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

Exercise 6.1

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

i. $\text{81}$

Ans:

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

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

ii. $\text{272}$

Ans:

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

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

iii. $\text{799}$

Ans:

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

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

iv. $\text{3853}$

Ans:

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

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

v. $\text{1234}$

Ans:

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

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

vi. $\text{26387}$

Ans:

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

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

vii. $\text{52698}$

Ans:

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

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

viii. $\text{99880}$

Ans:

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

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

ix. $\text{12796}$

Ans:

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

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

x. $\text{55555}$

Ans:

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

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

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

i. $\text{1057}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$. Also, a perfect square has an even number of zeroes at the end of it.

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

Hence, $1057$cannot be a perfect square.

ii. $\text{23453}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$Also, a perfect square has an even number of zeroes at the end of it.

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

Hence, $23453$ cannot be a perfect square.

iii. $\text{7928}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$Also, a perfect square has an even number of zeroes at the end of it.

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

Hence, $7928$ cannot be a perfect square.

iv. $\text{222222}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$Also, a perfect square has an even number of zeroes at the end of it.

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

Hence, $222222$ cannot be a perfect square.

v. $\text{64000}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$ Also, a perfect square has an even number of zeroes at the end of it.

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

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

vi. $\text{89722}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$Also, a perfect square has an even number of zeroes at the end of it.

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

Hence, $89722$ cannot be a perfect square.

vii. $\text{222000}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$Also, a perfect square has an even number of zeroes at the end of it.

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

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

viii. $\text{505050}$

Ans:

The square of numbers may end with any one of the digits $0$, $1$\[,4\], $5$, $6$, or $9$Also, a perfect square has an even number of zeros at the end of it.

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

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

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

i. $\text{243}$

Ans:

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

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

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

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

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

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

ii. $\text{256}$

Ans:

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

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

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

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

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

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

iii. $\text{72}$

Ans:

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

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

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

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

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

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

iv. $\text{675}$

Ans:

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

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

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

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

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

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

v. $\text{100}$

Ans:

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

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

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

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

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

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

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

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

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

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

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

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

Ans:

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

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

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

Hence,

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

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

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

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

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

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

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

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

Ans:

It can be observed from the given pattern that:

• the square of the numbers has odd number of digits 

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

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

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

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

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

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

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

Hence,

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

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

6. Find the missing numbers using the given pattern.

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

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

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

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

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

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

Ans:

It can be observed from the given pattern that:

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

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

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

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

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

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

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

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

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

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

7. Find the sum without adding.

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

Ans:

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

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

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

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

Ans:

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

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

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

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

Ans:

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

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

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

8.

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

Ans: 

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

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

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

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

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

Ans: 

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

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

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

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

9.

How many numbers lie between squares of the following numbers?

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

Ans: 

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

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

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

Ans: 

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

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

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

Ans: 

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

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

Exercise 6.2

1. Find the square of the following numbers.

i. \[\text{32}\]

Ans:

\[32=30+2\]

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

Since, (a+b)2=a2 +2ab+b2 

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

\[=900+120+4\]

\[=1024\]

ii. \[\text{35}\]

Ans:

\[35=30+5\]

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

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

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

\[=900+300+25\]

\[=1225\]

iii. \[\text{86}\]

Ans:

\[86=80+6\]

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

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

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

\[=6400+960+36\]

\[=7396\]

iv. \[\text{93}\]

Ans:

\[93=90+3\]

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

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

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

\[=8100+540+9\]

\[=8649\]

v. \[\text{71}\]

Ans:

\[71=70+1\]

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

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

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

\[=4900+140+1\]

\[=5041\]

vi. \[\text{46}\]

Ans:

\[46=40+6\]

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

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

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

\[=1600+480+36\]

\[=2116\]

2. Write a Pythagoras triplet whose one number is

i. \[\text{6}\]

Ans:

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

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

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

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

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

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

So let 2m\[=6\]

Then we get, m\[=3\]

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

\[=9-1\]

\[=8\]

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

\[=9+1\]

\[=10\]

