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# NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Concepts of Squares and Square Roots Class 8 Chapter 5 - FREE PDF Download

NCERT Chapter square and square roots for class 8, you will learn about the different techniques used to determine whether a given natural number is a perfect square or not.  The Chapter also covers various methods for finding the square roots of square numbers. If you are familiar with exponents, understanding the concept of square roots will be easy. It's recommended to review the Chapter before solving the NCERT Solutions for Class 8 Maths square and square roots. Class 8 Maths NCERT Solutions are prepared by Vedantu which contains various problems with square and square roots. Download the free NCERT Solutions for CBSE Class 8 Maths Syllabus to prepare well for exams.

Table of Content
1. NCERT Solutions for Concepts of Squares and Square Roots Class 8 Chapter 5 - FREE PDF Download
2. Glance on Maths Class 8 Chapter 5  Squares and Square Roots
3. Access Exercise wise NCERT Solutions for Chapter 5 Maths Class 8
4. Exercises Under NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots
5. Access NCERT Solutions for Class 8 Maths Chapter 5 – Squares and Square Roots
5.1Exercise 5.1
6. Overview of Deleted Syllabus for CBSE Class 8 Maths Chapter 5  Squares and Square Roots
7. Class 8 Maths Chapter 5: Exercises Breakdown
8. Other Study Material for CBSE Class 8 Maths Chapter 5
9. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

## Glance on Maths Class 8 Chapter 5  Squares and Square Roots

• Properties of Squares include :

• Understanding the characteristics and properties of square numbers.

• Recognizing patterns in squares, such as the ending digits of perfect squares.

• Perfect Squares involve :

• Identifying perfect squares from 1 to 100.

• Learning shortcuts and tricks to quickly determine if a number is a perfect square.

• Pythagorean Triplets

• Finding Square Roots:

• Prime Factorization Method: Breaking down a number into its prime factors to find its square root.

• This article contains Chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 5 -  Squares and Square Roots, which you can download as PDFs.

• There are four exercises (30 fully solved questions) in class 8th maths Chapter 5 Squares and Square Roots.

## Exercises Under NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots

• Exercise 5.1 Introduces students to square numbers, helping them identify and list perfect squares. The exercise includes basic problems that focus on recognizing and calculating the squares of various numbers. Students learn to distinguish between perfect and non-perfect squares, setting the foundation for more complex concepts.

• Exercise 5.2 Delves into finding square roots using the prime factorization method. It provides practice problems that require students to break down numbers into their prime factors, enabling them to calculate the square roots of given numbers accurately. This exercise reinforces the importance of understanding prime factors and their role in determining square roots.

• Exercise 5.3 Focuses on the long division method for finding square roots. Students are guided through detailed steps to use the long division method effectively, followed by a series of problems that necessitate applying this technique to find square roots. This exercise helps students grasp the procedural aspect of calculating square roots through long division.

• Exercise 5.4 Covers the properties and patterns of square numbers, along with an introduction to Pythagorean triplets. Students explore various properties and recognize patterns in square numbers, such as the ending digits of perfect squares. The exercise also includes problems involving the identification and application of Pythagorean triplets, linking the concept of squares to real-life scenarios and enhancing problem-solving skills.

## Access NCERT Solutions for Class 8 Maths Chapter 5 – Squares and Square Roots

### Exercise 5.1

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. The following numbers are obviously not perfect squares. Give reason.

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. The squares of which of the following would be odd numbers?

(i) 431 (ii) 2826 (iii) 7779 (iv) 82004

Ans: We know that the square of an odd number is odd and the square of an even number is even.

So,

(i) The square root of 431 is an odd number.

(ii) The square root of 2826 is an even number.

(iii) The square root of 7779 is an odd number.

(iv) The square root of 82004 is an even number.

4. Observe the following pattern and find the missing digits.

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

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

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

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

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

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. Observe the following pattern and supply the missing numbers.

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

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

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

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

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

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. Using the given pattern, find the missing numbers.

