RD Sharma Solutions Class 8 Cubes And Cube Roots Exercise 4.5

RD Sharma Solutions Class 8 Chapter 4 Exercise 4.5

RD Sharma Class 8 Solutions Chapter 4 Ex 4.5 PDF Free Download

RD Sharma Solutions Class 8 Chapter 4 Exercise 4.5

Making use of the cube root table, find the table, find the cube roots of the following (correct to three decimal points):

  1. 7
  2. 70
  3. 700
  4. 7000
  5. 1100
  6. 780
  7. 7800
  8. 1346
  9. 940
  10. 5112            
  11. 9800
  12. 732
  13. 7342
  14. 133100
  15. 37800
  16. 0.27
  17. 8.6
  18. 0.86
  19. 8.65
  20. 7532   
  21. 833     
  22. 34.2

Answer:

 

Q1. 7

Answer :

Because 7 lies between 1 and 100, we will look at the row containing 7 in the column of x.

By the cube root table, we have:

\(\sqrt[3]{7}\) = 1.913

Thus, the answer is 1.913.

 

 

Q2. 70

Because 70 lies between 1 and 100, we will look at the row containing 70 in the column of x.

By the cube root table, we have:

\(\sqrt[3]{70}\) = 4.121

Thus, the answer is 4.121

 

Q3. We have:

700 = 70 x 10

Cube root of 700 will be in the column of \(\sqrt[3]{10x}\) against 70.

By the cube root table, we have:

\(\sqrt[3]{700}\) = 8.879

Thus, the answer is 8.879

 

Q4. We have:

7000 = 70 x 100

\(\sqrt[3]{7000}\) = \(\sqrt[3]{7 \times 1000}\) = \(\sqrt[3]{7}\) x \(\sqrt[3]{1000}\)

By the cube root table, we have:

\(\sqrt[3]{7}\) = 1.913 and \(\sqrt[3]{1000}\) = 10

\(\sqrt[3]{7000}\) = \(\sqrt[3]{7}\) = \(\sqrt[3]{1000}\) = 1.913 x 10 = 19.13

Thus, the answer is 19.13

 

Q5. We have:

1100 = 11 x 100

Therefore,

\(\sqrt[3]{1100}\) = \(\sqrt[3]{11 \times 100}\) = \(\sqrt[3]{11}\) x \(\sqrt[3]{100}\)

By the cube root table, we have:

\(\sqrt[3]{11}\) = 2.224 and \(\sqrt[3]{100}\) = 4.642

\(\sqrt[3]{1100}\) = \(\sqrt[3]{11}\)\(\sqrt[3]{100}\) = 2.224 x 4.642 = 10.323 (Up to three decimal places)

Thus, the answer is 10.323.

 

Q6. We have:

780 = 78 x 10

Therefore, Cube root of 780 will be in the column of \(\sqrt[3]{10x}\) against 78.

By the cube root table, we have:

\(\sqrt[3]{780}\) = 9.025

Thus, the answer is 9.025

 

Q7. 7800

7800 = 78 x 100

\(\sqrt[3]{7800}\) = \(\sqrt[3]{78 \times 100}\) = \(\sqrt[3]{78}\)\(\sqrt[3]{100}\)

By the cube root table, we have:

\(\sqrt[3]{78}\) = 4.273 and  \(\sqrt[3]{100}\) = 4.642

\(\sqrt[3]{7800}\) = \(\sqrt[3]{78}\)\(\sqrt[3]{100}\) = 4.273 x 4.642 = 19.835 (up to three decimal places)

Thus, the answer is 19.835

 

Q8. 1346

Answer:

By prime factorisation, we have:

1346 = 2 x 673 => \(\sqrt[3]{1346}\) = \(\sqrt[3]{2}\) x \(\sqrt[3]{673}\)

Also 670 < 673 < 680 => \(\sqrt[3]{670}\) < \(\sqrt[3]{673}\) < \(\sqrt[3]{680}\)

From the cube root table, we have:

\(\sqrt[3]{670}\) = 8.750 and \(\sqrt[3]{680}\) = 8.794

For the difference (680 – 670), i.e., 10, the difference in the values

= 8.794 – 8.750 = 0. 044

For the difference of (673 – 670), i.e., 3, the difference in the values

=  \(\frac{0.044 \times 3}{10}\) = 0.0132 = 0.013  (up to three decimal places)

= 8.750 + 0.013 = 8.763

Now,

\(\sqrt[3]{1346}\) = \(\sqrt[3]{2}\) x \(\sqrt[3]{8.763}\) = 1.260 x 8.763 = 11.041 (up to three decimal places)

Thus, the answer is 11.041

Q 9. 940

Answer :

We have:

250 = 25 x 100

Cube root of 250 would be in the column of \(\sqrt[3]{10x}\) against 25.

By the cube root table, we have:

\(\sqrt[3]{250}\) = 6.3

Thus, the required cube root is 6.3.