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

ii. \[\text{14}\]

Ans:

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

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

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

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

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

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

So let 2m=14 

Then we get, m=7 

Now, m2-1=72-1

\[=49-1\]

\[=48\]

Similarly, m2+1=72+1 

\[=49+1\]

\[=50\]

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

iii. \[\text{16}\]

Ans:

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

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

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

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

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

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

So let 2m=16 

Then we get, m=8 

Now, m2-1=82-1

\[=64-1\]

\[=63\]

Similarly, m2+1=82+1

\[=64+1\]

\[=65\]

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

iv. \[\text{18}\]

Ans:

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

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

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

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

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

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

So let 2m=18 

Then we get, m=9 

Now, m2-1=92-1

\[=81-1\]

\[=80\]

Similarly, m2+1=92-1

\[=81+1\]

\[=82\]

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

Exercise 6.3

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

i. \[\text{9801}\]

Ans: We know that the one’s digit of the square root of the number ending with \[1\] can be \[1\] or \[9\].

Thus, the possible one’s digit of the square root of \[9801\] is either \[1\] or \[9\].

ii. \[\text{99856}\]

Ans: We know that the one’s digit of the square root of the number ending with  \[6\] can be \[6\] or \[4\].

Thus, the possible one’s digit of the square root of \[99856\] is either \[6\] or \[4\].

iii. \[\text{998001}\]

Ans: We know that the one’s digit of the square root of the number ending with  \[1\] can be \[1\] or \[9\].

Thus, the possible one’s digit of the square root of \[998001\] is either \[1\] or \[9\].

iv. \[\text{657666025}\]

Ans: We know that the one’s digit of the square root of the number ending with  \[5\] will be \[5\] .

Thus, the only possible one’s digit of the square root of \[657666025\] is \[5\] .

2. Find the numbers which are surely not perfect squares without doing any calculations.

i. \[\text{153}\]

Ans:

The perfect square of numbers may end with any one of the digits $0$, $1$ ,\[4\], $5$, $6$, or $9$. Also, a perfect square has even number of zeroes at the end of it, if any.

We can see that \[153\] has its unit place digit as \[3\].

Hence, \[153\]cannot be a perfect square.

ii. \[\text{257}\]

Ans:

The perfect square of numbers may end with any one of the digits $0$, $1$ ,\[4\], $5$, $6$, or $9$. Also, a perfect square has even number of zeroes at the end of it, if any.

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

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

iii. \[\text{408}\]

Ans:

The perfect square of numbers may end with any one of the digits $0$, $1$ ,\[4\], $5$, $6$, or $9$. Also, a perfect square has even number of zeroes at the end of it, if any.

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

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

iv. \[\text{441}\]

Ans:

The perfect square of numbers may end with any one of the digits $0$, $1$ ,\[4\], $5$, $6$, or $9$. Also, a perfect square has even number of zeroes at the end of it, if any.

We can see that \[441\] has its unit place digit as \[1\].

Hence, \[441\]is a perfect square.

3. Find the square roots of \[\text{100}\] and \[\text{169}\] by the method of repeated subtraction.

Ans:

It is already known to us that the sum of the first n odd natural numbers is n2.

For \[\sqrt{100}\]

1. \[100-1=99\]

2. \[99-3=96\]

3. \[96-5=91\]

4. \[91-7=84\]

5. \[84-9=75\]

6. \[75-11=64\]

7. \[64-13=51\]

8. \[51-15=36\]

9. \[36-17=19\]

10. \[19-19=0\]

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

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

For \[\sqrt{169}\]

1. \[169-1=168\]

2. \[168-3=165\]

3. \[165-5=160\]

4. \[160-7=153\]

5. \[153-9=144\]

6. \[144-11=133\]

7. \[133-13=120\]

8. \[120-15=105\]

9. \[105-17=88\]

10. \[88-19=69\]

11.  \[69-21=48\]

12.  \[48-23=25\]

13.  \[25-25=0\]

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

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

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

i. \[\text{729}\]

Ans:

