Find here step-by-step NCERT Maths Solution Class 8 Exercise 6.3 | Square Roots | Chapter 5 Square And Square Roots from CBSE NCERT textbook.

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## NCERT Maths Solution Class 8 Exercise 6.3

Problems in Class 8 Ex 6.3 require the knowledge Square Roots and their Properties.

Square root of a number is the value which when multiplied by itself gives that number, it is the inverse operation of squaring.

Every perfect square has two integral square roots, one is positive and other is its negative.

• Negative square root of a number is denoted by the symbol -√ .

• Positive square root of a number is denoted by the symbol √ .

Square Root of a number can be found by two methods:

• Prime Factorisation Method

• Division Method

### Exercise 6.3

Sauare Roots

**QUESTION 1**

What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801 (ii) 99856

(iii) 998001 (iv) 657666025

**ANSWER**

(i) Since square of numbers having unit digit 1 and 9 ends in 1,

Therefore, the possible one’s digits of the square root of 9801 are 1 and 9.

(ii) Since square of numbers having unit digit 4 and 6 ends in 6,

Therefore, the possible one’s digits of the square root of 99856 are 4 and 6.

(iii) Since square of numbers having unit digit 1 and 9 ends in 1,

Therefore, the possible one’s digits of the square root of 998001 are 1 and 9.

(iv) Since only square of numbers having unit digit 5 ends in 5,

Therefore, the possible one’s digit of the square root of 657666025 is five.

**QUESTION 2**

Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257

(iii) 408 (iv) 441

**ANSWER**

A perfect square ends in either one of the following digits at units place:

0 (even), 1, 5, 6 and 9,

therefore, (i), (ii) and (iii) are surely not perfect squares.

[ Find here step-by-step NCERT Maths Solution Class 8 Exercise 6.3 ]

**QUESTION 3**

Find the square roots of 100 and 169 by the method of repeated subtraction.

