Squares and Square Roots — Exercise 3.3

Find the square root of the following number by prime factorisation method:

784

Answer

Finding prime factors of 784

$\begin{array}{l|r} 2 & 784 \\ \hline 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow 784 = (2 \times 2) \times (2 \times 2) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{784} = 2 \times 2 \times 7\\[1em] \Rightarrow \sqrt{784} = 28$

Hence, $\mathbf{\sqrt{784} = 28}$

Find the square root of the following number by prime factorisation method:

441

Answer

Finding prime factors of 441

$\begin{array}{l|r} 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow 441 = (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{441} = 3 \times 7\\[1em] \Rightarrow \sqrt{441} = 21$

Hence, $\mathbf{\sqrt{441} = 21}$

Find the square root of the following number by prime factorisation method:

1849

Answer

Finding prime factors of 1849

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

Solving,

⇒ 1849 = (43 × 43)

$\Rightarrow \sqrt{1849} = 43$

Hence, $\mathbf{\sqrt{1849} = 43}$

Find the square root of the following number by prime factorisation method:

4356

Answer

Finding prime factors of 4356

$\begin{array}{l|r} 2 & 4356 \\ \hline 2 & 2178 \\ \hline 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow 4356 = (2 \times 2) \times (3 \times 3) \times (11 \times 11)\\[1em] \Rightarrow \sqrt{4356} = 2 \times 3 \times 11\\[1em] \Rightarrow \sqrt{4356} = 66$

Hence, $\mathbf{\sqrt{4356} = 66}$

Find the square root of the following number by prime factorisation method:

6241

Answer

Finding prime factors of 6241

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

Solving,

⇒ 6241 = (79 x 79)

$\Rightarrow \sqrt{6241} = 79$

Hence, $\mathbf{\sqrt{6241} = 79}$

Find the square root of the following number by prime factorisation method:

8836

Answer

Finding prime factors of 8836

$\begin{array}{l|r} 2 & 8836 \\ \hline 2 & 4418 \\ \hline 47 & 2209 \\ \hline 47 & 47 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow 8836 = (2 \times 2) \times (47 \times 47)\\[1em] \Rightarrow \sqrt{8836} = 2 \times 47\\[1em] \Rightarrow \sqrt{8836} = 94$

Hence, $\mathbf{\sqrt{8836} = 94}$

Find the square root of the following number by prime factorisation method:

8281

Answer

Finding prime factors of 8281

$\begin{array}{l|r} 7 & 8281 \\ \hline 7 & 1183 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow 8281 = (7 \times 7) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{8281} = 7 \times 13\\[1em] \Rightarrow \sqrt{8281} = 91$

Hence, $\mathbf{\sqrt{8281} = 91}$

Find the square root of the following number by prime factorisation method:

9025

Answer

Finding prime factors of 9025

$\begin{array}{l|r} 5 & 9025 \\ \hline 5 & 1805 \\ \hline 19 & 361 \\ \hline 19 & 19 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow 9025 = (5 \times 5) \times (19 \times 19)\\[1em] \Rightarrow \sqrt{9025} = 5 \times 19\\[1em] \Rightarrow \sqrt{9025} = 95$

Hence, $\mathbf{\sqrt{9025} = 95}$

Find the square root of the following number by prime factorisation method:

$9\dfrac{67}{121}$

Answer

Converting mixed fraction to improper fraction:

$9\dfrac{67}{121} = \dfrac{1156}{121}$

Finding prime factors of 1156 and 121

$\begin{array}{l|r} 2 & 1156 \\ \hline 2 & 578 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow \sqrt{9\dfrac{67}{121}}\\[1em] \Rightarrow \sqrt{\dfrac{1156}{121}}\\[1em] \Rightarrow \dfrac{\sqrt{1156}}{\sqrt{121}}\\[1em] \Rightarrow \dfrac{\sqrt{(2 \times 2) \times (17 \times 17)}}{\sqrt{(11 \times 11)}}\\[1em] \Rightarrow \dfrac{2 \times 17}{11}\\[1em] \Rightarrow \dfrac{34}{11}\\[1em] \Rightarrow 3\dfrac{1}{11}$

