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Chapter 5 Class 8 Squares and Square Roots

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Get NCERT Solutions of Chapter 5 Class 8 Squares and Square Roots free at Teachoo. All NCERT Exercise questions and examples have been solved in an easy to understand way with detailed explanation of each step. 

In this chapter, we will learn

• What are Square Numbers
• Unit digit of square numbers
• Number of zeroes in square of a number
• Number of numbers between two square numbers
• Sum of first n odd numbers is n 2
• Patterns in Square numbers
• Finding Square of a number without actual multiplication
• What is a Pythagorean Triplet, and how to find them
• What are square roots
• One's digit of square roots
• Finding square root through repeated subtraction
• Finding square root through prime factorization
• Finding square root by division method - of both whole numbers and decimals

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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.

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. $\text{81}$

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. $\text{272}$

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. $\text{799}$

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. $\text{3853}$

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. $\text{1234}$

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. $\text{26387}$

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. $\text{52698}$

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. $\text{99880}$

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. $\text{12796}$

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. $\text{55555}$

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. $\text{1057}$

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. $\text{23453}$

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. $\text{7928}$

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

Hence, $7928$ cannot be a perfect square.

iv. $\text{222222}$

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

Hence, $222222$ cannot be a perfect square.

v. $\text{64000}$

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. $\text{89722}$

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

Hence, $89722$ cannot be a perfect square.

vii. $\text{222000}$

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. $\text{505050}$

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. $\text{243}$

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. $\text{256}$

$256$ 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. $\text{72}$

$72$ 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. $\text{675}$

$675$ 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. $\text{100}$

$100$ 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.

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

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

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

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

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

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 $2$.

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

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

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

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

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

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

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

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

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

6. Find the missing numbers using the given pattern.

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

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

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

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

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

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

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

Since, the sum of first n odd natural numbers is n 2 .

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

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

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

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

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

Ans:  

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.

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

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

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

How many numbers lie between squares of the following numbers?

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

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 \[\text{26}\]

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

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

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. \[\text{32}\]

\[32=30+2\]

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

Since, (a+b) 2 =a 2 +2ab+b 2  

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

\[=900+120+4\]

ii. \[\text{35}\]

\[35=30+5\]

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

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

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

\[=900+300+25\]

iii. \[\text{86}\]

\[86=80+6\]

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

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

\[=6400+960+36\]

iv. \[\text{93}\]

\[93=90+3\]

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

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

\[=8100+540+9\]

v. \[\text{71}\]

\[71=70+1\]

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

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

\[=4900+140+1\]

vi. \[\text{46}\]

\[46=40+6\]

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

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

\[=1600+480+36\]

2. Write a Pythagoras triplet whose one number is

i. \[\text{6}\]

We know that 2m, m 2 -1, m 2 +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 m 2 -1=6, then we get m 2 \[=7\]

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

Similarly, if we take m 2 +1=6, then we get m 2 \[=5\]

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

So let 2m\[=6\]

Then we get, m\[=3\]

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

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

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

ii. \[\text{14}\]

We know that 2m, m 2 -1, m 2 +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 m 2 -1=14, then we get m 2 -1=6

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

Similarly, if we take m 2 +1=14, then we get m 2 +1=14

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

So let 2m=14 

Then we get, m=7 

Now, m 2 -1=7 2 -1

Similarly, m 2 +1=7 2 +1 

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

iii. \[\text{16}\]

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

If we take m 2 -1=16, then we get m 2 =17

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

Similarly, if we take m 2 +1=16, then we get m 2 =15

So let 2m=16 

Then we get, m=8 

Now, m 2 -1=8 2 -1

Similarly, m 2 +1=8 2 +1

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

iv. \[\text{18}\]

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

If we take m 2 -1=18, then we get m 2 =19

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

Similarly, if we take m 2 +1=18, then we get m 2 =17

So let 2m=18 

Then we get, m=9 

Now, m 2 -1=9 2 -1

Similarly, m 2 +1=9 2 -1

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. \[\text{153}\]

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. \[\text{257}\]

We can see that \[257\] has its unit place digit as \[7\].

Hence, \[257\]cannot be a perfect square.

iii. \[\text{408}\]

We can see that \[408\] has its unit place digit as \[8\].

Hence, \[408\]cannot be a perfect square.

iv. \[\text{441}\]

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.

It is already known to us that the sum of the first n odd natural numbers is n 2 .

For \[\sqrt{100}\]

\[100-1=99\]

\[99-3=96\]

\[96-5=91\]

\[91-7=84\]

\[84-9=75\]

\[75-11=64\]

\[64-13=51\]

\[51-15=36\]

\[36-17=19\]

\[19-19=0\]

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

Hence, \[\sqrt{100}=10\]

For \[\sqrt{169}\]

\[169-1=168\]

\[168-3=165\]

\[165-5=160\]

\[160-7=153\]

\[153-9=144\]

\[144-11=133\]

\[133-13=120\]

\[120-15=105\]

\[105-17=88\]

\[88-19=69\]

 \[69-21=48\]

 \[48-23=25\]

 \[25-25=0\]

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

Hence, \[\sqrt{169}=13\]

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

i. \[\text{729}\]

The factorization of \[729\] is as follows:

\[729=\underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}\]

\[\sqrt{729}=3\times 3\times 3\]

So, \[\sqrt{729}=27\]

ii. \[\text{400}\]

The factorization of \[400\] is as follows:

\[400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}\]

\[\sqrt{400}=2\times 2\times 5\]

So, \[\sqrt{400}=20\]

iii. \[\text{1764}\]

The factorization of \[1764\] is as follows:

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

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

So, \[\sqrt{1764}=42\]

iv. \[\text{4096}\]

The factorization of \[4096\] is as follows:

\[4096=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\]

\[\sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2\]

So, \[\sqrt{4096}=64\]

v. \[\text{7744}\]

The factorization of \[7744\] is as follows:

\[7744=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{11\times 11}\]

\[\sqrt{7744}=2\times 2\times 2\times 11\]

So, \[\sqrt{7744}=88\]

vi. \[\text{9604}\]

The factorization of \[9604\] is as follows:

\[9604=\underline{2\times 2}\times \underline{7\times 7}\times \underline{7\times 7}\]

\[\sqrt{9604}=2\times 7\times 7\]

So, \[\sqrt{9604}=98\]

vii. \[\text{5929}\]

The factorization of \[5929\] is as follows:

\[5929=\underline{7\times 7}\times \underline{11\times 11}\]

\[\sqrt{5929}=7\times 11\]

So, \[\sqrt{5929}=77\]

viii. \[\text{9216}\]

The factorization of \[9216\] is as follows:

\[9216=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\]

\[\sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3\]

So, \[\sqrt{9216}=96\]

ix. \[\text{529}\]

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

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

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

x. \[\text{8100}\]

The factorization of \[8100\] is as follows:

\[8100=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}\]

\[\sqrt{8100}=2\times 3\times 3\times 5\]

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. \[\text{252}\]

The factorization of \[252\] is as follows:

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

ii. \[\text{180}\]

The factorization of \[180\] is as follows:

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. \[\text{1008}\]

The factorization of \[1008\] is as follows:

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

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. \[\text{2028}\]

The factorization of \[2028\] is as follows:

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. \[\text{1458}\]

The factorization of \[1458\] is as follows:

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. \[\text{768}\]

The factorization of \[768\] is as follows:

Here, \[768=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times 3\]

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.

