# NCERT Solutions Class 8 Maths Chapter 6

## NCERT Solutions for Class 8 Mathematics Chapter 6: Square and Square Roots

Mathematics is a subject which requires a deeper understanding of the concepts and problem solving skills . In addition, it also has complex theorems and formulas. Thus, students of class 8 have to prepare with more focus to gain an advantage over the subject. For students interested in this subject, it opens many opportunities in Engineering and Science. Therefore, it is a critical learning stage for them, as it prepares them for the upcoming opportunities in life.

Class 8 Mathematics Chapter 6: Square and Square Roots class 8 is an essential chapter that introduces concepts like finding squares of big numbers by expansion techniques and other patterns observed in square numbers. . It sheds light on square numbers and square roots, their properties, and some exciting patterns.

Extramarks’ NCERT Solutions for Class 8 Mathematics Chapter 6 are available for the students on the website or app of Extramarks. It covers detailed and accurate solutions to questions given in the NCERT textbook, which will help them understand how to solve the different kinds of problems in a step-by- step manner . It consists of 30 questions, of which 21 are medium to hard, four are moderately easy, and 5 are long answer-type questions. Our solution guide revolves around many essential questions that can also appear in the entrance exams.

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For the latest updates and exam-related news, students can visit our Extramarks’ website. Further, they can also refer to other class solutions such as NCERT Solutions Class 10, NCERT Solutions Class 11, and NCERT Solutions Class 12.

## Key Topics Covered in NCERT Solutions for Class 8 Mathematics Chapter 6:

The Squares and Square Roots Class 8 – discusses the properties of square numbers. Here, students will learn how to find the sum and product of consecutive even and odd natural numbers and also about  Pythagorean triplets. Our NCERT Solutions for Class 8 Mathematics Chapter 6 outline all the formulas for finding square roots with the help of repeated subtraction and through prime factorization and long division method to find the square roots of numbers.

Some of the key topics featured in NCERT Solution for Class 8 Mathematics Chapter 6 are:

### Square of a number

If any number is raised to exponent two, it is known as the square of the base. Hence, 25 is referred to as the square of 5, while 64 is the square of 8. One can find the square of a number by multiplying the given number two times. For example, 5 Square = 5 x 5 = 25, whereas 8 Square = 8 x 8 = 64.

If we find the square of a whole number, the resultant number is a perfect square. Numbers such as perfect squares are 4, 9, 16, 25, 36, 49, 64, and so on. Therefore, a square of a number is always a positive number.

To understand the applications of the square of a number, students can refer to NCERT Solutions for Class 8 Mathematics Chapter 6.

### Perfect Squares

A perfect square is a number expressed as the product of an integer itself or as the second exponent of an integer. For example, if 25 is a perfect square, it is the product of integer five by itself, 5 x 5 = 25. On the other hand, a number such as 21 is not a perfect square number because it is not eligible to be expressed as the product of two exact integers.

Therefore, a perfect square is a positive integer. It can be obtained by multiplying an integer by itself. In simpler words, one can say that perfect squares are numbers that are the products of integers. Therefore, one can express a perfect square as x2, where x is an integer and 2 is the value of a perfect square.

### Properties of Square Numbers

Properties of square numbers explained in NCERT Solutions for Class 8 Mathematics Chapter 6 are:

• Numbers such as 0, 1, 4, 5, 6, or 9 in the unit’s place may or may not be a square.
• Numbers such as 2, 3, 7, or 8 are in their unit place then it is not a square number.
• The number has 1 or 9 in the unit’s place; its square ends in 1.
• If a square number ends in 6, the number will have 4 or 6 in the unit place.

### Square Roots

The inverse operation of squaring a number is known as the square root of a number. The value is obtained when we multiply the number by itself.

We can calculate the square root of a number by finding a number that delivers the original number when it is squared. For example, if ‘a’ is the square root of ‘b’, it means that a x a = b. It is said that the square of any number is always a positive number. Hence, every number has two square roots, one of a positive value and one of a negative value.

There are few example elaborated in NCERT Solutions for Class 8 Mathematics Chapter 6:

Numbers such as 2 and -2 are square roots of 4. So in most places, only the positive value is written in the square root of a number.

### Square Root of a Negative Number

A square root of a negative number does not give a negative result. The answer is either positive or zero. Besides, complex numbers have a solution to finding a negative number’s square root. The square root of -x is: √(-x)= i√x. Therefore, i is the square root of -1.

