# Important Questions Square and Square roots class 8

## Important Questions Class 8 Mathematics Chapter 6 – Squares And Square Roots

Chapter 6 of Class 8 Mathematics is about ‘Squares and Square Roots’. In this chapter Squares and square roots, both concepts are exactly opposite to each other. Squares are the numbers produced after multiplying a value by itself. At the same time, the square root of a number is a value that, on getting multiplied by itself, gives its original value. Hence, both are converse and inverse methods. For example, the square of 2 is 4, and at the same time, the square root of 4 is 2.

If n is a particular number, then its square is represented by n raised to the power 2, i.e., n2 and its square root is expressed as ‘√n’, where ‘√’ is called radical. The value under the root symbol is said to be radicand.

The square numbers are widely explained in terms of the area of a square shape. The shape of a square has all its sides equal. Therefore, the area of the square is equal to (side x side) or side2. Hence, if the square’s side length is 3cm, its area is 32= 9 sq. cm.

Properties of Square Numbers

The square numbers are the values produced when we multiply a number by itself. Some of the properties are:

• A square of 1 is equal to 1
• Square of positive numbers are positive
• A square of negative numbers is also positive. For example, (-3)2 = 9
• The square of zero is zero
• The square of a number’s root equals the value under the root. For example, (√3)2 = 3
• The unit place of the square of any even number will have an even number only.
• If a number has 1 or 9 in the unit’s place, its square ends in 1.
• If a number has 4 or 6 in the unit’s place, its square ends in 6.

Extramarks is one of the best learning platforms for students to achieve excellent scores. It helps students by providing them with important questions from the CBSE syllabus. The Extramarks team refers to NCERT books, CBSE sample papers, CBSE revision notes, CBSE past years’ questions, etc., and provides you with the solutions to important questions. Students can get a good hold on this chapter by solving our Mathematics Class 8 Chapter 6 important questions.

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Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

### Important Questions Class 8 Mathematics Chapter 6 – With Solutions

Given below are a few of the questions and their answers from our Chapter 6 Class 8 Mathematics important questions.

 CBSE Class 8 Maths Important Questions Sr No. Chapters Chapters Name 1 Chapter 1 Rational Numbers 2 Chapter 2 Linear Equations in One Variable 3 Chapter 3 Understanding Quadrilaterals 4 Chapter 4 Practical Geometry 5 Chapter 5 Data Handling 6 Chapter 6 Squares and Square Roots 7 Chapter 7 Cubes and Cube Roots 8 Chapter 8 Comparing Quantities 9 Chapter 9 Algebraic Expressions and Identities 10 Chapter 10 Visualising Solid Shapes 11 Chapter 11 Mensuration 12 Chapter 12 Exponents and Powers 13 Chapter 13 Direct and Inverse Proportions 14 Chapter 14 Factorisation 15 Chapter 15 Introduction to Graphs 16 Chapter 16 Playing with Numbers

Question 1:Show that the sum of two consecutive natural numbers is 13².

Answer 1:Let 2n + 1 = 13

So, n = 6

So, ( 2n + 1)² = 4n² + 4n + 1

= (2n² + 2n) + (2n² + 2n + 1)

Substitute n = 6,

(13)² = ( 2 x 6² + 2 x 6) + (2 x 6² + 2 x 6 + 1)

= (72 + 12) + (72 + 12 + 1)

= 84 + 85

Question 2: What would be the square root of the number 625 using the identity

(a +b)² = a² + b² + 2ab?

= (600 + 25)²

= 600² + 2 x 600 x 25 +25²

= 360000 + 30000 + 625

= 390625

Question 4:Use the following identity and find the square of 189.

(a – b)² = a² – 2ab + b²

Answer 4: 189 = (200 – 11) 2

= 40000 – 2 x 200 x 11 + 112

= 40000 – 4400 + 121

= 35721

Question 5: Find the smallest whole number from which 1008 should be multiplied in order to obtain a perfect square number. Also, find out the square root of the square number so obtained.

Let us factorise the number 1008. 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

= ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 ) × 7

Here, 7 cannot be paired.