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. \[\text{2925}\]

The factorization of \[2925\] is as follows:

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. \[\text{396}\]

The factorization of \[396\] is as follows:

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 \[396\div 11=36\]

iv. \[\text{2645}\]

The factorization of \[2645\] is as follows:

Here, \[2645=5\times \underline{23\times 23}\]

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

v. \[\text{2800}\]

The factorization of \[2800\] is as follows:

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

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

vi. \[\text{1620}\]

The factorization of \[1620\] is as follows:

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.

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 \[\sqrt{2401}\]

\[\sqrt{2401}=\sqrt{\underline{7\times 7}\times \underline{7\times 7}}\]

\[=7\times 7\]

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.

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.

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 \[=\]\[\sqrt{2025}\]

\[\sqrt{2025}=\sqrt{\underline{5\times 5}\times \underline{3\times 3}\times \underline{3\times 3}}\]

\[=5\times 3\times 3\]

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 \[\text{10}\].

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 

L.C.M\[=2\times 2\times 3\times 3\times 5\]

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 \[\text{20}\].

We know that the number that is perfectly divisible by each one of \[8,15\] and \[20\]is their L.C.M

L.C.M\[=2\times 2\times 2\times 3\times 5\]

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. \[\text{2304}\]

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

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

ii. \[\text{4489}\]

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

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

iii. \[\text{3481}\]

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

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

iv. \[\text{529}\]

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

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

v. \[\text{3249}\]

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

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

vi. \[\text{1369}\]

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

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

vii. \[\text{5776}\]

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

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

viii. \[\text{7921}\]

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

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

ix. \[\text{576}\]

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

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

x. \[\text{1024}\]

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

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

xi. \[\text{3136}\]

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

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

xii. \[\text{900}\]

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

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

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

i. \[\text{64}\]

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. \[\text{144}\]

\[144=\overline{1}\overline{44}\] 

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

iii. \[\text{4489}\]

\[4489=\overline{44}\overline{89}\] 

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

iv. \[\text{27225}\]

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

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

v. \[\text{390625}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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. \[\text{402}\]

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

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. \[\text{1989}\]

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

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.\[\text{3250}\]

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

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. \[\text{825}\]

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

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. \[\text{4000}\]

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

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. \[\text{525}\]

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

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

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. \[\text{1750}\]

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

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. \[\text{252}\]

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

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. \[\text{1825}\]

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

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. \[\text{6412}\]

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

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 \[\text{441}{{\text{m}}^{\text{2}}}\] .

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

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

We get x\[\times \]

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

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 

It is given that triangle ABC is right angled at B

So, on applying Pythagoras Theorem, we get

AC 2 =AB 2 +BC 2

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

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

Thus, AC=10 cm.

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

AB 2 =AC 2 -BC 2  

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

\[=169-25\]

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. 

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:

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? 

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.

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.

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 = 6 2 = (2 x 3) 2 = 2 2 x 3 2 .

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 = 2 2 ,       

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 n 2 . 

Ex: 1 + 3 + 5 = (sum of first 3 odd natural numbers = 9 = 3 2 .

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. 5 2 = 25 = 12 + 13, 9 2 = 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  a 2 + b 2 = c 2

If n be any natural number m > 1, then (2n, n 2 - 1, n 2 + 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

[sqrtpq = √p/√q.

There are two integral square roots of a perfect square number n + viz. √n & -√n. Ex: 6 2 = 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: 31 2 = 30 2 + 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

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.

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.

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. 

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. 

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

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

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Related Questions

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

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

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

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

Find the square of: 

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

Find the value of 

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

Find the perfect square numbers between

(i) 30 and 40

(ii) 50 and 60.

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

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

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.

NCERT Solutions for Class 8 Maths

Ncert solutions for class 8.

NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots

The NCERT solutions for Class 8 maths Chapter 6 Squares and Square Roots sheds light on square numbers and square roots, their properties, and some interesting patterns between them. We know that when a number is multiplied by itself, the product that we get is known as the square of that particular number. For example, 4 is square of 2, 9 is square of 3, and so on.

Concepts like finding squares of big numbers by expansion techniques and other patterns observed in square numbers are also covered in the NCERT solutions Class 8 maths Chapter 6. All square numbers will end with 0, 1, 4, 5, 6, or 9 at units' place. And square numbers will have an even number of zeroes in the end. These are a few of the important facts which will help the students throughout the Class 8 maths NCERT solutions Chapter 6 for the understanding of the examples and the exercise questions asked. The pdf file of this chapter can be found below and also you can find some of these in the exercises given below.

• NCERT Solutions Class 8 Maths Chapter 6 Ex 6.1
• NCERT Solutions Class 8 Maths Chapter 6 Ex 6.2
• NCERT Solutions Class 8 Maths Chapter 6 Ex 6.3
• NCERT Solutions Class 8 Maths Chapter 6 Ex 6.4

NCERT Solutions for Class 8 Maths Chapter 6 PDF

The concept of square root forms the basis for many higher math topics ahead. Sufficient practice on the same is required to build up a quick mental math skill which will help students in examinations too while saving enough time on calculations. A brief glance at the exercises in the NCERT solutions Class 8 maths Chapter 6 can be observed below :

☛ Download Class 8 Maths NCERT Solutions Chapter 6

NCERT Class 8 Maths Chapter 6   Download PDF

NCERT Solutions for Class 8 Maths Chapter 6

Square and square roots form a major part of mathematics. Almost in every other topic, one can encounter them, whether it's algebra , geometry , or calculus ; hence, one must know how to calculate squares and square roots, in order to understand the concepts better and faster. An exercise-wise detailed analysis of NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots can be found in the pdf below :

• Class 8 Maths Chapter 6 Ex 6.1 - 9 Questions
• Class 8 Maths Chapter 6 Ex 6.2 - 2 Questions
• Class 8 Maths Chapter 6 Ex 6.3 - 10 Questions
• Class 8 Maths Chapter 6 Ex 6.4 - 9 Questions

☛ Download Class 8 Maths Chapter 6 NCERT Book

Topics Covered: The class 8 maths NCERT solutions Chapter 6 discusses the properties of square numbers, how to find the sum and product of consecutive even and odd natural numbers , what are Pythagorean triplets, finding square root with the help of repeated subtraction , and through prime factorization as well as the long division method to find square roots of numbers .

Total Questions: The class 8 maths Chapter 6 Squares and Square Roots consists of a total of 30 questions out of which 21 are easy, 4 are moderately easy and, 5 are long answer type questions.

List of Formulas in NCERT Solutions Class 8 Maths Chapter 6

The NCERT solutions Class 8 maths Chapter 6 talks about some of the important facts while dealing with square and square roots of numbers. We know that if a natural number ‘x’ is written as ‘y 2 ’, where y is a natural number, then ‘x’ denotes the square number. The important facts revolving around the above concept can be found in the NCERT solutions for Class 8 maths Chapter 6 some of which are given below :

• The ‘ √ ‘ symbol represents the positive square root of a number
• The square root can be said to be the inverse operation of the square.

