Skip to content
Related Articles

Related Articles

Class 8 RD Sharma Solutions – Chapter 3 Squares and Square Roots – Exercise 3.1 | Set 1
  • Last Updated : 06 Apr, 2021

Question 1. Which of the following numbers are perfect squares?

(i) 484

Solution:

Prime factorization of 484 is

484 = 2×2×11×11

By grouping prime factors



= (2×2) × (11×11)

Proper pairing of all the factors 

Therefore, 484 is a perfect square.

(ii) 625

Solution:

Prime factorization of 625

625 = 5×5×5×5

By grouping the prime factors 

= (5×5) × (5×5)

Proper pairing of all the factors 

Therefore, 625 is a perfect square.

(iii) 576

Solution:

Prime factorization of 576

576 = 2×2×2×2×2×2×3×3

By grouping the prime factors 

= (2×2) × (2×2) × (2×2) × (3×3)

Proper pairing of all the factors 



Therefore, 576 is a perfect square.

(iv) 941

Solution:

Prime factorization of 941

941 = 941 × 1

941 itself is a prime factor.

Therefore, 941 is not a perfect square.

(v) 961

Solution:

Prime factorization of 961

961 = 31×31

By grouping the prime factors 

= (31×31)

Proper pairing of all factors

Therefore, 961 is a perfect square.

(vi) 2500

Solution:

Prime factorization of

2500 = 2×2×5×5×5×5

By grouping the prime factors 

= (2×2) × (5×5) × (5×5)

Proper pairing of all factors

Therefore, 2500 is a perfect square.

Question 2. Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number in each case:

(i) 1156

Solution:

Prime factorization of 1156

1156 = 2×2×17×17

By grouping prime factors

= (2×2) × (17×17)

Proper pairing of all factors

Therefore,1156 is a perfect square.

To find whose square the given number is

1156 = (2×17) × (17×2)

= 34 × 34

= (34)2

Therefore,1156 is a square of 34.

(ii) 2025

Solution:

Prime factorization of 2025

2025 = 3×3×3×3×5×5

By grouping prime factors 

= (3×3) × (3×3) × (5×5)

Proper pairing of all factors

Therefore, 2025 is a perfect square.

To find whose square the given number is

2025 = (3×3×5) × (3×3×5)

= 45 × 45

= (45)2

Therefore, 2025 is a square of 45.

(iii) 14641

Solution:

Prime factorization of 14641

14641 = 11×11×11×11

By grouping prime factors

= (11×11) × (11×11)

Proper pairing of all factors

Therefore,14641 is a perfect square.

To find whose square the given number is

14641 = (11×11) × (11×11)

= 121 × 121

= (121)2

Therefore,14641 is a square of 121.

(iv) 4761

Solution:

Prime factorization of 4761

4761 = 3×3×23×23

By grouping the prime factors 

= (3×3) × (23×23)

Proper pairing of all factors

Therefore, 4761 is a perfect square.

To find whose square the given number is

4761 = (3×23) × (3×23)

= 69 × 69

= (69)2

Therefore, 4761 is a square of 69.

Question 3. Find the smallest number by which the given number must be multiplied so that the product is a perfect square:

(i) 23805

Solution:

Prime factorization of 23805

23805 = 3×3×23×23×5

By grouping the prime factors

= (3×3) × (23×23) × 5

Prime factor 5 is left out.

Multiplying by 5,

23805 × 5 = (3×3) × (23×23) × (5×5)

= (3×5×23) × (3×5×23)

= 345 × 345

= (345)2

Therefore, product is the square of 345.

(ii) 12150

Solution:

Prime factorization of 12150

12150 = 2×3×3×3×3×3×5×5

By grouping the prime factors 

= 2×3 × (3×3) × (3×3) × (5×5)

Prime factor 2 and 3 are left out.

Multiplying by 2×3 = 6 

12150 × 6 = 2×3 × (3×3) × (3×3) × (5×5) × 2 × 3

= (2×3×3×3×5) × (2×3×3×3×5)

= 270 × 270

= (270)2

Therefore, product is the square of 270.

(iii) 7688

Solution:

Prime factorization of 7688

7688 = 2×2×31×31×2

By grouping the prime factors

= (2×2) × (31×31) × 2

Prime factor 2 is left out.

Multiplying by 2 

7688 × 2 = (2×2) × (31×31)× (2×2)

= (2×31×2) × (2×31×2)

= 124 × 124

= (124)2

Therefore, product is the square of 124.

