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

CBSE Class 8 is regarded as a pivotal point in a student’s academic career. It is a class where students must master all the principles that will help them score higher grades in the future. It is a very significant level for a student since it teaches them all the fundamental principles that they will need in their future studies. Students in CBSE Level 8 should master the fundamentals of each subject because the following higher class is similar but in a more thorough format

In Chapter 6 of CBSE Class 8 Maths, students learn about squares and square roots. Students will learn everything there is to know about squaring a number, including how to calculate the square root value of a number, what perfect squares are, the qualities of square numbers, and so on. To ensure a thorough understanding of the subject, students should complete the NCERT Maths Exercise 6.3 Class 8 Maths questions. If students have completed the preceding chapter on exponents, they will have no trouble understanding the questions in NCERT Maths Class 8 Chapter 6 Exercise 6.3. Students are encouraged to take help from the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. They can download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, from the website and mobile application of Extramarks.

Squares and square roots are one of the most significant topics in Class 8 Mathematics because the chapter contributes not only to the advanced mathematics that students will learn in the future but also to other Science disciplines. Squares and square roots are among the few mathematical formulas that can be used to solve algebraic, geometric, and trigonometric problems. Square and Square Roots Class 8 Exercise 6.3 must be thoroughly

practised by students. Students can use the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3,to practise questions. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, contain the most accurate answers for all the exercises.

The Class 8 Mathematics syllabus is lengthy, with many essential chapters that are important to prepare for term-end exams. Chapter 6, Square and Square Roots, is one of the most crucial chapters in the Class 8 Maths syllabus. The chapter contains a large set of questions that make up a significant portion of the exam. As a result, if students study the chapter thoroughly, square roots can account for a significant portion of their grade. If students wish to improve their skills at solving square root problems, they may always solve the NCERT exercises and if the NCERT exercises are difficult to solve, then students can use the resources provided by Extramarks. Students who find Exercise 6.3 of Chapter 6 Class 8 difficult, can use the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, will guide them in solving all the exercise questions. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, are available on the website and mobile application of Extramarks. Students can completely trust the resources provided by Extramarks.

Mathematics is an essential subject for students to study. Learning Mathematics is fun as it teaches various new concepts and their applications to students. It also improves their cognitive and thinking skills. Students can improve their analytical thinking skills by practising Mathematics. Mathematics is a great tool to develop mental discipline. Mathematics has numerous applications. Mathematics is employed in everyday lives in some form or another. Therefore, Mathematics is introduced to students at a very young age.

Mathematical ability is very helpful in different professions. Mathematical concepts are applied in professions and subjects such as Computer Science, Engineering, Astronomy, Statistics, Data Science, and others. Students who are interested in pursuing higher education in these fields should concentrate on Class 8 Mathematics. Class 8 Maths is quite helpful in understanding the theories and concepts of other subjects as well.

Mathematics is generally perceived as difficult by students. To make studying Maths easy and simple, students must practice a lot of questions. Moreover, for effective understanding, consistency is required in Mathematics. Students must solve questions of varied difficulty levels to be fully prepared for Mathematics examinations. When beginning the chapter, students should start by solving the easy questions first and then move on to the questions that require them to apply their reasoning and analytical skills. Often in exams, students will find questions of a nature where analytical skills are required, so students must be prepared for all kinds of questions. With ample practice, students will be able to solve questions quickly within the stipulated time period. For the most accurate and to-the-point answers for Class 8 Chapter 6 Exercise 6.3, students should refer to the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. By practising questions with the help of the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, students will be able to find the most simple solutions for all the questions, which will help them to finish their paper on time. Therefore, students must download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, from the website and mobile application of Extramarks. It is recommended that students download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, in PDF format. In PDF format, students will be able to access the NCERT solutions whenever they want.

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

Chapter 6 of Class 8 Mathematics is on Squares and Square Roots. To ace in Mathematics, students must practise a lot of questions. Students can practise the questions of Class 8 Chapter 6 from the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, are offered by Extramarks for the benefit of students. Students can download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, from the website and mobile application of Extramarks. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, can be downloaded in PDF format for offline access.

