CBSE Class 10 Maths Chapter 6 — Triangles: NCERT Solutions, Theorems and Important Questions 2027

Last Updated: April 2026

CBSE Class 10 Maths Chapter 6 — Triangles is one of the most important chapters for board exams, contributing 6-10 marks in every CBSE board paper. This chapter covers the criteria for similarity of triangles, the Basic Proportionality Theorem (Thales’ Theorem), Pythagoras’ theorem, and area ratios of similar triangles. These complete NCERT solutions and notes are aligned with the CBSE Class 10 syllabus for 2027.

Chapter Overview — Board Exam Importance

Section 1: Similar Triangles — Basic Concepts

What is Similarity?

Two figures are similar if they have the same shape but not necessarily the same size. For triangles, similarity means:

• Corresponding angles are equal (Angle-Angle)
• Corresponding sides are in the same ratio (proportional)

Symbol for similarity: △ABC ~ △DEF (read: triangle ABC is similar to triangle DEF)

Note: If △ABC ~ △DEF, then:
∠A = ∠D, ∠B = ∠E, ∠C = ∠F AND AB/DE = BC/EF = CA/FD

Section 2: Basic Proportionality Theorem (Thales’ Theorem)

Statement (Theorem 6.1)

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.

Given: In △ABC, DE ∥ BC where D is on AB and E is on AC

To prove: AD/DB = AE/EC

Proof (Board Exam Must-Know)

Draw DM ⊥ AC and EN ⊥ AB. Draw BE and CD.

ar(ADE) = (1/2) × AD × EN … (i)

ar(BDE) = (1/2) × DB × EN … (ii)

From (i) ÷ (ii): ar(ADE)/ar(BDE) = AD/DB … (iii)

Similarly: ar(ADE) = (1/2) × AE × DM

ar(CDE) = (1/2) × EC × DM

ar(ADE)/ar(CDE) = AE/EC … (iv)

Since BDE and CDE share the same base DE and lie between the same parallel lines DE and BC: ar(BDE) = ar(CDE) … (v)

From (iii), (iv), (v): AD/DB = AE/EC (Proved)

Converse of BPT (Theorem 6.2)

If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.

Section 3: Criteria for Similarity of Triangles

Theorem 6.3: AA (Angle-Angle) Similarity Criterion

If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

If ∠A = ∠D and ∠B = ∠E, then △ABC ~ △DEF (and ∠C = ∠F automatically)

Theorem 6.4: SSS (Side-Side-Side) Similarity Criterion

If the corresponding sides of two triangles are in the same ratio, the triangles are similar.

If AB/DE = BC/EF = CA/FD, then △ABC ~ △DEF

Theorem 6.5: 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 in the same ratio, the triangles are similar.

If ∠A = ∠D and AB/DE = AC/DF, then △ABC ~ △DEF

Section 4: Areas of Similar Triangles

Theorem 6.6

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If △ABC ~ △DEF, then:

ar(△ABC)/ar(△DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)²

Important Result

If the ratio of corresponding sides = m:n, then ratio of areas = m²:n²

Example: If △ABC ~ △PQR and AB:PQ = 3:5, then ar(△ABC):ar(△PQR) = 9:25

Section 5: Pythagoras’ Theorem

Theorem 6.8 (Pythagoras’ Theorem)

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

If ∠B = 90° in △ABC, then AC² = AB² + BC²

Proof Using Similarity

Given: △ABC with ∠ABC = 90°. BD ⊥ AC (altitude to hypotenuse)

Prove: AC² = AB² + BC²

In △ADB and △ABC: ∠A = ∠A (common), ∠ADB = ∠ABC = 90°

∴ △ADB ~ △ABC (AA similarity)

∴ AD/AB = AB/AC → AB² = AD × AC … (i)

Similarly: △BDC ~ △ABC → BC² = DC × AC … (ii)

Adding (i) and (ii): AB² + BC² = AC(AD + DC) = AC × AC = AC² (Proved)

Converse of Pythagoras’ Theorem (Theorem 6.9)

If in a triangle, square of one side is equal to sum of squares of the other two sides, then the angle opposite to the first side is a right angle.

Important Solved Examples

Example 1: In △ABC, D and E are points on AB and AC respectively such that DE ∥ BC. If AD = 3 cm, DB = 5 cm and AE = 6 cm, find EC.

Solution: By BPT: AD/DB = AE/EC → 3/5 = 6/EC → EC = 10 cm

Example 2: The areas of two similar triangles ABC and PQR are 25 cm² and 49 cm² respectively. If AB = 5 cm, find PQ.

Solution: ar(ABC)/ar(PQR) = AB²/PQ² → 25/49 = 25/PQ² → PQ² = 49 → PQ = 7 cm

Example 3: A ladder of length 13 m rests against a vertical wall. The foot of the ladder is 5 m from the wall. Find the height at which the ladder touches the wall.

Solution: By Pythagoras: height² + 5² = 13² → height² = 169 – 25 = 144 → height = 12 m

CBSE Board Practice Questions

Q1. (1 mark) If △ABC ~ △PQR, AB = 6 cm, PQ = 9 cm, then ar(△ABC)/ar(△PQR) = ?
Answer: (6/9)² = 4/9

Q2. (2 marks) In the given figure, if LM ∥ CB and LN ∥ CD, prove that AM/MB = AN/ND.
Answer: In △ABC: LM ∥ BC → AM/MB = AL/LC (BPT) … (i). In △ACD: LN ∥ CD → AN/ND = AL/LC (BPT) … (ii). From (i) and (ii): AM/MB = AN/ND. (Proved)

Q3. (3 marks) In a △ABC, D is the mid-point of BC and AL ⊥ BC. Prove that AC² = AD² + BC·DL + (BC/2)².
Answer: Using Pythagoras in △ALC: AC² = AL² + LC². LC = LD + DC = LD + BC/2. Expand and substitute AL² = AD² – DL² (from △ADL). Simplify to get the result.

Q4. (3 marks) State and prove the BPT (Basic Proportionality Theorem).
Answer: [State theorem as given above, then provide the area-ratio proof as detailed in Section 2 of these notes.]

Common Mistakes to Avoid in Board Exams

1. Confusing similarity notation: △ABC ~ △DEF means A↔D, B↔E, C↔F — always match vertices in the STATED order
2. BPT ratio direction: AD/DB = AE/EC (smaller to full is WRONG); the ratio is the divided part to the remaining part, consistently on both sides
3. Areas vs sides: Area ratio is SQUARE of sides ratio — a very common calculation error
4. Proof of Pythagoras: Show both triangles similar to the original, not just to each other
5. Not stating theorems: In proofs, clearly state which theorem/criterion you are using

Frequently Asked Questions

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