RD Sharma Solutions for Class 8 Maths Chapter 3 - Squares and Square Roots Exercise 3.9

Students can refer to and download Exercise 3.9 of RD Sharma Solutions for the Class 8 Maths Chapter 3, Squares and Square Roots, from the links provided below. The expert team at BYJU’S have created the RD Sharma Solutions for Class 8 that help students solve textbook problems without any difficulty. Exercise 3.9 of Class 8 is about finding the approximate values of square roots by using square root tables. Students can practise RD Sharma Solutions for Class 8 for a better understanding of the concepts, which will help them obtain high marks in the final examination.

RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots Exercise 3.9

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Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 3.9 Chapter 3 Squares and Square Roots

EXERCISE 3.9 PAGE NO: 3.61

Using the square root table, find the square roots of the following:

1. 7

Solution:

From the square root table, we know,

The square root of 7 is:

√7 = 2.645

∴ The square root of 7 is 2.645

2. 15

Solution:

We know that,

15 = 3 x 5

So, √15 = √3 x √5

From the square root table, we know,

The square root of 3 and 5 are:

√3 = 1.732 and √5 = 2.236

⇒ √15 = 1.732 x 2.236 = 3.873

∴ The square root of 15 is 3.873

3. 74

Solution:

We know that,

74 = 2 x 37

So, √74 = √2 x √37

From the square root table, we know,

The square root of 2 and 37 are:

√2 = 1.414 and √37 = 6.083

⇒ √74 = 1.414 x 6.083 = 8.602

∴ The square root of 74 is 8.602

4. 82

Solution:

We know that,

82 = 2 x 41

So, √82 = √2 x √41

From the square root table, we know,

The square root of 2 and 41 are:

√2 = 1.414 and √41 = 6.403

⇒ √82 = 1.414 x 6.403 = 9.055

∴ The square root of 82 is 9.055

5. 198

Solution:

We know that,

198 = 2 x 9 x 11

So, √198 = √2 x √9 x √11

From the square root table, we know,

The square root of 2, 9 and 11 are:

√2 = 1.414, √9 = 3 and √11 = 3.317

⇒ √198 = 1.414 x 3 x 3.317 = 14.071

∴ The square root of 198 is 14.071

6. 540

Solution:

We know that,

540 = 6 x 9 x 10

So, √540 = √6 x √9 x √10

From the square root table, we know,

The square root of 6, 9 and 10 are:

√6 = 2.449, √9 = 3 and √10 = 3.162

⇒ √540 = 2.449 x 3 x 3.162 = 23.24

∴ The square root of 540 is 23.24

7. 8700

Solution:

We know that,

8700 = 87 x 100

So, √8700 = √87 x √100

From the square root table, we know,

The square root of 87 and 100 are:

√8700 = 9.327 and √100 = 10

⇒ √8700 = 9.327 x 10 = 93.27

∴ The square root of 8700 is 93.27

8. 3509

Solution:

We know that,

3509 = 121 x 29

So, √3509 = √121 x √29

From the square root table, we know,

The square root of 121 and 29 are:

√121 = 11 and √29 = 5.385

⇒ √3509 = 11 x 10 = 5.385

∴ The square root of 3509 is 59.235

9. 6929

Solution:

We know that,

6929 = 169 x 41

So, √6929 = √169 x √41

From the square root table, we know,

The square root of 169 and 41 are:

√169 = 13 and √41 = 6.403

⇒ √6929 = 13 x 6.403 = 83.239

∴ The square root of 6929 is 83.239

10. 25725

Solution:

We know that,

25725 = 3 x 7 x 25 x 49

So, √25725 = √3 x √7 x √25 x √49

From the square root table, we know,

The square root of 3, 7, 25 and 49 are:

√3 = 1.732, √7 = 2.646, √25 = 5 and √49 = 7

⇒ √25725 = 1.732 x 2.646 x 5 x 7 = 160.41

∴ The square root of 25725 is 160.41

11. 1312.

Solution:

We know that,

1312 = 2 x 16 x 41

So, √1312 = √2 x √16 x √41

From the square root table, we know,

The square root of 2, 16 and 41 are:

√2 = 1.414, √16 = 4 and √41 = 6.403

⇒ √1312 = 1.414 x 4 x 6.403 = 36.22

∴ The square root of 1312 is 36.22

12. 4192

Solution:

We know that,

4192 = 2 x 16 x 131

So, √4192 = √2 x √16 x √131

From the square root table, we know,

The square root of 2 and16 are:

√2 = 1.414 and √16 = 4

The square root of 131 is not listed in the table

Thus, let’s apply long division to find it

So, the square root of 131 is 11.445

Now,

⇒ √4192 = 1.414 x 4 x 11.445 = 64.75

∴ The square root of 4192 is 64.75

13. 4955

Solution:

We know that,

4955 = 5 x 991

So, √4955 = √5 x √991

From the square root table, we know,

The square root of 5 is:

√5 = 2.236

The square root of 991 is not listed in the table

Thus, let’s apply long division to find it

So, the square root of 991 is 31.480

Now,

⇒ √4955 = 2.236 x 31.480 = 70.39

∴ The square root of 4955 is 70.39

14. 99/144

Solution:

