Chapter 15: Similarity (With Applications to Maps & Models) - ICSE Class 10 Mathematics Master Notes

Official ICSE Syllabus & Scope of Examination

Prescribed Syllabus: Axioms of similarity of triangles (SAS, AA / AAA, SSS); Basic Proportionality Theorem (BPT / Thales Theorem) and its applications; Altitude drawn to the hypotenuse in a right-angled triangle; Theorem on the ratio of areas of similar triangles; Applications of similarity to Maps and Scale Models (Scale Factor $k$, $k^2$ for Areas, $k^3$ for Volumes).

Core Theorems Covered with Complete Proofs:
1. Similarity Axioms: Equal corresponding angles and proportional corresponding sides ($\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$).
2. Basic Proportionality Theorem (BPT): Line parallel to one side of a triangle divides the other two sides proportionally ($\frac{AD}{DB} = \frac{AE}{EC}$ and $\frac{AD}{AB} = \frac{DE}{BC}$).
3. Altitude to Hypotenuse: $\triangle DBA \sim \triangle DAC \sim \triangle ABC \implies AD^2 = BD \times CD, AB^2 = BD \times BC, AC^2 = CD \times BC$.
4. Area Ratio of Similar Triangles: $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{\text{Altitude}_1}{\text{Altitude}_2}\right)^2 = \left(\frac{\text{Median}_1}{\text{Median}_2}\right)^2$.
5. Size Transformations & Maps/Models: Linear scale $k$, Area scale $k^2$, Volume scale $k^3$, and unit conversions ($1\text{ km} = 10^5\text{ cm}$, $1\text{ km}^2 = 10^{10}\text{ cm}^2$).

PART (A): Similarity vs Congruency & Axioms of Similar Triangles

1. Similarity vs Congruency of Geometric Figures

Property	Similar Figures ($\sim$)	Congruent Figures ($\cong$)
Shape	Same Shape (All corresponding angles are equal).	Same Shape (All corresponding angles are equal).
Size	May Differ in Size (Corresponding sides are in proportion).	Exactly Same Size (Corresponding sides are strictly equal).
Relation	Similar figures are not necessarily congruent.	Congruent figures are ALWAYS similar.

ICSE Conceptual Rules on Polygons Similarity (True/False Checklist)

All Circles are Similar: Any two circles have identical shape ($360^\circ$), differing only by radius $r$.
All Squares are Similar: All angles are $90^\circ$ and all four sides are equal, so side ratios are always $1 : 1$.
All Equilateral Triangles are Similar: All angles are strictly $60^\circ$ (by AAA postulate).
All Isosceles-Right Triangles are Similar: Angles are always $45^\circ, 45^\circ, 90^\circ$ (by AAA postulate).
Rectangles & Rhombuses: Two rectangles or two rhombuses are NOT necessarily similar because their angles (for rhombus) or side aspect ratios (for rectangle) may differ!
Transitive Property of Similarity: Two geometric figures which are similar to the same figure are similar to each other: $$ \text{If } \triangle A \sim \triangle B \quad \text{and} \quad \triangle B \sim \triangle C \implies \mathbf{\triangle A \sim \triangle C} $$

2. Similar Triangles & Order of Vertices

SIMILARITY DEFINITION & SIDES PROPORTIONALITY

Two triangles $\triangle ABC$ and $\triangle PQR$ are similar (written $\triangle ABC \sim \triangle PQR$) if and only if:

$$ \text{(i) } \angle A = \angle P, \quad \angle B = \angle Q, \quad \angle C = \angle R $$ $$ \text{(ii) } \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} $$

Corresponding sides are opposite to equal corresponding angles: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$.

Rules for Writing Corresponding Parts & Vertex Ordering

Rule of Corresponding Sides: Sides opposite to equal angles are corresponding sides.
Rule of Corresponding Angles: Angles opposite to proportional sides are corresponding angles and are equal.
Vertex Sequence Strictness: If vertex $A \leftrightarrow D$, $B \leftrightarrow E$, and $C \leftrightarrow F$, we must write $\triangle ABC \sim \triangle DEF$. Writing $\triangle ABC \sim \triangle DFE$ is mathematically incorrect!

3. Postulates / Criteria for Similarity

Similarity Postulate	Statement / Condition	Mathematical Form
1. AA (or AAA) Postulate	If two angles of one triangle are respectively equal to two angles of another triangle, the triangles are similar. (Third angles are automatically equal by angle-sum property $180^\circ$).	If $\angle A = \angle D$ and $\angle B = \angle E \implies \mathbf{\triangle ABC \sim \triangle DEF}$.
2. SAS Postulate	If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar.	If $\angle A = \angle D$ and $\frac{AB}{DE} = \frac{AC}{DF} \implies \mathbf{\triangle ABC \sim \triangle DEF}$.
3. SSS Postulate	If all three corresponding sides of two triangles are in the same ratio, the triangles are similar.	If $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \implies \mathbf{\triangle ABC \sim \triangle DEF}$.

