# NCERT Solutions Class 8 Maths Chapter 6 Exercise 6.4

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**NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.4**

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**Access NCERT Solutions for Mathematics Chapter 6 – Squares and Square Roots Exercise 6.4**

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**Maths NCERT Solutions Class 8 Chapter 6 Exercise 6.4**

Class 8 Maths Chapter 6 Exercise 6.4, covers several important topics such as Finding Square Roots by Division Method, Square Roots of Decimals, and Estimating Square Roots. These topics are essential to learning to find the square root of a number. Students should learn and follow all the methods step by step to be able to solve questions based on these topics. Along with that, they need to practice as many questions as possible. Students need access to a variety of questions and well-written solutions before they appear for their exams. Most question banks base their content primarily on the exercises and examples from NCERT textbooks. It is simpler for students to respond to questions when they have access to the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4. It helps them understand the topics more quickly. A student feels more prepared for the examinations and gains confidence in grasping the concepts more quickly. This gives them more time to study while maintaining their mental focus on other facets of their education. Understanding the topics covered in the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4, will help students advance in their studies. Having access to the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4, which is adjusted to students’ needs, can be a helpful study aid, despite the fact that the chapter’s content can occasionally be difficult and perplexing. Using these study tools will make learning the concepts related to square roots much easier. Students who want to progress in their preparation should consider practising with the NCERT questions. The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4, by Extramarks, focus on a particular chapter. For students who want to do better on their exams, there are many study resources available at Extramarks, including chapter-wise worksheets, mock tests and important questions. This makes Extramarks the one-stop destination for studying, practising, and testing. As Extramarks’ NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4 are written and edited by subject specialists, all students are ensured of a smooth educational experience.

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**NCERT Solutions for Class 8**

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**Q.1 **Find the square root of each of the following numbers by Division method.

(i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369 (vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900

**Ans**

\begin{array}{l}\text{(i) Consider}\sqrt{2304}\\ \\ \begin{array}{cc}& 48\\ 4& \begin{array}{l}\text{}\overline{23}\text{}\overline{04}\\ -16\end{array}\\ 88& \begin{array}{l}\text{}704\\ -704\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{2304}=48\\ \\ \text{(ii) Consider}\sqrt{4489}\\ \\ \begin{array}{cc}& 67\\ 6& \begin{array}{l}\text{}\overline{44}\text{}\overline{89}\\ -36\end{array}\\ 127& \begin{array}{l}\text{}889\\ -889\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{4489}=67\end{array}

\begin{array}{l}\text{(iii) Consider}\sqrt{3481}\\ \\ \begin{array}{cc}& 59\\ 5& \begin{array}{l}\text{}\overline{34}\text{}\overline{81}\\ -25\end{array}\\ 109& \begin{array}{l}\text{}981\\ -981\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{3481}=59\\ \\ \text{(iv) Consider}\sqrt{529}\\ \\ \begin{array}{cc}& 23\\ 2& \begin{array}{l}\text{}\overline{5}\text{}\overline{29}\\ -4\end{array}\\ 43& \begin{array}{l}\text{}129\\ -129\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{529}=23\end{array}

\begin{array}{l}\text{(v) Consider}\sqrt{3249}\\ \\ \begin{array}{cc}& 57\\ 5& \begin{array}{l}\text{}\overline{32}\text{}\overline{49}\\ -25\end{array}\\ 107& \begin{array}{l}\text{}749\\ -749\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{3249}=57\\ \\ \text{(vi) Consider}\sqrt{1369}\\ \\ \begin{array}{cc}& 37\\ 3& \begin{array}{l}\text{}\overline{13}\text{}\overline{69}\\ -9\end{array}\\ 67& \begin{array}{l}\text{}469\\ -469\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{1369}=37\end{array}

\begin{array}{l}\text{(vii) Consider}\sqrt{5776}\\ \\ \begin{array}{cc}& 76\\ 7& \begin{array}{l}\text{}\overline{57}\text{}\overline{76}\\ -49\end{array}\\ 146& \begin{array}{l}\text{876}\\ -876\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{5776}=76\\ \\ \text{(viii) Consider}\sqrt{7921}\\ \\ \begin{array}{cc}& 89\\ 8& \begin{array}{l}\text{}\overline{79}\text{}\overline{21}\\ -64\end{array}\\ 169& \begin{array}{l}\text{1521}\\ -1521\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{7921}=89\end{array}

