In this chapter, we provide RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5 Maths pdf, free RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5 Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 8 |

Subject | Maths |

Chapter | Chapter 3 |

Chapter Name | Squares and Square Roots |

Exercise | Ex 3.5 |

**RD Sharma Solutions for Class 8 Chapter 3 Squares and Square Roots Ex 3.5 Download PDF**

**Question 1.**

**Find the square root of each of 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**

**(jcv) 20421361**

**(xvi) 62504836**

**(xvii) 82264900**

**(xviii) 3226694416**

**(xix)6407522209**

**(xx) 3915380329**

**Solution:**

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**Question 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) 16160**

**(v) 4401624**

**Solution:**

(i) 2361

Finding the square root of 2361

We get 48 as quotient and remainder = 57

∴ To make it a perfect square, we have to subtract 57 from 2361

∴ Least number to be subtracted = 57

(ii) 194491

Finding the square root of 194491

We get 441 as quotient and remainder = 10

∴ To make it a perfect square, we have to subtract 10 from 194491

∴ Least number to be subtracted = 10

(iii) 26535

Finding the square root of 26535

We get 162 as quotient and 291 as remainder

∴ To make it a perfect square, we have to subtract 291 from 26535

∴ Least number to be subtracted = 291

(iv)16160

Finding the square root of 16160

We get 127 as quotient and 31 as remainder

∴ To make it a perfect square, we have to subtract 31 from 16160

∴ Least number to be subtracted = 31

(v) 4401624

Find the square root of 4401624

We get 2098 as quotient and 20 as remainder

∴ To make it a perfect square, we have to subtract 20 from 4401624

∴ Least number to be subtracted = 20

**Question 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

Finding the square root of 5607, we see that 74

^{2}= 5607- 131 =5476 and 75

^{2}= 5625

∴ 5476 < 5607 < 5625

∴ 5625 – 5607 = 18 is to be added to get a perfect square

∴ Least number to be added = 18

(ii) 4931

Finding the square root of 4931, we see that 70

^{2}= 4900

∴ 71

^{2}= 5041 4900 <4931 <5041

∴ 5041 – 4931 = 110 is to be added to get a perfect square.

∴ Least number to be added =110

(iii) 4515600

Finding the square root of 4515600, we see

that 2124

^{2}= 4511376

and 2 1 25

^{2}= 45 1 56 25

∴ 4511376 <4515600 <4515625

∴ 4515625 – 4515600 = 25 is to be added to get a perfect square.

∴ Least number to be added = 25

(iv) 37460

Finding the square root of 37460

that 193

^{2}= 37249, 194

^{2}= =37636

∴ 37249 < 37460 < 37636

∴ 37636 – 37460 = 176 is to be added to get a perfect square.

∴ Least number to be added =176

(v) 506900

Finding the square root of 506900, we see that

711

^{2}= 505521, 712

^{2}= 506944

∴ 505521 < 506900 < 506944

∴ 506944 – 506900 = 44 is to be added to get a perfect square.

∴ Least number to be added = 44

**Question 4.**Find the greatest number of 5 digits which is a perfect square.

**Solution:**

Greatest number of 5-digits = 99999 Finding square root, we see that 143 is left as remainder

∴ Perfect square = 99999 – 143 = 99856 If we add 1 to 99999, it will because a number of 6 digits

∴ Greatest square 5-digits perfect square = 99856

**Question 5.**Find the least number of four digits which is a perfect square.

**Solution:**

Least number of 4-digits = 10000

Finding square root of 1000

We see that if we subtract 39

From 1000, we get three digit number

∴ We shall add 124 – 100 = 24 to 1000 to get a

perfect square of 4-digit number

∴ 1000 + 24 = 1024

∴ Least number of 4-digits which is a perfect square = 1024

**Question 6.**Find the least number of six-digits which is a perfect square.

**Solution:**

Least number of 6-digits = 100000

Finding the square root of 100000, we see that if we subtract 544, we get a perfect square of 5-digits.

So we shall add

4389 – 3900 = 489

to 100000 to get a perfect square

Past perfect square of six digits= 100000 + 489 =100489

**Question 7.**Find the greatest number of 4-digits which is a perfect square.

**Solution:**

Greatest number of 4-digits = 9999

Finding the square root, we see that 198 has been left as remainder

∴ 4-digit greatest perfect square = 9999 – 198 = 9801

**Question 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 be 8160, find the number of soldiers in each row.

**Solution:**

Total number of soldiers = 8160 Soldiers left after arranging them in a square = 60

∴ Number of soldiers which are standing in a square = 8160 – 60 = 8100

**Question 9.**The area of a square field is 60025 m

^{2}. A man cycle along its boundry at 18 km/hr. In how much time will be return at the starting point.

**Solution:**

Area of a square field = 60025 m

^{2}

**Question 10.**The cost of levelling and turfing a square lawn at Rs. 250 per m

^{2}is Rs. 13322.50. Find the cost of fencing it at Rs. 5 per metre ?

**Solution:**

Cost of levelling a square field = Rs. 13322.50

Rate of levelling = Rs. 2.50 per m

^{2}

and perimeter = 4a = 4 x 73 = 292 m Rate of fencing the field = Rs. 5 per m

∴ Total cost of fencing = Rs. 5 x 292 = Rs. 1460

**Question 11.**Find the greatest number of three digits which is a perfect square.

**Solution:**

3-digits greatest number = 999

Finding the square root, we see that 38 has been left

∴ Perfect square = 999 – 38 = 961

∴ Greatest 3-digit perfect square = 961

**Question 12.**Find the smallest number which must be added to 2300 so that it becomes a perfect square.

**Solution:**

Finding the square root of 2300

We see that we have to add 704 – 700 = 4 to 2300 in order to get a perfect square

∴ Smallest number to be added = 4

**All Chapter RD Sharma Solutions For Class 8 Maths**

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**All Subject NCERT Solutions For Class 8**

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