NCERT Class 8 Maths Squares and Square Roots

# NCERT Class 8 Maths Squares and Square Roots

The chapter 6 begins with an introduction to square numbers and then explains the properties of square numbers.Some interesting patterns associated with adding triangular numbers, numbers between square numbers, adding odd numbers and so on.The next section of the chapter deals with Pythagorean Triplets and the procedure to find the similar sets of triplets.The next topic of discussion is square roots and the procedure to find square roots through repeated subtraction, division method. Square roots of decimals and the estimation of square roots is discussed in the last part of the chapter.

## Chapter 6 Ex.6.1 Question 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) $$55555$$

### Solution

What is known?

Numbers

What is unknown?

Unit digit of the square of numbers

Reasoning 1:

If a number has $$1$$ or $$9$$ in its unit digit, then it’s square ends with $$1$$.

$$\left( 1\times 1=1 \right)$$

Steps:

Since $$81$$ has $$1$$ as its unit digit, $$1$$ will be the unit digit of its square.

Similar Examples

$$91, 721, 4321$$

Reasoning 2:

If a number has either $$2$$ or $$8$$ as its unit digit, then it’s square ends with $$4$$.

Steps:

Since $$272$$ has $$2$$ as its unit digit, then its square number ends with $$4$$.

$$\left( {2 \times 2 = 4} \right)$$

Similar Examples

$$22, 2432, 147322$$

Reasoning 3:

If a number has $$1$$ or $$9$$ in its unit digit, then it’s square ends with $$1$$

Steps:

Since $$799$$ has $$9$$ as its unit digit, $$1$$ will be the unit digit of its square.

$$\left( 9\times 9=81 \right)$$

Reasoning 4:

If a number has either $$3$$ or $$7$$ as its unit digit, then its square number ends with $$9$$.

Steps:

Since, $$3853$$ has $$3$$ as its unit digit, $$9$$ will be the unit digit of its square.

$$\left( {3 \times 3 = 9} \right)$$

Similar Examples

$$13, 433, 63$$

Reasoning 5:

If a number has either $$4$$ or $$6$$ as its unit digit, then its square ends with $$6$$.

Steps:

Since, $$1234$$ has $$4$$ as its unit digit, $$6$$ will be the unit digit of its square.

$$\left( {4 \times 4 = 16} \right)$$

Similar Examples

$$14, 114, 484, 1594$$

Reasoning 6:

If a number has either $$3$$ or $$7$$ as its unit digit, then its square number ends with $$9$$.

Steps:

Since, $$26387$$ has $$7$$ as its unit digit, $$9$$ will be the unit digit of its square.

$$\left( {7 \times 7 = 49} \right)$$

Reasoning 7:

If a number has either $$2$$ or $$8$$ as its unit digit, then it’s square ends with $$4$$.

Steps:

Since $$52698$$ has $$8$$ as its unit digit, then its square number ends with $$4$$.

$$\left( {8 \times 8 = 64} \right)$$

Reasoning 8:

If a number has $$0$$ as its unit digit, then its square ends with $$0$$.

Steps:

Since, $$99880$$ has $$0$$ as its unit digit, $$0$$ will be the unit digit of its square.

Similar Examples

$$190, 1240, 167850$$

Reasoning 9:

If a number has either $$4$$ or $$6$$ as its unit digit, then its square ends with $$6$$.

Steps:

Since, $$12796$$ has $$6$$ as its unit digit, $$6$$ will be the unit digit of its square.

$$\left( {6 \times 6 = 36} \right)$$

Reasoning 10:

If a number has $$5$$ as its unit digit, then its square ends with $$5$$.

$$\left( 5\times 5=25 \right)$$

Steps:

Since, $$55555$$ has $$5$$ as its unit digit, $$5$$ will be the unit digit of its square.

Similar Examples

$$105, 85, 3425$$

## Chapter 6 Ex.6.1 Question 2

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

(i) $$1057$$

(ii) $$23453$$

(iii) $$7928$$

(iv) $$222222$$

(v) $$64000$$

(vi) $$89722$$

(vii) $$222000$$

(viii) $$505050$$

### Solution

What is known?

Numbers whhich are perfect square

What is unknown?

Why these number are not perfect square

Reasoning:

The square of a number has  $$0,1,4,5,6$$ or $$9$$ at its unit’s place are perfect squares.

Also, square of a number can only have even number of zeros at the end.

Steps:

In the above question unit digit of numbers are $$7, 3, 8, 2, 000, 2, 000, 0$$ respectively, so these number are obviously not perfect square.

## Chapter 6 Ex.6.1 Question 3

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

(i) $$431$$

(ii) $$2826$$

(iii) $$7779$$

(iv) $$82004$$

### Solution

What is known?

Numbers

What is unknown?

Square of which number would be odd

Reasoning:

Square of an odd number is always odd and square of an even number is always even.

Steps:

Since, $$431$$ and $$7779$$ are odd number so, squares of these numbers will be odd.

## Chapter 6 Ex.6.1 Question 4

Observe the following pattern and find the missing digits.

\begin{align}{11^2} &= 121\\{101^2} &= 10201\\{1001^2} &= 1002001\\{100001^2} &= 1 \ldots .2....1\\{10000001^2} &= \ldots ..........\end{align}

### Solution

What is known?

Pattern

What is uknown?

Missing number in the pattern

Reasoning:

The square of the given number has the same number of zeros before and after the digit $$2$$ as it has in the original number.