**ANSWER**

Subtracting successive odd

numbers from 100 starting from 1,

(i) 100 – 1 = 99 (ii) 99 – 3 = 96

(iii) 96 – 5 = 91 (iv) 91 – 7 = 84

(v) 84 – 9 = 75 (vi) 75 – 11 = 64

(vii) 64 – 13 = 51 (viii) 51 – 15 = 36

(ix) 36 – 17 = 19 (x) 19 – 19 = 0

Since we obtained 0 at 10th step, therefore

Similarly, subtracting successive odd numbers from 169 starting from 1,

(i) 169 – 1 = 168 (ii) 168 – 3 = 165

(iii) 165 – 5 = 160 (iv) 160 – 7 = 153

(v) 153 – 9 = 144 (vi) 144 – 11 = 133

(vii) 133 – 13 = 120 (viii) 120 – 15 = 105

(ix) 105 – 17 = 88 (x) 88 – 19 = 69

(xi) 69 – 21 = 48 (xii) 48 – 23 = 25

(xiii) 25 – 25 = 0

Since we obtained 0 at 13th step, therefore

**QUESTION 4**

Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729 (ii) 400 (iii) 1764

(iv) 4096 (v) 7744 (vi) 9604

(vii) 5929 (viii) 9216 (ix) 529

(x) 8100

**ANSWER**

(i) Prime factorisation of 729

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 |

729 = 3×3×3×3×3×3

(ii) Prime factorisation of 400

2 | 400 |

2 | 200 |

2 | 100 |

2 | 50 |

5 | 25 |

5 |

729 = 2×2×2×2×5×5

(iii) Prime factorisation of 1764

2 | 1764 |

2 | 882 |

3 | 441 |

3 | 147 |

7 | 49 |

7 |

729 = 2×2×3×3×7×7

(iv) Prime factorisation of 4096

2 | 4096 |

2 | 2048 |

2 | 1024 |

2 | 512 |

2 | 256 |

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 |

4096 = 2×2×2×2×2×2×2×2×2×2×2×2

(v) Prime factorisation of 7744

2 | 7744 |

2 | 3872 |

2 | 1936 |

2 | 968 |

2 | 484 |

2 | 242 |

11 | 121 |

11 |

7744 = 2×2×2×2×2×2×11×11

(vi) Prime factorisation of 9604

2 | 9604 |

2 | 4802 |

7 | 2401 |

7 | 343 |

7 | 49 |

7 |

9604 = 2×2×7×7×7×7

(vii) Prime factorisation of 5929

7 | 5929 |

7 | 847 |

11 | 121 |

3 | 11 |

5929 = 7×7×11×11

(viii) Prime factorisation of 9216

2 | 9216 |

2 | 4608 |

2 | 2304 |

2 | 1152 |

2 | 576 |

2 | 288 |

2 | 144 |

2 | 72 |

2 | 36 |

2 | 18 |

3 | 9 |

3 |

9216 = 2×2×2×2×2×2×3×3

(ix) Prime factorisation of 529

23 | 529 |

23 |

529 = 23×23

(x) Prime factorisation of 8100

2 | 8100 |

2 | 4050 |

3 | 2025 |

3 | 675 |

3 | 225 |

3 | 75 |

5 | 25 |

5 |

8100 = 2×2×3×3×3×3×5×5

[ NCERT Maths Solution Class 8 Exercise 6.3 ]

**QUESTION 5**

For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252 (ii) 180 (iii) 1008

(iv) 2028 (v) 1458 (vi) 768

**ANSWER**

(i) Prime factorisation of 252

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 |

252 = 2×2×3×3×7

Here, prime factor 7 is not paired, therefore the smallest whole number the number should be multiplied with to get a perfect square is 7,

252×7 = 2×2×3×3×7×7 = 1764 is the required square number,

= 2×3×7 = 42.

(ii) Prime factorisation of 180

2 | 180 |

2 | 90 |

3 | 45 |

3 | 15 |

5 |

180 = 2×2×3×3×5

Here, prime factor 5 is not paired, therefore the smallest whole number the number should be multiplied with to get a perfect square is 5,

180×5 = 2×2×3×3×5×5 = 900 is the required square number,

= 2×3×5 = 30.

(iii) Prime factorisation of 1008

2 | 1008 |

2 | 504 |

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 |

1008 = 2×2×2×2×3×3×7,

Here, prime factor 7 is not paired, therefore the smallest whole number the number should be multiplied with to get a perfect square is 7,

1008×7 = 2×2×2×2×3×3×7×7 = 7056 is the required square number,

= 2×2×3×7 = 84.

(iv) Prime factorisation of 2028

2 | 2028 |

2 | 1014 |

3 | 507 |

13 | 169 |

13 |

2028 = 2×2×3×13×13

Here, prime factor 3 is not paired, therefore the smallest whole number the number should be multiplied with to get a perfect square is 3,

2028×3 = 2×2×3×3×13×13 = 6084 is the required square number,

= 2×3×13 = 78.

(v) Prime factorisation of 1458

2 | 1458 |

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 |

1458 = 2×3×3×3×3×3×3

Here, prime factor 2 is not paired, therefore the smallest whole number the number should be multiplied with to get a perfect square is 2,

1458×2 = 2×3×3×3×3×3×3 = 2916 is the required square number,

= 2×3×3×3 = 54.

(vi) Prime factorisation of 768

2 | 768 |

2 | 384 |

2 | 192 |

2 | 96 |

2 | 48 |

2 | 24 |

2 | 12 |

2 | 6 |

3 |

768 = 2×2×2×2×2×2×2×2×3

Here, prime factor 3 is not paired, therefore the smallest whole number the number should be multiplied with to get a perfect square is 3,

768×3 = 2×2×2×2×2×2×2×2×3×3 = 2304 is the required square number,

= 2×2×2×2×3 = 48.

[ NCERT Maths Solution Class 8 Exercise 6.3 ]

**QUESTION 6**

For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252 (ii) 2925 (iii) 396

(iv) 2645 (v) 2800 (vi) 1620

**ANSWER**

(i) Prime factorisation of 252

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 |

252 = 2×2×3×3×7

Here, prime factor 7 is not paired, therefore the smallest whole number the number should be divided by to get a perfect square is 7,

252 ÷ 7 = 2×2×3×3×7 ÷ 7 = 36 is the required square number,

= 2×3 = 6.

