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WBB • Class X • Mathematics • Ch 18

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Similarity

Chapter 18 of WBBSE Class 10 Mathematics Ganit Prakash introduces the transformative Euclidean theory of Similarity (সদৃশতা). While congruent figures are identical in both shape and size (scale factor k = 1), similar figures share the exact same shape while their sizes can differ by a constant scale factor. Two rectilinear polygons of the same number of sides are defined as similar if their corresponding angles are equal and the lengths of their corresponding sides are proportional. For triangles, the chapter proves that these two conditions are mutually interdependent. The theoretical cornerstone of the chapter is Theorem 48, commonly known as Thales Theorem or the Basic Proportionality Theorem (BPT), which proves that if a straight line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those two sides in the exact same ratio (AD/DB = AE/EC). Its converse proves that a line dividing two sides proportionally must be parallel to the third side. The curriculum formalizes the three similarity criteria: AAA (or the AA corollary), SAS, and SSS similarity. A major highlight of the chapter is the Right-Angled Triangle Altitude Theorem, which proves that the altitude drawn from the right angle to the hypotenuse generates two sub-triangles that are similar to each other and to the parent triangle. This yields three indispensable geometric mean formulas: BD squared equals AD times DC, AB squared equals AD times AC, and BC squared equals CD times AC. Finally, the chapter establishes that the ratio of the areas of two similar triangles equals the square of the ratio of their corresponding sides, and applies these principles to solve crucial Madhyamik board riders.

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Have You Ever Wondered?

How was the Greek mathematician Thales of Miletus able to measure the exact height of the Great Pyramid of Giza over 2,500 years ago, simply by waiting for the moment when his own shadow equaled his height and then measuring the shadow of the pyramid? And how do modern architects scale down a 50-story skyscraper onto a sheet of blueprint paper without distorting a single angle? The secret lies in the profound Euclidean concept of Similarity (সদৃশতা)—the geometry of proportional scaling, identical angles, and the master theorems that form the bedrock of trigonometry and modern cartography.

Why This Chapter Matters

Similarity is one of the most practically pervasive concepts in all of human science, technology, art, and commerce. Without similarity, maps, architectural blueprints, satellite photographs, engineering scale models, and digital screens could not function, as they all depend on scaling dimensions proportionally while preserving angular geometry. In surveying and navigation, similarity provides the principle of indirect measurement, allowing engineers to measure the heights of mountains, the widths of raging rivers, and the distances of celestial bodies without physically traversing them. In modern computer graphics and animation, 3D rendering engines use 4x4 transformation matrices that apply uniform similarity scaling to models as they move closer to or further from the virtual camera. Furthermore, similarity is the mathematical bridge that gave birth to trigonometry: the fact that right triangles with equal angles have proportional side ratios is what makes trigonometric ratios like sine, cosine, and tangent universal constants. For West Bengal Madhyamik candidates, Chapter 18 carries immense weight, appearing regularly as 5-mark theorem proofs (such as Theorem 48), 3-mark geometric riders, and compulsory 2-mark short problems.

Before You Begin (Prerequisites)

Concept of ratio and proportion (a/b = c/d) and basic algebraic operations from Chapter 5.
Concepts of congruence of triangles (SSS, SAS, ASA, RHS) where shapes are identical in both shape and size.
Parallel lines and properties of alternate and corresponding angles cut by a transversal line.
Formula for the area of a triangle: Area = (1/2) * Base * Altitude.

What You Will Learn (Core Objectives)

Distinguish clearly between Congruence (সর্বসমতা - same shape and same size) and Similarity (সদৃশতা - same shape, proportional size).
Formulate the formal conditions for similarity of polygons: corresponding angles are equal and corresponding sides are in the same ratio.
State and prove Theorem 48 (Thales' Theorem / Basic Proportionality Theorem): A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (AD/DB = AE/EC) using triangle area ratios.
State and apply the Converse of Thales' Theorem to determine when a line segment is parallel to a side of a triangle.
Master the three fundamental criteria for similarity of triangles: AAA (and the AA corollary), SAS, and SSS similarity.
State and prove the Right-Angled Triangle Altitude Theorem: The altitude to the hypotenuse divides the triangle into two triangles similar to each other and to the whole triangle, yielding the geometric mean relations (BD² = AD · DC, AB² = AD · AC, BC² = CD · AC).
Prove and apply the Area Ratio Theorem: The ratio of the areas of two similar triangles equals the square of the ratio of their corresponding sides.
Solve high-frequency Madhyamik board examination riders involving proportions, shadows, and intersecting transversals.

