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

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

Solution:

Let the side of square field as x

x2 = 5184 m2

x = âˆš5184m

x = 2 Ã— 2 Ã—2 Ã— 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 Ã— 48

= 96m

Area of rectangle = l Ã— b

Area of rectangle = 96 Ã— 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 Ã— 32 Ã— 5 (Pairing of 2 and 3)

5 is left out

Multiplying the number with 5

180 Ã— 5 = 22 Ã— 32 Ã— 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 Ã— 32 Ã—2 Ã— 5

2 and 5 are left out

Multiplying the number with 2 Ã— 5 = 10

360 Ã— 10 = 22 Ã— 32 Ã— 52 Ã— 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 â€“ 1 = 120

120 â€“ 3 = 117

117 â€“ 5 = 112

112 â€“ 7 = 105

105 â€“ 9 = 96

96 â€“ 11 = 85

85 â€“ 13 = 72

72 â€“ 15 = 57

57 â€“ 17 = 40

40 â€“ 19 = 21

21 â€“ 21 = 0

11 times subtraction operation is carried out

Therefore, âˆš121 = 11

169 â€“ 1 = 168

168 â€“ 3 = 165

165 â€“ 5 = 160

160 â€“ 7 = 153

153 â€“ 9 = 144

144 â€“ 11 = 133

133 â€“ 13 = 120

120 â€“ 15 = 105

105 â€“ 17 = 88

88 â€“ 19 = 69

69 â€“ 21 = 48

48 â€“ 23 = 25

25 â€“ 25 = 0

13 times subtraction operation is carried out

Therefore, âˆš169 = 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 Ã— 22 Ã— 22 Ã— 112

Therefore, the square root of 7744 is

âˆš7744 = 2 Ã— 2 Ã— 2 Ã— 11

= 88

(ii) 9604

Solution:

Prime factorization of 9604 is

9604 = 22 Ã— 72 Ã— 72

Therefore, the square root of 9604 is

âˆš9604 = 2 Ã— 7 Ã— 7

= 98

(iii) 5929

Solution:

Prime factorization of 5929 is

5929 = 112 Ã— 72

Therefore, the square root of 5929 is

âˆš5929 = 11 Ã— 7

= 77

(iv) 7056

Solution:

Prime factorization of 7056 is

7056 = 22 Ã— 22 Ã— 72 Ã— 32

Therefore, the square root of 7056 is

âˆš7056 = 2 Ã— 2 Ã— 7 Ã— 3

= 84

### Question 15. The students of class VIII of a school donated Rs 2401 for PMâ€™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 Ã— x rupees = 2401

x2 = 2401

x = âˆš2401

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 = âˆš5929

x = 77

Therefore, total number of rows are 77.

Previous
Next