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NCERT Solutions for Class 8 Maths Exercise 6.4 Class 8 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 8 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

**NCERT solutions for Class 8 Maths Squares and Square Roots Download as PDF**

## NCERT Solutions for Class 8 Maths Squares and Square Roots

**Class –VIII Mathematics (Ex. 6.4)**

**NCERT SOLUTION**

**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 =

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

Therefore, number of digits in square root =

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

Therefore, number of digits in square root =

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

Therefore, number of digits in square root =

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

Therefore, number of digits in square root =

NCERT Solutions for Class 8 Maths Exercise 6.4

**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.

NCERT Solutions for Class 8 Maths Exercise 6.4

**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 53must 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.

NCERT Solutions for Class 8 Maths Exercise 6.4

**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 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 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 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 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 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.

NCERT Solutions for Class 8 Maths Exercise 6.4

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

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

Area of square

According to question,

= 441

=

=

= 21 m

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

NCERT Solutions for Class 8 Maths Exercise 6.4

**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. **(a) Using Pythagoras theorem,

= 36 + 84 = 100

AC = 10 cm

(b) Using Pythagoras theorem,

= 169 – 25

= 144

AB= 12 cm

NCERT Solutions for Class 8 Maths Exercise 6.4

**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 required 24 more plants.

NCERT Solutions for Class 8 Maths Exercise 6.4

**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 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.

## NCERT Solutions for Class 8 Maths Exercise 6.4

NCERT Solutions Class 8 Mathematics PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 8 Mathematics includes text book solutions from Class 8 Maths Book . NCERT Solutions for CBSE Class 8 Maths have total 16 chapters. 8 Maths NCERT Solutions in PDF for free Download on our website. Ncert class 8 solutions PDF and Maths ncert class 8 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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