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Class 8 RD Sharma Solutions – Chapter 3 Squares and Square Roots – Exercise 3.6
  • Last Updated : 28 Dec, 2020

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.

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