For example: If we make a perfect square number like 16. Then, the square root of -16 will not have the real square root of -16. √(-16)= √16 × √(-1) = 4i (as, √(-1)= i), ‘i’ is denoted as the square root of -1.

### Finding the square of a number using identity

It is not that difficult to find the square root of a number which is a perfect square. These are the squares whose positive numbers can be expressed as the product of a number itself. In simple words, perfect squares are numbers expressed as any integer’s value of power 2. There are four methods explained in NCERT Solutions for Class 8 Mathematics Chapter 6 are as follows:

• Repeated Subtraction Method of Square Root
• Square Root by Prime Factorization Method
• Square Root by Estimation Method
• Square Root by Long Division Method

The first three methods are usually used for perfect squares, while the fourth method, the long division method, is mainly used for any number, whether it is a perfect square or not. To understand how these different methods are used to find the square of a number, students can refer to NCERT Solutions for Class 8 Mathematics Chapter 6.

### Interesting Patterns

There are many exciting patterns observed in square and perfect square roots:

• Numbers between square numbers
• A sum of consecutive natural numbers
• Product of two consecutive even or odd natural numbers

### NCERT Solutions for Class 8 Mathematics Chapter 6: Exercise & Solutions

The solutions are prepared by subject experts who have years of experience in teaching. All the answers are stated stepwise for quick retention.

Every answer of every chapter in NCERT Solutions for Class 10 Science is written as per the CBSE guidelines. As the explanations are comprehensive, the fundamentals of the chapter are understood by the students in a better way. The answers in NCERT solutions are explained in detail, which gives students an idea of how to attempt a question in the board exam in the right manner

NCERT Class 8 Mathematics Chapter 6 introduces 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 introduces the various methods for finding the square roots of square numbers. Students can click the links below for specific questions and solutions:

Along with this, students can also refer to other solutions for primary and secondary classes:

## NCERT Exemplar for Class 8 Mathematics:

Mathematics interests many students, but it is equally challenging as it involves complex problem solving. Many students face difficulty in remembering the concepts and theories. Thus, it is beneficial for them to solve different questions of different a difficulty levels. In addition, it helps students remember the formulas and their use at various stages of problems.

NCERT Exemplar consists of all solved textual questions  that are explained in easy to understand language. Going through these questions and answers helps in building  a strong foundation on all the concepts. Most of the questions asked in the annual examination are from NCERT books. Further, exemplars offer multiple-choice questions, descriptive-type questions, and objective-type questions.

### Key Features of NCERT Solutions for Class 8 Mathematics Chapter 6:

Class 8 Mathematics discusses the properties of square numbers and odd natural numbers. Some of the essential topics such as some interesting patterns, Pythagorean triples, square roots, and square roots of decimals. Our Extramarks NCERT Solutions for Class 8 Mathematics Chapter 6 cover all essential topics and concepts.

The key benefits of NCERT Solutions for Class 8 Mathematics Chapter 6 are listed here:

• The solutions have well-curated examples and solutions to exercise questions covering all the essential concepts of the topic.
• Students can first  solve the problems and then cross check their answers from the given solutions.
• Students can solve questions of different difficulty levels, hence fully preparing themselves for the exams.
• By accessing other NCERT study material like important questions from the chapter, frequently asked questions, MCQs, past years’ papers, expert tips & tricks to grasp answers on Extramarks app or website, students can leverage higher marks in their examination.
• The additional questions provided would enable students to practise thoroughly.

Q.1 What will be the unit digit of the squares of the following numbers?(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555

Ans

1. 81- Here, the number ends with 1.So its square will end with (1×1=1).Therefore, the unit digit of the square of 81 will be 1.
2. 272 – Here, the number ends with 2.So its square will end with (2×2=4).Therefore, the unit digit of the square of 272 will be 4.
3. 799 – Here, the number ends with 9.So its square will end with (9×9= 81).Therefore, the unit digit of the square of 799 will be 1.
4. 3853- Here, the number ends with 3.So its square will end with (3×3=9).Therefore, the unit digit of the square of 3853 will be 9.
5. 1234 – Here, the number ends with 4. So its square will end with (4×4=16).Therefore, the unit digit of the square of 1234 will be 6.
6. 26387- Here, the number ends with 7. So its square will end with (7×7=49).Therefore, the unit digit of the square of 23387 will be 9.
7. 52698 – Here, the number ends with 8. So its square will end with (8×8=64).Therefore, the unit digit of the square of 52698 will be 4
8. 99880- Here, the number ends with 0. So its square will end with (0×0=0).Therefore, the unit digit of the square of 99880 will be 0.
9. 12796 – Here, the number ends with 6. So its square will end with (6×6=36).Therefore, the unit digit of the square of 12796 will be 6.
10. 55555- Here, the number ends with 5. So its square will end with (5×5=25). Therefore, the unit digit of the square of 55555 will be 5.

