Squares and Cube Roots Class 8 RS Aggarwal Exe-3D Goyal Brothers ICSE Maths Solutions

Squares and Cube Roots Class 8 RS Aggarwal Exe-3D Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3. We provide step by step Solutions of cisce prescribed publication / textbook  to develop skill and confidence among students. In this articles you would learn to find out squire roots of decimal numbers. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.

Squares and Cube Roots Class 8 RS Aggarwal Exe-3D Goyal Brothers ICSE Maths Solutions

Squares and Cube Roots Class 8 RS Aggarwal Exe-3D Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3

Board ICSE
Publications Goyal Brothers Prakshan
Subject Maths
Class 8th
Writer RS Aggarwal
Book Name Foundation
Ch-3 Squires and Squire roots , Cubes and Cube roots
Exe-3D squire roots of decimal numbers
Edition 2024-2025

Squire Roots of Decimal Numbers

(Squares and Cube Roots Class 8 RS Aggarwal Exe-3D Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3)

Page- 51,52

Exercise- 3D

Squire Roots of Decimal Numbers

Que-1: Find the square root of :

(i) 2.89     (ii) 33.64      (iii) 156.25 (iv) 75.69    (v) 5.4289        (vi) 18.1476

Solution- (i) √2.89 = 1.7
(ii) √33.64 = 5.8
(iii) √156.25 = 12.5
(iv) √75.69 = 8.7
(v) √5.429 = 2.33
(vi) √18.1476 = 4.26

Que-2: Evaluate √2 up to two places of decimal.

Solution- √2 = 1.414
⇒ √2 = 1.41 Ans.

Evaluate √2 up to two places of decimal

Que-3: Evaluate √0.4 correct up to two places of decimal.

Solution- Since, 0.4 = 4/10
Taking square root from both sides.
√0.4 = √(4/10)
⇒ √0.4 = √[(4/10) × (10/10)]
⇒ √0.4 = √(40/100)
⇒ √0.4 = 1/10 √40                (∴√40 = 6.3245553)
⇒ √0.4 = 1/10 × 6.3245553
⇒ √0.4 = 0.63245553
⇒ √0.4 = 0.63 Ans.

Que-4: Evaluate √2.8 correct up to two places of decimal.

Solution- √(2.8) = 1.673
⇒ √(2.8) = 1.67 (correct up to two decimal places) Ans.

Evaluate √2.8 correct up to two places of decimal.

Que-5: Find the length of each side of a square whose area is equal to the area of rectangle of length 13,6 metres and breadth 3.4 metres.

Solution-Length of rectangle (l) = 13.6 m
and width (b) = 3.4 m
∴ Area = l × b = 13.6 × 3.4
= 46.24 m²
Now, area of square = a
∴ a² = 46.24 m²
⇒ a = √46.24 m = 6.8m Ans.

Que-6: If √(1 + 27/169) = 1 + x/13, find the value of x.

Solution-  Given, √(1+27/169) = (1+x/13)
⇒ √(169+27)/169 = (1+x/13)
⇒ √(196/169) = (1+x/13)
⇒ 14/13 = (1+x/13)
⇒ 1/13 = x/13
⇒ x = 1 Ans.

Que-7: Find the square root of :

(i) 3*(13/36)   (ii) 4*(73/324)   (iii) 3*(942/2209) (iv) 1089/4624

Solution-(i) 3*(13/36) = (3×36+13)/36 = (108+13)/36 = 121/36
​Now, we’ll find the square root of this fractions.
√121/36 = √121/√36 = 11/6 = 1*(5/6) Ans.

(ii) 4*(73/324) ​= (4×324+73)/324 ​= (1296+73)/324 ​= 1369/324
√1369/324 = √1369/√324 = 37/18 = 2*(1/18) Ans.

(iii) 3*(942/2209) ​= (3×2209+942​)/2209 = (6627+942)/2209 ​= 7569/2209
√7569/2209 = √7569/√2209 = 87/47 = 1*(40/47) Ans.

