RD Sharma solution class 8 chapter 3 Square and Square roots Exercise 3.9

Exercise 3.9

Page-3.61

Question 1:

Using square root table, find the square root
7

Answer 1:

From the table, we directly find that the square root of 7 is 2.646.

Question 2:

Using square root table, find the square root
15

Answer 2:

Using the table to find $\sqrt{3 }\mathrm{and} \sqrt{5}$
$$\begin{gathered} \sqrt{15}=\sqrt{3}\times \sqrt{5} \\ =1.732\times 2.236 \\ =3.873 \end{gathered}$$
              
       

Question 3:

Using square root table, find the square root
74

Answer 3:

  Using the table to find $\sqrt{2} \mathrm{and} \sqrt{37}$

$$\begin{gathered} \sqrt{74}=\sqrt{2}\times \sqrt{37} \\ =1.414\times 6.083 \\ =8.602 \end{gathered}$$
               
         

Question 4:

Using square root table, find the square root
82

Answer 4:

Using the table to find $\sqrt{2} \mathrm{and} \sqrt{41}$

$$\begin{gathered} \sqrt{82}=\sqrt{2}\times \sqrt{41} \\ =1.414\times 6.403 \\ =9.055 \end{gathered}$$
                  
         

Question 5:

Using square root table, find the square root
198

Answer 5:

Using the table to find $\sqrt{2} \mathrm{and} \sqrt{11}$

$$\begin{gathered} \sqrt{198}=\sqrt{2}\times \sqrt{9}\times \sqrt{11} \\ =1.414\times 3\times 3.317 \\ = 14.070 \end{gathered}$$
                   
           

Question 6:

Using square root table, find the square root
540

Answer 6:

Using the table to find $\sqrt{3} \mathrm{and} \sqrt{5}$

$$\begin{gathered} \sqrt{540}=\sqrt{54}\times \sqrt{10} \\ =2\times 3\sqrt{3}\times \sqrt{5} \\ =2\times 3\times 1.732\times 2.2361 \\ =23.24 \end{gathered}$$
            
                

Question 7:

Using square root table, find the square root
8700

Answer 7:

Using the table to find $\sqrt{3} \mathrm{and} \sqrt{29}$

$$\begin{gathered} \sqrt{8700}=\sqrt{3}\times \sqrt{29}\times \sqrt{100} \\ =1.7321\times 5.385\times 10 \\ =93.27 \end{gathered}$$
                      
              

Question 8:

Using square root table, find the square root
3509

Answer 8:

Using the table to find $\sqrt{29}$

$$\begin{gathered} \sqrt{3509}=\sqrt{121}\times \sqrt{29} \\ =11\times 5.3851 \\ =59.235 \end{gathered}$$
                         
              

Question 9:

Using square root table, find the square root
6929

Answer 9:

Using the table to find $\sqrt{41}$

$$\begin{gathered} \sqrt{6929}=\sqrt{169}\times \sqrt{41} \\ =13 \times 6.4031 \\ =83.239 \end{gathered}$$
                       
              

Question 10:

Using square root table, find the square root
25725

Answer 10:

Using the table to find $\sqrt{3} \mathrm{and} \sqrt{7}$

$$\begin{gathered} \sqrt{25725}=\sqrt{3\times 5\times 5\times 7\times 7\times 7} \\ =\sqrt{3}\times 5\times 7\times \sqrt{7} \\ =1.732 \times 5\times 7\times 2.646 \\ =160.41 \end{gathered}$$
                          
                          
                 

Question 11:

Using square root table, find the square root
1312

Answer 11:

Using the table to find $\sqrt{2} \mathrm{and} \sqrt{41}$

$$\begin{gathered} \sqrt{1312}=\sqrt{2\times 2\times 2\times 2\times 2\times 41} \\ =2\times 2\sqrt{2}\times \sqrt{41} \\ =2\times 2\times 1.414\times 6.4031 \\ =36.222 \end{gathered}$$
              
                       
              

Question 12:

Using square root table, find the square root
4192

Answer 12:

$$\begin{gathered} \sqrt{4192}=\sqrt{2\times 2\times 2\times 2\times 2\times 131} \\ =2\times 2\sqrt{2}\times \sqrt{131} \end{gathered}$$
              
The square root of 131 is not listed in the table. Hence, we have to apply long division to find it.