Hence, $\mathbf{\sqrt{9\dfrac{67}{121}} = 3\dfrac{1}{11}}$

Find the square root of the following number by prime factorisation method:

$17\dfrac{13}{36}$

Answer

Converting mixed fraction to improper fraction:

$17\dfrac{13}{36} = \dfrac{625}{36}$

Finding prime factors of 625 and 36

$\begin{array}{l|r} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow \sqrt{17\dfrac{13}{36}}\\[1em] \Rightarrow \sqrt{\dfrac{625}{36}}\\[1em] \Rightarrow \dfrac{\sqrt{625}}{\sqrt{36}}\\[1em] \Rightarrow \dfrac{\sqrt{(5 \times 5) \times (5 \times 5)}}{\sqrt{(2 \times 2) \times (3 \times 3)}}\\[1em] \Rightarrow \dfrac{5 \times 5}{2 \times 3}\\[1em] \Rightarrow \dfrac{25}{6}\\[1em] \Rightarrow 4\dfrac{1}{6}$

Hence, $\mathbf{\sqrt{17\dfrac{13}{36}} = 4\dfrac{1}{6}}$

Find the square root of the following number by prime factorisation method:

1.96

Answer

Converting decimal to fraction:

$1.96 = \dfrac{196}{100}$

Finding prime factors of 196 and 100

$\begin{array}{l|r} 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow \sqrt{1.96}\\[1em] \Rightarrow \sqrt{\dfrac{196}{100}}\\[1em] \Rightarrow \dfrac{\sqrt{196}}{\sqrt{100}}\\[1em] \Rightarrow \dfrac{\sqrt{(2 \times 2) \times (7 \times 7)}}{\sqrt{(2 \times 2) \times (5 \times 5)}}\\[1em] \Rightarrow \dfrac{2 \times 7}{2 \times 5}\\[1em] \Rightarrow \dfrac{14}{10}\\[1em] \Rightarrow 1.4$

Hence, $\mathbf{\sqrt{1.96} = 1.4}$

Find the square root of the following number by prime factorisation method:

0.0064

Answer

Converting decimal to fraction:

$0.0064 = \dfrac{64}{10000}$

Finding prime factors of 64 and 10000

$\begin{array}{l|r} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 10000 \\ \hline 2 & 5000 \\ \hline 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow \sqrt{0.0064}\\[1em] \Rightarrow \sqrt{\dfrac{64}{10000}}\\[1em] \Rightarrow \dfrac{\sqrt{64}}{\sqrt{10000}}\\[1em] \Rightarrow \dfrac{\sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2)}}{\sqrt{(2 \times 2) \times (2 \times 2) \times (5 \times 5) \times (5 \times 5)}}\\[1em] \Rightarrow \dfrac{2 \times 2 \times 2}{2 \times 2 \times 5 \times 5}\\[1em] \Rightarrow \dfrac{8}{100}\\[1em] \Rightarrow 0.08$

Hence, $\mathbf{\sqrt{0.0064} = 0.08}$

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

588

Answer

Finding prime factors of 588

$\begin{array}{l|r} 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

588 = (2 × 2) × (7 × 7) × 3

Since the prime factor 3 is left unpaired, so 588 should be multiplied by 3.

$\Rightarrow 588 \times 3 = 1764\\[1em] \Rightarrow 1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{1764} = 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{1764} = 42$

Hence, 588 should be multiplied by 3 and $\mathbf{\sqrt{1764} = 42}$.

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

720

Answer

Finding prime factors of 720

$\begin{array}{l|r} 2 & 720 \\ \hline 2 & 360 \\ \hline 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

720 = (2 × 2) × (2 × 2) × (3 × 3) × 5

Since the prime factor 5 is left unpaired, so 720 should be multiplied by 5.

$\Rightarrow 720 \times 5 = 3600\\[1em] \Rightarrow 3600 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (5 \times 5)\\[1em] \Rightarrow \sqrt{3600} = 2 \times 2 \times 3 \times 5\\[1em] \Rightarrow \sqrt{3600} = 60$

Hence, 720 should be multiplied by 5 and $\mathbf{\sqrt{3600} = 60}$.