(iv) √1089/4624 = √1089/√4624 = 33/68 Ans.

Que-8: Find the value of :

(i) √243/√867    (ii) √1183/√2023

Solution- (i) We have
243 = 3×3×3×3×3
867 = 3×17×17
Now, √243/√867 = √(3×3×3×3×3)/√(3×17×17)
= √(3×3×3×3)/√(17×17)
= (3×3)/17 = 9/17 Ans.

(ii) Factorise:
1183 = 7×13×13
2023 = 7×17×17
√1183/√2023 = √(7×13×13)/√(7×17×17)
= √(13×13)/√(17×17)
= 13/17 Ans.

Que-9: Find the value of √15625 and hence evaluate √156.25 + √1.5625.

Solution-√15625​ = 125 Ans.
Now, let’s use this to evaluate √156.25 + √1.5625.
√156.25 = √15625/100 = √15625/√100 = 125/10 = 12.5.
Similarly,
√1.5625 = √156.25/100 = √156.25/√100 = 12.5/10=1.25.
So,
√156.25 + √1.5625 = 12.5 + 1.25 = 13.75. Ans.

Que-10- Evaluate : (i) √99 x √396      (ii) √147 x √243

Solution- (i) First we write the prime factorization of the given numbers
99 = 3×3×11
396 = 2×2×3×3×11
Therefore, √99 × √396 = √(99×396)
= √(3×3×11×2×2×3×3×11)
= 3×2×3×11 = 198 Ans.

(ii) √143 = √3 × √49 = √3 × 7
√243 = √3 × √81 = √3 × 9
√143 × √243 = 3×9×7 = 189 Ans.

Que-11- Evaluate :

(i) √(0.289/0.00121)     (ii) √(48.4/0.289) (iii) √(0.01 + √0.0064)   (iv) √0.01 + √0.81 + √1.21 + √0.0009 (v) √[41 – √{21 + √(19 – √9)}]   (vi) √[10 + √{25 + √(108 + √(154 + √225))}]

Solution- (i) √0.289/0.00121
= √0.28900/0.00121
= √28900/121
=170/11 Ans.

(ii) √48.4/0.289 ​​= √484/289​​
Now, we can simplify the fraction 484/289​:
484/289 = 22²/17²​
Taking the square root of both the numerator and denominator:
√484/√289 = √22²/√17²​​
= 22/17 Ans.

(iii) √0.0064 = 0.08
0.01 + 0.08 = 0.09
Now, taking the square root of 0.09:
√0.09 = 0.3 Ans.

(iv) √0.01 = 0.1
√0.81 = 0.9
√1.21 = 1.1
√0.0009 = 0.03
Now, let’s add them up:
0.1 + 0.9 + 1.1 + 0.03 = 2.13 Ans.

(v) Start with the innermost square root: √9 = 3
Substitute √9 into the expression: √(19 – 3) = √16 = 4
Now we have: √{41 – √(21 + 4)}
Simplify inside the inner square root: √(21 + 4) = √25 = 5
Substitute √(21 + 4) into the expression: √{41 – 5}
Simplify the expression inside the outer square root: 41 – 5 = 36
Now we have: √36
Simplify the square root of 36: √36 = 6
So, the simplified expression is 6. Ans.

(vi) Start from the innermost expression: √225 = 15.
Next, √(154 + 15) = √169 = 13.
Continuing, √(108 + 13) = √121 = 11.
Then, √(25 + 11) = √36 = 6.
Finally, √(10 + 6) = √16 = 4.
So, the result is 4. Ans.

Que-12: Three-fifths of the square of a certain number is 126.15. Find the number.

Solution- Let the number be x.
Then, 3/5 x²= 126.15
⇔ x² = (126.15×5/3)
= 210.25
⇔ x = √210.25=14.5. Ans.

— : End of Squares and Cube Roots Class 8 RS Aggarwal Exe-3D Goyal Brothers Maths Solutions :–

Return to- ICSE Class -8 RS Aggarwal Goyal Brothers Math Solutions

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