With the help of the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 students will be able to solve the Class 8 Maths Chapter 6 Exercise 6.3 thoroughly. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 are written in a stepwise manner for the clarity of students. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 can help students to self study and enhance their Mathematics skills.

**Access NCERT Solutions for Class 8 Maths Chapter 6 – Squares and Square Roots**

Students can access the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, from the website and mobile application of Extramarks. Students are advised to download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, in PDF format. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, will help students score well in the exams. Students must download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 for effective preparation.

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

Class 8 Mathematics chapter 6 is a very important chapter. Students learn new Mathematical concepts in Class 8.

- Calculating square roots

Calculation of square roots is a very basic concept in Mathematics. Students can prepare for this topic with the help of the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3.

The square of a number is the value obtained by multiplying it by itself, whereas the square root of a number is the factor of a number that, when multiplied by itself, yields the original number. If ‘x’ is the square root of ‘y,’ then x*x = y. Students can download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, from the website and mobile application of Extramarks. Further, the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, can be downloaded in PDF format for the benefit of students.

It is easy to find the square root of a perfect square number. Perfect squares are positive numbers that can be stated as the product of two numbers. Perfect squares are positive numbers that can be stated as the product of a single number. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 will guide students to solve the exercise questions accurately.

- 2. Obtaining the square root by repeated subtraction

In this procedure, we subtract the consecutive odd numbers from the number we wish to determine the square root of until the resultant value is 0. The number of times we subtract will be the same as the square root of the supplied number. This approach only works for perfect square numbers. Students can learn more about this method by solving the exercise questions. Students are encouraged to take help from the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, if they face any challenges. NCERT exercises must be solved completely to score well in exams. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3

Will benefit students a lot and help them solve NCERT exercises.

- Using prime factorization to find the square root

A number is prime-factored when it is expressed as a product of prime factors. The following steps must be taken to get the square root of a given number using the prime factorization method:

- Divide the given integer into its prime factors first.
- Then, make pairings of similar factors so that both factors in each pair are equal.
- Take one of the factors from each pair.
- Then compute the product of the factors obtained by selecting one component from each pair.
- The value of the product is equal to the square root of the provided number.

To solve questions related to prime factorization, students must take help from NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. Students can use the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, to match their solved answers with the correct answers. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, are solved by expert Mathematics teachers and can be relied upon by students. All the solutions mentioned in the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, are accurate and are proofread regularly. Students must download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, from the website and mobile application of Extramarks.

**PDF Solutions of Class 8 Maths Chapter 6 Exercise 6.3**

Students can download the PDF version of the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, from the website and mobile application of Extramarks. The PDF format of the solutions will make it convenient for students to study at their own pace. They can also use the Extramarks platform to download PDF solutions to other exercises in Class 8 Chapter 6 Maths. With the use of the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, and other study materials offered by Extramarks, students can gain knowledge and enhance their performance. Therefore, students must take help from the resources offered by Extramarks.

**NCERT Maths Class 8 Chapter 6 Exercise 6.3 Solutions makes exam preparations easier**

The NCERT solutions for Class 8 Maths offered by Extramarks are very beneficial for students as:

- Students will find the most precise and easy-to-understand solutions for all the exercise questions.
- All the NCERT Solutions provided by Extramarks are solved by expert Mathematics teachers. Moreover, all the resources provided are updated regularly.
- The NCERT solutions are written in a stepwise manner for better understanding.
- The solutions are provided for all exercises in Chapter 6 Class 8.
- The study resources provided by Extramarks cover the entire syllabus holistically.

Therefore, to make their exam preparation easy, students must refer to the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3.

**NCERT Solution Class 8 Maths Chapter 6 Other Exercises**

In Class 8 Maths Chapter 6 there are four exercises. All the exercises are important to gain knowledge regarding the concept and to score high marks in the Mathematics exam. Extramarks provides NCERT solutions for all the NCERT exercises. To prepare for Exercise 6.3, students must refer to the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, are available on the website and mobile application of Extramarks. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, can be downloaded in PDF format by students. Students can find the solutions for Exercise 6.1, Exercise 6.2, Exercise 6.4 on the website and mobile application of Extramarks. All the study resources offered by Extramarks are very useful for exam preparation.

