Model Textbook of Mathematics Grade 7 (FBISE / NBF)
Class 7 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 7 (FBISE / NBF)

Class 7 Mathematics Ch 2 Mastery Guide: Square and Square Roots, Primes, Factorization & Real-World Applications (FBISE)

📖 Chapter 2: Square and Square Roots 📅 Updated: Sep 10, 2026
Teacher & Student Roadmap Grade 7 Mathematics • FBISE / National Curriculum (NBF)

Instructional Guide: Unit 02 Square and Square Roots

Target Learning Outcomes (SLOs)
  • Differentiate between Even and Odd numbers based on divisibility by 2 and their unit digits.
  • Define Prime numbers (exactly two distinct factors) and Composite numbers (more than two factors), recognizing that $1$ is neither prime nor composite.
  • Construct and apply the Sieve of Eratosthenes to identify all 25 prime numbers between 1 and 100.
  • Identify Twin Primes (primes differing by 2) and the unique Prime Triplet $(3, 5, 7)$.
  • Express numbers as the sum of two or three prime numbers (Goldbach-style decompositions).
  • Compute the square of numbers up to three digits, proper fractions, and decimal numbers.
  • Recognize that the square of a proper fraction or a decimal less than $1$ is smaller than the number itself ($\left(\frac{3}{5}\right)^2 = \frac{9}{25} < \frac{3}{5}$).
  • Understand the concept of Square Root as the inverse operation of squaring ($\sqrt{x} = x^{1/2}$).
  • Find the square root of natural numbers, common fractions, and decimal numbers using the prime factorization method.
  • Verify and apply the fundamental square root properties: $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$.
  • Solve real-world word problems involving area of squares, perimeters, dimensions of rectangular rooms, grouping soldiers/students in square formations, and fencing/cementing costs.
Prerequisites & Bridge Concepts
  • Class 5 & 6 Factors & Multiples: Prime factorization using factor trees and division ladders.
  • Class 6 Exponents: Understanding base and index notation ($x^2 = x \times x$).
  • Geometry Basics: Area of a square ($\text{side} \times \text{side} = s^2$) and perimeter ($4 \times s$).
Common Misconceptions & Pitfalls
  • Squaring vs. Multiplying by 2: Squaring means multiplying a number by itself ($5^2 = 5 \times 5 = 25$), NOT multiplying by $2$ ($5 \times 2 = 10$).
  • Square of Fractions: While squaring a whole number $>1$ produces a larger number ($4^2 = 16 > 4$), squaring a proper fraction produces a smaller number ($\left(\frac{1}{2}\right)^2 = \frac{1}{4} < \frac{1}{2}$).
  • The Status of 1: 1 is neither prime nor composite because it has only one factor (itself), whereas primes require exactly two distinct factors.
  • Square Root of Sums Fallacy: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. For instance, $\sqrt{9 + 16} = \sqrt{25} = 5$, which is NOT equal to $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$!
  • Mixed Numbers Before Rooting: You CANNOT take the square root of whole and fractional parts separately! In $\sqrt{1\frac{7}{9}}$, you must first convert to the improper fraction $\sqrt{\frac{16}{9}} = \frac{4}{3} = 1\frac{1}{3}$.
Pedagogical Flow & Practical Teaching Tips

Begin with visual dot arrays: arrange 4, 9, 16, and 25 counters into physical squares. When students see a $4 \times 4$ grid makes 16 dots, the connection between "square shape" and "square number" becomes intuitive. Introduce square root as "finding the length of the side of that square". Connect prime factorization to pairing socks: every factor must have an identical twin to emerge from the radical sign $\sqrt{\cdot}$.

Kid-Friendly Rhymes & Golden Rules

The Perfect Square Chant:
"Multiply a number by its very own face,
You make a perfect square in its rightful place!
$3 \times 3$ is nine, $4 \times 4$ sixteen,
The neatest square grids that you've ever seen!"
The Square Root Detective:
"The radical sign $\sqrt{\cdot}$ is a detective in disguise,
Looking for pairs right before your eyes!
Two identical factors inside the door,
One steps outside, and the radical is no more!"
Twin Primes Friendship:
"Prime numbers with just two between,
Are the sweetest twins that you've ever seen!
$(3, 5), (5, 7), (11, 13)$ too,
Differ by two and stay forever true!"

Key Definitions & Properties Reference Bank

Concept Mathematical Definition Key Formula / Rule Example
Even Number Divisible by 2; ends in 0, 2, 4, 6, 8. $2k$ where $k \in \mathbb{N}$ 142, 7134, 911756
Odd Number Not divisible by 2; ends in 1, 3, 5, 7, 9. $2k + 1$ where $k \in \mathbb{W}$ 33, 1111, 75667
Prime Number Has exactly 2 distinct factors: 1 and itself. Factors of $p = \{1, p\}$ 2, 3, 5, 7, 11, 13, 97
Composite Number Has more than 2 distinct factors. Smallest composite number is 4 4, 6, 8, 9, 10, 12, 18
Twin Primes Two prime numbers with a difference of 2. $p_2 - p_1 = 2$ (3, 5), (5, 7), (11, 13), (41, 43), (71, 73)
Prime Triplet Three consecutive primes differing by 2. Only known triplet is $(3, 5, 7)$ $5 - 3 = 2$ and $7 - 5 = 2$
Perfect Square Product of a whole number with itself. $n = x^2$ $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121$
Square Root Number which when multiplied by itself gives $y$. $\sqrt{y} = y^{1/2}$ $\sqrt{36} = 6$ because $6 \times 6 = 36$
Product Rule of Roots Root of a product equals product of roots. $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ $\sqrt{25 \times 16} = 5 \times 4 = 20$
Quotient Rule of Roots Root of a fraction equals root of num / root of den. $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ $\sqrt{\frac{64}{81}} = \frac{8}{9}$

Fast Recall: First 25 Squares ($1^2$ to $25^2$)

$1^2 = 1$
$2^2 = 4$
$3^2 = 9$
$4^2 = 16$
$5^2 = 25$
$6^2 = 36$
$7^2 = 49$
$8^2 = 64$
$9^2 = 81$
$10^2 = 100$
$11^2 = 121$
$12^2 = 144$
$13^2 = 169$
$14^2 = 196$
$15^2 = 225$
$16^2 = 256$
$17^2 = 289$
$18^2 = 324$
$19^2 = 361$
$20^2 = 400$
$21^2 = 441$
$22^2 = 484$
$23^2 = 529$
$24^2 = 576$
$25^2 = 625$

Active Study Cues & Critical Thinking Prompts

Why is 2 the only even prime?

Any even number greater than 2 is divisible by 2 in addition to 1 and itself, so it must have at least 3 factors and is therefore composite!

Can a square number end in 2, 3, 7, or 8?

No! Every perfect square ends ONLY in 0, 1, 4, 5, 6, or 9. If a number ends in 2, 3, 7, or 8, it can never be a perfect square.

Why $\sqrt{x}$ can't be split over addition?

Because geometrically, the side of a square of area $(a+b)$ is NOT equal to the sum of sides of separate squares: $\sqrt{9+16} = \sqrt{25} = 5 \neq 3+4=7$.

Concept Explorations & Comprehensive Theory

Section 1: Number Types, Primes, Sieve of Eratosthenes & Twin Primes

Mathematics builds upon the fundamental classification of natural numbers $\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$:

  • Even Numbers: Natural numbers divisible by 2 without remainder (multiples of 2). A number is even if its units digit is $0, 2, 4, 6,$ or $8$. Examples: $2, 4, 6, 8, 10, 56798, 911756$.
  • Odd Numbers: Natural numbers not divisible by 2. A number is odd if its units digit is $1, 3, 5, 7,$ or $9$. Examples: $1, 3, 5, 7, 9, 798125, 3219$.

Algebraic Properties of Even and Odd Numbers:

Sum Properties:
• $\text{Even} + \text{Even} = \text{Even}$ (e.g. $8 + 4 = 12$)
• $\text{Odd} + \text{Odd} = \text{Even}$ (e.g. $5 + 9 = 14$)
• $\text{Even} + \text{Odd} = \text{Odd}$ (e.g. $9 + 12 = 21$)
Product Properties:
• $\text{Even} \times \text{Even} = \text{Even}$ (e.g. $8 \times 4 = 32$)
• $\text{Odd} \times \text{Odd} = \text{Odd}$ (e.g. $7 \times 3 = 21$)
• $\text{Even} \times \text{Odd} = \text{Even}$ (e.g. $4 \times 5 = 20$)

Prime & Composite Numbers:

A natural number greater than 1 that has only two distinct factors (1 and itself) is called a prime number. Examples: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31$.
A natural number that has more than two different factors is called a composite number. Examples: $4, 6, 8, 9, 10, 12, 14, 15, 16, 18$.

Crucial Cardinal Fact: The number $1$ has only one factor (itself). Therefore, $1$ is neither prime nor composite! $2$ is the smallest prime and the only even prime number. The smallest composite number is $4$.

The Sieve of Eratosthenes (276 B.C. – 194 B.C.):

To find all primes up to 100: 1. Write numbers 1 to 100 in 10 rows. Cross out 1.
2. Circle 2 (prime), then cross out all multiples of 2 ($4, 6, 8, \dots$).
3. Circle 3 (prime), then cross out all multiples of 3 ($6, 9, 12, \dots$).
4. Circle 5 (prime), cross out all multiples of 5; circle 7, cross out all multiples of 7.
5. The remaining circled numbers are the 25 prime numbers between 1 and 100:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Twin Primes & Prime Triplets:

  • Twin Primes: Prime numbers that differ by exactly 2. Examples: $(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)$.
  • Prime Triplet: Three consecutive prime numbers that differ by 2. The set $\mathbf{(3, 5, 7)}$ is the only known prime triplet in all of mathematics!
  • Goldbach's Representation: Every even number greater than 4 can be written as the sum of two odd primes (e.g. $12 = 5 + 7$, $36 = 17 + 19$, $60 = 29 + 31$).

Section 2: Squares of Numbers, Proper Fractions & Decimals

When a number $x$ is multiplied by itself, the result $x \times x = x^2$ is called the square of $x$. Geometrically, $x^2$ is the area of a square whose side length is $x$.

  • Perfect Square of Natural Numbers: An integer is a perfect square if its prime factors can be grouped into exact pairs of equal factors. For example, $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = (2^2) \times (2^2) \times (3^2) = (2 \times 2 \times 3)^2 = 12^2$, so 144 is a perfect square. But $80 = 2^4 \times 5$ has an unpaired factor 5, so 80 is NOT a perfect square.
  • Square of Even & Odd Numbers:
    • The square of an even number is always even: $2^2 = 4$, $6^2 = 36$, $14^2 = 196$, $20^2 = 400$.
    • The square of an odd number is always odd: $1^2 = 1$, $5^2 = 25$, $11^2 = 121$, $17^2 = 289$.
  • Square of Proper Fractions: When squaring a proper fraction $\frac{a}{b}$ ($a < b$): $$\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$$ Notice that $\frac{a^2}{b^2} < \frac{a}{b}$! For example: $\left(\frac{3}{5}\right)^2 = \frac{9}{25} = 0.36$, whereas $\frac{3}{5} = 0.60$. Since $0.36 < 0.60$, the square of a proper fraction is always less than the fraction itself!
  • Square of Decimals Less than 1: Similarly, for any decimal number $0 < d < 1$, $d^2 < d$. For example, $(0.2)^2 = 0.04 < 0.2$; $(0.5)^2 = 0.25 < 0.5$; $(0.07)^2 = 0.0049 < 0.07$.

Section 3: Square Roots & The Prime Factorization Method

The square root of a number $y$ is a number $x$ which, when multiplied by itself, produces $y$: $$x^2 = y \iff x = \sqrt{y} = y^{1/2}$$ The symbol $\sqrt{\cdot}$ is called the radical sign, and the number beneath it is called the radicand.

While $(-4) \times (-4) = 16$ and $4 \times 4 = 16$, every positive real number has two square roots ($+4$ and $-4$). In Grade 7, we study the principal (positive) square root: $\sqrt{16} = 4$.

The 4-Step Prime Factorization Algorithm for Square Roots:

  1. Step 1: Find prime factors: Divide the number by prime numbers ($2, 3, 5, 7, 11, \dots$) until the quotient is 1.
  2. Step 2: Group identical factors into pairs: Write the factors in exponential form $p_1^2 \times p_2^2 \times \dots$.
  3. Step 3: Apply the radical sign: Take one factor from each pair outside the radical sign: $\sqrt{p^2} = p$.
  4. Step 4: Multiply the extracted factors: The product gives the exact square root!
Worked Example: Find $\sqrt{11025}$.
Prime factorization: $11025 = 3 \times 3 \times 5 \times 5 \times 7 \times 7 = 3^2 \times 5^2 \times 7^2$.
Taking root: $\sqrt{11025} = \sqrt{3^2 \times 5^2 \times 7^2} = 3 \times 5 \times 7 = \mathbf{105}$.

The Fundamental Product Property:

$$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$$ Example: $\sqrt{25 \times 16} = \sqrt{400} = 20$. Separately: $\sqrt{25} \times \sqrt{16} = 5 \times 4 = 20$. Both sides are strictly equal!

Section 4: Square Roots of Common and Mixed Fractions

To find the square root of a common fraction $\frac{a}{b}$, apply the Quotient Rule: $$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$ Find the prime factorization of the numerator $a$ and denominator $b$ independently, group factors into pairs, and take square roots separately.

Golden Rule for Mixed Numbers:
Always convert the mixed number to an improper fraction FIRST before taking any square root!
Example: $\sqrt{5\frac{41}{64}} = \sqrt{\frac{(5 \times 64) + 41}{64}} = \sqrt{\frac{320 + 41}{64}} = \sqrt{\frac{361}{64}} = \frac{\sqrt{361}}{\sqrt{64}} = \frac{19}{8} = \mathbf{2\frac{3}{8}}$.
Never attempt $\sqrt{5} + \sqrt{\frac{41}{64}}$, which is completely mathematically invalid.

Section 5: Square Roots of Decimal Numbers

To find the square root of a decimal fraction by prime factorization:

  1. Convert the decimal into a common fraction: Remove the decimal point and place 1 followed by as many zeros in the denominator as there are decimal digits (e.g. $1.69 = \frac{169}{100}$, $0.0256 = \frac{256}{10000}$).
  2. Take square root of numerator and denominator: $\sqrt{\frac{169}{100}} = \frac{\sqrt{169}}{\sqrt{100}} = \frac{13}{10}$.
  3. Convert back to decimal form: $\frac{13}{10} = \mathbf{1.3}$.
Notice the Decimal Place Rule: When taking the square root, the number of decimal places is halved! • $1.69$ (2 decimal places) $\implies \sqrt{1.69} = 1.3$ (1 decimal place).
• $0.0256$ (4 decimal places) $\implies \sqrt{0.0256} = 0.16$ (2 decimal places).
• $0.000081$ (6 decimal places) $\implies \sqrt{0.000081} = 0.009$ (3 decimal places).

Section 6: Real-World Applications & Word Problem Formulas

Square roots are ubiquitous in construction, land measurement, and physical arrangements:

Application Scenario Given Condition Mathematical Formula
Side of Square from Area Area $A$ of square region $\text{Side } s = \sqrt{A}$
Perimeter of Square from Area Area $A$ of square $\text{Perimeter } P = 4 \times s = 4 \times \sqrt{A}$
Square Formation of People/Soldiers Number of rows $=$ Number of students per row $\text{Number of rows} = \sqrt{\text{Total Students}}$
Equal Society Contribution Each member pays as many Rs as total members $\text{Members} = \sqrt{\text{Total Collection}}$
Rectangular Field (Length $= 2 \times$ Width) Area $= (2w)(w) = 2w^2$ $w = \sqrt{\frac{\text{Area}}{2}}$, $\text{Length } l = 2w$, $P = 2(l+w)$
Fencing / Cementing Cost Rate per metre for boundary $\text{Total Cost} = \text{Perimeter} \times \text{Rate} = 4\sqrt{A} \times \text{Rate}$

Complete Solved Exercises (Step-by-Step Textbook Mastery)

Exercise 2.1: Even, Odd, Prime, Composite & Twin Primes
Question 1: Separate even and odd numbers from the following:
$33, 50, 142, 7134, 1111, 3219, 523, 625, 75667, 911756, 778$.
Solution:
Look at the units digit (last digit) of each number:
• Even numbers end in $0, 2, 4, 6, 8$: $\mathbf{50, 142, 7134, 911756, 778}$.
• Odd numbers end in $1, 3, 5, 7, 9$: $\mathbf{33, 1111, 3219, 523, 625, 75667}$.
Question 2: List the prime numbers between 20 and 60.
Solution:
Prime numbers have only two factors (1 and itself). Checking numbers between 20 and 60:
$\mathbf{23, 29, 31, 37, 41, 43, 47, 53, 59}$.
Question 3: A prime number when added to 101 gives an odd prime number. Find it.
Solution:
Let the prime number be $p$.
$101$ is an odd prime. The sum of two odd numbers is always even. If $p$ were odd, $101 + p$ would be an even number greater than 2, which cannot be prime!
Therefore, $p$ must be the only even prime, which is $2$.
Check: $101 + 2 = 103$, and $103$ is indeed an odd prime number.
Answer: $\mathbf{2}$.
Question 4: One of the twin primes is 7. Find other if sum of both numbers is an even number.
Solution:
Twin primes differ by 2. The primes differing by 2 from 7 are $7 - 2 = 5$ and $7 + 2 = 9$.
Since 9 is composite ($3 \times 3 = 9$), the only prime twin of 7 is 5.
Check sum: $5 + 7 = 12$, which is an even number.
Answer: $\mathbf{5}$.
Question 5: Find the greatest two digit prime number which, when added with 4, gives the smallest three digit prime number.
Solution:
The smallest three-digit prime number is $101$.
Let the two-digit prime be $p$. Then:
$p + 4 = 101 \implies p = 101 - 4 = 97$.
Check: $97$ is indeed the greatest two-digit prime number.
Answer: $\mathbf{97}$.
Question 6: Express each of the following as sum of two prime numbers.
(a) 34    (b) 33    (c) 42    (d) 60    (e) 54
Solution:
(a) 34: $17 + 17 = 34$ (or $3 + 31 = 34$ or $5 + 29 = 34$).
(b) 33: Since 33 is odd, one prime must be the even prime $2$: $31 + 2 = 33$.
(c) 42: $37 + 5 = 42$ (or $23 + 19 = 42$ or $11 + 31 = 42$).
(d) 60: $23 + 37 = 60$ (or $29 + 31 = 60$ or $17 + 43 = 60$).
(e) 54: $43 + 11 = 54$ (or $31 + 23 = 54$ or $47 + 7 = 54$ or $37 + 17 = 54$).
Question 7: Find pairs of twin primes between 40 and 70.
Solution:
List primes between 40 and 70: $41, 43, 47, 53, 59, 61, 67$.
Pairs differing by 2:
1. $43 - 41 = 2 \implies \mathbf{(41, 43)}$
2. $61 - 59 = 2 \implies \mathbf{(59, 61)}$
Answer: $\mathbf{(41, 43)}$ and $\mathbf{(59, 61)}$.
Question 8: Write the numbers as the sum of twin primes.
(a) 36    (b) 84    (c) 144    (d) 60
Solution:
(a) 36: $36 / 2 = 18$. The primes adjacent to 18 differing by 2 are 17 and 19. $17 + 19 = \mathbf{36}$.
(b) 84: $84 / 2 = 42$. Adjacent twin primes are 41 and 43. $41 + 43 = \mathbf{84}$.
(c) 144: $144 / 2 = 72$. Adjacent twin primes are 71 and 73. $71 + 73 = \mathbf{144}$.
(d) 60: $60 / 2 = 30$. Adjacent twin primes are 29 and 31. $29 + 31 = \mathbf{60}$.
Question 9: List 10 composite numbers between 41 and 55.
Solution:
Composite numbers between 41 and 55 (excluding primes 43, 47, 53):
$\mathbf{42, 44, 45, 46, 48, 49, 50, 51, 52, 54}$ (exactly 10 composite numbers).
Question 10: List 5 consecutive composite numbers between 60 and 70.
Solution:
Primes in this range are 61 and 67. The composite numbers lying between 61 and 67 are:
$\mathbf{62, 63, 64, 65, 66}$ (5 consecutive composite numbers).
Exercise 2.2: Squares of Numbers, Fractions, Decimals & Perfect Squares
Question 1: Find the square of the following numbers.
(i) 7   (ii) 9   (iii) 10   (iv) 13   (v) 16   (vi) 19   (vii) 30   (viii) 100   (ix) 150   (x) 21   (xi) 23   (xii) 25
Solution:
• (i) $7^2 = 7 \times 7 = \mathbf{49}$
• (ii) $9^2 = 9 \times 9 = \mathbf{81}$
• (iii) $10^2 = 10 \times 10 = \mathbf{100}$
• (iv) $13^2 = 13 \times 13 = \mathbf{169}$
• (v) $16^2 = 16 \times 16 = \mathbf{256}$
• (vi) $19^2 = 19 \times 19 = \mathbf{361}$
• (vii) $30^2 = 30 \times 30 = \mathbf{900}$
• (viii) $100^2 = 100 \times 100 = \mathbf{10{,}000}$
• (ix) $150^2 = 150 \times 150 = \mathbf{22{,}500}$
• (x) $21^2 = 21 \times 21 = \mathbf{441}$
• (xi) $23^2 = 23 \times 23 = \mathbf{529}$
• (xii) $25^2 = 25 \times 25 = \mathbf{625}$
Question 2: Find the square of the following numbers (fractions and decimals).
(i) $\frac{2}{3}$   (ii) $\frac{16}{13}$   (iii) $\frac{9}{16}$   (iv) $\frac{7}{23}$   (v) $0.3$   (vi) $0.5$   (vii) $0.07$   (viii) $0.09$
Solution:
• (i) $\left(\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \mathbf{\frac{4}{9}}$
• (ii) $\left(\frac{16}{13}\right)^2 = \frac{16^2}{13^2} = \mathbf{\frac{256}{169}}$
• (iii) $\left(\frac{9}{16}\right)^2 = \frac{9^2}{16^2} = \mathbf{\frac{81}{256}}$
• (iv) $\left(\frac{7}{23}\right)^2 = \frac{7^2}{23^2} = \mathbf{\frac{49}{529}}$
• (v) $(0.3)^2 = 0.3 \times 0.3 = \mathbf{0.09}$
• (vi) $(0.5)^2 = 0.5 \times 0.5 = \mathbf{0.25}$
• (vii) $(0.07)^2 = 0.07 \times 0.07 = \mathbf{0.0049}$
• (viii) $(0.09)^2 = 0.09 \times 0.09 = \mathbf{0.0081}$
Question 3: Tell which of the following are perfect squares?
(i) 16   (ii) 27   (iii) 48   (iv) 121   (v) 64   (vi) 72   (vii) 1000   (viii) 625
Solution:
Check prime factorization for paired factors:
• (i) $16 = 2^4 = (4)^2 \implies$ Perfect square.
• (ii) $27 = 3^3 = 3^2 \times 3$ (unpaired 3) $\implies$ Not a perfect square.
• (iii) $48 = 2^4 \times 3$ (unpaired 3) $\implies$ Not a perfect square.
• (iv) $121 = 11^2 \implies$ Perfect square.
• (v) $64 = 8^2 = 2^6 \implies$ Perfect square.
• (vi) $72 = 2^3 \times 3^2 = 2^2 \times 3^2 \times 2$ (unpaired 2) $\implies$ Not a perfect square.
• (vii) $1000 = 10^3 = 10^2 \times 10$ (odd number of zeros) $\implies$ Not a perfect square.
• (viii) $625 = 5^4 = (25)^2 \implies$ Perfect square.
Answer: $\mathbf{16, 121, 64, 625}$ are perfect squares.
Question 4: Find all the perfect squares between 1 and 101.
Solution:
The perfect squares strictly between 1 and 101 are the squares of natural numbers $2, 3, 4, 5, 6, 7, 8, 9, 10$:
$2^2 = 4$, $3^2 = 9$, $4^2 = 16$, $5^2 = 25$, $6^2 = 36$, $7^2 = 49$, $8^2 = 64$, $9^2 = 81$, $10^2 = 100$.
Answer: $\mathbf{4, 9, 16, 25, 36, 49, 64, 81, 100}$.
Exercise 2.3: Square Roots by Factorization & Real-World Word Problems
Question 1: Find the square root of the following.
(i) 196    (ii) 225    (iii) 144    (iv) 625
Solution:
• (i) $196 = 2^2 \times 7^2 \implies \sqrt{196} = 2 \times 7 = \mathbf{14}$
• (ii) $225 = 3^2 \times 5^2 \implies \sqrt{225} = 3 \times 5 = \mathbf{15}$
• (iii) $144 = 2^4 \times 3^2 = (2^2 \times 3)^2 \implies \sqrt{144} = 4 \times 3 = \mathbf{12}$
• (iv) $625 = 5^4 = (5^2)^2 \implies \sqrt{625} = 25 = \mathbf{25}$
Question 2: Compute the following.
(i) $\sqrt{81}$    (ii) $\sqrt{169}$    (iii) $\sqrt{289}$    (iv) $\sqrt{400}$    (v) $\sqrt{2^2 \times 3^2}$    (vi) $\sqrt{7^2 \times 8^2}$
Solution:
• (i) $\sqrt{81} = \sqrt{9^2} = \mathbf{9}$
• (ii) $\sqrt{169} = \sqrt{13^2} = \mathbf{13}$
• (iii) $\sqrt{289} = \sqrt{17^2} = \mathbf{17}$
• (iv) $\sqrt{400} = \sqrt{20^2} = \mathbf{20}$
• (v) $\sqrt{2^2 \times 3^2} = 2 \times 3 = \mathbf{6}$
• (vi) $\sqrt{7^2 \times 8^2} = 7 \times 8 = \mathbf{56}$
Question 3: Find the square root of the following by factorization.
(i) 36   (ii) 64   (iii) 196   (iv) 324   (v) 441   (vi) 484   (vii) 625   (viii) 841   (ix) 961   (x) 1089   (xi) 1225   (xii) 1521   (xiii) 1024   (xiv) 1764   (xv) 2025   (xvi) 4096   (xvii) 4225   (xviii) 2116   (xix) 4356   (xx) 6400   (xxi) 9801   (xxii) 8100   (xxiii) 12100   (xxiv) 102400
Solution:
• (i) $36 = 2^2 \times 3^2 \implies \sqrt{36} = 2 \times 3 = \mathbf{6}$
• (ii) $64 = 2^6 = (2^3)^2 \implies \sqrt{64} = 2^3 = \mathbf{8}$
• (iii) $196 = 2^2 \times 7^2 \implies \sqrt{196} = 2 \times 7 = \mathbf{14}$
• (iv) $324 = 2^2 \times 3^4 \implies \sqrt{324} = 2 \times 3^2 = 2 \times 9 = \mathbf{18}$
• (v) $441 = 3^2 \times 7^2 \implies \sqrt{441} = 3 \times 7 = \mathbf{21}$
• (vi) $484 = 2^2 \times 11^2 \implies \sqrt{484} = 2 \times 11 = \mathbf{22}$
• (vii) $625 = 5^4 \implies \sqrt{625} = 5^2 = \mathbf{25}$
• (viii) $841 = 29^2 \implies \sqrt{841} = \mathbf{29}$
• (ix) $961 = 31^2 \implies \sqrt{961} = \mathbf{31}$
• (x) $1089 = 3^2 \times 11^2 \implies \sqrt{1089} = 3 \times 11 = \mathbf{33}$
• (xi) $1225 = 5^2 \times 7^2 \implies \sqrt{1225} = 5 \times 7 = \mathbf{35}$
• (xii) $1521 = 3^2 \times 13^2 \implies \sqrt{1521} = 3 \times 13 = \mathbf{39}$
• (xiii) $1024 = 2^{10} \implies \sqrt{1024} = 2^5 = \mathbf{32}$
• (xiv) $1764 = 2^2 \times 3^2 \times 7^2 \implies \sqrt{1764} = 2 \times 3 \times 7 = \mathbf{42}$
• (xv) $2025 = 3^4 \times 5^2 \implies \sqrt{2025} = 3^2 \times 5 = 9 \times 5 = \mathbf{45}$
• (xvi) $4096 = 2^{12} \implies \sqrt{4096} = 2^6 = \mathbf{64}$
• (xvii) $4225 = 5^2 \times 13^2 \implies \sqrt{4225} = 5 \times 13 = \mathbf{65}$
• (xviii) $2116 = 2^2 \times 23^2 \implies \sqrt{2116} = 2 \times 23 = \mathbf{46}$
• (xix) $4356 = 2^2 \times 3^2 \times 11^2 \implies \sqrt{4356} = 2 \times 3 \times 11 = \mathbf{66}$
• (xx) $6400 = 64 \times 100 = 8^2 \times 10^2 \implies \sqrt{6400} = 8 \times 10 = \mathbf{80}$
• (xxi) $9801 = 3^4 \times 11^2 \implies \sqrt{9801} = 3^2 \times 11 = 9 \times 11 = \mathbf{99}$
• (xxii) $8100 = 81 \times 100 = 9^2 \times 10^2 \implies \sqrt{8100} = 9 \times 10 = \mathbf{90}$
• (xxiii) $12100 = 121 \times 100 = 11^2 \times 10^2 \implies \sqrt{12100} = 11 \times 10 = \mathbf{110}$
• (xxiv) $102400 = 1024 \times 100 = 32^2 \times 10^2 \implies \sqrt{102400} = 32 \times 10 = \mathbf{320}$
Question 4: Verify the property $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$.
(i) $\sqrt{25 \times 16} = \sqrt{25} \times \sqrt{16}$
(ii) $\sqrt{36 \times 4} = \sqrt{36} \times \sqrt{4}$
(iii) $\sqrt{49 \times 81} = \sqrt{49} \times \sqrt{81}$
(iv) $\sqrt{100 \times 121} = \sqrt{100} \times \sqrt{121}$
(v) $\sqrt{144 \times 169} = \sqrt{144} \times \sqrt{169}$
Solution:
(i): $\text{LHS} = \sqrt{400} = 20$. $\text{RHS} = 5 \times 4 = 20$. $\text{LHS} = \text{RHS}$. Verified.
(ii): $\text{LHS} = \sqrt{144} = 12$. $\text{RHS} = 6 \times 2 = 12$. $\text{LHS} = \text{RHS}$. Verified.
(iii): $\text{LHS} = \sqrt{3969} = 63$. $\text{RHS} = 7 \times 9 = 63$. $\text{LHS} = \text{RHS}$. Verified.
(iv): $\text{LHS} = \sqrt{12100} = 110$. $\text{RHS} = 10 \times 11 = 110$. $\text{LHS} = \text{RHS}$. Verified.
(v): $\text{LHS} = \sqrt{24336} = 156$. $\text{RHS} = 12 \times 13 = 156$. $\text{LHS} = \text{RHS}$. Verified.
Question 5: Area of a square region is $121\text{ m}^2$. Find the length of its side.
Solution:
$\text{Area of square} = s^2 = 121\text{ m}^2$.
Taking square root on both sides: $s = \sqrt{121} = \sqrt{11^2} = \mathbf{11\text{ m}}$.
Answer: Length of side is $\mathbf{11\text{ m}}$.
Question 6: The area of a square photo frame is $256\text{ cm}^2$. Find the perimeter of the photo frame.
Solution:
$\text{Area} = s^2 = 256\text{ cm}^2 \implies s = \sqrt{256} = 16\text{ cm}$.
$\text{Perimeter} = 4 \times s = 4 \times 16 = \mathbf{64\text{ cm}}$.
Answer: Perimeter is $\mathbf{64\text{ cm}}$.
Question 7: Product of a number and its half is 72. Find the number.
Solution:
Let the number be $x$. Its half is $\frac{x}{2}$.
According to the condition: $x \times \frac{x}{2} = 72 \implies \frac{x^2}{2} = 72 \implies x^2 = 72 \times 2 = 144$.
Taking square root: $x = \sqrt{144} = \mathbf{12}$.
Answer: The number is $\mathbf{12}$.
Question 8: Area of a square field is $2500\text{ m}^2$. Find the perimeter of the square field.
Solution:
$\text{Side } s = \sqrt{2500} = \sqrt{25 \times 100} = 5 \times 10 = 50\text{ m}$.
$\text{Perimeter} = 4 \times s = 4 \times 50 = \mathbf{200\text{ m}}$.
Answer: Perimeter is $\mathbf{200\text{ m}}$.
Question 9: If 5 litres of paint will cover $16\text{ m}^2$ and you are painting a square pattern, what is the length of square the paint will cover?
Solution:
The total area covered by the paint is $16\text{ m}^2$.
Since the pattern is a square: $\text{Area} = s^2 = 16\text{ m}^2$.
Taking square root: $s = \sqrt{16} = \mathbf{4\text{ m}}$.
Answer: The length of the square pattern is $\mathbf{4\text{ m}}$.
Question 10: Area of a square field is $900\text{ m}^2$. Find the cost of cementing boundary wall at a rate of Rs. 50 per metre.
Solution:
1. Find side of field: $s = \sqrt{900} = 30\text{ m}$.
2. Find perimeter (length of boundary): $P = 4 \times s = 4 \times 30 = 120\text{ m}$.
3. Calculate cost: $\text{Cost} = 120\text{ m} \times \text{Rs. } 50/\text{m} = \mathbf{\text{Rs. } 6000}$.
Answer: Total cost is $\mathbf{\text{Rs. } 6000}$.
Question 11: Length of rectangular region is double of its width. Find perimeter if its area is $128\text{ cm}^2$.
Solution:
Let the width be $w\text{ cm}$. Then length $l = 2w\text{ cm}$.
$\text{Area} = l \times w = (2w)(w) = 2w^2 = 128\text{ cm}^2$.
$w^2 = \frac{128}{2} = 64 \implies w = \sqrt{64} = 8\text{ cm}$.
Length $l = 2 \times 8 = 16\text{ cm}$.
$\text{Perimeter} = 2(l + w) = 2(16 + 8) = 2(24) = \mathbf{48\text{ cm}}$.
Answer: Perimeter is $\mathbf{48\text{ cm}}$.
Exercise 2.4: Square Roots of Common and Mixed Fractions
Questions 1 to 15: Find the square root of the following fractions:
1. $\frac{64}{81}$: $\sqrt{\frac{64}{81}} = \frac{\sqrt{64}}{\sqrt{81}} = \frac{\sqrt{8^2}}{\sqrt{9^2}} = \mathbf{\frac{8}{9}}$.

2. $\frac{361}{529}$: $\sqrt{\frac{361}{529}} = \frac{\sqrt{19^2}}{\sqrt{23^2}} = \mathbf{\frac{19}{23}}$.

3. $\frac{256}{625}$: $\sqrt{\frac{256}{625}} = \frac{\sqrt{16^2}}{\sqrt{25^2}} = \mathbf{\frac{16}{25}}$.

4. $\frac{1225}{169}$: $\sqrt{\frac{1225}{169}} = \frac{\sqrt{35^2}}{\sqrt{13^2}} = \frac{35}{13} = \mathbf{2\frac{9}{13}}$ (or $\mathbf{\frac{35}{13}}$).

5. $\frac{400}{121}$: $\sqrt{\frac{400}{121}} = \frac{\sqrt{20^2}}{\sqrt{11^2}} = \frac{20}{11} = \mathbf{1\frac{9}{11}}$ (or $\mathbf{\frac{20}{11}}$).

6. $\frac{144}{49}$: $\sqrt{\frac{144}{49}} = \frac{\sqrt{12^2}}{\sqrt{7^2}} = \frac{12}{7} = \mathbf{1\frac{5}{7}}$ (or $\mathbf{\frac{12}{7}}$).

7. $\frac{3025}{4096}$: Factorization gives $3025 = 5^2 \times 11^2 = 55^2$, and $4096 = 64^2$.
$\sqrt{\frac{3025}{4096}} = \frac{55}{64} = \mathbf{\frac{55}{64}}$.

8. $\frac{1296}{324}$: Method A (Simplify first): $\frac{1296}{324} = 4 \implies \sqrt{4} = 2$.
Method B (Separate roots): $\frac{\sqrt{1296}}{\sqrt{324}} = \frac{36}{18} = \mathbf{2}$ (or $\mathbf{\frac{36}{18}}$).

9. $\frac{196}{729}$: $\sqrt{\frac{196}{729}} = \frac{\sqrt{14^2}}{\sqrt{27^2}} = \mathbf{\frac{14}{27}}$.

10. $\frac{4761}{841}$: $4761 = 69^2$, $841 = 29^2$. $\sqrt{\frac{4761}{841}} = \frac{69}{29} = \mathbf{2\frac{11}{29}}$ (or $\mathbf{\frac{69}{29}}$).

11. $3\frac{22}{49}$: Convert to improper fraction: $\frac{(3 \times 49) + 22}{49} = \frac{147 + 22}{49} = \frac{169}{49}$.
$\sqrt{\frac{169}{49}} = \frac{13}{7} = \mathbf{1\frac{6}{7}}$.

12. $3\frac{13}{81}$: Convert to improper: $\frac{(3 \times 81) + 13}{81} = \frac{243 + 13}{81} = \frac{256}{81}$.
$\sqrt{\frac{256}{81}} = \frac{16}{9} = \mathbf{1\frac{7}{9}}$.

13. $5\frac{41}{64}$: Convert to improper: $\frac{(5 \times 64) + 41}{64} = \frac{320 + 41}{64} = \frac{361}{64}$.
$\sqrt{\frac{361}{64}} = \frac{19}{8} = \mathbf{2\frac{3}{8}}$.

14. $40\frac{41}{64}$: Convert to improper: $\frac{(40 \times 64) + 41}{64} = \frac{2560 + 41}{64} = \frac{2601}{64}$.
Since $2601 = 51^2$: $\sqrt{\frac{2601}{64}} = \frac{51}{8} = \mathbf{6\frac{3}{8}}$.

15. $26\frac{25}{36}$: Convert to improper: $\frac{(26 \times 36) + 25}{36} = \frac{936 + 25}{36} = \frac{961}{36}$.
Since $961 = 31^2$: $\sqrt{\frac{961}{36}} = \frac{31}{6} = \mathbf{5\frac{1}{6}}$.
Exercise 2.5: Square Roots of Decimals & Multi-Step Evaluations
Question 1: Find the square root by factorization for the following decimals:
• (i) $0.09 = \frac{9}{100} \implies \sqrt{\frac{9}{100}} = \frac{3}{10} = \mathbf{0.3}$
• (ii) $0.16 = \frac{16}{100} \implies \sqrt{\frac{16}{100}} = \frac{4}{10} = \mathbf{0.4}$
• (iii) $12.25 = \frac{1225}{100} \implies \sqrt{\frac{1225}{100}} = \frac{35}{10} = \mathbf{3.5}$
• (iv) $1.44 = \frac{144}{100} \implies \sqrt{\frac{144}{100}} = \frac{12}{10} = \mathbf{1.2}$
• (v) $0.0256 = \frac{256}{10000} \implies \sqrt{\frac{256}{10000}} = \frac{16}{100} = \mathbf{0.16}$
• (vi) $0.1936 = \frac{1936}{10000} \implies \sqrt{\frac{1936}{10000}} = \frac{44}{100} = \mathbf{0.44}$ (official textbook key: $\mathbf{0.4}$)
• (vii) $19.36 = \frac{1936}{100} \implies \sqrt{\frac{1936}{100}} = \frac{44}{10} = \mathbf{4.4}$
• (viii) $11.56 = \frac{1156}{100} \implies \sqrt{\frac{1156}{100}} = \frac{34}{10} = \mathbf{3.4}$
• (ix) $30.25 = \frac{3025}{100} \implies \sqrt{\frac{3025}{100}} = \frac{55}{10} = \mathbf{5.5}$
• (x) $0.0441 = \frac{441}{10000} \implies \sqrt{\frac{441}{10000}} = \frac{21}{100} = \mathbf{0.21}$
• (xi) $0.1225 = \frac{1225}{10000} \implies \sqrt{\frac{1225}{10000}} = \frac{35}{100} = \mathbf{0.35}$
• (xii) $0.0064 = \frac{64}{10000} \implies \sqrt{\frac{64}{10000}} = \frac{8}{100} = \mathbf{0.08}$
• (xiii) $2.89 = \frac{289}{100} \implies \sqrt{\frac{289}{100}} = \frac{17}{10} = \mathbf{1.7}$
• (xiv) $0.000081 = \frac{81}{1000000} \implies \sqrt{\frac{81}{1000000}} = \frac{9}{1000} = \mathbf{0.009}$
Question 2: Area of a square region is $43.56\text{ km}^2$. Find the length of its side.
Solution:
$\text{Side } s = \sqrt{43.56} = \sqrt{\frac{4356}{100}} = \frac{\sqrt{4356}}{\sqrt{100}} = \frac{66}{10} = \mathbf{6.6\text{ km}}$.
Answer: Length of the side is $\mathbf{6.6\text{ km}}$.
Question 3: Find the positive number which when multiplied by itself gives 110.25.
Solution:
Let the number be $x$. Then $x^2 = 110.25$.
$x = \sqrt{110.25} = \sqrt{\frac{11025}{100}} = \frac{105}{10} = \mathbf{10.5}$.
Answer: The positive number is $\mathbf{10.5}$.
Question 4: What is the length of a side of a square whose area is $23.04\text{ cm}^2$?
Solution:
$\text{Side } s = \sqrt{23.04} = \sqrt{\frac{2304}{100}} = \frac{48}{10} = \mathbf{4.8\text{ cm}}$.
Answer: Length of side is $\mathbf{4.8\text{ cm}}$.
Question 5: Evaluate the following expressions:
(i) $\left(\frac{1}{2}\right)^2 + \sqrt{0.25}$
(ii) $\sqrt{(0.5)^2 + (1.2)^2}$
(iii) $\sqrt{(0.5)^2 - (0.4)^2}$
Solution:
(i): $\left(\frac{1}{2}\right)^2 = 0.25 = \frac{1}{4}$. $\sqrt{0.25} = 0.5 = \frac{1}{2}$.
Sum $= 0.25 + 0.5 = \mathbf{0.75}$ (or $\frac{3}{4}$).

(ii): $(0.5)^2 = 0.25$, and $(1.2)^2 = 1.44$.
Sum $= 0.25 + 1.44 = 1.69$.
$\sqrt{1.69} = \mathbf{1.3}$.

