RD Sharma Solutions for Class 8 Maths Exercise 3.4 Chapter 3, Squares and Square Roots are given below. The expert faculty at BYJU’S have solved the questions in Exercise 3.4, which will help students in solving this exercise of Class 8 Maths without any problems. Students who find it challenging to solve tough Maths problems can refer to these RD Sharma Class 8 Maths solutions to get better guidance. Solving this exercise will ensure that students excel in annual exams. In this exercise, they will study the definition of square roots and some properties of square roots. By practising the Solutions of RD Sharma Class 8, students will be able to grasp the concepts thoroughly, and they can attain good marks in their examinations.
RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots Exercise 3.4
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EXERCISE 3.4 PAGE NO: 3.38
1. Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i) 9801
We know that the unit digit of 9801 is 1
Unit digit of square root = 1 or 9
Since the number is odd, the square root is also odd
(ii) 99856
We know that the unit digit of 99856 = 6
Unit digit of square root = 4 or 6
Since the number is even, the square root is also even
(iii) 998001
We know that the unit digit of 998001 = 1
Unit digit of square root = 1 or 9
Since the number is odd, the square root is also odd
(iv) 657666025
We know that the unit digit of 657666025 = 5
Unit digit of square root = 5
Since the number is odd, the square root is also odd
2. Find the square root of each of the following by prime factorization.
(i) 441 (ii) 196
(iii) 529 (iv) 1764
(v) 1156 (vi) 4096
(vii) 7056 (viii) 8281
(ix) 11664 (x) 47089
(xi) 24336 (xii) 190969
(xiii) 586756 (xiv) 27225
(xv) 3013696
Solution:
(i) 441
Firstly let’s find the prime factors for
441 = 3×3×7×7
= 32×72
√441 = 3×7
= 21
(ii) 196
Firstly let’s find the prime factors for
196 = 2×2×7×7
= 22×72
√196 = 2×7
= 14
(iii) 529
Firstly let’s find the prime factors for
529 = 23×23
= 232
√529 = 23
(iv) 1764
Firstly let’s find the prime factors for
1764 = 2×2×3×3×7×7
= 22×32×72
√1764 = 2×3×7
= 42
(v) 1156
Firstly let’s find the prime factors for
1156 = 2×2×17×17
= 22×172
√1156 = 2×17
= 34
(vi) 4096
Firstly let’s find the prime factors for
4096 = 2×2×2×2×2×2×2×2×2×2×2×2
= 212
√4096 = 26
= 64
(vii) 7056
Firstly let’s find the prime factors for
7056 = 2×2×2×2×21×21
= 22×22×212
√7056 = 2×2×21
= 84
(viii) 8281
Firstly let’s find the prime factors for
8281 = 91×91
= 912
√8281 = 91
(ix) 11664
Firstly let’s find the prime factors for
11664 = 2×2×2×2×3×3×3×3×3×3
= 22×22×32×32×32
√11664 = 2×2×3×3×3
= 108
(x) 47089
Firstly let’s find the prime factors for
47089 = 217×217
= 2172
√47089 = 217
(xi) 24336
Firstly let’s find the prime factors for
24336 = 2×2×2×2×3×3×13×13
= 22×22×32×132
√24336 = 2×2×3×13
= 156
(xii) 190969
Firstly let’s find the prime factors for
190969 = 23×23×19×19
= 232×192
√190969 = 23×19
= 437
(xiii) 586756
Firstly let’s find the prime factors for
586756 = 2×2×383×383
= 22×3832
√586756 = 2×383
= 766
(xiv) 27225
Firstly let’s find the prime factors for
27225 = 5×5×3×3×11×11
= 52×32×112
√27225 = 5×3×11
= 165
(xv) 3013696
Firstly let’s find the prime factors for
3013696 = 2×2×2×2×2×2×217×217
= 26×2172
√3013696 = 23×217
= 1736
3. Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.
Solution:
Firstly let’s find the prime factors for
180 = (2 × 2) × (3 × 3) × 5
=22 × 32 × 5
To make the unpaired 5 into paired, multiply the number with 5
180 × 5 = 22 × 32 × 52
∴ Square root of √ (180 × 5) = 2 × 3 × 5
= 30
4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Firstly let’s find the prime factors for
147 = (7 × 7) × 3
=72 × 3
To make the unpaired 3 into paired, multiply the number with 3
147 × 3 = 72 × 32
∴ Square root of √ (147 × 3) = 7 × 3
= 21
5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.
Solution:
Firstly let’s find the prime factors for
3645 = (3 × 3) × (3 × 3) × (3 × 3) × 5
=32 × 32 × 32 × 5
To make the unpaired 5 into paired, the number 3645 has to be divided by 5
3645 ÷ 5 = 32 × 32 × 32
∴ Square root of √ (3645 ÷ 5) = 3 × 3 × 3
= 27
6. Find the smallest number by which 1152 must be divided so that it becomes a square. Also, find the square root of the number so obtained.