${{\text{1}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}\text{=}{{\text{3}}^{\text{2}}}$

${{\text{2}}^{\text{2}}}\text{+}{{\text{3}}^{\text{2}}}\text{+}{{\text{6}}^{\text{2}}}\text{=}{{\text{7}}^{\text{2}}}$

${{\text{3}}^{\text{2}}}\text{+}{{\text{4}}^{\text{2}}}\text{+1}{{\text{2}}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}$

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

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

${{\text{6}}^{\text{2}}}\text{+}{{\text{7}}^{\text{2}}}\text{+}{{...}^{\text{2}}}\text{=}{{...}^{\text{2}}}$

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. Without adding, find the sum.

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 5.2

1. Find the square of the following numbers.

i. $\text{32}$

Ans:

$32=30+2$

Squaring on Both sides

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

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:

$35=30+5$

Squaring on Both sides

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

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:

$86=80+6$

Squaring on Both sides

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

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:

$93=90+3$

Squaring on Both sides

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

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:

$71=70+1$

Squaring on Both sides

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

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:

$46=40+6$

Squaring on Both sides

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

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 5.3

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. Without doing any calculations, find the numbers which are surely not perfect squares

i. $\text{153}$

Ans:

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

We can see that $153$ has its unit place digit as $3$.

Hence, $153$cannot be a perfect square.

ii. $\text{257}$

Ans:

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

We can see that $257$ has its unit place digit as $7$.

Hence, $257$cannot be a perfect square.

iii. $\text{408}$

Ans:

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

We can see that $408$ has its unit place digit as $8$.

Hence, $408$cannot be a perfect square.

iv. $\text{441}$

Ans:

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

We can see that $441$ has its unit place digit as $1$.

Hence, $441$is a perfect square.

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

Ans:

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

For $\sqrt{100}$

1. $100-1=99$

2. $99-3=96$

3. $96-5=91$

4. $91-7=84$

5. $84-9=75$

6. $75-11=64$

7. $64-13=51$

8. $51-15=36$

9. $36-17=19$

10. $19-19=0$

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

Hence, $\sqrt{100}=10$

For $\sqrt{169}$

1. $169-1=168$

2. $168-3=165$

3. $165-5=160$

4. $160-7=153$

5. $153-9=144$

6. $144-11=133$

7. $133-13=120$

8. $120-15=105$

9. $105-17=88$

10. $88-19=69$

11.  $69-21=48$

12.  $48-23=25$

13.  $25-25=0$

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

Hence, $\sqrt{169}=13$

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

i. $\text{729}$

Ans:

The factorization of $729$ is as follows:

 $3$ $729$ $3$ $243$ $3$ $81$ $3$ $27$ $3$ $9$ $3$ $3$ $1$

$729=\underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$

$\sqrt{729}=3\times 3\times 3$

So, $\sqrt{729}=27$

ii. $\text{400}$

Ans:

The factorization of $400$ is as follows:

 $2$ $400$ $2$ $200$ $2$ $100$ $2$ $50$ $5$ $25$ $5$ $5$ $1$

$400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}$

$\sqrt{400}=2\times 2\times 5$

So, $\sqrt{400}=20$

iii. $\text{1764}$

Ans:

The factorization of $1764$ is as follows:

 $2$ $1764$ $2$ $882$ $3$ $441$ $3$ $147$ $7$ $49$ $7$ $7$ $1$

$1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$

$\sqrt{1764}=2\times 3\times 7$

So, $\sqrt{1764}=42$

iv. $\text{4096}$

Ans:

The factorization of $4096$ is as follows:

 $2$ $4096$ $2$ $2048$ $2$ $1024$ $2$ $512$ $2$ $256$ $2$ $128$ $2$ $64$ $2$ $32$ $2$ $16$ $2$ $8$ $2$ $4$ $2$ $2$ $1$

$4096=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}$

$\sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2$

So, $\sqrt{4096}=64$

v. $\text{7744}$

Ans:

The factorization of $7744$ is as follows:

 $2$ $7744$ $2$ $3872$ $2$ $1936$ $2$ $968$ $2$ $484$ $2$ $242$ $11$ $121$ $11$ $11$ $1$

$7744=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{11\times 11}$

$\sqrt{7744}=2\times 2\times 2\times 11$

So, $\sqrt{7744}=88$

vi. $\text{9604}$

Ans:

