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

### Chapter 3 Squares and Square Roots – Exercise 3.4 | Set 1

**Question 11. The area of a square field is 5184 m**^{2}. 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.

^{2}. 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

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x

^{2}= 5184 m^{2}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 m

^{2}= 4608 m

^{2}

**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 = 2

^{2}× 3^{2}× 5 (Pairing of 2 and 3)5 is left out

Multiplying the number with 5

180 × 5 = 2

^{2}× 3^{2}× 5^{2}= 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 = 2

^{2}× 3^{2}×2 × 52 and 5 are left out

Multiplying the number with 2 × 5 = 10

360 × 10 = 2

^{2}× 3^{2}× 5^{2}× 2^{2}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 = 2

^{2}× 2^{2}× 2^{2}× 11^{2}Therefore, the square root of 7744 is

√7744 = 2 × 2 × 2 × 11

= 88

**(ii) 9604**

**Solution:**

Prime factorization of 9604 is

9604 = 2

^{2}× 7^{2}× 7^{2}Therefore, the square root of 9604 is

√9604 = 2 × 7 × 7

= 98

**(iii) 5929**

**Solution:**

Prime factorization of 5929 is

5929 = 11

^{2}× 7^{2}Therefore, the square root of 5929 is

√5929 = 11 × 7

= 77

**(iv) 7056**

**Solution:**

Prime factorization of 7056 is

7056 = 2

^{2}× 2^{2}× 7^{2}× 3^{2}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

x

^{2}= 2401x = √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 = x

^{2}71 students are left out

Total students x

^{2}+ 71 = 6000x

^{2}= 5929x = √5929

x = 77

Therefore, total number of rows are 77.