\begin{array}{l}\text{(ix) Consider}\sqrt{576}\\ \\ \begin{array}{cc}& 24\\ 2& \begin{array}{l}\text{}\overline{5}\text{}\overline{76}\\ -4\end{array}\\ 44& \begin{array}{l}\text{176}\\ -176\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{576}=24\\ \\ \text{(x) Consider}\sqrt{1024}\\ \\ \begin{array}{cc}& 32\\ 3& \begin{array}{l}\text{}\overline{10}\text{}\overline{24}\\ -9\end{array}\\ 62& \begin{array}{l}\text{124}\\ -124\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{1024}=32\end{array}

$\begin{array}{l}\text{(xi) Consider}\sqrt{3136}\\ \\ \begin{array}{cc}& 56\\ 5& \begin{array}{l}\text{}\overline{31}\text{}\overline{36}\\ -25\end{array}\\ 106& \begin{array}{l}\text{636}\\ -636\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{3136}=56\\ \\ \text{(xii) Consider}\sqrt{900}\\ \\ \begin{array}{cc}& 30\\ 3& \begin{array}{l}\text{}\overline{9}\text{}\overline{00}\\ -9\end{array}\\ 60& \begin{array}{l}\text{00}\\ 00\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{900}=30\end{array}$ \begin{array}{l}\end{array}

**Q.2** Find the number of digits in the square root of each of the following numbers (without any calculation).

(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625

**Ans**

We can find the number of digits in the square root of each of the given numbers by placing bars on them.

(i) By placing bars, we obtain

$64=\overline{64}$

Since there is only one bar, the square root of 64 will have only one digit in it.

(ii) By placing bars, we obtain

$144=\overline{1}\overline{44}$

Here, we have two bars, so the square root of 144 will have 2 digits in it.

(iii) By placing bars, we obtain

$4489=\overline{44}\overline{89}$

Since there are two bars, the square root of 4489 will have 2 digits in it.

(iv) By placing bars, we obtain

$27225=\overline{2}\overline{72}\overline{25}$

Here we have three bars, so the square root of 27225 will have three digits in it.

$390625=\overline{39}\overline{06}\overline{25}$

Here, also we have three bars; the square root of 390625 will have 3 digits in it.

**Q.3 **Find the square root of the following decimal numbers.

(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36

**Ans**

\begin{array}{l}\text{(i) Consider}\sqrt{2.56}\\ \\ \begin{array}{cc}& 1.6\\ 1& \begin{array}{l}\text{}\overline{2}\text{}\text{.}\overline{56}\\ -1\end{array}\\ 26& \begin{array}{l}\text{156}\\ -156\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{2.56}=1.6\end{array}

\begin{array}{l}\text{(ii)Consider}\sqrt{7.29}\\ \\ \begin{array}{cc}& 2.7\\ 2& \begin{array}{l}\text{}\overline{7}\text{}\text{.}\overline{29}\\ -4\end{array}\\ 47& \begin{array}{l}\text{329}\\ -329\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{7.29}=2.7\\ \\ \\ \text{(iii)Consider}\sqrt{51.84}\\ \\ \begin{array}{cc}& 7.2\\ 7& \begin{array}{l}\text{}\overline{51}\text{}\text{.}\overline{84}\\ -49\end{array}\\ 142& \begin{array}{l}\text{284}\\ -284\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{51.84}=7.2\end{array}

\begin{array}{l}\text{(iv)Consider}\sqrt{42.25}\\ \\ \begin{array}{cc}& 6.5\\ 6& \begin{array}{l}\text{}\overline{42}\text{}\text{.}\overline{25}\\ -36\end{array}\\ 125& \begin{array}{l}\text{625}\\ -625\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{42.25}=6.5\\ \\ \\ \text{(v)Consider}\sqrt{31.36}\\ \\ \begin{array}{cc}& 5.6\\ 5& \begin{array}{l}\text{}\overline{31}\text{}\text{.}\overline{36}\\ -25\end{array}\\ 106& \begin{array}{l}\text{636}\\ -636\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{31.36}=5.6\end{array}