Q.2 The following numbers are obviously not perfect squares. Give reason.

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

Ans

A perfect square of any number ends with any of the digits 0, 1,4,5,6 or 9 at its unit’s place. Also, a perfect square will end with even number of zeroes.

1. 1057: Since the unit digit of 1057 is 7.Therefore, it is not a perfect square.
2. 23453: Since the unit digit of 23453 is 3.Therefore, it is not a perfect square.
3. 7928: Since the unit digit of 7928 is 8.Therefore, it is not a perfect square.
4. 222222: Since the unit digit of 222222 is 2. Therefore, it is not a perfect square.
5. 64000: A perfect square should have even number of zeroes at the end but 64000 is having three zeroes at the end which is not an even number. Therefore, it is not a perfect square.
6. 89722 : Since the unit digit of 89722 is 2.Therefore,it is not a perfect square.
7. 222000 : A perfect square should have even number of zeroes at the end but 222000 is having three zeroes at the end which is not an even number. Therefore, it is not a perfect square.
8. A perfect square should have even number of zeroes at the end but 505050 is having one zero at the end which is not an even number. Therefore, it is not a perfect square.

Q.3 The squares of which of the following would be odd numbers?

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

Ans

Here, 431 and 7779 are odd numbers and 2826 and 82004 are even numbers

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

Thus, the squares of 431 and 7779 will be odd numbers.

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

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1 ……… 2 ……… 1

100000012 = ………………………

Ans

We can observe that, the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,

1000012 = 10000200001

100000012 = 100000020000001

Q.5 Observe the following pattern and supply the missing numbers.

112 = 1 2 1

1012 = 1 0 2 0 1

101012 = 102030201

10101012 = ………………………

…………2 = 10203040504030201

Ans

We can observe that, the squares of the given numbers have the same number of zeroes after every digit as it was in the original number.

Therefore,

10101012 = 1020304030201

1010101012 = 10203040504030201

Hence, the missing numbers are:

1020304030201, 1010101012

Q.6 Using the given pattern, find the missing numbers.

12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + _2 = 212

52 + _2 + 302 = 312

62 + 72 + _2 = __2

Ans

Clearly, the third number is the product of the first two numbers and the fourth number can be obtained by adding 1 to the third number.

Thus, the missing numbers in the pattern will be as follows.

$\begin{array}{l}{\text{4}}^{\text{2}}+{\text{5}}^{\text{2}}+\underset{¯}{{20}^{2}}={\text{21}}^{\text{2}}\\ {\text{5}}^{\text{2}}+\underset{¯}{{6}^{2}}+\text{3}{0}^{\text{2}}={\text{31}}^{\text{2}}\\ {\text{6}}^{\text{2}}+{\text{7}}^{\text{2}}+\underset{¯}{{42}^{2}}=\underset{¯}{{43}^{2}}\end{array}$

Q.7 Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Ans

1. Since, the sum of first n odd natural numbers is n2.Therefore,1 + 3 + 5 + 7 + 9 = (5)2 = 25.

2. Since, the sum of first n odd natural numbers is n2.Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100.

3. Since, the sum of first n odd natural numbers is n2.Therefore,1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)2 = 144

Q.8 (i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

Ans

(i) We know that the sum of first n odd natural numbers is n2.Therefore, 49 is the sum of first 7 odd natural numbers.

49 = (7)2

49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) We know that the sum of first n odd natural numbers is n2.Therefore, 121 is the sum of first 11 odd natural numbers.

121 = (11)2

121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Q.9 How many numbers lie between squares of the following numbers?

(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Ans

We know that there will be 2n numbers in between the squares of the numbers n and (n + 1).

(i) Between 122 and 132, there will be 2 × 12 = 24 numbers

(ii) Between 252 and 262, there will be 2 × 25 = 50 numbers

(iii) Between 992 and 1002, there will be 2 × 99 = 198 numbers

Q.10 Find the square of the following numbers.

