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NCERT Solutions for Class 8 Maths Chapter 6 - Squares And Square Roots

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NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots - Free PDF

Introduction 

NCERT Solutions for Class 8 Maths Chapter 6, Square and Square Roots  provided by Vedantu, is developed as per the latest syllabus and under strict guidelines set by CBSE board. To understand the concept, students must go through the solutions and the notes related to the solution. You can download the free pdf format of the NCERT Solutions Chapter 6 from the official website of Vedantu. NCERT Solutions for other subjects for other classes are also available on Vedantu. You can reach out to us if you need extra help relating to any subject.


Class:

NCERT Solutions For Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 6 - Squares and Square Roots

Content Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes


In this chapter, you will learn the different techniques used to determine whether a given natural number is a perfect square number or not. These techniques are demonstrated by the properties and patterns followed by square numbers. This chapter also deals with the various methods for finding the square roots of square numbers. If you have knowledge about exponents then understanding the concept of Square roots will be easy. Please go through a quick review of the chapter before solving the NCERT Solutions for Chapter 6. You can also find NCERT Solutions for Class 8 Science on Vedantu.

Access NCERT Solutions for Maths Chapter 6 - Squares and Square Roots

Exercise 6.1

1. What will be the unit digit of the square of the following numbers?

i. $\text{81}$

Ans:

It is known to us that, the square of the number having a unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is $1$,its square will end with the unit digit of the multiplication \[\left( 1\times 1=1 \right)\] i.e., 1.


ii. $\text{272}$

Ans:

It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’..

Now, in the given number, the unit’s place digit is 2, its square will end with the unit digit of the multiplication \[\left( 2\times 2=4 \right)\]i.e., 4.


iii. $\text{799}$

Ans:

It is known to us that, the square of the number having a unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is 9, its square will end with the unit digit of the multiplication \[\left( 9\times 9=81 \right)\]i.e., 1.


iv. $\text{3853}$

Ans:

It is known to us that, the square of the number having a unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is 3, its square will end with the unit digit of the multiplication \[\left( 3\times 3=9 \right)\] i.e., 9.


v. $\text{1234}$

Ans:

It is known to us that, the square of the number having a unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is $4$,its square will end with the unit digit of the multiplication $\left( 4\times 4=16 \right)$ i.e., 6.


vi. $\text{26387}$

Ans:

It is known to us that, the square of the number having a unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is 7, its square will end with the unit digit of the multiplication $\left( 7\times 7=49 \right)$ i.e., 9.


vii. $\text{52698}$

Ans:

It is known to us that, the square of the number having a unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is 8, its square will end with the unit digit of the multiplication $\left( 8\times 8=64 \right)$ i.e., 4.


viii. $\text{99880}$

Ans:

It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is $0$,its square will have two zeroes at the end. Hence, the unit digit of the square of the given number is $0$.


ix. $\text{12796}$

Ans:

It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is 6, its square will end with the unit digit of the multiplication $\left( 6\times 6=36 \right)$ i.e., 6.


x. $\text{55555}$

Ans:

It is known to us that, the square of the number having unit place digit as ‘a’, will end up with the unit digit of ‘a\[\times \]a’.

Now, in the given number, the unit’s place digit is 5, its square will end with the unit digit of the multiplication $\left( 5\times 5=25 \right)$ i.e., 5.


2. Give a reason why the following numbers are not perfect squares.

i. $\text{1057}$

Ans:

The 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.

We can see that $1057$has its unit place digit as $7$.

Hence, $1057$cannot be a perfect square.


ii. $\text{23453}$

Ans:

The 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.

We can see that $23453$ has its unit place digit as $3$.

Hence, $23453$ cannot be a perfect square.


iii. $\text{7928}$

Ans:

The 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.

We can see that $7928$ has its unit place digit as $8$.

Hence, $7928$ cannot be a perfect square.


iv. $\text{222222}$

Ans:

The 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.

We can see that $222222$ has its unit place digit as $2$.

Hence, $222222$ cannot be a perfect square.


v. $\text{64000}$

Ans:

The 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.

We can see that $64000$ has three zeroes at the end of it.

Since a perfect square cannot end with an odd number of zeroes, therefore, $64000$ is not a perfect square.


vi. $\text{89722}$

Ans:

The 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.

We can see that $89722$ has its unit place digit as $2$.

Hence, $89722$ cannot be a perfect square.


vii. $\text{222000}$

Ans:

The 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.

We can see that $222000$ has three zeroes at the end of it.

Since a perfect square cannot end with an odd number of zeroes, therefore, $222000$ is not a perfect square.


viii. $\text{505050}$

Ans:

The 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 zeros at the end of it.

We can see that $505050$ has three zeroes at the end of it.

Since a perfect square cannot end with an odd number of zeroes, therefore, $505050$ is not a perfect square.


3. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

i. $\text{243}$

Ans:

First, we’ll break the number into its factors.

$243$ can be written as $243=\underline{3\times 3\times 3}\times 3\times 3$

Here, two $3$s are left which are not in a triplet. So, we need one more $3$to make $243$ a cube.

If $243$ is multiplied by $3$, then we get,

$243\times 3=\underline{3\times 3\times 3}\times \underline{3\times 3\times 3}=729$ (which is a perfect cube).

Thus, $3$ is the smallest number by which $243$ must be multiplied to obtain a perfect cube.


ii. $\text{256}$

Ans:

First, we’ll break the number into its factors.

$256$ can be written as $256=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 2\times 2$

Here, two $2$s are left which are not in a triplet. So, we need one more $2$to make $256$ a cube.

If $256$ is multiplied by $2$, then we get,

$256\times 2=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}=512$ (which is a perfect cube).

