Class 8 Mathematics

Chapter 1

Rational Numbers

Definition

A rational number is a number that can be written as $\dfrac{p}{q}$ where $p, q$ are integers and $q \neq 0$.

Properties of Operations on Rational Numbers

Property	Addition	Multiplication
Closure	$\dfrac{a}{b}+\dfrac{c}{d}$ is rational	$\dfrac{a}{b}\times\dfrac{c}{d}$ is rational
Commutativity	$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{c}{d}+\dfrac{a}{b}$	$\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{c}{d}\times\dfrac{a}{b}$
Associativity	$(a+b)+c=a+(b+c)$	$(a\times b)\times c=a\times(b\times c)$
Identity	$a+0=0+a=a$	$a\times1=1\times a=a$
Inverse	$a+(-a)=0$	$a\times\dfrac{1}{a}=1\;(a\neq0)$

Distributive Property

$$a \times (b + c) = a \times b + a \times c$$

Rational Number Between Two Rationals

Between any two rational numbers $a$ and $b$, there exists a rational number:
$$\text{Mean} = \frac{a + b}{2}$$ In fact, infinitely many rationals lie between any two rationals.

Proof — Rational Mean lies between a and b

Let $a < b$. We must show $a < \dfrac{a+b}{2} < b$.

Since $a < b$, we have $a + a < a + b$, so $2a < a+b$, thus $a < \dfrac{a+b}{2}$.

Similarly $a+b < b+b = 2b$, so $\dfrac{a+b}{2} < b$. ∴ Mean lies strictly between $a$ and $b$. ∎

Additive Inverse

The additive inverse of $\dfrac{p}{q}$ is $-\dfrac{p}{q}$, since $\dfrac{p}{q} + \left(-\dfrac{p}{q}\right) = 0$

Multiplicative Inverse (Reciprocal)

The multiplicative inverse of $\dfrac{p}{q}$ is $\dfrac{q}{p}$ (where $p, q \neq 0$), since $\dfrac{p}{q} \times \dfrac{q}{p} = 1$

Subtraction and division are NOT commutative for rational numbers: $a - b \neq b - a$ and $a \div b \neq b \div a$ in general.

Chapter 2

Linear Equations in One Variable

Standard Form and Solution

General Linear Equation

$$ax + b = 0 \quad (a \neq 0) \implies x = -\frac{b}{a}$$

Equation with Variable on Both Sides

$$ax + b = cx + d \implies x = \frac{d - b}{a - c} \quad (a \neq c)$$

Cross Multiplication (Proportions)

$$\frac{a}{b} = \frac{c}{d} \implies ad = bc$$

Proof — Cross Multiplication

Given $\dfrac{a}{b} = \dfrac{c}{d}$. Multiply both sides by $bd$:

$\dfrac{a}{b} \times bd = \dfrac{c}{d} \times bd \implies ad = bc$ ∎

Key Application Formulas

Speed, Distance, Time

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}, \quad D = S \times T, \quad T = \frac{D}{S}$$

Transposition Rule

Any term can be moved from one side to the other by changing its sign:
$ax + b = c \implies ax = c - b \implies x = \dfrac{c-b}{a}$

Chapter 3

Understanding Quadrilaterals

Angle Sum of Polygons

Sum of Interior Angles of a Polygon

$$S = (n - 2) \times 180°$$ where $n$ = number of sides of the polygon.

Proof — Polygon Angle Sum Formula

A polygon with $n$ sides can be divided into $(n-2)$ triangles by drawing diagonals from one vertex.

Each triangle has angle sum = 180°.

∴ Total angle sum = $(n-2) \times 180°$ ∎

Hexagon divided into (n−2) = 4 triangles from vertex A. Each triangle contributes 180°.