**NCERT Solutions for Class 8**

In Class 8 students study a lot of subjects. The core subjects of Class 8 are Mathematics, Social Science, Science, Sanskrit/Hindi, English. All the subjects are important and must be thoroughly studied by students. Students can prepare for these subjects by using the resources provided by Extramarks. Extramarks provide resources such as Past year’s paper solutions, NCERT solutions and Revision notes. Students must take help from the resources provided to enhance their Class 8 marks. The resources offered by Extramarks are very helpful while self-studying. Students can trust the resources provided by Extramarks. All the resources are curated by expert subject teachers and are proofread regularly. Moreover, the study materials are updated regularly to meet the latest CBSE guidelines.

Students must prepare for Class 8 effectively. Preparing thoroughly for Class 8 will ensure that the basic concepts of the major subjects are understood clearly by students. Good knowledge of Class 8 subjects will help students to study efficiently in higher classes. Students must solve all the exercise questions in Class 8 and also carry out the additional activities mentioned in the NCERT books. In this manner, students will be able to get practical knowledge of subjects. Students must refer to the resources provided by Extramarks in case they face any challenges.

A few tips for students to prepare for Class 8 are:

Syllabus: Before beginning exam preparation, students should be familiar with their specific syllabus. The Class 8 CBSE syllabus for all courses will assist pupils in understanding what they will learn during the academic year. Knowing the syllabus will help Class 8 learners study well for their exam, and it is developed by the CBSE board in accordance with the requirements.

Past year’s papers: Solve past years’ question papers and sample papers on a regular basis to improve their test scores. Solving the CBSE example problems for Class 8 may aid students in their exam preparation, and some of the questions contained in the sample papers may appear on the final exam question paper. It is usually a good idea to complete these practice papers prior to the exam. These mock examinations help learners assess their own preparation. Students are also encouraged to take unit exams after finishing a chapter. They should try to answer questions to determine how well they grasp the chapter. Unit tests must be analysed in order to discover weak areas that students can subsequently work on. Students should also prepare for the term-end exam by taking full-length mock exams. The full mock exams will help them manage their time on exam day. Mock examinations should be taken in an exam-like setting so that students are not overly nervous on exam day. Students are encouraged to use the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, to analyse the mock examinations. Students can refer to the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, to find the most accurate and precise way to solve different questions. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, contain solutions along with clear explanations of the formulas used for the students’ understanding.

Revision : Revision is the most important component of exam preparation. Revision helps students remember a large amount of information for a longer period of time. Students should be able to recall the majority of the formulas from their Class 8 Mathematics textbooks with proper revision. Revisions must be carried out on a regular basis. Students should review their short notes as well as their mock tests and unit tests. After several revisions, students will gain confidence. The most effective way to revise mathematics is to practise questions on a regular basis. Regular question practice will decrease the time it takes students to solve each question, allowing them to complete their paper on time. Students can revise the NCERT Solutions for Exercise 6.3 from the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, are available on the website and mobile application of Extramarks. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 are very beneficial for students to prepare efficiently for Class 8 Mathematics Chapter 6.

Solving NCERT Exercises: The most important resource for Class 8 school examinations are NCERT questions. NCERT questions provide students with an idea of the sorts of questions they may face in the examinations. NCERT activities offer questions of varied levels of difficulty. Students must practise the NCERT questions since questions in the CBSE exams and school term-end exams are usually very similar to the NCERT exercises. Sometimes direct questions also appear in NCERT exercises. If students face any difficulties in solving the NCERT exercises or if they want to find simple solutions, they can use the study resources offered by Extramarks. Extramarks’ materials are reliable and trustworthy. NCERT practise questions can also help students prepare for competitive examinations. To solve questions from Exercise 6.3 Chapter 6 Class 8 Maths, students can refer to the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, are available to download in PDF format on the mobile application and website of Extramarks.

Students should have a thorough comprehension of the concepts and topics covered in their individual classes. Students must prepare in accordance with the syllabus outlined in the course textbook. Students should study all the concepts carefully so that they can write precise answers in the exam.

**Q.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

**Ans**

(i) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) The one’s digit of the given number is 6. If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) The one’s digit of the given number is 5. If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.