Thus, $2$ is the smallest number by which $256$ must be multiplied to obtain a perfect cube.


iii. $\text{72}$

Ans:

First, we’ll break the number into its factors.

$72$ can be written as $72=\underline{2\times 2\times 2}\times 3\times 3$

Here, two $3$s are left which are not in a triplet. So, we need one more $3$to make $72$ a cube.

If $72$ is multiplied by $3$, then we get,

$72\times 3=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}=216$ (which is a perfect cube).

Thus, $3$ is the smallest number by which $72$ must be multiplied to obtain a perfect cube.


iv. $\text{675}$

Ans:

First, we’ll break the number into its factors.

$675$ can be written as $675=\underline{3\times 3\times 3}\times 5\times 5$

Here, two $5$s are left which are not in a triplet. So, we need one more $5$to make $675$ a cube.

If $675$ is multiplied by $5$, then we get,

$675\times 5=\underline{3\times 3\times 3}\times \underline{5\times 5\times 5}=3375$ (which is a perfect cube).

Thus, $5$ is the smallest number by which $675$ must be multiplied to obtain a perfect cube.


v. $\text{100}$

Ans:

First, we’ll break the number into its factors.

$100$ can be written as $100=2\times 2\times 5\times 5$

Here, two $2$s and two $5$s are left which are not in a triplet. So, we need one more $2$ and one more $5$to make $100$ a cube.

If $100$ is multiplied by $2$ and $5$, then we get,

$100\times 2\times 5=\underline{2\times 2\times 2}\times \underline{5\times 5\times 5}=1000$ (which is a perfect cube).

Thus, $2\times 5$$=10$ is the smallest number by which $100$ must be multiplied to obtain a perfect cube.


4. Find the missing digits after observing the following pattern.

$\text{1}{{\text{1}}^{\text{2}}}\text{=121}$

$\text{10}{{\text{1}}^{\text{2}}}\text{=10201}$

$\text{100}{{\text{1}}^{\text{2}}}\text{=1002001}$

$\text{10000}{{\text{1}}^{\text{2}}}\text{=1}...\text{2}...\text{1}$

$\text{1000000}{{\text{1}}^{\text{2}}}\text{=}...$

Ans:

It can be observed from the given pattern that after doing the square of the number, there are a same number of zeroes before the digit and a same number of zeroes after the digit as there are in the original number.

So, the square of the number $100001$ will have four zeroes before $2$ and four zeroes after $2$.

Similarly, the square of the number $10000001$ will have six zeroes before $2$ and six zeroes after $2$.

Hence,

${{100001}^{2}}=10000200001$

${{10000001}^{2}}=100000020000001$


5.  Find the missing number after observing the following pattern.

$\text{1}{{\text{1}}^{\text{2}}}\text{=121}$

$\text{10}{{\text{1}}^{\text{2}}}\text{=10201}$

$\text{1010}{{\text{1}}^{\text{2}}}\text{=102030201}$

$\text{101010}{{\text{1}}^{\text{2}}}\text{=}...$

${{...}^{\text{2}}}\text{=10203040504030201}$

Ans:

It can be observed from the given pattern that:

  • the square of the numbers has odd number of digits 

  • the first and the last digit of the square of the numbers is $1$

  • the square of the numbers is symmetric about the middle digit

Since there are four  $1$ in $1010101$, so the square of this number will have natural numbers up to $4$ with $0$ in between every consecutive number and then making the number symmetric about $4$

That is, ${{1010101}^{2}}=1020304030201$

Now, \[10203040504030201\] has natural numbers up to \[5\]and the number is symmetric about.

So, the number whose square is \[10203040504030201\], is \[101010101\]

That is, \[{{101010101}^{2}}=10203040504030201\]

Hence,

${{1010101}^{2}}=1020304030201$

\[{{101010101}^{2}}=10203040504030201\]


6. Find the missing numbers using the given pattern.

\[{{\text{1}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}\text{=}{{\text{3}}^{\text{2}}}\]

\[{{\text{2}}^{\text{2}}}\text{+}{{\text{3}}^{\text{2}}}\text{+}{{\text{6}}^{\text{2}}}\text{=}{{\text{7}}^{\text{2}}}\]

\[{{\text{3}}^{\text{2}}}\text{+}{{\text{4}}^{\text{2}}}\text{+1}{{\text{2}}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}\]

\[{{\text{4}}^{\text{2}}}\text{+}{{\text{5}}^{\text{2}}}\text{+}{{...}^{\text{2}}}\text{=2}{{\text{1}}^{\text{2}}}\]

\[{{\text{5}}^{\text{2}}}\text{+}{{...}^{\text{2}}}\text{+3}{{\text{0}}^{\text{2}}}\text{=3}{{\text{1}}^{\text{2}}}\]

\[{{\text{6}}^{\text{2}}}\text{+}{{\text{7}}^{\text{2}}}\text{+}{{...}^{\text{2}}}\text{=}{{...}^{\text{2}}}\]

Ans:

It can be observed from the given pattern that:

  • The third number in the addition is the product of first two numbers.

  • The R.H.S can be obtained by adding to the third number.

That is, in the first three patterns, it can be observed that

${{1}^{2}}+{{2}^{2}}+{{\left( 1\times 2 \right)}^{2}}={{\left( 2+1 \right)}^{2}}$ 

${{2}^{2}}+{{3}^{2}}+{{\left( 2\times 3 \right)}^{2}}={{\left( 6+1 \right)}^{2}}$ 

$ {{3}^{2}}+{{4}^{2}}+{{\left( 3\times 4 \right)}^{2}}={{\left( 12+1 \right)}^{2}}$

Hence, according to the pattern, the missing numbers are as follows:

${{4}^{2}}+{{5}^{2}}+{{\underline{20}}^{2}}={{21}^{2}}$

${{5}^{2}}+{{\underline{6}}^{2}}+{{30}^{2}}={{31}^{2}}$ 

${{6}^{2}}+{{7}^{2}}+{{\underline{42}}^{2}}={{\underline{43}}^{2}}$


7. Find the sum without adding.

i. \[\text{1+3+5+7+9}\]

Ans:

Since, the sum of first n odd natural numbers is n2.