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

2178

Answer

Finding prime factors of 2178

$\begin{array}{l|r} 2 & 2178 \\ \hline 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

2178 = (3 × 3) × (11 × 11) × 2

Since the prime factor 2 is left unpaired, so 2178 should be multiplied by 2.

$\Rightarrow 2178 \times 2 = 4356\\[1em] \Rightarrow 4356 = (2 \times 2) \times (3 \times 3) \times (11 \times 11)\\[1em] \Rightarrow \sqrt{4356} = 2 \times 3 \times 11\\[1em] \Rightarrow \sqrt{4356} = 66$

Hence, 2178 should be multiplied by 2 and $\mathbf{\sqrt{4356} = 66}$.

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

3042

Answer

Finding prime factors of 3042

$\begin{array}{l|r} 2 & 3042 \\ \hline 3 & 1521 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

3042 = (3 × 3) × (13 × 13) × 2

Since the prime factor 2 is left unpaired, so 3042 should be multiplied by 2.

$\Rightarrow 3042 \times 2 = 6084\\[1em] \Rightarrow 6084 = (2 \times 2) \times (3 \times 3) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{6084} = 2 \times 3 \times 13\\[1em] \Rightarrow \sqrt{6084} = 78$

Hence, 3042 should be multiplied by 2 and $\mathbf{\sqrt{6084} = 78}$.

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

6300

Answer

Finding prime factors of 6300

$\begin{array}{l|r} 2 & 6300 \\ \hline 2 & 3150 \\ \hline 3 & 1575 \\ \hline 3 & 525 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

6300 = (2 × 2) × (3 × 3) × (5 × 5) × 7

Since the prime factor 7 is left unpaired, so 6300 should be multiplied by 7.

$\Rightarrow 6300 \times 7 = 44100\\[1em] \Rightarrow 44100 = (2 \times 2) \times (3 \times 3) \times (5 \times 5) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{44100} = 2 \times 3 \times 5 \times 7\\[1em] \Rightarrow \sqrt{44100} = 210$

Hence, 6300 should be multiplied by 7 and $\mathbf{\sqrt{44100} = 210}$.

For the following number, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also find the square root of the square number so obtained:

1872

Answer

Finding prime factors of 1872

$\begin{array}{l|r} 2 & 1872 \\ \hline 2 & 936 \\ \hline 2 & 468 \\ \hline 2 & 234 \\ \hline 3 & 117 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

1872 = (2 × 2) × (2 × 2) × (3 × 3) × 13

Since the prime factor 13 is left unpaired, so 1872 should be divided by 13.

$\Rightarrow 1872 \div 13 = 144\\[1em] \Rightarrow 144 = (2 \times 2) \times (2 \times 2) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{144} = 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{144} = 12$

Hence, 1872 should be divided by 13 and $\mathbf{\sqrt{144} = 12}$.

For the following number, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also find the square root of the square number so obtained:

2592

Answer

Finding prime factors of 2592

$\begin{array}{l|r} 2 & 2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

2592 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2

Since the prime factor 2 is left unpaired, so 2592 should be divided by 2.

$\Rightarrow 2592 \div 2 = 1296\\[1em] \Rightarrow 1296 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{1296} = 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt{1296} = 36$

Hence, 2592 should be divided by 2 and $\mathbf{\sqrt{1296} = 36}$.

For the following number, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also find the square root of the square number so obtained:

3380

Answer

Finding prime factors of 3380

$\begin{array}{l|r} 2 & 3380 \\ \hline 2 & 1690 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

3380 = (2 × 2) × (13 × 13) × 5

Since the prime factor 5 is left unpaired, so 3380 should be divided by 5.

$\Rightarrow 3380 \div 5 = 676\\[1em] \Rightarrow 676 = (2 \times 2) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{676} = 2 \times 13\\[1em] \Rightarrow \sqrt{676} = 26$

Hence, 3380 should be divided by 5 and $\mathbf{\sqrt{676} = 26}$.

For the following number, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also find the square root of the square number so obtained:

16224

Answer

Finding prime factors of 16224

$\begin{array}{l|r} 2 & 16224 \\ \hline 2 & 8112 \\ \hline 2 & 4056 \\ \hline 2 & 2028 \\ \hline 2 & 1014 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

16224 = (2 × 2) × (2 × 2) × (13 × 13) × 2 × 3

Since the prime factors 2 and 3 are left unpaired, so 16224 should be divided by 2 × 3 = 6.

