Class 8 RD Sharma Solutions Chapter 3 Squares and Square Roots Exercise 3.6 (original) (raw)

Last Updated : 23 Jul, 2025

Exercise 3.6 of RD Sharma's Class 8 Maths textbook is based on the concept of square roots of rational numbers. It focuses on rational numbers expressed in fractional form.

In this article, we will provide the solution for all the questions mentioned in Exercise 3.6 of Class 8 RD Sharma Book.:

Exercise 3.6 in Chapter 3 of RD Sharma's Class 8 Mathematics textbook focuses on advanced applications of squares and square roots. This exercise likely builds upon the concepts introduced in previous sections, challenging students with more complex problems involving square roots. The problems in this set may include a mix of numerical calculations, algebraic manipulations, and word problems that require students to apply their knowledge of squares and square roots in various contexts.

How to Find the Square Root of a Rational Number?

Below are the steps to find the square root of a rational number:

**Simplify the fraction: Reduce the given fraction to its simplest form.
**Find the square root of the numerator and denominator: Determine the square roots of both the numerator and the denominator separately.
**Simplify the result: If possible, simplify the resulting fraction further.

**Question 1. Find the square root of :

****(i) 441/961**

****(ii) 324/841**

****(iii) 4 29⁄29**

****(iv) 2 14⁄25**

****(v) 2 137⁄196**

****(vi) 23 26⁄121**

****(vii) 25 544⁄729**

****(viii) 75 46⁄49**

****(ix) 3 942⁄2209**

****(x) 3 334⁄3025**

****(xi) 21 2797⁄3364**

****(xii) 38 11⁄25**

****(xiii) 23 394⁄729**

****(xiv) 21 51⁄169**

****(xv) 10 151⁄225**

**Solution:

****(i) 441/961**

The square root of √441/961 = 21/31

****(ii) 324/841**

The square root of √324/841= 18/29

****(iii) 4 29⁄29**

The square root of √(4 29⁄29) = √(225/49) = 15/7

****(iv) 2 14⁄25**

The square root of √(2 14⁄25) = √(64/25) = 8/5

****(v) 2 137⁄196**

The square root of √2 137⁄196 = √ (529/196) = 23/14

****(vi) 23 26⁄121**

The square root of √(23 26⁄121) = √(2809/121) = 53/11

****(vii) 25 544⁄729**

The square root of √(25 544⁄729) = √(18769/729) = 137/27

****(viii) 75 46⁄49**

The square root of √(75 46⁄49) = √(3721/49) = 61/7

****(ix) 3 942⁄2209**

The square root of √(3 942⁄2209) = √(7569/2209) = 87/47

****(x) 3 334⁄3025**

The square root of √(3 334⁄3025) = √(9409/3025) = 97/55

****(xi) 21 2797⁄3364**

The square root of √(21 2797⁄3364) = √(73441/3364) = 271/58

****(xii) 38 11⁄25**

The square root of √(38 11⁄25) = √(961/25) = 31/5

****(xiii) 23 394⁄729**

The square root of √(23 394⁄729) = √(17161/729) = 131/27 = 4 23/27

****(xiv) 21 51⁄169**

The square root of √(21 51⁄169) = √(3600/169) = 60/13 = 4 8/13

****(xv) 10 151⁄225**

The square root of √(10 151⁄225) = √(2401/225) = 49/15 = 3 4/15

**Question 2. Find the value of:

****(i) √80/√405**

****(ii) √441/√625**

****(iii) √1587/√1728**

****(iv) √72 × √338**

****(v) √45 × √20**

**Solution:

****(i) √80/√405** = √16/√81 = 4/9

****(ii) √441/√625** = 21/25

****(iii) √1587/√1728** = √529/√576 = 23/24

****(iv) √72 ×√338**

= √(2×2×2×3×3) ×√(2×13×13)

As we know the formula **√a × √b = √(a×b)

= √(2×2×2×3×3×2×13×13) = 22 × 3 × 13 = 156

****(v) √45 × √20** = √(5×3×3) × √(5×2×2)

As we know the formula **√a × √b = √(a×b)

= √(5×3×3×5×2×2) = 5 × 3 × 2 = 30

**Question 3. The area of a square field is 80 244⁄729 square metres. Find the length of each side of the field.

**Solution:

Given that,

Area of square field = 80 244⁄729 m2 = 58564/729 m2

Let's assume L is length of each side then,

L2 = 58564/729

L = √ (58564/729) = √58564/√729

= 242/27 = 8 26⁄27

The Length of each side of field is 8 26⁄27 m.

**Question 4. The area of a square field is 30 1⁄4m2. Calculate the length of the side of the square.

**Solution:

Given that,

Area of square field = 30 1⁄4 m2 = 121/4 m2

Let's assume L is length of each side then,

L2 = 121/4

L = √(121/4) = √121/√4 = 11/2

The Length of each side of field is 11/2 m.

**Question 5. Find the length of a side of a square playground whose area is equal to the area of a rectangular field of dimensions 72m and 338 m.

**Solution :

Given that,

l = 72m , b = 338m

As we know that Area of rectangular field = l × b

= 72 × 338 m2

= 24336 m2

Area of square = L2 = 24336 m2

L = √24336 = 156 m

The length of a side of a square playground 156 m.

Practice Questions

1. If √(x + 14) + √(x - 14) = 10, find the value of x.

2. Simplify: (√3 + √2)(√3 - √2) + (√5 + √3)(√5 - √3)

3. A square field has a diagonal of 20√2 meters. Find its area and perimeter.

4. Solve for x: √(2x + 3) + √(2x - 3) = 4

5. The sum of a number and its square root is 132. Find the number.

6. If a² + b² = 25 and ab = 12, find the value of (a + b)².

7. A rectangular paper of length 18 cm is rolled to form a cylinder of radius 3 cm. Find the height of the cylinder.

8. Simplify: (√7 + √3)² + (√7 - √3)²

9. The area of a rectangular field is 1800 square meters. If the length is 15 meters more than the breadth, find the dimensions of the field.

10. If x = √3 + 1 and y = √3 - 1, find the value of x² + y² and xy.

Summary

Exercise 3.6 of RD Sharma's Class 8 Mathematics textbook provides a comprehensive set of advanced problems on squares and square roots. This exercise is designed to challenge students and deepen their understanding of these fundamental mathematical concepts. The problems likely cover a wide range of applications, including:

Solving equations involving square roots
Simplifying complex expressions with square roots
Applying square roots in geometry (areas, perimeters, diagonals)
Word problems requiring the use of squares and square roots
Algebraic manipulations involving squares and square roots

By working through these problems, students will develop stronger problem-solving skills and gain a more intuitive understanding of how squares and square roots behave in various mathematical contexts. This exercise set helps prepare students for more advanced mathematical topics they will encounter in higher grades.