So, the sum of the first five odd natural numbers is \[{{\left( 5 \right)}^{2}}=25\]

Thus, \[1+3+5+7+9={{\left( 5 \right)}^{2}}=25\]


ii. \[\text{1+3+5+7+9+11+13+15+17+19}\]

Ans:

Since, the sum of first n odd natural numbers is n2.

So, the sum of the first ten odd natural numbers is \[{{\left( 10 \right)}^{2}}=100\]

Thus, \[1+3+5+7+9+11+13+15+17+19={{\left( 10 \right)}^{2}}=100\]


iii. \[\text{1+3+5+7+9+11+13+15+17+19+21+23}\]

Ans:

Since, the sum of first n odd natural numbers is n2.

So, the sum of the first twelve odd natural numbers is \[{{\left( 12 \right)}^{2}}=144\]

Thus, \[1+3+5+7+9+11+13+15+17+19+21+23={{\left( 12 \right)}^{2}}=144\]


8.

i. Express \[\text{49}\] as the sum of \[\text{7}\] odd numbers.

Ans: 

Since, the sum of first n odd natural numbers is n2.

We know that \[49={{\left( 7 \right)}^{2}}\]

\[49=\] Sum of \[7\]odd natural numbers

Hence, \[49=1+3+5+7+9+11+13\]


ii. Express \[\text{121}\] as the sum of \[\text{11}\]odd numbers.

Ans: 

Since, the sum of first n odd natural numbers is n2.

We know that \[121={{\left( 11 \right)}^{2}}\]

\[121=\] Sum of \[11\] odd natural numbers

Hence, \[121=1+3+5+7+9+11+13+15+17+19+21\]


9.

How many numbers lie between squares of the following numbers?

i. \[\text{12}\] and \[\text{13}\]

Ans: 

Between the squares of the numbers n and (n+1), there will be 2n numbers.

So, there will be \[2\times 12=24\] numbers between \[{{\left( 12 \right)}^{2}}\]and \[{{\left( 13 \right)}^{2}}\].


ii. \[\text{25}\] and \[\text{26}\]

Ans: 

Between the squares of the numbers n and (n+1), there will be 2n numbers.

So, there will be \[2\times 25=50\] numbers between \[{{\left( 25 \right)}^{2}}\]and \[{{\left( 26 \right)}^{2}}\] .


iii. \[\text{99}\] and \[\text{100}\]

Ans: 

Between the squares of the numbers n and (n+1), there will be 2n numbers.

So, there will be \[2\times 99=198\] numbers between \[{{\left( 99 \right)}^{2}}\] and \[{{\left( 100 \right)}^{2}}\] .


Exercise 6.2

1. Find the square of the following numbers.

i. \[\text{32}\]

Ans:

\[32=30+2\]

\[{{\left( 32 \right)}^{2}}={{\left( 30+2 \right)}^{2}}\]

Since, (a+b)2=a2 +2ab+b2 

So, \[{{\left( 30+2 \right)}^{2}}={{30}^{2}}+2\times 30\times 2+{{2}^{2}}\]

\[=900+120+4\]

\[=1024\]


ii. \[\text{35}\]

Ans:

\[35=30+5\]

\[{{\left( 35 \right)}^{2}}={{\left( 30+5 \right)}^{2}}\]

Since, (a+b)2=a2 +2ab+b2

So, \[{{\left( 30+5 \right)}^{2}}={{30}^{2}}+2\times 30\times 5+{{5}^{2}}\]

\[=900+300+25\]

\[=1225\]


iii. \[\text{86}\]

Ans:

\[86=80+6\]

\[{{\left( 86 \right)}^{2}}={{\left( 80+6 \right)}^{2}}\]

Since, (a+b)2=a2 +2ab+b2

So, \[{{\left( 80+6 \right)}^{2}}={{80}^{2}}+2\times 80\times 6+{{6}^{2}}\]

\[=6400+960+36\]

\[=7396\]


iv. \[\text{93}\]

Ans:

\[93=90+3\]

\[{{\left( 93 \right)}^{2}}={{\left( 90+3 \right)}^{2}}\]

Since, (a+b)2=a2 +2ab+b2

So, \[{{\left( 90+3 \right)}^{2}}={{90}^{2}}+2\times 90\times 3+{{3}^{2}}\]

\[=8100+540+9\]

\[=8649\]


v. \[\text{71}\]

Ans:

\[71=70+1\]

\[{{\left( 71 \right)}^{2}}={{\left( 70+1 \right)}^{2}}\]

Since, (a+b)2=a2 +2ab+b2

So, \[{{\left( 70+1 \right)}^{2}}={{70}^{2}}+2\times 70\times 1+{{1}^{2}}\]

\[=4900+140+1\]

\[=5041\]


vi. \[\text{46}\]

Ans:

\[46=40+6\]

\[{{\left( 46 \right)}^{2}}={{\left( 40+6 \right)}^{2}}\]

Since, (a+b)2=a2 +2ab+b2

So, \[{{\left( 40+6 \right)}^{2}}={{40}^{2}}+2\times 40\times 6+{{6}^{2}}\]

\[=1600+480+36\]

\[=2116\]


2. Write a Pythagoras triplet whose one number is

i. \[\text{6}\]

Ans:

We know that 2m, m2-1, m2+1 is the Pythagoras triplet for any natural number m >1 

It is given that one number in the triplet is \[6\].