**Q.4 **Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000

**Ans**

\begin{array}{l}\text{(i) Consider}\sqrt{402}\\ \\ \begin{array}{cc}& 20\\ 2& \begin{array}{l}\text{}\overline{4}\text{}\overline{02}\\ -4\end{array}\\ 40& \begin{array}{l}\text{02}\\ 00\end{array}\\ & 2\end{array}\\ \\ \text{Here,the remainder is 2}.\text{Therefore},\text{if we subtract 2 from 402,we}\\ \text{will get a perfect square}\text{.}\\ \text{Therefore},\text{required perfect square}=\text{4}0\text{2}-\text{2}=\text{4}00\\ \therefore \sqrt{400}=20\end{array}

\begin{array}{l}\text{(ii) Consider}\sqrt{1989}\\ \\ \begin{array}{cc}& 44\\ 4& \begin{array}{l}\text{}\overline{19}\text{}\overline{89}\\ -16\end{array}\\ 84& \begin{array}{l}\text{389}\\ 336\end{array}\\ & 53\end{array}\\ \\ \text{Here,the remainder is 53}.\text{Therefore},\text{if we subtract 53 from 1989,we}\\ \text{will get a perfect square}\text{.}\\ \text{Therefore},\text{required perfect square}=1989-53=1936\\ \therefore \sqrt{1936}=44\\ \\ \text{(iii) Consider}\sqrt{3250}\\ \\ \begin{array}{cc}& 57\\ 5& \begin{array}{l}\text{}\overline{32}\text{}\overline{50}\\ -25\end{array}\\ 107& \begin{array}{l}\text{750}\\ 749\end{array}\\ & 1\end{array}\\ \\ \text{Here,the remainder is 1}.\text{Therefore},\text{if we subtract 1 from 3250,we}\\ \text{will get a perfect square}\text{.}\\ \text{Therefore},\text{required perfect square}=3250-1=3249\\ \therefore \sqrt{3249}=57\end{array}

\begin{array}{l}\text{(iv) Consider}\sqrt{825}\\ \\ \begin{array}{cc}& 28\\ 2& \begin{array}{l}\text{}\overline{8}\text{}\overline{25}\\ -4\end{array}\\ 48& \begin{array}{l}\text{425}\\ 384\end{array}\\ & 41\end{array}\\ \\ \text{Here,the remainder is 41}.\text{Therefore},\text{if we subtract 41 from 825,we}\\ \text{will get a perfect square}\text{.}\\ \text{Therefore},\text{required perfect square}=825-41=784\\ \therefore \sqrt{784}=28\\ \\ \text{(v) Consider}\sqrt{4000}\\ \\ \begin{array}{cc}& 63\\ 6& \begin{array}{l}\text{}\overline{40}\text{}\overline{00}\\ -36\end{array}\\ 123& \begin{array}{l}\text{400}\\ 369\end{array}\\ & 31\end{array}\\ \\ \text{Here,the remainder is 31}.\text{Therefore},\text{if we subtract 31 from 4000,we}\\ \text{will get a perfect square}\text{.}\\ \text{Therefore},\text{required perfect square}=4000-31=3969\\ \therefore \sqrt{3969}=63\end{array}

**Q.5 **Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412

**Ans**

\begin{array}{l}\text{(i) Consider}\sqrt{525}\\ \\ \begin{array}{cc}& 22\\ 2& \begin{array}{l}\text{}\overline{5}\text{}\overline{25}\\ -4\end{array}\\ 42& \begin{array}{l}\text{125}\\ 84\end{array}\\ & 41\end{array}\\ \\ \text{Here,the remainder is 41}.\text{}\text{}\text{This means that square of 22 is less than 525}\text{.}\\ \text{The next number is 23 and the square of 23 is 529}\text{.}\\ \\ {\text{Hence, the number to be added to 525 = 23}}^{\text{2}}-525=529-525=4\\ \\ \text{Therefore},\text{required perfect square}=525+4=529\\ \therefore \sqrt{529}=23\end{array}