Steps:

\begin{align}{11^2} &= 121\\{101^2} &= 10201\\{1001^2}& = 1002001\\{100001^2} &= 10000200001\\{10000001^2}& = 100000020000001\end{align}

## Chapter 6 Ex.6.1 Question 5

Observe the following pattern and supply the missing numbers.

\begin{align}{11^2} &= 121\\{101^2} &= 10201\\{10101^2} &= 102030201\\{1010101^2} &= ?\\{?^2} &= 10203040504030201\end{align}

### Solution

What is known?

Pattern

What is unknown?

Missing number in the pattern

Reasoning:

Start with 1 followed by a zero and go up to as many number as there are number of 1s given, follow the same pattern in reverse order.

Solution:

\begin{align}{11^2} &= 121\\{101^2} &= 10201\\{10101^2} &= 102030201\\{1010101^2} &= 1020304030201\\{101010101^2} &= 10203040504030201\end{align}

## Chapter 6 Ex.6.1 Question 6

Using the given pattern, find the missing numbers.

\begin{align}{1^2} + {2^2} + {2^2} &= {3^2}\\{2^2} + {3^2} + {6^2} &= {7^2}\\{3^2} + {4^2} + {{12}^2} &= {{13}^2}\\{4^2} + {5^2} + { - ^2} &= {{21}^2}\\{5^2} + { - ^2} + {{30}^2} &= {{31}^2}\\{6^2} + {7^2} + \_\_{\_^2} &= \_\_\_{\_^2}\end{align}

### Solution

What is known?

Pattern

What is uknown?

Missing Number in the pattern

Reasoning:

The third number is the product of the first two numbers and the fourth number is obtained by adding $$1$$ to the third number

Steps:

\begin{align}{1^2} + {2^2} + {2^2} &= {3^2}\\{2^2} + {3^2} + {6^2} &= {7^2}\\{3^2} + {4^2} + {12}^2 &= {13}^2\\{4^2} + {5^2} + {20}^2 &= {21}^2\\{5^2} + {6^2} + {30}^2 &= {31}^2\\{6^2} + {7^2} + {42}^2 &= {43}^2\end{align}

## Chapter 6 Ex.6.1 Question 7

(i)

$$1 + 3 + 5 + 7 + 9$$

(ii)

\left[ \begin{align} &1 + 3 + 5 + 7 + 9 + 11 + \\ &13 + 15 + 17 + 19 \\ \end{align} \right]

(iii)

\left[ \begin{align} &1 + 3 + 5 + 7 + 9 + 11 + 13 + \\ &15 + 17 + 19 + 21 + 23 \\ \end{align} \right]

### Solution

What is known?

Consecutive odd numbers.

What is uknown?

Sum of these consecutive odd number without adding.

Reasoning:

Sum of the first n odd natural numbers is $$n^2$$ .

Steps:

(i) $$1 + 3 + 5 + 7 + 9$$

Here number of term $$(n)$$ is $$5$$

$${\mathop{\rm Sum}\nolimits} = {(5)^2} = 25$$

(ii)

\left[ \begin{align} &1 + 3 + 5 + 7 + 9 + 11 + \\ &13 + 15 + 17 + 19 \\ \end{align} \right]

Here number of term $$(n)$$ is $$10$$

$${\mathop{\rm Sum}\nolimits} = {(10)^2} = 100$$

(iii)

\left[ \begin{align} &1 + 3 + 5 + 7 + 9 + 11 + 13 + \\ &15 + 17 + 19 + 21 + 23 \\ \end{align} \right]

Here number of term $$(n)$$ is $$12$$

$${\mathop{\rm Sum}\nolimits} = {(12)^2} = 144$$

## Chapter 6 Ex.6.1 Question 8

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

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

### Solution

What is known?

Sum of $$7$$ odd number is $$49$$

Sum of $$11$$ odd number is $$121$$

What is uknown?

Express $$49$$ as the sum of $$7$$ odd numbers and $$121$$ as the sum of 11 odd numbers.

Reasoning:

The sum of successive odd natural numbers is $$n^2$$.

Steps:

(i) $$49 = {\left( 7 \right)^2}$$

Therefore ,$$49$$ is the sum of first $$7$$ odd natural numbers

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

(ii) $$121={\left(11 \right)^2}$$

Therefore ,$$121$$ is the sum of first $$11$$ odd natural numbers

$$121 = \left[ \begin{array}{l} 1 + 3 + 5 + 7 + 9 + 11 + \\13 + 15 + 17 + 19 + 21 \\ \end{array} \right]$$

## Chapter 6 Ex.6.1 Question 9

How many numbers lie between squares of the following numbers?

(i) $$12\; \text{ and} \; 13$$

(ii) $$25\; \text{and} \; 26$$

(iii) $$99\; \text{and} \; 100$$

### Solution

What is known?

Numbers

What is uknown?

How many number lie between squares of given number.

Reasoning:

There lies “$$2n$$” number between the square of the number $$n$$ and $$(n+1)$$

Steps:

(i) $$12\; \text{ and} \; 13$$

Here,$$n=12$$ gives $$2 \times 12 = 24$$, i.e., $$24$$ numbers between $${\left( {12} \right)^2}$$ and $${\left( {13} \right)^2}$$

(ii) $$25\; \text{and} \; 26$$

Here, $$n=25$$ gives $$2 \times 25 = 50$$, i.e., $$50$$ numbers. between $${\left( {25} \right)^2}$$ and $${\left( {26} \right)^2}$$

(iii) $$99\; \text{and} \; 100$$

Here, $$n=99$$ gives $$2 \times 99 = 198$$, i.e., $$198$$ numbers between $${\left( {99} \right)^2}$$ and $${\left( {100} \right)^2}$$

Squares and Square Roots | NCERT Solutions
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