If we take m2-1=6, then we get m2 \[=7\]

And m\[=\sqrt{7}\] which is not an integer.

Similarly, if we take m2+1=6, then we get m2 \[=5\]

And m\[=\sqrt{5}\] which is not an integer.

So let 2m\[=6\]

Then we get, m\[=3\]

Now, m2\[-1={{3}^{2}}-1\]

\[=9-1\]

\[=8\]

Similarly, m2\[+1={{3}^{2}}+1\]

\[=9+1\]

\[=10\]

Therefore \[\left( 6,8,10 \right)\] is the Pythagoras triplet.


ii. \[\text{14}\]

Ans:

We know that 2m, m2-1, m2+1 is the Pythagoras triplet for any natural number m >1

It is given that one number in the triplet is \[14\].

If we take m2-1=14, then we get m2-1=6

And m\[=\sqrt{15}\] which is not an integer.

Similarly, if we take m2+1=14, then we get m2+1=14

And m\[=\sqrt{13}\] which is not an integer.

So let 2m=14 

Then we get, m=7 

Now, m2-1=72-1

\[=49-1\]

\[=48\]

Similarly, m2+1=72+1 

\[=49+1\]

\[=50\]

Therefore \[\left( 14,48,50 \right)\] is the Pythagoras triplet.


iii. \[\text{16}\]

Ans:

We know that 2m, m2-1, m2+1 is the Pythagoras triplet for any natural number m >1

It is given that one number in the triplet is \[16\].

If we take m2-1=16, then we get m2=17

And m\[=\sqrt{17}\] which is not an integer.

Similarly, if we take m2+1=16, then we get m2=15

And m\[=\sqrt{15}\] which is not an integer.

So let 2m=16 

Then we get, m=8 

Now, m2-1=82-1

\[=64-1\]

\[=63\]

Similarly, m2+1=82+1

\[=64+1\]

\[=65\]

Therefore \[\left( 16,63,65 \right)\] is the Pythagoras triplet.


iv. \[\text{18}\]

Ans:

We know that 2m, m2-1, m2+1 is the Pythagoras triplet for any natural number m >1

It is given that one number in the triplet is \[18\].

If we take m2-1=18, then we get m2=19

And m\[=\sqrt{19}\] which is not an integer.

Similarly, if we take m2+1=18, then we get m2=17

And m\[=\sqrt{17}\] which is not an integer.

So let 2m=18 

Then we get, m=9 

Now, m2-1=92-1

\[=81-1\]

\[=80\]

Similarly, m2+1=92-1

\[=81+1\]

\[=82\]

Therefore \[\left( 18,80,82 \right)\] is the Pythagoras triplet.


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 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 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 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 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 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. \[\text{9216}\]

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. \[\text{529}\]

The factorization of \[529\] is as follows:

\[23\]

\[529\]

\[23\]

\[23\]


\[1\]


\[529=\underline{23\times 23}\]

So, \[\sqrt{529}=23\]


x. \[\text{8100}\]

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 \[252\times 7=1764\]

\[1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}\] which is a perfect square

\[\sqrt{1764}=2\times 3\times 7\]

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 , 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.

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. 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 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 \[120\times 2\times 3\times 5\]\[=3600\]


Exercise 6.4

1. Find the square root of each of the following numbers by the division method.

i. \[\text{2304}\]

Ans:

The square root of \[2304\] by division method is calculated as follows:


\[48\]

\[4\]

$\overline{23}\overline{04} $ 

 $-16$

\[88\]

704 

704



\[0\]

Hence, \[\sqrt{2304}=48\]


ii. \[\text{4489}\]

Ans:

The square root of \[4489\] by division method is calculated as follows:


\[67\]

\[6\]

$\overline{44}\overline{89} $

$-36$ 

\[127\]

889 

889 


\[0\]

Hence, \[\sqrt{4489}=67\]


iii. \[\text{3481}\]

Ans:

The square root of \[3481\] by division method is calculated as follows:


\[59\]

\[5\]

$\overline{34}\overline{81}$ 

$ -25 $

\[109\]

981  

981 


\[0\]

Hence, \[\sqrt{3481}=59\]


iv. \[\text{529}\]

Ans:

The square root of \[529\] by division method is calculated as follows:


\[23\]

\[2\]

$ \overline{5}\overline{29}$

 $-4$

\[43\]

129 

129  


\[0\]

Hence, \[\sqrt{529}=23\]


v. \[\text{3249}\]

Ans:

The square root of \[3249\] by division method is calculated as follows:


\[57\]

\[5\]

$\overline{32}\overline{49}$ 

$ -25$

\[107\]

749  

749 


\[0\]

Hence, \[\sqrt{3249}=57\]


vi. \[\text{1369}\]

Ans:

The square root of \[1369\] by division method is calculated as follows:


\[37\]

\[3\]

$\overline{13}\overline{69}$ 

 $ -9 $

\[67\]

469 

469


\[0\]

Hence, \[\sqrt{1369}=37\]


vii. \[\text{5776}\]

The square root of \[5776\] by division method is calculated as follows:


\[76\]

\[7\]

$\overline{57}\overline{76}$ 

 -49 

\[146\]

$overline{57}\overline{76}$ 

-49 


\[0\]

Hence, \[\sqrt{5776}=76\]


viii. \[\text{7921}\]

The square root of \[7921\] by division method is calculated as follows:


\[89\]

\[8\]

$\overline{79}\overline{21}$ 

-64  

\[169\]

1521 

 1521 


\[0\]

Hence, \[\sqrt{7921}=89\]


ix. \[\text{576}\]

The square root of \[576\] by division method is calculated as follows:


\[24\]

\[2\]

