# NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.3) Exercise 6.3

## NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.3) Exercise 6.3 NCERT Solutions for Class 8 Maths are provided by Vedantu. Our updated PDF solutions guide answers all the questions in the latest NCERT exercise book for the subject. You will find broadly described and simplified steps in the solution guide that help you understand the problems instead of just providing the solution. The PDF can be freely downloaded from Vedantu’s website and can go a long way to improve your chances for scoring well in your CBSE Class 8 Maths exam.Vedantu is a platform that provides free NCERT Solution and other study materials for students. Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.

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Exercise 6.3

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

i. $\text{9801}$

Ans: We know that the one’s digit of the square root of the number ending with  $1$ can be $1$ or $9$.

Thus, the possible one’s digit of the square root of $9801$ is either $1$ or $9$.

ii. $\text{99856}$

Ans: We know that the one’s digit of the square root of the number ending with  $6$ can be $6$ or $4$.

Thus, the possible one’s digit of the square root of $99856$ is either $6$ or $4$.

iii. $\text{998001}$

Ans: We know that the one’s digit of the square root of the number ending with  $1$ can be $1$ or $9$.

Thus, the possible one’s digit of the square root of $998001$ is either $1$ or $9$.

iv. $\text{657666025}$

Ans: We know that the one’s digit of the square root of the number ending with $5$ will be $5$.

Thus, the only possible one’s digit of the square root of $657666025$ is $5$.

2. Find the numbers which are surely not perfect squares without doing any calculations.

i. $\text{153}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $153$ has its unit place digit as $3$.

Hence, $153$ cannot be a perfect square.

ii. $\text{257}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $257$ has its unit place digit as $7$.

Hence, $257$ cannot be a perfect square.

iii. $\text{408}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $408$ has its unit place digit as $8$.

Hence, $408$ cannot be a perfect square.

iv. $\text{441}$

Ans: The perfect square of numbers may end with any one of the digits $0$, $1$, $4$, $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it, if any.

We can see that $441$ has its unit place digit as $1$.

Hence, $441$ is a perfect square.

3. Find the square roots of $\text{100}$ and $\text{169}$ by the method of repeated subtraction.

Ans: It is already known to us that the sum of the first n odd natural numbers is n2.

For $\sqrt{100}$

1. $100-1=99$

2. $99-3=96$

3. $96-5=91$

4. $91-7=84$

5. $84-9=75$

6. $75-11=64$

7. $64-13=51$

8. $51-15=36$

9. $36-17=19$

10.  $19-19=0$

After subtracting successive odd numbers from $1$ to $100$ , we are getting a $0$ at the 10th step.

Hence, $\sqrt{100}=10$

For $\sqrt{169}$

1. $169-1=168$

2. $168-3=165$

3. $165-5=160$

4. $160-7=153$

5. $153-9=144$

6. $144-11=133$

7. $133-13=120$

8. $120-15=105$

9. $105-17=88$

10. $88-19=69$

11.  $69-21=48$

12.  $48-23=25$

13.  $25-25=0$

After subtracting successive odd numbers from $1$ to $169$, we are getting a $0$ at the 13th step.

Hence, $\sqrt{169}=13$

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

i. $\text{729}$

Ans:

The factorization of $729$ is as follows:

 $3$ $729$ $3$ $243$ $3$ $81$ $3$ $27$ $3$ $9$ $3$ $3$ $1$

$729=\underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$

$\sqrt{729}=3\times 3\times 3$

So, $\sqrt{729}=27$

ii. $\text{400}$

Ans: The factorization of $400$ is as follows:

 $2$ $400$ $2$ $200$ $2$ $100$ $2$ $50$ $5$ $25$ $5$ $5$ $1$

$400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}$

$\sqrt{400}=2\times 2\times 5$

So, $\sqrt{400}=20$

iii. $\text{1764}$

Ans:

The factorization of $1764$ is as follows:

 $2$ $1764$ $2$ $882$ $3$ $441$ $3$ $147$ $7$ $49$ $7$ $7$ $1$

$1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$

$\sqrt{1764}=2\times 3\times 7$

So, $\sqrt{1764}=42$

iv. $\text{4096}$

Ans: The factorization of $4096$ is as follows:

 $2$ $4096$ $2$ $2048$ $2$ $1024$ $2$ $512$ $2$ $256$ $2$ $128$ $2$ $64$ $2$ $32$ $2$ $16$ $2$ $8$ $2$ $4$ $2$ $2$ $1$

$4096=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}$

$\sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2$

So, $\sqrt{4096}=64$

v. $\text{7744}$

Ans: The factorization of $7744$ is as follows:

 $2$ $7744$ $2$ $3872$ $2$ $1936$ $2$ $968$ $2$ $484$ $2$ $242$ $11$ $121$ $11$ $11$ $1$