**Q.2 **Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

**Ans**

A perfect square of any number ends with any of the digits 0, 1,4,5,6 or 9 at its unit’s place. Also, a perfect square will end with even number of zeroes.

(i) Since the number 153 has 3 in its unit’s place, so it is not a perfect square.

(ii) Since the number 257 has 7 in its unit’s place, so it is not a perfect square.

(iii) Since the number 408 has 8 in its unit’s place, so it is not a perfect square.

(iv) Since the number 441 has 1 in its unit’s place and a perfect square number may end with 1, so it is a perfect square.

**Q.3 **Find the square roots of 100 and 169 by the method of repeated subtraction.

**Ans**

\begin{array}{l}{\text{We know that the sum of the first n odd natural numbers is n}}^{\text{2}}\text{.}\\ \text{The square root of 169 can be obtained by the method of}\\ \text{repeated subtraction as follows}.\end{array}

\begin{array}{l}\text{Consider:}\\ \\ \left(\text{i}\right)\text{100}\text{}-\text{1}\text{}\text{= 99}\\ \left(\text{ii}\right)\text{99}\text{}-\text{}\text{3}\text{}\text{= 96}\\ \left(\text{iii}\right)\text{96}\text{}-\text{}\text{5}\text{}\text{= 91}\\ \left(\text{iv}\right)\text{91}\text{}-\text{}\text{7}\text{}\text{= 84}\\ \left(\text{v}\right)\text{84}\text{}-\text{}\text{9}\text{}\text{= 75}\\ \left(\text{vi}\right)\text{75}\text{}-\text{}\text{11}\text{}\text{= 64}\\ \left(\text{vii}\right)\text{64}\text{}-\text{13 = 51}\\ \left(\text{viii}\right)\text{51}\text{}-\text{15}\text{}\text{= 36}\\ \left(\text{ix}\right)\text{36}\text{}-\text{17}\text{}\text{= 19}\\ \left(\text{x}\right)\text{19}\text{}-\text{19}\text{}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting from}\\ {\text{1 to 100, and obtained 0 at 10}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\sqrt{\text{100}}\text{=10}\end{array}

$\begin{array}{l}\text{The square root of 169 can be obtained by the method}\\ \text{of repeated subtraction as follows:}\\ \\ \text{Consider:}\\ \\ \text{(i) 169}-\text{1}\text{= 168}\\ \left(\text{ii}\right)\text{168}-\text{3}\text{= 165}\\ \left(\text{iii}\right)\text{165}-\text{5}\text{= 160}\\ \left(\text{iv}\right)\text{160}-\text{7}\text{= 153}\\ \left(\text{v}\right)\text{153}-19\text{}\text{= 144}\\ \left(\text{vi}\right)\text{144}-\text{11}\text{= 133}\\ \left(\text{vii}\right)\text{133}-\text{13}\text{= 120}\\ \left(\text{viii}\right)\text{120}-\text{15 = 105}\\ \left(\text{ix}\right)\text{105}-\text{17}\text{= 88}\\ \left(\text{x}\right)\text{88}-\text{19}\text{= 69}\\ \left(\text{xi}\right)\text{69}-\text{21}\text{= 48}\\ \left(\text{xii}\right)\text{48}-\text{23}\text{= 25}\\ \left(\text{xiii}\right)\text{25}-\text{25}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting}\\ {\text{from 1 to 169, and obtained 0 at 13}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{\text{169}}\text{=13}\end{array}$

**Q.4 **Find the square roots of the following numbers by the Prime Factorization Method.

(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100

**Ans**

\begin{array}{l}\text{(i)}729\text{can be factorised as follows:}\\ \\ \begin{array}{cc}3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ \text{729}=\underset{\xaf}{\text{3}\times \text{3}}\times \underset{\xaf}{\text{3}\times \text{3}}\times \underset{\xaf}{\text{3}\times \text{3}}\\ \therefore \sqrt{729}=3\times 3\times 3=\text{27}\end{array}

\begin{array}{l}\text{(ii)400}\\ \\ \begin{array}{cc}2& 400\\ 2& 200\\ 2& 100\\ 5& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \text{400}=\underset{\xaf}{\text{2}\times \text{2}}\times \underset{\xaf}{\text{2}\times \text{2}}\times \underset{\xaf}{\text{5}\times \text{5}}\\ \therefore \sqrt{400}=2\times 2\times 5=\text{20}\end{array}