The factorization of $9604$ is as follows:

 $2$ $9604$ $2$ $4802$ $7$ $2401$ $7$ $343$ $7$ $49$ $7$ $7$ $1$

$9604=\underline{2\times 2}\times \underline{7\times 7}\times \underline{7\times 7}$

$\sqrt{9604}=2\times 7\times 7$

So, $\sqrt{9604}=98$

vii. $\text{5929}$

Ans:

The factorization of $5929$ is as follows:

 $7$ $5929$ $7$ $847$ $11$ $121$ $11$ $11$ $1$

$5929=\underline{7\times 7}\times \underline{11\times 11}$

$\sqrt{5929}=7\times 11$

So, $\sqrt{5929}=77$

viii. $\text{9216}$

The factorization of $9216$ is as follows:

 $2$ $9216$ $2$ $4608$ $2$ $2304$ $2$ $1152$ $2$ $576$ $2$ $288$ $2$ $144$ $2$ $72$ $2$ $36$ $2$ $18$ $3$ $9$ $3$ $3$ $1$

$9216=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}$

$\sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3$

So, $\sqrt{9216}=96$

ix. $\text{529}$

The factorization of $529$ is as follows:

 $23$ $529$ $23$ $23$ $1$

$529=\underline{23\times 23}$

So, $\sqrt{529}=23$

x. $\text{8100}$

The factorization of $8100$ is as follows:

 $2$ $8100$ $2$ $4050$ $3$ $2025$ $3$ $675$ $3$ $225$ $3$ $75$ $5$ $25$ $5$ $5$ $1$

$8100=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}$

$\sqrt{8100}=2\times 3\times 3\times 5$

So, $\sqrt{8100}=90$

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

i. $\text{252}$

Ans:

The factorization of $252$ is as follows:

 $2$ $252$ $2$ $126$ $3$ $63$ $3$ $21$ $7$ $7$ $1$

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

We can see that $7$is not paired

So, we have to multiply $252$ by $7$ to get a perfect square.

The new number will be $252\times 7=1764$

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

$\sqrt{1764}=2\times 3\times 7$

So, $\sqrt{1764}=42$

ii. $\text{180}$

Ans:

The factorization of $180$ is as follows:

 $2$ $180$ $2$ $90$ $3$ $45$ $3$ $15$ $5$ $5$ $1$

Here, $180=\underline{2\times 2}\times \underline{3\times 3}\times 5$

We can see that $5$is not paired

So, we have to multiply $180$ by $5$ to get a perfect square.

The new number will be $180\times 5=900$

$900=\underline{2\times 2}\times \underline{3\times 3}\times \underline{5\times 5}$ which is a perfect square

$\sqrt{900}=2\times 3\times 5$

So, $\sqrt{900}=30$

iii. $\text{1008}$

Ans:

The factorization of $1008$ is as follows:

 $2$ $1008$ $2$ $504$ $2$ $252$ $2$ $126$ $3$ $63$ $3$ $21$ $7$ $7$ $1$

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

We can see that $7$is not paired

So, we have to multiply $1008$ by $7$ to get a perfect square.

The new number will be $1008\times 7=7056$

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

$\sqrt{7056}=2\times 2\times 3\times 7$

So, $\sqrt{7056}=84$

iv. $\text{2028}$

Ans:

The factorization of $2028$ is as follows:

 $2$ $2028$ $2$ $1014$ $3$ $507$ $13$ $169$ $13$ $13$ $1$

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

We can see that $3$is not paired

So, we have to multiply $2028$ by $3$ to get a perfect square.

The new number will be $2028\times 3=6084$

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

$\sqrt{6084}=2\times 3\times 13$

So, $\sqrt{6084}=78$

v. $\text{1458}$

Ans:

The factorization of $1458$ is as follows:

 $2$ $1458$ $3$ $729$ $3$ $243$ $3$ $81$ $3$ $27$ $3$ $9$ $3$ $3$ $1$

Here, $1458=2\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$

We can see that $2$is not paired

So, we have to multiply $1458$ by $2$ to get a perfect square.