4. Proportionality of Altitudes, Medians, Angle Bisectors & Perimeters

CORRESPONDING LINEAR ELEMENTS THEOREM

If two triangles are similar ($\triangle ABC \sim \triangle PQR$), then the ratio of their corresponding sides is equal to the ratio of their:

$$ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \mathbf{\frac{\text{Altitude } AM}{\text{Altitude } PN}} = \mathbf{\frac{\text{Median } AX}{\text{Median } PY}} = \mathbf{\frac{\text{Angle Bisector } AL}{\text{Angle Bisector } PZ}} = \mathbf{\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle PQR)}} $$

4. Classroom Practice Problems (Similarity Axioms & Sides Calculation)

✍ PRACTICE PROBLEM A1: PERPENDICULARS ON BASE (AA SIMILARITY)

Q. In $\triangle ABC$, $AB \perp BC$ and $DE \perp BC$. If $AB = 9\text{ cm}$, $DE = 3\text{ cm}$, and $AC = 24\text{ cm}$, calculate the lengths of $DC$ and $AD$.

Step 1: Prove Similarity of $\triangle ABC$ and $\triangle DEC$:

$\angle ABC = \angle DEC = 90^\circ$ (Given).
$\angle C = \angle C$ (Common angle).
$\therefore \mathbf{\triangle ABC \sim \triangle DEC}$ by AA Postulate.

Step 2: Set up Proportionality Ratio: $$ \frac{AC}{DC} = \frac{AB}{DE} \implies \frac{24}{DC} = \frac{9}{3} = 3 $$ $$ DC = \frac{24}{3} = \mathbf{8\text{ cm}} $$

Step 3: Calculate $AD$: $$ AD = AC - DC = 24 - 8 = \mathbf{16\text{ cm}} $$

✍ PRACTICE PROBLEM A2: INTERSECTING CHORDS / TRANSVERSALS IN TRAPEZIUM

Q. In a trapezium $ABCD$, side $AB \parallel DC$. The diagonals $AC$ and $BD$ intersect each other at $P$.
(i) Prove that $\triangle APB \sim \triangle CPD$.
(ii) If $AP = 3.6\text{ cm}$, $PC = 2.4\text{ cm}$, and $PB = 4.0\text{ cm}$, find the length of $PD$.

(i) Proof of Similarity ($\triangle APB \sim \triangle CPD$):

$\angle PAB = \angle PCD$ (Alternate interior angles since $AB \parallel DC$).
$\angle PBA = \angle PDC$ (Alternate interior angles since $AB \parallel DC$).
$\angle APB = \angle CPD$ (Vertically opposite angles).
$\therefore \mathbf{\triangle APB \sim \triangle CPD}$ by AA Postulate. (Hence Proved!)

(ii) Calculate $PD$: Corresponding sides are proportional: $$ \frac{PA}{PC} = \frac{PB}{PD} \implies \frac{3.6}{2.4} = \frac{4.0}{PD} $$ $$ PD = \frac{4.0 \times 2.4}{3.6} = \frac{9.6}{3.6} = \mathbf{2.67\text{ cm}} \quad \left(\text{or } \frac{8}{3}\text{ cm}\right) $$

PART (B): Basic Proportionality Theorem (BPT / Thales Theorem)

1. Theorem Statement & Formal Statement-Reason Proof

THEOREM 1: BASIC PROPORTIONALITY THEOREM (BPT)

Statement: A line drawn parallel to one side of a triangle dividing the other two sides in distinct points divides the two sides in the same ratio.

Basic Proportionality Theorem: $DE \parallel BC \implies \frac{AD}{DB} = \frac{AE}{EC}$.

Given: In $\triangle ABC$, line $DE \parallel BC$, intersecting $AB$ at $D$ and $AC$ at $E$.

To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.

Statement	Reason
1. In $\triangle ABC$ and $\triangle ADE$: $\angle ADE = \angle ABC$ $\angle AED = \angle ACB$ $\angle A = \angle A$	Corresponding angles ($DE \parallel BC$). Corresponding angles ($DE \parallel BC$). Common angle.
2. $\therefore \triangle ADE \sim \triangle ABC$	By AAA Postulate.
3. $\frac{AB}{AD} = \frac{AC}{AE}$	Corresponding sides of similar triangles are proportional.
4. $\frac{AD + DB}{AD} = \frac{AE + EC}{AE} \implies 1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}$	Splitting $AB = AD + DB$ and $AC = AE + EC$.
5. $\frac{DB}{AD} = \frac{EC}{AE} \implies \mathbf{\frac{AD}{DB} = \frac{AE}{EC}}$	Subtracting $1$ from both sides and taking reciprocal. (Hence Proved!)