Chapter Roadmap & Progression

1 Module 1: Congruence versus Similar...

2 Module 2: Theorem 48 (Thales' Theor...

3 Module 3: Criteria for Similarity o...

4 Module 4: The Right Triangle Altitu...

5 Module 5: Ratio of Areas of Similar...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Congruence versus Similarity and Similar Polygons

1.1 Congruence (সর্বসমতা) versus Similarity (সদৃশতা)

In plane geometry, two geometric figures can relate to each other in two distinct ways:

Congruent Figures (সর্বসম চিত্র): Have the same shape and the same size. If one figure is placed over the other, they coincide completely. Their ratio of magnification (scale factor) is $k = 1$. Symbol: $\cong$.
Similar Figures (সদৃশ চিত্র): Have the same shape, but not necessarily the same size. One is an enlarged or reduced scale copy of the other. Symbol: $\sim$.

Fundamental Golden Rule: All congruent figures are similar, but all similar figures are not necessarily congruent! For example, all circles are similar to each other; all equilateral triangles are similar to each other; all squares are similar to each other.

1.2 Formal Conditions for Similarity of Polygons

Two polygons with the same number of sides are defined to be similar if and only if both of the following conditions are simultaneously satisfied:

All corresponding angles are equal: $$ngle A = ngle A', \quad ngle B = ngle B', \quad ngle C = ngle C', \quad \dots$$
All corresponding sides are proportional (in the same ratio): $$rac{AB}{A'B'} = rac{BC}{B'C'} = rac{CD}{C'D'} = \dots = k \quad ( ext{Scale Factor / অনুপাত})$$

Crucial Note for Polygons: For polygons with $n \ge 4$ sides, satisfying only one condition does NOT guarantee similarity. For instance, a square and a rectangle have all corresponding angles equal ($90^\circ$), but their sides are not proportional, so they are not similar. A square and a rhombus have all sides proportional, but their angles are not equal, so they are not similar. However, for triangles ($n = 3$), either condition automatically implies the other!

Module 2: Theorem 48 (Thales' Theorem / Basic Proportionality Theorem)

2.1 Statement of Theorem 48 (উপপাদ্য ৪৮ - থ্যালিসের উপপাদ্য)

Theorem 48: If a straight line is drawn parallel to any side of a triangle, it divides the other two sides (or their extensions) in the same ratio (কোনো ত্রিভুজের যেকোনো বাহুর সমান্তরাল সরলরেখা অপর দুটি বাহুকে বা তাদের বর্ধিতাংশকে সমানুপাতে বিভক্ত করে)।

2.2 Complete Euclidean Proof using Triangle Area Ratios

Given: In triangle $ riangle ABC$, the straight line $DE$ is drawn parallel to side $BC$ ($DE \parallel BC$), intersecting side $AB$ at point $D$ and side $AC$ at point $E$.

To Prove: $\mathbf{rac{AD}{DB} = rac{AE}{EC}}$

Construction: Join $B, E$ and $C, D$. From vertex $E$, draw perpendicular $EN \perp AB$. From vertex $D$, draw perpendicular $DM \perp AC$.