Important Questions for Class 8 Maths NCERT Solutions Chapter 6

Ncert solutions for class 8 maths video chapter 6, faqs on ncert solutions class 8 maths chapter 6, why are ncert solutions class 8 maths chapter 6 important.

The NCERT Solutions Class 8 Maths Chapter 6 covers important concepts regarding the squares of numbers and the derivation of square roots. These concepts are very helpful in attempting big and lengthy calculations as they make the problem-solving process much easier and smooth. The NCERT team has made sure to cover all of these concepts in a gradual manner through appropriate explanations of examples, making these solutions a very important resource of learning math.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 8 Maths Squares and Square Roots?

If one needs to excel at any subject, regular and consistent practice is a must. And especially with mathematics, this becomes all the more important. Hence the students must make sure to practice all the examples step by step and they must try to solve all the exercise questions in order to explore all the challenging aspects of the topic making them confident in approaching any math problem.

What are the Important Topics Covered in NCERT Solutions Class 8 Maths Chapter 6?

The NCERT Solutions Class 8 Maths Chapter 6 covers the properties of square numbers, process of finding the sum and product of consecutive even and odd natural numbers, description of Pythagorean triplets, methods to find square root with the help of repeated subtraction, and through prime factorization, land ong division technique to find square roots of numbers with the help of relevant examples.

How Many Questions are there in Class 8 Maths NCERT Solutions Chapter 6 Squares and Square Roots?

The NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots consists of a total of 30 questions. The students will be able to solve 21 of them very easily as they follow the direct logic of finding the squares and square roots, 4 would require a bit of logical thinking while the remaining 5 can be categorized as long-form questions as they involve lengthier calculations.

How CBSE Students can utilize NCERT Solutions Class 8 Maths Chapter 6 effectively?

The NCERT Solutions Class 8 Maths Chapter 6 has well-curated examples, and exercise questions covering all the important concepts of the topic. Hence, the students must first read the text of the book properly to gain a good understanding and thereby practice all the examples along with the exercise questions. The highlights presented in between the chapter should also be focussed on and learned well in order to utilize these solutions effectively.

Why Should I Practice Class 8 Maths NCERT Solutions Squares and Square Roots Chapter 6?

The CBSE board highly recommends studying from the NCERT Solutions Class 8 Maths Squares and Square Roots Chapter 6, which in itself proves the importance of the knowledge that is compiled in the book. The team at NCERT makes sure to do enough research and then comes up with the appropriate easy explanations of all the topics through real-life examples and exercise questions. Hence, the students must practice these solutions to gain a deeper understanding of the subject in an easy and strategic way.

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Class 8 Mathematics Squares and Square Roots Assignments

We have provided below free printable Class 8 Mathematics Squares and Square Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 8 Mathematics Squares and Square . These Assignments for Grade 8 Mathematics Squares and Square cover all important topics which can come in your standard 8 tests and examinations. Free printable Assignments for CBSE Class 8 Mathematics Squares and Square , school and class assignments, and practice test papers have been designed by our highly experienced class 8 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Squares and Square Class 8 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Squares and Square Class 8. Students can click on the links below and download all Pdf Assignments for Mathematics Squares and Square class 8 for free. All latest Kendriya Vidyalaya Class 8 Mathematics Squares and Square Assignments with Answers and test papers are given below.

Mathematics Squares and Square Class 8 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 8 Mathematics Squares and Square . Students and teachers can download and save all free Mathematics Squares and Square assignments in Pdf for grade 8th. Our expert faculty have covered Class 8 important questions and answers for Mathematics Squares and Square as per the latest syllabus for the current academic year. All test papers and question banks for Class 8 Mathematics Squares and Square and CBSE Assignments for Mathematics Squares and Square Class 8 will be really helpful for standard 8th students to prepare for the class tests and school examinations. Class 8th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 8 Mathematics Squares and Square Download in Pdf

Advantages of Class 8 Mathematics Squares and Square Assignments

• As we have the best and largest collection of Mathematics Squares and Square assignments for Grade 8, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Mathematics Squares and Square textbooks for Class 8 .
• All Mathematics Squares and Square assignments for Class 8 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 8 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Mathematics Squares and Square chapter wise worksheets and assignments for free in Pdf
• Class 8 Mathematics Squares and Square question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 8 Mathematics Squares and Square students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Mathematics Squares and Square Roots Class 8 which you can download in Pdf

We provide here Standard 8 Mathematics Squares and Square Roots chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Mathematics Squares and Square Roots in Grade 8, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

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Class viii math, class-8 square and square roots, introduction to square, square numbers / perfect squares, facts of perfect squares, properties of perfect square, number patterns of square numbers, square roots, methods of finding square roots, prime factorisation method, division method.

Square And Square Roots Test

Square And Square Roots Worksheets

Answer Sheet

As we know, area of a square is the product of it's two sides i.e. square of sides (side) 2 . If we consider the side of a square is 'a' unit, then Area of square = (a × a ) square unit = a 2 sq. unit. From this we conclude that, square of a number is the product of the number by the same number.

Example 1. 2 2 = 2 × 2

Example 2. ( 1 ⁄ 3 ) 2 = 1 ⁄ 3 × 1 ⁄ 3

Square numbers also known as Perfect square. For example, 9, 16, 25, 36, 49, 64, 81, ... are perfect squares.

All the natural numbers are not perfect square, but square of natural numbers are perfect squares.

If a natural number is N, and it is expressed as M 2 . Here M is natural number and N is a perfect square. But all the natural numbers do not form perfect square. From the below examples, we can know whether a given natural number is a perfect square.

Example 1. 25 is a square number because it is expressed by multiplying two same natural numbers of 5. 25 = 5 × 5

Example 2. 18 is not a square number because it can be expressed by multiplying different natural numbers like 18 = 3 × 6 = 2 × 9

A perfect square can always form by multiplying the pairs of equal prime factors.

When pairs of equal prime factors are multiplied with each other they form a perfect square. But when the prime factors are multiplied without pairing, they do not form a perfect square. Let's see some examples.

Example 1. 36 = 2 × 2 × 3 × 3 . (Here 36 is perfect square)

Example 2. 108 = 2 × 2 × 3 × 3 × 3. (Here 108 is not a perfect square as 3 cannot paired)

When a smallest natural number is multiplied by a non-perfect square to make it a perfect square.

Example 1. Find the smallest natural number that multiplied by 108 to make it a perfect square. Solution. Given number is 108. First we must calculate its prime factors by using prime factorisation method. Then arrange the prime factors as the product of pairs of equal prime number and unpaired prime number as shown below.         108 = 2 × 2 × 3 × 3 × 3 We can see, 3 is left unpaired. So, to make 108 a perfect square another 3 is required. Thus 108 is multiplied by 3 to get the perfect square. Hence 3 is the smallest number.

When a smallest natural number is divided by a non-perfect square to make it perfect square.

Example 1. Find the smallest natural number that will divide 48 to make it a perfect square? Solution. Here the given number is 48. First we must calculate its prime factors by using prime factorisation method. Then arrange the prime factors as the product of pairs of equal prime number and unpaired prime number as shown below.         48 = 2 × 2 × 2 × 2 × 3 Here number 3 is left unpaired. So to make 48 a perfect square we divide it by 3. Hence 3 is the smallest number.

The natural number ends with 0, 1, 4, 5, 6, or 9 at its unit's place are not square numbers. But a square number always ends with 0, 1, 4, 5, 6, or 9 at it's unit place. A number having 2,3,7,or 8 at its unit's place is never a square number.

Example 1. Check if 2978 is a perfect square number. Solution. 2978 is not a perfect square, because it ends with 8 at its unit's place.

Example 2. Check if 76302 is a perfect square number. Solution. 76302 is not a perfect square, because it ends with 2 at its unit's place.

Example 3. Check if 289 is a perfect square number. Solution. 289 is a perfect square number, because it ends with 9 at its unit's place.

Example 4. Check if 3253 is a perfect square number. Solution. 3253 is not a perfect square number, because it ends with 3 at its unit's place.

A number which has odd number of zeros at the end will never be a perfect square. Perfect square always ends with even number of zeros.

Example 1. Check if 90000 is a perfect square number. Solution. 300 2 = 9 0000 We can notice 4 zeros are present in 90000, hence it is a perfect square.

Example 2. Check if 2500 is a perfect square number. Solution. 50 2 = 25 00 We can notice 2 zeros are present in 2500, hence it is a perfect square.

Example 3. Check if 400000 is a perfect square number. Solution. We can notice 5 zeros are present in 4 00000 , hence it is not a perfect square.

Example 4. Check if 2000 is a perfect square number. Solution. We can notice 3 zeros are present in 2 000 , hence it is not a perfect square.

Example 5. Check if 40 is a perfect square number. Solution. We can notice only 1 zero is present in 4 0 , hence it is not a perfect square.

Square of odd natural numbers always get odd natural number.

Example 1. 379 2 = 1,43,641 is odd because 379 is an odd natural number. Example 2. 137 2 = 18,769 is odd because 137 is an odd natural number.

Square of even natural numbers is always getting even number.

Example 1. 64 2 = 4,096 is even because 64 is an even natural number. Example 2. 32 2 = 1,024 is even because 32 is an even natural number.

The square unit place number depends upon the number present in unit's place of natural number.

If a number ends with 1 or 9 in the unit's place, then its square unit's place will have 1.         9 2 = 81, 31 2 = 961

If a number ends with 0 in the unit's place, then its square ends in 0.         10 2 = 100

If a number ends with 5 in the unit's place, then its square ends in 5.         25 2 = 625

If a number ends with 3 or 7 in the unit's place, then its square ends in 9.         13 2 = 169, 17 2 = 289

If a number ends with 2 or 8 in the unit's place, then its square ends in 4         12 2 = 144, 8 2 = 64

If a number ends with 4 or 6 in the unit's place, then its square ends in 6.         14 2 = 196, 16 2 = 256.

Non square numbers between two consecutive square numbers

Adding consecutive odd numbers to express perfect square number, square of odd number greater than one is expressed by adding two consecutive natural number, square of numbers with digits 5 at unit's place, pythagorean triplet.

If we consider two consecutive numbers (p) and (p+1), then there are 2p non square numbers present in between p 2 and ( p+1) 2 . Let's see some examples.

Example 1. How many non-square natural numbers are present between 5 2 and 6 2 ? Solution. Two perfect square numbers are 5 2 = 25 and 6 2 = 36 As per the formula we will be having 10 (2 × 5) non-square natural numbers. In order to prove this, we have to find out the numbers present between 25 and 36. And the numbers are 26, 27, 28, 29, 30, 31, 32, 33, 34, and 35. So, the total count of numbers are 10.

To get perfect square number, consecutive odd natural numbers should be added starting from 1. Let's see some examples.

Example 1. Express 64 as the sum of the numbers of odd number. Solution. 64 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 2 Here we add first 8 odd natural number and get square of 8 i.e. 64.

If a natural number not expressed as a sum of consecutive odd natural numbers starting from 1, then that number is not a perfect square. In other word, to know a Number is a perfect square, we must subtract all first odd natural numbers (starting from 1) from the given square number untill the result is zero. Let's see some examples.

Example 1. Check if number 65 is perfect square. Solution. Consecutively subtract 1, 3, 5, 7, ... from number 65 untill we get zero. If the answer comes out to be zero then we can conclude the number to be perfect square. 65 − 1 = 64, 64 − 3 = 61, 61 − 5 = 56, 56 − 7 = 49, 49 − 9 = 40, 40 − 11 = 29, 29 − 13 = 16, 16 − 17 = -1. Here, we can see the answer turns out to be −1. So, we can conclude that number 65 is not a perfect square.

Example 2. Check if number 16 is perfect square. Solution. Consecutively subtract 1, 3, 5, 7, ... from number 16 untill we get zero. 16 − 1 = 15, 15 − 3 = 12, 12 − 5 = 7, 7 − 7 = 0 Here, we can see the answer turns out to be 0. So, we can conclude that number 16 is a perfect square.

Square of any odd number greater than 1 is expressed by summing of consecutive natural numbers. But sum of any two consecutive integers does not always form perfect square.

There is a formula to calculate first number and second number. Let's assume perfect square number is n 2 . First number is calculated by using formula n 2 -1 ⁄ 2 Second number is n 2 +1 ⁄ 2

Example 1. Find the two consecutive natural numbers whose summation will be equal to 13 2 . Solution. First number = 13 2 -1 ⁄ 2 = 84 Second number = 13 2 +1 ⁄ 2 = 85 13 2 = 169 = 84 + 85.

Example 2. Find the two consecutive natural numbers whose summation will be equal to 39 2 . Solution. First number = 39 2 -1 ⁄ 2         = (1521-1) ⁄ 2 = 760 Second number = 39 2 +1 ⁄ 2         = (1521+1) ⁄ 2 = 761 So, 39 2 = 760 + 761.

Here is the formula to find out square of numbers having 5 at the unit place. (m5) 2 = m(m + 1)hundreds + 25 = m(m + 1) × 100 + 25 Here m is digits before the units place. Let's see some examples.

Example 1. Find the square of 45. Solution. As per the above mentioned formual here m is 4. 45 2 = 4(4+1)hundred + 25         = {(4 × 5) × 100} + 25         = 20 × 100 + 25         = 2000 + 25         = 2025

Example 2. Find the square of 235. Solution. Here m is 23. 235 2 = 23(23+1)hundred + 25         = {(23 × 24) × 100} + 25         = 552 × 100 + 25         = 55200 + 25         = 55225

If we consider P, Q and R are three natural numbers and they are following the Pythagorean theory that is P 2 + Q 2 = R 2 then these three natural numbers form Pythagorean triplet. For any natural number P > 1, then (2P) 2 + (P 2 − 1) 2 = (P 2 + 1) 2 . So here( 2P, P 2 −1, P 2 +1 ) are formed Pythagorean triplet. Let's see some examples.

Example 1. Find Pythagorean triplet, whose smallest number is 16. Solution. Let us consider 2P = 16.         => P = 16 ⁄ 2         => P = 8 Now we put the value of P in P 2 −1 = 8 2 −1 = 63 Now we put the value of P in P 2 + 1 = 8 2 + 1 = 65 The Pythagorean triplets are (16, 63, and 65).

Example 2. Find Pythagorean triplet with one number is 18. Solution. Let us first consider P 2 − 1 = 18 , P 2 = 18 − 1 = 17. So, here value of P will not a natural number. Now, we consider P 2 + 1 = 18, P 2 = 18 + 1 = 19. So, here value of P will not a natural number. Now, consider 2P = 18, P = 18 ⁄ 2 = 9. So, now we can put the value of P = 9 in P 2 − 1 = 9 2 − 1 = 80, and P 2 + 1 = 9 2 + 1 = 82. Then the collection of numbers (18, 80, 82) is Pythagorean triplet with one number is 18.

Square root, is a factor of a number, that when multiplied by itself gives the original number. If M is the square root of N, then M × M = N The square root of number N is shown as √  N  . So, √  N  = M. NOTE - finding the square root is the inverse operation of finding square.

• Prime factorisation method
• Division method

To find square root of natural numbers Step 1. The given number is expressed into product of prime factors. Step 2. Make the prime numbers in pair of same prime number. Step 3. Consider one prime number from each pair and multiply them together. Step 4. The product is known as square root of the given number. Let's see some examples.

Example 1. Find the square root of 256 by prime factorisation. Solution. First convert 256 into prime factors by prime factorisation method. √  256  = √  2×2×2×2×2×2×2×2          = √   2×2 × 2×2 × 2×2 × 2×2           = 2×2×2×2 (Taking one prime number from each pair) So, √  256  = 16.

To find square root fraction First, we should calculate the square root of numerator and then denominator of the fraction. Square root of fraction = Square root of numerator/Square root of denominator.

Example 1. Find the square root of 196 ⁄ 144 . Solution. √( 196 ⁄ 144 ) = √  196  ⁄ √  144          = √  (2×2×7×7)  ⁄ √  (2×2×2×2×3×3)          = √(2 2 ×7 2 ) ⁄ √(2 2 ×2 2 ×3 2 )         = (2×7) ⁄ (2×2×3)         = 14 ⁄ 12         = 7 ⁄ 6

To find square root of decimal number We should first convert it to fraction, then express it to lowest term if needed then calculate square root as method used for fraction above.

Example 1. Find the square root of 2.89. Solution. First convert the decimal number 2.89 to fraction.         2.89 = 289 ⁄ 100 Then calculate square root of 289 ⁄ 100 √( 289 ⁄ 100 ) = √  289  ⁄ √  100          = √  17×17  ⁄ √  10×10          = 17 ⁄ 10         = 1.7 Hence square root of 2.89 = 1.7.

To find square root of mixed fraction number First, we must convert the mixed fraction into improper fraction. Then find out the square root using normal fraction method we dicuseed above.

Example 1. Find the square root of 4 261 ⁄ 25 Solution. First convert mixed fraction into improper fraction number.         4 261 ⁄ 25 = 361 ⁄ 25 Now, calculate the square root of 361 ⁄ 25         = √( 361 ⁄ 25 )         = √  361  ⁄ √  25          = √  (19×19)  ⁄ √  (5×5)          = √(19) 2 ⁄ √(5) 2         = 19 ⁄ 5         = 3 4 ⁄ 5 Hence, square root of 4 261 ⁄ 25 is 3 4 ⁄ 5 .

This method is used to calculate square root of large number. It is also called long division method. Let's see some examples to understand the process.

• First put bars over the pair of digits from right to left starting from unit's place.
• If number of digits is odd, bar must be put over remaining single digit that is present extreme left.
• Then first consider the first pair of digit or single digit number.
• Here, first digit is 5 now find out greatest number whose square is less than or equal to 5.
• Here, we get that number is 2.Then write 2 in the place of divisor and on the top in the quotient.
• Then subtract 4 from 5 and we get remainder as 1.
• Next pair of digits 29 is brought down and written near to 1 and we get new dividend as 129
• Add divisor with its quotient, here we get 4 and it became a new divisor for new dividend as written above.
• Keep some space right to 4.
• Then find the largest digit to fill that blank and it also become new digit in the quotient.
• The new divisor is multiplied by new digit in the quotient, then product is less than or equal to the dividend.
• So, here 43 multiplied by 3 and we get 129, that means new digit we choose is 3.
• Then subtract 129 from 129 and we get zero as shown above.
• Hence, √  529  = 23.

• Put bar over the pair of digits from right to left starting from unit's place.
• Then consider the first pair of digits i.e. 73.
• Find out greatest number whose square is less than or equal to 73.
• We get the number as 8. Then write 8 in the place of divisor and in the place of quotient as shown here.
• Subtract 64 from 73, and we get remainder as 9.
• Next pair of digits 96 is brought down and written near to 9 and we get new dividend as 996
• Add the divisor with its quotient, and we get 16 as our new divisor for new dividend 996.
• Keep some space right to 16.
• In this case, 166 multiplied with 6 and we get 996. That means, new digit we choose is 6.
• Subtract 996 from 996 and get zero as shown above.
• Hence, √  7396  = 86

Square roots of Decimal by long division method

Following method is used to calculate square root of decimal number. Let's see the below given example.

• Put bars on the pair of integer present before decimal point from right to left and put bars over the pair of integers from left to right after decimal point.
• First consider the first pair of digits i.e. 19.
• Now find out greatest number whose square is less than or equal to 19. And the number is 4.
• Write 4 in the place of divisor and on the top in the place of quotient.
• Subtract 16 from 19 and we get remainder as 3.
• As next pair of digits come after decimal point, so we should write the decimal point in the quotient.
• Then next pair of digits 98 is brought down and written next to 3 and we get new dividend as 398
• Add divisor with its quotient, we get 8 and it became a new divisor for new dividend as written above.
• Space must be blank right to 8.
• Find the largest digit to fill that blank and it also become new digit in the quotient.
• So, here 84 multiplied with 4 and get 336 that means new digit we choose is 4.
• Subtract 336 from 398, we get 62 as shown above.
• Then next pair of digits 09 is brought down and written next to 62, we get new dividend as 6209.
• Add divisor 84 with its quotient 4, we get 88 and it became a new divisor for the new dividend 6209.
• Space must be blank right to 88.
• So, here 887 multiplied by 7 and we get 6209 that means new digit we choose is 7.
• Subtract 6209 from 6209 and we get zero as shown above.
• Hence, √  19.9809  = 4.47

Square root of a fraction using long division method

In this case, we should first convert the fraction into decimal form and then find out the square root of the decimal number using the method described above.

Class-8 Square And Square Roots Test

Square And Square Roots - 1

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Class-8 Square And Square Roots Worksheets

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Revision Notes on Squares and Square Roots

Square number.

Any natural number ‘p’ which can be represented as y 2 , where y is a natural number, then ‘p’ is called a Square Number .

Where 2, 3, 4 are the natural numbers and 4, 9, 16 are the respective square numbers.

Such types of numbers are also known as Perfect Squares .

Some of the Square Numbers

Properties of Square Numbers

We can see that the square numbers are ending with 0, 1, 4, 5, 6 or 9 only . None of the square number is ending with 2, 3, 7 or 8.

Any number having 1 or 9 in its one’s place will always have a square ending with 1 .

Any number which has 4 or 6 in its unit’s place, its square will always end with 6 .

Any number which has 0 in its unit’s place, its square will always have an even number of zeros at the end .

Some More Interesting Patterns

1. adding triangular numbers.

If we could arrange the dotted pattern of the numbers in a triangular form then these numbers are called Triangular Numbers . If we add two consecutive triangular numbers then we can get the square number.

2. Numbers between Square Numbers

If we take two consecutive numbers n and n + 1, then there will be (2n) non-perfect square numbers between their squares numbers.

Let’s take n = 5 and 5 2 = 25

n + 1 = 5 + 1 = 6 and 6 2 = 36

2n = 2(5) = 10

There must be 10 numbers between 25 and 36.

The numbers are 26, 27, 28, 29, 30, 31, 32, 33, 34, 35.

3. Adding Odd Numbers

Sum of first n natural odd numbers is n 2 .

Any square number must be the sum of consecutive odd numbers starting from 1.

And if any natural number which is not a sum of successive odd natural numbers starting with 1, then it will not be a perfect square.

4. A Sum of Consecutive Natural Numbers

Every square number is the summation of two consecutive positive natural numbers.

If we are finding the square of n the to find the two consecutive natural numbers we can use the formula

12 + 13 = 25

Likewise, you can check for other numbers like

11 2 = 121 = 60 + 61

5. The Product of Two Consecutive Even or Odd Natural Numbers

If we have two consecutive odd or even numbers (a + 1) and (a -1) then their product will be (a 2 - 1)

Let take two consecutive odd numbers 21 and 23.

21 × 23 = (20 - 1) × (20 + 1) = 20 2 - 1

6. Some More Interesting Patterns about Square Numbers

Finding the Square of a Number

To find the square of any number we needed to divide the number into two parts then we can solve it easily.

If number is ‘x’ then x = (p + q) and x 2 = (p + q) 2

You can also use the formula (p + q) 2 = p 2 + 2pq + q 2

Find the square of 53.

Divide the number in two parts.

53 = 50 + 3

53 2 = (50 + 3) 2

= (50 + 3) (50 + 3)

= 50(50 + 3) +3(50 + 3)

= 2500 + 150 + 150 + 9

1. Other pattern for the number ending with 5

For numbers ending with 5 we can use the pattern

(a5) 2 = a × (a + 1)100 + 25

25 2 = 625 = (2 × 3) 100 + 25

45 2 = 2025 = (4 × 5) 100 + 25

95 2 = 9025 = (9 × 10) 100 + 25

125 2 = 15625 = (12 × 13) 100 + 25

2. Pythagorean Triplets

If the sum of two square numbers is also a square number, then these three numbers form a Pythagorean triplet.

For any natural number p >1, we have (2p)  2  + (p 2  -1) 2  = (p 2  + 1) 2 . So, 2p, p 2 -1 and p 2 +1 forms a Pythagorean triplet.

Write a Pythagorean triplet having 22 as one its member.

p 2  + 1 = 10

p 2  - 1 = 8.

Thus, the Pythagorean triplet is 6, 8 and 10.

6 2 + 8 2 = 10 2

36 + 64 = 100

Square Roots

The square root is the inverse operation of squaring. To find the number with the given square is called the Square Root .

 2 2 = 4, so the square root of 4 is 2

10 2 = 100, therefore square root of 100 is 10

There are two square roots of any number. One is positive and other is negative.

The square root of 100 could be 10 or -10.

Symbol of Positive Square Root

Finding Square Root

1. through repeated subtraction.

As we know that every square number is the sum of consecutive odd natural numbers starting from 1, so we can find the square root by doing opposite because root is the inverse of the square.

We need to subtract the odd natural numbers starting from 1 from the given square number until the remainder is zero to get its square root.

The number of steps will be the square root of that square number.

Calculate the square root of 64 by repeated addition.

64 – 1 = 63

63 – 3 = 60

60 – 5 = 55

55 – 7 = 48

28 – 13 = 15

48 – 9 = 39

15 – 15 = 0

39 – 11 = 28

2. Prime Factorization

In this method, we need to list the prime factors of the given number and then make the pair of two same numbers.

Then write one number for each pair and multiply to find the square root.

Calculate the square root of 784 using prime factorization method.

List the prime factors of 784.

784 = 2 × 2 × 2 × 2 × 7 × 7

√784 = 2 × 2 × 7 = 28

3. Division Method

Steps to find the square root by division method

Step 1: First we have to start making the pair of digits starting from the right and if there are odd number of digits then the single digit left over at the left will also have bar .

Step 2: Take the largest possible number whose square is less than or equal to the number which is on the first bar from the left. Write the same number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide to get the remainder.

Step 3: Like a normal division process bring the digits in next bar down and write next to the remainder.

Step 4: In next part the quotient will get double and we will right in next line with a blank on its right.

Step 5: Now we have to take a number to fill the blank so that the if we take it as quotient then the product of the new divisor and the new digit in quotient is less than or equal to the dividend.

Step 6: If there are large number of digits then you can repeat the steps 3, 4, 5 until the remainder does not become 0.

 Calculate the square root of √729 using division method.

Thus, √729 = 27.

Square Roots of Decimals

To find the square root of a decimal number we have to put bars on the primary part of the number in the same manner as we did above. And for the digits on the right of the decimal we have to put bars starting from the first decimal place. Rest of the method is same as above. We just need to put the decimal in between when the decimal will come in the division.

 Find √7.29 using division method.

Thus, √7.29 = 2.7

Remark: To put the bar on a number like 174.241, we will put a bar on 74 and a bar on 1 as it is a single digit left. And in the numbers after decimal, we will put a bar on 24 and put zero after 1 to make it double-digit.

174 . 24 10

Estimating Square Root

Sometimes we have to estimate the square root of a number if it’s not possible to calculate the exact square root.

Estimate the square root of 300.

 We know that, 300 comes between 100 and 400 i.e. 100 < 300 < 400. Now, √100 = 10 and √400 = 20.

 So, we can say that

10 < √300 < 20.

We can further estimate the numbers as we know that 17 2  = 289 and 18 2  = 324. Thus, we can say that the square root of √300 = 17 as 289 is much closer to 300 than 324.

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Assignments For Class 8 Mathematics Squares and Square Roots

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Squares and Square Roots Class 8 Worksheet pdf give practice questions based on square roots or squares, their properties and examples, Pythagorean trios, different procedures of finding the square roots of natural numbers, square roots of fractions and decimals, and various techniques for determining square roots such as prime factorization method, repeated subtraction, and long division method.

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Class 8 (Old)

Course: class 8 (old)   >   unit 5.

• Intro to square roots
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Square roots

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• Your answer should be
• an integer, like 6 ‍  
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• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

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Maths assignment class viii | quadrilateral, lesson plan math class 8 (ch-1) | square & square roots.

  E- LESSON PLAN   SUBJECT MATHEMATICS    CLASS- 8

E-LESSON PLAN MATHEMATICS CLASS-VIII CHAPTER-1 SQUARE & SQUARE ROOTS

• General Tables from 2 to 20
• Square table from 1 to 30
• Cubic table from 1 to 20
• To introduce the concept of square and square root to students.
• To teach students how to calculate the square and square root of a number.
• To help students understand the practical applications of squares and square roots.
• Method of finding the Pythagorean Triplet.
• Repeated subtraction method of finding the square.
• Prime factorization method of finding the square root.
• Long division method of finding the square root.
• Method of estimation for finding the square root of a number.
• Pythagorean Triplet.
• Repeated subtraction method of finding the square root of a perfect square number.
• Factorization method of finding the square root of a number.
• Long Division Method of finding the square root of a number.
• Method of estimation of finding the square of a number.

Arrange the students either in groups or individuals and plan an activity of writing the squares of natural numbers from 1 to 30. Also help students to find the square of other important numbers like 35, 40, 45, 50, 55 ….. etc.

Teacher may also suggest some shortcut ways of finding square of such numbers

Help the students to find the square root by using different methods like

• Repeated subtraction method of finding the square root of a number.
• Long division method of finding the square root of a number.
• Method of estimation of finding the square root of a number.

Application of Squares and Square Roots

Mind map / concept map.

Assessment:

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DAV Class 8 Maths Chapter 1 Worksheet 5 Solutions

The DAV Class 8 Maths Book Solutions and DAV Class 8 Maths Chapter 1 Worksheet 5  Solutions of Squares and Square Roots offer comprehensive answers to textbook questions.

DAV Class 8 Maths Ch 1 WS 5 Solutions

Question 1. Find the square root of the following fractions:

(i) \(\frac{324}{361}\)

(ii) \(\frac{441}{961}\)

(iii) \(5 \frac{19}{25}\)

(iv) \(21 \frac{51}{169}\)

(v) \(\frac{5625}{441}\)

(vi) \(7 \frac{18}{49}\)

(vii) \(23 \frac{394}{729}\)

(viii)\(35 \frac{85}{1444}\)

Solution: (i) \(\sqrt{\frac{324}{361}}=\frac{\sqrt{324}}{\sqrt{361}}=\frac{\sqrt{18^2}}{\sqrt{19^2}}=\frac{18}{19}\)

(ii) \(\sqrt{\frac{441}{961}}=\frac{\sqrt{441}}{\sqrt{961}}=\frac{\sqrt{21^2}}{\sqrt{31^2}}=\frac{21}{31}\)

(iii) \(\sqrt{5 \frac{19}{25}}=\sqrt{\frac{144}{25}}=\frac{\sqrt{144}}{\sqrt{25}}\) = \(\frac{\sqrt{12^2}}{\sqrt{5^2}}=\frac{12}{5}=2 \frac{2}{5}\)

(iv) \(\sqrt{21 \frac{51}{169}}=\sqrt{\frac{3600}{169}}=\frac{\sqrt{3600}}{\sqrt{169}}\) = \(\frac{\sqrt{60^2}}{\sqrt{13^2}}=\frac{60}{13}=4 \frac{8}{13}\)

(v) \(\sqrt{\frac{5625}{441}}=\frac{\sqrt{5625}}{\sqrt{441}}=\frac{\sqrt{75^2}}{\sqrt{21^2}}\) = \(\frac{75}{21}=3 \frac{12}{21}\)

(vi) \(\sqrt{7 \frac{18}{49}}=\sqrt{\frac{361}{49}}=\sqrt{\frac{19^2}{7^2}}\) = \(\frac{19}{7}=2 \frac{5}{7}\)

(vii) \(\sqrt{23 \frac{394}{729}}=\sqrt{\frac{17161}{729}}=\sqrt{\frac{131^2}{27^2}}\)

= \(\frac{131}{27}=4 \frac{23}{27}\)

(viii) \(\sqrt{35 \frac{85}{1444}}=\sqrt{\frac{50625}{1444}}=\sqrt{\frac{225^2}{38^2}}\)

= \(\frac{225}{38}=5 \frac{35}{38}\)

Question 2. (i) \(\sqrt{0.0009}\) (ii) \(\sqrt{0.0081}\) (iii) \(\sqrt{0.012321}\) (iv) \(\sqrt{7.29}\) Solution: (i) \(\sqrt{0.0009}\)

Hence \(\sqrt{0.0009}\) = 0.03

(ii) \(\sqrt{0.0081}\)

Hence \(\sqrt{0.0081}\) = 0.09

(iii) \(\sqrt{0.012321}\)

Hence \(\sqrt{0.012321}\) = 0.111

(iv) \(\sqrt{7.29}\)

Hence \(\sqrt{7.29}\) = 2.7.

Question 3. Find the square root of (i) 0.053361 (ii) 0.00053361 (iii) 150.0625 (iv) 0.374544 (v) 610.09 Solution: (i) 0.053361

Hence \(\sqrt{0.053361}\) = 0.231

(ii) 0.00053361

Hence \(\sqrt{0.00053361}\) = 0.231

(iii) 150.0625

Hence \(\sqrt{150.0625}\) = 12.25

(iv) 0.374544

Hence \(\sqrt{0.374544}\) = 0.612

Hence \(\sqrt{610.09}\) = 24.7

Question 4. Find the square root of the following (correct to three decimal places). (i) 7 (ii) 2.5 (iii) \(2 \frac{1}{12}\) (iv) 367 \(\frac{2}{7}\) Solution: (i) 7

Hence \(\sqrt{7}\) = 2.6457 = 2.646 (correct to 3 places of decimal)

Hence \(\sqrt{2.5}\) = 1.5811 = 1.581 (correct to 3 places of decimal)

(iii) \(\sqrt{2 \frac{1}{12}}=\sqrt{\frac{25}{12}}=\sqrt{\frac{5^2}{12}}=\frac{5}{\sqrt{12}}\)

Hence, \(\sqrt{2 \frac{1}{12}}=\frac{5}{3.464}\) = 1.443

(iv) 367 \(\frac{2}{7}\) \(\sqrt{367 \frac{2}{7}}=\sqrt{\frac{2571}{7}}\)

Hence, \(\sqrt{367 \frac{2}{7}}=\frac{50.705}{2.646}\) = 19.163 (correct to three places of decimal)

Question 5. Estimate the value of the following to the nearest one decimal place. (i) \(\sqrt{90}\) (ii) \(\sqrt{150}\) (iii) \(\sqrt{600}\) (iv) \(\sqrt{1000}\) Solution: (i) The perfect square near 90 are 81 and 100. ∴ 81 < 90 < 100 ⇒ 9 2 < 90 < 10 2 So we can estimate the value between 9 and 10. Let us try with 9.5. ∴ 9.5 × 9.5 = 90.25 > 90 ∴ 9 2 < 90 < (9.5) 2 Now 9.4 × 9.4 = 88.36 < 90 Hence, the estimated value of \(\sqrt{90}\) is 9.48.