2. Corollaries & Three Parallel Lines Intercepts

Crucial Corollaries of BPT

Full Side Proportionality: $$ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} \quad \text{and} \quad \frac{DB}{AB} = \frac{EC}{AC} $$
Converse of BPT: If a line divides any two sides of a triangle in the same ratio ($\frac{AD}{DB} = \frac{AE}{EC}$), then the line is parallel to the third side ($DE \parallel BC$).
Three Parallel Lines Intercepted by Two Transversals: If three parallel lines $l \parallel m \parallel n$ are intersected by transversals at $A, B, C$ and $P, Q, R$, then: $$ \frac{AB}{BC} = \frac{PQ}{QR} $$

3. Classroom Practice Problems (BPT & Corollaries)

✍ PRACTICE PROBLEM B1: SEGMENTS CALCULATION USING BPT

Q. In $\triangle ABC$, $DE \parallel BC$. $D$ divides $AB$ in the ratio $AD : DB = 2 : 3$. If $BC = 7.5\text{ cm}$, calculate:
(i) $\frac{AE}{EC}$, (ii) $\frac{AE}{AC}$, (iii) Length of $DE$.

(i) Find $\frac{AE}{EC}$: By BPT: $$ \frac{AE}{EC} = \frac{AD}{DB} = \mathbf{\frac{2}{3}} $$

(ii) Find $\frac{AE}{AC}$: $$ \frac{AE}{AC} = \frac{AE}{AE + EC} = \frac{2}{2 + 3} = \mathbf{\frac{2}{5}} $$

(iii) Calculate $DE$: Since $\triangle ADE \sim \triangle ABC$: $$ \frac{DE}{BC} = \frac{AD}{AB} = \frac{2}{5} \implies DE = \frac{2}{5} \times 7.5 = \mathbf{3.0\text{ cm}} $$

✍ PRACTICE PROBLEM B2: TESTING PARALLELISM USING CONVERSE OF BPT

Q. In $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively. Check whether $DE \parallel BC$ in each case:
(a) $AD = 3\text{ cm}, DB = 4.5\text{ cm}, AE = 4\text{ cm}, AC = 10\text{ cm}$.
(b) $AB = 7\text{ cm}, BD = 4.5\text{ cm}, AE = 3.5\text{ cm}, CE = 5.6\text{ cm}$.

Case (a): $EC = AC - AE = 10 - 4 = 6\text{ cm}$. $$ \frac{AD}{DB} = \frac{3}{4.5} = \frac{2}{3} \quad \text{and} \quad \frac{AE}{EC} = \frac{4}{6} = \frac{2}{3} $$ Since $\frac{AD}{DB} = \frac{AE}{EC} \implies \mathbf{DE \parallel BC}$ (By Converse of BPT).

Case (b): $AD = AB - BD = 7 - 4.5 = 2.5\text{ cm}$. $$ \frac{AD}{DB} = \frac{2.5}{4.5} = \frac{5}{9} \quad \text{and} \quad \frac{AE}{CE} = \frac{3.5}{5.6} = \frac{5}{8} $$ Since $\frac{5}{9} \ne \frac{5}{8} \implies \mathbf{DE \text{ is NOT parallel to } BC}$.

PART (C): Altitude to Hypotenuse in a Right-Angled Triangle

1. Theorem & The Three Right-Triangle Similarities

RIGHT TRIANGLE ALTITUDE THEOREM

Theorem Statement: If a perpendicular is drawn from the vertex of the right angle of a right-angled triangle to its hypotenuse, then the two triangles on each side of the perpendicular are similar to the whole triangle and to each other.

Altitude $AD \perp BC$ gives $\triangle DBA \sim \triangle DAC \sim \triangle ABC$.

