Important Questions for CBSE Class 8 Maths Chapter 5 - Squares and Square Roots

Very Short Answer Questions (1 Mark)

1. A number ending in __, __, __ or __ is never a perfect square.

Ans: \[2,3,7,8\]

2. If a number divided by 3 leaves a remainder 2, it is not a perfect square. Say true or false.

Ans: True

3. \[{{\mathbf{8}}^{\mathbf{2}}}\]

Ans: ${{8}^{2}}=8\times 8=64$

4. The square of even number is always ___.

Ans: Even

5. The square of odd number is always ___.

Ans: Odd

6. If a number when divided by 4 leaves a remainder 2 or 3, then it is a perfect square. Say true or false.

Ans: False

7. The sum of first $n$ odd natural numbers is ___.

Ans: ${{n}^{2}}$

8. The difference of two perfect squares is a perfect square. Say true or false.

Ans: False

9. The product of two perfect square is a perfect square. Say true or false.

Ans: True

Short Answer Questions (2 Marks)

10. Is \[\mathbf{496}\] a perfect square?

Ans: 

$\begin{align}& 2\left| \!{\underline {\,496 \,}} \right.  \\ & 2\left| \!{\underline {\,148 \,}} \right.  \\  & 2\left| \!{\underline {\,124 \,}} \right.  \\ & 2\left| \!{\underline {\,62 \,}} \right.  \\  & \text{   }31 \\  & \Rightarrow 496=2\times 2\times 2\times 2\times 31 \\ \end{align}$

Hence, the number cannot be expressed as pairs of factors.

11.\[~\mathbf{4000}\] is never a perfect square. Why? Ans: The number $4000$ has three zeros. Hence, it can never be a perfect square.

12. \[\mathbf{{{\left( n+1 \right)}^{2}}-{{n}^{2}}=}\]?

Ans: 

\[\begin{align}  & {{\left( n+1 \right)}^{2}}-{{n}^{2}} \\  & =\left( n+1+n \right)\left( n+1-n \right) \\  & =\left\{ \left( n+1 \right)+n \right\} \\ \end{align}\]

13. Write the general form of Pythagorean triplet using m, n, p.

Ans: For every natural number $m>1$, we have $\left( 2m,{{m}^{2}}-1,{{m}^{2}}+1 \right)$ as Pythagorean triplet.

14. State whether the square of given number is even or odd.

1. \[\mathbf{524}\]

2. \[\mathbf{655}\]

Ans:

1. The square of $524$ is even.

2. The square of $655$ is odd.

15. Write a Pythagorean triplet whose smallest number is \[\mathbf{9}\].

Ans: 

$\begin{align}& {{a}^{2}}=b+c \\ & c-b=1 \\ & \Rightarrow {{9}^{2}}=81=40+41 \\ \end{align}$

So, the Pythagorean triplet is $\left( 9,40,41 \right)$.

Short Answer Questions (3 Marks)

16. Is \[\mathbf{5292}\] a perfect square?

Ans:

$\begin{align}& 2\left| \!{\underline {\,5292 \,}} \right.  \\ & 2\left| \!{\underline {\,2646 \,}} \right.  \\ & 2\left| \!{\underline {\,1323 \,}} \right.  \\  & 3\left| \!{\underline {\, 441 \,}} \right.  \\ & 3\left| \!{\underline {\,147 \,}} \right.  \\ & 7\left| \!{\underline {\,49 \,}} \right.  \\ & \text{   }7 \\ & \Rightarrow 5292=2\times 2\times 2\times 3\times 3\times 7\times 7 \\ \end{align}$

The number cannot be expressed as pairs of factors.

Hence, it is not a perfect square.

17. Solve \[~{{\left( \mathbf{36} \right)}^{\mathbf{2}}}-{{\left( \mathbf{35} \right)}^{\mathbf{2}}}\].

Ans:

${{36}^{2}}-{{35}^{2}}=36+35=71$

18. Without adding, find the sum of \[\mathbf{1}+\mathbf{3}+\mathbf{5}+\mathbf{7}+\mathbf{9}+\mathbf{11}+\mathbf{13}+\mathbf{15}+\mathbf{17}+\mathbf{19}+\mathbf{21}\].

Ans:

$\text{Sum of first }11\text{ odd numbers}={{11}^{2}}=121$

19. \[{{\left( \mathbf{509} \right)}^{\mathbf{2}}}=\]?