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

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

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

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

ii. \[\text{400}\]

Ans:

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

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

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

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

iii. \[\text{1764}\]

Ans:

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

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

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

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

iv. \[\text{4096}\]

Ans:

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

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

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

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

v. \[\text{7744}\]

Ans:

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

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

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

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

vi. \[\text{9604}\]

Ans:

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

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

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

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

vii. \[\text{5929}\]

Ans:

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

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

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

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. \[\text{252}\]

Ans: 

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

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

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

ii. \[\text{180}\]

Ans:

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. \[\text{1008}\]

Ans:

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

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

We can see that \[7\]is not paired

So, we have to multiply \[1008\] by \[7\] to get a perfect square.

The new number will be \[1008\times 7=7056\]

\[7056=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}\] which is a perfect square

\[\sqrt{7056}=2\times 2\times 3\times 7\]

So, \[\sqrt{7056}=84\]

iv. \[\text{2028}\]

Ans:

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

Here, \[2028=\underline{2\times 2}\times 3\times \underline{13\times 13}\]

We can see that \[3\]is not paired

So, we have to multiply \[2028\] by \[3\] to get a perfect square.

The new number will be \[2028\times 3=6084\]

\[6084=\underline{2\times 2}\times \underline{3\times 3}\times \underline{13\times 13}\] which is a perfect square

\[\sqrt{6084}=2\times 3\times 13\]

So, \[\sqrt{6084}=78\]

v. \[\text{1458}\]

Ans:

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

Here, \[1458=2\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}\]

We can see that \[2\]is not paired

So, we have to multiply \[1458\] by \[2\] to get a perfect square.

The new number will be \[1458\times 2=2916\]

\[2916=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}\] which is a perfect square

\[\sqrt{2916}=2\times 3\times 3\times 3\]

So, \[\sqrt{2916}=54\]

vi. \[\text{768}\]

Ans:

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

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

We can see that \[3\]is not paired

So, we have to multiply \[768\] by \[3\] to get a perfect square.

The new number will be \[768\times 3=2304\]

\[2304=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\] which is a perfect square

\[\sqrt{2304}=2\times 2\times 2\times 2\times 3\]

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

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

i. \[\text{252}\]

Ans:

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

Here, \[252=\underline{2\times 2}\times \underline{3\times 3}\times 7\]

We can see that \[7\]is not paired

So, we have to divide \[252\] by \[7\] to get a perfect square.

The new number will be \[252\div 7=36\]

\[36=\underline{2\times 2}\times \underline{3\times 3}\] which is a perfect square

\[\sqrt{36}=2\times 3\]

So, \[\sqrt{36}=6\]

ii. \[\text{2925}\]

Ans:

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

Here, \[2925=\underline{3\times 3}\times \underline{5\times 5}\times 13\]

We can see that \[13\]is not paired

So, we have to divide \[2925\] by \[13\] to get a perfect square.

The new number will be \[2925\div 13=225\]

\[225=\underline{3\times 3}\times \underline{5\times 5}\] which is a perfect square

\[\sqrt{225}=3\times 5\]

So, \[\sqrt{225}=15\]

iii. \[\text{396}\]

Ans:

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

Here, \[396=\underline{2\times 2}\times \underline{3\times 3}\times 11\]

We can see that \[11\]is not paired

So, we have to divide \[396\] by \[11\] to get a perfect square.

The new number will be \[396\div 11=36\]

\[36=\underline{2\times 2}\times \underline{3\times 3}\] which is a perfect square

\[\sqrt{36}=2\times 3\]

So, \[\sqrt{36}=6\]

iv. \[\text{2645}\]

Ans:

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

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

We can see that \[5\]is not paired

So, we have to divide \[2645\] by \[5\] to get a perfect square.

The new number will be \[2645\div 5=529\]

\[529=\underline{23\times 23}\] which is a perfect square

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

v. \[\text{2800}\]

Ans:

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

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

We can see that \[7\]is not paired

So, we have to divide \[2800\] by \[7\] to get a perfect square.