Therefore, we will multiply 1008 by 7 to get a perfect square.

New number so obtained = 1008 ×7 = 7056

Now, let us find the square root of 7056 7056 = 2 × 2 × 2 × 2 × 3 × 3× 7 × 7

7056 = (2 ×  2 ) × ( 2 × 2 ) × ( 3 × 3 ) ×( 7 × 7 )

7056 = 2² × 2² × 3² × 7²

7056 = (2 × 2 × 3 × 7)²

Therefore;

√7056 = 2×2×3×7 = 84

Question 6: The sides of a right triangle whose hypotenuse is 17cm are _________ and _________.

For each natural number, m > 1, 2m, m2 –1 and m2 + 1 form a Pythagorean triplet.

Now,

m² + 1 = (2m)² + (m2 – 1)²

Where,

m² + 1 = 17

m² = 17 – 1

m² = 16

m = √16

m = 4

Then,

2m = 2 × 4

= 8

And,

m² – 1 = 4² – 1

= 16 -1

= 15

Question 7: √(1.96) = _________.

= √(1.96)

= √(196/100)

= √((14 × 14)/(10 × 10))

= √(142 / 102)

= 14/10

= 1.4

Question 8:If m is the required square of a natural number given by n, then n is

(a) the square of m

(b) greater than m

(c) equal to m

(d) √m

n² = m

Then,

= n = √m

Question 9:There are _________ perfect squares between 1 and 100.

Answer 9: There are 8 perfect squares between 1 and 100.

2 × 2 = 4

3 × 3 = 9

4 × 4 = 16

5 × 5 = 25

6 × 6 = 36

7 × 7 = 49

8 × 8 = 64

9 × 9 = 81

Question 10: By what least number should the given number be divided to get a perfect square number? In each of the following cases, find the number whose square is the new number 1575.

Answer 10: A method for determining the prime factors of a given number, such as a composite number, is known as prime factorisation.

Given 1575,

Resolve 1575 into prime factors, we get

1575 = 3 X 3 X 5 X 5 X 7 = (3² X 5² X 7)

To obtain a perfect square, we have to divide the above equation by 7

Then we get, 3380 = 3 X 3 X 5 X 5

New number = (9 X 25) = (3² X 5² )

Taking squares on both sides of the above equation, we get

∴ New number = (3 X 5)² = (15)²

Therefore, the new number is a square of 15

Question 11:Show that each of the numbers is a perfect square. In each case, find the number whose square is the given number:

7056

A perfect square is always expressed as a product of pairs of prime factors.

Resolving 7056 into prime factors, we obtain

7056 = 11 X 539

= 12 X 588

= 12 X 7 X 84

= 84 X 84

= (84)²

Thus, 84 is the number whose square is 5929

Therefore,7056 is a perfect square.

Question 12: Without adding, find the sum of the following:

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

As per the given property of perfect square, for any natural number n, we

have some of the first n odd natural numbers=n²

But here n=12

By applying the above the law, we get

thus, (1+3+5+7+9+11+13+15+17+19+21+23) = 12² = 144

Question 13: Using the formula (a – b)²=(a² – 2ab + b²), evaluate:

(196)²

We have (a – b)² = (a² – 2ab + b²)

(196)² can be written as 200-4

So here, a=200 and b=4

Using the formula,

(200 – 4)² = (200² – 2 X 200 X 4 + 4² )

On simplifying, we get

(200 – 4)² = (40000 – 1600 +16)

(196)² =38416

Question 14: By what least number should the number be divided to obtain a number with a perfect square? In this, in each case, find the number whose square is the new number 4851.

Answer 14: The number is a perfect square if and only if the prime factorization creates pairs; it is not exactly a perfect square if it is not paired up.

Given 4851,

Resolving 4851 into prime factors, we obtain

4851 = 3 X 3 X 7 X 7 X 11

= (32 X 72 X 11)

To obtain a perfect square, we need to divide the above equation by

11

we obtain, 9075 = 3 X 3 X 7 X 7

The new number = (9 X 49)

= (3² X 7² )

Taking squares on both sides from the above equation, we obtain

∴ The new number = (3 X 7)²

= (21)²

Therefore, the new number is a square of 21

Question 15: By what least number should the number be divided to obtain a perfect square number? In such a case, find the number whose square is the new number 4500.