\begin{array}{l}\text{(ii) Consider}\sqrt{1750}\\ \\ \begin{array}{cc}& 41\\ 4& \begin{array}{l}\text{}\overline{17}\text{}\overline{50}\\ -16\end{array}\\ 81& \begin{array}{l}\text{150}\\ 81\end{array}\\ & 69\end{array}\\ \\ \text{Here,the remainder is 69}.\text{}\text{}\text{This means that square of 41 is less than 1750}\text{.}\\ \text{The next number is 42 and the square of 42 is 1764}\text{.}\\ {\text{Hence, the number to be added to 1750 = 42}}^{\text{2}}-1750=1764-1750=14\\ \text{Therefore},\text{required perfect square}=1750+14=1764\\ \therefore \sqrt{1764}=42\\ \\ \text{(iii) Consider}\sqrt{252}\\ \\ \begin{array}{cc}& 15\\ 1& \begin{array}{l}\text{}\overline{2}\text{}\overline{52}\\ -1\end{array}\\ 25& \begin{array}{l}\text{152}\\ 125\end{array}\\ & 27\end{array}\\ \\ \text{Here,the remainder is 27}.\text{}\text{}\text{This means that square of 15 is less than 252}\text{.}\\ \text{The next number is 16 and the square of 16 is 256}\text{.}\\ {\text{Hence, the number to be added to 252 = 16}}^{\text{2}}-252=256-252=4\\ \text{Therefore},\text{required perfect square}=252+4=256\\ \therefore \sqrt{256}=16\end{array}

\begin{array}{l}\text{(iv) Consider}\sqrt{1825}\\ \\ \begin{array}{cc}& 42\\ 4& \begin{array}{l}\text{}\overline{18}\text{}\overline{25}\\ -16\end{array}\\ 82& \begin{array}{l}\text{225}\\ 164\end{array}\\ & 61\end{array}\\ \\ \text{Here,the remainder is 61}.\text{}\text{}\text{This means that square of 42 is less than 1825}\text{.}\\ \text{The next number is 43 and the square of 43 is 1849}\text{.}\\ {\text{Hence, the number to be added to 1825 = 43}}^{\text{2}}-1825=1849-1825=24\\ \text{Therefore},\text{required perfect square}=1825+24=1849\\ \therefore \sqrt{1849}=43\\ \\ \text{(v) Consider}\sqrt{6412}\\ \\ \begin{array}{cc}& 80\\ 8& \begin{array}{l}\text{}\overline{64}\text{}\overline{12}\\ -64\end{array}\\ 160& \begin{array}{l}\text{012}\\ 0\end{array}\\ & 12\end{array}\\ \\ \text{Here,the remainder is 12}.\text{}\text{}\text{This means that square of 80 is less than}6412.\\ \text{The next number is 81 and the square of 81 is 6561}\text{.}\\ \text{Hence, the number to be added to}6412{\text{= 81}}^{\text{2}}-6412=6561-6412=149\\ \text{Therefore},\text{required perfect square}=6412+149=6561\\ \therefore \sqrt{6561}=81\end{array}

**Q.6** Find the length of the side of a square whose area is 441 m^{2}.

**Ans**

$\begin{array}{l}\text{Let the length of the side of the square be}\mathrm{x}\text{metres.}\\ \text{Area of the square}=441\text{}{\mathrm{m}}^{2}\text{}\\ \text{i.e.}{\mathrm{x}}^{2}=441\text{}{\mathrm{m}}^{2}\text{}\\ \Rightarrow \mathrm{x}=\sqrt{441}\\ \\ \text{The square root of 441 can be obtained as follows:}\\ \begin{array}{cc}& 21\\ 2& \begin{array}{l}\text{}\overline{4}\text{}\overline{41}\\ -4\end{array}\\ 41& \begin{array}{l}\text{41}\\ 41\end{array}\\ & 0\end{array}\\ \\ \therefore \mathrm{x}=21\text{m}\\ \text{Therefore},\text{the length of the side of the square be}21\text{metres.}\end{array}$

**Q.7 **In a right triangle ABC, ∠B = 90°.

(a) If AB = 6 cm, BC = 8 cm, find AC (b) If AC = 13 cm, BC = 5 cm, find AB.