Three Similar Pairs & Core Algebraic Formulas:

$$ \text{1. } \triangle DBA \sim \triangle ABC \implies \frac{AB}{BC} = \frac{BD}{AB} \implies \mathbf{AB^2 = BD \times BC} $$ $$ \text{2. } \triangle DAC \sim \triangle ABC \implies \frac{AC}{BC} = \frac{CD}{AC} \implies \mathbf{AC^2 = CD \times BC} $$ $$ \text{3. } \triangle DBA \sim \triangle DAC \implies \frac{AD}{CD} = \frac{BD}{AD} \implies \mathbf{AD^2 = BD \times CD} \quad (\text{Geometric Mean Theorem}) $$ $$ \text{4. Ratio of Squares of Sides: } \mathbf{\frac{AB^2}{AC^2} = \frac{BD \times BC}{CD \times BC} = \frac{BD}{CD}} \quad \text{and} \quad \mathbf{\frac{BC^2}{AC^2} = \frac{BD}{AD}} $$

2. Classroom Practice Problems (Right-Triangle Altitude Models)

✍ PRACTICE PROBLEM C1: ALTITUDE & SIDES CALCULATION

Q. In right-angled $\triangle ABC$, $\angle A = 90^\circ$ and $AD \perp BC$. If $BD = 3.6\text{ cm}$ and $CD = 6.4\text{ cm}$, calculate:
(i) The length of altitude $AD$, (ii) The length of side $AB$, (iii) The length of side $AC$, (iv) The ratio $\frac{AB^2}{AC^2}$.

Total Hypotenuse $BC$: $$ BC = BD + CD = 3.6 + 6.4 = 10.0\text{ cm} $$

(i) Calculate Altitude $AD$ (Geometric Mean): $$ AD^2 = BD \times CD = 3.6 \times 6.4 = 23.04 \implies AD = \sqrt{23.04} = \mathbf{4.8\text{ cm}} $$

(ii) Calculate Side $AB$: $$ AB^2 = BD \times BC = 3.6 \times 10 = 36 \implies AB = \sqrt{36} = \mathbf{6.0\text{ cm}} $$

(iii) Calculate Side $AC$: $$ AC^2 = CD \times BC = 6.4 \times 10 = 64 \implies AC = \sqrt{64} = \mathbf{8.0\text{ cm}} $$

(iv) Ratio $\frac{AB^2}{AC^2}$: $$ \frac{AB^2}{AC^2} = \frac{BD}{CD} = \frac{3.6}{6.4} = \mathbf{\frac{9}{16}} $$

PART (D): Relation Between Areas of Two Similar Triangles

1. Area Theorem & Complete Proof

THEOREM 2: AREA RATIO OF SIMILAR TRIANGLES

Statement: The ratio of the areas of two similar triangles is proportional to the squares of their corresponding sides.

$$ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2 = \left(\frac{AM}{DN}\right)^2 = \left(\frac{\text{Perimeter}_1}{\text{Perimeter}_2}\right)^2 $$

Area ratio is proportional to the square of corresponding sides and altitudes: $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AM}{DN}\right)^2$.

Statement	Reason
1. $\text{Area}(\triangle ABC) = \frac{1}{2} BC \times AM$ $\text{Area}(\triangle DEF) = \frac{1}{2} EF \times DN$ $\implies \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \frac{BC}{EF} \times \frac{AM}{DN}$ … (I)	$\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{altitude}$.
2. In $\triangle ABM$ and $\triangle DEN$: $\angle B = \angle E$ $\angle AMB = \angle DNE = 90^\circ$ $\therefore \triangle ABM \sim \triangle DEN \implies \frac{AM}{DN} = \frac{AB}{DE}$ … (II)	Given ($\triangle ABC \sim \triangle DEF$). By construction ($AM \perp BC, DN \perp EF$). By AA Postulate.
3. Since $\triangle ABC \sim \triangle DEF \implies \frac{AB}{DE} = \frac{BC}{EF}$ … (III)	Corresponding sides of similar triangles are in proportion.
4. From (II) and (III): $\frac{AM}{DN} = \frac{BC}{EF}$	Transitive equality.
5. Substituting into (I): $$ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \frac{BC}{EF} \times \frac{BC}{EF} = \mathbf{\frac{BC^2}{EF^2}} $$	Hence Proved!

The Trapezium Diagonals Area Quad-Property

In any trapezium $ABCD$ where $AB \parallel DC$ and diagonals $AC$ and $BD$ intersect at point $P$:

$\triangle APB \sim \triangle CPD \implies \mathbf{\frac{\text{Area}(\triangle APB)}{\text{Area}(\triangle CPD)} = \left(\frac{AB}{DC}\right)^2 = \left(\frac{AP}{PC}\right)^2}$.
Equal Side-Triangles Area: $\mathbf{\text{Area}(\triangle APD) = \text{Area}(\triangle BPC)}$ (because $\triangle ADC$ and $\triangle BDC$ lie on the same base $DC$ between the same parallels $AB \parallel DC$, so $\text{Area}(\triangle ADC) = \text{Area}(\triangle BDC)$. Subtracting $\triangle DPC$ from both gives $\text{Area}(\triangle APD) = \text{Area}(\triangle BPC)$).
Linear Base Ratio: $\frac{\text{Area}(\triangle APD)}{\text{Area}(\triangle APB)} = \frac{PD}{PB} = \frac{DC}{AB}$ (Common altitude from $A$).

