Triangles
This chapter is about similar triangles — figures with the same shape but not necessarily the same size. You will learn the Basic Proportionality Theorem (Thales' theorem), the criteria that prove two triangles similar, how the areas of similar triangles compare, and the Pythagoras theorem. Similarity proofs are a regular feature of the CBSE Class 10 board exam.
Learning objectives
- Understand when two figures are similar.
- State and apply the Basic Proportionality Theorem.
- Use the AA, SSS and SAS criteria to prove triangles similar.
- Relate the areas of two similar triangles to their sides.
- Apply the Pythagoras theorem and its converse.
Key concepts
Similar figures
Two figures are similar if they have the same shape — their corresponding angles are equal and corresponding sides are in the same ratio. All circles are similar, and all squares are similar; triangles are similar only under certain conditions.
Basic Proportionality Theorem (Thales)
If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then it divides those two sides in the same ratio. So if DE ∥ BC in triangle ABC, then AD/DB = AE/EC.
Criteria for similarity
Two triangles are similar by AA (two pairs of equal angles), SSS (all three pairs of sides in the same ratio), or SAS (one pair of equal angles between two pairs of proportional sides).
Areas and the Pythagoras theorem
The ratio of the areas of two similar triangles equals the ratio of the squares of their corresponding sides. In a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides (Pythagoras theorem).
Important formulas
Basic Proportionality Theorem
If DE ∥ BC, then AD/DB = AE/EC
Ratio of areas (similar triangles)
ar(ABC)/ar(PQR) = (AB/PQ)²
Pythagoras theorem
hypotenuse² = base² + height²
Key definitions
- Similar triangles
- Triangles with equal corresponding angles and proportional corresponding sides (same shape).
- Congruent triangles
- Triangles that are similar AND the same size (sides equal, ratio 1:1).
- Hypotenuse
- The side opposite the right angle in a right-angled triangle — the longest side.
Solved examples
Q1. In triangle ABC, DE ∥ BC with AD = 2 cm, DB = 3 cm and AE = 4 cm. Find EC.
Solution: By the Basic Proportionality Theorem, AD/DB = AE/EC ⇒ 2/3 = 4/EC ⇒ EC = (4 × 3)/2 = 6 cm.
Q2. Two similar triangles have areas 36 cm² and 81 cm². Find the ratio of their corresponding sides.
Solution: Ratio of areas = (ratio of sides)². So ratio of sides = √(36/81) = 6/9 = 2/3.
Q3. Find the hypotenuse of a right triangle with legs 6 cm and 8 cm.
Solution: hypotenuse² = 6² + 8² = 36 + 64 = 100 ⇒ hypotenuse = 10 cm.
Common mistakes to avoid
- Matching the wrong corresponding sides when writing the similarity ratio.
- Using the side ratio for areas instead of the square of the ratio.
- Confusing similarity (same shape) with congruence (same shape and size).
- Applying Pythagoras to a non-right triangle.
Triangles — MCQ Quiz
10 questions with instant feedback. Use number keys 1–4 to answer.
The Basic Proportionality Theorem is also known as:
Practice questions
Short answer
State the AA similarity criterion.
If two angles of one triangle are equal to two angles of another, the triangles are similar.
Are all squares similar? Why?
Yes — all their angles are 90° and corresponding sides are in equal ratio.
A right triangle has legs 5 cm and 12 cm. Find the hypotenuse.
√(25 + 144) = √169 = 13 cm.
Long answer
In triangle ABC, D and E are points on AB and AC such that DE ∥ BC. If AD = 4 cm, DB = 6 cm and AC = 15 cm, find AE.
By BPT, AD/DB = AE/EC. Let AE = x, then EC = 15 − x. So 4/6 = x/(15 − x) ⇒ 4(15 − x) = 6x ⇒ 60 − 4x = 6x ⇒ 60 = 10x ⇒ x = 6 cm.
The areas of two similar triangles are 100 cm² and 49 cm². If a side of the larger triangle is 20 cm, find the corresponding side of the smaller one.
Ratio of sides = √(100/49) = 10/7. So the corresponding side = 20 × 7/10 = 14 cm.
HOTS (Higher Order Thinking)
A vertical pole 6 m high casts a shadow 4 m long. At the same time, a tower casts a 28 m shadow. Find the height of the tower.
The triangles formed are similar, so height/shadow is constant: 6/4 = h/28 ⇒ h = 6 × 28/4 = 42 m.
Prove that the diagonal of a square of side a has length a√2.
The diagonal forms a right triangle with the two sides as legs. By Pythagoras, diagonal² = a² + a² = 2a², so diagonal = a√2.
Quick revision
Revision notes
- Similar = same shape; equal angles, proportional sides.
- BPT: DE ∥ BC ⇒ AD/DB = AE/EC.
- Similarity criteria: AA, SSS, SAS.
- Area ratio = (side ratio)².
- Pythagoras: hyp² = base² + height².
Key takeaways
- Identify corresponding vertices carefully before writing ratios.
- Square the side ratio for areas — a very common exam trap.
- Similarity is the basis of heights-and-distances problems.
Frequently asked questions
What is the focus of the Class 10 Triangles chapter?
Mainly similarity of triangles — the Basic Proportionality Theorem and the AA, SSS and SAS criteria.
What is the difference between similar and congruent triangles?
Similar triangles have the same shape (proportional sides); congruent triangles also have the same size (sides equal, ratio 1:1).
How are areas of similar triangles related?
Their areas are in the ratio of the squares of their corresponding sides.