$\overline{5}\overline{76}$

  -4 

\[44\]

176

176


\[0\]

Hence, \[\sqrt{576}=24\]


x. \[\text{1024}\]

The square root of \[1024\] by division method is calculated as follows:


\[32\]

\[3\]

$\overline{10}\overline{24}$ 

-9 

\[62\]

124

124


\[0\]

Hence, \[\sqrt{1024}=32\]


xi. \[\text{3136}\]

The square root of \[3136\] by division method is calculated as follows:


\[56\]

\[5\]

$\overline{31}\overline{36}$ 

-25

\[106\]

636

636


\[0\]

Hence, \[\sqrt{3136}=56\]


xii. \[\text{900}\]

The square root of \[900\] by division method is calculated as follows:


\[30\]

\[3\]

$\overline{9}\overline{00}$

-9  

\[60\]

00

00


\[0\]

Hence, \[\sqrt{900}=30\]


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

i. \[\text{64}\]

Ans:

In order to find the number of digits in the square root, without any calculation, we have to place the bars on the given number

After placing bars, we get

\[64=\overline{64}\] 

We can see that there is only one bar. So, the square root of  \[64\] will have only one digit.


ii. \[\text{144}\]

Ans:

In order to find the number of digits in the square root, without any calculation, we have to place the bars on the given number

After placing bars, we get

\[144=\overline{1}\overline{44}\] 

We can see that there are two bars. So, the square root of  \[144\] will have two digits.


iii. \[\text{4489}\]

Ans:

In order to find the number of digits in the square root, without any calculation, we have to place the bars on the given number

After placing bars, we get

\[4489=\overline{44}\overline{89}\] 

We can see that there are two bars. So, the square root of  \[4489\] will have two digits.


iv. \[\text{27225}\]

Ans:

In order to find the number of digits in the square root, without any calculation, we have to place the bars on the given number

After placing bars, we get

\[27225=\overline{2}\overline{72}\overline{25}\] 

We can see that there are three bars. So, the square root of  \[27225\] will have three digits.


v. \[\text{390625}\]

Ans:

In order to find the number of digits in the square root, without any calculation, we have to place the bars on the given number

After placing bars, we get

\[390625=\overline{39}\overline{06}\overline{25}\] 

We can see that there are three bars. So, the square root of  \[390625\] will have three digits.


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

i. \[\text{2}\text{.56}\]

Ans:

The square root of \[2.56\] by division method is calculated as follows:


\[1.6\]

\[1\]

$\overline{2}.\overline{56} $ 

  -1  

\[26\]

156

156


\[0\]

Hence, \[\sqrt{2.56}=1.6\]


ii. \[\text{7}\text{.29}\]

Ans:

The square root of \[7.29\] by division method is calculated as follows:


\[2.7\]

\[2\]

$\overline{7}.\overline{29}$ 

 -4 

\[47\]

329

329


\[0\]

Hence, \[\sqrt{7.29}=2.7\]


iii. \[\text{51}\text{.84}\]

Ans:

The square root of \[51.84\] by division method is calculated as follows:


\[7.2\]

\[7\]

$\overline{51}.\overline{84}$ 

-49 

\[142\]

284

284


\[0\]

Hence, \[\sqrt{51.84}=7.2\]


iv. \[\text{42}\text{.25}\]

Ans:

The square root of \[42.25\] by division method is calculated as follows:


\[6.5\]

\[6\]

$\overline{42}.\overline{25} $ 

-36

\[125\]

625

625


\[0\]

Hence, \[\sqrt{42.25}=6.5\]


v. \[\text{31}\text{.36}\]

Ans:

The square root of \[31.36\] by division method is calculated as follows:


\[5.6\]

\[5\]

$\overline{31}.\overline{36}$ 

-25 

\[106\]

636

636


\[0\]

Hence, \[\sqrt{31.36}=5.6\]


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. \[\text{402}\]

Ans:

The square root of \[402\] by division method is calculated as follows:


\[20\]

\[2\]

$\overline{4}\overline{02}$

-4 


\[40\]

02

00


\[2\]


We are getting a remainder \[2\].

This means that the square of 20 is less than 402 by 2. 

So, we must subtract \[2\] from 402 in order to get a perfect square.

Hence, the required perfect square is \[402-2=400\]

The square root of the perfect square obtained is \[\sqrt{400}=20\]


ii. \[\text{1989}\]

Ans:

The square root of \[1989\] by division method is calculated as follows:


\[44\]

\[4\]

$\overline{19}\overline{89}$

-16 

\[84\]

389

336


\[53\]


We are getting a remainder \[53\].

This means that the square of 44 is less than 1989 by 53. 

So, we must subtract \[53\] from 1989 in order to get a perfect square.

Hence, the required perfect square is \[1989-53=1936\]

The square root of the perfect square obtained is \[\sqrt{1936}=44\]


iii.\[\text{3250}\]

Ans:

The square root of \[3250\] by division method is calculated as follows:


\[57\]

\[5\]

$\overline{32}\overline{50}$ 

-25 

\[107\]

750

749


\[1\]


We are getting a remainder \[1\] .

This means that the square of 57 is less than 3250 by \[1\]. 

So, we must subtract \[1\] from 3250 in order to get a perfect square.

Hence, the required perfect square is \[3250-1=3249\]

The square root of the perfect square obtained is \[\sqrt{3249}=57\]


iv. \[\text{825}\]

Ans:

The square root of \[825\] by division method is calculated as follows:


\[28\]

\[2\]

$\overline{8}\overline{25}$

-4 

\[48\]

425

384


\[41\]


We are getting a remainder \[41\].

This means that the square of 28 is less than 825 by \[41\]. 

