# Excercise 6.1 Squares and Square Roots- NCERT Solutions Class 8

Go back to  'Squares and Square Roots'

## 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 $${\left( {100} \right)^2}$$

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