The new number will be $1458\times 2=2916$

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

$\sqrt{2916}=2\times 3\times 3\times 3$

So, $\sqrt{2916}=54$

vi. $\text{768}$

Ans:

The factorization of $768$ is as follows:

 $2$ $768$ $2$ $384$ $2$ $192$ $2$ $96$ $2$ $48$ $2$ $24$ $2$ $12$ $2$ $6$ $3$ $3$ $1$

Here, $768=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times 3$

We can see that $3$is not paired

So, we have to multiply $768$ by $3$ to get a perfect square.

The new number will be $768\times 3=2304$

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

$\sqrt{2304}=2\times 2\times 2\times 2\times 3$

So, $\sqrt{2304}=48$

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

i. $\text{252}$

Ans:

The factorization of $252$ is as follows:

 $2$ $252$ $2$ $126$ $3$ $63$ $3$ $21$ $7$ $7$ $1$

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

We can see that $7$is not paired

So, we have to divide $252$ by $7$ to get a perfect square.

The new number will be $252\div 7=36$

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

$\sqrt{36}=2\times 3$

So, $\sqrt{36}=6$

ii. $\text{2925}$

Ans:

The factorization of $2925$ is as follows:

 $3$ $2925$ $3$ $975$ $5$ $325$ $5$ $65$ $13$ $13$ $1$

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

We can see that $13$is not paired

So, we have to divide $2925$ by $13$ to get a perfect square.

The new number will be $2925\div 13=225$

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

$\sqrt{225}=3\times 5$

So, $\sqrt{225}=15$

iii. $\text{396}$

Ans:

The factorization of $396$ is as follows:

 $2$ $396$ $2$ $198$ $3$ $99$ $3$ $33$ $11$ $11$ $1$

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

We can see that $11$is not paired

So, we have to divide $396$ by $11$ to get a perfect square.

The new number will be $396\div 11=36$

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

$\sqrt{36}=2\times 3$

So, $\sqrt{36}=6$

iv. $\text{2645}$

Ans:

The factorization of $2645$ is as follows:

 $5$ $2645$ $23$ $529$ $23$ $23$ $1$

Here, $2645=5\times \underline{23\times 23}$

We can see that $5$is not paired

So, we have to divide $2645$ by $5$ to get a perfect square.

The new number will be $2645\div 5=529$

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

So, $\sqrt{529}=23$

v. $\text{2800}$

Ans:

The factorization of $2800$ is as follows:

 $2$ $2800$ $2$ $1400$ $2$ $700$ $2$ $350$ $5$ $175$ $5$ $35$ $7$ $7$ $1$

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

We can see that $7$is not paired

So, we have to divide $2800$ by $7$ to get a perfect square.

The new number will be $2800\div 7=400$

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

$\sqrt{400}=2\times 2\times 5$

So, $\sqrt{400}=20$

vi. $\text{1620}$

Ans:

The factorization of $1620$ is as follows:

 $2$ $1620$ $2$ $810$ $3$ $405$ $3$ $135$ $3$ $45$ $3$ $15$ $5$ $5$ $1$

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

We can see that $5$ is not paired

So, we have to divide $1620$ by $5$ to get a perfect square.

The new number will be $1620\div 5=324$

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

$\sqrt{324}=2\times 3\times 3$

So, $\sqrt{324}=18$

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

Ans:

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

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

Total amount donated by students is Rs. $2401$

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

$\sqrt{2401}=\sqrt{\underline{7\times 7}\times \underline{7\times 7}}$

$=7\times 7$

$=49$

Thus, there are total $49$ students in the class.

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

Ans:

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

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

Hence,

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

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

The number of rows $\times$ The number of rows $=$ $2025$

The number of rows $=$$\sqrt{2025}$

$\sqrt{2025}=\sqrt{\underline{5\times 5}\times \underline{3\times 3}\times \underline{3\times 3}}$

$=5\times 3\times 3$

$=45$

Thus, the number of rows $=45$ and the number of plants in each row $=45$.

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

Ans:

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

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

 $2$ $4,9,10$ $2$ $2,9,5$ $3$ $1,9,5$ $3$ $1,3,5$ $5$ $1,1,5$ $1,1,1$

L.C.M$=2\times 2\times 3\times 3\times 5$

$=180$

It can be clearly seen that $5$ cannot be paired.

