Exercise 3.1 | Set 1 of Chapter 3 (Squares and Square Roots) in RD Sharma's Class 8 mathematics textbook introduces students to the fundamental concepts of squares and square roots. This exercise focuses on identifying perfect squares, understanding the properties of square numbers, and developing techniques to calculate squares and square roots efficiently. Students learn to recognize patterns in square numbers and use this knowledge to simplify calculations. The problems in this set are designed to build a strong foundation in working with squares and square roots, which is essential for more advanced mathematical concepts.
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.
Summary
Exercise 3.1 | Set 1 of Chapter 3 in RD Sharma's Class 8 mathematics textbook lays the groundwork for understanding squares and square roots. It covers a range of topics including identifying perfect squares, calculating squares of numbers, finding square roots, and understanding the properties of square numbers. The problems in this set are designed to help students develop mental math skills, recognize patterns in square numbers, and apply various techniques to simplify calculations. This exercise is crucial in building number sense and preparing students for more complex mathematical concepts involving quadratic equations and geometry. By mastering these fundamental skills, students gain confidence in working with squares and square roots, which is essential for success in higher-level mathematics.