Answer 15: The number is exactly a perfect square if and only if the prime factorization creates pairs; or else, it is not a perfect square number.

As per the given 4500,

Resolving 4500 into prime factors, we obtain

3380 = 2 X 2 X 3 X 3 X 5 X 5 X 5

= (2² X 3² X 5² X 5)

To obtain a perfect square, we need to divide the above equation by 5

Then we obtain, 4500 = 2 X 2 X 3 X 3 X 5 X 5

The new number = (4 X 9 X 25)

= (2² X 3² X 5² )

Taking squares on both sides from the above equation, we obtain

∴ The new number = (2 X 3 X 5)²

= (30)²

Therefore, the new number is a square of 30

Question 16: Write a Pythagorean triplet whose one member is:

(i) 6

(ii) 14

(iii) 16

(iv) 18

Answer 16: Any natural number m, 2m, m2–1, m2+1 is a Pythagorean triplet.

(i) 2m = 6

m = 6/2

m = 3

m²–1= 3² – 1 = 9–1 = 8

m²+1= 3²+1 = 9+1 = 10

Thus, (6, 8, 10) is a Pythagorean triplet.

(ii) 2m = 14

⇒ m = 14/2 = 7

m²–1= 7²–1 = 49–1 = 48

m²+1 = 7²+1 = 49+1 = 50

Thus, (14, 48, 50) is not a Pythagorean triplet.

(iii) 2m = 16

⇒ m = 16/2 = 8

m²–1 = 8²–1 = 64–1 = 63

m²+ 1 = 8²+1 = 64+1 = 65

Thus, (16, 63, 65) is a Pythagorean triplet.

(iv) 2m = 18

⇒ m = 18/2 = 9

m²–1 = 9²–1 = 81–1 = 80

m²+1 = 9²+1 = 81+1 = 82

Thus, (18, 80, 82) is a Pythagorean triplet.

Question 17: How many numbers lie between the

squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100

Answer 17:  As we know, between n² and (n+1)², the number of non–perfect square numbers are 2n.

(i) Between 122 and 132, there are 2×12 = 24 natural numbers.

(ii) Between 252 and 262, there are 2×25 = 50 natural numbers.

(iii) Between 992 and 1002, there are 2×99 =198 natural numbers.

Question 18: 2025 plants are to be planted in a garden in a way that each of the rows contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Let the number of rows be x.

Thus, the number of plants in each row = x.

Total many contributed by all the students = x × x = x²

Given, x² = Rs.2025

x2 = 3×3×3×3×5×5

⇒ x2 = (3×3)×(3×3)×(5×5)

⇒ x2 = (3×3×5)×(3×3×5)

⇒ x2 = 45×45

⇒ x = √(45×45)

⇒ x = 45

Therefore,

Number of rows = 45

Number of plants in each row = 45

Question 19: The digit at the one’s place of the number 572 is _________.

Answer 19: The digit at the one’s place of the number 572 is 9.

We see that 3 or 7 at the unit’s place ends in 9.

= 572

= 57 × 57

Question 20: Give a reason to show that the number given below is a perfect square:

5963

Answer 20:The unit digit of the square numbers will be 0, 1, 4, 5, 6, or 9 if we examine the squares of numbers from 1 to 10. Thus, the unit digit for all perfect squares will be 0, 1, 4, 5, 6, or 9, and none of the square numbers will end in 2, 3, 7, or 8.

Given 5963

We have the property of a perfect square, i.e. a number ending in 3 is never

a perfect square.

Therefore the given number 5963 is not a perfect square.

### Benefits Of Solving Important Questions Class 8 Mathematics Chapter 6

Mathematics requires a lot of practice. The Mathematics of Classes 8, 9, and 10 forms the basic  knowledge for Class 11 and Class 12 Mathematics. We suggest students access our platform Extramarks to access important questions Class 8 Mathematics Chapter 6. By rigorously solving questions and going through all the necessary solutions, students will eventually instil confidence in themselves to resolve any complex problems that come their way from this particular chapter of square and square roots.