 

Q10.   5112

Answer :

By prime factorisation, we have:

5112 = 23 x 32 x 71 => \(\sqrt[3]{}\) = 2 x \(\sqrt[3]{9}\) x \(\sqrt[3]{71}\)

By the cube root table, we have:

\(\sqrt[3]{9}\) = 2.080 and \(\sqrt[3]{71}\) = 4.141

\(\sqrt[3]{5112}\) = 2 x \(\sqrt[3]{9}\) x \(\sqrt[3]{71}\)= 2 x 2.080 x 4.141 = 17.227 (up to three decimal places)

Thus, the required cube root is 17.227.

 

Q11. We have:

9800 = 98 x 100

\(\sqrt[3]{9800}\) = \(\sqrt[3]{98 \times 100}\) = \(\sqrt[3]{98}\) x \(\sqrt[3]{100}\)

By the cube root table, we have:

\(\sqrt[3]{98}\) = 4.610 and \(\sqrt[3]{100}\) = 4.642

\(\sqrt[3]{9800}\) = \(\sqrt[3]{98}\) x \(\sqrt[3]{100}\) = 4.610 x 4.642 = 21.40  (up to three decimal places)

Thus, the required cube root is 21.40.

 

Q12. 732

Answer :

We have:

730 < 732 < 740 => \(\sqrt[3]{730}\) < \(\sqrt[3]{732}\) < \(\sqrt[3]{740}\)

From cube root table, we have:

\(\sqrt[3]{730}\) = 9.004 and \(\sqrt[3]{740}\) = 9.045

For the difference (740 – 730), i.e., 10, the difference in values

= 9.045 – 9.004 = 0.041

For the difference of (732 – 730), i.e., 2, the difference in values

\(\frac{0.044 \times 2}{10}\) = 0.0082

\(\sqrt[3]{732}\)  = 9.004 + 0.008 = 9.012

 

Q13. 7342

Answer:

We have:

7300 < 7342 < 7400 => \(\sqrt[3]{7300}\) < \(\sqrt[3]{7342}\) < \(\sqrt[3]{7400}\)

From the cube root table, we have:

\(\sqrt[3]{7300}\) = 19.39 and \(\sqrt[3]{7400}\) = 19.48

For the difference (7400 – 7300), i.e., 100, the difference in values

= 19.48 – 19.39 = 0.09

For the difference of (7342 – 7300), i.e., 42, the difference in the values

= \(\frac{0.09 \times 42}{100}\) = 0.0378 = 0.037

\(\sqrt[3]{7342}\) = 19.39 + 0.037 = 19.427

 

Q14. We have:

133100 = 1331 x 100 => \(\sqrt[3]{133100}\) = \(\sqrt[3]{1331 \times 100}\) = 11 x \(\sqrt[3]{100}\)

From the cube root table, we have:

\(\sqrt[3]{100}\) = 4.642

\(\sqrt[3]{133100}\) = 11 x \(\sqrt[3]{100}\) = 11 x 4.642 = 51.062

 

Q15. We have,

37800 = \(2 ^{3} \times 3 ^{3} \times 175 => \sqrt[3]{37800} = \sqrt[3]{2 ^{3} \times 3 ^{3} \times 175} = 6 \times \sqrt[3]{175}\)

Also 170 < 175 < 180 => \(\sqrt[3]{170}\) < \(\sqrt[3]{175}\) < \(\sqrt[3]{180}\)

From cube root table, we have:

\(\sqrt[3]{170}\) = 5.540 and \(\sqrt[3]{180}\) = 5.646

For the difference (180 – 170), i.e., 10, the difference in values

= 5.646 – 5.540 = 0.106

For the difference of (175 – 170), i.e., 5, the difference in values

\(\frac{0.106 \times 5}{10}\) = 0.053

\(\sqrt[3]{175}\) = 5.540 + 0.053 = 5.593

Now 37800 = 6 x \(\sqrt[3]{175}\) = 6 x 5.593 = 33.558

Thus, the required cube root is 33.558.

 

Q16. 0.27

The number 0.27 can be written as \(\frac{27}{100}\)

Now,

\(\sqrt[3]{0.27} = \sqrt[3]{\frac{27}{100}} = \frac{\sqrt[3]{27}}{\sqrt[3]{100}} = \frac{3}{\sqrt[3]{100}}\)

From cube root table, we have:

\(\sqrt[3]{100}\) = 4.642

\(\sqrt[3]{0.27}\) = \(\frac{3}{\sqrt[3]{100}} = \frac{3}{4.642} = 0.646\)

Thus, the required cube root is 0.646.

Q17. 8.6

The number 8.6 can be written as \(\frac{86}{10}\)

Now

\(\sqrt[3]{8.6} = \sqrt[3]{\frac{86}{10}} = \frac{\sqrt[3]{86}}{\sqrt[3]{10}} \\ From \; cube \; root \; table, \; we \; have: \\ = \sqrt[3]{86} = 4.414 \; and \; \sqrt[3]{10} = 2.154 \\ = \sqrt[3]{8.6} = \frac{\sqrt[3]{86}}{\sqrt[3]{10}} = \frac{4.414}{2.154} = 2.049\)

Thus, the required cube root is 2.049.