⚠️ CRITICAL EXAM DISTINCTION: SIMILAR VS COMMON VERTEX TRIANGLES

Similar Triangles ($\triangle ADE \sim \triangle ABC$): Areas are proportional to the SQUARE of sides: $$ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left(\frac{DE}{BC}\right)^2 = \left(\frac{AD}{AB}\right)^2 $$
Triangles with Common Vertex / Collinear Bases: Areas are proportional directly to the RATIO OF BASES (Linear, NOT squared!): $$ \frac{\text{Area}(\triangle ABE)}{\text{Area}(\triangle AEC)} = \frac{BE}{EC} $$

2. Classroom Practice Problems (Area Theorem & Trapezium Ratios)

✍ PRACTICE PROBLEM D1: PARALLEL LINE AREA & TRAPEZIUM SUBTRACTION

Q. In $\triangle ABC$, $DE \parallel BC$. Given that $AD = \frac{1}{2} BD$ and $BC = 4.5\text{ cm}$:
(i) Calculate the length of $DE$.
(ii) Find the ratio $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)}$.
(iii) Find the ratio $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\text{trapezium } BCED)}$.

(i) Calculate $DE$: Given $AD = \frac{1}{2}BD \implies \frac{AD}{BD} = \frac{1}{2}$. $$ \frac{AD}{AB} = \frac{AD}{AD + BD} = \frac{1}{1 + 2} = \frac{1}{3} $$ Since $\triangle ADE \sim \triangle ABC$: $$ \frac{DE}{BC} = \frac{AD}{AB} = \frac{1}{3} \implies DE = \frac{1}{3} \times 4.5 = \mathbf{1.5\text{ cm}} $$

(ii) Find $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)}$: $$ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left(\frac{DE}{BC}\right)^2 = \left(\frac{1}{3}\right)^2 = \mathbf{\frac{1}{9}} $$

(iii) Find Area Ratio with Trapezium: $$ \text{Area}(\text{trapezium } BCED) = \text{Area}(\triangle ABC) - \text{Area}(\triangle ADE) = 9 - 1 = 8 \text{ units} $$ $$ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\text{trapezium } BCED)} = \mathbf{\frac{1}{8}} $$

✍ PRACTICE PROBLEM D2: LINE PARALLEL TO BASE BISECTING TRIANGLE AREA

Q. In $\triangle ABC$, a line segment $PQ \parallel BC$ divides $\triangle ABC$ into two parts equal in area. Find the ratio $\frac{BP}{AB}$.

Step 1: Set up Area Ratio: Given: $\text{Area}(\triangle APQ) = \text{Area}(\text{trapezium } PBCQ) = \frac{1}{2}\text{Area}(\triangle ABC)$. $$ \frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle ABC)} = \frac{1}{2} $$

Step 2: Relate to Corresponding Sides: Since $PQ \parallel BC \implies \triangle APQ \sim \triangle ABC$: $$ \left(\frac{AP}{AB}\right)^2 = \frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle ABC)} = \frac{1}{2} \implies \frac{AP}{AB} = \frac{1}{\sqrt{2}} $$

Step 3: Solve for $\frac{BP}{AB}$: $$ \frac{BP}{AB} = \frac{AB - AP}{AB} = 1 - \frac{AP}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \mathbf{\frac{2 - \sqrt{2}}{2}} $$

PART (E): Applications to Maps, Models & Size Transformations

1. Similarity as a Size Transformation

Enlargement & Reduction About Center $P$

When a geometric figure is transformed such that every point moves radially from a fixed center $P$ by a constant ratio $k$:

Scale Factor $k > 1$: The transformation is an Enlargement (Image is larger than pre-image).
Scale Factor $k < 1$: The transformation is a Reduction (Image is smaller than pre-image).
Scale Factor $k = 1$: The transformation is an Identity Transformation (Image is congruent to pre-image).
Preservation: Angles are preserved; collinear lines remain lines; the image of an $n$-sided polygon is a similar $n$-sided polygon.

