Students can click on the links given below to access RD Sharma Solutions for Class 8 Maths Exercise 3.6 Chapter 3, Squares and Square Roots. The questions present in this exercise have been solved by BYJU’S experts in Maths in a comprehensive manner, which will help students in solving Exercise 3.6 of Class 8 easily and quickly. Exercise 3.6 of Class 8 helps students to understand the concept of finding the square root of rational numbers in the form of fractions.
By practising the RD Sharma Solutions, Class 8, students can gain confidence in attempting the annual examination, which plays a vital role in scoring good marks.
RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots Exercise 3.6
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 3.6 Chapter 3 Squares and Square Roots
EXERCISE 3.6 PAGE NO: 3.48
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
2. Find the value of:
(i) √80/√405
(ii) √441/√625
(iii) √1587/√1728
(iv) √72 ×√338
(v) √45 × √20
Solution:
(i) √80/√405
√80/√405 = √16/√81 = 4/9
(ii) √441/√625
√441/√625 = 21/25
(iii) √1587/√1728
√1587/√1728 = √529/√576 = 23/24
(iv) √72 ×√338
√72 ×√338 = √(2×2×2×3×3) ×√(2×13×13)
By using the formula √a × √b = √(a×b)
= √(2×2×2×3×3×2×13×13)
= 22 × 3 × 13
= 156
(v) √45 × √20
√45 × √20 = √(5×3×3) × √(5×2×2)
By using the formula √a × √b = √(a×b)
= √(5×3×3×5×2×2)
= 5 × 3 × 2
= 30
3. The area of a square field is 80 244/729 square metres. Find the length of each side of the field.
Solution:
We know that the given area = 80 244/729 m2
= 58564/729 m2
If L is the length of each side
L2 = 58564/729
L = √ (58564/729) = √58564/√729
= 242/27
= 8 26/27
∴ Length is 8 26/27
4. The area of a square field is 30 1/4m2. Calculate the length of the side of the square.
Solution:
We know that the given area = 30 1/4 m2
= 121/4 m2
If L is the length of each side, then,
L2 = 121/4
L = √(121/4) = √121/√4
= 11/2
∴ Length is 11/2
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:
By using the formula
Area of rectangular field = l × b
= 72 × 338 m2
= 24336 m2
Area of the square, L2 = 24336 m2
L = √24336
= 156 m
∴ The length of the side of the square playground is 156 m.
RD Sharma Solutions for Class 8 Maths Exercise 3.6 Chapter 3 – Squares and Square Roots
Exercise 3.6 of RD Sharma Solutions for Chapter 3 Squares and Square Roots, mainly deals with the concepts related to the square roots of rational numbers in the form of fractions.
RD Sharma Class 8 Solutions can help the students practise and learn each and every concept, as it provides solutions to all questions in the RD Sharma textbook. Students who aim to secure high marks in Maths in the final exam are advised to practise all the questions present in the textbook. Very lengthy questions that require many steps can also be easily remembered by students by practising RD Sharma Solutions.
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