Given below are a few benefits of regularly solving the questions from our question bank of Important Questions Class 8 Mathematics Chapter 6:

• Our expert Mathematics subject teachers have carefully assembled the most Important Questions in Class 8 Mathematics Chapter 6 by analysing the past years’  questions.
• The questions and solutions provided are based on the NCERT books and strictly follow the  latest CBSE syllabus and guidelines. So the students can completely count on it.
• The questions covered in our set of Important Questions in Class 8 Mathematics Chapter 6 are based on various topics covered in square and square roots. It is recommended that students go  through it and clear their confusion or doubts before solving these important questions.
• By solving our Important Questions Class 8 Mathematics Chapter 6, students will get a thorough idea of how the paper will be set. Solving questions similar to the exam questions would help the students to definitely perform better in their upcoming exams, and get excellent scores. .

Extramarks believes in incorporating joyful learning experiences through its own repository of resources. Extramarks provides a range of learning solutions for students from Class 1 to Class 12. We have other study resources on our website, along with important questions and solutions. Students can also click on the below-mentioned links and access some of these resources:

• NCERT books
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• Important formulas
• CBSE extra questions

Q.1 What is the value of 942?

Marks:1

A. 8836

B. 10000

C. 9836

D. 7836

Ans

8836

Q.2 Find the square root of 0.9 correct up to three places of decimal.

Marks:4
Ans

$\begin{array}{l}\begin{array}{cc}& 0.9486\\ \begin{array}{l}9\\ 9\end{array}& \begin{array}{l}0.90000000\\ \text{}81\end{array}\\ \begin{array}{l}18\\ \text{}8\end{array}& \begin{array}{l}900\\ 736\end{array}\\ \begin{array}{l}26\\ \text{}6\end{array}& \begin{array}{l}\text{}16400\\ \text{}15104\end{array}\\ \begin{array}{l}32\\ \text{}2\end{array}& \begin{array}{l}\text{}129600\\ \text{}113796\end{array}\\ & \text{}15804\end{array}\\ \sqrt{0.9}=0.9486\dots \\ \mathrm{Upto}\text{three dec}\mathrm{i}\text{mal places}\\ \sqrt{0.9}=0.949\end{array}$

Q.3 Find the square root of 3136 by division method.

Marks:2
Ans ## 1. Are there 2 answers to a square root?

The square root of a number always has two answers, one is positive and the other is negative.

## 2. What are the important chapters in Class 8 Mathematics Chapter 6?

The NCERT Mathematics book has 16 chapters. When it comes to grasping the fundamentals and taking the test, each chapter is equally important. Additionally, CBSE does not specify chapter weightage, and each chapter must be completely understood  in order to receive a good grade.

## 3. Where can I get important questions for Class 8 Mathematics Chapter 6 online?

On the Extramarks website, you can find all of the important questions for Class 8 Mathematics Chapter 6, along with their answers. On the website, you can also find important questions and NCERT solutions for all classes. This is true for other subjects as well.

## 4. Give a list of all chapters from Class 8 Mathematics.

Class 8 Mathematics covers important topics from Arithmetic,  Algebra and Geometry.  Here is a list of all 16 chapters from Class 8 Mathematics.

• Chapter 1- Rational Numbers
• Chapter 2 – Linear Equations in One Variable
• Chapter 3 – Understanding Quadrilaterals
• Chapter 4 – Practical Geometry
• Chapter 5 – Data Handling
• Chapter 6 – Square and Square Roots
• Chapter 7 – Cube and Cube Roots
• Chapter 8 – Comparing Quantities
• Chapter 9 – Algebraic Expressions and Identities
• Chapter 10 – Visualising Solid Shapes
• Chapter 11- Mensuration
• Chapter 12 – Exponents and Powers
• Chapter 13 – Direct and Inverse Proportions
• Chapter 14 – Factorisation
• Chapter 15 – Introduction to Graphs
• Chapter 16 – Playing with Numbers