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

Squares and Square Roots

Exercise 6.1

## 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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

Without adding, find the sum.

(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**

**Video 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**

**Video 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**

**Video 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 \({\left( {100} \right)^2}\)