Students can download the RD Sharma Solutions for Exercise 3.5 Class 8 Maths Chapter 3, Squares and Square Roots, from the link accessible here. The subject-matter experts at BYJU’S have solved the questions in Exercise 3.5 in Chapter 3 Maths in a precise manner, which will help students in solving this exercise without any difficulty. By practising regularly using the RD Sharma Solutions, the students can obtain high scores in Class 8 final exams. This exercise will help students to understand the concept of finding the square root of a perfect square by the long-division method.
By practising the RD Sharma Class 8 Solutions, students will be able to understand the concepts clearly, which in turn helps in gaining confidence, which plays a vital role in their examinations.
RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots Exercise 3.5
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 3.5 Chapter 3 Squares and Square Roots
EXERCISE 3.5 PAGE NO: 3.43
1. Find the square root of each of the following by the long division method:
(i) 12544 (ii) 97344
(iii) 286225 (iv) 390625
(v) 363609 (vi) 974169
(vii) 120409 (viii) 1471369
(ix) 291600 (x) 9653449
(xi) 1745041 (xii) 4008004
(xiii) 20657025 (xiv) 152547201
(xv) 20421361 (xvi)62504836
(xvii) 82264900 (xviii) 3226694416
(xix) 6407522209 (xx) 3915380329
Solution:
(i) 12544
By using the long division method
∴ the square root of 12544
√12544 = 112
(ii) 97344
By using the long division method
∴ the square root of 97344
√97344 = 312
(iii) 286225
By using the long division method
∴ the square root of 286225
√286225 = 535
(iv) 390625
By using the long division method
∴ the square root of 390625
√390625 = 625
(v) 363609
By using the long division method
∴ the square root of 363609
√36369 = 603
(vi) 974169
By using the long division method
∴ the square root of 974169
√974169 = 987
(vii) 120409
By using long division method
∴ the square root of 120409
√120409 = 347
(viii) 1471369
By using the long division method
∴ the square root of 1471369
√1471369 = 1213
(ix) 291600
By using the long division method
∴ the square root of 291600
√291600 = 540
(x) 9653449
By using the long division method
∴ the square root of 9653449
√9653449 = 3107
(xi) 1745041
By using the long division method
∴ the square root of 1745041
√1745041 = 1321
(xii) 4008004
By using the long division method
∴ the square root of 4008004
√4008004 = 2002
(xiii) 20657025
By using the long division method
∴ the square root of 20657025
√20657025 = 4545
(xiv) 152547201
By using the long division method
∴ the square root of 152547201
√152547201 = 12351
(xv) 20421361
By using the long division method
∴ the square root of 20421361
√20421361 = 4519
(xvi) 62504836
By using the long division method
∴ the square root of 62504836
√62504836 = 7906
(xvii) 82264900
By using the long division method
∴ the square root of 82264900
√82264900 = 9070
(xviii) 3226694416
By using the long division method
∴ the square root of 3226694416
√3226694416 = 56804
(xix) 6407522209
By using the long division method
∴ the square root of 6407522209
√6407522209 = 80047
(xx) 3915380329
By using the long division method
∴ the square root of 3915380329
√3915380329 = 62573
2. Find the least number which must be subtracted from the following numbers to make them a perfect square:
(i) 2361
(ii) 194491
(iii) 26535
(iv) 161605
(v) 4401624
Solution:
(i) 2361
By using the long division method
∴ 57 has to be subtracted from 2361 to get a perfect square.
(ii) 194491
By using the long division method
∴ 10 has to be subtracted from 194491 to get a perfect square.
(iii) 26535
By using long division method
∴ 291 has to be subtracted from 26535 to get a perfect square.
(iv) 161605
By using the long division method
∴ 1 has to be subtracted from 161605 to get a perfect square.
(v) 4401624
By using the long division method
∴ 20 has to be subtracted from 4401624 to get a perfect square.