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

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

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

Ans:

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

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

 $2$ $8,15,20$ $2$ $4,15,10$ $2$ $2,15,5$ $3$ $1,15,5$ $5$ $1,5,5$ $1,1,1$

L.C.M$=2\times 2\times 2\times 3\times 5$

$=120$

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

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

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

Exercise 5.4

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

i. $\text{2304}$

Ans:

The square root of $2304$ by division method is calculated as follows:

 $48$ $4$ $\overline{23}\overline{04}$  $-16$ $88$ 704 704 $0$

Hence, $\sqrt{2304}=48$

ii. $\text{4489}$

Ans:

The square root of $4489$ by division method is calculated as follows:

 $67$ $6$ $\overline{44}\overline{89}$$-36$ $127$ 889 889 $0$

Hence, $\sqrt{4489}=67$

iii. $\text{3481}$

Ans:

The square root of $3481$ by division method is calculated as follows:

 $59$ $5$ $\overline{34}\overline{81}$ $-25$ $109$ 981  981 $0$

Hence, $\sqrt{3481}=59$

iv. $\text{529}$

Ans:

The square root of $529$ by division method is calculated as follows:

 $23$ $2$ $\overline{5}\overline{29}$ $-4$ $43$ 129 129 $0$

Hence, $\sqrt{529}=23$

v. $\text{3249}$

Ans:

The square root of $3249$ by division method is calculated as follows:

 $57$ $5$ $\overline{32}\overline{49}$ $-25$ $107$ 749  749 $0$

Hence, $\sqrt{3249}=57$

vi. $\text{1369}$

Ans:

The square root of $1369$ by division method is calculated as follows:

 $37$ $3$ $\overline{13}\overline{69}$  $-9$ $67$ 469 469 $0$

Hence, $\sqrt{1369}=37$

vii. $\text{5776}$

The square root of $5776$ by division method is calculated as follows:

 $76$ $7$ $\overline{57}\overline{76}$  -49 $146$ $overline{57}\overline{76}$ -49 $0$

Hence, $\sqrt{5776}=76$

viii. $\text{7921}$

The square root of $7921$ by division method is calculated as follows:

 $89$ $8$ $\overline{79}\overline{21}$ -64 $169$ 1521  1521 $0$

Hence, $\sqrt{7921}=89$

ix. $\text{576}$

The square root of $576$ by division method is calculated as follows:

 $24$ $2$ $\overline{5}\overline{76}$  -4 $44$ 176176 $0$

Hence, $\sqrt{576}=24$

x. $\text{1024}$

The square root of $1024$ by division method is calculated as follows:

 $32$ $3$ $\overline{10}\overline{24}$ -9 $62$ 124124 $0$

Hence, $\sqrt{1024}=32$

xi. $\text{3136}$

The square root of $3136$ by division method is calculated as follows:

 $56$ $5$ $\overline{31}\overline{36}$ -25 $106$ 636636 $0$

Hence, $\sqrt{3136}=56$

xii. $\text{900}$

The square root of $900$ by division method is calculated as follows:

 $30$ $3$ $\overline{9}\overline{00}$-9 $60$ 0000 $0$

Hence, $\sqrt{900}=30$

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

i. $\text{64}$

Ans:

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

After placing bars, we get

$64=\overline{64}$

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

ii. $\text{144}$

Ans:

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

After placing bars, we get

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

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

iii. $\text{4489}$

Ans:

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

After placing bars, we get

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

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

iv. $\text{27225}$

Ans:

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

After placing bars, we get

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

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

v. $\text{390625}$

Ans:

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

After placing bars, we get

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

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

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

i. $\text{2}\text{.56}$

Ans:

The square root of $2.56$ by division method is calculated as follows:

 $1.6$ $1$ $\overline{2}.\overline{56}$   -1 $26$ 156156 $0$

Hence, $\sqrt{2.56}=1.6$

ii. $\text{7}\text{.29}$

Ans:

The square root of $7.29$ by division method is calculated as follows:

