Hence, (204)² = 41616.

Using the identity $(a + b)^2 = (a^2 + 2ab + b^2)$, evaluate:

(307)²

Answer

Solving,

⇒ (307)²

⇒ (300 + 7)²

⇒ (300)² + 2 × 300 × 7 + (7)²

⇒ 90000 + 4200 + 49

⇒ 94249

Hence, (307)² = 94249.

Using the identity $(a + b)^2 = (a^2 + 2ab + b^2)$, evaluate:

(430)²

Answer

Solving,

⇒ (430)²

⇒ (400 + 30)²

⇒ (400)² + 2 × 400 × 30 + (30)²

⇒ 160000 + 24000 + 900

⇒ 184900

Hence, (430)² = 184900.

Using the identity $(a - b)^2 = (a^2 - 2ab + b^2)$, evaluate:

(94)²

Answer

Solving,

⇒ (94)²

⇒ (100 - 6)²

⇒ (100)² - 2 × 100 × 6 + (6)²

⇒ 10000 - 1200 + 36

⇒ 8836

Hence, (94)² = 8836.

Using the identity $(a - b)^2 = (a^2 - 2ab + b^2)$, evaluate:

(192)²

Answer

Solving,

⇒ (192)²

⇒ (200 - 8)²

⇒ (200)² - 2 × 200 × 8 + (8)²

⇒ 40000 - 3200 + 64

⇒ 36864

Hence, (192)² = 36864.

Using the identity $(a - b)^2 = (a^2 - 2ab + b^2)$, evaluate:

(389)²

Answer

Solving,

⇒ (389)²

⇒ (400 - 11)²

⇒ (400)² - 2 × 400 × 11 + (11)²

⇒ 160000 - 8800 + 121

⇒ 151321

Hence, (389)² = 151321.

Evaluate:

59 × 61

Answer

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 59 × 61

⇒ (60 - 1) × (60 + 1)

⇒ (60)² - (1)²

⇒ 3600 - 1

⇒ 3599

Hence, 59 × 61 = 3599.

Evaluate:

88 × 92

Answer

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 88 × 92

⇒ (90 - 2) × (90 + 2)

⇒ (90)² - (2)²

⇒ 8100 - 4

⇒ 8096

Hence, 88 × 92 = 8096.

Evaluate:

94 × 106

Answer

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 94 × 106

⇒ (100 - 6) × (100 + 6)

⇒ (100)² - (6)²

⇒ 10000 - 36

⇒ 9964

Hence, 94 × 106 = 9964.

Evaluate:

48 × 52

Answer

Using the identity $a^2 - b^2 = (a + b) (a - b)$,

⇒ 48 × 52

⇒ (50 - 2) × (50 + 2)

⇒ (50)² - (2)²

⇒ 2500 - 4

⇒ 2496

Hence, 48 × 52 = 2496.

State whether each of the following statements is true or false:

(i) A perfect square number always has an even number of digits.

(ii) Every number ending in an even number of zeros is a perfect square.

(iii) The square of a prime number is prime.

(iv) There are 10 perfect square numbers from 1 to 100.

(v) The sum of two perfect squares is a perfect square.

(vi) The difference of two perfect squares is a perfect square.

(vii) The product of two perfect squares is a perfect square.

Answer

(i) The perfect squares 4, 9, 121 and 625 have one, one, three and three digits respectively.

Hence, the given statement is false.

(ii) A perfect square can only have an even number of zeros at the end but the converse is not true.

For example, 300, 1200 and 50000 end with an even number of zeros but these are not perfect squares.

Hence, the given statement is false.

(iii) The square of a prime number $p$ is $p \times p$, which has $p$ as a factor other than 1 and itself.

For example, 2 is a prime number but 2² = 4, which is not a prime number.

Hence, the given statement is false.

(iv) The perfect squares from 1 to 100 are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100, which are 10 in number.

Hence, the given statement is true.

(v) The numbers 9 and 16 are perfect squares.

9 + 16 = 25 = 5², which is a perfect square.

But, the numbers 4 and 9 are also perfect squares and 4 + 9 = 13, which is not a perfect square.

Hence, the given statement is false.

(vi) The numbers 25 and 9 are perfect squares.

25 - 9 = 16 = 4², which is a perfect square.

But, the numbers 9 and 4 are also perfect squares and 9 - 4 = 5, which is not a perfect square.

Hence, the given statement is false.

(vii) Let the two perfect squares be $a^2$ and $b^2$.

$a^2 \times b^2 = (ab)^2$, which is a perfect square.

For example, 4 × 9 = 36 = 6², which is a perfect square.

Hence, the given statement is true.