(iii): $(0.5)^2 = 0.25$, and $(0.4)^2 = 0.16$.
Difference $= 0.25 - 0.16 = 0.09$.
$\sqrt{0.09} = \mathbf{0.3}$.
Review Exercise 2: Comprehensive Review & Mastery Assessment
Question 1: Encircle the correct answer for the following questions (13 MCQs):
(i) Twin primes differ by: (a) 1, (b) 2, (c) 3, (d) 4. → Correct: (b) 2.
(ii) A number has two factors one odd and other even. The number is: (a) even, (b) odd, (c) infinite, (d) prime. → Correct: (a) even (the only such number is 2, whose factors are 1 and 2, which is even).
(iii) A composite number has: (a) 1 factor, (b) 2 factors, (c) more than 2 factors, (d) none of these. → Correct: (c) more than 2 factors.
(iv) The square of an odd positive integer is: (a) odd, (b) even, (c) odd or even, (d) negative number. → Correct: (a) odd.
(v) The square of an even positive integer is: (a) odd, (b) even, (c) odd or even, (d) negative number. → Correct: (b) even.
(vi) $\sqrt{5 \times 5 \times 7 \times 7}$ is equal to: (a) 5, (b) 7, (c) 12, (d) 35. → Correct: (d) 35 ($5 \times 7 = 35$).
(vii) $\sqrt{121} \times \sqrt{144}$ is equal to: (a) 11, (b) 12, (c) 132, (d) 23. → Correct: (c) 132 ($11 \times 12 = 132$).
(viii) $\sqrt{2\frac{1}{4}}$ is equal to: (a) $\frac{2}{3}$, (b) $\frac{3}{2}$, (c) $\frac{3}{4}$, (d) $\frac{4}{3}$. → Correct: (b) $\frac{3}{2}$ ($\sqrt{9/4} = 3/2$).
(ix) $\sqrt{1.21}$ is equal to: (a) 1.1, (b) 2.1, (c) 1.2, (d) 0.1. → Correct: (a) 1.1.
(x) $\sqrt{\frac{169}{196}}$ is equal to: (a) $\frac{12}{13}$, (b) $\frac{13}{15}$, (c) $\frac{13}{14}$, (d) $\frac{14}{13}$. → Correct: (c) $\frac{13}{14}$.
(xi) $\sqrt{1\frac{7}{9}}$ is equal to: (a) $\frac{4}{5}$, (b) $\frac{4}{3}$, (c) $\frac{3}{4}$, (d) $\frac{5}{3}$. → Correct: (b) $\frac{4}{3}$ ($\sqrt{16/9} = 4/3$).
(xii) $\left(\frac{9}{4}\right)^2$ is equal to: (a) $\frac{81}{16}$, (b) $\frac{81}{25}$, (c) $\frac{100}{16}$, (d) $\frac{100}{16}$. → Correct: (a) $\frac{81}{16}$.
(xiii) $\sqrt{5^2 \times 6^2}$ is equal to: (a) 30, (b) 36, (c) 25, (d) 24. → Correct: (a) 30 ($5 \times 6 = 30$).
Question 2: Find the number, which is factor of every number but is neither prime nor composite.
Solution:
The number is $\mathbf{1}$. It divides every natural number without remainder ($a = 1 \times a$), and it has exactly one factor (itself), making it neither prime nor composite.
Question 3: Sum of two twin primes is 144 and they lie between 70 and 80. Find them.
Solution:
Primes between 70 and 80 are 71, 73, and 79.
Twin primes are 71 and 73 (difference $73 - 71 = 2$).
Their sum: $71 + 73 = 144$.
Answer: $\mathbf{71}$ and $\mathbf{73}$.
Question 4: Express each of the following as sum of three prime numbers:
(a) 38    (b) 31    (c) 41
Solution:
(a) 38: $19 + 17 + 2 = \mathbf{38}$ (or $31 + 5 + 2 = 38$).
(b) 31: $11 + 17 + 3 = \mathbf{31}$ (or $23 + 5 + 3 = 31$).
(c) 41: $13 + 17 + 11 = \mathbf{41}$ (or $31 + 7 + 3 = 41$ or $29 + 7 + 5 = 41$).
Question 5: Evaluate $\sqrt{\frac{2809}{4096}}$.
Solution:
$2809 = 53^2$, and $4096 = 64^2$.
$\sqrt{\frac{2809}{4096}} = \frac{\sqrt{2809}}{\sqrt{4096}} = \mathbf{\frac{53}{64}}$.
Question 6: Calculate $\sqrt{2\frac{1337}{3844}}$.
Solution:
Convert to improper fraction:
$2\frac{1337}{3844} = \frac{(2 \times 3844) + 1337}{3844} = \frac{7688 + 1337}{3844} = \frac{9025}{3844}$.
Factorize: $9025 = 95^2$ ($5^2 \times 19^2$), and $3844 = 62^2$ ($2^2 \times 31^2$).
$\sqrt{\frac{9025}{3844}} = \frac{95}{62} = \mathbf{1\frac{33}{62}}$ (or $\mathbf{\frac{95}{62}}$).
Question 7: Find square root of 4900 using factorization.
Solution:
$4900 = 49 \times 100 = 7^2 \times 10^2 = (2^2 \times 5^2 \times 7^2)$.
$\sqrt{4900} = 2 \times 5 \times 7 = \mathbf{70}$.
Question 8: Find square root of 167281 using factorization.
Solution:
Testing prime factors: $167281$ is divisible by $409$ ($409 \times 409 = 167281$).
Since $409$ is a prime number: $167281 = 409^2$.
$\sqrt{167281} = \mathbf{409}$.
Question 9: Find square root of 19.36.
Solution:
$19.36 = \frac{1936}{100}$.
$\sqrt{\frac{1936}{100}} = \frac{\sqrt{1936}}{\sqrt{100}} = \frac{44}{10} = \mathbf{4.4}$.
Question 10: The product of two positive numbers is 2400. One of them is six times the other. Find the numbers.
Solution:
Let the smaller number be $x$. Then the other number is $6x$.
$x \times 6x = 2400 \implies 6x^2 = 2400 \implies x^2 = \frac{2400}{6} = 400$.
$x = \sqrt{400} = 20$.
Second number $= 6 \times 20 = 120$.
Answer: The numbers are $\mathbf{20}$ and $\mathbf{120}$.
Question 11: 3600 students are asked to stand in different rows. Every row has as many students as there are rows. Find the number of rows.
Solution:
Let the number of rows be $r$. Then number of students per row is also $r$.
Total students $= r \times r = r^2 = 3600$.
$r = \sqrt{3600} = \sqrt{36 \times 100} = 6 \times 10 = \mathbf{60}$.
Answer: There are $\mathbf{60\text{ rows}}$.
Question 12: Find the perimeter of a square whose area is $6889\text{ m}^2$.
Solution:
1. Side $s = \sqrt{6889}$. Factorizing: $6889 = 83^2 \implies s = 83\text{ m}$.
2. Perimeter $P = 4 \times s = 4 \times 83 = \mathbf{332\text{ m}}$.
Answer: Perimeter is $\mathbf{332\text{ m}}$.
Question 13: A society collected Rs. 8836. Each member contributed as many rupees as there were members. Find the number of members of the society.
Solution:
Let the number of members be $m$. Then each member paid Rs. $m$.
Total collection $= m \times m = m^2 = 8836$.
$m = \sqrt{8836}$. Factorizing: $8836 = 2^2 \times 47^2 \implies m = 2 \times 47 = \mathbf{94}$.
Answer: There are $\mathbf{94\text{ members}}$ in the society.
Brain Buster Challenge: $1 + \sqrt{7 - 2\sqrt{4}}$ is equal to:
(a) $\sqrt{6} + 1$    (b) $\sqrt{6} - 1$    (c) $\sqrt{3} + 1$    (d) $\sqrt{3} - 1$
Solution:
1. Evaluate innermost root: $\sqrt{4} = 2$.
2. Substitute: $7 - 2(2) = 7 - 4 = 3$.
3. Expression becomes: $1 + \sqrt{3} = \sqrt{3} + 1$.
Answer: Correct option is (c) $\mathbf{\sqrt{3} + 1}$.

Active Recall Knowledge Checks (Self-Testing)

1. Why is 1 neither prime nor composite?
A prime number must have exactly two distinct positive factors (1 and itself). A composite number has more than two factors. Since 1 has only one positive factor (1), it fits neither definition!
2. How can you tell if a number cannot be a perfect square just by looking at its last digit?
The square of any integer must end in $0, 1, 4, 5, 6,$ or $9$. Therefore, any number ending in $2, 3, 7,$ or $8$ can NEVER be a perfect square.
3. Is the square of a fraction always smaller than the fraction? Explain.
Only for proper fractions (where numerator $<$ denominator, value $< 1$). For example, $\left(\frac{1}{2}\right)^2 = \frac{1}{4} < \frac{1}{2}$. For improper fractions ($> 1$), the square is larger: $\left(\frac{3}{2}\right)^2 = \frac{9}{4} > \frac{3}{2}$.
4. Why does $\sqrt{0.000081}$ have 3 decimal places when $0.000081$ has 6?
Because $0.000081 = \frac{81}{10^6}$. The square root is $\frac{\sqrt{81}}{\sqrt{10^6}} = \frac{9}{10^3} = \frac{9}{1000} = 0.009$. The exponent of 10 in the denominator is halved from 6 to 3!
5. Why is $\sqrt{9 + 16}$ equal to 5 and not 7?
Order of operations requires computing inside the radical first: $9 + 16 = 25$, and $\sqrt{25} = 5$. The radical sign does NOT distribute over addition: $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$.

Synthesis Summary: The Core Pillars of Unit 02

Unit 02 bridges foundational arithmetic and higher algebra through the complementary operations of Squaring ($x \mapsto x^2$) and Extracting Roots ($x^2 \mapsto x$). Mastering the prime factorization method empowers students to systematically decompose large numbers, verify perfect squares by matching twin factors, and seamlessly evaluate fractions and decimals. Remember the golden rules: (1) $1$ is unique (neither prime nor composite), (2) squares of proper fractions shrink while squares of integers grow, (3) roots distribute over multiplication ($\sqrt{ab} = \sqrt{a}\sqrt{b}$) and division ($\sqrt{a/b} = \sqrt{a}/\sqrt{b}$) but NEVER over addition, and (4) mixed numbers must always be converted to improper fractions before rooting!

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