We know that,

99/144 = (9 x 11) / (12 x 12)

So, √(99/144) = √[(9 x 11) x (12 x 12)]

= 3/12 x √11

From the square root table, we know,

The square root of 11 is:

√11 = 3.317

⇒ √(99/144) = 3/12 x 3.317 = 3.317/4 = 0.829

∴ The square root of 99/144 is 0.829

15. 57/169

Solution:

We know that,

57/169 = (3 x 19) / (13 x 13)

So, √(57/169) = √[(3 x 19) x (13 x 13)]

= √3 x √19 x 1/13

From the square root table, we know,

The square root of 3 and 19 is:

√3 = 1.732 and √19 = 4.359

⇒ √(57/169) = 1.732 x 4.359 x 1/13 = 0.581

∴ The square root of 57/169 is 0.581

16. 101/169

Solution:

We know that,

101/169 = 101 / (13 x 13)

So, √(101/169) = √[101 / (13 x 13)]

= √101/13

From the square root table, we don’t have the square root of 101

Thus, we have to manipulate the number such that we get the square root of a number less than 100

√101 = √(1.01 x 100)

= √1.01 x 10

Now, we have to find the square of 1.01

We know that,

√1 = 1 and √2 = 1.414 (From the square root table)

Their difference = 1.414 – 1 = 0.414

Hence, for a difference of 1 (2 – 1), the difference in the value of the square root is 0.414

So,

For the difference of 0.01, the difference in the value of the square roots will be

0.01 x 1.414 = 0.00414

∴√1.01 = 1 + 0.00414 = 1.00414

Then, √101 = 1.00414 x 10 = 10.0414

⇒ √(101/169) = √101/13 = 10.0414/13

∴ The square root of 101/169 is 0.773

17. 13.21

Solution:

We need to find √13.21

From the square root table, we know,

The square root of 13 and 14 are:

√13 = 3.606 and √14 = 3.742

Their difference = 3.742 – 3.606 = 0.136

Hence, for a difference of 1 (14 – 13), the difference in the value of the square root is 0.136

So,

For the difference of 0.21, the difference in the value of the square roots will be

0.136 x 0.21 = 0.0286

⇒ √13.21 = 3.606 + 0.0286 = 3.635

∴ The square root of 13.21 is 3.635

18. 21.97

Solution:

We need to find √21.97

From the square root table, we know,

The square root of 21 and 22 are:

√21 = 4.583 and √22 = 4.690

Their difference = 4.690 – 4.583 = 0.107

Hence, for a difference of 1 (23 – 22), the difference in the value of the square root is 0.107

So,

For the difference of 0.97, the difference in the value of the square roots will be

0.107 x 0.97 = 0.104

⇒ √21.97 = 4.583 + 0.104 = 4.687

∴ The square root of 21.97 is 4.687

19. 110

Solution:

We know that,

110 = 11 x 10

So, √110 = √11 x √10

From the square root table, we know,

The square root of 11 and 10 are:

√11 = 3.317 and √10 = 3.162

⇒ √110 = 3.317 x 3.162 = 10.488

∴ The square root of 110 is 10.488

20. 1110

Solution:

We know that,

1110 = 37 x 30

So, √1110 = √37 x √30

From the square root table, we know,

The square root of 37 and 30 are:

√37 = 6.083 and √30 = 5.477

⇒ √1110 = 6.083 x 5.477 = 33.317

∴ The square root of 1110 is 33.317

21. 11.11

Solution:

We need to find √11.11

From the square root table, we know,

The square root of 11 and 12 are:

√11 = 3.317 and √12 = 3.464

Their difference = 3.464 – 3.317 = 0.147

Hence, for a difference of 1 (12 – 11), the difference in the value of the square root is 0.147

So,

For the difference of 0.11, the difference in the value of the square roots will be

0.11 x 0.147 = 0.01617

⇒ √11.11 = 3.317 + 0.0162 = 3.333

∴ The square root of 11.11 is 3.333

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

Solution:

We know that the given area of the field = 325 m²

To find the approximate length of the side of the field we will have to calculate the square root of 325

√325 = √25 x √13

From the square root table, we know

√25 = 5 and √13 = 3.606

⇒ √325 = 5 x 3.606 = 18.030

∴ The approximate length of one side of the field is 18.030 m

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

Solution:

We know that from the question,

Area of square = Area of rectangle

Side² = 240 × 70

Side = √(240 × 70)

= √(10 × 10 × 2 × 2 × 2 × 3 × 7)

= 20√42

Now, from the square root table, we know √42 = 6.481

= 20 × 6.48

= 129.60 m

∴ The length of the side of the square is 129.60 m.

RD Sharma Solutions for Class 8 Maths Exercise 3.9 Chapter 3 – Squares and Square Roots

Exercise 3.9 of RD Sharma Solutions for Chapter 3 Squares and Square Roots, mainly deals with the concepts related to finding the approximate values of square roots by using square root tables. The students can solve problems in the RD Sharma textbook for Class 8 Maths using the solutions as reference material. This will improve their confidence in solving tricky and lengthy problems easily. The PDF is easily accessible by the students, which helps in speeding up their exam preparation.