Each Interior Angle of a Regular n-gon

$$\text{Each interior angle} = \frac{(n-2) \times 180°}{n}$$

Sum of Exterior Angles of Any Polygon

$$\text{Sum of exterior angles} = 360°$$ (regardless of number of sides)

Each Exterior Angle of a Regular n-gon

$$\text{Each exterior angle} = \frac{360°}{n}$$ Also: Interior angle + Exterior angle = 180°

Polygon Angle Sum Table

Polygon	Sides (n)	Sum of Interior Angles	Each Angle (regular)
Triangle	3	180°	60°
Quadrilateral	4	360°	90°
Pentagon	5	540°	108°
Hexagon	6	720°	120°
Heptagon	7	900°	128.6°
Octagon	8	1080°	135°

Properties of Special Quadrilaterals

Shape	Key Properties
Parallelogram	Opposite sides equal & parallel; opposite angles equal; consecutive angles supplementary; diagonals bisect each other
Rectangle	All properties of parallelogram; all angles = 90°; diagonals are equal
Rhombus	All properties of parallelogram; all sides equal; diagonals are perpendicular bisectors of each other
Square	All properties of rectangle + rhombus; all sides equal; all angles = 90°; diagonals equal and perpendicular
Trapezium	Exactly one pair of parallel sides (called bases)
Kite	Two pairs of adjacent equal sides; diagonals perpendicular; one diagonal bisects the other

Chapter 4

Data Handling

Arithmetic Mean

$$\text{Mean} = \bar{x} = \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{\sum x_i}{n}$$

Range

$$\text{Range} = \text{Maximum value} - \text{Minimum value}$$

Probability of an Event

$$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$$

Complementary Probability

$$P(E) + P(\bar{E}) = 1 \implies P(\bar{E}) = 1 - P(E)$$ where $P(\bar{E})$ is the probability that event $E$ does NOT occur.

Proof — Probability Range

Favourable outcomes $n(E)$ satisfy: $0 \leq n(E) \leq n(S)$.

Dividing by $n(S)$: $0 \leq P(E) \leq 1$.

So probability always lies between 0 and 1 (inclusive). ∎

Central Angle in Pie Chart

$$\text{Central angle} = \frac{\text{Value of component}}{\text{Sum of all values}} \times 360°$$

Chapter 5

Squares and Square Roots

Sum of First n Odd Numbers = n²

$$1 + 3 + 5 + \cdots + (2n-1) = n^2$$

Geometric Proof — n² as sum of odd numbers

An $n \times n$ square can be built by adding L-shaped borders (gnomons) to progressively larger squares. The $k$-th gnomon contains $2k-1$ unit squares.

$1^2 = 1$; $2^2 = 1+3$; $3^2 = 1+3+5$; $n^2 = 1+3+5+\cdots+(2n-1)$ ∎

Each new "L-shaped gnomon" adds the next odd number, building n² step by step.

Pythagorean Triplet Formula

For any integer $m > 1$, the set $(2m,\; m^2-1,\; m^2+1)$ is a Pythagorean triplet.
Example: $m=2 \Rightarrow (4, 3, 5)$; $m=3 \Rightarrow (6, 8, 10)$; $m=4 \Rightarrow (8, 15, 17)$

Proof — Pythagorean Triplet

We must show $(2m)^2 + (m^2-1)^2 = (m^2+1)^2$.

LHS $= 4m^2 + m^4 - 2m^2 + 1 = m^4 + 2m^2 + 1$

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

LHS = RHS ∎

Square Root Properties

$$\sqrt{ab} = \sqrt{a} \times \sqrt{b}, \quad a,b > 0$$ $$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, \quad a,b > 0$$ $$\sqrt{a^2} = |a| = a \text{ (when } a \geq 0\text{)}$$

Key Square Root Facts

Numbers ending in 2, 3, 7, 8 are NEVER perfect squares. Perfect squares end in 0, 1, 4, 5, 6, or 9.

Chapter 6

Cubes and Cube Roots

Cube of a Number

$$n^3 = n \times n \times n$$ Geometrically: the volume of a cube with side $n$.

Cube = Sum of Consecutive Odd Numbers

$$1^3 = 1$$ $$2^3 = 3 + 5$$ $$3^3 = 7 + 9 + 11$$ $$4^3 = 13 + 15 + 17 + 19$$ The $k$-th group starts at $(k^2 - k + 1)$-th odd number and has $k$ terms.

Cube Root Properties

$$\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$$ $$\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}} \quad (b \neq 0)$$

A number is a perfect cube only if every prime factor appears in groups of exactly three in its prime factorization.