Proof:

Recall the area of a triangle: $ ext{Area} = rac{1}{2} imes ext{Base} imes ext{Altitude}$.
Considering side $AB$ as base: $$ ext{Area}( riangle ADE) = rac{1}{2} imes AD imes EN$$ $$ ext{Area}( riangle BDE) = rac{1}{2} imes DB imes EN$$ Dividing the two areas: $$rac{ ext{Area}( riangle ADE)}{ ext{Area}( riangle BDE)} = rac{rac{1}{2} imes AD imes EN}{rac{1}{2} imes DB imes EN} = \mathbf{rac{AD}{DB}} \quad ext{--- (Equation 1)}$$
Similarly, considering side $AC$ as base: $$ ext{Area}( riangle ADE) = rac{1}{2} imes AE imes DM$$ $$ ext{Area}( riangle CDE) = rac{1}{2} imes EC imes DM$$ Dividing the two areas: $$rac{ ext{Area}( riangle ADE)}{ ext{Area}( riangle CDE)} = rac{rac{1}{2} imes AE imes DM}{rac{1}{2} imes EC imes DM} = \mathbf{rac{AE}{EC}} \quad ext{--- (Equation 2)}$$
Now consider $ riangle BDE$ and $ riangle CDE$. Both triangles lie on the same base $DE$ and between the same parallel lines $DE$ and $BC$ ($DE \parallel BC$).
Triangles on the same base and between the same parallels have equal areas (WBBSE Class 9 Theorem): $$\mathbf{ ext{Area}( riangle BDE) = ext{Area}( riangle CDE)} \quad ext{--- (Equation 3)}$$
Comparing Equations 1, 2, and 3, the left-hand sides are identical. Therefore, their right-hand sides must be equal: $$\mathbf{rac{AD}{DB} = rac{AE}{EC}}$$ (Hence Proved - 5 Marks Compulsory Theorem in Madhyamik)

2.3 Useful Corollaries of Thales' Theorem

By applying componendo ($rac{a+b}{b}$) to the basic ratio $rac{AD}{DB} = rac{AE}{EC}$:

Corollary 1: $\mathbf{rac{AB}{AD} = rac{AC}{AE}}$ or $\mathbf{rac{AD}{AB} = rac{AE}{AC}}$
Corollary 2: $\mathbf{rac{AB}{DB} = rac{AC}{EC}}$ or $\mathbf{rac{DB}{AB} = rac{EC}{AC}}$
Full Proportionality with third side: $\mathbf{rac{AD}{AB} = rac{AE}{AC} = rac{DE}{BC}}$

2.4 Converse of Thales' Theorem

Converse: If a straight line divides any two sides of a triangle in the same ratio ($rac{AD}{DB} = rac{AE}{EC}$), then the straight line is strictly parallel to the third side ($\mathbf{DE \parallel BC}$).

Module 3: Criteria for Similarity of Triangles (AAA, SAS, SSS)

3.1 The Three Triangle Similarity Criteria

To prove that two triangles $ riangle ABC$ and $ riangle DEF$ are similar, it is not necessary to verify all six conditions (3 angles and 3 side ratios). Any one of the following three criteria is sufficient:

Criterion	Condition Required	Conclusion & Significance
AAA (or AA) Similarity	$ngle A = ngle D$, $ngle B = ngle E$ (The third angle $ngle C = ngle F$ is automatically equal by $180^\circ$ sum)	$ riangle ABC \sim riangle DEF \implies rac{AB}{DE} = rac{BC}{EF} = rac{AC}{DF}$. Most frequently used criterion!
SSS Similarity	All three pairs of corresponding sides are proportional: $rac{AB}{DE} = rac{BC}{EF} = rac{AC}{DF}$	$ riangle ABC \sim riangle DEF \implies$ All corresponding angles are equal.
SAS Similarity	One angle of a triangle is equal to one angle of another ($ngle A = ngle D$) AND the sides including these angles are proportional: $rac{AB}{DE} = rac{AC}{DF}$	$ riangle ABC \sim riangle DEF \implies$ The remaining sides and angles are proportional.

The AA Corollary: If two angles of one triangle are respectively equal to two angles of another triangle, the two triangles are similar ($ riangle ABC \sim riangle DEF$). In board problems, finding just two pairs of equal angles is the fastest path to similarity!