So, we must subtract \[41\] from 825 in order to get a perfect square.

Hence, the required perfect square is \[825-41=784\]

The square root of the perfect square obtained is \[\sqrt{784}=28\]


v. \[\text{4000}\]

Ans:

The square root of \[4000\] by division method is calculated as follows:


\[63\]

\[6\]

$\overline{40}\overline{00} $

-36

\[123\]

400  

369 


\[31\]


We are getting a remainder \[31\].

This means that the square of 63 is less than 4000 by \[31\]. 

So, we must subtract \[31\] from 4000 in order to get a perfect square.

Hence, the required perfect square is \[4000-31=3969\]

The square root of the perfect square obtained is \[\sqrt{3969}=63\]


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. \[\text{525}\]

Ans:

The square root of \[4000\] by division method is calculated as follows:


\[22\]

\[2\]

$\overline{5}\overline{25}$

-4  

\[42\]

125

84


\[41\]


Since after finding the square root of 525, we will have 41 as the remainder.

Also, it can be observed that $(22)^2<525<(23)^2$

i.e.,

525 is greater than $(22)^2$ but less than $(23)^2$

Since $(23)^2$ = $529$.

Therefore, to have a perfect square, we need to add 4 to 525.

So, $525+4=529$

And the square root of $529$ = $23$


ii. \[\text{1750}\]

Ans:

The square root of \[1750\] by division method is calculated as follows:


\[41\]

\[4\]

$\overline{17}\overline{50}$ 

-16 

\[81\]

150

81


\[69\]


We are getting a remainder \[69\].

This means that the square of 41 is less than 1750.

The next number is 42 and its square is  \[{{42}^{2}}=1764\]

So, the number that should be added to 1750 is \[{{42}^{2}}-1750=1764-1750=14\] 

Hence, the required perfect square is \[1750+14=1764\]

The square root of the perfect square obtained is \[\sqrt{1764}=42\]


iii. \[\text{252}\]

Ans:

The square root of \[252\] by division method is calculated as follows:


\[15\]

\[1\]

$\overline{2}\overline{52}$

-1 

\[25\]

152

125


\[27\]


We are getting a remainder \[27\].

This means that the square of 15 is less than \[252\].

The next number is 16 and its square is  \[{{16}^{2}}=256\]

So, the number that should be added to 252 is \[{{16}^{2}}-252=256-252=4\] 

Hence, the required perfect square is \[252+4=256\]

The square root of the perfect square obtained is \[\sqrt{256}=16\]


iv. \[\text{1825}\]

Ans:

The square root of \[1825\] by division method is calculated as follows:


\[42\]

\[4\]

$\overline{18}\overline{25}$

-16 

\[82\]

225

164


\[61\]


We are getting a remainder \[61\].

This means that, the square of 42 is less than \[1825\]

The next number is 43 and its square is  \[{{43}^{2}}=1849\]

So, the number that should be added to \[1825\] is \[{{43}^{2}}-1825=1849-1825=24\] 

Hence, the required perfect square is \[1825+24=1849\]

The square root of the perfect square obtained is \[\sqrt{1849}=43\]


v. \[\text{6412}\]

Ans:

The square root of \[6412\] by division method is calculated as follows:


\[80\]

\[8\]

$\overline{64}\overline{12}$

-64 

\[160\]

012

0


\[12\]

We are getting a remainder \[12\].

This means that the square of 80 is less than \[6412\]

The next number is 81 and its square is  \[{{81}^{2}}=6561\]

So, the number that should be added to \[6412\] is \[{{81}^{2}}-6412=6561-6412=149\] 

Hence, the required perfect square is \[6412+149=6561\]

The square root of the perfect square obtained is \[\sqrt{6561}=81\]


6.Find the length of the side of a square whose area is \[\text{441}{{\text{m}}^{\text{2}}}\].

Ans:

Let us consider that the side of the square be x m in length

Then the area of the square is (x)2=441 m2 

We get x\[\times \]

Calculating the square root of 441 using long division method as follows:


\[21\]

\[2\]

$\overline{4}\overline{41}$

-4 

\[41\]

041

41


\[0\]


We get x=21 m 

Therefore, the length of the side of the square is 21 m.


7. In a right triangle ABC, \[\angle \text{B=9}{{\text{0}}^{^{O}}}\]

a) Find AC if AB=6 cm, BC=8 cm 

Ans:

It is given that triangle ABC is right angled at B

So, on applying Pythagoras Theorem, we get

AC2=AB2+BC2

AC2\[={{6}^{2}}+{{8}^{2}}\]

\[=36+64\]

\[=100\]

AC\[=\sqrt{10\times 10}\]

Thus, AC=10 cm.


b) Find AB if AC=13 cm, BC=5 cm

Ans:

It is given that triangle ABC is right angled at B

So, on applying Pythagoras Theorem, we get

AC2=AB2+BC2

AB2=AC2-BC2 

AB2\[={{13}^{2}}-{{5}^{2}}\]

\[=169-25\]

\[=144\]

AB\[=\sqrt{12\times 12}\]

Thus, AB=12 cm. 


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 the same. Find the minimum number of plants he needs more for this. 

Ans:

According to the gardener has 1000 plants and the number of rows and columns are the same.

Our aim is to find the minimum number of plants that he needs so that after planting them, the number of rows and columns are the same.

This means that we have to find the number that should be added to 1000 to make it a perfect square.

The square root of \[1000\] by long division method is calculated as follows:


\[31\]

\[3\]

$\overline{10}\overline{00} $

-9 

\[61\]

100

61


\[39\]


We are getting a remainder \[39\].

This means that the square of 31 is less than \[1000\]

The next number is 32 and its square is  \[{{32}^{2}}=1024\]

So, the number that should be added to \[1000\] is \[{{32}^{2}}-1000=1024-1000=24\] 

Hence, the required number of plants is 24.