The new number will be \[2800\div 7=400\]

\[400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}\] which is a perfect square

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

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

vi. \[\text{1620}\]

Ans:

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

Here, \[1620=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times 5\]

We can see that \[5\] is not paired

So, we have to divide \[1620\] by \[5\] to get a perfect square.

The new number will be \[1620\div 5=324\]

\[324=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\] which is a perfect square

\[\sqrt{324}=2\times 3\times 3\]

So, \[\sqrt{324}=18\]

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Ans:

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

We can find the number of students in the class by doing the square root of the total amount donated by the students of Class VIII.

Total amount donated by students is Rs. $2401$

Then, the number of students in the class will be \[\sqrt{2401}\]

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

\[=7\times 7\]

\[=49\]

Thus, there are total \[49\] students in the class.

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Ans:

According to the plants are being planted in a garden in such a way that each row contains as many plants as the number of rows.

So, the number of rows will be equal to the number of plants in each row.

Hence,

The number of rows \[\times \] Number of plants in each row \[=\]Total number of plants

The number of rows \[\times \] Number of plants in each row \[=\] \[2025\]

The number of rows \[\times \] The number of rows \[=\] \[2025\]

The number of rows \[=\]\[\sqrt{2025}\]

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

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

\[=45\]

Thus, the number of rows \[=45\] and the number of plants in each row \[=45\].

9. Find the smallest square number that is divisible by each of the numbers \[\text{4,9}\] and \[\text{10}\].

Ans:

We know that the number that is perfectly divisible by each one of \[4,9\] and \[10\] is their L.C.M

So, taking the L.C.M of these numbers 

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

\[=180\]

It can be clearly seen that \[5\] cannot be paired.

Therefore, we have to multiply \[180\] by \[5\] in order to get a perfect square.

Thus, the smallest square number divisible by \[4,9\] and \[10\]\[=180\times 5=900\]

10. Find the smallest square number that is divisible by each of the numbers \[\text{8,15}\] and \[\text{20}\].

Ans:

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

So, taking the L.C.M of these numbers 

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

\[=120\]

It can be clearly seen that the prime factors \[2\],\[3\] and \[5\] cannot be paired.

Therefore, we have to multiply \[120\] by \[2\],\[3\] and \[5\] in order to get a perfect square.

Thus, the smallest square number divisible by \[8,15\] and \[20\]is \[120\times 2\times 3\times 5\]\[=3600\]

Exercise 6.4

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

i. \[\text{2304}\]

Ans:

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

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

ii. \[\text{4489}\]

Ans:

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

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

iii. \[\text{3481}\]

Ans:

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

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

iv. \[\text{529}\]

Ans:

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

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

v. \[\text{3249}\]

Ans:

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

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

vi. \[\text{1369}\]

Ans:

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. \[\text{64}\]

Ans:

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

After placing bars, we get

\[64=\overline{64}\] 

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

ii. \[\text{144}\]

Ans:

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

After placing bars, we get

\[144=\overline{1}\overline{44}\] 

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

iii. \[\text{4489}\]

Ans:

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

After placing bars, we get

\[4489=\overline{44}\overline{89}\] 

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

iv. \[\text{27225}\]

Ans:

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

After placing bars, we get

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

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

v. \[\text{390625}\]

Ans:

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

After placing bars, we get

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

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

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

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

Ans:

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

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

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

Ans:

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

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

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

Ans:

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

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

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

Ans:

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

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

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

Ans:

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

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

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

i. \[\text{402}\]

Ans:

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

We are getting a remainder \[2\].

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

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

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

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

ii. \[\text{1989}\]

Ans:

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

We are getting a remainder \[53\].

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

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

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

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

iii.\[\text{3250}\]

Ans:

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

We are getting a remainder \[1\] .

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

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

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

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

iv. \[\text{825}\]

Ans:

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

We are getting a remainder \[41\].

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

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

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

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

v. \[\text{4000}\]

Ans:

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

We are getting a remainder \[31\].

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

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

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

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

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

i. \[\text{525}\]

Ans:

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

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

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

i.e.,

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