Module 4: The Right Triangle Altitude Theorem and Geometric Means

4.1 The Right-Angled Triangle Altitude Theorem

Theorem: In a right-angled triangle, if an altitude is drawn from the vertex containing the right angle to the hypotenuse, then the triangles on both sides of the altitude are similar to the whole triangle and to each other (সমকোণী ত্রিভুজের সমকৌণিক বিন্দু থেকে অতিভুজের উপর লম্ব অঙ্কন করলে যে দুটি ত্রিভুজ উৎপন্ন হয়, তারা প্রত্যেকে মূল ত্রিভুজের সাথে সদৃশ এবং নিজেরাও পরস্পর সদৃশ)।

4.2 Three Fundamental Geometric Mean Formulas

Let $ riangle ABC$ be a right-angled triangle with $ngle B = 90^\circ$. Let altitude $BD \perp AC$ meet the hypotenuse $AC$ at point $D$. This generates three similar triangles: $$\mathbf{ riangle ADB \sim riangle BDC \sim riangle ABC}$$

From their corresponding side proportions, we obtain three master relations:

Altitude is Geometric Mean of Hypotenuse Segments: $$ riangle ADB \sim riangle BDC \implies rac{AD}{BD} = rac{BD}{DC} \implies \mathbf{BD^2 = AD \cdot DC} \implies BD = \sqrt{AD \cdot DC}$$
Leg AB is Geometric Mean of Adjacent Segment and Whole Hypotenuse: $$ riangle ADB \sim riangle ABC \implies rac{AD}{AB} = rac{AB}{AC} \implies \mathbf{AB^2 = AD \cdot AC}$$
Leg BC is Geometric Mean of Adjacent Segment and Whole Hypotenuse: $$ riangle BDC \sim riangle ABC \implies rac{CD}{BC} = rac{BC}{AC} \implies \mathbf{BC^2 = CD \cdot AC}$$

Direct Proof of Pythagoras' Theorem: Adding relations (2) and (3): $$AB^2 + BC^2 = AD \cdot AC + CD \cdot AC = (AD + CD) \cdot AC = AC \cdot AC = \mathbf{AC^2}$$ This provides the most elegant and famous similarity proof of Pythagoras' Theorem ($a^2 + b^2 = c^2$)!

Module 5: Ratio of Areas of Similar Triangles and Madhyamik Riders

5.1 Area Ratio Theorem for Similar Triangles

Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides (দুটি সদৃশ ত্রিভুজের ক্ষেত্রফলের অনুপাত তাদের অনুরূপ বাহুগুলির বর্গের অনুপাতের সমান)।

$$ ext{If } riangle ABC \sim riangle DEF, ext{ then:}$$ $$\mathbf{rac{ ext{Area}( riangle ABC)}{ ext{Area}( riangle DEF)} = \left(rac{AB}{DE} ight)^2 = \left(rac{BC}{EF} ight)^2 = \left(rac{AC}{DF} ight)^2}$$

Furthermore, this quadratic scaling law extends to all linear elements of similar triangles:

Ratio of Altitudes (উচ্চতা): $rac{ ext{Area}_1}{ ext{Area}_2} = \left(rac{h_1}{h_2} ight)^2$
Ratio of Medians (মধ্যমা): $rac{ ext{Area}_1}{ ext{Area}_2} = \left(rac{m_1}{m_2} ight)^2$
Ratio of Perimeters (পরিসীমা): $\mathbf{rac{ ext{Perimeter}_1}{ ext{Perimeter}_2} = rac{s_1}{s_2}}$ (Perimeter scales linearly, while Area scales quadratically!).

Key Formulas, Identities & Theorems

Thales' Theorem Ratio

$$\frac{AD}{DB} = \frac{AE}{EC}$$

Full Triangle Proportionality

$$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$$

Altitude to Hypotenuse Geometric Mean

$$BD^2 = AD \cdot DC$$

Leg to Hypotenuse Relation 1

$$AB^2 = AD \cdot AC$$

Leg to Hypotenuse Relation 2

$$BC^2 = CD \cdot AC$$

Ratio of Areas of Similar Triangles

$$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{s_1}{s_2}\right)^2$$

Ratio of Perimeters

$$\frac{\text{Perimeter}_1}{\text{Perimeter}_2} = \frac{s_1}{s_2}$$

Conceptual Solved Examples & Case Studies

Example 1

In triangle ABC, DE is parallel to BC. If AD = x, DB = x - 2, AE = x + 2, and EC = x - 1, find the value of x.