2. Mathematical Foundation of Scale Factor ($k$) in Maps & Models

SCALE FACTOR ($k$) & TRANSFORMATION MATRIX $$ k = \frac{\text{Linear Dimension of Model / Map}}{\text{Corresponding Linear Dimension of Actual Object}} $$

Dimension	Model vs Actual Relation	Scale Ratio Formula
1D: Length / Height / Perimeter	$\text{Length}_{\text{model}} = k \times \text{Length}_{\text{actual}}$	$\frac{\text{Length}_{\text{model}}}{\text{Length}_{\text{actual}}} = \mathbf{k}$
2D: Surface Area / Plot Area	$\text{Area}_{\text{model}} = k^2 \times \text{Area}_{\text{actual}}$	$\frac{\text{Area}_{\text{model}}}{\text{Area}_{\text{actual}}} = \mathbf{k^2}$
3D: Volume / Capacity / Mass	$\text{Volume}_{\text{model}} = k^3 \times \text{Volume}_{\text{actual}}$	$\frac{\text{Volume}_{\text{model}}}{\text{Volume}_{\text{actual}}} = \mathbf{k^3}$

Essential Unit Conversion Factors

Length: $1\text{ km} = 1000\text{ m} = 100,000\text{ cm} = 10^5\text{ cm}$.
Area: $1\text{ km}^2 = (10^5\text{ cm})^2 = 10^{10}\text{ cm}^2 = 10^6\text{ m}^2$.
Area: $1\text{ m}^2 = (100\text{ cm})^2 = 10,000\text{ cm}^2 = 10^4\text{ cm}^2$.
Volume / Capacity: $1\text{ m}^3 = (100\text{ cm})^3 = 10^6\text{ cm}^3 = 1000\text{ liters}$.
Volume: $1\text{ liter} = 1000\text{ cm}^3$.

3. Classroom Practice Problems (Maps & Models)

✍ PRACTICE PROBLEM E1: MAP SCALE & ACTUAL PLOT AREA

Q. The scale of a map is $1 : 50,000$. On the map, a triangular plot $ABC$ has dimensions $AB = 2\text{ cm}, BC = 3.5\text{ cm}, \angle B = 90^\circ$. Calculate:
(i) The actual length of side $BC$ in kilometers.
(ii) The actual area of the plot in $\text{km}^2$.

Scale Factor: $k = \frac{1}{50,000}$.

(i) Actual Length of $BC$: $$ \text{Actual } BC = 3.5\text{ cm} \times 50,000 = 175,000\text{ cm} = \frac{175,000}{100,000}\text{ km} = \mathbf{1.75\text{ km}} $$

(ii) Actual Area of Plot: $$ \text{Area on Map} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 2 \times 3.5 = 3.5\text{ cm}^2 $$ $$ \text{Actual Area} = \frac{\text{Map Area}}{k^2} = 3.5 \times (50,000)^2\text{ cm}^2 = 3.5 \times 2.5 \times 10^9\text{ cm}^2 = 8.75 \times 10^9\text{ cm}^2 $$ Convert to $\text{km}^2$ (divide by $10^{10}$): $$ \text{Actual Area} = \frac{8.75 \times 10^9}{10^{10}} = \mathbf{0.875\text{ km}^2} $$

✍ PRACTICE PROBLEM E2: SHIP MODEL (LENGTH, AREA & VOLUME)

Q. A model of a ship is made to a scale of $1 : 200$.
(i) If the length of the model is $4\text{ m}$, calculate the actual length of the ship.
(ii) The area of the deck of the ship is $160,000\text{ m}^2$. Find the area of the deck of the model.
(iii) The volume of the model is $200\text{ liters}$. Calculate the volume of the ship in $\text{m}^3$.

Scale Factor: $k = \frac{1}{200}$.

(i) Actual Length: $$ \text{Actual Length} = \frac{\text{Model Length}}{k} = 4\text{ m} \times 200 = \mathbf{800\text{ m}} $$

(ii) Model Deck Area: $$ \text{Area}_{\text{model}} = k^2 \times \text{Area}_{\text{actual}} = \left(\frac{1}{200}\right)^2 \times 160,000 = \frac{160,000}{40,000} = \mathbf{4\text{ m}^2} $$

(iii) Actual Volume in $\text{m}^3$: $$ \text{Volume}_{\text{actual}} = \frac{\text{Volume}_{\text{model}}}{k^3} = 200\text{ liters} \times (200)^3 = 200 \times 8,000,000 = 1.6 \times 10^9\text{ liters} $$ Since $1\text{ m}^3 = 1000\text{ liters}$: $$ \text{Actual Volume} = \frac{1.6 \times 10^9}{1000} = \mathbf{1.6 \times 10^6\text{ m}^3} $$

✍ PRACTICE PROBLEM E3: AEROPLANE MODEL & SURFACE PAINTING COST

Q. An aeroplane is $30\text{ m}$ long and its model is $15\text{ cm}$ long. If the total outer surface area of the model is $150\text{ cm}^2$, find the cost of painting the outer surface of the aeroplane at the rate of $\text{Rs. } 120\text{ per m}^2$, given that $50\text{ m}^2$ of the surface is left for windows.