\begin{array}{l}\text{(iii)1764}\\ \begin{array}{cc}2& 1764\\ 2& 882\\ 3& 441\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \text{1764}=\underset{\xaf}{\text{2}\times \text{2}}\times \underset{\xaf}{\text{3}\times \text{3}}\times \underset{\xaf}{\text{7}\times \text{7}}\\ \therefore \sqrt{1764}=2\times 3\times 7=\text{42}\end{array}

\begin{array}{l}\text{(iv)}4096\text{can be factorised as follows:}\\ \\ \begin{array}{cc}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\end{array}\\ \\ \text{4096}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{\text{2}\times \text{2}}\times \underset{\xaf}{\text{2}\times \text{2}}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\\ \therefore \sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2=\text{64}\\ \end{array}

\begin{array}{l}\text{(vi)}9604\text{can be factorised as follows:}\\ \\ \begin{array}{cc}2& 9604\\ 2& 4802\\ 7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ \text{9604}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{\text{7}\times \text{7}}\times \underset{\xaf}{\text{7}\times \text{7}}\\ \therefore \sqrt{9604}=2\times 7\times 7=\text{98}\\ \\ \text{(vii)5929}\\ \\ \begin{array}{cc}7& 5929\\ 7& 847\\ 11& 121\\ 11& 11\\ & 1\end{array}\\ \\ \text{5929}=\underset{\xaf}{7\times 7}\times \underset{\xaf}{\text{11}\times \text{11}}\\ \therefore \sqrt{5929}=7\times 11=\text{77}\end{array}

\begin{array}{l}\text{(viii) 9216}\\ \\ \begin{array}{cc}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\end{array}\\ \text{9216}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{\text{3}\times 3}\\ \therefore \sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3=\text{96}\\ \\ \text{(ix) 529}\\ \\ \begin{array}{cc}23& 529\\ 23& 23\\ & 1\end{array}\\ \\ \text{529}=\underset{\xaf}{23\times 23}\\ \therefore \sqrt{529}=\text{23}\end{array}

\begin{array}{l}\text{(x)}8100\\ \\ \begin{array}{cc}2& 8100\\ 2& 4050\\ 3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 8100=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{5\times 5}\\ \therefore \sqrt{8100}=2\times 3\times 3\times 5=90\\ \end{array}

**Q.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

**Ans**

$\begin{array}{l}\text{(i) 252 can be factorised as:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 252 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ 252\times 7=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{7\times 7}\\ \\ \therefore \text{252\xd77 = 1764 is a perfect square}\text{.}\\ \\ \therefore \sqrt{1764}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{7\times 7}=2\times 3\times 7=42\\ \end{array}$

$\begin{array}{l}\text{(ii)180 can be factorised as:}\\ \\ \begin{array}{cc}2& 180\\ 2& 90\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ 180=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we multiply the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 180 has to be multiplied with 5}\\ \text{to obtain a perfect square}\text{.}\\ \text{180}\times 5=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{5\times 5}\\ \therefore \text{180\xd75 = 900 is a perfect square}\text{.}\\ \therefore \sqrt{900}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{5\times 5}=2\times 3\times 5=30\\ \\ \text{(iii)1008 can be factorised as:}\\ \\ \begin{array}{cc}2& 1008\\ 2& 504\\ 2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\end{array}$

\begin{array}{l}1008=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 1008 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ \text{1008}\times 7=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{7\times 7}\\ \therefore \text{1008\xd77 = 7056 is a perfect square}\text{.}\\ \therefore \sqrt{7056}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{7\times 7}=2\times 2\times 3\times 7=84\\ \\ \text{(iv) 2028 can be factorised as follows:}\\ \begin{array}{cc}2& 2028\\ 2& 1014\\ 3& 507\\ 13& 169\\ 13& 13\\ & 1\end{array}\\ 2028=\underset{\xaf}{2\times 2}\times \underset{\xaf}{13\times 13}\times 3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2028 has to be multiplied with 3}\\ \text{to obtain a perfect square}\text{.}\\ \text{2028}\times 3=\underset{\xaf}{2\times 2}\times \underset{\xaf}{13\times 13}\times \underset{\xaf}{3\times 3}\\ \therefore \text{2028}\times 3\text{= 6084 is a perfect square}\text{.}\\ \therefore \sqrt{6084}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{13\times 13}\times \underset{\xaf}{3\times 3}=2\times 3\times 13=78\end{array}