(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{}{32}^{2}={\left(30\text{}+\text{}2\right)}^{2}\\ =\left(30\text{}+\text{}2\right)\left(30\text{}+\text{}2\right)\\ =30\left(30\text{}+\text{}2\right)+2\left(30\text{}+\text{}2\right)\\ =30×30+2×30+2×30+\text{}2×2\\ =900+120+4\\ =\text{}1024\\ \\ \left(\mathrm{ii}\right){35}^{2}={\left(30+5\right)}^{2}\\ =\left(30+5\right)\left(30+5\right)\\ =30\left(30+5\right)+5\left(30+5\right)\\ =30×30+5×30+5×30+\text{}5×5\\ =900+150+150+25\\ =\text{}1225\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{}{86}^{2}={\left(80+6\right)}^{2}\\ =\left(80+6\right)\left(80+6\right)\\ =80\left(80+6\right)+6\left(80+6\right)\\ =80×80+6×80+6×80+\text{}6×6\\ =6400+480+480+36\\ =7396\\ \text{}\\ \left(\mathrm{iv}\right)\text{}{93}^{2}={\left(90+3\right)}^{2}\\ =\left(90+3\right)\left(90+3\right)\\ =90\left(90+3\right)+3\left(90+3\right)\\ =90×90+3×90+3×90+3×3\\ =8100+270+270+9\\ =8649\\ \\ \left(\mathrm{v}\right)\text{}{71}^{2}\text{}={\left(70+1\right)}^{2}\\ =\left(70+1\right)\left(70+1\right)\\ =70\left(70+1\right)+1\left(70+1\right)\\ =70×70+1×70+1×70+1×1\\ =4900+70+70+1\\ =5041\end{array}$

$\begin{array}{l}\left(\mathrm{vi}\right)\text{}{46}^{2}={\left(40+6\right)}^{2}\\ =\left(40+6\right)\left(40+6\right)\\ =40\left(40+6\right)+6\left(40+6\right)\\ =40×40+6×40+6×40+6×6\\ =1600+240+240+36\\ =\text{}2116\end{array}$

Q.11 Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14 (iii) 16 (iv) 18

Ans

$\begin{array}{l}\text{For any natural number}\mathrm{m}>\text{1},\\ \text{2}\mathrm{m},{\mathrm{m}}^{\text{2}}-\text{1},{\mathrm{m}}^{\text{2}}+\text{1 forms a Pythagorean triplet}.\\ \left(\text{i}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{6},\text{then}{\mathrm{m}}^{\text{2}}=\text{5}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\end{array}$

$\begin{array}{l}\text{If we take}{m}^{\text{2}}-\text{1}=\text{6},\text{then}{m}^{\text{2}}=\text{7}\\ \text{Again the value of}m\text{is not an integer}.\\ {\text{So, we try to take m}}^{\text{2}}\text{+ 1 = 6}\text{.}\\ {\text{Again m}}^{\text{2}}\text{= 5 will not give an integer value for m}\text{.}\\ \text{Let 2}m=\text{6}⇒m=\text{3}\\ \text{Therefore},\text{the Pythagorean triplets are}\\ \text{2}×\text{3},{\text{3}}^{\text{2}}-\text{1},{\text{3}}^{\text{2}}+\text{1 or 6},\text{8},\text{and 1}0.\end{array}$

$\begin{array}{l}\left(\text{ii}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{14},\text{then}{\mathrm{m}}^{\text{2}}=\text{13}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{14},\text{then}{\mathrm{m}}^{\text{2}}=\text{15.}\\ \text{Again the value of}\mathrm{m}\text{is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{14}⇒\mathrm{m}=\text{7}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{49}-\text{1}=\text{48 and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{49}+\text{1}=\text{5}0\\ \text{Therefore},\text{the required triplet is 14},\text{48},\text{and 5}0.\end{array}$

$\begin{array}{l}\left(\text{iii}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{16},\text{then}{\mathrm{m}}^{\text{2}}=\text{15}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{16},\text{then}{\mathrm{m}}^{\text{2}}=\text{17}\\ \text{Again the value of}\mathrm{m}\text{\hspace{0.17em}is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{16}⇒\mathrm{m}=\text{8}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{64}-\text{1}=\text{63 and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{64}+\text{1}=\text{65}\\ \text{Therefore},\text{the Pythagorean triplet is 16},\text{63},\text{and 65}.\end{array}$

$\begin{array}{l}\left(\text{iv}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{18},{\mathrm{m}}^{\text{2}}=\text{17}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{18},\text{then}{\mathrm{m}}^{\text{2}}=\text{19}\\ \text{Again the value of}\mathrm{m}\text{\hspace{0.17em}\hspace{0.17em}is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{18}⇒\mathrm{m}=\text{9}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{81}-\text{1}=\text{8}0\text{and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{81}+\text{1}=\text{82}\\ \text{Therefore},\text{the Pythagorean triplet is 18},\text{8}0,\text{and 82}.\end{array}$

Q.12 What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

Ans

(i) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) The one’s digit of the given number is 6. If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) The one’s digit of the given number is 5. If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.

