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Triangles - Concept of Similarity

Grade 10CBSE

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

🔑Concepts

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Definition of Similarity: Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional). Visually, similar triangles have the same shape but can differ in size, much like an original photograph and its enlargement.

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Basic Proportionality Theorem (BPT) or Thales Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Imagine a large triangle ΔABC\Delta ABC with a line segment DEDE drawn inside it such that DD is on ABAB, EE is on ACAC, and DEDE is visually parallel to the base BCBC.

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Converse of BPT: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. In a diagram, if you measure ADDB\frac{AD}{DB} and AEEC\frac{AE}{EC} and find them equal, the line DEDE will never meet BCBC no matter how far they are extended.

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AA (Angle-Angle) Similarity Criterion: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This is because the third angle must also be equal due to the angle sum property. Visually, if you stack the smaller triangle inside the larger one at a shared corner, their opposite sides will appear parallel.

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SSS (Side-Side-Side) Similarity Criterion: If the corresponding sides of two triangles are proportional, then their corresponding angles are equal, and the triangles are similar. This means if you scale every side of a triangle by the same factor kk, the resulting shape remains identical in proportion.

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SAS (Side-Angle-Side) Similarity Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar. Visually, this fixes the 'spread' of the two sides and their relative lengths, forcing the third side to fall into the same proportion.

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Similarity in Right Triangles: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other. This creates three nested right-angled triangles of different sizes but identical shapes.

📐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.