RS Aggarwal solution class 8 chapter 3 Squares and Square roots Exercise 3D

Exercise 3D

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Q1 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL  | myhelper

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Question 1:

Find the square root of number by using the method of prime factorisation:
225

Answer 1:

By prime factorisation method:

225=3×3×5×5225=3×5=15


Q2 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 2:

Find the square root of number by using the method of prime factorisation:
441

Answer 2:

By prime factorisation:

      441=3×3×7×7  441=3×7=21


Q3 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 3:

Find the square root of number by using the method of prime factorisation:
729

Answer 3:

Resolving into prime factors:

729=3×3×3×3×3×3

729=3×3×3=27


Q4 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 4:

Find the square root of number by using the method of prime factorisation:
1296

Answer 4:

Resolving into prime factors:

1296=2×2×2×2×3×3×3×3

1296=2×2×3×3=36


Q5 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 5:

Find the square root of number by using the method of prime factorisation:
2025

Answer 5:

Resolving into prime factors:

2025=3×3×3×3×5×5

2025=3×3×5=45


Q6 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 6:

Find the square root of number by using the method of prime factorisation:
4096

Answer 6:

Resolving into prime factors:
4096=2×2×2×2×2×2×2×2×2×2×2×2

4096=2×2×2×2×2×2=64


Q7 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 7:

Find the square root of number by using the method of prime factorisation:
7056

Answer 7:

Resolving into prime factors:

7056=2×2×2×2×3×3×7×7

7056=2×2×3×7=84


Q8 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 8:

Find the square root of number by using the method of prime factorisation:
8100

Answer 8:

Resolving into prime factors:

8100=2×2×3×3×3×3×5×5

8100=2×3×3×5=90


Q9 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 9:

Find the square root of number by using the method of prime factorisation:
9216

Answer 9:

Resolving into prime factors:

9216=2×2×2×2×2×2×2×2×2×2×3×3

9216=2×2×2×2×2×3=96


Q10 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 10:

Find the square root of number by using the method of prime factorisation:
11025

Answer 10:

Resolving into prime factors:

11025=3×3×5×5×7×7

11025=3×5×7=105


Q11 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 11:

Find the square root of number by using the method of prime factorisation:
15876

Answer 11:

Resolving into prime factors:

15876=2×2×3×3×3×3×7×7

15876=2×3×3×7=126


Q12 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 12:

Find the square root of number by using the method of prime factorisation:
17424

Answer 12:

Resolving into prime factors:

17424=2×2×2×2×3×3×11×11

17424=2×2×3×11=132


Q13 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 13:

Find the smallest number by which 252 must be multiplied to get a perfect square. Also, find the square root of the perfect square so obtained.

Answer 13:

Resolving into prime factors:

252=2×2×3×3×7

Thus, the given number must be multiplied by 7 to get a perfect square.

New number = 252×7=1764

1764=2×3×7=42


Q14 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 14:

Find the smallest number by which 2925 must be divided to obtain a perfect square. Also, find the square root of the perfect square so obtained.

Answer 14:

Resolving into prime factors:

2925=3×3×5×5×13

13 is the smallest number by which the given number must be divided to make it a perfect square.

New number = 2925÷13=225

225=3×5=15


Q15 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 15:

1225 Plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Answer 15:

Let the number of rows be x.
Therefore, the number of plants in each row is also x.
Total number of plants =x × x=x2=1225
x2=1225=5×5×7×7x=1225=5×7=35

Thus, the number of rows is 35 and the number of plants in each row is 35.


Q16 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 16:

The students of a class arranged a picnic. Each student contributed as many rupees as the number of students in the class. If the total contribution is Rs 1156, find the strength of the class.

Answer 16:

Let the number of students be x.
Hence, the amount contributed by each student is Rs x.

Total amount contributed =x×x=x2=1156

1156=2×2×17×17x=1156=2×17=34

Thus, the strength of the class is 34.


Q17 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 17:

Find the least square number which is exactly divisible by each of the numbers 6, 9, 15 and 20.

Answer 17:

The smallest number divisible by each of these numbers is their L.C.M.
L.C.M. of 6, 9, 15, 20 = 180

Resolving into prime factors: 
180=2×2×3×3×5
To make it a perfect square, we multiply it with 5.

Required number = 180×5=900


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Q18 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL | myhelper

Question 18:

Find the least square number which is exactly divisible by each of the numbers 8, 12, 15 and 20.

Answer 18:

The smallest number divisible by each of these numbers is their L.C.M.
L.C.M. of 8, 12, 15, 20 = 120

Resolving into prime factors: 
120=2×2×2×3×5

To make this into a perfect square, we need to multiply the number with 2×3×5=30.

Required number = 120×30=3600

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