(ii) Prime factorisation of 2925

3 | 2925 |

3 | 975 |

5 | 325 |

5 | 65 |

13 |

2925 = 3×3×5×5×13

Here, prime factor 13 is not paired, therefore the smallest whole number the number should be divided by to get a perfect square is 13,

2925 ÷ 13 = 3×3×5×5×13 ÷ 13 = 225 is the required square number,

= 3×5 = 15.

(iii) Prime factorisation of 396

2 | 396 |

2 | 198 |

3 | 99 |

3 | 33 |

11 |

396 = 2×2×3×3×11

Here, prime factor 11 is not paired, therefore the smallest whole number the number should be divided by to get a perfect square is 11,

396 ÷ 11 = 2×2×3×3×11 ÷ 11 = 36 is the required square number,

= 2×3 = 6.

(iv) Prime factorisation of 2645

5 | 2645 |

23 | 529 |

23 |

2645 = 5×23×23

Here, prime factor 5 is not paired, therefore the smallest whole number the number should be divided by to get a perfect square is 5,

2645 ÷ 5 = 5×23×23 ÷ 5 = 529 is the required square number,

.

(v) Prime factorisation of 2800

2 | 2800 |

2 | 1400 |

2 | 700 |

2 | 350 |

5 | 175 |

5 | 35 |

7 |

2800 = 2×2×2×2×5×5×7

Here, prime factor 7 is not paired, therefore the smallest whole number the number should be divided by to get a perfect square is 7,

2800 ÷ 7 = 2×2×2×2×5×5×7 ÷ 7 = 400 is the required square number,

= 2×2×5 = 20.

(vi) Prime factorisation of 1620

2 | 1620 |

2 | 810 |

3 | 405 |

3 | 135 |

3 | 45 |

3 | 15 |

5 |

1620 = 2×2×3×3×3×3×5

Here, prime factor 5 is not paired, therefore the smallest whole number the number should be divided by to get a perfect square is 5,

1620 ÷ 3 = 2×2×3×3×3×3×5 ÷ 5 = 324 is the required square number,

= 2×3×3 = 18.

[ NCERT Maths Solution Class 8 Exercise 6.3 ]

**QUESTION 7**

The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

**ANSWER**

Let, the number of students in the class be *x*,

So, amount donated by each student be ₹ *x*,

A/Q,

Taking prime factorisation on RHS

Taking square root on both sides,

Therefore, the number of students in the class = *x* = 49.

**QUESTION 8**

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

**ANSWER**

Let, the number of rows be *x*,

So, the number of plants be *x* too,

A/Q,

Taking prime factorisation on RHS

Taking square root on both sides,

Therefore, the number of rows and the number of plants in each row = *x* = 45.

[ NCERT Maths Solution Class 8 Exercise 6.3 ]

**QUESTION 9**

Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

**ANSWER**

Smallest number that is divisible by each of the numbers 4, 9 and 10 is their LCM,

2 | 4, 9, 10 |

3 | 2, 9, 5 |

2, 3, 5 |

LCM of 4, 9 and 10 = 2×2×3×3×5 = 180,

If we multiply 180 with 5, the number we get is a square number that is divisible by 4, 9 and 10,

180 × 5 = 900,

Thus, 900 is the smallest square number that is divisible by each of the numbers 4, 9 and 10.

**QUESTION 10**

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

**ANSWER**

Smallest number that is divisible by each of the numbers 8, 15 and 20 is their LCM,

2 | 8, 15, 20 |

2 | 4, 15, 10 |

5 | 2, 15, 5 |

2, 3, 1 |

LCM of 8, 15 and 20 = 2×2×2×3×5 = 120,

If we multiply 120 with unpaired factors i.e., 2×3×5, the number we get is a square number that is divisible by 8, 15 and 20,

120 × 2×3×5 = 3600,

Thus, 3600 is the smallest square number that is divisible by each of the numbers 8, 15 and 20.

We hope NCERT Maths Solution Class 8 Exercise 6.3 has helped you understand how to find Square Roots and their applications.

Please write in the comment section for any error or any solution related queries from the exercise. [ NCERT Maths Solution Class 8 Exercise 6.3 ]

Check NCERT Solution of other exercises from Class 8 Maths Chapter 6 Squares And Sauare Roots by clicking the links given below.