$7744=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{11\times 11}$

$\sqrt{7744}=2\times 2\times 2\times 11$

So, $\sqrt{7744}=88$

vi. $\text{9604}$

Ans: The factorization of $9604$ is as follows:

 $2$ $9604$ $2$ $4802$ $7$ $2401$ $7$ $343$ $7$ $49$ $7$ $7$ $1$

$9604=\underline{2\times 2}\times \underline{7\times 7}\times \underline{7\times 7}$

$\sqrt{9604}=2\times 7\times 7$

So, $\sqrt{9604}=98$

vii. $\text{5929}$

Ans: The factorization of $5929$ is as follows:

 $7$ $5929$ $7$ $847$ $11$ $121$ $11$ $11$ $1$

$5929=\underline{7\times 7}\times \underline{11\times 11}$

$\sqrt{5929}=7\times 11$

So, $\sqrt{5929}=77$

viii. $\text{9216}$

Ans: The factorization of $9216$ is as follows:

 $2$ $9216$ $2$ $4608$ $2$ $2304$ $2$ $1152$ $2$ $576$ $2$ $288$ $2$ $144$ $2$ $72$ $2$ $36$ $2$ $18$ $3$ $9$ $3$ $3$ $1$

$9216=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}$

$\sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3$

So, $\sqrt{9216}=96$

ix. $\text{529}$

Ans: The factorization of $529$ is as follows:

 $23$ $529$ $23$ $23$ $1$

$529=\underline{23\times 23}$

So, $\sqrt{529}=23$

x. $\text{8100}$

Ans: The factorization of $8100$ is as follows:

 $2$ $8100$ $2$ $4050$ $3$ $2025$ $3$ $675$ $3$ $225$ $3$ $75$ $5$ $25$ $5$ $5$ $1$

$8100=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}$

$\sqrt{8100}=2\times 3\times 3\times 5$

So, $\sqrt{8100}=90$

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. $\text{252}$

Ans: The factorization of $252$ is as follows:

 $2$ $252$ $2$ $126$ $3$ $63$ $3$ $21$ $7$ $7$ $1$

Here, $252=\underline{2\times 2}\times \underline{3\times 3}\times 7$

We can see that $7$ is not paired

So, we have to multiply $252$ by $7$ to get a perfect square.

The new number will be $252\times 7=1764$

$1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$

$\sqrt{1764}=2\times 3\times 7$

So, $\sqrt{1764}=42$

ii. $\text{180}$

Ans: The factorization of $180$ is as follows:

 $2$ $180$ $2$ $90$ $3$ $45$ $3$ $15$ $5$ $5$ $1$

Here, $180=\underline{2\times 2}\times \underline{3\times 3}\times 5$

We can see that $5$ is not paired

So, we have to multiply $180$ by $5$ to get a perfect square.

The new number will be $180\times 5=900$

$900=\underline{2\times 2}\times \underline{3\times 3}\times \underline{5\times 5}$ which is a perfect square

$\sqrt{900}=2\times 3\times 5$

So, $\sqrt{900}=30$

iii. $\text{1008}$

Ans: The factorization of $1008$ is as follows:

 $2$ $1008$ $2$ $504$ $2$ $252$ $2$ $126$ $3$ $63$ $3$ $21$ $7$ $7$ $1$

Here, $1008=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times 7$

We can see that $7$ is not paired

So, we have to multiply $1008$ by $7$ to get a perfect square.

The new number will be $1008\times 7=7056$

$7056=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ which is a perfect square

$\sqrt{7056}=2\times 2\times 3\times 7$

So, $\sqrt{7056}=84$

iv. $\text{2028}$

Ans: The factorization of $2028$ is as follows:

 $2$ $2028$ $2$ $1014$ $3$ $507$ $13$ $169$ $13$ $13$ $1$

Here, $2028=\underline{2\times 2}\times 3\times \underline{13\times 13}$

We can see that $3$ is not paired

So, we have to multiply $2028$ by $3$ to get a perfect square.

The new number will be $2028\times 3=6084$

$6084=\underline{2\times 2}\times \underline{3\times 3}\times \underline{13\times 13}$ which is a perfect square

$\sqrt{6084}=2\times 3\times 13$

So, $\sqrt{6084}=78$

v. $\text{1458}$

Ans: The factorization of $1458$ is as follows:

 $2$ $1458$ $3$ $729$ $3$ $243$ $3$ $81$ $3$ $27$ $3$ $9$ $3$ $3$ $1$

Here, $1458=2\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$

We can see that $2$ is not paired

So, we have to multiply $1458$ by $2$ to get a perfect square.