Cubes of First 10 Integers

n	1	2	3	4	5	6	7	8	9	10
n³	1	8	27	64	125	216	343	512	729	1000

Chapter 7

Comparing Quantities

Percentages

Percentage Formula

$$x\% \text{ of } y = \frac{x \times y}{100}$$ $$\text{Percentage increase} = \frac{\text{Increase}}{\text{Original value}} \times 100$$ $$\text{Percentage decrease} = \frac{\text{Decrease}}{\text{Original value}} \times 100$$

Profit and Loss

Profit and Loss Formulas

$$\text{Profit} = \text{SP} - \text{CP} \quad \text{(when SP > CP)}$$ $$\text{Loss} = \text{CP} - \text{SP} \quad \text{(when CP > SP)}$$ $$\text{Profit\%} = \frac{\text{Profit}}{\text{CP}} \times 100, \quad \text{Loss\%} = \frac{\text{Loss}}{\text{CP}} \times 100$$ $$\text{SP} = \text{CP} \times \frac{100 + \text{Profit\%}}{100} \quad \text{(profit case)}$$ $$\text{SP} = \text{CP} \times \frac{100 - \text{Loss\%}}{100} \quad \text{(loss case)}$$ $$\text{CP} = \frac{\text{SP} \times 100}{100 + \text{Profit\%}} \quad \text{CP} = \frac{\text{SP} \times 100}{100 - \text{Loss\%}}$$

Discount

Discount Formulas

$$\text{Discount} = \text{Marked Price (MP)} - \text{Sale Price (SP)}$$ $$\text{Discount\%} = \frac{\text{Discount}}{\text{MP}} \times 100$$ $$\text{SP} = \text{MP} \times \frac{100 - \text{Discount\%}}{100}$$

Tax (GST / VAT)

Tax Formula

$$\text{Tax amount} = \text{Cost price} \times \frac{\text{Tax\%}}{100}$$ $$\text{Bill amount} = \text{CP} + \text{Tax} = \text{CP} \times \frac{100 + \text{Tax\%}}{100}$$

Simple Interest

$$\text{SI} = \frac{P \times R \times T}{100}$$ $$\text{Amount } A = P + \text{SI} = P\left(1 + \frac{RT}{100}\right)$$ where $P$ = Principal, $R$ = Rate% per annum, $T$ = Time in years.

Compound Interest

Compound Interest — Annually

$$A = P\left(1 + \frac{R}{100}\right)^n$$ $$\text{CI} = A - P = P\left[\left(1 + \frac{R}{100}\right)^n - 1\right]$$ where $n$ = number of years.

Compound Interest — Half Yearly

$$A = P\left(1 + \frac{R}{200}\right)^{2n}$$ Rate becomes $R/2$ per period; periods become $2n$ per year.

Compound Interest — Quarterly

$$A = P\left(1 + \frac{R}{400}\right)^{4n}$$

Growth and Depreciation

$$\text{Final Value} = P\left(1 + \frac{R}{100}\right)^n \quad \text{(growth)}$$ $$\text{Final Value} = P\left(1 - \frac{R}{100}\right)^n \quad \text{(depreciation / decay)}$$

Derivation — Why CI Formula Uses Powers

Year 1: Amount $= P + P\cdot\frac{R}{100} = P\left(1+\frac{R}{100}\right)$. Call this $P_1$.

Year 2: Amount $= P_1\left(1+\frac{R}{100}\right) = P\left(1+\frac{R}{100}\right)^2$.

After $n$ years: $A = P\left(1+\frac{R}{100}\right)^n$ ∎

Difference from SI: CI is always greater than SI (for same P, R, T with $n > 1$), since interest is earned on interest.

Chapter 8

Algebraic Expressions and Identities

The Four Standard Algebraic Identities

Identity 1

$$(a + b)^2 = a^2 + 2ab + b^2$$

Algebraic Proof — (a+b)²

$(a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2$ ∎

Geometric proof: A square of side (a+b) splits into four regions with areas a², ab, ab, b².

Identity 2

$$(a - b)^2 = a^2 - 2ab + b^2$$

Algebraic Proof — (a−b)²

$(a-b)^2 = (a-b)(a-b) = a(a-b) - b(a-b) = a^2 - ab - ab + b^2 = a^2 - 2ab + b^2$ ∎

Also: $(a-b)^2 = (a+b)^2 - 4ab$ (useful identity)

Rearranging the big square (a²) minus two (a−b)×b strips gives (a−b)² = a²−2ab+b².