Q.13 Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Ans

A perfect square of any number ends with any of the digits 0, 1,4,5,6 or 9 at its unit’s place. Also, a perfect square will end with even number of zeroes.

(i) Since the number 153 has 3 in its unit’s place, so it is not a perfect square.

(ii) Since the number 257 has 7 in its unit’s place, so it is not a perfect square.

(iii) Since the number 408 has 8 in its unit’s place, so it is not a perfect square.

(iv) Since the number 441 has 1 in its unit’s place and a perfect square number may end with 1, so it is a perfect square.

Q.14 Find the square roots of 100 and 169 by the method of repeated subtraction.

Ans

$\begin{array}{l}{\text{We know that the sum of the first n odd natural numbers is n}}^{\text{2}}\text{.}\\ \text{The square root of 169 can be obtained by the method of}\\ \text{repeated subtraction as follows}.\end{array}$

$\begin{array}{l}\text{Consider:}\\ \\ \left(\text{i}\right)\text{100}\text{}-\text{1}\text{}\text{= 99}\\ \left(\text{ii}\right)\text{99}\text{}-\text{}\text{3}\text{}\text{= 96}\\ \left(\text{iii}\right)\text{96}\text{}-\text{}\text{5}\text{}\text{= 91}\\ \left(\text{iv}\right)\text{91}\text{}-\text{}\text{7}\text{}\text{= 84}\\ \left(\text{v}\right)\text{84}\text{}-\text{}\text{9}\text{}\text{= 75}\\ \left(\text{vi}\right)\text{75}\text{}-\text{}\text{11}\text{}\text{= 64}\\ \left(\text{vii}\right)\text{64}\text{}-\text{13 = 51}\\ \left(\text{viii}\right)\text{51}\text{}-\text{15}\text{}\text{= 36}\\ \left(\text{ix}\right)\text{36}\text{}-\text{17}\text{}\text{= 19}\\ \left(\text{x}\right)\text{19}\text{}-\text{19}\text{}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting from}\\ {\text{1 to 100, and obtained 0 at 10}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\sqrt{\text{100}}\text{=10}\end{array}$

$\begin{array}{l}\text{The square root of 169 can be obtained by the method}\\ \text{of repeated subtraction as follows:}\\ \\ \text{Consider:}\\ \\ \text{(i) 169}-\text{1}\text{= 168}\\ \left(\text{ii}\right)\text{168}-\text{3}\text{= 165}\\ \left(\text{iii}\right)\text{165}-\text{5}\text{= 160}\\ \left(\text{iv}\right)\text{160}-\text{7}\text{= 153}\\ \left(\text{v}\right)\text{153}-19\text{}\text{= 144}\\ \left(\text{vi}\right)\text{144}-\text{11}\text{= 133}\\ \left(\text{vii}\right)\text{133}-\text{13}\text{= 120}\\ \left(\text{viii}\right)\text{120}-\text{15 = 105}\\ \left(\text{ix}\right)\text{105}-\text{17}\text{= 88}\\ \left(\text{x}\right)\text{88}-\text{19}\text{= 69}\\ \left(\text{xi}\right)\text{69}-\text{21}\text{= 48}\\ \left(\text{xii}\right)\text{48}-\text{23}\text{= 25}\\ \left(\text{xiii}\right)\text{25}-\text{25}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting}\\ {\text{from 1 to 169, and obtained 0 at 13}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{\text{169}}\text{=13}\end{array}$

Q.15 Find the square roots of the following numbers by the Prime Factorization Method.

(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100

Ans

$\begin{array}{l}\text{(i)}729\text{can be factorised as follows:}\\ \\ \begin{array}{cc}3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ \text{729}=\underset{¯}{\text{3}×\text{3}}×\underset{¯}{\text{3}×\text{3}}×\underset{¯}{\text{3}×\text{3}}\\ \therefore \sqrt{729}=3×3×3=\text{27}\end{array}$

$\begin{array}{l}\text{(ii)400}\\ \\ \begin{array}{cc}2& 400\\ 2& 200\\ 2& 100\\ 5& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \text{400}=\underset{¯}{\text{2}×\text{2}}×\underset{¯}{\text{2}×\text{2}}×\underset{¯}{\text{5}×\text{5}}\\ \therefore \sqrt{400}=2×2×5=\text{20}\end{array}$

