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NCERT Solutions for Class 8 Maths Chapter 6 (Ex 6.4) Squares and Square Roots

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 NameSquares and Square Roots
ChapterChapter 6
ExerciseExercise 6.4
ClassClass 8
SubjectMaths NCERT Solutions
BoardCBSE
TEXTBOOKCBSE NCERT
CategoryNCERT 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.