The new number will be $1458\times 2=2916$

$2916=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{2916}=2\times 3\times 3\times 3$

So, $\sqrt{2916}=54$

vi. $\text{768}$

Ans: The factorization of $768$ is as follows:

 $2$ $768$ $2$ $384$ $2$ $192$ $2$ $96$ $2$ $48$ $2$ $24$ $2$ $12$ $2$ $6$ $3$ $3$ $1$

Here, $768=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times 3$

We can see that $3$ is not paired

So, we have to multiply $768$ by $3$ to get a perfect square.

The new number will be $768\times 3=2304$

$2304=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{2304}=2\times 2\times 2\times 2\times 3$

So, $\sqrt{2304}=48$

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. $\text{252}$

Ans: The factorization of $252$ is as follows:

 $2$ $252$ $2$ $126$ $3$ $63$ $3$ $21$ $7$ $7$ $1$

Here, $252=\underline{2\times 2}\times \underline{3\times 3}\times 7$

We can see that $7$ is not paired

So, we have to divide $252$ by $7$ to get a perfect square.

The new number will be $252\div 7=36$

$36=\underline{2\times 2}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{36}=2\times 3$

So, $\sqrt{36}=6$

ii. $\text{2925}$

Ans: The factorization of $2925$ is as follows:

 $3$ $2925$ $3$ $975$ $5$ $325$ $5$ $65$ $13$ $13$ $1$

Here, $2925=\underline{3\times 3}\times \underline{5\times 5}\times 13$

We can see that $13$ is not paired

So, we have to divide $2925$ by $13$ to get a perfect square.

The new number will be $2925\div 13=225$

$225=\underline{3\times 3}\times \underline{5\times 5}$ which is a perfect square

$\sqrt{225}=3\times 5$

So, $\sqrt{225}=15$

iii. $\text{396}$

Ans: The factorization of $396$ is as follows:

 $2$ $396$ $2$ $198$ $3$ $99$ $3$ $33$ $11$ $11$ $1$

Here, $396=\underline{2\times 2}\times \underline{3\times 3}\times 11$

We can see that $11$ is not paired

So, we have to divide $396$ by $11$ to get a perfect square.

The new number will be $396\div 11=36$

$36=\underline{2\times 2}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{36}=2\times 3$

So, $\sqrt{36}=6$

iv. $\text{2645}$

Ans: The factorization of $2645$ is as follows:

 $5$ $2645$ $23$ $529$ $23$ $23$ $1$

Here, $2645=5\times \underline{23\times 23}$

We can see that $5$ is not paired

So, we have to divide $2645$ by $5$ to get a perfect square.

The new number will be $2645\div 5=529$

$529=\underline{23\times 23}$ which is a perfect square

So, $\sqrt{529}=23$

v. $\text{2800}$

Ans: The factorization of $2800$ is as follows:

 $2$ $2800$ $2$ $1400$ $2$ $700$ $2$ $350$ $5$ $175$ $5$ $35$ $7$ $7$ $1$

Here, $2800=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}\times 7$

We can see that $7$ is not paired

So, we have to divide $2800$ by $7$ to get a perfect square.

The new number will be $2800\div 7=400$

$400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}$ which is a perfect square

$\sqrt{400}=2\times 2\times 5$

So, $\sqrt{400}=20$

vi. $\text{1620}$

Ans: The factorization of $1620$ is as follows:

 $2$ $1620$ $2$ $810$ $3$ $405$ $3$ $135$ $3$ $45$ $3$ $15$ $5$ $5$ $1$

Here, $1620=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times 5$

We can see that $5$ is not paired

So, we have to divide $1620$ by $5$ to get a perfect square.

The new number will be $1620\div 5=324$

$324=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}$ which is a perfect square

$\sqrt{324}=2\times 3\times 3$

So, $\sqrt{324}=18$

7. The students of Class VIII of a school donated Rs 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.

Ans: According to the question, each student donated as many rupees as the number of students in the class.

We can find the number of students in the class by doing the square root of the total amount donated by the students of Class VIII.

The total amount donated by students is Rs. $2401$

Then, the number of students in the class will be $\sqrt{2401}$

$\sqrt{2401}=\sqrt{\underline{7\times 7}\times \underline{7\times 7}}$

$=7\times 7$

$=49$

Thus, there are total $49$ students in the class.

8. Around 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.

Ans: According to the question, the plants are being planted in a garden in such a way that each row contains as many plants as the number of rows.

So, the number of rows will be equal to the number of plants in each row.