Ans:

$\begin{align}& {{509}^{2}} \\ & ={{\left( 500+9 \right)}^{2}} \\ & ={{500}^{2}}+{{9}^{2}}+2\times 500\times 9 \\ & =250000+81+9000 \\ & =259081 \\ \end{align}$

20. Evaluate ${{\left( \mathbf{691} \right)}^{\mathbf{2}}}$.

Ans:

 $\begin{align}& {{691}^{2}} \\ & ={{\left( 700-9 \right)}^{2}} \\ & ={{700}^{2}}+{{9}^{2}}-2\times 700\times 9 \\  & =490000+81-12600 \\  & =477481 \\ \end{align}$

21. Evaluate \[\mathbf{39}\times \mathbf{41}\].

Ans:

 $\begin{align}& 39\times 41 \\ & =\left( 40-1 \right)\left( 40+1 \right) \\ & ={{40}^{2}}-{{1}^{2}} \\ & =1600-1 \\ & =1599 \\ \end{align}$

22. Express \[\mathbf{121}\] as sum of \[\mathbf{11}\] odd numbers.

Ans: We know that,

$\begin{align}& \text{Sum of first }n\text{ odd numbers}={{n}^{2}} \\ & \text{Here, }{{11}^{2}}=121 \\ \end{align}$

Hence, expressing $121$ as sum of $11$ odd numbers,

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

Long Answer Questions (4-5 Marks)

23. By what least number should 720 be multiplied to get a perfect square number? Also, find the number whose square is the new number.

Ans:

 $\begin{align}& 2\left| \!{\underline {\,720 \,}} \right.  \\ & 2\left| \!{\underline {\,360 \,}} \right.  \\ & 2\left| \!{\underline {\,180 \,}} \right.  \\ & 2\left| \!{\underline {\,90 \,}} \right.  \\ & 3\left| \!{\underline {\,45 \,}} \right.  \\ & 3\left| \!{\underline {\,15 \,}} \right.  \\ & \text{    }3 \\  & \Rightarrow 720=2\times 2\times 2\times 2\times 3\times 3\times 5 \\ \end{align}$

Here, $5$ doesn’t have a pair. If the number is multiplied by $5$, $720$ becomes a perfect square.

Hence, the new number formed after multiplying is $720\times 5=3600$.

24. By what least number should \[\mathbf{2527}\] be divided to get a perfect square number? Find the number whose square is the new number.

Ans: 

\[\begin{align}& 7\left| \!{\underline {\,2527 \,}} \right.  \\ & 19\left| \!{\underline {\,361 \,}} \right.  \\  & \text{      }19 \\  & \Rightarrow 2527=7\times 19\times 19 \\ \end{align}\]

Here, $7$ doesn’t have a pair. If the number is divided by $7$, $2527$ becomes a perfect square.

Hence, the new number formed after multiplying is $2527\div 7=361$.

25. Find the largest number of \[\mathbf{4}\] digits which is a perfect square.

Ans: The largest $4$digit number is $9999$.

${{99}^{2}}$ is less than $9999$ by $198$.

Hence, the required number is $9999-198=9801$.

26. Find the sum of \[\mathbf{49}\] by column method.

Ans: Given number$=49$

$a=4,b=9$

$\begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ a^{2} & 2 \times a \times b & b^{2} \\ 4^{2} & 2 \times 4 \times 9 & 9^{2} \\ 16 & 72 & 81 \\+8 & +8 & \\ \hline 24 & {80} \end{array}$

Hence, ${{49}^{2}}=2401$.

27. Find the squares of \[\mathbf{41}\] using diagonal method.

Ans:

Hence, ${{41}^{2}}=1681$

28. Find the square root of \[\mathbf{441}\].

Ans: 

$\begin{align}& 3\left| \!{\underline {\,441 \,}} \right.  \\ & 3\left| \!{\underline {\,147 \,}} \right.  \\ & 7\left| \!{\underline {\,49 \,}} \right.  \\  & \text{    }7 \\ & \Rightarrow \sqrt{441}=\sqrt{3\times 3\times 7\times 7} \\ & =3\times 7 \\  & =21 \\ \end{align}$

29. Find the square root of \[\mathbf{11025}\].

Ans: 

$\begin{align}& 3\left| \!{\underline {\,11025 \,}} \right.  \\ & 3\left| \!{\underline {\,3675 \,}} \right.  \\ & 5\left| \!{\underline {\,1225 \,}} \right.  \\ & 5\left| \!{\underline {\,245 \,}} \right.  \\ & 7\left| \!{\underline {\,49 \,}} \right.  \\ & \text{    }7 \\ & \Rightarrow \sqrt{11025}=\sqrt{3\times 3\times 5\times 5\times 7\times 7} \\ & =3\times 5\times 7 \\  & =105 \\ \end{align}$

30. In a movie theatre, the number of rows is equal to the number of chairs in each row. If the capacity of the theatre is 2025. Find the number of chairs in each row.

