Similarity of Triangles
Introduction
Two triangles are similar if they have the same shape but not necessarily the same size. In ICSE Class 10, you learn the criteria for similarity, the Basic Proportionality Theorem, and applications involving ratios of areas and map scales.
Key Definition
Two triangles are similar if:
- Their corresponding angles are equal.
- Their corresponding sides are in the same ratio (proportional).
Symbol: ΔABC ~ ΔPQR
Similarity Criteria
1. AA (Angle-Angle) Criterion
If two angles of one triangle are respectively equal to two angles of another triangle, the triangles are similar.
If ∠A = ∠P and ∠B = ∠Q, then ΔABC ~ ΔPQR.
Note: The third angle automatically becomes equal (angle sum property).
2. SSS (Side-Side-Side) Criterion
If the corresponding sides of two triangles are in the same ratio, the triangles are similar.
If AB/PQ = BC/QR = CA/RP, then ΔABC ~ ΔPQR.
3. SAS (Side-Angle-Side) Criterion
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 AB/PQ = BC/QR and ∠B = ∠Q, then ΔABC ~ ΔPQR.
Basic Proportionality Theorem (Thales' Theorem)
If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those two sides in the same ratio.
In ΔABC, if DE ∥ BC (D on AB, E on AC), then:
AD / DB = AE / EC
Corollary
AD / AB = AE / AC = DE / BC
Ratio of Areas of Similar Triangles
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
If ΔABC ~ ΔPQR with scale factor k = AB/PQ, then:
Area(ΔABC) / Area(ΔPQR) = k² = (AB/PQ)² = (BC/QR)² = (CA/RP)²
Map Scales
A map scale is a ratio of distances on the map to actual distances on the ground. This is an application of similarity.
Map distance : Actual distance = 1 : k
Example: If the map scale is 1 : 50,000, then 1 cm on the map represents 50,000 cm = 0.5 km on the ground.
Worked Examples
Example 1: Proving Similarity (AA)
In ΔABC and ΔDEF, ∠A = 40°, ∠B = 60°, ∠D = 40°, ∠E = 80°. Are the triangles similar?
Solution: ∠C = 180° − (40° + 60°) = 80° ∠F = 180° − (40° + 80°) = 60°
∠A = ∠D = 40° and ∠B = ∠F = 60°
Therefore, by AA criterion: ΔABC ~ ΔDFE ✓
Example 2: Basic Proportionality Theorem
In ΔABC, DE ∥ BC. AD = 3 cm, DB = 6 cm, AE = 4 cm. Find EC.
Solution: By BPT: AD / DB = AE / EC 3 / 6 = 4 / EC 1 / 2 = 4 / EC EC = 8 cm
Example 3: Ratio of Areas
ΔABC ~ ΔPQR. AB : PQ = 3 : 5. If area of ΔABC is 36 cm², find the area of ΔPQR.
Solution: Area(ΔABC) / Area(ΔPQR) = (AB/PQ)² = (3/5)² = 9/25 36 / Area(ΔPQR) = 9/25 Area(ΔPQR) = 36 × 25 / 9 = 100 cm²
Example 4: Map Scale Application
On a map, the distance between two cities is 8 cm. The actual distance is 40 km. Find the map scale.
Solution: 8 cm represents 40 km = 40 × 1,00,000 cm = 40,00,000 cm Scale = 8 : 40,00,000 = 1 : 5,00,000
Map scale = 1 : 5,00,000
Comparison: Congruence vs Similarity
| Feature | Congruent Triangles | Similar Triangles |
|---|---|---|
| Shape | Same | Same |
| Size | Same | May differ |
| Corresponding sides | Equal | Proportional |
| Corresponding angles | Equal | Equal |
| Ratio of areas | 1 : 1 | k² : 1 |
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| Applying BPT without parallel condition | DE must be ∥ BC for AD/DB = AE/EC |
| Confusing similarity and congruence criteria | AA/SSS/SAS for similarity; SSS/SAS/ASA/AAS/RHS for congruence |
| Using corresponding sides in wrong order | Match vertices in the same order |
| Incorrectly squaring the scale factor for area | Area ratio = (side ratio)² |
ICSE Exam Focus
Similarity carries 10–14 marks in ICSE exams — one of the most heavily weighted geometry topics.
- Proving similarity using criteria.
- Applying BPT to find unknown lengths.
- Area ratio problems.
- Map scale problems.
- Proof-based questions.
Marks Blueprint:
| Topic | Marks |
|---|---|
| Identifying similarity (AA/SSS/SAS) | 3 |
| Basic Proportionality Theorem | 4 |
| Area ratio problems | 3 |
| Map scale problems | 2 |
| Proof-based questions | 4–6 |
Self-Test Questions
-
In ΔABC, DE ∥ BC. AD = 4.5 cm, DB = 9 cm, EC = 12 cm. Find AE.
-
ΔABC ~ ΔPQR. AB = 6 cm, BC = 8 cm, PQ = 9 cm. Find QR.
-
The areas of two similar triangles are 64 cm² and 144 cm². Find the ratio of their corresponding sides.
-
State and prove the Basic Proportionality Theorem.
-
On a map with scale 1 : 2,00,000, two towns are 12 cm apart. Find the actual distance in km.
-
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
In ICSE, the BPT proof is frequently asked as a 4-mark question. Practice writing it with clear steps and diagram reference.