3. Find the least number which must be added to the following numbers to make them a perfect square:
(i) 5607
(ii)4931
(iii) 4515600
(iv) 37460
(v) 506900
Solution:
(i) 5607
By using the long division method
The remainder is 131
Since (74)2 < 5607
We take the next perfect square number, i.e., (75)2
(75)2 = 5625 > 5607
So, the number to be added = 5625 – 5607 = 18
(ii) 4931
By using the long division method
The remainder is 31
Since, (70)2 < 4931
We take the next perfect square number, i.e., (71)2
(71)2 = 5041 > 4931
So, the number to be added = 5041 – 4931 = 110
(iii) 4515600
By using the long division method
The remainder is 4224
Since, (2124)2 < 4515600
We take the next perfect square number, i.e., (2125)2
(2125)2 = 4515625 > 4515600
So, the number to be added = 4515625 – 4515600 = 25
(iv) 37460
By using the long division method
The remainder is 211
Since, (193)2 < 37460
We take the next perfect square number, i.e., (194)2
(194)2 = 37636 > 37460
So, the number to be added = 37636 – 37460 = 176
(v) 506900
By using the long division method
The remainder is 1379
Since, (711)2 < 506900
We take the next perfect square number, i.e., (712)2
(712)2 = 506944 > 506900
So, the number to be added = 506944 – 506900 = 44
4. Find the greatest number of 5 digits, which is a perfect square.
Solution:
We know that the greatest 5 digit number is 99999
By using long division method
The remainder is 143
So, the greatest 5 digit perfect square number is:
99999 – 143 = 99856
∴ 99856 is the required greatest 5 digit perfect square number.
5. Find the least number of 4 digits, which is a perfect square.
Solution:
We know that the least 4 digit number is 1000
By using long division method
The remainder is 39
Since, (31)2 < 1000
We take the next perfect square number, i.e., (32)2
(32)2 = 1024 > 1000
∴ 1024 is the required least number 4 digit number which is a perfect square.
6. Find the least number of six digits, which is a perfect square.
Solution:
We know that the least 6 digit number is 100000
By using the long division method
The remainder is 144
Since, (316)2 < 100000
We take the next perfect square number, i.e., (317)2
(317)2 = 100489 > 100000
∴ 100489 is the required least number 6 digit number which is a perfect square.
7. Find the greatest number of 4 digits, which is a perfect square.
Solution:
We know that the greatest 4 digit number is 9999
By using the long division method
The remainder is 198
So, the greatest 4 digit perfect square number is:
9999 – 198 = 9801
∴ 9801 is the required greatest 4 digit perfect square number.
8. A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers is 8160, find the number of soldiers in each row
Solution:
We know that the total number of soldiers = 8160
Number of soldiers left out = 60
Number of soldiers arranged in rows to form a perfect square = 8160 – 60 = 8100
∴ number of soldiers in each row = √8100
= √ (9×9×10×10)
= 9×10
= 90
9. The area of a square field is 60025m2. A man cycles along its boundary at 18 Km/hr. In how much time will he return at the starting point?
Solution:
We know that the area of the square field = 60025 m2
Speed of cyclist = 18 km/h
= 18 × (1000/60×60)
= 5 m/s2
Area = 60025 m2
Side2 = 60025
Side = √60025
= 245
We know, Total length of the boundary = 4 × Side
= 4 × 245
= 980 m
∴ Time taken to return to the starting point = 980/5
= 196 seconds
= 3 minutes 16 seconds
10. The cost of levelling and turning a square lawn at Rs 2.50 per m2 is Rs13322.50 Find the cost of fencing it at Rs 5 per metre.
Solution:
We know that the cost of levelling and turning a square lawn = 2.50 per m2
The total cost of levelling and turning = Rs. 13322.50
Total area of square lawn = 13322.50/2.50
= 5329 m2
Side2 = 5329
Side of square lawn = √5329
= 73 m
So, the total length of the lawn = 4 × 73
= 292 m
∴ Cost of fencing the lawn at Rs 5 per metre = 292 × 5
= Rs. 1460
11. Find the greatest number of three digits which is a perfect square.
Solution:
We know that the greatest 3 digit number is 999
By using the long division method
The remainder is 38
So, the greatest 3 digit perfect square number is:
999 – 38 = 961
∴ 961 is the required greatest 3 digit perfect square number.
12. Find the smallest number which must be added to 2300 so that it becomes a perfect square.
Solution:
By using the long division method, let’s find the square root of 2300
The remainder is 91
Since, (47)2 < 2300
We take the next perfect square number, i.e., (48)2
(48)2 = 2304 > 2300
∴ The smallest number required to be added to 2300 to get a perfect square is
2304 – 2300 = 4
RD Sharma Solutions for Class 8 Maths Exercise 3.5 Chapter 3 – Squares and Square Roots
Exercise 3.5 of RD Sharma Solutions for Chapter 3 Squares and Square Roots, primarily deals with the concepts related to the long division method, which provides the means for detecting the square root of a perfect square by the long division 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. The solutions are solved in a precise manner which is easily understandable by the students to fetch good marks in the final exams.
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