Identity 3

$$(a + b)(a - b) = a^2 - b^2$$

Algebraic Proof — (a+b)(a−b) = a²−b²

$(a+b)(a-b) = a(a-b)+b(a-b) = a^2 - ab + ab - b^2 = a^2 - b^2$ ∎

The L-shaped region (a²−b²) can be rearranged into a rectangle with sides (a+b) and (a−b).

Identity 4

$$(x + a)(x + b) = x^2 + (a + b)x + ab$$

Algebraic Proof — Identity 4

$(x+a)(x+b) = x(x+b) + a(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab$ ∎

Summary of Identities

Identity	Expression
1. Square of a sum	$(a+b)^2 = a^2 + 2ab + b^2$
2. Square of a difference	$(a-b)^2 = a^2 - 2ab + b^2$
3. Product of sum & difference	$(a+b)(a-b) = a^2 - b^2$
4. Product of binomials	$(x+a)(x+b) = x^2+(a+b)x+ab$

Useful derived result: $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$ and $(a+b)^2 - (a-b)^2 = 4ab$

Chapter 9

Mensuration

Two-Dimensional Figures

Trapezium (Trapezoid)

$$\text{Area} = \frac{1}{2}(a + b) \times h$$ where $a, b$ are the parallel sides and $h$ is the perpendicular height.

Proof — Area of Trapezium

Take two identical trapeziums and join them to form a parallelogram.

The parallelogram has base $= a+b$ and height $= h$.

Area of parallelogram $= (a+b) \times h$.

Since two trapeziums were used: Area of one trapezium $= \dfrac{1}{2}(a+b)h$ ∎

Two congruent trapeziums form a parallelogram with base (a+b) and height h.

Rhombus

$$\text{Area} = \frac{1}{2} \times d_1 \times d_2$$ where $d_1$ and $d_2$ are the two diagonals.

Proof — Area of Rhombus

The diagonals of a rhombus bisect each other at right angles.

This creates 4 right triangles, each with legs $d_1/2$ and $d_2/2$.

Area $= 4 \times \dfrac{1}{2} \times \dfrac{d_1}{2} \times \dfrac{d_2}{2} = \dfrac{1}{2}d_1 d_2$ ∎

Diagonals of rhombus meet at 90°, creating 4 equal right triangles.

General Quadrilateral

$$\text{Area} = \frac{1}{2} \times d \times (h_1 + h_2)$$ where $d$ is one diagonal and $h_1, h_2$ are perpendiculars from the other two vertices to that diagonal.

Three-Dimensional Figures

Cuboid (Box): length l, breadth b, height h

Lateral Surface Area (LSA) $= 2h(l+b)$
Total Surface Area (TSA) $= 2(lb + bh + hl)$
Volume $= l \times b \times h$
Face diagonal $= \sqrt{l^2+b^2}$, $\sqrt{b^2+h^2}$, $\sqrt{l^2+h^2}$
Space diagonal $= \sqrt{l^2+b^2+h^2}$

Cuboid with dimensions l (length), b (breadth), h (height).

Cube: all sides = a

Lateral Surface Area (LSA) $= 4a^2$
Total Surface Area (TSA) $= 6a^2$
Volume $= a^3$
Face diagonal $= a\sqrt{2}$
Space (body) diagonal $= a\sqrt{3}$

Proof — TSA of Cube

A cube has 6 identical square faces, each of area $a^2$.

TSA $= 6 \times a^2 = 6a^2$ ∎

LSA: Only 4 side faces contribute $\Rightarrow$ LSA $= 4a^2$ ∎

Right Circular Cylinder: radius r, height h

Curved Surface Area (CSA) $= 2\pi r h$
Total Surface Area (TSA) $= 2\pi r(r + h) = 2\pi r^2 + 2\pi rh$
Volume $= \pi r^2 h$

Proof — CSA of Cylinder

Unroll the curved surface of a cylinder into a flat rectangle.

Width of rectangle = circumference of base $= 2\pi r$.

Height of rectangle $= h$.

CSA $= 2\pi r \times h = 2\pi rh$ ∎

TSA $=$ CSA $+ 2 \times$ base area $= 2\pi rh + 2\pi r^2 = 2\pi r(r+h)$ ∎

Complete Formula Book — Chapterwise with Proofs & Diagrams

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