Step 1: Calculate Scale Factor $k$: $$ k = \frac{\text{Length of Model}}{\text{Length of Aeroplane}} = \frac{15\text{ cm}}{30\text{ m}} = \frac{15\text{ cm}}{3000\text{ cm}} = \frac{1}{200} $$

Step 2: Calculate Actual Outer Surface Area: $$ \text{Area of Aeroplane} = \frac{\text{Area of Model}}{k^2} = 150 \times (200)^2\text{ cm}^2 = 150 \times 40,000\text{ cm}^2 = 6,000,000\text{ cm}^2 $$ Convert to $\text{m}^2$ (divide by $10,000$): $$ \text{Actual Area} = \frac{6,000,000}{10,000} = 600\text{ m}^2 $$

Step 3: Area to be Painted & Total Cost: $$ \text{Area to be painted} = 600 - 50 = 550\text{ m}^2 $$ $$ \text{Total Cost} = 550 \times \text{Rs. } 120 = \mathbf{\text{Rs. } 66,000} \quad (\text{₹ } 66,000) $$

PART (F): ICSE Board Examination Masterclass (High-Yield Multi-Concept PYQs)

✍ BOARD EXEMPLAR F1: THREE PERPENDICULARS ON A LINE (1/x + 1/y = 1/z)

Q. In the given figure, $AB \perp BF$, $CD \perp BF$, and $EF \perp BF$. If $AB = x$, $CD = z$, and $EF = y$, prove that: $$ \frac{1}{x} + \frac{1}{y} = \frac{1}{z} $$

Step 1: In $\triangle ABF$, $CD \parallel AB$: Since $CD \perp BF$ and $AB \perp BF \implies CD \parallel AB$.
$\therefore \triangle FDC \sim \triangle FBA \implies \frac{CD}{AB} = \frac{DF}{BF} \implies \frac{z}{x} = \frac{DF}{BF}$ … (I)

Step 2: In $\triangle EBF$, $CD \parallel EF$: Since $CD \perp BF$ and $EF \perp BF \implies CD \parallel EF$.
$\therefore \triangle BDC \sim \triangle BFE \implies \frac{CD}{EF} = \frac{BD}{BF} \implies \frac{z}{y} = \frac{BD}{BF}$ … (II)

Step 3: Add Equations (I) and (II): $$ \frac{z}{x} + \frac{z}{y} = \frac{DF}{BF} + \frac{BD}{BF} = \frac{DF + BD}{BF} = \frac{BF}{BF} = 1 $$ Divide the entire equation by $z$: $$ \mathbf{\frac{1}{x} + \frac{1}{y} = \frac{1}{z}} $$ Hence Proved! (Standard ICSE 4-mark question).

✍ BOARD EXEMPLAR F2: MEDIANS OF TRIANGLE & CENTROID 2:1 RATIO

Q. The medians $BD$ and $CE$ of $\triangle ABC$ intersect at $G$. Prove that:
(i) $\triangle EGD \sim \triangle CGB$,
(ii) $BG = 2 GD$ (Centroid divides median in $2 : 1$).

Step 1 (Mid-Point Theorem on $ED$): $E$ is mid-point of $AB$, and $D$ is mid-point of $AC$.
By Mid-point Theorem: $ED \parallel BC$ and $ED = \frac{1}{2} BC \implies \frac{ED}{BC} = \frac{1}{2}$.

Step 2 (Similarity of $\triangle EGD$ and $\triangle CGB$):

$\angle DEG = \angle BCG$ (Alternate interior angles since $ED \parallel BC$).
$\angle EDG = \angle CBG$ (Alternate interior angles since $ED \parallel BC$).
$\angle EGD = \angle CGB$ (Vertically opposite angles).
$\therefore \mathbf{\triangle EGD \sim \triangle CGB}$ by AA Postulate.

Step 3 (Proof of $BG = 2GD$): By ratio of corresponding sides: $$ \frac{GD}{GB} = \frac{ED}{BC} = \frac{1}{2} \implies \mathbf{BG = 2 GD} $$ Hence Proved!

✍ BOARD EXEMPLAR F3: RIGHT TRIANGLE WITH INSCRIBED TRANSVERSAL (ICSE 2014)

Q. In $\triangle ABC$, $\angle ABC = \angle DAC$. $AB = 8\text{ cm}, AC = 4\text{ cm}, AD = 5\text{ cm}$.
(i) Prove that $\triangle ACD \sim \triangle BCA$.
(ii) Find the lengths of $BC$ and $CD$.
(iii) Find $\frac{\text{Area}(\triangle ACD)}{\text{Area}(\triangle BCA)}$.

(i) Proof of $\triangle ACD \sim \triangle BCA$:

$\angle CAD = \angle CBA$ (Given: $\angle DAC = \angle ABC$).
$\angle C = \angle C$ (Common angle to both triangles).
$\therefore \mathbf{\triangle ACD \sim \triangle BCA}$ by AA Postulate.