Hence,

The number of rows $\times$ Number of plants in each row $=$Total number of plants

The number of rows $\times$ Number of plants in each row $=$ $2025$

The number of rows $\times$ The number of rows $=$ $2025$

The number of rows $=$$\sqrt{2025} \sqrt{2025}=\sqrt{\underline{5\times 5}\times \underline{3\times 3}\times \underline{3\times 3}} =5\times 3\times 3 =45 Thus, the number of rows =45 and the number of plants in each row =45. 9. Find the smallest square number that is divisible by each of the numbers \text{4,9} and \text{10}. Ans: We know that the number that is perfectly divisible by each one of 4,9 and 10 is their L.C.M So, taking the L.C.M of these numbers  2 4,9,10 2 2,9,5 3 1,9,5 3 1,3,5 5 1,1,5 1,1,1 L.C.M=2\times 2\times 3\times 3\times 5 =180 It can be clearly seen that 5 cannot be paired. Therefore, we have to multiply 180 by 5 in order to get a perfect square. Thus, the smallest square number divisible by 4,9 and 10$$=180\times 5=900$

10. Find the smallest square number that is divisible by each of the numbers $\text{8,15}$ and $\text{20}$.

Ans: We know that the number that is perfectly divisible by each one of $8,15$ and $20$ is their L.C.M

So, taking the L.C.M of these numbers

 $2$ $8,15,20$ $2$ $4,15,10$ $2$ $2,15,5$ $3$ $1,15,5$ $5$ $1,5,5$ $1,1,1$

L.C.M$=2\times 2\times 2\times 3\times 5$

$=120$

It can be clearly seen that the prime factors $2$, $3$ and $5$ cannot be paired.

Therefore, we have to multiply $120$ by $2$, $3$ and $5$ in order to get a perfect square.

Thus, the smallest square number divisible by $8, 15$ and $20$ is $120\times 2\times 3\times 5 = 3600$

## What do you learn from Exercise 6.3 Class 8 Maths?

CBSE Class 8 Maths introduces students to squares and square roots in the 6th lesson. In this chapter students get to learn everything about squaring a number, how to derive the square root value of a number, perfect squares, what are perfect squares, properties of square numbers and so on. To get a good grasp over the lesson students may proceed to complete the NCERT Maths Class 8, ex. 6.3 exercises. If you have completed the previous chapter on exponents, then it may not be difficult to understand the problems in exercise 6.3 Class 8 NCERT.

Squares and square roots are one of the most important lessons in Class 8 Mathematics because the chapter not only contributes to advanced Mathematics that you shall study in the future but other Science subjects as well. Squares and square roots are one of the few Mathematical formulae that are applied in algebraic, geometric and trigonometric problems too. Ex 6.3 Class 8 Maths covers some of the most important patterns in square and square root problems. This section covers common and uncommon problems related to the sub-sections and theorems mentioned above.

For instance, a problem from section 6.3 may ask you to find out which among a group of numbers is a perfect square. The NCERT exercise book contains ten sums on the subject, under section 6.3. Students who wish to score well in their CBSE Class 8 Maths paper and wish to score well must try to solve the NCERT exercise on their own and revise it several times.

In the event you are unable to do so on your own, then do not fret, Vedantu is there to help you. Vedantu’s PDF solution guide to NCERT Class 8 Maths, Chapter 6, ex. 6.3 provides accurate, and simplified, step-by-step solutions that can aid in your understanding of Squares and Square Roots. All you need to do is visit Vedantu’s site and download it.

The syllabus is a long-winded one, with numerous important chapters that are highly anticipated in the CBSE test papers. Chapter 6.3, Square and Square Roots happen to be one of the most critically important chapters in the Class 8 Maths syllabus. The chapter presents a long list of problems that form a good percentage of the question paper. So, provided you study the chapter well, square roots can account for a lot of your marks. If you want to get better at solving square root problems, you can always resort to solving the NCERT exercises; and, if the NCERT exercises seem daunting, then you can always count on Vedantu for NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.3.

### Vedantu’s PDF Solutions of Class 8 Maths Chapter 6 Exercise 6.3

Vedantu’s NCERT Solution guide can benefit you like it has done for countless students in the past, and continues to do so regularly. Vedantu’s NCERT Solutions for the Class 8, 6th chapter, exercise 6.3 has detailed step-by-step answers that appeal to your understanding of the lesson in a broader, simplified sense. As CBSE students must be aware, their exam papers are modeled quite similarly to the NCERT question-answer model; practicing the sums from the NCERT exercise with the help of the solution and then on your own can make a substantial difference in your understanding of the subject.

If you wish to clear all of your doubts associated with Squares and Square Roots, class 8 Maths Exercise 6.3 solutions you have to study the Vedantu solution guide properly. Every solution is given outlines how numerical problems of different patterns can be tackled and simplified for the deduction. If you are not convinced how much of a difference the solution guide may make, simply attempt the previous ten years’ questions after you are done consulting the steps illustrated in the solution guide. The results shall manifest themselves and help you appear for your CBSE Class 8 Maths paper more confidently.

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