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 a number of columns. How many children would be left out in this arrangement? 

Ans:

According to the given question, there are 500 children in the school and they have to stand for a PT drill in such a way that the number of rows and columns are equal.

We have to calculate the number of children that are left out in this arrangement.

This means that we have to find the number that should be subtracted from 500 in order to make it a perfect square.

The square root of \[1000\] by long division method is calculated as follows:


\[22\]

\[2\]

$\overline{5}\overline{00}$

-4

\[42\]

100

84


\[16\]


We are getting a remainder \[16\].

This means that the square of 22 is less than \[500\] by \[16\]

So, we must subtract \[16\]  from \[500\] in order to get a perfect square.

Hence, the required perfect square is \[500-16=484\]

Thus, 16 children will be left out of this arrangement.


NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots - PDF Download

Important Concepts Covered Under NCERT Class 8 Maths Chapter 6 Squares and Square Roots

Squares and Square Roots is the sixth chapter of Class 8 Maths NCERT. It is an important chapter of NCERT Class 8 Maths and is also the foundation chapter for Maths in higher classes. Therefore, students must understand the concepts of Squares and Square Roots thoroughly before they jump to solving questions. 


This is why we have provided an insight into the topics of Class 8 Maths Chapter 6 to help students know what they can expect from the chapter. Check them out here.

Sl. No.

Important Topics

1.

Properties of Square Numbers

2.

Some Interesting Patterns

  • Adding triangular numbers

  • Numbers between square numbers

  • Adding odd numbers

  • A sum of consecutive natural numbers

  • Product of two consecutive even or odd natural numbers

  • Some more patterns in square numbers

3.

Finding the Square of a Number

  • Other patterns in squares

  • Pythagorean triplets

4.

Square Roots

  • Finding square roots

  • Finding square root through repeated subtraction

  • Finding square root through prime factorisation

  • Finding square root by division method

5.

Square Roots of Decimals

6.

Estimating Square Root


Facts

  • When a natural number is raised to the power 2 then it is called the square of the number.

  • If a natural number is multiplied by itself, the result is a perfect square number which is also a natural number.

  • Any natural number may not be a perfect square.

  • A square number can be expressed as the product of equal prime factors.

Ex: 36 = 62 = (2 x 3)2 = 22 x 32.

  • A square number will always have either 0, 1, 4, 5 or 9 at the unit's place but the converse is not true. Ex: 21 is not a square.

  • If a natural number has either 2, 3, 7 or 8 at its unit place then it cannot be a square number. 

  • The square of an even number is always even while the square of an odd number is always odd. 

  • If two consecutive triangle numbers are added, we always obtain a square number.

Ex: 1, 3, 6, 10, …… is the set of triangular numbers and 1 + 3 = 4 = 22,       

  • If we take squares of two consecutive natural numbers  n & n + 1, we will find 2n non-square natural numbers between them.  

  • The sum of the first natural odd natural numbers is n2

Ex: 1 + 3 + 5 = (sum of first 3 odd natural numbers = 9 = 32.

  • If a natural number is not the sum of consecutive odd natural numbers, starting from 1 then it cannot be a perfect square.

  • The square of any odd natural number greater than 1 is always the sum of two consecutive natural numbers.

i.e. 52 = 25 = 12 + 13, 92 = 81 = 40 + 41, etc.

  • The square of any number having 5 in the unit’s place can be formed by using the formula (a5)2 = a(a + 1) x 100 + 25.

  • A set of three natural numbers a, b, c are said to form a Pythagorean triplet (a, b, c) if  a2 + b2 = c2

  • If n be any natural number m > 1, then (2n, n2 - 1, n2 + 1) will form a Pythagorean triplet. 

  • The square root of any number n is that number which when multiplied by itself gives the original number n.

  • The square root is the inverse function of finding square. pq

  • [sqrt

  • p

  • q

  • [sqrtpq
    = √p/√q.

  • There are two integral square roots of a perfect square number n + viz. √n & -√n. Ex: 62 = 36 => √36 = ± 6 i.e. +6 & -6

  • If the square of a natural number is known then the square of the next natural number can be formed out by the sum of the square, the number & the next natural number.

Ex: 312 = 302 + 30 + 31 = 961.


 Applications of a Square Root

  • Square roots are used to determine the length of the diagonal of a square or a triangle.

  • When we need to calculate the third side of a triangle where the measurement of the other two sides of the triangle are known then square roots are used. 

  • Square roots are used to calculate the standard deviation.

  • In order to solve a quadratic equation, square roots are used.


Methods of Finding Square Root

  1. By Prime Factorization Method: To find the square root of a number by prime factorization method, first, we need to derive the number and then determine the prime factors of the number by successive division. Then pair the derived prime factors in such a way that both the factors are equal in the pair. Calculate the product of one factor taken from each pair to obtain the desired square root.

  2. By Repeated Subtraction: This method is used when the natural numbers are small. To calculate the number of odd numbers whose sum is the given natural number, we keep subtracting 1, 3, 5, 7, 9, …… till we arrive at 0. Then we count the number of times the subtraction is performed to arrive at 0. The count becomes the square root of the number.

  3. Long Division Method: When the square numbers are big then we use the long division method because the prime factorization method becomes difficult. In this method, we place the bar on each pair of the given number starting from its unit’s place. Then divide the leftmost digit of the given number by a number whose square root is less than or equal to the leftmost number of the given number. 

  4. Divide the number and write the quotient and the remainder. Then pull the next paired digits next to the remainder and continue the division method till you arrive at 0. 