Substituting the values:
            =   $2\times 2\times 11.4455$          (using the table to find $\sqrt{2}$)
              = 64.75

Question 13:

Using square root table, find the square root
4955

Answer 13:

On prime factorisation:
4955 is equal to 5 $\times$ 991, which means that $\sqrt{4955}=\sqrt{5}\times \sqrt{11}$.
The square root of 991 is not listed in the table; it lists the square roots of all the numbers below 100.
Hence, we have to manipulate the number such that we get the square root of a number less than 100. This can be done in the following manner:
$\sqrt{4955}=\sqrt{49.55\times 100}=\sqrt{49.55}\times 10$
Now, we have to find the square root of 49.55.
We have: $\sqrt{49}=7 \mathrm{and} \sqrt{50}=7.071$ .
Their difference is 0.071.
Thus, for the difference of 1 (50 $-$ 49), the difference in the values of the square roots is 0.071.
For the difference of 0.55, the difference in the values of the square roots is:
0.55 $\times$ 0.0701 = 0.03905
$\therefore$ $\sqrt{49.55}=7+0.03905=7.03905$

Finally, we have:
$\sqrt{4955}=\sqrt{49.55}\times 10=7.03905\times 10=70.3905$

Question 14:

Using square root table, find the square root
$\frac{99}{144}$

Answer 14:

$\sqrt{\frac{99}{144}}=\frac{\sqrt{3\times 3\times 11}}{\sqrt{144}}$
             = $\frac{3\sqrt{11}}{12}$             
           
            =$\frac{3\times 3.3166}{12}$                    (using the square root table to find $\sqrt{11}$)
             =0.829
       
             

Question 15:

Using square root table, find the square root
$\frac{57}{169}$

Answer 15:

$\sqrt{\frac{57}{169}}=\frac{\sqrt{3}\times \sqrt{19}}{\sqrt{169}}$
             $\frac{1.732\times 4.3589}{13}$          (using the square root table to find $\sqrt{3}$ and $\sqrt{19}$)
              $0.581$

Question 16:

Using square root table, find the square root
$\frac{101}{169}$

Answer 16:

$\sqrt{\frac{101}{169}}=\frac{\sqrt{101}}{\sqrt{169}}$
The square root of 101 is not listed in the table. This is because the table lists the square roots of all the numbers below 100.
Hence, we have to manipulate the number such that we get the square root of a number less than 100. This can be done in the following manner:
$\sqrt{101}=\sqrt{1.01\times 100}=\sqrt{1.01}\times 10$
Now, we have to find the square root of 1.01.

We have:
 $\sqrt{1}=1 \mathrm{and} \sqrt{2}=1.414$
Their difference is 0.414.
Thus, for the difference of 1 (2 $-$ 1), the difference in the values of the square roots is 0.414.
For the difference of 0.01, the difference in the values of the square roots is:
0.01 $\times$ 0.414 = 0.00414
$\therefore$ $$\begin{gathered} \sqrt{1.01}=1+0.00414=1.00414 \\ \sqrt{101}=\sqrt{1.01}\times 10=1.00414\times 10=10.0414 \end{gathered}$$

Finally, $\sqrt{\frac{101}{169}}=\frac{\sqrt{101}}{1313}=\frac{10.0414}{13}=0.772$
This value is really close to the one from the key answer.