Question 4. Find the smallest number by which the given number must be divided so that the resulting number is a perfect square:

(i) 14283

Solution:

Prime factorization of 14283

14283 = 3×3×3×23×23

By grouping the prime factors 

= (3×3) × (23×23) × 3

Prime factor 3 is left out.

Dividing by 3 

14283/3 = (3×3) × (23×23)

= (3×23) × (3×23)

= 69 × 69

= (69)2

Therefore, resultant is the square of 69.

(ii) 1800

Solution:

Prime factorization of 1800

1800 = 2×2×5×5×3×3×2

By grouping the prime factors

= (2×2) × (5×5) × (3×3) × 2

Prime factor 2 is left out.

Dividing by 2 

1800/2 = (2×2) × (5×5) × (3×3)

= (2×5×3) × (2×5×3)

= 30 × 30

= (30)2

Therefore, resultant is the square of 30.
 

(iii) 2904

Solution:

Prime factorization of 2904

2904 = 2×2×11×11×2×3

By grouping the prime factors

= (2×2) × (11×11) × 2 × 3

Prime factor 2 and 3 are left out.

Dividing by 6 

2904/6 = (2×2) × (11×11)

= (2×11) × (2×11)

= 22 × 22

= (22)2

Therefore, resultant is the square of 22.

Question 5. Which of the following numbers are perfect squares?

11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121

Solution:

11 = 1×11

11 is a prime number

Therefore, 11 is not a perfect square.

12 = 2×2×3

3 is left out in pairing of factors

Therefore, 12 is not a perfect square.

16 = 2×2×2×2

= (2×2) × (2×2)

Proper pairing of factors

Therefore,16 is a perfect square.

32 = 2×2×2×2×2

= (2×2) × (2×2) × 2

2 is left out in pairing of factors

Therefore, 32 is not a perfect square.

36 = 2×2×3×3

Proper pairing of factors

Therefore, 36 is a perfect square.

50 = 2×5×5

2 is left out in pairing of factors

Therefore, 50 is not a perfect square.

64 = 2×2×2×2×2×2

Proper pairing of factors

Therefore, 64 is a perfect square.

79 = 1×79

79 is a prime number

79 is not a perfect square.

81 = 3×3×3×3

Proper pairing of factors 

Therefore, 81 is a perfect square.

111 = 1×111

111 it is a prime number.

Therefore, 111 is not a perfect square.

121 = 11×11

Proper pairing of factors

Therefore, 121 is a perfect square.

Question 6. Using prime factorization method, find which of the following numbers are perfect squares?

189, 225, 2048, 343, 441, 2961, 11025, 3549

Solution:

189 prime factors are

189 = 3×3×3×7

As, proper pairing of factors is not there.

Therefore, it is not a perfect square

225 prime factors are

225 = (5×5) × (3×3)

Proper pairing of factors is there.

 Therefore, it is a perfect square.

2048 prime factors are

2048 = (2×2) × (2×2) × (2×2) × (2×2) × (2×2) × 2

2 is left out from pairing of factors.

Therefore, 2048 is not a perfect square.

343 prime factors are

343 = (7×7) × 7

7 is left out from proper pairing of factors.

Therefore, 343 is not a perfect square.

441 prime factors are

441 = (7×7) × (3×3)

Proper pairing of factors is there.

 Therefore, 441 is a perfect square.

2961 prime factors are

2961 = (3×3) × (3×3) × (3×3) × (2×2)

Proper pairing of factors is there.

Therefore, 2961 is a perfect square.

11025 prime factors are

11025 = (3×3) × (5×5) × (7×7)

Proper pairing of factors is there. 

Therefore, 11025 is a perfect square.

3549 prime factors are

3549 = (13×13) × 3 × 7

Proper pairing of factors is not there.

Therefore, 3549 is not a perfect square.

Question 7. By what number should each of the following numbers by multiplied to get a perfect square in each case? Also find the number whose square is the new number.

(i) 8820

Solution:

Prime factorization of 8820

8820 = 2×2×3×3×7×7×5

By grouping the prime factors 

= (2×2) × (3×3) × (7×7) × 5

Prime factor 5 is left out.

Multiplying by 5 

8820 × 5 = (2×2) × (3×3) × (7×7) × (5×5)

= (2×3×7×5) × (2×3×7×5)

= 210 × 210

= (210)2

Therefore, new number is square of 210.