(ii) The perfect square number near 150 are 144 and 169. ∴ 144 < 150 < 169 ⇒ 12 2 < 150 < 13 2 So we can estimate the value of \(\sqrt{150}\) between 12 and 13. Let us try for 12.5. ∴ 12.5 × 12.5 = 156.25 > 150 and 12.4 < 12.4 = 153.76 12.3 × 12.3 = 151.29 > 150 12.2 × 12.2 = 148.84 < 150 ∴ 12.2 2 < 150 < 12.4 2 ∴ 12 2 < 150 < 12.5 2 ∴ 12 2 < 150 < 12.4 2 ∴ 12 2 < 150 < 151.29 ∴ 148.84 < 150 < 151.29 ∴ Estimated value of \(\sqrt{150}\) = 12.25.

(iii) The perfect squares near 600 are 576 and 625. ∴ 576 < 600 < 625 ⇒ 24 2 < 600 < 25 2 Let us try for 24.5. ∴ 24.5 × 24.5 = 600.25 > 600 ∴ 24 2 < 600 < 24.5 2 24.495 2 = 600.005 Hence, the estimated value of \(\sqrt{600}\) = 24.495

(iv) The perfect square numbers near 1000 are 961 and 1024. ∴ 961 < 1000 < 1024 ⇒ 31 2 < 1000 < 32 2 Let us try for 31.5 ∴ 31.5 × 31.5 = 992.25 < 1000 and 31.6 × 31.6 = 998.56 < 1000 then 31.7 × 31.7 = 1004.89 > 1000 ∴ 31.6 2 < 1000 < 31.7 2 31.62 2 = 999.82 < 1000 Hence, the estimated value of \(\sqrt{1000}\) is 31.63.

Question 6. Devika has a square piece of cloth of area 9 m 2 and she wants to make 16 square-shaped scarves of equal size out of it. What should possibly be the length of the side of the scarf that can be made out of this piece? Solution: Area of 1 square shapped scarf = \(\frac{9}{16}\) m 2

side of the scarf = \(\sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}}\) = \(\frac{\sqrt{3^2}}{\sqrt{4^2}}=\frac{3}{4}\) m.

Hence, the length of the scarf = \(\frac {3}{4}\) m.

Question 7. The area of a square plot is 800 m 2 . Find the estimated length of the side of the plot. Solution: The perfect square numbers near to 800 are 784 and 841. ∴ 784 < 800 < 841 ⇒ 28 2 < 800 < 29 2 Let us check for 28.5. ∴ 28.5 × 28.5 = 812.25 > 800 ∴ 28 2 < 800 < 28.5 2 28.4 × 28.4 = 806.56 < 800 ∴ 28 2 < 800 < 28.4 28.3 × 28.3 = 800.89 ∴ 28 2 < 800 < (28.3) 2 28.2 × 28.2 = 79.524 < 800 ∴ 28.2 2 < 800 < 28.3 2 ∴ Estimated value of \(\sqrt{800}\) = 28.25 Hence, the length of the side of the plate = 28.25 m

DAV Class 8 Maths Chapter 1 Value Based Questions

Question 1. Priya wants to wish her teacher on Teacher’s Day by giving her a self-made greeting card. She chooses a pink coloured square sheet of paper. A side of that paper measures 19.5 cm. (i) Find the area of paper she chooses for the card. (ii) What act of Priya did you like? Solution: (i) ∵ Side of square sheet of paper she chooses for the card = 19.5 cm ∴ Area of the card = (side) 2 = (19.5 cm) 2 = 19.5 cm x 19.5 cm = 380.25 cm 2

(ii) Respect, gratitude and thankful nature.

Question 2. The students of Class-VIII B of a school donated 2,304 for the Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. (i) Find the number of students in VIII B. (ii) What quality of the students do you appreciate here? Solution: (i) Total donation = 2,304 ∴ No. of students in class VIII B = \(\sqrt{2,304}\) = \(\sqrt{48 \times 48}\) = 48.

(ii) The students exhibit responsibility and duty towards the nation.

DAV Class 8 Maths Chapter 1 Worksheet 5 Notes

Square root of a rational number:

(i) \(\sqrt{a \times b}=\sqrt{a} \times \sqrt{b}\) (ii) \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\), b ≠ 0

Example 1. Find \(\sqrt{36 \times 81}\) and show that \(\sqrt{36 \times 81}=\sqrt{36} \times \sqrt{81} \) Solution: \(\sqrt{36 \times 81}\) = \(\sqrt{6^2 \times 9^2}\) =\(\sqrt{(6 \times 9)^2}\) = \(\sqrt{(54)^2}\) = 54 Now \(\sqrt{36} \times \sqrt{81}=\sqrt{6^2} \times \sqrt{9^2}\) = 6 × 9 = 54 Hence, \(\sqrt{36 \times 81}=\sqrt{36} \times \sqrt{81} \)

Example 2. Find \(\sqrt{\frac{49}{121}}\) and show that \(\sqrt{\frac{49}{121}}=\sqrt{\frac{49}{121}}\). Solution: \(\sqrt{\frac{49}{121}}\) = \(\sqrt{\frac{7^2}{11^2}}\) = \(\sqrt{\left(\frac{7}{11}\right)^2}\) = \(\frac{7}{11}\) Now \(\frac{\sqrt{49}}{\sqrt{121}}=\frac{\sqrt{7^2}}{\sqrt{11^2}}\) = \(\frac{7}{11}\) Hence, \(\sqrt{\frac{49}{121}}=\sqrt{\frac{49}{121}}\)

Example 3. Find the square root of 1\(\frac{123}{361}\). Solution: \(\sqrt{1 \frac{123}{361}}\) = \(\sqrt{\frac{484}{361}}\) = \(\frac{\sqrt{484}}{\sqrt{361}}\) = \(\frac{\sqrt{22^2}}{\sqrt{19^2}}\) = \(\frac{22}{19}\) [∵ \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)]

Squares of terminating decimals:

Example 4. Find the squares of the following rational numbers: (i) 0.32 (ii) 0.2 (iii) 0.008 Solution: (i) (0.32) 2 = 0.1024 (ii) (0.25) 2 = 0.0625 (iii) (0.008) 2 = 0.000064 (iv) (0.0001) 2 = 0.0000001

Square roots of numbers in decimal form (which are perfect squares):

Example 5. Find the square root of the following decimal numbers: (i) 17.64 (ii) 147.1369 Solution:

Hence, \(\sqrt{17.64}\) = 4.2

Hence, \(\sqrt{147.1369}\) = 12.13.

Approximate value of the square root of natural numbers (Which are not perfect Squares)

Example 6. Find the square root of 5 upto three decimal places. Solution:

Hence, √5 = 2.236 upto 3 decimal place.

Example 7. Find the square root of 3\(\frac{3}{5}\) correct to two places of decimals. Solution: 3\(\frac{3}{5}\) = \(\frac{18}{5}\) = 3.6 ∴ \(\sqrt{3 \frac{3}{5}}=\sqrt{3.6}\)

Hence, \(\sqrt{3 \frac{3}{5}}\) = 1.897 = 1.90