 $2.7$ $2$ $\overline{7}.\overline{29}$  -4 $47$ 329329 $0$

Hence, $\sqrt{7.29}=2.7$

iii. $\text{51}\text{.84}$

Ans:

The square root of $51.84$ by division method is calculated as follows:

 $7.2$ $7$ $\overline{51}.\overline{84}$ -49 $142$ 284284 $0$

Hence, $\sqrt{51.84}=7.2$

iv. $\text{42}\text{.25}$

Ans:

The square root of $42.25$ by division method is calculated as follows:

 $6.5$ $6$ $\overline{42}.\overline{25}$ -36 $125$ 625625 $0$

Hence, $\sqrt{42.25}=6.5$

v. $\text{31}\text{.36}$

Ans:

The square root of $31.36$ by division method is calculated as follows:

 $5.6$ $5$ $\overline{31}.\overline{36}$ -25 $106$ 636636 $0$

Hence, $\sqrt{31.36}=5.6$

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

i. $\text{402}$

Ans:

The square root of $402$ by division method is calculated as follows:

 $20$ $2$ $\overline{4}\overline{02}$-4 $40$ 0200 $2$

We are getting a remainder $2$.

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

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

Hence, the required perfect square is $402-2=400$

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

ii. $\text{1989}$

Ans:

The square root of $1989$ by division method is calculated as follows:

 $44$ $4$ $\overline{19}\overline{89}$-16 $84$ 389336 $53$

We are getting a remainder $53$.

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

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

Hence, the required perfect square is $1989-53=1936$

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

iii.$\text{3250}$

Ans:

The square root of $3250$ by division method is calculated as follows:

 $57$ $5$ $\overline{32}\overline{50}$ -25 $107$ 750749 $1$

We are getting a remainder $1$ .

This means that the square of 57 is less than 3250 by $1$.

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

Hence, the required perfect square is $3250-1=3249$

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

iv. $\text{825}$

Ans:

The square root of $825$ by division method is calculated as follows:

 $28$ $2$ $\overline{8}\overline{25}$-4 $48$ 425384 $41$

We are getting a remainder $41$.

This means that the square of 28 is less than 825 by $41$.

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

Hence, the required perfect square is $825-41=784$

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

v. $\text{4000}$

Ans:

The square root of $4000$ by division method is calculated as follows:

 $63$ $6$ $\overline{40}\overline{00}$-36 $123$ 400  369 $31$

We are getting a remainder $31$.

This means that the square of 63 is less than 4000 by $31$.

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

Hence, the required perfect square is $4000-31=3969$

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

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

i. $\text{525}$

Ans:

The square root of $4000$ by division method is calculated as follows:

 $22$ $2$ $\overline{5}\overline{25}$-4 $42$ 12584 $41$

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

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

i.e.,

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

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

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

So, $525+4=529$

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

ii. $\text{1750}$

Ans:

The square root of $1750$ by division method is calculated as follows:

 $41$ $4$ $\overline{17}\overline{50}$ -16 $81$ 15081 $69$

We are getting a remainder $69$.

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

The next number is 42 and its square is  ${{42}^{2}}=1764$

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

Hence, the required perfect square is $1750+14=1764$

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

iii. $\text{252}$

Ans:

The square root of $252$ by division method is calculated as follows:

 $15$ $1$ $\overline{2}\overline{52}$-1 $25$ 152125 $27$

We are getting a remainder $27$.

This means that the square of 15 is less than $252$.

The next number is 16 and its square is  ${{16}^{2}}=256$

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

Hence, the required perfect square is $252+4=256$

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

iv. $\text{1825}$

Ans:

The square root of $1825$ by division method is calculated as follows:

 $42$ $4$ $\overline{18}\overline{25}$-16 $82$ 225164 $61$

We are getting a remainder $61$.

This means that, the square of 42 is less than $1825$

The next number is 43 and its square is  ${{43}^{2}}=1849$

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

Hence, the required perfect square is $1825+24=1849$

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

v. $\text{6412}$

Ans:

The square root of $6412$ by division method is calculated as follows:

 $80$ $8$ $\overline{64}\overline{12}$-64 $160$ 0120 $12$

We are getting a remainder $12$.