$\Rightarrow 16224 \div 6 = 2704\\[1em] \Rightarrow 2704 = (2 \times 2) \times (2 \times 2) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{2704} = 2 \times 2 \times 13\\[1em] \Rightarrow \sqrt{2704} = 52$

Hence, 16224 should be divided by 6 and $\mathbf{\sqrt{2704} = 52}$.

For the following number, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also find the square root of the square number so obtained:

61347

Answer

Finding prime factors of 61347

$\begin{array}{l|r} 3 & 61347 \\ \hline 11 & 20449 \\ \hline 11 & 1859 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

61347 = (11 × 11) × (13 × 13) × 3

Since the prime factor 3 is left unpaired, so 61347 should be divided by 3.

$\Rightarrow 61347 \div 3 = 20449\\[1em] \Rightarrow 20449 = (11 \times 11) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{20449} = 11 \times 13\\[1em] \Rightarrow \sqrt{20449} = 143$

Hence, 61347 should be divided by 3 and $\mathbf{\sqrt{20449} = 143}$.

Find the smallest square number that is divisible by the following numbers:

3, 6, 10, 15

Answer

LCM of 3, 6, 10 and 15 = 2 × 3 × 5 = 30.

So, the smallest number divisible by each one of 3, 6, 10 and 15 is 30.

Prime Factorisation of 30:

$\begin{array}{l|r} 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

30 = 2 × 3 × 5

Since the prime factors 2, 3 and 5 are left unpaired, so 30 should be multiplied by 2 × 3 × 5 = 30.

Required square number = 30 × 30

= 900

Hence, the smallest square number divisible by 3, 6, 10 and 15 is 900.

Find the smallest square number that is divisible by the following number:

6, 9, 27, 36

Answer

LCM of 6, 9, 27 and 36 = 2 × 2 × 3 × 3 × 3 = 108.

So, the smallest number divisible by each one of 6, 9, 27 and 36 is 108.

Prime Factorisation of 108:

$\begin{array}{l|r} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

108 = (2 × 2) × (3 × 3) × 3

Since the prime factor 3 is left unpaired, so 108 should be multiplied by 3.

Required square number = 108 × 3

= 324

Hence, the smallest square number divisible by 6, 9, 27 and 36 is 324.

Find the smallest square number that is divisible by the following number:

4, 7, 8, 16

Answer

LCM of 4, 7, 8 and 16 is 2 × 2 × 2 × 2 × 7 = 112.

So, the smallest number divisible by each one of 4, 7, 8 and 16 is 112.

Prime Factorisation of 112:

$\begin{array}{l|r} 2 & 112 \\ \hline 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

112 = (2 × 2) × (2 × 2) × 7

Since the prime factor 7 is left unpaired, so 112 should be multiplied by 7.

Required square number = 112 × 7

= 784

Hence, the smallest square number divisible by 4, 7, 8 and 16 is 784.

4225 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.

Answer

Let the number of rows be $x$.

The number of plants in each row = the number of rows = $x$

Finding prime factors of 4225

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

Total number of plants = 4225

$\Rightarrow x \times x = 4225\\[1em] \Rightarrow x^2 = 4225\\[1em] \Rightarrow x = \sqrt{4225}\\[1em] \Rightarrow x = \sqrt{(5 \times 5) \times (13 \times 13)}\\[1em] \Rightarrow x = 5 \times 13\\[1em] \Rightarrow x = 65$

Hence, the number of rows is 65 and the number of plants in each row is 65.

The area of a rectangle is 1936 sq. m. If the length of the rectangle is 4 times its breadth, find the dimensions of the rectangle.

Answer

Let the breadth of the rectangle be $x$ m.

Length of the rectangle = 4 × breadth = $4x$ m

Area of rectangle = length × breadth

Area of rectangle = 1936

$\Rightarrow 4x \times x = 1936\\[1em] \Rightarrow 4x^2 = 1936\\[1em] \Rightarrow x^2 = \dfrac{1936}{4}\\[1em] \Rightarrow x^2 = 484\\[1em] \Rightarrow x = \sqrt{484}\\[1em]$

Finding prime factors of 484

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

Solving,

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

Breadth of the rectangle = 22 m

Length of the rectangle = 4 × 22 = 88 m

Hence, the length of the rectangle is 88 m and its breadth is 22 m.

