RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1

Other Exercises

Question 1.
Which of the following numbers are perfect squares ?
(i)484
(ii) 625
(iii) 576
(iv) 941
(v) 961
(vi) 2500
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 1
Grouping the factors in pairs, we have left no factor unpaired
∴ 484 is a perfect square of 22
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 2
∴ Grouping the factors in pairs, we have left no factor unpaired
∴ 625 is a perfect square of 25.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 3
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 576 is a perfect square of 24
(iv) 941 has no prime factors
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 4
∴ 941 is not a perfect square.
(v) 961 =31 x 31
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 961 is a perfect square of 31
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 5
Grouping the factors in pairs, we see that no factor is left impaired
∴ 2500 is a perfect square of 50 .

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
(ii) 2025
(iii) 14641
(iv) 4761
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 6
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 1156 is a perfect square of 2 x 17 = 34
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 7
Grouping the factors in pairs, we see that no factor is left unpaired
2025 is a perfect square of 3 x 3 x 5 =45
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 8
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 14641 is a perfect square of 11×11 = 121
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 9
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 4761 is a perfect square of 3 x 23 = 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
(ii) 12150
(iii) 7688
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 10
Grouping the factors in pairs of equal factors, we see that 5 is left unpaird
∴ In order to complete the pairs, we have to multiply 23805 by 5, then the product will be the perfect square.
Requid smallest number = 5
(ii) 12150 = 2 x 3 x 3×3 x 3×3 x 5×5
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 11
Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired
∴ In order to complete the pairs, we have to multiply 12150 by 2 x 3 =6 i.e., then the product will be the complete square.
∴ Required smallest number = 6
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 12
Grouping the factors in pairs of equal factors, we see that factor 2 is left unpaired
∴ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square
∴ Required smallest number = 2

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
(ii) 1800
(iii) 2904
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 13
Grouping the factors in pairs of equal factors, we see that factors we see that 3 is left unpaired
Deviding by 3, the quotient will the perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 14
Grouping the factors in pair of equal factors, we see that 2 is left unpaired.
∴ Dividing by 2, the quotient will be the perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 15
Grouping the factors in pairs of equal factors, we see that 2 x 3 we left unpaired
∴ Dividing by 2 x 3 = 6, the quotient will be the perfect square.

Question 5.
Which of the following numbers are perfect squares ?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121
Solution:
11 is not a perfect square as 11 = 1 x 11
12 is not a perfect square as 12 = 2×2 x 3
16 is a perfect square as 16 = 2×2 x 2×2
32 is not a perfect square as 32 = 2×2 x 2×2 x 2
36 is a perfect square as 36 = 2×2 x 3×3
50 is not a perfect square as 50 = 2 x 5×5
64 is a perfect square as 64 = 2×2 x 2×2 x 2×2
79 is not a perfect square as 79 = 1 x 79
81 is a perfect square as 81 = 3×3 x 3×3
111 is not a perfect square as 111 = 3 x 37
121 is a perfect square as 121 = 11 x 11
Hence 16, 36, 64, 81 and 121 are perfect squares.

Question 6.
Using prime factorization method, find which of the following numbers are perfect squares ?
∴ 189,225,2048,343,441,2916,11025,3549
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 16
Grouping the factors in pairs, we see that are 3 and 7 are left unpaired
∴ 189 is not a perfect square
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 17
Grouping the factors in pairs, we see no factor left unpaired
∴ 225 is a perfect square
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 18
Grouping the factors in pairs, we see no factor left unpaired
∴ 2048 is a perfect square
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 19
Grouping the factors in pairs, we see that one 7 is left unpaired
∴ 343 is not a perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 20
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 441 is a perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 21
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 2916 is a perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 22
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 11025 is a perfect square.
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 23
Grouping the factors in pairs, we see that 3, no factor 7 are left unpaired
∴ 3549 is a perfect square.

Question 7.
By what number should each of the following numbers be multiplied to get a perfect square in each case ? Also, find the number whose square is the new number.
(i) 8820
(ii) 3675
(iii) 605
(iv) 2880
(v) 4056
(vi) 3468
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 24
Grouping the factors in pairs, we see that 5 is left unpaired
∴ By multiplying 8820 by 5, we get the perfect square and square root of product will be
= 2 x 3 x 5 x 7 = 210
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 25
Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3675 by 3, we get a perfect square and square of the product will be
= 3 x 5 x 7 = 105
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 26
Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 605 by 5, we get a perfect square and square root of the product will be
= 5 x 11 =55
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 27
Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 2880 by 5, we get the perfect square.
Square rooi of product will be = 2 x 2 * 2 – 3 x 5 = 120
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 28
Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 4056 by 2 x 3 i.e., 6, we get the perfect square.
and square root of the product will be
= 2 x 2 x 3 x 13 = 156
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 29
Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3468 by 3 we get a perfect square, and square root of the product will be 2 x 3 x 17 = 102
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 30
Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 7776 by 2 x 3 or 6 We get a perfect square and square root of the product will be
= 2 x 2 x 2 x 3 x 3 x 3 = 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
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 31
Grouping the factors in pairs, we see that 2 is left unpaired
∴ Dividing by 2, we get the perfect square and square root of the quotient will be 7 x 13 = 91
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 32
Grouping the factors in pairs, we see that 2 is left unpaired,
∴ Dividing 3698 by 2, the quotient is a perfect square
and square of quotient will be = 43
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 33
Grouping the factors in pairs, we see that 7 is left unpaired
∴ Dividing 5103 by 7, we get the quotient a perfect square.
and square root of the quotient will be 3 x 3 x 3 = 27
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 34
Grouping the factors iq pairs, we see that 2 and 3 are left unpaired
∴ Dividing 3174 by 2 x 3 i.e. 6, the quotient will be a perfect square and square root of the quotient will be = 23
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 35
Grouping the factors in pairs, we find that 7 is left unpaired i
∴ Dividing 1575 by 7, the quotient is a perfect square
and square root of the quotient will be = 3 x 5 = 15

Question 9.
Find the greatest number of two digits which is a perfect square.
Solution:
The greatest two digit number = 99 We know, 92 = 81 and 102 = 100 But 99 is in between 81 and 100
∴ 81 is the greatest two digit number which is a perfect square.

Question 10.
Find the least number of three digits which is perfect square.
Solution:
The smallest three digit number =100
We know that 92 = 81, 102 = 100, ll2 = 121
We see that 100 is the least three digit number which is a perfect square.

Question 11.
Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.
Solution:
By factorization:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 36
Grouping the factors in pairs, we see that 11 is left unpaired
∴ The least number is 11 by which multiplying 4851, we get a perfect square.

Question 12.
Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.
Solution:
By factorization,
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 37
Grouping the factors in pairs, we see that 13 is left unpaired
∴ Dividing 28812 by 3, the quotient will be a perfect square.

Question 13.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.
Solution:
By factorization,
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1 38
Grouping the factors in pairs, we see that one 2 is left unpaired.
∴ Dividing 1152 by 2, we get the perfect square and square root of the resulting number 576, will be 2 x 2 x 2 x 3 = 24

 

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