$\begin{array}{l}\text{(iii)1764}\\ \begin{array}{cc}2& 1764\\ 2& 882\\ 3& 441\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \text{1764}=\underset{¯}{\text{2}×\text{2}}×\underset{¯}{\text{3}×\text{3}}×\underset{¯}{\text{7}×\text{7}}\\ \therefore \sqrt{1764}=2×3×7=\text{42}\end{array}$

$\begin{array}{l}\text{(iv)}4096\text{can be factorised as follows:}\\ \\ \begin{array}{cc}2& 4096\\ 2& 2048\\ 2& 1024\\ 2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\\ \\ \text{4096}=\underset{¯}{2×2}×\underset{¯}{\text{2}×\text{2}}×\underset{¯}{\text{2}×\text{2}}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}\\ \therefore \sqrt{4096}=2×2×2×2×2×2=\text{64}\\ \end{array}$

$\begin{array}{l}\text{(vi)}9604\text{can be factorised as follows:}\\ \\ \begin{array}{cc}2& 9604\\ 2& 4802\\ 7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ \text{9604}=\underset{¯}{2×2}×\underset{¯}{\text{7}×\text{7}}×\underset{¯}{\text{7}×\text{7}}\\ \therefore \sqrt{9604}=2×7×7=\text{98}\\ \\ \text{(vii)5929}\\ \\ \begin{array}{cc}7& 5929\\ 7& 847\\ 11& 121\\ 11& 11\\ & 1\end{array}\\ \\ \text{5929}=\underset{¯}{7×7}×\underset{¯}{\text{11}×\text{11}}\\ \therefore \sqrt{5929}=7×11=\text{77}\end{array}$

$\begin{array}{l}\text{(viii) 9216}\\ \\ \begin{array}{cc}2& 9216\\ 2& 4608\\ 2& 2304\\ 2& 1152\\ 2& 576\\ 2& 288\\ 2& 144\\ 2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \text{9216}=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{\text{3}×3}\\ \therefore \sqrt{9216}=2×2×2×2×2×3=\text{96}\\ \\ \text{(ix) 529}\\ \\ \begin{array}{cc}23& 529\\ 23& 23\\ & 1\end{array}\\ \\ \text{529}=\underset{¯}{23×23}\\ \therefore \sqrt{529}=\text{23}\end{array}$

$\begin{array}{l}\text{(x)}8100\\ \\ \begin{array}{cc}2& 8100\\ 2& 4050\\ 3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 8100=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{3×3}×\underset{¯}{5×5}\\ \therefore \sqrt{8100}=2×3×3×5=90\\ \end{array}$

Q.16 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) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768

Ans

$\begin{array}{l}\text{(i) 252 can be factorised as:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{¯}{2×2}×\underset{¯}{3×3}×7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 252 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ 252×7=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{7×7}\\ \\ \therefore \text{252×7 = 1764 is a perfect square}\text{.}\\ \\ \therefore \sqrt{1764}=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{7×7}=2×3×7=42\\ \end{array}$

$\begin{array}{l}\text{(ii)180 can be factorised as:}\\ \\ \begin{array}{cc}2& 180\\ 2& 90\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ 180=\underset{¯}{2×2}×\underset{¯}{3×3}×5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we multiply the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 180 has to be multiplied with 5}\\ \text{to obtain a perfect square}\text{.}\\ \text{180}×5=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{5×5}\\ \therefore \text{180×5 = 900 is a perfect square}\text{.}\\ \therefore \sqrt{900}=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{5×5}=2×3×5=30\\ \\ \text{(iii)1008 can be factorised as:}\\ \\ \begin{array}{cc}2& 1008\\ 2& 504\\ 2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\end{array}$

$\begin{array}{l}1008=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{3×3}×7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 1008 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ \text{1008}×7=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{7×7}\\ \therefore \text{1008×7 = 7056 is a perfect square}\text{.}\\ \therefore \sqrt{7056}=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{7×7}=2×2×3×7=84\\ \\ \text{(iv) 2028 can be factorised as follows:}\\ \begin{array}{cc}2& 2028\\ 2& 1014\\ 3& 507\\ 13& 169\\ 13& 13\\ & 1\end{array}\\ 2028=\underset{¯}{2×2}×\underset{¯}{13×13}×3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2028 has to be multiplied with 3}\\ \text{to obtain a perfect square}\text{.}\\ \text{2028}×3=\underset{¯}{2×2}×\underset{¯}{13×13}×\underset{¯}{3×3}\\ \therefore \text{2028}×3\text{= 6084 is a perfect square}\text{.}\\ \therefore \sqrt{6084}=\underset{¯}{2×2}×\underset{¯}{13×13}×\underset{¯}{3×3}=2×3×13=78\end{array}$

