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# NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3- Squares and Square Roots

Last updated date: 12th Sep 2024
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## NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 - FREE PDF Download

NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 are provided by Vedantu. Our updated PDF solutions guide answers all the questions in the latest NCERT exercise book for the subject. Class 8 Math 5.3 dives deeper into the concepts introduced earlier in the chapter, helping students understand how to find square roots using various methods. This exercise is crucial as it builds foundational knowledge that is applicable in higher-level math and various real-life scenarios. The PDF can be freely downloaded from Vedantu’s website and can go a long way to improve your chances of scoring well in your CBSE Class 8 Maths exam.

Table of Content
1. NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 5 Exercise 5.3 Class 8 | Vedantu
3. Access NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 - Squares and Square Roots
4. Class 8 Maths Chapter 5: Exercises Breakdown
5. CBSE Class 8 Maths Chapter 5 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 5 Exercise 5.3 Class 8 | Vedantu

• Prime factorization method which involves breaking down the number into its prime factors.

• Division method which is also known as long division method and is used to find the square roots of larger numbers.

• Estimation method by using the nearest perfect squares method, this method approximates the square root.

• This exercise is about developing a better intuition for squares and square roots, which are essential building blocks in mathematics

• Exercise 5.3 Class 8 Maths NCERT Solutions has overall 10 fully solved questions.

## Access NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 - Squares and Square Roots

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 $120\times 2\times 3\times 5 = 3600$

## Conclusion

NCERT Solutions for Class 8 Maths Chapter Square and Square Roots Exercise 5.3 by Vedantu helps students master various methods of finding square roots, including prime factorization, division, and estimation methods. This exercise is crucial for developing a strong foundation in understanding the properties of numbers and improving problem-solving skills. Students should focus on mastering each method and practicing a variety of problems to gain confidence. Vedantu's detailed, step-by-step solutions ensure that students can follow along easily, making complex concepts more understandable and aiding in thorough exam preparation.

## Class 8 Maths Chapter 5: Exercises Breakdown

 Exercise Number of Questions Exercise 5.1 9 Questions and Solutions Exercise 5.2 2 Questions and Solutions Exercise 5.4 9 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3- Squares and Square Roots

1. How can we Find square roots Class 8 Maths Chapter 5 Squares and Square Roots (EX 5.3) Exercise 5.3?

The square root of a number is the factor that results from multiplying the original number by itself. If "x" equals the square root of "y," then X × X = Y.It is easy to determine the square root of a perfect square integer. Positive integers that may be expressed as the sum of two numbers are called "perfect squares." Positive numbers that can be expressed as the product of a single number are called perfect squares. In layman's terms, perfect squares are those numbers that can be written as the value of any number's power 2.

2. What is the use of repeated subtraction to calculate the square root  Class 8 Maths Chapter 5 Squares and Square?

In the repeated subtraction method, the square root of the number is calculated by subtracting each succeeding odd integer until the result equals zero. The number of times we subtract will be equal to the square root of the supplied number. This method can only be applied to perfect square numbers.

3. How Using prime factorization, we can find the square root in Class 8 Maths Chapter 5 Squares and Square?

The following steps must be taken to use the prime factorization method to determine the square root of a given number:

1. Primarily, determine the prime factors of the given number.

2. Next, group related factors into pairs by ensuring that each pair's two components are equal.

3. Select one element from each pair.

4. Next, calculate the product of the factors you got by taking one from each pair of factors.

5. The square root of the specified number is equal to the product value.

4. How does Vedantu NCERT Maths Class 8 Chapter 5 Exercise 5.3 Solutions make exam preparations easier?

Here are some of the ways that Vedantu NCERT Maths Class 8 Chapter 5 Exercise 5.3 Solutions make exam preparation easier:

1. In the world of Indian school education, Vedantu is a well-known brand. Vedantu's There are many reasons why CBSE students can depend on Vedantu for help with their academics.

2. The internal tutors at Vedantu are some of the most knowledgeable individuals in their respective fields. They have extensive knowledge of the CBSE question format and have years of experience in their field of study.

3. The Vedantu NCERT Solutions are precise and easily understood. The explanations given for the examples in students' school textbooks are frequently found wanting.

5. Why are Class 8 Maths Chapter 5 Squares and Square Roots so important?

It is important in the formula for the roots of a quadratic equation. Based on square roots, quadratic fields and rings of quadratic integers are important in algebra and have been used in geometry. Square roots are frequently used in foreign mathematical and scientific formulas. These concepts simplify and streamline problem-solving, which is especially useful when performing extensive and time-consuming calculations.

6. What topics are covered in Class 8 ex 5.3 Maths?

Exercise 5.3 focuses on different methods to find the square roots of numbers. This includes the prime factorization method, the division method, and the estimation method.

7. How do you find the square root of a perfect square using prime factorization in Class 8 Math Exercise 5.3?

To find the square root of a perfect square using prime factorization, break down the number into its prime factors and then pair the factors. For example, for 36, the prime factors are $2 \times 2 \times 3 \times 3$  Pair the factors to get 2 \times 3 = 6, so the square root of 36 is 6.