(ii) 3675

Solution:

Prime factorization of 3675

3675 = 5×5×7×7×3

By grouping the prime factors 

= (5×5) × (7×7) × 3

Prime factor 3 is left out.

Multiplying by 3 

3675 × 3 = (5×5) × (7×7) × (3×3)

= (5×7×3) × (5×7×3)

= 105 × 105

= (105)2

Therefore, new number is the square of 105.

(iii) 605

Solution:

Prime factorization of 605

605 = 5×11×11

By grouping the prime factors 

= (11×11) × 5

Prime factor 5 is left out.

Multiply by 5 

605 × 5 = (11×11) × (5×5)

= (11×5) × (11×5)

= 55 × 55

= (55)2

Therefore, new number is the square of 55.

(iv) 2880

Solution:

Prime factorization of 2880

2880 = 5×3×3×2×2×2×2×2×2

By grouping the prime factors 

= (3×3) × (2×2) × (2×2) × (2×2) × 5

Prime factor 5 is left out.

Multiply by 5 

2880 × 5 = (3×3) × (2×2) × (2×2) × (2×2) × (5×5)

= (3×2×2×2×5) × (3×2×2×2×5)

= 120 × 120

= (120)2

Therefore, new number is the square of 120.
 

(v) 4056

Solution:

Prime factorization of 4056

4056 = 2×2×13×13×2×3

By grouping the prime factors 

= (2×2) × (13×13) × 2 × 3

Prime factors 2 and 3 are left out.

Multiplying by 6 we get,

4056 × 6 = (2×2) × (13×13) × (2×2) × (3×3)

= (2×13×2×3) × (2×13×2×3)

= 156 × 156

= (156)2

Therefore, new number is the square of 156.

(vi) 3468

Solution:

Prime factorization of 3468

3468 = 2×2×17×17×3

By grouping the prime factors

= (2×2) × (17×17) × 3

Prime factor 3 is left out.

Multiplying by 3 we get,

3468 × 3 = (2×2) × (17×17) × (3×3)

= (2×17×3) × (2×17×3)

= 102 × 102

= (102)2

Therefore, new number is the square of 102.

(vii) 7776

Solution:

Prime factorization of 7776

7776 = 2×2×2×2×3×3×3×3×2×3

By grouping the prime factors 

= (2×2) × (2×2) × (3×3) × (3×3) × 2 × 3

Prime factors 2 and 3 are left out.

Multiplying by 6 we get,

7776 × 6 = (2×2) × (2×2) × (3×3) × (3×3) × (2×2) × (3×3)

= (2×2×3×3×2×3) × (2×2×3×3×2×3)

= 216 × 216

= (216)2

Therefore, the new number is square of 216.

Question 8. By What numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.

(i) 16562

Solution:

Prime factorization of16562

16562 = 7×7×13×13×2

By grouping the prime factors 

= (7×7) × (13×13) × 2

Prime factor 2 is left out.

Dividing by 2 

16562/2 = (7×7) × (13×13)

= (7×13) × (7×13)

= 91 × 91

= (91)2

Therefore, the new number is square of 91.
 

(ii) 3698

Solution:

Prime factorization of 3698

3698 = 2×43×43

By grouping the prime factors 

= (43×43) × 2

Prime factor 2 is left out.

Dividing by 2 

3698/2 = (43×43)

= (43)2

Therefore, the new number is square of 43.
 

(iii) 5103

Solution:

Prime factorization of 5103

5103 = 3×3×3×3×3×3×7

By grouping the prime factors 

= (3×3) × (3×3) × (3×3) × 7

Prime factor 7 is left out.

Dividing by 7 

5103/7 = (3×3) × (3×3) × (3×3)

= (3×3×3) × (3×3×3)

= 27 × 27

= (27)2

Therefore, new number is the square of 27.

(iv) 3174

Solution:

Prime factorization of 3174

3174 = 2×3×23×23

By grouping the prime factors

= (23×23) × 2 × 3

Prime factor 2 and 3 are left out.

Dividing by 6 

3174/6 = (23×23)

= (23)2

Therefore, new number is the square of 23.

(v) 1575

Solution:

Prime factorization  for 1575

1575 = 3×3×5×5×7

By grouping the prime factors

= (3×3) × (5×5) × 7

Prime factor 7 is left out.

 Dividing by 7

1575/7 = (3×3) × (5×5)

= (3×5) × (3×5)

= 15 × 15

= (15)2

Therefore, new number is square of 15.

Chapter 3 Squares and Square Roots – Exercise 3.1 | Set 2

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :