Model Textbook of Mathematics Grade 8 (FBISE / NBF)
Class 8 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 8 (FBISE / NBF)

Class 8 Mathematics - Ch 3: Square & Square Roots, Cubes & Cube Roots Mastery Guide (FBISE)

📖 Chapter 3: Square and Square Roots, Cubes and Cube Roots 📅 Updated: Sep 13, 2026
Teacher & Student Roadmap Grade 8 Mathematics • FBISE / National Curriculum (NBF)

Instructional Guide: Unit 03 Square and Square Roots, Cubes and Cube Roots (مربع اور جذر المربع، مکعب اور جذر المکعب)

Target Student Learning Outcomes (SLOs)
  • Find squares of whole numbers up to 4 digits and understand patterns of squared numbers.
  • Compute square roots of natural numbers, common fractions, and decimals by both Prime Factorization and Division Method.
  • Calculate square roots of non-perfect square numbers and irrational decimals up to specified decimal places.
  • Determine cubes of natural numbers and recognize perfect cubes.
  • Find cube roots ($\sqrt[3]{\cdot}$) using prime factor triplet grouping.
  • Solve real-world geometric word problems involving square areas, circular fields, perimeters, and cube volumes.
Common Student Misconceptions & Traps
  • Decimals Pairing Direction: Pairing digits incorrectly. In decimal numbers, integral part pairs from right to left, but the fractional part pairs from left to right (e.g., $0.9 \to 0.\overline{90}$, $\sqrt{0.9} \approx 0.948$, NOT $0.3$!).
  • Division Method Step: Forgetting to double the existing quotient or add the last digit to form the next trial divisor.
  • Square Root vs Cube Root Grouping: For square root, group prime factors in pairs ($2^2 \to 2$); for cube root, group prime factors in triplets ($2^3 \to 2$).
  • Units in Real-Life Problems: Confusing units: area is $\text{m}^2$ / $\text{cm}^2$, perimeter and side length are $\text{m}$ / $\text{cm}$, volume is $\text{m}^3$ / $\text{cm}^3$.
3-Step Concept Mastery Strategy
  1. Step 1: Form Pairs / Triplets: Place bars over digit pairs starting from decimal point. For cube roots, write complete prime factorization.
  2. Step 2: Division / Extraction Cycle: Find largest digit $x$ such that $x^2 \le \text{group}$, subtract, bring down next pair, double quotient for next divisor.
  3. Step 3: Verification & Unit Check: Multiply result by itself (square or cube) to verify original radicand and check dimensional units.
Curriculum Note: This unit strictly aligns with FBISE Grade 8 Mathematics (NBF - Single National Curriculum). All decimal rounding, division ladders, algebraic proofs, and 100% solved exercises are presented with step-by-step mathematical working.

Unit 3: Square and Square Roots, Cubes and Cube Roots

A complete conceptual mastery guide covering perfect squares, long division square roots, fractional and decimal roots, approximations, cubes, and prime factorization cube roots with real-life applications.

Geometric Anatomy: 2D Square vs 3D Cube

Understanding how Side Length, Area ($s^2$), and Volume ($s^3$) relate to Square Roots ($\sqrt{A}$) and Cube Roots ($\sqrt[3]{V}$).

2D Square (مربع) Area = s² Side = s s Square: s² = s × s Square Root: s = √Area Perimeter = 4s 3D Cube (مکعب) V = s³ Length = s Cube: s³ = s × s × s Cube Root: s = ∛Volume Face Area = s²

1. Perfect Squares & Patterns of Squared Numbers

When a number is multiplied by itself, the resulting product is called the square of that number. If the square of a natural number $n$ is $x$ ($n^2 = x$), then $x$ is called a perfect square (کامل مربع).

Geometric Interpretation:

If a square has side length $17\text{ cm}$, its area is calculated as:
$$\text{Area} = \text{Length} \times \text{Length} = 17\text{ cm} \times 17\text{ cm} = 17^2\text{ cm}^2 = 289\text{ cm}^2$$ Thus, $289$ is the square of $17$, written as $17^2 = 289$.

Reference Table: First 20 Perfect Squares

$n$ $n^2$ $n$ $n^2$ $n$ $n^2$ $n$ $n^2$
116361112116256
247491214417289
398641316918324
4169811419619361
525101001522520400

Triangular Sum Pattern of Squared Numbers

Every perfect square $n^2$ equals the symmetric consecutive sum from $1$ up to $n$ and back down to $1$:

1 = 1² = 1
1 + 2 + 1 = 2² = 4
1 + 2 + 3 + 2 + 1 = 3² = 9
1 + 2 + 3 + 4 + 3 + 2 + 1 = 4² = 16
1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5² = 25
...
1 + 2 + 3 + ... + n + ... + 3 + 2 + 1 = n²
Fundamental Properties of Perfect Squares:
  • Never Negative: The square of any real number is always non-negative ($x^2 \ge 0$).
  • Parity Rule: The square of an even number is always even (e.g., $6^2 = 36$); the square of an odd number is always odd (e.g., $7^2 = 49$).
  • Ending Digit Rule: A perfect square never ends in $2, 3, 7, \text{ or } 8$. If a number ends in $2, 3, 7,$ or $8$, it can never be a perfect square.
  • Non-Perfect Squares: Numbers like $87, 123, 195, 326, 778, 1345, 1856, 2458$ are not perfect squares because their square roots are not whole numbers.
Check Point (Textbook Page 33)

Can you find the squares of 14, 16, 17, 18, 19, and 20?

• $14^2 = 14 \times 14 = \mathbf{196}$
• $16^2 = 16 \times 16 = \mathbf{256}$
• $17^2 = 17 \times 17 = \mathbf{289}$
• $18^2 = 18 \times 18 = \mathbf{324}$
• $19^2 = 19 \times 19 = \mathbf{361}$
• $20^2 = 20 \times 20 = \mathbf{400}$

2. Square Root: Prime Factorization & Division Method

The square root (جذر المربع) of a number $x$ is the value that, when multiplied by itself, gives $x$. It is denoted by the radical symbol $\sqrt{x}$.
Since $6 \times 6 = 36$, $\sqrt{36} = 6$. Since $7 \times 7 = 49$, $\sqrt{49} = 7$.

Long Division Method for Square Roots: Step-by-Step Anatomy

8 2 5 (Quotient = √680625) 68 06 25 ← Step 1: Pair from right 8 64 Step 2: 8² = 64 ≤ 68 16 2 4 06 3 24 Step 3: Double 8 → 16; 162 × 2 = 324 164 5 82 25 82 25 0 (Remainder) Step 4: 162 + 2 = 164; 1645 × 5 = 8225
Theorem: Number of Digits in Square Root:

If a perfect square number has $n$ digits:
• If $n$ is even, its square root has exactly $\frac{n}{2}$ digits (e.g., $1444$ has 4 digits $\implies 4/2 = 2$ digits in $\sqrt{1444}=38$).
• If $n$ is odd, its square root has exactly $\frac{n+1}{2}$ digits (e.g., $680625$ has 6 digits $\implies 3$ digits in $825$; $196$ has 3 digits $\implies (3+1)/2 = 2$ digits).

Textbook Worked Examples (Pages 33–35)

Example 1 (Page 33): Find the length of the side of a square whose area is $680625\text{ cm}^2$.
Given: $\text{Area of square} = 680625\text{ cm}^2$
Formula: $\text{Length of side} = \sqrt{\text{Area}} = \sqrt{680625\text{ cm}^2}$
By division method: $\sqrt{680625} = \mathbf{825\text{ cm}}$.
Example 2 (Page 34): Area of square field is $61504\text{ cm}^2$. Find the length of a side and the perimeter of the square field.
Given: $\text{Area} = 61504\text{ cm}^2$
Length of side: $s = \sqrt{61504} = \mathbf{248\text{ cm}}$
Perimeter of square: $P = 4 \times s = 4(248\text{ cm}) = \mathbf{992\text{ cm}}$.
Example 3 (Page 34): Find the least number that must be subtracted from $734\text{ cm}$ to get a perfect square.
Applying long division to $734$:
$2^2 = 4 \implies 7 - 4 = 3$; bring down $34 \to 334$.
Divisor $47 \times 7 = 329 \implies \text{Remainder} = 334 - 329 = \mathbf{5}$.
Therefore, the least number to subtract is $\mathbf{5}$ (resulting in $734 - 5 = 729 = 27^2$).
Example 4 (Pages 34–35): Some students of grade VIII contributed as many rupees as the number of students. If the total collection was Rs. 27225, find the number of students and the amount contributed by each.
Let the number of students $= x$. Amount contributed by each $= \text{Rs } x$.
$\text{Total collection} = x \times x = x^2 = 27225$
$x = \sqrt{27225} = \mathbf{165}$
Answer: Number of students $= \mathbf{165}$, Amount contributed by each $= \mathbf{\text{Rs. } 165}$.
Example 5 (Page 35): The product of two positive numbers is 84500. One of them is 5 times the other. Find the numbers.
Let first number $= y$, then second number $= 5y$.
Product $= y \times 5y = 5y^2 = 84500$
$y^2 = \frac{84500}{5} = 16900 \implies y = \sqrt{16900} = 130$
$1^{\text{st}}\text{ number} = y = \mathbf{130}$, $2^{\text{nd}}\text{ number} = 5y = 5(130) = \mathbf{650}$.

3. Exercise 3.1 — 100% Solved Step-by-Step Solutions

Question 1: Which of the following are perfect squares?

(i) 196:
Prime factorization: $196 = 2 \times 2 \times 7 \times 7 = (2 \times 7)^2 = 14^2$.
✔ Yes, 196 is a perfect square ($14^2$).
(ii) 1296:
Prime factorization: $1296 = 2^4 \times 3^4 = (2^2 \times 3^2)^2 = (4 \times 9)^2 = 36^2$.
✔ Yes, 1296 is a perfect square ($36^2$).
(iii) 325:
Prime factorization: $325 = 5 \times 5 \times 13 = 5^2 \times 13$. Factor 13 is not paired.
✘ No, 325 is NOT a perfect square.
(iv) 6561:
Prime factorization: $6561 = 3^8 = (3^4)^2 = 81^2$.
✔ Yes, 6561 is a perfect square ($81^2$).
(v) 4097:
$64^2 = 4096$, $65^2 = 4225$. 4097 lies between $64^2$ and $65^2$.
✘ No, 4097 is NOT a perfect square.

Question 2: Find the square root by Division Method.

(i) 841:
Pairs: $\overline{8}\;\overline{41}$
$2^2 = 4 \implies 8-4 = 4$, bring down $41 \to 441$.
Divisor: $49 \times 9 = 441 \implies \text{Remainder} = 0$.
$\sqrt{841} = \mathbf{29}$
(ii) 7921:
Pairs: $\overline{79}\;\overline{21}$
$8^2 = 64 \implies 79-64 = 15$, bring down $21 \to 1521$.
Divisor: $169 \times 9 = 1521 \implies \text{Remainder} = 0$.
$\sqrt{7921} = \mathbf{89}$
(iii) 1296:
Pairs: $\overline{12}\;\overline{96}$
$3^2 = 9 \implies 12-9 = 3$, bring down $96 \to 396$.
Divisor: $66 \times 6 = 396 \implies \text{Remainder} = 0$.
$\sqrt{1296} = \mathbf{36}$
(iv) 9801:
Pairs: $\overline{98}\;\overline{01}$
$9^2 = 81 \implies 98-81 = 17$, bring down $01 \to 1701$.
Divisor: $189 \times 9 = 1701 \implies \text{Remainder} = 0$.
$\sqrt{9801} = \mathbf{99}$
(v) 42025:
Pairs: $\overline{4}\;\overline{20}\;\overline{25}$
$2^2 = 4 \implies 0$, bring down $20 \to 20$. Divisor $40 \times 0 = 0 \to 20$, bring down $25 \to 2025$.
Divisor: $405 \times 5 = 2025 \implies \text{Remainder} = 0$.
$\sqrt{42025} = \mathbf{205}$
(vi) 49284:
Pairs: $\overline{4}\;\overline{92}\;\overline{84}$
$2^2 = 4 \implies 0$, bring down $92 \to 92$. Divisor $42 \times 2 = 84 \implies 92-84 = 8$, bring down $84 \to 884$.
Divisor: $442 \times 2 = 884 \implies \text{Remainder} = 0$.
$\sqrt{49284} = \mathbf{222}$
(vii) 46225:
Pairs: $\overline{4}\;\overline{62}\;\overline{25}$
$2^2 = 4 \implies 0$, bring down $62 \to 62$. Divisor $41 \times 1 = 41 \implies 62-41 = 21$, bring down $25 \to 2125$.
Divisor: $425 \times 5 = 2125 \implies \text{Remainder} = 0$.
$\sqrt{46225} = \mathbf{215}$
(viii) 78961:
Pairs: $\overline{7}\;\overline{89}\;\overline{61}$
$2^2 = 4 \implies 3$, bring down $89 \to 389$. Divisor $48 \times 8 = 384 \implies 389-384 = 5$, bring down $61 \to 561$.
Divisor: $561 \times 1 = 561 \implies \text{Remainder} = 0$.
$\sqrt{78961} = \mathbf{281}$
(ix) 119025:
Pairs: $\overline{11}\;\overline{90}\;\overline{25}$
$3^2 = 9 \implies 11-9 = 2$, bring down $90 \to 290$. Divisor $64 \times 4 = 256 \implies 290-256 = 34$, bring down $25 \to 3425$.
Divisor: $685 \times 5 = 3425 \implies \text{Remainder} = 0$.
$\sqrt{119025} = \mathbf{345}$
(x) 167281:
Pairs: $\overline{16}\;\overline{72}\;\overline{81}$
$4^2 = 16 \implies 0$, bring down $72 \to 72$. Divisor $80 \times 0 = 0 \to 72$, bring down $81 \to 7281$.
Divisor: $809 \times 9 = 7281 \implies \text{Remainder} = 0$.
$\sqrt{167281} = \mathbf{409}$
(xi) 1522756:
Pairs: $\overline{1}\;\overline{52}\;\overline{27}\;\overline{56}$
$1^2 = 1 \implies 0$, bring down $52 \to 52$. Divisor $22 \times 2 = 44 \implies 8$, bring down $27 \to 827$.
Divisor $243 \times 3 = 729 \implies 827-729 = 98$, bring down $56 \to 9856$.
Divisor $2464 \times 4 = 9856 \implies \text{Remainder} = 0$.
$\sqrt{1522756} = \mathbf{1234}$
(xii) 4227136:
Pairs: $\overline{4}\;\overline{22}\;\overline{71}\;\overline{36}$
$2^2 = 4 \implies 0$, bring down $22 \to 22$. Divisor $40 \times 0 = 0 \to 22$, bring down $71 \to 2271$.
Divisor $405 \times 5 = 2025 \implies 2271-2025 = 246$, bring down $36 \to 24636$.
Divisor $4106 \times 6 = 24636 \implies \text{Remainder} = 0$.
$\sqrt{4227136} = \mathbf{2056}$

Question 3: Find the least numbers which must be subtracted from the following numbers to make them perfect square.

(i) 1299:
$3^2 = 9 \implies 12-9=3$, bring down $99 \to 399$.
Divisor $66 \times 6 = 396 \implies \text{Remainder} = 399 - 396 = \mathbf{3}$.
$\therefore$ Subtract 3 ($1299 - 3 = 1296 = 36^2$).
(ii) 1854:
$4^2 = 16 \implies 18-16=2$, bring down $54 \to 254$.
Divisor $83 \times 3 = 249 \implies \text{Remainder} = 254 - 249 = \mathbf{5}$.
$\therefore$ Subtract 5 ($1854 - 5 = 1849 = 43^2$).
(iii) 9806:
$9^2 = 81 \implies 98-81=17$, bring down $06 \to 1706$.
Divisor $189 \times 9 = 1701 \implies \text{Remainder} = 1706 - 1701 = \mathbf{5}$.
$\therefore$ Subtract 5 ($9806 - 5 = 9801 = 99^2$).
(iv) 42029:
$2^2 = 4 \implies 0$, bring down $20 \to 20$ (divisor $40 \times 0 = 0$), bring down $29 \to 2029$.
Divisor $405 \times 5 = 2025 \implies \text{Remainder} = 2029 - 2025 = \mathbf{4}$.
$\therefore$ Subtract 4 ($42029 - 4 = 42025 = 205^2$).
Question 4: The area of a square field is $19600\text{ m}^2$. Find the length of the side of the square.
Solution: $\text{Area} = s^2 = 19600\text{ m}^2 \implies s = \sqrt{19600} = \sqrt{196 \times 100} = 14 \times 10 = \mathbf{140\text{ m}}$.
Question 5: Area of a circular field is $74536\text{ m}^2$. Find the circumference of the circle. (Take $\pi \approx \frac{22}{7}$)
Solution:
$\text{Area} = \pi r^2 = 74536 \implies \frac{22}{7} r^2 = 74536 \implies r^2 = \frac{74536 \times 7}{22} = 3388 \times 7 = 23716$
$r = \sqrt{23716} = 154\text{ m}$
$\text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 154 = 2 \times 22 \times 22 = \mathbf{968\text{ m}}$.
Question 6: The area of a square field is $1449616\text{ sqm}$. Find the perimeter.
Solution:
$\text{Side length } s = \sqrt{1449616} = 1204\text{ m}$
$\text{Perimeter} = 4 \times s = 4 \times 1204\text{ m} = \mathbf{4816\text{ m}}$.
Question 7: Find the least number of four digits which is a perfect square.
Solution:
The smallest four-digit number is $1000$.
Taking square root: $31^2 = 961 < 1000$ (3 digits), while $32^2 = 1024$ (4 digits).
$\therefore$ The least 4-digit perfect square is $32^2 = \mathbf{1024}$.
Question 8: Find the least number which must be subtracted from $3151$ to make it a perfect square.
Solution:
Applying division method on $3151$: Pairs $\overline{31}\;\overline{51}$.
$5^2 = 25 \implies 31-25 = 6$, bring down $51 \to 651$.
Divisor $106 \times 6 = 636 \implies \text{Remainder} = 651 - 636 = \mathbf{15}$.
$\therefore$ The least number that must be subtracted is $\mathbf{15}$ (since $3151 - 15 = 3136 = 56^2$).
Question 9: The product of two positive numbers is 230496. One of the number is 6 times the other. Find the numbers.
Solution:
Let the smaller number be $x$, then the other number is $6x$.
$x \times 6x = 230496 \implies 6x^2 = 230496 \implies x^2 = \frac{230496}{6} = 38416$
$x = \sqrt{38416} = 196$
$\text{First number} = \mathbf{196}$, $\text{Second number} = 6 \times 196 = \mathbf{1176}$.
Question 10: For a charity show each student of a class contributed as many rupees as the number of the students. If the total collection was Rs. 22500, find the number of students and amount contributed by each.
Solution:
Let number of students $= n$. Contribution per student $= \text{Rs } n$.
$\text{Total Collection} = n \times n = n^2 = 22500$
$n = \sqrt{22500} = \sqrt{225 \times 100} = 15 \times 10 = 150$
Answer: Number of students $= \mathbf{150}$, Contribution per student $= \mathbf{\text{Rs. } 150}$.

4. Square Roots of Fractions and Decimals

Properties of Square Roots:
  • Multiplication Law: $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$
  • Division Law: $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$   $(b \ne 0)$
  • Mixed Fractions: Always convert $w\frac{p}{q}$ into improper fraction $\frac{wq+p}{q}$ first.
Decimal Pairing Rules:
  • Integral Part (Left of '.'): Group in pairs from right to left ($\leftarrow$).
  • Decimal Part (Right of '.'): Group in pairs from left to right ($\rightarrow$).
  • Add a trailing zero if the last decimal group has an odd digit count.
Check Point (Textbook Page 37)
1. Find $\sqrt{\frac{361}{529}}$:
$\sqrt{\frac{361}{529}} = \frac{\sqrt{361}}{\sqrt{529}} = \frac{\sqrt{19^2}}{\sqrt{23^2}} = \mathbf{\frac{19}{23}}$

2. Find square root of $0.005329$:
Pairs: $0.\overline{00}\;\overline{53}\;\overline{29}$
First pair $00 \implies 0$; pair $53 \implies 7^2 = 49$, rem $4$, bring down $29 \to 429$. Divisor $143 \times 3 = 429 \implies \mathbf{0.073}$.

5. Exercise 3.2 — 100% Solved Step-by-Step Solutions

Question 1: Find the square root of the following fractions.

(i) $\frac{25}{49}$: $\sqrt{\frac{25}{49}} = \frac{\sqrt{5^2}}{\sqrt{7^2}} = \mathbf{\frac{5}{7}}$
(ii) $\frac{225}{169}$: $\sqrt{\frac{225}{169}} = \frac{\sqrt{15^2}}{\sqrt{13^2}} = \mathbf{\frac{15}{13} = 1\frac{2}{13}}$
(iii) $\frac{1681}{841}$: $\sqrt{\frac{1681}{841}} = \frac{\sqrt{41^2}}{\sqrt{29^2}} = \mathbf{\frac{41}{29} = 1\frac{12}{29}}$
(iv) $\frac{361}{625}$: $\sqrt{\frac{361}{625}} = \frac{\sqrt{19^2}}{\sqrt{25^2}} = \mathbf{\frac{19}{25}}$
(v) $\frac{1296}{1225}$: $\sqrt{\frac{1296}{1225}} = \frac{\sqrt{36^2}}{\sqrt{35^2}} = \mathbf{\frac{36}{35} = 1\frac{1}{35}}$
(vi) $\frac{3025}{729}$: $\sqrt{\frac{3025}{729}} = \frac{\sqrt{55^2}}{\sqrt{27^2}} = \mathbf{\frac{55}{27} = 2\frac{1}{27}}$
(vii) $\frac{2116}{2601}$: $\sqrt{\frac{2116}{2601}} = \frac{\sqrt{46^2}}{\sqrt{51^2}} = \mathbf{\frac{46}{51}}$
(viii) $\frac{2025}{1444}$: $\sqrt{\frac{2025}{1444}} = \frac{\sqrt{45^2}}{\sqrt{38^2}} = \mathbf{\frac{45}{38} = 1\frac{7}{38}}$

Question 2: Simplify the following mixed fractions.

(i) $\sqrt{4\frac{29}{49}}$:
$4\frac{29}{49} = \frac{4 \times 49 + 29}{49} = \frac{196 + 29}{49} = \frac{225}{49}$
$\sqrt{\frac{225}{49}} = \frac{15}{7} = \mathbf{2\frac{1}{7}}$
(ii) $\sqrt{40\frac{41}{64}}$:
$40\frac{41}{64} = \frac{40 \times 64 + 41}{64} = \frac{2560 + 41}{64} = \frac{2601}{64}$
$\sqrt{\frac{2601}{64}} = \frac{51}{8} = \mathbf{6\frac{3}{8}}$
(iii) $\sqrt{10\frac{6}{25}}$:
$10\frac{6}{25} = \frac{10 \times 25 + 6}{25} = \frac{256}{25}$
$\sqrt{\frac{256}{25}} = \frac{16}{5} = \mathbf{3\frac{1}{5}}$
(iv) $\sqrt{10\frac{151}{225}}$:
$10\frac{151}{225} = \frac{10 \times 225 + 151}{225} = \frac{2250 + 151}{225} = \frac{2401}{225}$
$\sqrt{\frac{2401}{225}} = \frac{49}{15} = \mathbf{3\frac{4}{15}}$
(v) $\sqrt{9\frac{67}{121}}$:
$9\frac{67}{121} = \frac{9 \times 121 + 67}{121} = \frac{1089 + 67}{121} = \frac{1156}{121}$
$\sqrt{\frac{1156}{121}} = \frac{34}{11} = \mathbf{3\frac{1}{11}}$
(vi) $\sqrt{21\frac{51}{169}}$:
$21\frac{51}{169} = \frac{21 \times 169 + 51}{169} = \frac{3549 + 51}{169} = \frac{3600}{169}$
$\sqrt{\frac{3600}{169}} = \frac{60}{13} = \mathbf{4\frac{8}{13}}$

Question 3: Find the square root of the following decimals.

(i) 0.16: $\sqrt{0.16} = \mathbf{0.4}$
(ii) 20.25: $\sqrt{20.25} = \mathbf{4.5}$
(iii) 46.24: $\sqrt{46.24} = \mathbf{6.8}$
(iv) 0.1296: $\sqrt{0.1296} = \mathbf{0.36}$
(v) 9.8596: $\sqrt{9.8596} = \mathbf{3.14}$
(vi) 42.5104: $\sqrt{42.5104} = \mathbf{6.52}$
(vii) 0.000225: $\sqrt{0.000225} = \mathbf{0.015}$
(viii) 727.9204: $\sqrt{727.9204} = \mathbf{26.98}$
(ix) 207.0721: $\sqrt{207.0721} = \mathbf{14.39}$
(x) 460.1025: $\sqrt{460.1025} = \mathbf{21.45}$
(xi) 7260.7441: $\sqrt{7260.7441} = \mathbf{85.21}$
(xii) 0.00001296: $\sqrt{0.00001296} = \mathbf{0.0036}$
Question 4: The area of a square lawn of a school is $42025\text{ m}^2$. If you complete $2\frac{1}{5}$ rounds of the square lawn, how much distance you traveled?
Solution:
$\text{Length of side } s = \sqrt{\text{Area}} = \sqrt{42025\text{ m}^2} = 205\text{ m}$
$\text{Perimeter of 1 round} = 4 \times 205\text{ m} = 820\text{ m}$
$\text{Total distance} = 2\frac{1}{5} \times 820 = \frac{11}{5} \times 820 = 11 \times 164 = \mathbf{1804\text{ m}}$.
Question 5: The length of a rectangular field is $2\frac{1}{2}$ times of width. If the area of rectangular field is $50.625\text{ m}^2$, find the length and width of the rectangular field.
Solution:
Let width $= w$. Then length $l = 2\frac{1}{2} w = 2.5 w = \frac{5}{2} w$.
$\text{Area} = l \times w = 2.5 w^2 = 50.625\text{ m}^2$
$w^2 = \frac{50.625}{2.5} = 20.25 \implies w = \sqrt{20.25} = 4.5\text{ m}$
$\text{Length } l = 2.5 \times 4.5 = \mathbf{11.25\text{ m}}$, $\text{Width } w = \mathbf{4.5\text{ m}}$.

6. Exercise 3.3 — Square Roots of Non-Perfect Squares & Decimal Approximations

To find the square root of a non-perfect square or fraction to $n$ decimal places, add sufficient pairs of zeros after the decimal point and calculate to $(n+1)$ decimal places (or inspect remainder vs half divisor) to round off correctly.

Question 1: Find the square roots of the following numbers upto three places of decimal.

(i) 3: $\sqrt{3.000000} \approx 1.73205 \implies \mathbf{1.732}$
(ii) 5: $\sqrt{5.000000} \approx 2.23606 \implies \mathbf{2.236}$
(iii) 7: $\sqrt{7.000000} \approx 2.64575 \implies \mathbf{2.646}$
(iv) 2.5: $\sqrt{2.500000} \approx 1.58113 \implies \mathbf{1.581}$
(v) 13: $\sqrt{13.000000} \approx 3.60555 \implies \mathbf{3.606}$
(vi) 1.1: $\sqrt{1.100000} \approx 1.04880 \implies \mathbf{1.049}$
(vii) 20: $\sqrt{20.000000} \approx 4.47213 \implies \mathbf{4.472}$
(viii) 1.7: $\sqrt{1.700000} \approx 1.30384 \implies \mathbf{1.304}$

Question 2: Find the square roots of the following numbers upto two places of decimal.

(i) 0.9: Pairs $0.\overline{90}\;\overline{00}\dots$
$\sqrt{0.9000} \approx 0.9486 \implies \mathbf{0.95}$
(ii) $2\frac{1}{12}$: $\frac{25}{12} \approx 2.083333$
$\sqrt{2.083333} \approx 1.4433 \implies \mathbf{1.44}$
(iii) $\frac{13}{7}$: $\frac{13}{7} \approx 1.857142$
$\sqrt{1.857142} \approx 1.3627 \implies \mathbf{1.36}$
(iv) 9573.853: $9573.8530$
$\sqrt{9573.8530} \approx 97.846 \implies \mathbf{97.85}$
(v) 654.69: $654.6900$
$\sqrt{654.6900} \approx 25.5869 \implies \mathbf{25.59}$
(vi) $\frac{1}{10}$: $\frac{1}{10} = 0.1 = 0.1000$
$\sqrt{0.1000} \approx 0.3162 \implies \mathbf{0.32}$
Question 3: The area of a square photo frame is $250\text{ cm}^2$. Find the perimeter of the photo frame, leaving your answer correct to 1 decimal place.
Solution:
$\text{Side } s = \sqrt{250} \approx 15.811388\text{ cm}$
$\text{Perimeter} = 4 \times s = 4 \times 15.811388 = 63.24555\dots\text{ cm}$
Rounding to 1 d.p.: $\mathbf{63.2\text{ cm}}$.
Question 4: Area of a one face of a cube is $997\text{ cm}^2$, find the length of the face of the cube correct to one decimal place.
Solution:
$\text{Length of side } l = \sqrt{\text{Face Area}} = \sqrt{997} \approx 31.5753\text{ cm}$
Rounding to 1 d.p.: $\mathbf{31.6\text{ cm}}$.

Prime Factorization: Square Roots (Pairs) vs Cube Roots (Triplets)

Square Root: Groups of 2 (Pairs) √144 = √(2 × 2 × 2 × 2 × 3 × 3) 2 × 2 2 × 2 3 × 3 Result = 2 × 2 × 3 = 12 Cube Root: Groups of 3 (Triplets) ∛216 = ∛(2 × 2 × 2 × 3 × 3 × 3) 2 × 2 × 2 (2³) 3 × 3 × 3 (3³) Result = 2 × 3 = 6

7. Cubes, Cube Roots & Exercise 3.4

When a number $x$ is multiplied by itself three times, the product is called the cube (مکعب) of $x$:
$$x \times x \times x = x^3$$ Conversely, the cube root (جذر المکعب) of $y$ is denoted as $\sqrt[3]{y} = y^{1/3}$.

Key Reference: Cubes of Common Natural Numbers

$n$ $n^3$ $n$ $n^3$ $n$ $n^3$
116216111331
287343121728
3278512132197
4649729142744
5125101000153375
Check Point (Textbook Page 41)

Can you find the cubes of 12, 13, 15, 16, 17, 18, 19, 21?

• $12^3 = 12 \times 12 \times 12 = \mathbf{1728}$
• $13^3 = 13 \times 13 \times 13 = \mathbf{2197}$
• $15^3 = 15 \times 15 \times 15 = \mathbf{3375}$
• $16^3 = 16 \times 16 \times 16 = \mathbf{4096}$
• $17^3 = 17 \times 17 \times 17 = \mathbf{4913}$
• $18^3 = 18 \times 18 \times 18 = \mathbf{5832}$
• $19^3 = 19 \times 19 \times 19 = \mathbf{6859}$
• $21^3 = 21 \times 21 \times 21 = \mathbf{9261}$

Exercise 3.4 — 100% Solved Solutions

Question 1: Find the cubes of the following numbers.

(i) 5: $5^3 = 5 \times 5 \times 5 = \mathbf{125}$
(ii) 8: $8^3 = 8 \times 8 \times 8 = \mathbf{512}$
(iii) 15: $15^3 = 15 \times 15 \times 15 = \mathbf{3375}$
(iv) 20: $20^3 = 20 \times 20 \times 20 = \mathbf{8000}$
(v) 25: $25^3 = 25 \times 25 \times 25 = \mathbf{15625}$

Question 2: Find the cube root of each of the following by prime factorization.

(i) 64:
$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3$
$\sqrt[3]{64} = 2 \times 2 = \mathbf{4}$
(ii) 729:
$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^3 \times 3^3$
$\sqrt[3]{729} = 3 \times 3 = \mathbf{9}$
(iii) 2197:
$2197 = 13 \times 13 \times 13 = 13^3$
$\sqrt[3]{2197} = \mathbf{13}$
(iv) 3375:
$3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 = 3^3 \times 5^3$
$\sqrt[3]{3375} = 3 \times 5 = \mathbf{15}$
(v) 2744:
$2744 = 2 \times 2 \times 2 \times 7 \times 7 \times 7 = 2^3 \times 7^3$
$\sqrt[3]{2744} = 2 \times 7 = \mathbf{14}$
Question 3: The volume of a room is $1331\text{ ft}^3$.
(i) Find the length of the room: $l = \sqrt[3]{1331} = \mathbf{11\text{ ft}}$ (since $11^3 = 1331$).
(ii) Find the area of the floor: $\text{Area} = l \times l = 11\text{ ft} \times 11\text{ ft} = \mathbf{121\text{ sq ft}}$ (or $\text{ft}^2$).
Question 4: The volume of a box in the shape of a cube is $2197\text{ cm}^3$.
(i) Find the length of the box: $l = \sqrt[3]{2197} = \mathbf{13\text{ cm}}$ (since $13^3 = 2197$).
(ii) Find the area of one face of the box: $\text{Face Area} = l^2 = 13\text{ cm} \times 13\text{ cm} = \mathbf{169\text{ cm}^2}$.
Question 5: Given that the prime factorization of $9261$ is $3^3 \times 7^3$. Find $\sqrt[3]{9261}$ without using calculator.
Solution:
$\sqrt[3]{9261} = \sqrt[3]{3^3 \times 7^3} = (3^3 \times 7^3)^{1/3} = (3^3)^{1/3} \times (7^3)^{1/3} = 3 \times 7 = \mathbf{21}$.

8. Review Exercise 3 — 100% Solved Solutions

Question 1: Choose the correct option.

(i) $\frac{\sqrt{3 \times 49}}{\sqrt{64 \times 3}} = \dots$
$\frac{\sqrt{3} \times \sqrt{49}}{\sqrt{64} \times \sqrt{3}} = \frac{7}{8}$
Correct Option: (b) $\frac{7}{8}$
(ii) $\frac{\sqrt{6}}{\sqrt{3}}$ is an:
$\sqrt{\frac{6}{3}} = \sqrt{2} \approx 1.414\dots$ (non-terminating, non-repeating)
Correct Option: (c) irrational number
(iii) The square root of 1444 consists of … digits:
1444 has 4 digits ($n=4$, even) $\implies 4/2 = 2$ digits.
Correct Option: (b) two
(iv) $\sqrt{1\frac{11}{25}}$ is equal to:
$\sqrt{\frac{36}{25}} = \frac{6}{5} = 1.2$
Correct Option: (a) 1.2
(v) $\frac{\sqrt{7.29}}{2}$ is equal to:
$\frac{2.7}{2} = 1.35$
Correct Option: (d) 1.35
(vi) $\frac{\sqrt{625} \div 25 \times 5}{\sqrt{625} \div 5}$ is equal to:
$\frac{25 \div 25 \times 5}{25 \div 5} = \frac{1 \times 5}{5} = \frac{5}{5} = 1$
Correct Option: (d) 1
(vii) The area of the square region is $1296\text{ m}^2$, the length of each side is:
$s = \sqrt{1296} = 36\text{ m}$
Correct Option: (b) 36m
(viii) $\frac{2}{\sqrt{4}} \times \sqrt{\frac{16}{9}}$ is equal to:
$\frac{2}{2} \times \frac{4}{3} = 1 \times \frac{4}{3} = \frac{4}{3}$
Correct Option: (c) $\frac{4}{3}$
(ix) The volume of a box is 216. The length of the box is:
$l = \sqrt[3]{216} = 6$
Correct Option: (b) 6
(x) Cube root of $(24 \div 8) \times 9$ is:
$\sqrt[3]{3 \times 9} = \sqrt[3]{27} = 3$
Correct Option: (c) 3

Questions 2 – 7: Descriptive & Numerical Solutions

Question 2: Find $\sqrt{1\frac{19}{81}}$.
Solution:
$1\frac{19}{81} = \frac{1 \times 81 + 19}{81} = \frac{100}{81}$
$\sqrt{\frac{100}{81}} = \frac{\sqrt{100}}{\sqrt{81}} = \frac{10}{9} = \mathbf{1\frac{1}{9}}$.
Question 3: Calculate $\sqrt{1+0.69} \times \sqrt{2-0.04}$.
Solution:
$\sqrt{1.69} \times \sqrt{1.96} = 1.3 \times 1.4 = \mathbf{1.82}$.
Question 4: Evaluate $\sqrt{\frac{7+\frac{1}{5}}{9+\frac{4}{5}}}$.
Solution:
$\text{Numerator} = 7 + \frac{1}{5} = \frac{36}{5}$
$\text{Denominator} = 9 + \frac{4}{5} = \frac{49}{5}$
$\sqrt{\frac{36/5}{49/5}} = \sqrt{\frac{36}{49}} = \frac{\sqrt{36}}{\sqrt{49}} = \mathbf{\frac{6}{7}}$.
Question 5: A square field has an area of $289\text{ m}^2$. Find its perimeter.
Solution:
$\text{Side length } s = \sqrt{289\text{ m}^2} = 17\text{ m}$
$\text{Perimeter} = 4 \times s = 4 \times 17\text{ m} = \mathbf{68\text{ m}}$.
Question 6: Volume of a cube is $4913\text{ cm}^3$. Find length of one side and also area of two faces.
Solution:
Length of one side: $s = \sqrt[3]{4913} = \mathbf{17\text{ cm}}$ (since $17^3 = 4913$)
Area of one face: $s^2 = 17 \times 17 = 289\text{ cm}^2$
Area of two faces: $2 \times s^2 = 2 \times 289\text{ cm}^2 = \mathbf{578\text{ cm}^2}$.
Question 7: Estimate the value of the following:
(i) $\sqrt{66}$: Since $8^2 = 64 < 66 < 81 = 9^2$, $\sqrt{66} \approx \mathbf{8.1}$ (exact $\approx 8.12$).
(ii) $\sqrt{80}$: Since $8^2 = 64 < 80 < 81 = 9^2$ (very close to 81), $\sqrt{80} \approx \mathbf{8.9}$ (exact $\approx 8.94$).
(iii) $\sqrt[3]{218}$: Since $6^3 = 216 < 218 < 343 = 7^3$ (very close to 216), $\sqrt[3]{218} \approx \mathbf{6.0}$ (exact $\approx 6.02$).

9. Chapter Summary & Rapid Review Flashcards

Square vs Cube • Square: $x^2 = x \times x$, Root: $\sqrt{x}$
• Cube: $x^3 = x \times x \times x$, Root: $\sqrt[3]{x}$
Square Properties • A perfect square is never negative.
• Square of even is even; square of odd is odd.
• Never ends in 2, 3, 7, or 8.
Fraction / Product Rules • $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$
• $\sqrt{a/b} = \sqrt{a}/\sqrt{b}$
• $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$
Words Board (Textbook Page 44) • Square (مربع)
• Square Root (جذر المربع)
• Cube (مکعب)
• Cube Root (جذر المکعب)
• Division Method (تقسیمی طریقہ)
1. Why is $\sqrt{0.9} \approx 0.95$ and NOT $0.3$?
Because $0.3 \times 0.3 = 0.09$, NOT $0.9$. To find $\sqrt{0.9}$, we group decimals starting from the decimal point to the right as $0.\overline{90}$. Since $9^2 = 81 < 90$, the root begins with $0.948\dots \approx 0.95$.
2. How do you find the least number to subtract to make a number a perfect square?
Perform standard long division square root on the given number. The remainder at the end of the division is the exact least number that must be subtracted from the original number.
3. What is the difference between finding side length from area vs from volume?
For a 2D square with area $A$, the side length is the square root: $s = \sqrt{A}$. For a 3D cube with volume $V$, the side length is the cube root: $s = \sqrt[3]{V}$.

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