Ans: 

$\begin{align}& 5\left| \!{\underline {\,2025 \,}} \right.  \\ & 5\left| \!{\underline {\,405 \,}} \right.  \\ & 3\left| \!{\underline {\,81 \,}} \right.  \\ & 3\left| \!{\underline {\,27 \,}} \right.  \\ & 3\left| \!{\underline {\,9 \,}} \right.  \\ & 3\left| \!{\underline {\,3 \,}} \right.  \\ & \text{  1} \\ & \Rightarrow \sqrt{2025}=\sqrt{5\times 5\times 3\times 3\times 3\times 3} \\ & =5\times 3\times 3 \\ & =45 \\ \end{align}$

31. Find the square root of \[\mathbf{55696}\] by long division method.

Ans: 

Hence, $\sqrt{55696}=236$

32. \[\sqrt{\mathbf{12}.\mathbf{25}}\] and $\sqrt{\frac{\mathbf{625}}{\mathbf{729}}}$.

Ans:

$\sqrt{12.25}$ can be written as,

$\sqrt{12.25}=\sqrt{\frac{1225}{100}}$

For numerator,

$\begin{align}& 5\left| \!{\underline {\,1225 \,}} \right.  \\ & 5\left| \!{\underline {\,245 \,}} \right.  \\ & 7\left| \!{\underline {\,49 \,}} \right.  \\ & \text{     7} \\ & \Rightarrow \sqrt{1225}=\sqrt{5\times 5\times 7\times 7} \\ & =5\times 7 \\  & =35 \\ \end{align}$

For denominator,

$\begin{align}& 2\left| \!{\underline {\,100 \,}} \right.  \\ & 2\left| \!{\underline {\,50 \,}} \right.  \\ & 5\left| \!{\underline {\,25 \,}} \right.  \\ & 5\left| \!{\underline {\, 5 \,}} \right.  \\ & \text{  1} \\ & \Rightarrow \sqrt{100}=\sqrt{2\times 2\times 5\times 5} \\ & =2\times 5 \\ & =10 \\ \end{align}$

Hence, 

$\begin{align}& \sqrt{12.25}=\sqrt{\frac{1225}{100}} \\ & =\frac{\sqrt{1225}}{\sqrt{100}} \\ & =\frac{35}{10} \\ & =3.5 \\ \end{align}$

Square root of $\sqrt{\frac{625}{729}}$,

For numerator,

$\begin{align}& 5\left| \!{\underline {\,625 \,}} \right.  \\ & 5\left| \!{\underline {\,125 \,}} \right.  \\ & 5\left| \!{\underline {\,25 \,}} \right.  \\ & \text{     5} \\ & \Rightarrow \sqrt{625}=\sqrt{5\times 5\times 5\times 5} \\ & =5\times 5 \\ & =25 \\ \end{align}$

For denominator,

$\begin{align}& 3\left| \!{\underline {\,729 \,}} \right.  \\ & 3\left| \!{\underline {\,243 \,}} \right.  \\ & 3\left| \!{\underline {\,81 \,}} \right.  \\ & 3\left| \!{\underline {\,27 \,}} \right.  \\ & 3\left| \!{\underline {\,9 \,}} \right.  \\ & 3\left| \!{\underline {\,3 \,}} \right.  \\ & \text{  1} \\ & \Rightarrow \sqrt{729}=\sqrt{3\times 3\times 3\times 3\times 3\times 3} \\ & =3\times 3\times 3 \\ & =27 \\ \end{align}$

Hence, 

\[\begin{align}& \sqrt{\frac{625}{729}}=\frac{\sqrt{625}}{\sqrt{729}} \\ & =\frac{25}{27} \\ \end{align}\]

Chapter 5 Square and Square Roots Important Questions PDF

In this chapter, you will learn the concepts of square and square roots. There are many important questions given for the chapter in the PDF version free that will give you an idea of what kind of question you can expect in exams. You can use the pdf of the solutions for your extensive revision for exams.

Square and Square Roots

In the earlier classes, you have learnt that the area of a square = side x side. That means the area of a square = (side)2 i.e square of side.

If the side of a square is given a units, then

Area of square = (a x a)sq.units = a2 sq.units

Therefore, the square of a number is the product of the number by itself.