This means that the square of 80 is less than $6412$

The next number is 81 and its square is  ${{81}^{2}}=6561$

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

Hence, the required perfect square is $6412+149=6561$

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

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

Ans:

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

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

We get x$\times$

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

 $21$ $2$ $\overline{4}\overline{41}$-4 $41$ 04141 $0$

We get x=21 m

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

7. In a right triangle ABC, $\angle {\text{B = }}{90^ \circ }$

a) Find AC if AB=6 cm, BC=8 cm

Ans:

It is given that triangle ABC is right angled at B

So, on applying Pythagoras Theorem, we get

AC2=AB2+BC2

AC2$={{6}^{2}}+{{8}^{2}}$

$=36+64$

$=100$

AC$=\sqrt{10\times 10}$

Thus, AC=10 cm.

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

Ans:

It is given that triangle ABC is right angled at B

So, on applying Pythagoras Theorem, we get

AC2=AB2+BC2

AB2=AC2-BC2

AB2$={{13}^{2}}-{{5}^{2}}$

$=169-25$

$=144$

AB$=\sqrt{12\times 12}$

Thus, AB=12 cm.

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

Ans:

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

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

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

The square root of $1000$ by long division method is calculated as follows:

 $31$ $3$ $\overline{10}\overline{00}$-9 $61$ 10061 $39$

We are getting a remainder $39$.

This means that the square of 31 is less than $1000$

The next number is 32 and its square is  ${{32}^{2}}=1024$

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

Hence, the required number of plants is 24.

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

Ans:

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

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

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

The square root of $1000$ by long division method is calculated as follows:

 $22$ $2$ $\overline{5}\overline{00}$-4 $42$ 10084 $16$

We are getting a remainder $16$.

This means that the square of 22 is less than $500$ by $16$

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

Hence, the required perfect square is $500-16=484$

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

## Overview of Deleted Syllabus for CBSE Class 8 Maths Chapter 5  Squares and Square Roots

 Chapter Dropped Topics Squares and Square Roots Estimating Square Root

## Class 8 Maths Chapter 5: Exercises Breakdown

 Exercise Number of Questions Exercise 5.1 9 Question and Solutions Exercise 5.2 2 Questions and Solutions Exercise 5.3 10 Questions and Solutions Exercise 5.4 9 Questions and Solutions

## Conclusion

NCERT Class 8 maths Chapter 5 solutions concludes by highlighting how crucial it is to understand the ideas behind squares and square roots. Understanding how to calculate squares and square roots of numbers, both by hand and via prime factorization, is essential. To learn these skills, students should concentrate on working through a variety of issues. 8-12 questions from this Chapter have been asked on tests in the past. Students that practise these kinds of questions will gain confidence and perform well on tests.

## Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the Chapter-wise NCERT Solutions for Class 8 Maths. Go through these Chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots

1. What are the topics covered in Chapter 5 for class 8, Square and Square Roots?

In chapter 5 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.

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

There are four exercises in Class 8 Chapter 5. 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 5 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 5 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.

8. Where are square roots used in class 8 maths ch 5?

In squares and square roots class 8 solutions there are many uses for square roots in many different domains. Here are a few examples:

• Geometry: The Pythagorean theorem, which determines the length of a missing side in a right-angled triangle, depends on square roots.

• Distance: When computing the distance between two points, square roots are needed, particularly when using the distance formula.

• Areas and Volumes: When working with diagonals or irrational numbers—that is, numbers that cannot be stated as a straightforward fraction—square roots can be utilised to determine the areas and volumes of forms.

• Physics: Square roots can be found in many formulas related to physics, such as those that compute velocity, or speed with direction, or simple harmonic motion.

• Engineering: Square roots are utilised in many different engineering applications, like electrical circuit design and stress calculations in materials.

• Finance: Interest rates and the cumulative return on investments can be computed using square roots.

• Computer Science: Algorithms for database searches, data sorting, and computer graphics occasionally employ square roots.

9. Who was the first mathematician to use the symbol √ for square root?

Christoff Rudolff was the first mathematician to adopt the sign √ for square root in his book "Die Coß" (published in 1525). This book is regarded as the original algebraic text written in German.