 

Q18. 0.86

The number 0.86 can be written as \(\frac{86}{100}\)

Now

\(\sqrt[3]{0.86} = \sqrt[3]{\frac{86}{100}} = \frac{\sqrt[3]{86}}{\sqrt[3]{100}} \\ From \; cube \; root \; table, \; we \; have: \\ = \sqrt[3]{86} = 4.414 \; and \; \sqrt[3]{100} = 4.342 \\ = \sqrt[3]{0.86} = \frac{\sqrt[3]{86}}{\sqrt[3]{100}} = \frac{4.414}{4.642} = 0.951\)

Thus, the required cube root is 0.951.

 

Q19. 8.65

Answer :

The number 8.65 could be written as \(\frac{865}{100}\)

Now

\(\sqrt[3]{8.65} = \sqrt[3]{\frac{865}{100}} = \frac{\sqrt[3]{865}}{\sqrt[3]{100}} \\ Also, \; 860 \; < \; 865 \; < \; 870 => \sqrt[3]{860} \; < \; \sqrt[3]{865} \; < \; \sqrt[3]{870} \\ From \; cube \; root \; table, \; we \; have: \\ = \sqrt[3]{860} = 9.510 \; and \; \sqrt[3]{870} = 9.546 \\\)

For the difference (870 – 860), i.e., 10, the difference in values

= 9.546 – 9.510 = 0.036

For the difference of (865 – 860), i.e., 5, the difference in values

= \(\frac{0.036 \times 5}{10}\) = 0.018 (up to three decimal places)

\(\sqrt[3]{865}\) = 9.510 + 0.018 = 9.528 (up to three decimal places)

From cube root table, we also have:

\(\sqrt[3]{100}\) 4.642

\(= \sqrt[3]{8.65} = \frac{\sqrt[3]{865}}{\sqrt[3]{100}} = \frac{9.528}{4.642} = 2.053\) (up to three decimal places)

Thus, the required cube root is 2.053

 

Q20. We have, 7532

\(7500 \; < \; 7532 \; < \; 7600 => \sqrt[3]{7500} \; < \; \sqrt[3]{7532} \; < \; \sqrt[3]{7600} \\ From \; cube \; root \; table, \; we \; have: \\ = \sqrt[3]{7500} = 19.57 \; and \; \sqrt[3]{7600} = 19.66 \\\)

For the difference of (7600 – 7500), i.e., 100, the difference in values

= 19.66 – 19.57 = 0.09

For the difference of (7532 – 7500), i.e., 32, the difference in values,

= \(\frac{0.009 \times 32}{100}\) = 0.0288 = 0.029 (up to three decimal places)

\(\sqrt[3]{7532}\) = 19.57 + 0.029 = 19.599

Thus, the required cube root is 19.599

 

Q21. We have, 833

\(830 \; < \; 833 \; < \; 840 => \sqrt[3]{830} \; < \; \sqrt[3]{833} \; < \; \sqrt[3]{840} \\ From \; cube \; root \; table, \; we \; have: \\ = \sqrt[3]{830} = 9.398 \; and \; \sqrt[3]{840} = 9.435 \\\)

For the difference of (840 – 830), i.e., 10, the difference in values

= 9.435 – 9.398 = 0.037

For the difference of (833 – 830), i.e., 3, the difference in values

= \(\frac{0.037 \times 3}{10}\) = 0.0111 = 0.011 (up to three decimal places)

\(\sqrt[3]{833}\) = 9.398 + 0.011 = 9.409

Thus, the required cube root is 9.409

 

Q22. 34.2

The number 34.2 could be written as \(\frac{342}{10}\)

Now,

\(\sqrt[3]{34.2} = \sqrt[3]{\frac{342}{10}} = \frac{\sqrt[3]{342}}{\sqrt[3]{10}} \\ Also \\ 340 \; < \; 342 \; < \; 350 => \sqrt[3]{340} \; < \; \sqrt[3]{342} \; < \; \sqrt[3]{350} \\ From \; cube \; root \; table, \; we \; have: \\ = \sqrt[3]{340} = 6.980 \; and \; \sqrt[3]{350} = 7.047 \\\)

For the difference of (350 – 340), i.e., 10, the difference in values

= 7.047 – 6.980 = 0.067

For the difference of (342 – 340), i.e., 2, the difference in values

= \(\frac{0.067 \times 2}{10}\) (up to three decimal places)

From cube root table, we also have:

\(\sqrt[3]{10}\) = 2.154

\(\sqrt[3]{34.2}\) =  \(\frac{\sqrt[3]{342}}{\sqrt[3]{10}} = \frac{6.993}{2.154} = 3.246\)

Thus, the required cube root is 3.246


Practise This Question

Find the area of the triangle formed by joining the midpoints of the sides of the triangle formed with coordinates A (-4, 3), B (2, 3) and C (4, 5) in sq. units.