RS Aggarwal Solutions: Exercise 3D - Squares and Square Roots | Mathematics (Maths) Class 8 PDF Download

Q.1. Find the square root of number by using the method of prime factorisation: 225
Ans. By prime factorisation method:
225=3×3×5×5
√225 =3×5=15

Q.2. Find the square root of number by using the method of prime factorisation: 441
Ans. By prime factorisation:
441=3×3×7×7
∴ √441 = 3×7=21

Q.3. Find the square root of number by using the method of prime factorisation: 729
Ans. Resolving into prime factors: 729=3×3×3×3×3×3
∴ √729 = 3×3×3=27

Q.4. Find the square root of number by using the method of prime factorisation: 1296
Ans. Resolving into prime factors:
1296=2×2×2×2×3×3×3×3
∴√1296=2×2×3×3=36

Q.5. Find the square root of number by using the method of prime factorisation: 2025
Ans. Resolving into prime factors: 2025=3×3×3×3×5×5
∴√2025 =3×3×5=45

Q.6. Find the square root of number by using the method of prime factorisation: 4096
Ans. 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

Q.7. Find the square root of number by using the method of prime factorisation: 7056
Ans. Resolving into prime factors:
7056=2×2×2×2×3×3×7×7
∴√7056=2×2×3×7=84

Q.8. Find the square root of number by using the method of prime factorisation: 8100
Ans. Resolving into prime factors: 8100 = 2×2×3×3×3×3×5×5
∴√8100 = 2×3×3×5=90

Q.9. Find the square root of number by using the method of prime factorisation: 9216
Ans. 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

Q.10. Find the square root of number by using the method of prime factorisation: 11025
Ans. Resolving into prime factors:
11025=3×3×5×5×7×7
∴√11025=3×5×7=105

Q.11. Find the square root of number by using the method of prime factorisation: 15876
Ans. Resolving into prime factors:
15876=2×2×3×3×3×3×7×7
∴√15876=2×3×3×7=126

Q.12. Find the square root of number by using the method of prime factorisation: 17424
Ans. Resolving into prime factors: 17424=2×2×2×2×3×3×11×11
∴ √17424=2×2×3×11=132

Q.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.
Ans. 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

Q.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.
Ans. 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

Q.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.
Ans. 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 = 5×5×7×7
x= √1225 = 5×7=35
Thus, the number of rows is 35 and the number of plants in each row is 35.

Q.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.
Ans. 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=2×2×17×17
x= √1156=2×17=34
Thus, the strength of the class is 34.

Q.17. Find the least square number which is exactly divisible by each of the numbers 6, 9, 15 and 20.
Ans. 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

Q.18. Find the least square number which is exactly divisible by each of the numbers 8, 12, 15 and 20.
Ans. 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=302×3×5=30.
Required number = 120×30=3600