(ii) Calculate $BC$ and $CD$: $$ \frac{AC}{BC} = \frac{CD}{CA} = \frac{AD}{BA} \implies \frac{4}{BC} = \frac{CD}{4} = \frac{5}{8} $$

$\frac{4}{BC} = \frac{5}{8} \implies BC = \frac{32}{5} = \mathbf{6.4\text{ cm}}$.
$\frac{CD}{4} = \frac{5}{8} \implies CD = \frac{20}{8} = \mathbf{2.5\text{ cm}}$.

(iii) Area Ratio: $$ \frac{\text{Area}(\triangle ACD)}{\text{Area}(\triangle BCA)} = \left(\frac{AC}{BC}\right)^2 = \left(\frac{4}{6.4}\right)^2 = \left(\frac{5}{8}\right)^2 = \mathbf{\frac{25}{64}} $$

✍ BOARD EXEMPLAR F4: PARALLELOGRAM WITH MID-POINT & LINE PROOF (PE = 2PD)

Q. In a parallelogram $ABCD$, $M$ is the mid-point of side $BC$. Line $DM$ intersects diagonal $AC$ at $P$ and $AB$ produced at $E$. Prove that $PE = 2PD$.

Step 1: Prove Congruence of $\triangle DCM$ and $\triangle EBM$:

$MC = MB$ (Given: $M$ is mid-point of $BC$).
$\angle DMC = \angle EMB$ (Vertically opposite angles).
$\angle DCM = \angle EBM$ (Alternate angles since $DC \parallel AE$).
$\therefore \triangle DCM \cong \triangle EBM \implies DC = EB$.

Step 2: Total Length $AE$: $$ AE = AB + BE = DC + DC = 2 DC $$

Step 3: Similarity of $\triangle APE$ and $\triangle CPD$: Since $AE \parallel DC \implies \triangle APE \sim \triangle CPD$: $$ \frac{PE}{PD} = \frac{AE}{DC} = \frac{2 DC}{DC} = 2 \implies \mathbf{PE = 2 PD} $$ Hence Proved!

✍ BOARD EXEMPLAR F5: RHOMBUS & STRAIGHT LINE RATIO PROOF

Q. $ABCD$ is a rhombus. $DPR$ and $CBR$ are straight lines intersecting at $R$, and diagonal $AC$ intersects $DR$ at $P$. Prove that $DP \times CR = DC \times PR$.

Step 1: Identify Similar Triangles: In $\triangle DPC$ and $\triangle RPA$:

$\angle DPC = \angle RPA$ (Vertically opposite angles).
$\angle PCD = \angle PAR$ (Alternate angles since $AB \parallel DC$).
$\therefore \triangle DPC \sim \triangle RPA$ or considering parallel lines $DC \parallel AB$ in $\triangle RDC$: $$ \frac{DP}{PR} = \frac{DC}{CR} \implies \mathbf{DP \times CR = DC \times PR} $$

Hence Proved!

PART (G): Quick Revision Formula & Theorem Matrix

Theorem / Concept	Key Mathematical Formula	Crucial Exam Tips
Similar Triangles Definition	$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$	Always write vertices in correct corresponding order.
Linear Elements Proportionality	$\frac{\text{Side}_1}{\text{Side}_2} = \frac{\text{Altitude}_1}{\text{Altitude}_2} = \frac{\text{Median}_1}{\text{Median}_2} = \frac{\text{Perimeter}_1}{\text{Perimeter}_2}$	Medians, altitudes, and perimeters share the same linear ratio.
Basic Proportionality Theorem	$\frac{AD}{DB} = \frac{AE}{EC} \quad \text{and} \quad \frac{AD}{AB} = \frac{DE}{BC}$	Applies whenever a line is parallel to one triangle side.
Right Triangle Altitude	$AD^2 = BD \times CD, \ AB^2 = BD \times BC, \ \frac{AB^2}{AC^2} = \frac{BD}{CD}$	Altitude is the geometric mean of hypotenuse segments.
Similar Triangles Area Ratio	$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{side}_1}{\text{side}_2}\right)^2 = \left(\frac{\text{altitude}_1}{\text{altitude}_2}\right)^2$	Square the side ratio for similar triangles!
Trapezium Diagonal Areas	$\text{Area}(\triangle APD) = \text{Area}(\triangle BPC)$	Side non-parallel triangles in a trapezium have equal areas.
Maps & Scale Factor ($k$)	$\text{Length} \propto k, \ \text{Area} \propto k^2, \ \text{Volume} \propto k^3$	Always convert map units carefully ($1\text{ km} = 10^5\text{ cm}$).