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