$\begin{array}{l}\text{(v) 1458 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 1458=2×\underset{¯}{3×3}×\underset{¯}{3×3}×\underset{¯}{3×3}\\ \text{Here, the prime factor 2 does not have its pair}\text{.}\\ \text{If we multiply the given number by 2, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 1458 has to be multiplied with 2 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1458}×2=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{3×3}×\underset{¯}{3×3}\\ \\ \therefore \text{1458}×2\text{= 2916 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2916}=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{3×3}×\underset{¯}{3×3}=2×3×3×3=54\\ \\ \end{array}$

$\begin{array}{l}\text{(vi) 768 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 768\\ 2& 384\\ 2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}\\ \\ 768=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 768 has to be multiplied with 3 to obtain}\\ \text{a perfect square}\text{.}\\ \text{768}×3=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{3×3}\\ \\ \therefore \text{768}×3\text{= 2304 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2304}=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{3×3}=2×2×2×2×3=48\end{array}$

Q.17 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) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620

Ans

$\begin{array}{l}\text{(i) 252 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{¯}{2×2}×\underset{¯}{3×3}×7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 252 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{252}÷7=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{¯}{2×2}×\underset{¯}{3×3}\\ \therefore \sqrt{36}=\underset{¯}{2×2}×\underset{¯}{3×3}=2×3=6\end{array}$

$\begin{array}{l}\text{(ii) 2925 can be factorised as follows:}\\ \\ \begin{array}{cc}3& 2925\\ 3& 975\\ 5& 325\\ 5& 65\\ 13& 13\\ & 1\end{array}\\ \\ 2925=\underset{¯}{3×3}×\underset{¯}{5×5}×13\\ \text{Here, the prime factor 13 does not have its pair}\text{.}\\ \text{If we divide the given number by 13, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2925 has to be divided with 13 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2925}÷13=225\text{is a perfect square}\text{.}\\ \text{225 =}\underset{¯}{3×3}×\underset{¯}{5×5}\\ \therefore \sqrt{225}=\underset{¯}{3×3}×\underset{¯}{5×5}=3×5=15\\ \\ \text{(iii) 396 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 396\\ 2& 198\\ 3& 99\\ 3& 33\\ 11& 11\\ & 1\end{array}\\ \end{array}$

$\begin{array}{l}396=\underset{¯}{3×3}×\underset{¯}{2×2}×11\\ \text{Here, the prime factor 11 does not have its pair}\text{.}\\ \text{If we divide the given number by 11, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,396 has to be divided with 11 to obtain}\\ \text{a perfect square}\text{.}\\ \text{396}÷11=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{¯}{2×2}×\underset{¯}{3×3}\\ \therefore \sqrt{36}=\underset{¯}{2×2}×\underset{¯}{3×3}=2×3=6\\ \\ \text{(iv) 2645 can be factorised as follows:}\\ \\ \begin{array}{cc}5& 2645\\ 23& 529\\ 23& 23\\ & 1\end{array}\\ \\ 2645=\underset{¯}{23×23}×5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,2645 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2645}÷5=529\text{is a perfect square}\text{.}\\ \text{529 =}\underset{¯}{23×23}\\ \therefore \sqrt{529}=23\end{array}$

$\begin{array}{l}\text{(v) 2800 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 2800\\ 2& 1400\\ 2& 700\\ 2& 350\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}\\ \\ 2800=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{5×5}×7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2800 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2800}÷7=400\text{is a perfect square}\text{.}\\ \text{400 =}\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{5×5}\\ \therefore \sqrt{400}=\underset{¯}{2×2}×\underset{¯}{2×2}×\underset{¯}{5×5}=2×2×5=20\end{array}$

$\begin{array}{l}\text{(vi) 1620 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1620\\ 2& 810\\ 3& 405\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ \\ 1620=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{3×3}×5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,1620 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1620}÷5=324\text{is a perfect square}\text{.}\\ \text{324 =}\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{3×3}\\ \therefore \sqrt{324}=\underset{¯}{2×2}×\underset{¯}{3×3}×\underset{¯}{3×3}=2×3×3=18\end{array}$

Q.18 The students of Class VIII of a school donated ₹ 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 question, each student donated as many rupees as the number of students of the class.
Therefore, the number of students in the class will be the square root of the amount donated by the students of the class.
The total amount of donation is Rs 2401.