**Ans**

(a) Given: AB = 6 cm, BC = 8 cm and ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

\begin{array}{l}{\text{AC}}^{\text{2}}={\text{AB}}^{\text{2}}+{\text{BC}}^{\text{2}}\\ \Rightarrow {\text{AC}}^{\text{2}}={\left(\text{6 cm}\right)}^{\text{2}}+{\left(\text{8 cm}\right)}^{\text{2}}\\ \Rightarrow {\text{AC}}^{\text{2}}=\left(\text{36}+\text{64}\right){\text{cm}}^{\text{2}}=\text{1}00{\text{cm}}^{\text{2}}\\ \Rightarrow \text{AC}=\sqrt{\text{1}00\text{}}\text{cm}=\text{10 cm}\\ \therefore \text{AC}=\text{10cm}\end{array}

(b) Given: AC = 13 cm, BC = 5 cm and ΔABC is right-angled at B.

Therefore, by applying Pythagoras theorem, we obtain

\begin{array}{l}{\text{AC}}^{\text{2}}={\text{AB}}^{\text{2}}+{\text{BC}}^{\text{2}}\\ \Rightarrow {\left(\text{13cm}\right)}^{\text{2}}={\left(\text{AB}\right)}^{\text{2}}+{\left(\text{5 cm}\right)}^{\text{2}}\\ \Rightarrow {\text{AB}}^{\text{2}}=\left(\text{13}-5\right){\text{cm}}^{\text{2}}\\ \Rightarrow {\text{AB}}^{\text{2}}=\left(169-25\right){\text{cm}}^{\text{2}}=144{\text{cm}}^{\text{2}}\\ \Rightarrow \text{AB}=\sqrt{144}\text{cm}=\text{12 cm}\\ \therefore \text{AB}=\text{12 cm}\end{array}

**Q.8 **A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.

**Ans**

It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same.

We have to find the number of plants that should be more planted such that the number of rows and columns are same.

That means we have to find the number which should be added to 1000 to make it a perfect square

The square root of 1000 can be calculated as follows:

\begin{array}{l}\begin{array}{cc}& 31\\ 3& \begin{array}{l}\text{}\overline{10}\text{}\overline{00}\\ -9\end{array}\\ 61& \begin{array}{l}\text{100}\\ 61\end{array}\\ & 39\end{array}\\ \\ \text{The remainder is 39}.\text{It represents that the square of 31 is less than 1}000.\\ \text{The next number is 32 and the square of 32}=\text{1}0\text{24}\text{.}\\ \\ \text{Hence},\text{number to be added to 1}000\text{to make it a}\\ \text{perfect square}={\text{32}}^{\text{2}}-\text{1}000=\text{1}0\text{24}-\text{1}000=\text{24}\\ \text{Thus},\text{the required number of plants is 24}.\\ \end{array}

**Q.9 **There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?

**Ans**

It is given that there are 500 children in the school. They have to stand for a P.T. drill such that the number of rows is equal to the number of columns.

We have to find the number of children who will be left out in this arrangement.

That means we have to calculate the number which should be subtracted from 500 to make it a perfect square.

The square root of 500 can be calculated by long division method as follows.

\begin{array}{cc}& 22\\ 2& \begin{array}{l}\text{}\overline{5}\text{}\overline{00}\\ -4\end{array}\\ 42& \begin{array}{l}\text{100}\\ 84\end{array}\\ & 16\end{array}

$\begin{array}{l}\text{The remainder is 16}.\text{}\\ \therefore \text{the number to be subtracted from 500 to make it a}\\ \text{perfect square}=16\\ \text{So, the required number is}500-16=484\\ \text{Thus, the number of children who will be left out is 16}\text{.}\end{array}$

## FAQs (Frequently Asked Questions)

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### 2. Are questions from the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4 included in the Class 8 examinations?

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### 3. How many questions are resolved in the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4 offered by Extramarks?

The NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4 provide answers for all nine questions in Square And Square Roots Class 8 Exercise 6.4. Students can find all the steps necessary to solve the questions in the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4 offered by Extramarks in order to get the best possible grade on the exam.

### 4. Can one finish the Mathematics question paper in the allotted time by practising the NCERT Solutions For Class 8 Maths Chapter 6 Exercise 6.4 before the exam?

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