Solution:
Firstly let’s find the prime factors for
1152 = (2 × 2) × (2 × 2) × (2 × 2) × 2 × (3 × 3)
=22 × 22 × 22 × 32 × 2
To make the unpaired 2 into paired, the number 1152 has to be divided by 2
1152 ÷ 2 = 22 × 22 × 22 × 32
∴ Square root of √ (1152 ÷ 2) = 2 × 2 × 2 × 3
= 24
7. The product of two numbers is 1296. If one number is 16 times the other, find the numbers.
Solution:
Let us consider two numbers, a and b
So, we know that one of the number a = 16b
a × b = 1296
16b × b = 1296
16b2 = 1296
b2 = 1296/16 = 81
b = 9
a = 16b
= 16(9)
= 144
∴ a =144 and b =9
8. A welfare association collected Rs. 202500 as a donation from the residents. If each paid as many rupees as there were residents, find the number of residents.
Solution:
Let us consider the total residents as a
So, each paid Rs. a
Total collection = a (a) = a2
We know that the total Collection = 202500
a = √ 202500
a = √a(2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 5)
= 2 × 3 × 3 × 5 × 5a
= 450
∴ Total residents = 450
9. A society collected Rs 92.16. Each member collected as many paise as there were members. How many members were there, and how much did each contribute?
Solution:
Let us consider there were few members, each attributed a paise
a (a), i.e. total cost collected = 9216 paise
a2 = 9216
a = √9216
= 2 × 2 × 2 × 12
= 96
∴ There were 96 members in the society, and each contributed 96 paise
10. A society collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school?
Solution:
Let us consider number of school students as a
each student contributed a paise
Total money obtained = a2paise
= 2304 paise
a = √2304
a = √2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
a = 2 × 2 × 2 × 2 × 3
a = 48
∴ There were 48 students in the school
11. The area of a square field is 5184 m2. A rectangular field whose length is twice its breadth has a perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
Let us consider the side of the square field as a
a2 = 5184 m2
a = √5184m
a = 2 × 2 ×2 × 9
= 72 m
Perimeter of square = 4a
= 4(72)
= 288 m
Perimeter of rectangle = 2 (l + b) = perimeter of the square field
= 288 m
2 (2b + b) = 288
b = 48 and l = 96
Area of rectangle = 96 × 48 m2
= 4608 m2
12. Find the least square number, exactly divisible by each one of the numbers: (i) 6, 9, 15 and 20 (ii) 8, 12, 15 and 20
Solution:
(i) 6, 9, 15 and 20
Firstly take L.C.M for 6, 9, 15, and 20, which is 180
So the prime factors of 180 = 22 × 32 × 5
To make it a perfect square, we have to multiply the number with 5
180 × 5 = 22 × 32 × 52
∴ 900 is the least square number divisible by 6, 9, 15 and 20
(ii) 8, 2, 15 and 20
Firstly take L.C.M for 8, 2, 15, 20, which is 360
So the prime factors of 360 = 22 × 32 ×2 × 5
To make it a perfect square, we have to multiply the number by 2 × 5 = 10
360 × 10 = 22 × 32 × 52 × 22
∴ 3600 is the least square number divisible by 8, 12, 15 and 20
13. Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:
Let us find the square roots of 121 and 169 by the method of repeated subtraction
121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
Clearly, we have performed the operation 11 times
∴ √121 = 11
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0
Clearly, we have performed subtraction 13 times
∴ √169 = 13
14. Write the prime factorization of the following numbers and hence find their square roots.
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056
Solution:
(i) 7744
Prime factors of 7744 are
7744 = 22 × 22 × 22 × 112
∴ The square root of 7744 is
√7744 = 2 × 2 × 2 × 11
= 88
(ii) 9604
Prime factors of 9604 are
9604 = 22 × 72 × 72
∴ The square root of 9604 is
√9604 = 2 × 7 × 7
= 98
(iii) 5929
Prime factors of 5929 are
5929 = 112 × 72
∴ The square root of 5929 is
√5929 = 11 × 7
= 77
(iv) 7056
Prime factors of 7056 are
7056 = 22 × 22 × 72 × 32
∴ The square root of 7056 is
√7056 = 2 × 2 × 7 × 3
= 84
15. The students of class VIII of a school donated Rs. 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class, Find the number of students in the class.
Solution:
Let us consider number of students as a
Each student denoted a rupee
So, the total amount collected is a × a rupees = 2401
a2 = 2401
a = √2401
a = 49
∴ There are 49 students in the class.
16. A PT teacher wants to arrange a maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after the arrangement.
Solution:
Let us consider number of rows as a
No. of columns = a
Total number of students who sat in the field = a2
Total students a2 + 71 = 6000
a2 = 5929
a = √5929
a = 77
∴ The total number of rows is 77.
RD Sharma Solutions for Class 8 Maths Exercise 3.4 Chapter 3 – Squares and Square Roots
Exercise 3.4 of RD Sharma Solutions for Chapter 3 Squares and Square Roots, mainly deals with the definition and methods to find the square root of a perfect square by the prime factorisation method. The RD Sharma Solutions can help the students practise diligently while learning the fundamentals, as it provides the answers to all the questions in the RD Sharma textbook. Students are suggested to practise these solutions regularly, which will help them ace their exams and improve their overall percentage.
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