$\begin{array}{l}\text{Number of students in the class}=\sqrt{2400}\\ \\ \begin{array}{cc}7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ 2401=\underset{¯}{7×7}×\underset{¯}{7×7}\\ \therefore \sqrt{2401}=7×7=49\end{array}$

Hence, the number of students in the class is 49.

Q.19 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 question, each row contains as many plants as the number of rows.

Hence,

Number of rows = Number of plants in each row

Also, total number of plants = 2025

Since, the number of rows and number of plants in each row are equal, therefore (Number of rows)2 = 2025

$\begin{array}{l}\therefore \text{Number of rows =}\sqrt{2025}\\ \\ \begin{array}{cc}3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 2025=\underset{¯}{3×3}×\underset{¯}{3×3}×\underset{¯}{5×5}\\ \therefore \sqrt{2025}=3×3×5=45\end{array}$

Hence, the number of rows and the number of plants in each row is 45.

Q.20 Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Ans

The number that will be perfectly divisible by each one of 4, 9, and 10 is their LCM.

$\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{4},\text{9},\text{1}0\\ 2& \text{2},\text{9},\text{5}\\ 3& \text{1},\text{9},\text{5}\\ 3& \text{1},\text{3},\text{5}\\ 5& \text{1},\text{1},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}$

Therefore, LCM of 4, 9, 10 = 2 × 2 × 3 × 3 × 5 =180

Here, prime factor 5 does not have its pair. Therefore, 180 is not a perfect square. If we multiply 180 with 5, then the number will become a perfect square. Therefore, 180 should be multiplied with 5 to obtain a perfect square.

Hence, the required square number is 180 × 5 = 900.

Q.21 Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Ans

The number that will be perfectly divisible by each one of 8, 15, and 20 is their LCM.

$\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{8},\text{15},\text{2}0\\ 2& \text{4},\text{15},\text{1}0\\ 2& \text{2},\text{15},\text{5}\\ 3& \text{1},\text{15},\text{5}\\ 5& \text{1},\text{5},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}$

Therefore, LCM of 8, 15, and 20 = 2 × 2 ×2 × 3 × 5 =120

Here, prime factors 2, 3 and 5 do not have their pairs. Therefore, 120 is not a perfect square. If we multiply 120 with 2×3×5, then the number will become a perfect square. Therefore, 120 should be multiplied with 2×3×5 to obtain a perfect square.

Hence, the required square number is 120×2×3×5 = 3600.

Q.22 Find the square root of each of the following numbers by Division method.

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

Ans

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

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

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

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

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

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

Q.23 Find the number of digits in the square root of each of the following numbers (without any calculation).

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

Ans

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

(i) By placing bars, we obtain

$64=\overline{64}$

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

(ii) By placing bars, we obtain

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

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

(iii) By placing bars, we obtain

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

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

(iv) By placing bars, we obtain

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

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

(v) By placing the bars, we obtain

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

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

Q.24 Find the square root of the following decimal numbers.

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

Ans

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

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

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

Q.25 Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

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

Ans

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

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

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

Q.26 Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

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

Ans

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

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

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

Q.27 Find the length of the side of a square whose area is 441 m2.

Ans

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

Q.28 In a right triangle ABC, ∠B = 90°.

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

Ans

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

Therefore, by applying Pythagoras theorem, we obtain

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

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

Therefore, by applying Pythagoras theorem, we obtain

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

Q.29 A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.

Ans

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

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

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

The square root of 1000 can be calculated as follows:

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

Q.30 There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?

Ans

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

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

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

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

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

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

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### 1. Which are the essential topics under NCERT Solutions for Class 8 Mathematics Chapter 6?

There are various interesting topics in the chapter. Some of the concepts presented in the NCERT solution are listed below:

• Properties of square numbers
• Finding the square of a number
• Patterns in squares
• Pythagorean triplets
• Square roots of decimals
• Estimating square root

### 2. How can students utilise NCERT Solutions for Class 8 Mathematics Chapter 6 more efficiently?

The NCERT solutions offer well-curated solved  examples and exercise questions covering the topic’s essential concepts. Thus, students must first read the content of the chapter properly to gain a good understanding. The highlights given between the chapters should also be focused on. They should solve the exercises and then cross check their answers from  these solutions.

### 3. Are NCERT solutions important?

Yes, they are, as they cover numerous questions along with their answers to guide students with the concepts of   the squares of numbers and derivation of square roots. These concepts are beneficial in attempting extensive and lengthy calculations; they make problem-solving more effortless and smoother. Hence, NCERT solutions are vital in preparing for annual exams and entrance tests.