  5. The derived quotient becomes the square root of the given number.


From Exam Point of View

By going through the above notes, you will definitely be able to solve all the questions in NCERT Solutions provided by Vedantu. This chapter is very important chapter and also very long but there is no need to feel scared, Vedantu has a team of very experienced teachers from reputed institutions across the country who can help you overcome your fears for the subject and master the topic. There are a variety of sums given for the chapter in the NCERT Solutions pdf which will give you an idea of what kind of questions you can expect in exams. You can use the pdf of the solutions for your extensive practice for exams. 


We Cover All the Exercises of the Class 8 Maths Chapter 8 - Squares and Square Roots

Chapter 14 Squares and Square Roots Exercises in PDF Format

Exercise 6.1

9 Question and Solutions

Exercise 6.2

2 Questions and Solutions

Exercise 6.3

10 Questions and Solutions

Exercise 6.4

9 Questions and Solutions


Benefits of Using Vedantu’s NCERT Solutions

Countless students till today have benefitted from the NCERT Solutions provided by Vedantu. Given below are some of the benefits of using Vedantu’s NCERT Solutions for Class 8 Maths Chapter 6:

  • The NCERT Solutions for Class 8 Maths Chapter 6 provided on Vedantu are solved by our very experienced subject matter experts. 

  • Our subject experts have pprovided the solutions to every questions in a very simple and step-wise manner so that it becomes self-explanatory. 

  • We have provided solutions to every questions that are included under Class 8 Maths Chapter 6 with 100% accuracy. 

  • Solving these NCERT Solutions will boost your confidence and will also help you with time management during the exams. If you have any doubts related to the topic then you can completely rely on the NCERT Solutions provided by Vedantu. 

  • You can download the pdf format of Class 8 Maths Chapter 6 Solutions from Vedantu’s website which is available for free. You can download it on any device and practice at your convenient time. You can carry the pdf anywhere and anytime.


Related Questions

Question 1

A perfect square number can never have the digit …….. at the unit's place.

A.1

B.4

C.8

D.9

A.1B.4C.8D.9


Question 2

To what power should (-2) be raised to get 64?


Question 3

How many numbers lie in between the square of the following numbers?

A) 

12and13

12and13

B) 

25and26

25and26

C) 

99and100

99and100


Question 4

Find the square of: 

6.3

6.3

A.39.69

B.39.56

C.39.60

D.39.03

A.39.69B.39.56C.39.60D.39.03


Question 5

Find the value of 

3

−2

3−2


Question 6

Find the two consecutive positive integers, whose sum of squares is 365.


Question 7

Find the perfect square numbers between

(i) 30 and 40

(ii) 50 and 60.


Question 8

How many non-square numbers lie between the squares of 12 and 13?


Question 9

What is ten times the square root of the following decimal number?

7.29


Summary

Vedantu's provision of NCERT Solutions for Class 8 Maths Chapter 6 - "Squares and Square Roots" is a significant boon for students. These solutions offer clear and comprehensive explanations, demystifying the intricacies of squares and square roots. Vedantu's commitment to providing these resources enhances accessibility to quality educational materials, aiding students in their academic journey. By offering these solutions, Vedantu empowers students to not only comprehend the fundamentals but also excel in their mathematical studies. They serve as a reliable companion, promoting effective learning and helping students develop critical mathematical skills essential for problem-solving and future academic pursuits. Vedantu's contribution to education is invaluable.

FAQs on NCERT Solutions for Class 8 Maths Chapter 6 - Squares And Square Roots

1. What are the topics covered in Chapter 6 for class 8, Square and Square Roots?

In chapter 6 for class 8, Square and Square Roots, you will study square numbers or perfect squares, properties and patterns of square numbers, Pythagorean triplets, square roots and various methods to find square roots of natural numbers, decimals and fractions.

2. What are the applications of Square Roots?

Square roots are used to determine the length of the diagonal of a square or a triangle, and to determine the third side of the triangle if the other two sides are known. They are also used to calculate standard deviation and solve quadratic equations.

3. Why is Vedantu academic excellence?

Vedantu is one of the leading online education platforms in the country. The teachers in Vedantu have many years of experience in teaching students. They have prepared NCERT solutions and guides for CBSE Class 8 Maths after extensive research and as per the guidelines of the CBSE Board. Everything covered in the study guide as per the NCERT syllabus will help students to answer any question in unit tests, half-yearly exams and final exams. Experts and teachers have prepared the solutions in a very simple format so that students can easily understand. Experts have also included reference notes so that students can enhance their general knowledge.

4. Can we download pdf version of the NCERT solutions in any device?

Yes, the pdf version of the NCERT solutions can be downloaded on any device like mobile phone, Ipad, laptop and desktop.

5. How many exercises are there in the NCERT Solutions for Class 8 Maths Chapter 6?

There are four exercises in Class 8 Chapter 6. The first exercise covers nine questions. The second covers two, the third covers 10 and the last nine. These questions are offered along with their solutions. You can find NCERT Solutions for Class 8 Maths Chapter 6 on the Vedantu website for free and also on the Vedantu Mobile app.

6. What are square numbers?

If any natural number, let’s say ‘a’ can be expressed as n2, where n is also a natural number, then a can be called a square number. At the units' position, all square numbers end with zero, one, four, five, six, or nine. Only an even number of zeros can be found at the end of a square number. The inverse operation of a square is known as the square root. A perfect square number has two integral square roots.

7. How to get the maximum benefit from NCERT Solutions for Class 8 Maths?

The NCERT Solutions Class 8 Maths Chapter 6 contains well-curated examples and practice problems that cover all major ideas of the topics of the chapter. Hence, students must first study the book's content thoroughly to get a good grasp, and then practise all of the examples as well as the exercise problems. It's also worth paying attention to the highlights given in between chapters.