Question 17:

Using square root table, find the square root
13.21

Answer 17:

From the square root table, we have:
 $\sqrt{13}=3.606 \mathrm{and} \sqrt{14}=\sqrt{2}\times \sqrt{7}=3.742$
Their difference is 0.136.
Thus, for the difference of 1 (14 $-$ 13), the difference in the values of the square roots is 0.136.
For the difference of 0.21, the difference in the values of their square roots is:
$0.136\times 0.21=0.02856$
$\therefore$ $\sqrt{13.21}=3.606+0.02856\approx 3.635$

Question 18:

Using square root table, find the square root

Answer 18:

We have to find $\sqrt{21.97}$.
From the square root table, we have:
$\sqrt{21}=\sqrt{3}\times \sqrt{7}=4.583 \mathrm{and} \sqrt{22}=\sqrt{2}\times \sqrt{11}$=4.690
Their difference is 0.107.
Thus, for the difference of 1 (22 $-$ 21), the difference in the values of the square roots is 0.107.
For the difference of 0.97, the difference in the values of their square roots is:
$0.107\times 0.97=0.104$
$\therefore$ $\sqrt{21.97}=4.583+0.104\approx 4.687$

Question 19:

Using square root table, find the square root
110

Answer 19:

$$\begin{gathered} \sqrt{110}=\sqrt{2}\times \sqrt{5}\times \sqrt{11} \\ =1.414\times 2.236\times 3.317 (\mathrm{Using} \mathrm{the} \mathrm{square} \mathrm{root} \mathrm{table} \mathrm{to} \mathrm{find} \mathrm{all} \mathrm{the} \mathrm{square} \mathrm{roots}) \\ =10.488 \end{gathered}$$
                  
           

Question 20:

Using square root table, find the square root
1110

Answer 20:

$$\begin{gathered} \sqrt{1110}=\sqrt{2}\times \sqrt{3}\times \sqrt{5}\times \sqrt{37} \\ =1.414\times 1.732\times 2.236\times 6.083 (\mathrm{Using} \mathrm{the} \mathrm{table} \mathrm{to} \mathrm{find} \mathrm{all} \mathrm{the} \mathrm{square} \mathrm{roots} ) \\ =33.312 \end{gathered}$$
                          

Question 21:

Using square root table, find the square root
11.11

Answer 21:

We have:
$\sqrt{11}=3.317 \mathrm{and} \sqrt{12}=3.464$
Their difference is 0.1474.
Thus, for the difference of 1 (12 $-$ 11), the difference in the values of the square roots is 0.1474.
For the difference of 0.11, the difference in the values of the square roots is:
0.11 $\times$ 0.1474 = 0.0162
$\therefore$ $\sqrt{11.11}=3.3166+0.0162=3.328\approx 3.333$

Question 22:

The area of a square field is 325 m². Find the approximate length of one side of the field.

Answer 22:

The length of one side of the square field will be the square root of 325.
$\therefore$ $$\begin{gathered} \sqrt{325}=\sqrt{5\times 5\times 13} \\ =5\times \sqrt{13} \\ =5\times 3.605 \\ =18.030 \end{gathered}$$
           
Hence, the length of one side of the field is 18.030 m.

Question 23:

Find the length of a side of a sqiare, whose area is equal to the area of a rectangle with sides 240 m and 70 m.

Answer 23:

The area of the rectangle = 240 m $\times$ 70 m = 16800 m²
Given that the area of the square is equal to the area of the rectangle.
Hence, the area of the square will also be 16800 m².
The length of one side of a square is the square root of its area.
$\therefore$ $$\begin{gathered} \sqrt{16800}=\sqrt{2\times 2\times 2\times 2\times 2\times 3\times 5\times 5\times 7} \\ =2\times 2\times 5\sqrt{2\times 3\times 7} \\ =20\sqrt{42}\mathrm{m}=129.60 \mathrm{m} \end{gathered}$$
                 
                 
  Hence, the length of one side of the square is 129.60 m