In this chapter, you will study about square numbers or perfect squares, their properties and patterns, Pythagorean triplets, various techniques to determine square roots of natural numbers, square roots of decimals and fractions.

Methods of Finding Square Root

1. By Prime Factorization Method: We can use the prime factorization method to calculate the square root. In this method, first, we need to take out the number and then calculate the prime factors of the number by successive division. Then we need to pair the formulated prime factors in such a way that both the factors are equal in the pair. Determine the product of one factor taken from each pair to obtain the required square root.

2. By Repeated Subtraction: When the natural numbers are small then we use this method. We keep subtracting 1, 3, 5, 7, 9, …… till we arrive at 0 for calculating the number of odd numbers whose sum is the given natural number. Then we compute the number of times the subtraction is executed to reach at 0. The number of computations becomes the square root of the number.

3. Long Division Method: We use the long division method when the square numbers are big because the prime factorization method becomes difficult. We draw the bar on each pair of the given number starting from its unit’s place in this method and then divide the digit of the given number at the extreme left by a number whose square root is less than or equal to the leftmost number of the given number.

Write the quotient and the remainder after doing the division. Then we bring forward the next paired digits next to the remainder and continue the division method till you arrive at 0.

The quotient derived in the method becomes the square root of the given number.

Applications of a Square Root

• If we need to determine the length of the diagonal of a square or a triangle then we use techniques of square roots.

• Square roots are used when we need to determine the third side of a triangle where the measurement of the other two sides of the triangle is known.

• Standard deviation can be calculated by using the methods of Square roots.

• In order to solve a quadratic equation, square roots are used.

Important Facts

Following are the highlights of the chapter that will give you a quick glance of the entire chapter.

• When the exponent of a natural number is 2, the number derived is called a square number or a perfect square. Ex: 22, 32, 82, etc. are the square numbers or perfect square. 

• When a natural number is raised to the power 2 then it is called the square of the number.

• The product of a natural number is called a perfect square when the natural number is multiplied by itself.

• It is not necessary that all natural numbers are a perfect square.

• A square number can be defined as the product of equal prime factors.

Ex: 64 = 82 = (2 x 4)2 = 22 x 42.

• A square number must have 0, 1, 4, 5 or 9 at the unit's place. Opposite of the number does not exist.

• A natural number can never have 2, 3, 7 or 8 at its unit place then it cannot be a square number.

• An even number will always have a square number even while an odd number will always have a square number odd.

• If two consecutive triangle numbers are added, we always obtain a square number.

Ex: 1, 3, 6, 10, …… is the set of triangular numbers and 1 + 3 = 4 = 22.

• If we take squares of two consecutive natural numbers n & n + 1, we will find 2n non-square natural numbers between them.

• The total result of the first natural odd natural numbers is n2.

Ex: 1 + 3 + 5 = (sum of first 3 odd natural numbers = 9 = 32)

• A number can have a perfect square only when the number is not the sum of consecutive odd natural numbers, starting from 1.

• The sum of two consecutive natural numbers is equal to the square of an odd natural number greater than 1.

i.e. 52 = 25 = 12 + 13, 92 = 81 = 40 + 41, etc.

• The square of any number having 5 in the unit’s place can be formed by using the formula (a5)2 = a (a + 1) x 100 + 25.

• A set of three natural numbers a, b, c are said to form a Pythagorean triplet (a, b, c) if  a2 + b2 = c2

• If n be any natural number m > 1, then (2n, n2 - 1, n2 + 1) will form a Pythagorean triplet.

• The square root of any natural number, when multiplied by itself, gives the original number n.

• The square root is the inverse function of finding square. \[\sqrt{\frac{p}{q}}=\frac{\sqrt{p}}{\sqrt{q}}\].

• There are two important square roots of a perfect square number n + viz. \[\sqrt{n}\] & \[-\sqrt{n}\].

Ex: 62 = 36 = \[\sqrt{36}\] = ±6 i.e. +6 & -6

• If we know the square of a natural number then we can form the square of the next natural number from the sum of the square, the number & the next natural number.

Ex: 312 = 302 + 30 + 31 = 961.

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Chapterwise Important Questions for CBSE Class 8 Maths

• Chapter 1 - Rational Numbers

• Chapter 2 - Linear Equations in One Variable

• Chapter 3 - Understanding Quadrilaterals

• Chapter 4 - Practical Geometry

• Chapter 5 - Data Handling

• Chapter 7 - Cubes 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

Also Access, 

• Revision Notes for CBSE Class 8 Maths

• NCERT Solutions for Class 8 Maths

Conclusion

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