Key Concept: Square Numbers

a) 49

[Solution Description]

To determine if a number is a perfect square, we need to check if it can be expressed as the square of an integer. Let us evaluate each option:

1) 49: $7 \times 7 = 49$; thus, 49 is a perfect square.

2) 50: There is no integer $n$ such that $n^2 = 50$; thus, 50 is not a perfect square.

3) 72: There is no integer $n$ such that $n^2 = 72$; thus, 72 is not a perfect square.

4) 81: $9 \times 9 = 81$; thus, 81 is a perfect square.

From the above calculations, both 49 and 81 are perfect squares, but since only one option is correct, the correct answer is 49.

Key Concept: Square Numbers, Finding Perfect Squares Between Ranges

c) 36

[Solution Description]

To find the sum of all perfect square numbers between 30 and 40, we first identify the perfect squares in this range.

- The square of 5 is 25, which is below 30.

- The square of 6 is 36, which is between 30 and 40.

- The square of 7 is 49, which is above 40.

Therefore, the only perfect square between 30 and 40 is 36. The sum of all perfect squares in this range is 36.

Key Concept: Square Numbers, Perfect Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion (A) states that 144 is a perfect square because it can be expressed as $12^2$. This is true since $12 \times 12 = 144$. The reason (R) defines a perfect square as a natural number that can be expressed as the square of another natural number, which is also true. Furthermore, the reason correctly explains why 144 is a perfect square, as it fits the definition provided in the reason. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Square Numbers

b) 64

[Solution Description]

The next perfect square after 49 is found by calculating the square of the next integer.

- The square root of 49 is 7.

- The next integer is 8, and its square is $8^2 = 64$.

Therefore, the next perfect square after 49 is 64.

Key Concept: Identifying Square Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To determine if 49 is a square number, we need to check if it can be expressed as the square of a natural number. We know that $7 \times 7 = 49$, which means 49 is indeed a square number. Therefore, the Assertion is true. The Reason states that 49 can be expressed as the product of a natural number with itself, which is also true and correctly explains why 49 is a square number.

Key Concept: Identifying Square Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

A square number is a number that can be expressed as the product of an integer with itself. Here, 49 can be written as $7 \times 7$, which means it is indeed a square number. Therefore, both the Assertion and Reason are true, and the Reason correctly explains why 49 is a square number.

Key Concept: Perfect Squares

b) 11 cm

[Solution Description] The area of a square is given by the formula $Area = side \times side = side^2$. We need to find the side when the area is 121 cm². To find the side, we take the square root of the area:

$\text{Side} = \sqrt{121} = 11 \text{ cm}$

Key Concept: Perfect Squares, Units Place of Square Numbers

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion states that 144 is a perfect square because it can be expressed as $12^2$. This is true since $12 \times 12 = 144$.

The reason claims that any number ending with the digit 4 at the units place must be a perfect square. However, this is not always true. For example, the number 14 ends with the digit 4 but it is not a perfect square. Similarly, 24, 34, etc., are not perfect squares.

Therefore, while the assertion is true, the reason is false.

Key Concept: Square Numbers, Ending Digits

c) 5

[Solution Description]

To determine the unit digit of a number whose square ends with 5, we can look at the pattern of squares of numbers ending with different digits. For example, $5^2 = 25$, $15^2 = 225$, and $25^2 = 625$. In each case, the square ends with 5, and the original number ends with 5. Therefore, if the square of a number ends with 5, the original number must also end with 5.

Key Concept: Square Numbers

b) 144

[Solution Description]

To find the square of 12, multiply 12 by itself: $12 \times 12 = 144$.

Key Concept: Area of a Square

d) 49 cm$^2$

[Solution Description]

The area of a square is calculated using the formula: Area = side $\times$ side. Given the side length is 7 cm, the area is $7 \times 7 = 49$ cm$^2$. Therefore, the correct answer is 49 cm$^2$.

Key Concept: Square Numbers

b) 81

[Solution Description]

To find the square of 9, we multiply 9 by itself:

$9 \times 9 = 81$

Therefore, the square of 9 is 81.

Key Concept: Properties of Square Numbers

b) It might be a perfect square.

[Solution Description]

The number 1069 ends with the digit 9. According to the properties of square numbers, a number ending with 0, 1, 4, 5, 6, or 9 could be a perfect square. Since 1069 ends with 9, it might be a perfect square. However, we need further verification to confirm this.

Key Concept: Square Numbers, Properties of Units Place

c) 5

[Solution Description]

To find the digit in the units place of the square of $n = 12345$, we only need to consider the digit in the units place of $n$, which is 5. The square of 5 is $25$, so the digit in the units place of $n^2$ is 5.

Key Concept: Last Digits of Square Numbers

c) 7

[Solution Description]

According to the properties of square numbers, a perfect square can only end with 0, 1, 4, 5, 6, or 9. It cannot end with 2, 3, 7, or 8. Therefore, the number 7 cannot be the last digit of a perfect square.

Key Concept: Last Digits of Square Numbers

d) 9

[Solution Description]

The last digit of a square number depends on the last digit of the original number. If a number ends with 3, its square will end with 9 because $3^2 = 9$. Therefore, the last digit of the square of a number ending with 3 is 9.

Key Concept: Square Numbers and Zeros

b) 8

[Solution Description]

Let the number be $N$ which ends with 4 zeros. When we square $N$, the number of zeros at the end doubles because each zero in the original number contributes two zeros in the square. Therefore, if $N$ has 4 zeros, its square will have $4 \times 2 = 8$ zeros.

Key Concept: Square Numbers and Zeros

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

Let’s analyze the assertion and reason step by step. First, consider a number $N$ that ends with 3 zeros. This means $N$ can be written as $N = k \times 1000$, where $k$ is an integer. Squaring $N$ gives:

$N^2 = (k \times 1000)^2 = k^2 \times 1000000$

Here, $1000000$ has 6 zeros. Therefore, $N^2$ will indeed have 6 zeros at the end. This confirms that the assertion is true. Now, let’s analyze the reason. The reason states that the number of zeros at the end of a square number is always double the number of zeros in the original number. While this holds true for numbers ending with zeros, it is not universally applicable to all numbers. For example, the number 500 has 2 zeros, and its square 250000 has 4 zeros, which is double the number of zeros. However, this reasoning fails for numbers like 5, which has 0 zeros, and its square 25 also has 0 zeros, but doubling 0 still gives 0. Therefore, the reason is not a correct explanation of the assertion because it does not hold true for all cases.

Key Concept: Number of Zeros in Square

c) 4

[Solution Description]

To find the number of zeros at the end of the square of a number, we can use the property that if a number has \$n\$ zeros at the end, its square will have \$2n\$ zeros at the end.

The number 400 has 2 zeros at the end. Therefore, its square will have \$2 \times 2 = 4\$ zeros at the end.

So, the square of 400 will have 4 zeros at the end.

Key Concept: Properties of Square Numbers, Squares of Even and Odd Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let's analyze the assertion and reason step by step.

- **Assertion**: The square of an even number is always divisible by 4.

- **Reason**: An even number can be expressed as $2n$, where $n$ is an integer.

**Step 1**: Let’s consider an even number, which can be written as $2n$, where $n$ is an integer.

**Step 2**: Now, let's square this even number: $(2n)^2 = 4n^2$.

**Step 3**: The result, $4n^2$, clearly shows that it is divisible by 4, because it has a factor of 4.

**Conclusion**: Both the Assertion and Reason are true, and the Reason correctly explains why the Assertion is true. Therefore, the correct answer is option a).

Key Concept: Squares of Even and Odd Numbers

c) 727

[Solution Description]

The square of an odd number is always odd, while the square of an even number is always even. Among the given options, 727 is an odd number. Therefore, the square of 727 will be an odd number.

Key Concept: Properties of Square Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that 144 is a perfect square because it can be expressed as $12 \times 12$. This is true since $12^2 = 144$. The reason defines a perfect square as a number that can be expressed as the product of an integer with itself, which is also correct. Furthermore, the reason correctly explains why 144 is a perfect square. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Sum of Odd Numbers, Perfect Squares

b) The number is a perfect square

[Solution Description]

The sum of the first $n$ odd natural numbers is $n^2$. For $n = 15$, the sum is $15^2 = 225$. Since 225 is a perfect square, it can indeed be expressed as the sum of the first 15 odd natural numbers.

Key Concept: Sum of First n Odd Natural Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The sum of the first 5 odd natural numbers is calculated as follows: 1, 3, 5, 7, 9 are the first 5 odd natural numbers. Adding them together, we get:

1 + 3 = 4,

4 + 5 = 9,

9 + 7 = 16, and

16 + 9 = 25.

Therefore, the sum of the first 5 odd natural numbers is 25.

The formula for the sum of the first n odd natural numbers is $n^2$. For n = 5, this formula also gives $5^2 = 25$, which matches our calculation. Thus, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Adding Triangular Numbers

b) 55 and 66

[Solution Description]

Let the two consecutive triangular numbers be $T_n$ and $T_{n+1}$. The sum of two consecutive triangular numbers is a square number. Given that $T_n + T_{n+1} = 121$, and since $121$ is a perfect square ($11^2$), we can find $n$.

The formula for the $n^{th}$ triangular number is:

$T_n = \frac{n(n+1)}{2}$

Therefore,

$T_{n+1} = \frac{(n+1)(n+2)}{2}$

Adding them together:

$T_n + T_{n+1} = \frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2} = \frac{(n+1)(2n+2)}{2} = (n+1)^2$

Given $(n+1)^2 = 121$, we solve for $n$:

$n+1 = 11 \quad \Rightarrow \quad n = 10$

Therefore, the two triangular numbers are:

$T_{10} = \frac{10 \times 11}{2} = 55 \quad \text{and} \quad T_{11} = \frac{11 \times 12}{2} = 66$

So, the two numbers are 55 and 66.

Key Concept: Interesting Patterns in Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The first two consecutive triangular numbers are 1 and 3. According to the pattern mentioned in the syllabus, when two consecutive triangular numbers are combined, they form a square number. Here, $1 + 3 = 4$, which is equal to $2^2$. Therefore, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Adding Consecutive Triangular Numbers

b) 45 and 55

[Solution Description]

Let the pair be $T_n$ and $T_{n+1}$. Their sum is $(n+1)^2 = 100$. Solving for $n$, we get $n+1 = 10$ so $n = 9$. The pair is $T_9 = \frac{9(9+1)}{2} = 45$ and $T_{10} = \frac{10(10+1)}{2} = 55$. The correct pair is $45$ and $55$.

Key Concept: Adding Consecutive Triangular Numbers

b) 36

[Solution Description]

Let the smaller triangular number be $T_n = \frac{n(n+1)}{2}$. The next triangular number is $T_{n+1} = \frac{(n+1)(n+2)}{2}$. Their sum is $\frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2} = \frac{(n+1)(n+n+2)}{2} = \frac{(n+1)(2n+2)}{2} = (n+1)(n+1) = (n+1)^2 = 81$. Solving for $n$, we get $n+1 = 9$ so $n = 8$. The smaller triangular number is $T_8 = \frac{8(8+1)}{2} = 36$.

Key Concept: Numbers between square numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] According to the syllabus, the number of non-square numbers between two consecutive squares $n^2$ and $(n+1)^2$ is given by $2n$. Here, $n = 10$, so the number of non-square numbers between $10^2$ and $11^2$ should be $2 \times 10 = 20$. Therefore, the Assertion is true. The Reason is also true as it correctly states the general formula for the number of non-square numbers between two consecutive squares. Moreover, the Reason correctly explains the Assertion.

Key Concept: Number of Non-Square Numbers Between Two Square Numbers

c) 20

[Solution Description]

Using the formula $2n$ where $n = 10$:

$2n = 2 \times 10 = 20$

Thus, there are 20 non-square numbers between $10^2$ and $11^2$.

Key Concept: Sum of First n Odd Numbers

c) 30

[Solution Description]

A number can be expressed as the sum of successive odd numbers starting from 1 if and only if it is a perfect square. Among the options, 30 is not a perfect square (since $5^2 = 25$ and $6^2 = 36$), so it cannot be expressed in this form.

Key Concept: Sum of First n Odd Numbers

c) 36

[Solution Description]

A number is a perfect square if it can be expressed as the sum of consecutive odd numbers starting from 1. 36 is a perfect square because it can be expressed as $1 + 3 + 5 + 7 + 9 + 11$, and $6^2 = 36$.

Key Concept: Representing Square Numbers as a Sum of Two Consecutive Integers

b) 264 and 265

[Solution Description]

To express $23^2$ as the sum of two consecutive positive integers, we use the pattern for odd squares: If the square is of an odd number $n$, then it can be expressed as $\frac{n^2 - 1}{2} + \frac{n^2 + 1}{2}$.

Let's calculate:

$23^2 = 529 \\ \frac{529 - 1}{2} = 264 \\ \frac{529 + 1}{2} = 265$

Therefore, $23^2$ can be expressed as the sum of two consecutive positive integers, which are 264 and 265.

Key Concept: Representing Square Numbers as a Sum of Two Consecutive Integers

b) 144 and 145

[Solution Description]

To express $17^2$ as the sum of two consecutive positive integers, we use the formula: If the square is of an odd number $n$, then it can be expressed as $\frac{n^2 - 1}{2} + \frac{n^2 + 1}{2}$.

Let's calculate:

$17^2 = 289 \\ \frac{289 - 1}{2} = 144 \\ \frac{289 + 1}{2} = 145$

Therefore, $17^2$ can be expressed as the sum of two consecutive positive integers, which are 144 and 145.

Key Concept: Patterns in Square Numbers Ending with 5, Numerical and Analytical Difficulty

b) 42025

[Solution Description]

For the number 205, $a = 20$. Using the pattern $(a5)^2 = a(a + 1)$ hundreds + 25, we have $(205)^2 = 20 \times 21$ hundreds + 25. First, calculate $20 \times 21 = 420$. Then, add the hundreds place: $420$ hundreds = $42000$. Finally, add 25 to get $42000 + 25 = 42025$.

Key Concept: Patterns in Square Numbers Ending with 5, Real-World Application

b) 21025

[Solution Description]

The area of the plot is the square of the side length, which is 145 meters. Using the pattern for numbers ending with 5, where $a = 14$, we have $(145)^2 = 14 \times 15$ hundreds + 25. First, calculate $14 \times 15 = 210$. Then, add the hundreds place: $210$ hundreds = $21000$. Finally, add 25 to get $21000 + 25 = 21025$ square meters.

Key Concept: Pythagorean Triplets

a) 3, 4, 5

[Solution Description]

Given the formula for generating Pythagorean triplets: $(2m)^2 + (m^2 – 1)^2 = (m^2 + 1)^2$. Let's substitute $m = 2$.

Step 1: Calculate $2m$

$2m = 2 \times 2 = 4$

Step 2: Calculate $m^2 - 1$

$m^2 - 1 = (2)^2 - 1 = 4 - 1 = 3$

Step 3: Calculate $m^2 + 1$

$m^2 + 1 = (2)^2 + 1 = 4 + 1 = 5$

Therefore, the Pythagorean triplet is 3, 4, 5.

Key Concept: Pythagorean Triplets

b) 30

[Solution Description]

In a Pythagorean triplet, $a^2 + b^2 = c^2$. Here, $a = 16$ and $c = 34$.

We need to find $b$ such that $16^2 + b^2 = 34^2$.

First, calculate $16^2$ and $34^2$:

$16^2 = 256$

$34^2 = 1156$

Now, solve for $b^2$:

$256 + b^2 = 1156$

$b^2 = 1156 - 256$

$b^2 = 900$

$b = \sqrt{900}$

$b = 30$

So, the missing number is 30.

Key Concept: Other patterns in squares

b) 7225

[Solution Description] For numbers ending with 5, like 85, we use the formula $a(a + 1)$ hundreds + 25. Here, $a = 8$. So, we calculate:

$8 \times (8 + 1) = 8 \times 9 = 72$,

which gives $72$ hundreds + $25 = 7200 + 25 = 7225$.

Key Concept: Finding the Square of a Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To find the square of 45, we use the formula for numbers ending with 5: $(a5)^2 = a(a+1)$ hundreds + 25. Here, $a = 4$. So, $a(a+1) = 4 \times 5 = 20$. Therefore, $(45)^2 = 20$ hundreds + 25 = 2000 + 25 = 2025. The assertion is true, and the reason provides the correct explanation for finding the square of 45.

Key Concept: Finding the Square of a Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the square of 45 can be calculated using the formula $(40 + 5)^2 = 40^2 + 2 \times 40 \times 5 + 5^2$. This is correct because when we expand $(40 + 5)^2$ using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$, we get $40^2 + 2 \times 40 \times 5 + 5^2$. Calculating this, we have $1600 + 400 + 25 = 2025$, which is indeed the square of 45. The reason provided is also correct as it explains the derivation of the formula used in the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Squaring a Two-Digit Number

b) 225

[Solution Description]

To find the square of 15, we can use the pattern for numbers ending with 5.

For a number like $a5$, its square is $(a \times (a + 1))$ hundreds followed by 25.

Here $a = 1$. So, $15^2 = (1 \times 2)$ hundreds + 25 = 200 + 25 = 225.

Key Concept: Shortcut Methods for Squaring

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the square of a number ending with 5 can be calculated using the formula $a(a + 1)$ hundreds $+ 25$. The reason explains that this formula is derived from the expansion of $(10a + 5)^2$, which simplifies to $100a(a + 1) + 25$. Here, $a$ represents the digit(s) before the unit place in the number. The assertion is true because this shortcut method is mathematically valid and widely used. The reason is also true as it provides the correct derivation of the formula. Furthermore, the reason correctly explains why the assertion holds true.

Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Other patterns in squares

b) 7225

[Solution Description]

For numbers ending with 5, the square can be found using the pattern $a(a + 1)$ hundreds + 25. Here, $a = 8$.

Step 1: Calculate $a(a + 1)$, which is $8 \times 9 = 72$.

Step 2: Multiply by 100 to get the hundreds place: $72 \times 100 = 7200$.

Step 3: Add 25 to the result: $7200 + 25 = 7225$.

Therefore, the square of 85 is 7225.

Key Concept: Squares using identities, Multi-step reasoning

b) 2209

[Solution Description]

Let $a = 40$ and $b = 7$. Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, we substitute the values:

$(40 + 7)^2 = 40^2 + 2 \times 40 \times 7 + 7^2$

Calculate each term:

$40^2 = 1600$

$2 \times 40 \times 7 = 560$

$7^2 = 49$

Add them together:

$1600 + 560 + 49 = 2209$

Therefore, the square of 47 is 2209.

Key Concept: Squares using identities, Multi-step reasoning

b) 3969

[Solution Description]

Let $a = 70$ and $b = 7$. Using the identity $(a - b)^2 = a^2 - 2ab + b^2$, we substitute the values:

$(70 - 7)^2 = 70^2 - 2 \times 70 \times 7 + 7^2$

Calculate each term:

$70^2 = 4900$

$2 \times 70 \times 7 = 980$

$7^2 = 49$

Combine the results:

$4900 - 980 + 49 = 3969$

Therefore, the square of 63 is 3969.

Key Concept: Pattern in Squares

b) 7225

[Solution Description]

To find the square of 85 using the pattern for numbers ending with 5, follow these steps:

Step 1: Identify the number before the last digit 5, which is 8.

Step 2: Multiply this number by its successor: $8 \times 9 = 72$.

Step 3: Append 25 to the result obtained in step 2: $7225$.

Therefore, the square of 85 is 7225.

Key Concept: Other patterns in squares

b) 13225

[Solution Description]

To find the square of 115 using the given pattern, follow these steps:

1. Identify the number preceding 5 in 115, which is 11.

2. Multiply this number by its successor: $11 \times 12 = 132$.

3. The result represents the number of hundreds in the square: 132 hundreds = 13200.

4. Add 25 to this value: $13200 + 25 = 13225$.

Therefore, the square of 115 is 13225.

Key Concept: Square Roots, Repeated Subtraction

c) 12

[Solution Description]

To find the square root using repeated subtraction, we know that if a number is reduced to zero in $n$ steps by subtracting successive odd numbers starting from 1, then the square root of that number is $n$. Here, the number is reduced to zero in 12 steps, so the square root is 12.

Key Concept: Square Roots, Inverse Operations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

First, let's verify the assertion. We know that $39^2 = 1521$, so $\sqrt{1521} = 39$. Therefore, the assertion is true. Now, let's examine the reason. Squaring a number means multiplying it by itself, and finding the square root means finding a number that, when multiplied by itself, gives the original number. Hence, finding the square root is indeed the inverse operation of squaring. This means both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Square Roots, Positive Square Root, Repeated Subtraction

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To find the square root of 49 using repeated subtraction of successive odd numbers:

\begin{itemize}

Key Concept: Repeated Subtraction

c) 12

[Solution Description]

To find the square root of 144 using the repeated subtraction method, we subtract successive odd numbers starting from 1 until we reach 0.

Starting with 144, we subtract 1, then 3, then 5, and so on.

The process will be as follows:

144 - 1 = 143,

143 - 3 = 140,

140 - 5 = 135,

135 - 7 = 128,

128 - 9 = 119,

119 - 11 = 108,

108 - 13 = 95,

95 - 15 = 80,

80 - 17 = 63,

63 - 19 = 44,

44 - 21 = 23,

23 - 23 = 0.

It takes 12 steps to reach 0, so the square root of 144 is 12.

Therefore, the correct answer is 12.

Key Concept: Square Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the square root of 25 is 5, which is correct as $\sqrt{25} = 5$. The reason given is $5^2 = 25$, which is also true and correctly explains why the square root of 25 is 5. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Square Roots, Perfect Squares

b) 25

[Solution Description]

To find the smallest natural number whose square ends with the same last two digits as the original number, we need to find a number whose square ends with the same digits as itself. Let's denote the number as $x$. We need $x^2 \equiv x \pmod{100}$. This simplifies to $x(x - 1) \equiv 0 \pmod{100}$. The smallest natural number satisfying this condition is 25 because $25^2 = 625$, which ends with 25.

Key Concept: Square Root Concept

c) 12

[Solution Description]

Given $n^2 = 144$, we need to find the value of $n$. Since the square root of a number is the inverse operation of squaring, we can find $n$ by taking the square root of 144.

From the syllabus, we know that $12^2 = 144$. Therefore, the value of $n$ is 12.

Key Concept: Positive Square Root

b) 5

[Solution Description]

The positive square root of a number is denoted by the symbol $\sqrt{}$. We need to find the positive square root of 25.

From the syllabus, we know that $5^2 = 25$. Therefore, the positive square root of 25 is 5.

Key Concept: Square Root Calculation

c) 8

[Solution Description]

To find the positive square root of 64, we look for a number that, when squared, gives 64. We know that $8^2 = 64$. Therefore, the positive square root of 64 is 8.

Key Concept: Finding Square Root Through Repeated Subtraction

c) 8

[Solution Description]

To find the square root of 64 using repeated subtraction, we subtract successive odd numbers starting from 1 until we reach 0. The number of steps taken to reach 0 gives the square root.

Step 1: $64 - 1 = 63$

Step 2: $63 - 3 = 60$

Step 3: $60 - 5 = 55$

Step 4: $55 - 7 = 48$

Step 5: $48 - 9 = 39$

Step 6: $39 - 11 = 28$

Step 7: $28 - 13 = 15$

Step 8: $15 - 15 = 0$

We reached 0 at the 8th step. Therefore, the square root of 64 is 8.

Key Concept: Square Root Identification

c) 12

[Solution Description]

To find the square root of 144, we need to determine a number whose square equals 144. We know that $12^2 = 144$. Therefore, the square root of 144 is 12.

Key Concept: Square Roots, Pythagoras Theorem

b) 10 cm

[Solution Description]

Using the Pythagoras theorem, which states that in a right triangle, the square of the hypotenuse ($c$) is equal to the sum of the squares of the other two sides ($a$ and $b$):

$c^2 = a^2 + b^2$

Here, $a = 6$ cm and $b = 8$ cm. Plugging these values into the equation:

$c^2 = 6^2 + 8^2 = 36 + 64 = 100$

To find $c$, take the square root of both sides:

$c = \sqrt{100} = 10$

Therefore, the length of the hypotenuse is 10 cm.

Key Concept: Square Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that the square root of 25 is 5, which is correct because $5^2 = 25$. The reason explains that the symbol $\sqrt{}$ denotes only the positive square root, which is also correct. Moreover, the reason correctly explains why the square root of 25 is considered as 5 and not -5, even though both are mathematically valid. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Square Roots of Perfect Squares

c) 13

[Solution Description]

A perfect square number has two integral square roots, one positive and one negative. For 144, the square roots are 12 and -12 because $12^2 = 144$ and $(-12)^2 = 144$. Therefore, 13 is not a square root of 144.

Key Concept: Square Root Calculation

c) 16

[Solution Description]

First, perform the prime factorisation of 256.

$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Pairing the prime factors, we get:

$256 = (2 \times 2 \times 2 \times 2)^2$

Therefore, $\sqrt{256} = 2 \times 2 \times 2 \times 2 = 16$

Key Concept: Perfect Square

c) No, multiply by 2

[Solution Description]

First, perform the prime factorisation of 72.

$72 = 2 \times 2 \times 2 \times 3 \times 3$

The prime factor 2 does not occur in pairs. To make all factors appear in pairs, we need to multiply 72 by 2.

So, $72 \times 2 = 144$, which is a perfect square.

Therefore, the smallest number by which 72 should be multiplied to make it a perfect square is 2.

Key Concept: Finding square root through repeated subtraction

b) 5

[Solution Description]

Subtract successive odd numbers starting from 1 from 25 until you get 0. The number of steps taken to reach 0 is the square root of 25. Steps:

1. $25 – 1 = 24$

2. $24 – 3 = 21$

3. $21 – 5 = 16$

4. $16 – 7 = 9$

5. $9 – 9 = 0$. It takes 5 steps to reach 0, so $\sqrt{25} = 5$.

Key Concept: Square Roots, Repeated Subtraction

b) 49

[Solution Description]

The number of subtractions required to reach zero using repeated subtraction of successive odd numbers starting from 1 gives the square root of the number. Since it takes 7 steps, the square root of the number is 7, and the number itself is $7^2 = 49$.

Key Concept: Finding Square Root by Division Method

b) 89

[Solution Description]

Step 1: Place bars over every pair of digits starting from the one's place: $\overline{79} \overline{21}$. Since there are two bars, the square root will be a 2-digit number.

Step 2: Find the largest number whose square is less than or equal to 79. $8^2 = 64$ and $9^2 = 81$. So, 8 is the divisor and quotient. The remainder is $79 - 64 = 15$.

Step 3: Bring down the next pair (21) to the right of the remainder. New dividend is 1521.

Step 4: Double the quotient (8) to get 16, and leave a blank next to it.

Step 5: Guess the largest possible digit to fill the blank such that when multiplied by the new quotient, the product is $\leq$ 1521. $161 \times 9 = 1449$. So, the new digit is 9.

Step 6: The remainder is $1521 - 1449 = 72$, but since the process ends here, $\sqrt{7921} = 89$.

Key Concept: Division Method, Estimating Square Root

c) 3

[Solution Description]

To estimate the square root of 11664 using the division method:

Step 1: Place bars over every pair of digits starting from the right. So, $\overline{116}\overline{64}$.

Step 2: Find the largest number whose square is less than or equal to 116. The largest such number is 10 because $10^2 = 100$ and $11^2 = 121$ which is greater than 116. So, the divisor is 10 and the quotient is 10. Subtract $100$ from $116$ to get a remainder of $16$.

Step 3: Bring down the next pair of digits (64) to the right of the remainder to form the new dividend 1664.

Step 4: Double the current quotient (10) to get 20, and place it with a blank on the right.

Step 5: Guess the largest possible digit to fill the blank such that when the new divisor (20X) is multiplied by this digit, the product is less than or equal to 1664. Here, $208 \times 8 = 1664$. So, the new digit in the quotient is 8.

Step 6: Since the remainder is 0 and no digits are left, the square root of 11664 is 108. Since there are three bars, the number of digits in the square root is 3.

Key Concept: Repeated Subtraction Method

b) 11

[Solution Description]

In the repeated subtraction method, we subtract successive odd numbers starting from 1 until we reach 0. For 121:

Step 1: $121 - 1 = 120$

Step 2: $120 - 3 = 117$

Step 3: $117 - 5 = 112$

Step 4: $112 - 7 = 105$

Step 5: $105 - 9 = 96$

Step 6: $96 - 11 = 85$

Step 7: $85 - 13 = 72$

Step 8: $72 - 15 = 57$

Step 9: $57 - 17 = 40$

Step 10: $40 - 19 = 21$

Step 11: $21 - 21 = 0$

It takes 11 steps to reach 0. Therefore, $\sqrt{121} = 11$.

Key Concept: Finding Square Root through Prime Factorisation

c) 12

[Solution Description]

To find the square root of 144 using prime factorisation, we first find the prime factors of 144.

$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$

Now, for each prime factor, we take half of its exponent and multiply them together.

$\sqrt{144} = 2^{2/2} \times 3^{2/2} = 2 \times 3 = 6$

Therefore, the square root of 144 is 12.

Key Concept: Repeated Subtraction Method

b) 11

[Solution Description]

To find the square root of 121 using the repeated subtraction method, subtract successive odd numbers starting from 1 until the result is zero. The number of steps taken to reach zero gives the square root.

Step-by-step subtraction:

$121 - 1 = 120$

$120 - 3 = 117$

$117 - 5 = 112$

$112 - 7 = 105$

$105 - 9 = 96$

$96 - 11 = 85$

$85 - 13 = 72$

$72 - 15 = 57$

$57 - 17 = 40$

$40 - 19 = 21$

$21 - 21 = 0$

It took 11 steps to reach zero. Therefore, $\sqrt{121} = 11$.

Key Concept: Repeated Subtraction Method, Perfect Squares

b) 15 steps: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29$

[Solution Description]

To find the square root of 225 using the repeated subtraction method, we subtract successive odd numbers starting from 1 until we get 0. The number of steps gives the square root. For 225, the sequence of odd numbers subtracted is $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29$, totaling 15 steps. Therefore, $\sqrt{225} = 15$.

Key Concept: Prime Factorization Method

d) 36

[Solution Description]

First, perform the prime factorization of 1296:

$1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

Now, pair the prime factors:

$1296 = (2 \times 2 \times 3 \times 3)^2$

Therefore, the square root of 1296 is:

$\sqrt{1296} = 2 \times 2 \times 3 \times 3 = 36$

Key Concept: Prime Factorization, Square Root

c) 42

[Solution Description]

First, perform the prime factorization of 1764: $1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7$. Pairing the prime factors, we get $(2 \times 3 \times 7)^2 = 42^2$. Therefore, $\sqrt{1764} = 42$.

Key Concept: Long Division Method, Multi-step Solution

b) 96

[Solution Description]

To find the square root of 9216 using the long division method:

1. Place bars over every pair of digits: $92$ $16$.

2. The number of bars indicates that the square root will have 2 digits.

3. Find the largest number whose square is less than or equal to 92. Here, $9^2 = 81 < 92 < 10^2 = 100$. So, the first digit is 9.

4. Subtract 81 from 92 to get the remainder 11. Bring down the next pair (16) to make the new dividend 1116.

5. Double the quotient (9) and write it with a blank: 18_.

6. Guess the largest possible digit to fill the blank such that $184 \times 4 = 736 \leq 1116$. However, the correct digit is 6 because $186 \times 6 = 1116$.

7. Since the remainder is 0, the square root of 9216 is 96.

Key Concept: Long Division Method

b) 42

[Solution Description]

First, place bars over every pair of digits starting from the right: $17$ $64$. The largest number whose square is less than or equal to 17 is 4 ($4^2 = 16$). Subtract 16 from 17 to get a remainder of 1. Bring down the next pair of digits (64) to make the new dividend 164. Double the quotient (4) to get 8 and find a digit such that $8x \times x \leq 164$. The digit is 2 since $82 \times 2 = 164$. Therefore, $\sqrt{1764} = 42$.

Key Concept: Estimating the Square Root of a Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the number of digits in the square root of a perfect square, we place bars starting from the rightmost digit. Each bar covers two digits. For the number 14400, placing bars gives us three groups: 1, 44, and 00. Since there are three bars, the square root will have three digits. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Estimating the Square Root of a Number

b) 131

[Solution Description]

To find the least number that must be subtracted from 5607 to make it a perfect square, we use the long division method. When we divide 5607 by the largest possible perfect square, we get a remainder of 131. This means $74^2$ is less than 5607 by 131. Therefore, subtracting 131 from 5607 gives us a perfect square: $5607 - 131 = 5476$. The square root of 5476 is 74.

Key Concept: Finding the Smallest Number to be Subtracted to Get a Perfect Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine the smallest number to be subtracted from 5607 to make it a perfect square, we first find $\sqrt{5607}$ using the long division method. The remainder obtained is 131. This indicates that $74^2 = 5476$ is less than 5607 by 131. Therefore, subtracting 131 from 5607 gives us the perfect square 5476. Hence, the assertion is true. The reason is also true because the remainder obtained when finding $\sqrt{5607}$ is indeed 131, which explains why 131 must be subtracted to get a perfect square.

Key Concept: Finding the smallest number to be added, Square root by long division

b) 13

[Solution Description]

To find the least number to be added to 4212 to make it a perfect square, we first calculate $\sqrt{4212}$ using the long division method. Starting with 60, since $64^2 = 4096$, which is less than 4212, and $65^2 = 4225$, which is greater than 4212, we know that the square root lies between 64 and 65. Next, we find the difference between $65^2$ and 4212:

$4225 - 4212 = 13$

Therefore, the least number to be added is 13. The resulting perfect square is $4225$, and its square root is $65$.

Key Concept: Square Roots of Decimals

c) 5.5

[Solution Description]

To find the square root of $30.25$, follow these steps:

1. Place bars on the integral part (30) and the decimal part (25).

2. The leftmost bar is on 30. We know that $5^2 = 25 < 30 < 36 = 6^2$. So, we take 5 as the divisor and 30 as the dividend.

3. Divide 30 by 5 to get a quotient of 5 and a remainder of 5.

4. Bring down the next pair (25) to the right of the remainder to get 525.

5. Double the divisor (5) and write it with a blank on its right: 10_. Place a decimal point in the quotient after 5.

6. We need to find a digit such that $10x \times x$ is less than or equal to 525. The digit 5 satisfies this because $105 \times 5 = 525$.

7. Since the remainder is 0 and no bars are left, the square root of $30.25$ is $5.5$.

Key Concept: Square Roots of Decimals

b) 4.5

[Solution Description]

To find the square root of $20.25$, follow these steps:

1. Place bars on the integral part (20) and the decimal part (25).

2. The leftmost bar is on 20. We know that $4^2 = 16 < 20 < 25 = 5^2$. So, we take 4 as the divisor and 20 as the dividend.

3. Divide 20 by 4 to get a quotient of 4 and a remainder of 4.

4. Bring down the next pair (25) to the right of the remainder to get 425.

5. Double the divisor (4) and write it with a blank on its right: 8_. Place a decimal point in the quotient after 4.

6. We need to find a digit such that $8x \times x$ is less than or equal to 425. The digit 5 satisfies this because $85 \times 5 = 425$.

7. Since the remainder is 0 and no bars are left, the square root of $20.25$ is $4.5$.

Key Concept: Placing Bars on Integral and Decimal Parts

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the square root of a decimal number, we place bars on both the integral and decimal parts. For the integral part, we start from the unit's place close to the decimal and move towards the left. For example, in 176, the first bar is over 76, and the second bar is over 1. For the decimal part, we start from the decimal point and move towards the right. For example, in .341, the first bar is over 34, and to make pairs, we add a zero after 1, making it .3410. Both the Assertion and Reason are true, and the Reason correctly explains how bars are placed differently for the integral and decimal parts.

Key Concept: Placing Bars on the Integral and Decimal Parts

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

Placing bars on the decimal part of a number involves starting from the decimal point and moving towards the right. This method is used to group digits in pairs to simplify the calculation of the square root. However, while the assertion is correct, the reason provided does not fully explain why bars are placed in this manner. The primary purpose of placing bars is to group digits for easier calculation, but it is not solely about simplifying the process of finding the square root. Therefore, both the assertion and reason are true, but the reason does not correctly explain the assertion.

Key Concept: Square Roots of Decimals, Long Division Method

c) 4.8

[Solution Description]

To find $\sqrt{23.04}$, we start by placing bars on the integral part (23) and decimal part (04). The largest number whose square is less than 23 is 4 since $4^2 = 16$. Subtract 16 from 23 to get a remainder of 7. Bring down the next pair (04) to make it 704. Double the divisor (4) to get 8, and find a digit (d) such that $80d \times d \leq 704$. Here, $802 \times 2 = 1604$, which is too large, so $d = 8$. Thus, $\sqrt{23.04} = 4.8$.

Key Concept: Square Roots of Decimals

b) 3.5

[Solution Description]

To find the square root of $12.25$, we follow the long division method for decimals.

Step 1: Place bars on the integral part (12) and the decimal part (25) of the number.

Step 2: The left-most bar is on 12, and $3^2 = 9 < 12 < 16 = 4^2$. Take 3 as the divisor and 12 as the dividend. Divide and get the remainder.

Step 3: The remainder is 3. Write the number under the next bar (i.e., 25) to the right of this remainder to get 325.

Step 4: Double the divisor and enter it with a blank on its right. Since 25 is the decimal part, put a decimal point in the quotient.

Step 5: We know $65 \times 5 = 325$, therefore, the new digit is 5. Divide and get the remainder.

Step 6: Since the remainder is 0 and no bar is left, $\sqrt{12.25} = 3.5$.

Therefore, the correct answer is $3.5$.

Key Concept: Square Roots of Decimals

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The assertion states that the square root of 12.25 is 3.5. This is correct as $\sqrt{12.25} = 3.5$. The reason explains the general method to find the square root of decimal numbers, which includes steps like placing bars, dividing, and finding remainders. However, the reason does not directly explain why the square root of 12.25 is 3.5. Therefore, both the assertion and reason are true, but the reason is not a correct explanation of the assertion.

Key Concept: Square Roots of Decimals

c) 7.5

[Solution Description]

To find the square root of 56.25, we follow these steps:

Step 1: We know that $7^2 = 49$ and $8^2 = 64$. Since $49 < 56.25 < 64$, the square root lies between 7 and 8.

Step 2: Let us consider 7.5 as an approximation. Calculating $7.5^2$, we get $7.5 \times 7.5 = 56.25$.

Therefore, the square root of 56.25 is 7.5.

Key Concept: Pythagoras theorem, Square roots

c) 25 cm

[Solution Description]

To find the length of the hypotenuse, we use the Pythagoras theorem. Let $a = 7$ cm and $b = 24$ cm be the lengths of the legs, and $c$ be the length of the hypotenuse. According to the theorem:

$c^2 = a^2 + b^2$

Substituting the given values:

$c^2 = 7^2 + 24^2$

$c^2 = 49 + 576$

$c^2 = 625$

Taking the square root of both sides:

$c = \sqrt{625} = 25 \text{ cm}$

Key Concept: Square Root Application

c) 22.624

[Solution Description]

To find the diagonal of a square, first, we need to find the length of its side. The area of a square is given by $A = s^2$, where $s$ is the length of the side. Given that the area is 256 square units, we can write:

$256 = s^2$

To find $s$, we take the square root of both sides:

$s = \sqrt{256} = 16$

Now, using the Pythagorean theorem for a square, the diagonal $d$ can be found using:

$d = s\sqrt{2}$

Substituting the value of $s$:

$d = 16\sqrt{2}$

Calculating $\sqrt{2} \approx 1.414$, we get:

$d \approx 16 \times 1.414 = 22.624$

Therefore, the length of the diagonal is approximately 22.624 units.

Key Concept: Square Root in Geometry

c) $10\sqrt{2}$ cm

[Solution Description]

To find the diagonal of a square, we use the Pythagorean theorem. The diagonal divides the square into two right-angled triangles. Let the side of the square be $AB = BC = 10$ cm. The diagonal $AC$ can be calculated as:

$AC^2 = AB^2 + BC^2 = 10^2 + 10^2 = 100 + 100 = 200$

Taking the square root of both sides:

$AC = \sqrt{200} = 10\sqrt{2} \, \text{cm}$

Key Concept: Square Root in Geometry

b) $8\sqrt{2}$ units

[Solution Description]

First, find the side length of the square using the area formula. Let the side length be $s$. Then:

$s^2 = 64 \implies s = \sqrt{64} = 8 \, \text{units}$

Next, use the Pythagorean theorem to find the diagonal $d$ of the square:

$d^2 = s^2 + s^2 = 8^2 + 8^2 = 64 + 64 = 128$

Taking the square root of both sides:

$d = \sqrt{128} = 8\sqrt{2} \, \text{units}$

Key Concept: Pythagorean Triplets, Pythagorean Theorem

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To verify the assertion, we check if $7^2 + 24^2 = 25^2$. Calculating the squares: $7^2 = 49$, $24^2 = 576$, and $25^2 = 625$. Adding the squares of 7 and 24 gives $49 + 576 = 625$, which equals $25^2$. Hence, the assertion is true. The reason states that for any natural number $m > 1$, the numbers $2m$, $m^2 – 1$, and $m^2 + 1$ always form a Pythagorean triplet. This is also true as it is a general formula to generate Pythagorean triplets. However, the triplet (7, 24, 25) does not directly follow this formula since there is no integer $m$ such that $2m = 7$, $m^2 - 1 = 24$, or $m^2 + 1 = 25$. Therefore, the reason is not the correct explanation of the assertion.

Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Assertion is true, but Reason is false.

Assertion is false, but Reason is true.

Key Concept: Pythagorean Theorem, Square Roots

b) 10 meters

[Solution Description]

The ladder, the wall, and the ground form a right triangle. Let the length of the ladder be $L$ meters. Using the Pythagorean theorem:

$L^2 = \text{Base}^2 + \text{Height}^2 \\ \\ L^2 = 6^2 + 8^2 \\ \\ L^2 = 36 + 64 \\ \\ L^2 = 100 \\ \\ L = \sqrt{100} \\ \\ L = 10 \text{ meters}$

Therefore, the length of the ladder is 10 meters.

Key Concept: Finding the Length of a Square’s Side Using Its Area

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To find the side length of a square when the area is known, we use the formula for the area of a square: $Area = side^2$. Given that the area is 64 cm$^2$, we can set up the equation $64 = side^2$. Taking the square root of both sides, we get $side = \sqrt{64}$, which simplifies to $side = 8$ cm. Therefore, the Assertion is true. The Reason states that the area of a square is calculated by squaring the length of its side, which is also true and directly explains why the Assertion is correct.

Key Concept: Square Roots, Perfect Squares

a) 0

[Solution Description]

The total number of trees needed for equal rows and columns would form a perfect square. The smallest perfect square greater than or equal to 1225 needs to be found. The square root of 1225 is calculated as follows:

$\sqrt{1225} = 35$

Since 35 is an integer, 1225 itself is a perfect square. Therefore, no additional trees are required to arrange them in equal rows and columns.

Key Concept: Finding the Length of a Square’s Side Using Its Area

c) 8 cm

[Solution Description]

The area of a square is given by the formula $Area = side \times side$. To find the length of the side, we need to find the square root of the area. Here, the area is 64 cm². So, $side = \sqrt{64}$. We know that $8 \times 8 = 64$, so $\sqrt{64} = 8$. Therefore, the length of the side is 8 cm.

Key Concept: Finding the Number of Rows in a Square Formation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let the number of rows be $x$. Since the number of rows equals the number of columns, the total number of students is given by $x^2 = 1296$. Taking the square root of both sides, we get $x = \sqrt{1296} = 36$. Therefore, the number of rows is indeed 36. The reason correctly explains that the number of rows is equal to the square root of the total number of students.

Key Concept: Square Formation

c) 42

[Solution Description]

Let the number of rows be $x$. Since the formation is square, the number of columns is also $x$. Therefore, the total number of students is $x^2$.

Given that the total number of students is 1764, we have:

$x^2 = 1764$

To find $x$, take the square root of both sides:

$x = \sqrt{1764} = 42$

Thus, the number of rows is 42.

Key Concept: Square Root Calculation

c) 4.5

[Solution Description]

To find the square root of 20.25, we calculate $\sqrt{20.25} = 4.5$. Thus, the square root of 20.25 is $4.5$.