Step-by-Step Solution:

Given Data: In $\triangle ABC$, $DE \parallel BC$. $AD = x$, $DB = x - 2$, $AE = x + 2$, $EC = x - 1$. Step 1: Apply Thales' Theorem (Theorem 48) Since $DE \parallel BC$, by the Basic Proportionality Theorem: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Step 2: Substitute Expressions and Cross-Multiply $$\frac{x}{x - 2} = \frac{x + 2}{x - 1}$$ $$x(x - 1) = (x - 2)(x + 2)$$ Step 3: Expand and Solve the Equation $$x^2 - x = x^2 - 4$$ Subtract $x^2$ from both sides: $$-x = -4 \implies x = 4$$ Step 4: Check Validity of Dimensions $AD = 4 > 0$, $DB = 4 - 2 = 2 > 0$, $AE = 4 + 2 = 6 > 0$, $EC = 4 - 1 = 3 > 0$. Ratio $\frac{AD}{DB} = \frac{4}{2} = 2$ and $\frac{AE}{EC} = \frac{6}{3} = 2$. Ratios match perfectly! Final Answer: The value of $x$ is $\mathbf{4}$.

Example 2

A vertical pole of length 6 m casts a shadow 4 m long on the ground, and at the same time a vertical tower casts a shadow 28 m long on the ground. Find the height of the tower.

Step-by-Step Solution:

Given Data: Height of pole $h_1 = 6\text{ m}$, shadow of pole $s_1 = 4\text{ m}$. Height of tower $= H\text{ m}$, shadow of tower $s_2 = 28\text{ m}$. Step 1: Establish Triangle Similarity via Sun's Elevation At the exact same time of day, the sun's rays strike the ground at the same angle of elevation ($\theta$). Both the pole and the tower are vertical to the ground ($90^\circ$). Let $\triangle ABC$ represent the pole and its shadow, and $\triangle DEF$ represent the tower and its shadow. $$\angle B = \angle E = 90^\circ \quad \text{and} \quad \angle C = \angle F = \theta$$ By the AA Similarity Criterion: $$\triangle ABC \sim \triangle DEF$$ Step 2: Set Up Proportionality of Corresponding Sides $$\frac{\text{Height of Tower}}{\text{Height of Pole}} = \frac{\text{Shadow of Tower}}{\text{Shadow of Pole}}$$ $$\frac{H}{6} = \frac{28}{4}$$ $$\frac{H}{6} = 7$$ $$H = 7 \times 6 = 42\text{ m}$$ Final Answer: The height of the tower is $\mathbf{42\text{ m}}$.

Example 3

In a right-angled triangle ABC, ∠B = 90°, and BD ⊥ AC is the altitude to the hypotenuse. If AD = 4 cm and CD = 9 cm, find the lengths of: (i) altitude BD, (ii) side AB, and (iii) side BC.

Step-by-Step Solution:

Given Data: Right $\triangle ABC$ with $\angle B = 90^\circ$, $BD \perp AC$. $AD = 4\text{ cm}$, $CD = 9\text{ cm}$. Total hypotenuse $AC = AD + CD = 4 + 9 = 13\text{ cm}$. Step 1: Find Altitude BD using Geometric Mean Formula By the Right Triangle Altitude Theorem, $\triangle ADB \sim \triangle BDC$: $$BD^2 = AD \cdot CD$$ $$BD^2 = 4 \times 9 = 36$$ $$BD = \sqrt{36} = 6\text{ cm}$$ Step 2: Find Leg AB By similarity $\triangle ADB \sim \triangle ABC$: $$AB^2 = AD \cdot AC$$ $$AB^2 = 4 \times 13 = 52$$ $$AB = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}\text{ cm} \approx 7.21\text{ cm}$$ Step 3: Find Leg BC By similarity $\triangle BDC \sim \triangle ABC$: $$BC^2 = CD \cdot AC$$ $$BC^2 = 9 \times 13 = 117$$ $$BC = \sqrt{117} = \sqrt{9 \times 13} = 3\sqrt{13}\text{ cm} \approx 10.82\text{ cm}$$ Verification via Pythagoras: $$AB^2 + BC^2 = 52 + 117 = 169 = 13^2 = AC^2$$ (Matches perfectly!) Final Answer: (i) $BD = \mathbf{6\text{ cm}}$ (ii) $AB = \mathbf{2\sqrt{13}\text{ cm}}$ (iii) $BC = \mathbf{3\sqrt{13}\text{ cm}}$.

Example 4

The areas of two similar triangles ABC and DEF are 64 cm² and 121 cm² respectively. If EF = 15.4 cm, find the length of side BC.

Step-by-Step Solution:

Given Data: $\triangle ABC \sim \triangle DEF$. $\text{Area}(\triangle ABC) = 64\text{ cm}^2$. $\text{Area}(\triangle DEF) = 121\text{ cm}^2$. Side $EF = 15.4\text{ cm}$. Let $BC = x\text{ cm}$. Step 1: Apply the Area Ratio Theorem For similar triangles, the ratio of their areas equals the square of the ratio of their corresponding sides: $$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2$$ $$\frac{64}{121} = \left(\frac{x}{15.4}\right)^2$$ Step 2: Take Square Roots on Both Sides $$\sqrt{\frac{64}{121}} = \frac{x}{15.4}$$ $$\frac{8}{11} = \frac{x}{15.4}$$ Step 3: Solve for x $$x = \frac{8 \times 15.4}{11}$$ Notice that $15.4 / 11 = 1.4$: $$x = 8 \times 1.4 = 11.2\text{ cm}$$ Final Answer: The length of side $BC$ is $\mathbf{11.2\text{ cm}}$.

Example 5

In triangle ABC, line segment DE is drawn parallel to base BC such that it divides triangle ABC into two regions of equal area. Find the ratio BD / AB.

Step-by-Step Solution:

Given Data: In $\triangle ABC$, $DE \parallel BC$. Line segment $DE$ divides $\triangle ABC$ into two parts of equal area: $\text{Area}(\triangle ADE) = \text{Area}(\text{trapezium } BDEC)$. Step 1: Relate Areas of Sub-Triangle and Main Triangle $$\text{Area}(\triangle ABC) = \text{Area}(\triangle ADE) + \text{Area}(\text{trapezium } BDEC)$$ $$\text{Area}(\triangle ABC) = 2 \times \text{Area}(\triangle ADE)$$ $$\implies \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{1}{2}$$ Step 2: Apply Similarity and Area Ratio Theorem Since $DE \parallel BC$, $\angle ADE = \angle B$ and $\angle AED = \angle C$ (corresponding angles). By AA Similarity, $\triangle ADE \sim \triangle ABC$. $$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left(\frac{AD}{AB}\right)^2$$ $$\frac{1}{2} = \left(\frac{AD}{AB}\right)^2 \implies \frac{AD}{AB} = \frac{1}{\sqrt{2}}$$ Step 3: Determine the Ratio BD / AB Notice that $BD = AB - AD$: $$\frac{BD}{AB} = \frac{AB - AD}{AB} = 1 - \frac{AD}{AB}$$ $$\frac{BD}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$$ Rationalizing the denominator: $$\frac{BD}{AB} = \frac{(\sqrt{2} - 1)\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$$ Final Answer: The ratio $\frac{BD}{AB}$ is $\mathbf{\frac{\sqrt{2} - 1}{\sqrt{2}}}$ (or $\mathbf{\frac{2 - \sqrt{2}}{2}}$).

Common Misconceptions & Examiner Traps

Common Misconception

Writing Thales' theorem ratio as AD/DB = EC/AE (inverting one side).

Scientific Reality & Correction

Ensure exact correspondence in direction: Top/Bottom = Top/Bottom, i.e., AD/DB = AE/EC.

Common Misconception

Assuming ratio of areas of similar triangles equals ratio of sides (omitting the square).

Scientific Reality & Correction

Ratio of areas is proportional to the SQUARE of the sides: Area1/Area2 = (s1/s2)².

Common Misconception

Squaring perimeters when comparing similar triangles.

