## NCERT Solutions for Class 8 Chapter 6 Squares and Square Roots -Free PDF Download

Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 6 – Squares and Square Roots solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 6 – Squares and Square Roots Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.

Maths Revision Notes for Class 8

Chapter Name | Squares and Square Roots |

Chapter | Chapter 6 |

Exercise | Exercise 6.4 |

Class | Class 8 |

Subject | Maths NCERT Solutions |

Board | CBSE |

TEXTBOOK | CBSE NCERT |

Category | NCERT Solutions |

**NCERT SOLVED**

**1. Find the square roots of each of the following numbers by Division method:**

**(i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369**

**(vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900**

**Ans. (i)** 2304

Hence, the square root of 2304 is 48.

**(ii)** 4489

Hence, the square root of 4489 is 67.

**(iii)** 3481

Hence, the square root of 3481 is 59.

**(iv)** 529

Hence, the square root of 529 is 23.

**(v)** 3249

Hence, the square root of 3249 is 57.

**(vi)** 1369

Hence, the square root of 1369 is 37.

**(vii)** 5776

Hence, the square root of 5776 is 76.

**(viii)** 7921

Hence, the square root of 7921 is 89.

**(ix)** 576

Hence, the square root of 576 is 24.

**(x)** 1024

Hence, the square root of 1024 is 32.

**(xi)** 3136

Hence, the square root of 3136 is 56.

**(xii)** 900

Hence, the square root of 900 is 30.

**2. Find the number of digits in the square root of each of the following numbers (without any calculation):**

**(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625**

**Ans. (i)** Here, 64 contains two digits which is even.

Therefore, number of digits in square root = ( that is 8, which is single digit number)

**(ii)** Here, 144 contains three digits which is odd.

Therefore, number of digits in square root = (that is 12, which is a 2-digit number)

**(iii)** Here, 4489 contains four digits which is even.

Therefore, number of digits in square root = (that is 67, which is a 2-digit number)

**(iv)** Here, 27225 contains five digits which is odd.

Therefore, number of digits in square root = (that is 165, which is a 3-digit number)

**(v)** Here, 390625 contains six digits which is even.

Therefore, the number of digits in square root = (that is 625, which is a 3-digit number)

**3. Find the square root of the following decimal numbers:**

**(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36**

**Ans. (i)** 2.56

Hence, the square root of 2.56 is 1.6.

**(ii)** 7.29

Hence, the square root of 7.29 is 2.7.

**(iii)** 51.84

Hence, the square root of 51.84 is 7.2.

**(iv)** 42.25

Hence, the square root of 42.25 is 6.5.

**(v)** 31.36

Hence, the square root of 31.36 is 5.6.

**4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained:**

**(i) 402 **

**(ii) 1989**

**(iii) 3250 **

**(iv) 825**

**(v) 4000**

**Ans**. (i) 402

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 2. Therefore 2 must be subtracted from 402 to get a perfect square.

402 – 2 = 400

Hence, the square root of 400 is 20.

**(ii)** 1989

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 53. Therefore 53 must be subtracted from 1989 to get a perfect square.

1989 – 53 = 1936

Hence, the square root of 1936 is 44.

**(iii)** 3250

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 1. Therefore 1 must be subtracted from 3250 to get a perfect square.

3250 – 1 = 3249

Hence, the square root of 3249 is 57.

**(iv)** 825

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 41. Therefore 41 must be subtracted from 825 to get a perfect square.

825 – 41 = 784

Hence, the square root of 784 is 28.

**(v)** 4000

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 31. Therefore 31 must be subtracted from 4000 to get a perfect square.

4000 – 31 = 3969

Hence, the square root of 3969 is 63.

**5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained:**

**(i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412**

**Ans. (i)** 525

Since the remainder is 41.

Therefore

Next perfect square number = 529

Hence, number to be added

= 529 – 525 = 4

525 + 4 = 529

Hence, the square root of 529 is 23.

**(ii)** 1750

Since the remainder is 69.

Therefore

Next perfect square number = 1764

Hence, number to be added

= 1764 – 1750 = 14

1750 + 14 = 1764

Hence, the square root of 1764 is 42.

**(iii)** 252

Since the remainder is 27.

Therefore

Next perfect square number = 256

Hence, number to be added

= 256 – 252 = 4

252 + 4 = 256

Hence, the square root of 256 is 16.

**(iv)** 1825

Since the remainder is 61.

Therefore

Next perfect square number = 1849

Hence, number to be added = 1849 – 1825 = 24

1825 + 24 = 1849

Hence, the square root of 1849 is 43.

**(v)** 6412

Since the remainder is 12.

Therefore

Next perfect square number = 6561

Hence, number to be added

= 6561 – 6412 = 149

6412 + 149 = 6561

Hence, the square root of 6561 is 81.

**6. Find the length of the side of a square whose area is?**

**Ans. **Let the length of the side of a square be meter.

Area of square

According to question,

= 441

=

=

= 21 m

Hence, the length of the side of a square is 21 m.

**7. In a right triangle ABC, ****B = **

**(i) If AB = 6 cm, BC = 8 cm, find AC.**

**(ii) If AC = 13 cm, BC = 5 cm, find AB.**

**Ans. **(i) Using Pythagoras theorem,

= 36 + 84 = 100

AC = sqrt(100)sqrt(100)

AC = 10 cm

(ii) Using Pythagoras theorem,

= 169 – 25

= 144

AB = sqrt(144)sqrt(144)

AB= 12 cm

**8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and number of columns remain same. Find the minimum number of plants he needs more for this.**

**Ans. **Here, plants = 1000

Since remainder is 39.

Therefore

Next perfect square number = 1024

Hence, number to be added

= 1024 – 1000 = 24

1000 + 24 = 1024

Hence, the gardener requires 24 more plants.

**9. There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?**

**Ans. **Here, Number of children = 500

By getting the square root of this number, we get,

In each row, the number of children is 22.

And left out children are 16.