\begin{array}{l}\text{(v) 1458 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 1458=2\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\\ \text{Here, the prime factor 2 does not have its pair}\text{.}\\ \text{If we multiply the given number by 2, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 1458 has to be multiplied with 2 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1458}\times 2=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\\ \\ \therefore \text{1458}\times 2\text{= 2916 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2916}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}=2\times 3\times 3\times 3=54\\ \\ \end{array}

\begin{array}{l}\text{(vi) 768 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 768\\ 2& 384\\ 2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}\\ \\ 768=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times 3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 768 has to be multiplied with 3 to obtain}\\ \text{a perfect square}\text{.}\\ \text{768}\times 3=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\\ \\ \therefore \text{768}\times 3\text{= 2304 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2304}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}=2\times 2\times 2\times 2\times 3=48\end{array}

**Q.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

**Ans**

$\begin{array}{l}\text{(i) 252 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 252 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{252}\xf77=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\\ \therefore \sqrt{36}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}=2\times 3=6\end{array}$

$\begin{array}{l}\text{(ii) 2925 can be factorised as follows:}\\ \\ \begin{array}{cc}3& 2925\\ 3& 975\\ 5& 325\\ 5& 65\\ 13& 13\\ & 1\end{array}\\ \\ 2925=\underset{\xaf}{3\times 3}\times \underset{\xaf}{5\times 5}\times 13\\ \text{Here, the prime factor 13 does not have its pair}\text{.}\\ \text{If we divide the given number by 13, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2925 has to be divided with 13 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2925}\xf713=225\text{is a perfect square}\text{.}\\ \text{225 =}\underset{\xaf}{3\times 3}\times \underset{\xaf}{5\times 5}\\ \therefore \sqrt{225}=\underset{\xaf}{3\times 3}\times \underset{\xaf}{5\times 5}=3\times 5=15\\ \\ \text{(iii) 396 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 396\\ 2& 198\\ 3& 99\\ 3& 33\\ 11& 11\\ & 1\end{array}\\ \end{array}$

$\begin{array}{l}396=\underset{\xaf}{3\times 3}\times \underset{\xaf}{2\times 2}\times 11\\ \text{Here, the prime factor 11 does not have its pair}\text{.}\\ \text{If we divide the given number by 11, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,396 has to be divided with 11 to obtain}\\ \text{a perfect square}\text{.}\\ \text{396}\xf711=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\\ \therefore \sqrt{36}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}=2\times 3=6\\ \\ \text{(iv) 2645 can be factorised as follows:}\\ \\ \begin{array}{cc}5& 2645\\ 23& 529\\ 23& 23\\ & 1\end{array}\\ \\ 2645=\underset{\xaf}{23\times 23}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,2645 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2645}\xf75=529\text{is a perfect square}\text{.}\\ \text{529 =}\underset{\xaf}{23\times 23}\\ \therefore \sqrt{529}=23\end{array}$

$\begin{array}{l}\text{(v) 2800 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 2800\\ 2& 1400\\ 2& 700\\ 2& 350\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}\\ \\ 2800=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{5\times 5}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2800 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2800}\xf77=400\text{is a perfect square}\text{.}\\ \text{400 =}\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{5\times 5}\\ \therefore \sqrt{400}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{2\times 2}\times \underset{\xaf}{5\times 5}=2\times 2\times 5=20\end{array}$

$\begin{array}{l}\text{(vi) 1620 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1620\\ 2& 810\\ 3& 405\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ \\ 1620=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,1620 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1620}\xf75=324\text{is a perfect square}\text{.}\\ \text{324 =}\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\\ \therefore \sqrt{324}=\underset{\xaf}{2\times 2}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}=2\times 3\times 3=18\end{array}$

**Q.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.

**Ans**

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

Therefore, the number of students in the class will be the square root of the amount donated by the students of the class.