Scientific Reality & Correction

Perimeter is a one-dimensional linear measure: Perimeter1/Perimeter2 = s1/s2 (NO squaring).

Common Misconception

Misidentifying segments in right triangle altitude formula (writing BD² = AB · BC).

Scientific Reality & Correction

The square of the altitude to hypotenuse equals the product of the TWO SEGMENTS of the hypotenuse: BD² = AD · DC.

Common Misconception

Assuming any two rectangles or rhombuses are similar.

Scientific Reality & Correction

For polygons with 4 or more sides, BOTH corresponding angles must be equal AND corresponding sides must be proportional.

Geometric Architecture: Similarity Theorems & Right Triangle Altitude (WBBSE Class 10 Ganit Prakash)

Chapter Summary & 10 Key Takeaways

Takeaway 1

Congruence vs Similarity: Congruent shapes have same shape and size (k = 1); similar shapes have same shape with proportional size (scale factor k).

Takeaway 2

Polygon Similarity: Requires corresponding angles equal AND corresponding sides proportional simultaneously.

Takeaway 3

Theorem 48 (Thales / BPT): If DE ∥ BC in ΔABC, then AD/DB = AE/EC.

Takeaway 4

Corollaries of Thales: AD/AB = AE/AC = DE/BC and DB/AB = EC/AC.

Takeaway 5

Converse of Thales: If AD/DB = AE/EC, then DE is strictly parallel to BC.

Takeaway 6

Triangle Similarity Criteria: AAA (or AA), SSS, and SAS similarity.

Takeaway 7

Right Triangle Altitude Theorem: Altitude from right angle to hypotenuse yields ΔADB ~ ΔBDC ~ ΔABC.

Takeaway 8

Geometric Mean Relations: BD² = AD · DC, AB² = AD · AC, and BC² = CD · AC.

Takeaway 9

Area Ratio Theorem: Area1 / Area2 = (s1 / s2)² = (h1 / h2)² = (m1 / m2)².

Takeaway 10

Perimeter Ratio Law: Perimeter1 / Perimeter2 = s1 / s2 (linear scaling).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

In triangle ABC, D and E are points on AB and AC such that AD = 3 cm, DB = 4.5 cm, AE = 2 cm, and EC = 3 cm. Is DE parallel to BC?

Reveal Answer & Explanation

Answer: Calculate the ratios: AD / DB = 3 / 4.5 = 30 / 45 = 2 / 3. AE / EC = 2 / 3. Since AD / DB = AE / EC = 2 / 3, by the Converse of Thales' Theorem, DE is strictly parallel to BC.
Check if AD/DB equals AE/EC, then apply the Converse of Thales' Theorem.

The perimeters of two similar triangles ABC and PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, find the corresponding side AB.

Reveal Answer & Explanation

Answer: Ratio of perimeters equals ratio of corresponding sides: Perimeter(ABC) / Perimeter(PQR) = AB / PQ => 36 / 24 = AB / 10 => 3 / 2 = AB / 10 => AB = (3 * 10) / 2 = 15 cm.
Perimeters scale linearly with side lengths: P1 / P2 = s1 / s2.

In right triangle ABC, ∠B = 90° and BD ⊥ AC. If AD = 2 cm and CD = 8 cm, find altitude BD.

Reveal Answer & Explanation

Answer: BD² = AD * CD = 2 * 8 = 16 => BD = sqrt(16) = 4 cm.
Use the formula BD² = AD * CD.

The areas of two similar triangles are in the ratio 25 : 49. What is the ratio of their corresponding altitudes?

Reveal Answer & Explanation

Answer: Ratio of areas equals square of ratio of altitudes: Area1 / Area2 = (h1 / h2)² => 25 / 49 = (h1 / h2)² => h1 / h2 = sqrt(25 / 49) = 5 / 7. The ratio of altitudes is 5 : 7.
Take the square root of the area ratio.

State the AA similarity criterion for triangles.

Reveal Answer & Explanation

Answer: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar, because their third angles are automatically equal by the 180° angle sum property.
Think about why knowing two angles is sufficient to know the third angle.

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