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Triangles - Differentiate similar and congruent triangles using definitions and counterexamples

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Congruent Triangles are 'identical' in both shape and size. Two triangles are congruent if their corresponding sides are equal and their corresponding angles are equal. Symbol: ≅\cong.

Two congruent triangles showing identical size and shape.
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Similar Triangles have the same shape but not necessarily the same size. Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional). Symbol: ∼\sim.

A small triangle and a larger similar triangle.
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Key Differentiator: All congruent triangles are similar (ratio 1:11:1), but not all similar triangles are congruent. Similarity is a scaling transformation, while congruence is a rigid transformation.

Diagram
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Counterexample for Congruence: Two equilateral triangles with sides 3cm3 cm and 5cm5 cm respectively have equal angles (60∘60^{\circ} each), making them similar, but they are NOT congruent because their side lengths differ.

Two equilateral triangles of different sizes demonstrating similarity without congruence.

📐Formulae

If ΔABC∼ΔDEF\Delta ABC \sim \Delta DEF, then ∠A=∠D\angle A = \angle D, ∠B=∠E\angle B = \angle E, and ∠C=∠F\angle C = \angle F.

Side Ratio: ABDE=BCEF=ACDF\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}

Basic Proportionality Theorem: In ΔABC\Delta ABC, if DE∥BCDE \parallel BC, then ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}.

Corollary of BPT: ADAB=AEAC\frac{AD}{AB} = \frac{AE}{AC} and ABDB=ACEC\frac{AB}{DB} = \frac{AC}{EC}.

Area Ratio (for reference): Area(ΔABC)Area(ΔDEF)=(ABDE)2=(BCEF)2=(ACDF)2\frac{Area(\Delta ABC)}{Area(\Delta DEF)} = (\frac{AB}{DE})^2 = (\frac{BC}{EF})^2 = (\frac{AC}{DF})^2 (Note: Often used in advanced similarity problems).

💡Examples

Problem 1:

In ΔABC\Delta ABC, DE∥BCDE \parallel BC. If AD=xAD = x, DB=x−2DB = x - 2, AE=x+2AE = x + 2, and EC=x−1EC = x - 1, find the value of xx.

Solution:

  1. By Basic Proportionality Theorem (BPT), since DE∥BCDE \parallel BC, we have: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}
  2. Substitute the given values: xx−2=x+2x−1\frac{x}{x - 2} = \frac{x + 2}{x - 1}
  3. Cross-multiply: x(x−1)=(x+2)(x−2)x(x - 1) = (x + 2)(x - 2)
  4. Simplify: x2−x=x2−4x^2 - x = x^2 - 4
  5. Subtract x2x^2 from both sides: −x=−4-x = -4
  6. Therefore, x=4x = 4.

Explanation:

We apply the Thales Theorem which states that a line parallel to the base of a triangle divides the other two sides proportionally. We then solve the resulting algebraic equation for xx.

Problem 2:

A vertical pole of length 6m6 m casts a shadow 4m4 m long on the ground and at the same time a tower casts a shadow 28m28 m long. Find the height of the tower.

Solution:

  1. Let ABAB be the pole and BCBC be its shadow. Let PQPQ be the tower and QRQR be its shadow.
  2. In ΔABC\Delta ABC and ΔPQR\Delta PQR, ∠B=∠Q=90∘\angle B = \angle Q = 90^{\circ} (Vertical structures).
  3. ∠C=∠R\angle C = \angle R (Angle of elevation of the sun is the same for both at the same time).
  4. Therefore, ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR by AA similarity criterion.
  5. Thus, ABPQ=BCQR\frac{AB}{PQ} = \frac{BC}{QR}
  6. 6h=428\frac{6}{h} = \frac{4}{28}
  7. h=6×284=6×7=42mh = \frac{6 \times 28}{4} = 6 \times 7 = 42 m.

Explanation:

This problem uses the AA similarity criterion. Since the sun's rays hit the earth at the same angle for both objects, the triangles formed by the objects and their shadows are similar, allowing us to use side proportions.

Problem 3:

Check if ΔABC\Delta ABC with sides 3cm,4cm,5cm3 cm, 4 cm, 5 cm and ΔPQR\Delta PQR with sides 6cm,8cm,10cm6 cm, 8 cm, 10 cm are similar or congruent.

Two right-angled triangles with proportional sides (3-4-5 and 6-8-10).

Solution:

Ratioofcorrespondingsides:Ratio of corresponding sides: ABPQ=36=12\frac{AB}{PQ} = \frac{3}{6} = \frac{1}{2} BCQR=48=12\frac{BC}{QR} = \frac{4}{8} = \frac{1}{2} ACPR=510=12\frac{AC}{PR} = \frac{5}{10} = \frac{1}{2} Since all corresponding sides are in the same ratio 12\frac{1}{2}, the triangles are similar by SSS similarity criterion. However, since the sides are not equal (3≠63 \neq 6), they are not congruent.

Explanation:

Triangles are similar because their sides are proportional. They are not congruent because their sizes are different.

Problem 4:

In ΔLMN\Delta LMN, PQ∥MNPQ \parallel MN. If LP=2cmLP = 2 cm, PM=4cmPM = 4 cm, and PQ=3cmPQ = 3 cm, find the length of MNMN. Is ΔLPQ∼ΔLMN\Delta LPQ \sim \Delta LMN?

A triangle with a parallel line segment inside creating a smaller similar triangle.

Solution:

In ΔLPQ\Delta LPQ and ΔLMN\Delta LMN:

  1. ∠L=∠L\angle L = \angle L (Common)
  2. ∠LPQ=∠LMN\angle LPQ = \angle LMN (Corresponding angles as PQ∥MNPQ \parallel MN) Therefore, ΔLPQ∼ΔLMN\Delta LPQ \sim \Delta LMN (AA Similarity). By property of similarity: LPLM=PQMN\frac{LP}{LM} = \frac{PQ}{MN} LM=LP+PM=2+4=6cmLM = LP + PM = 2 + 4 = 6 cm 26=3MN\frac{2}{6} = \frac{3}{MN} 2×MN=182 \times MN = 18 MN=9cmMN = 9 cm

Explanation:

Parallel lines create corresponding angles, leading to similar triangles where side ratios are equal.