In a school a P.T. teacher wants to arrange 2000 students in the form of rows and columns for P.T. display. If the number of rows is equal to number of columns and 64 students could not be accommodated in this arrangement. Find the number of rows.

Answer

Let the number of rows be $x$.

The number of columns = the number of rows = $x$

Students arranged in the P.T. display = 2000 - 64 = 1936

Finding prime factors of 1936

$\begin{array}{l|r} 2 & 1936 \\ \hline 2 & 968 \\ \hline 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

Total number of students accommodated = 1936

$\Rightarrow x \times x = 1936\\[1em] \Rightarrow x^2 = 1936\\[1em] \Rightarrow x = \sqrt{1936}\\[1em] \Rightarrow x = \sqrt{(2 \times 2) \times (2 \times 2) \times (11 \times 11)}\\[1em] \Rightarrow x = 2 \times 2 \times 11\\[1em] \Rightarrow x = 44$

Hence, the number of rows is 44.

In a school, the students of class VIII collected ₹2304 for a picnic. Each student contributed as many rupees as the number of students in the class. Find the number of students in the class.

Answer

Let the number of students in the class be $x$.

Amount contributed by each student = number of students = ₹$x$

Finding prime factors of 2304

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

Total amount collected = ₹2304

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

Hence, the number of students in the class is 48.

The product of two numbers is 7260. If one number is 15 times the other number, find the numbers.

Answer

Let the first number be $x$.

The second number is 15 times the first number = $15x$

Product of two numbers = 7260

$\Rightarrow x \times 15x = 7260\\[1em] \Rightarrow 15x^2 = 7260\\[1em] \Rightarrow x^2 = \dfrac{7260}{15}\\[1em] \Rightarrow x^2 = 484\\[1em] \Rightarrow x = \sqrt{484}\\[1em]$

Finding prime factors of 484

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

Solving,

$\Rightarrow x = \sqrt{(2 \times 2) \times (11 \times 11)}\\[1em] \Rightarrow x = 22\\[1em] \Rightarrow 15x = 15 \times 22 = 330$

Hence, the numbers are 22 and 330.

Find three positive numbers in the ratio 2 : 3 : 5, the sum of whose squares is 950.

Answer

Let the three numbers be $2x$, $3x$ and $5x$.

$\Rightarrow (2x)^2 + (3x)^2 + (5x)^2 = 950\\[1em] \Rightarrow 4x^2 + 9x^2 + 25x^2 = 950\\[1em] \Rightarrow 38x^2 = 950\\[1em] \Rightarrow x^2 = \dfrac{950}{38}\\[1em] \Rightarrow x^2 = 25\\[1em] \Rightarrow x = \sqrt{25}\\[1em] \Rightarrow x = 5$

First number = 2 × 5 = 10

Second number = 3 × 5 = 15

Third number = 5 × 5 = 25

Hence, the three numbers are 10, 15 and 25.

The perimeters of two squares are 60 metres and 144 metres respectively. Find the perimeter of another square equal in area to the sum of the areas of first two squares.

Answer

Perimeter of a square = 4 × side

$\Rightarrow \text{Side of first square} = \dfrac{60}{4} = 15 \text{ m}\\[1em] \Rightarrow \text{Side of second square} = \dfrac{144}{4} = 36 \text{ m}$

Area of first square = 15 × 15 = 225 m²

Area of second square = 36 × 36 = 1296 m²

Sum of the areas of the two squares = 225 + 1296 = 1521 m²

Let the side of the required square be $x$ m.

Finding prime factors of 1521

$\begin{array}{l|r} 3 & 1521 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

Solving,

$\Rightarrow x^2 = 1521\\[1em] \Rightarrow x = \sqrt{1521}\\[1em] \Rightarrow x = \sqrt{(3 \times 3) \times (13 \times 13)}\\[1em] \Rightarrow x = 3 \times 13\\[1em] \Rightarrow x = 39$

Perimeter of the required square = 4 × 39 = 156 m

Hence, the perimeter of the required square is 156 m.