The total amount of donation is Rs 2401.

\begin{array}{l}\text{Number of students in the class}=\sqrt{2400}\\ \\ \begin{array}{cc}7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ 2401=\underset{\xaf}{7\times 7}\times \underset{\xaf}{7\times 7}\\ \therefore \sqrt{2401}=7\times 7=49\end{array}

Hence, the number of students in the class is 49.

**Q.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.

**Ans**

According to the question, each row contains as many plants as the number of rows.

Hence,

Number of rows = Number of plants in each row

Also, total number of plants = 2025

Since, the number of rows and number of plants in each row are equal, therefore (Number of rows)^{2} = 2025

$\begin{array}{l}\therefore \text{Number of rows =}\sqrt{2025}\\ \\ \begin{array}{cc}3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 2025=\underset{\xaf}{3\times 3}\times \underset{\xaf}{3\times 3}\times \underset{\xaf}{5\times 5}\\ \therefore \sqrt{2025}=3\times 3\times 5=45\end{array}$

Hence, the number of rows and the number of plants in each row is 45.

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

**Ans**

The number that will be perfectly divisible by each one of 4, 9, and 10 is their LCM.

\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{4},\text{9},\text{1}0\\ 2& \text{2},\text{9},\text{5}\\ 3& \text{1},\text{9},\text{5}\\ 3& \text{1},\text{3},\text{5}\\ 5& \text{1},\text{1},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}

Therefore, LCM of 4, 9, 10 = __2 × 2__ × __3 × 3__ × 5 =180

Here, prime factor 5 does not have its pair. Therefore, 180 is not a perfect square. If we multiply 180 with 5, then the number will become a perfect square. Therefore, 180 should be multiplied with 5 to obtain a perfect square.

Hence, the required square number is 180 × 5 = 900.

**Q.10** Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

**Ans**

The number that will be perfectly divisible by each one of 8, 15, and 20 is their LCM.

\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{8},\text{15},\text{2}0\\ 2& \text{4},\text{15},\text{1}0\\ 2& \text{2},\text{15},\text{5}\\ 3& \text{1},\text{15},\text{5}\\ 5& \text{1},\text{5},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}

Therefore, LCM of 8, 15, and 20 = __2 × 2__ ×2 × 3 × 5 =120

Here, prime factors 2, 3 and 5 do not have their pairs. Therefore, 120 is not a perfect square. If we multiply 120 with 2×3×5, then the number will become a perfect square. Therefore, 120 should be multiplied with 2×3×5 to obtain a perfect square.

Hence, the required square number is 120×2×3×5 = 3600.

## FAQs (Frequently Asked Questions)

### 1. Is Class 8 Mathematics tough?

It is true that Class 8 has an increased difficulty level in all subjects, but students can make it easier for themselves by studying regularly and practising NCERT exercises. Class 8 Mathematics must also be prepared in this manner. Students can solve NCERT exercises with the help of resources provided by Extramarks. To solve the questions of Chapter 6 Class 8, students can take the help of the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3.

### 2. From where can students download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 ?

Students can download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 from the website and mobile application of Extramarks.

### 3. Can the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 be downloaded in PDF format?

Yes, the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 are available in PDF format. Students are advised to download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 in PDF format for easy access.

### 4. Is Class 8 Mathematics syllabus important for Class 10 board examinations?

Yes, Class 8 Mathematics syllabus is very important for Class 10 board examinations as students learn the foundations in Class 8 and Class 9. If they understand the basics well, they will be able to learn the more complex topics easily. Moreover, Class 8, Class 9 and Class 10 are further very important for Class 11 and Class 12 Mathematics.

### 5. Will the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 help students to solve mock papers?

Yes, the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 will help students in solving the mock examination questions. They can also refer to Extramarks’ past years’ paper solutions to find authentic solutions. Students must download the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 for preparing chapter 6 efficiently.

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### 7. Are solutions of other Chapter 6 Exercises also provided by Extramarks?

Yes, along with the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3, solutions for Exercise 6.1, Exercise 6.2, Exercise 6.4 are available on the website and mobile application of Extramarks for the benefit of students. Students can refer to the solutions in case of any challenges.

### 8. Are the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.3 also provided in Hindi by Extramarks?

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