Class 8 RD Sharma Solutions - Chapter 3 Squares and Square Roots - Exercise 3.4 | Set 2

Exercise 3.4 Set 2 of RD Sharma's Class 8 Mathematics textbook continues the exploration of squares and square roots. This set likely focuses on more advanced applications and problem-solving techniques, building upon the concepts introduced in Set 1. Students can expect to encounter more complex calculations, multi-step problems, and real-world applications of square roots. Exercise 3.4 Set 2 of RD Sharma's Class 8 Mathematics textbook delves deeper into the intricate world of squares and square roots, building upon the foundation laid in previous exercises. This set is designed to challenge students' understanding and application of these fundamental mathematical concepts in more complex scenarios.

Question 11. The area of a square field is 5184 m². A rectangular field, whose length is twice its breadth has a perimeter equal to the perimeter of a square field. Find the area of the rectangular field.

Solution:

Let the side of square field as x 

x² = 5184 m²

x = √5184m

x = 2 × 2 ×2 × 9

  = 72 m

Perimeter of square = 4x

                                = 4(72)

                                = 288 m

Perimeter of rectangle = 2 (l + b) = perimeter of the square field

                                                     = 288 m

l = 2b

2 (2b + b) = 288

2(3b) = 288

6b = 288

b = 288/6 (Transposing 6)

b = 48m

l = 2 × 48

= 96m

Area of rectangle = l × b

Area of rectangle = 96 × 48 m²

                            = 4608 m²

Question 12. Find the least square number, exactly divisible by each one of the numbers:

 (i) 6, 9, 15 and 20 

Solution:

L.C.M of 6, 9, 15, 20 is 180

Prime factorization of 180 = 2² × 3² × 5 (Pairing of 2 and 3)

5 is left out

Multiplying the number with 5

180 × 5 = 2² × 3² × 5²

= 900

Therefore, 900 is the least square number divisible by 6, 9, 15 and 20

(ii) 8, 12, 15 and 20

Solution:

L.C.M of 8, 2, 15, 20 is 360

Prime factorization of 360 = 2² × 3² ×2 × 5

2 and 5 are left out

Multiplying the number with 2 × 5 = 10

360 × 10 = 2² × 3² × 5² × 2²

Therefore, 3600 is the least square number divisible by 8, 12, 15 and 20

Question 13. Find the square roots of 121 and 169 by the method of repeated subtraction.

Solution:

In repeated subtraction method, odd numbers are subtracted one by one from the previous result and number of times subtraction is carried out is the square root.

121 – 1 = 120

120 – 3 = 117

117 – 5 = 112

112 – 7 = 105

105 – 9 = 96

96 – 11 = 85

85 – 13 = 72

72 – 15 = 57

57 – 17 = 40

40 – 19 = 21

21 – 21 = 0

11 times subtraction operation is carried out

Therefore, √121 = 11

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 88

88 – 19 = 69

69 – 21 = 48

48 – 23 = 25

25 – 25 = 0

13 times subtraction operation is carried out

Therefore, √169 = 13

Question 14. Write the prime factorization of the following numbers and hence find their square roots.

(i) 7744

Solution:

Prime factorization of 7744 is

7744 = 2² × 2² × 2² × 11²

Therefore, the square root of 7744 is

√7744 = 2 × 2 × 2 × 11

           = 88

(ii) 9604

Solution:

Prime factorization of 9604 is

9604 = 2² × 7² × 7²

Therefore, the square root of 9604 is

√9604 = 2 × 7 × 7

           = 98

(iii) 5929

Solution:

Prime factorization of 5929 is

5929 = 11² × 7²

Therefore, the square root of 5929 is

√5929 = 11 × 7

           = 77

(iv) 7056

Solution:

Prime factorization of 7056 is

7056 = 2² × 2² × 7² × 3²

Therefore, the square root of 7056 is

√7056 = 2 × 2 × 7 × 3

           = 84

Question 15. The students of class VIII of a school donated Rs 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class, Find the number of students in the class.

Solution:

Let the number of students be x

Each student denoted x rupees

Total amount collected is x × x rupees = 2401

x² = 2401

x = √2401

x = 49

Therefore, there are 49 students in the class.

Question 16. A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.

Solution:

Let the number of rows be x

Number of columns = x

Total number of students in the arrangement = x²

71 students are left out

Total students x² + 71 = 6000

x² = 5929

x = √5929

x = 77

Therefore, total number of rows are 77.

Summary

Exercise 3.4 Set 2 provides students with challenging problems to deepen their understanding of squares and square roots. It likely includes a mix of computational exercises and word problems that require critical thinking and application of mathematical concepts. This set aims to enhance students' problem-solving skills and prepare them for more advanced mathematical topics.Exercise 3.4 Set 2 serves as a comprehensive exploration of advanced concepts related to squares and square roots, tailored for Class 8 students following the RD Sharma curriculum. This set of problems is meticulously designed to push students beyond basic computations, encouraging them to develop a more intuitive and analytical approach to mathematical problem-solving. The questions span a wide spectrum of difficulty levels and topics, including but not limited to: complex square root calculations using methods like long division; simplification and manipulation of expressions involving square roots; application of square root properties in geometric problems; solving equations where square roots play a central role; and tackling word problems that simulate real-life scenarios requiring the use of squares and square roots.